M4maths Previous Puzzles

27 Aug 2014 Profit or loss

A shopkeeper decides to give a discount of 20% + 30% on the printed price of a watch, but unfortunately the printed banner comes with the print as 30%+40% discount. If he gets a gain of only 5% of actual cost of watch in this new structure then how much profit in a actual cost did he decide to get in the earlier structure.

OPtion

1) 15%
2) 10%
3) 55%
4) 46%
5) 35%
6) 45%
7) 40%
8) 30%
9) 20%
10)None of these

Solution

Let printed cost = x and actual cost = y
then according to given condition
x(1-.3)(1-.4) - y = .05 y
x = (1.05/.42)*y

so x*(1-.2)*(1-.3) = (1.05/.42) * .8 * .7y
x * .8 * .7 = (1.05*8/6)*y
= (1.05/3)*4y = 1.4 y

x*.8*.7 - y = .4y
so profit = 40%

option 7)

Correct Option: 7 Best Solution (60)

Neel Patel   10 years AGO

In cade 1. 20%+ 30% discount =20+30-(20*30)/100
=44%discount
if marked price is 100x then selling price is 56x


In case 2. 30%+40%=30+40-30*40/100
=58% discount
if marked prire is 100x then selling price is 42x ,
on which the shopkeeper gains 5%
do costprise is=42x*(100/105)=40x

so in 1st case selling price =56xand cost price= 40x
profit precentage=(56-40/40)*100
=40%

Like? Yes (14) | No (4)  

Rahul Sah   10 years AGO

Let M.P=Rs.100
After discount of 20%+30%(100-->80-->56) S.P=Rs.56
After discount of 30%+40%(100-->70-->42) S.P=Rs.42

When discount of 30%+40%,profit=5%,S.P=Rs.42
=>S.P=105% of C.P
=>42=(105/100)*C.P
=>C.P=(42*100)/105

Original discount of 20%+30%,profit=?,S.P=Rs.56
=>S.P=(100+%profit)% of C.P
=>56=((100+%profit)/100) * (42*100)/105
=>%profit=40

Option 7)

Like? Yes (5) | No (1)  

Nischay Jain   10 years AGO

Let Or. cost=100
Later discount = 30% +40%
so S.P = 42
he gains a profit of 5% on selling at Rs 42
(105*x)/100=42
Therefore, the C.p is Rs 40
now if sold the watch at 20% + 30%
Then S.P = 56
now profit % = profit/c.p *100
16/40 *100= 40%
Therefore the ans is (7) 40%

Like? Yes (3) | No   

SUNITEE GARIMA   10 years AGO

Let Marked price be Rs100
On giving successive discount of 20% and 30% on MP,total discount is 44%.
so SP=(100-44)*100/100=Rs 56
In second case ,on giving successive disount of 30% and 40%,total discount offered is 58%.
So ,new SP=(100-58)*100/100=Rs 42.
A/q,profit earned is 5%.
So CP=(100/105)*42=Rs 40
So actual profit =56-40=Rs16
profit %=(16/40)*100=40%

Like? Yes (3) | No (1)  

Sourojit Dhua   10 years AGO

Let marked price be 100.
Sp of the item for 1st condition comes out as rs. 56.
Sp of the item for 2nd cnditn= rs.42
now, cp= x+x/20=42
so, cp=rs.40
now desired profit= (56-40)/40x100=40%

Like? Yes (2) | No   

swarup kumar prusti   10 years AGO

100-20% dis=80
80-30%dis=56
100-30% dis=70
70-40% dis=42
in 42 he have 5% profit
in 56 he will get 40% profit

Like? Yes (2) | No   

suraj Khanduri   10 years AGO

let printed price 100.
let actual price x.
after 20%+30% discount price= 56.
after 30%+40% discount price= 42
on selling 42 he gain 5%.
x+(x.5/100)=42
x=40. i.e. actual price.
according previous structure he would gain 56-40=16.
in term of %: (16/40)*100= 40
so ans=40%.

Like? Yes (1) | No   

Dr.Dipin Singh   10 years AGO

If C is the cost & P is printed price, then
sp in new structure = 0.7*0.6*P = 1.05C
SP in eralier structure = 0.8*0.7*P= 0.56P= C*0.56*1.05/0.42= 1.40C
Hence profit % in earlier structure = 40%

Like? Yes (3) | No (2)  

vikash kumar   10 years AGO

let the market price is 100 ,he want to give the discount of 20% first and then 30% means the desired sp will be 56rs but his actual sp will be 42rs after giving the discount of 30% and 40% and he has gained 5% from his actual cp and after calculating his actual cp was 40rs means his desired profit is (56-40)*100/40%=40%

Like? Yes (1) | No   

ADARSH GAURAV   10 years AGO

say MP =100
after giving disc. of 30% and 40% respectively, SP becomes 70 and 42 respectively
so when SP=42 and gain=5%
so ((SP-CP)/CP)*100=gain%
CP=40
again consider MP=100
after giving disc. of 20% and 30% respectively, SP becomes 80 and 56 respectively
gain%=((56-40)/40)*100=40%




40% is the answer.

Like? Yes (1) | No   

ANAND KUMAR   10 years AGO

let s.p=100
100-30%+40% of 100=42
so c.p=42*100/105
=40
now,100-20%+30% of 100=56
hence profit=16*100/40
=40%

Like? Yes (1) | No   

kumar aryan   10 years AGO

Let x be marked price after 30-40%it become23x/50 which is equal to selling price .now cost price is equal to 100*sp/100+%profit which comes 2x/5. Now after 20-30% selling price will come 14x/25 now profit is14x/25-2x/5=4x/25 %profit is4x/25*100/2x/5=40%.

Like? Yes (1) | No   

deepak kumar   10 years AGO

ON 42 GAIN WAS 5%, THEN ON 56 GAIN WILL BE 40%

Like? Yes (1) | No   

SAYAN CHATTERJEE   10 years AGO

let the printed price be x & cost price be c
so x*70/100*60/100-c=c*5/100
=> x=5*c/2
at desired price, profit= ( x*80/100*70/100-c)/c*100%
=40%. (7)

Like? Yes (1) | No   

Kapil Shirsath   10 years AGO

Let the marked price be 100Rs
When discount is 30%+40%,net discount=58%. So the selling price is 42Rs and profit earned is 5%
So, the cost price is 40Rs
Now when discount is 20%+30%,net discount=44%. So the selling price is 56Rs
Profit intended to be earned is 56-40=16rs
So the profit percentage=(100*16)/40=40%

Like? Yes (1) | No   

Rajat sadhu   10 years AGO

let the printed price=x then
selling price due to misprinting=0.56x (due to 30%+40%discount)
he gains 5% at this s.p. so the actual price is=95%of s.p.
=0.399x
earlier selling price=30% of (20% of x)=.56x
profit on actual price=(0.56x-0.399x)*100/(0.399x)=40%

Like? Yes (1) | No   

lalit   10 years AGO

consider initially MP=100
intial plan sp of shopkeeper=((100*20/100)*30)/100=56
but actual sp of shopkeeper=((100*30/100)*40)/100=42
acc to question 1.05cp=42
cp=40
profit=56-40=16=40%

Like? Yes (1) | No   

Pranoy Bej   10 years AGO

20%+30% = 44%
30%+40% = 58%
let printed price = 100
so SP= 42
then CP = 40
so the profit for earlier structure will be = (16/40)*100 = 40%

Like? Yes (1) | No   

manish kumar singh   10 years AGO

Let M.P = 100 Rs.
On giving Successive discount of 20% and 30% the S.P will be 56 Rs.

Also, giving Successive discounts of 30% and 40% the new S.P will be 42 Rs.
as given he gets 5% profit
so ((42-x)/x)*100=5
from here we get c.p=40
which is denoted as x
so ((56-40)/40)*100)
from here we get 40% as answer

Like? Yes (1) | No   

ANKIT SHARMA   10 years AGO

let cost be 100
5% profit
then 105
105/.6=175
175/.7=250
then
250-(250*.2)=200
200-(200*.3)=140
HENCE 40% ans

Like? Yes (1) | No   

venkat suhas   10 years AGO

In the new discount, the cost of the watch is 42% of the marked price.
It gives him 5% profit.
So,the original price is 40% of marked price.
By the initial discount, it is the 56% of the marked price
His profit is 16% on 40% which is equal to 40%.
So, the answer is 40%

Like? Yes (1) | No   

sujeet kumar   10 years AGO

let printed price=x.
selling price at (30%+40%)=x(100-58)/100=42x/100.
cost price at 5% profit= 42x/105.
selling price at(20%+30%)=56x/100.
profit=(56x/100-42x/105)=(0.56-0.4)x=0.16x.
%profit=(0.16x*100)/(42x/105)=40%

Like? Yes (1) | No   

shree   10 years AGO

40% is

Like? Yes (1) | No   

Jeevanantham.V.M   10 years AGO

Let M.P = 100 Rs.
On giving Successive discount of 20% and 30% the S.P will be 56 Rs.

Also, giving Successive discounts of 30% and 40% the new S.P will be 42 Rs.

According to the Question ,he gets only 4% profit
Therefore C.P will be 40 Rs.

So, the actual Profit = (56-40)*100 / 40 = 16*100/40 = 40 %

Like? Yes (1) | No   

pawan kumar chhipa   10 years AGO

let s.p is 100 then
30%+40% is 42
and it gain 5% on it
so c.p is 42*100/105 =40

and 20%+30% is 56

so its gain is 56-40/40*100
40%
ans is 7) 40%

Like? Yes (1) | No   

Akash Kumar   10 years AGO

Let assume the printed pricr =100 rs. after 30% and 40% off consiqutively then the sold price will be 70*60/100=42(as per new structure). on it 5% profit means actual price was=(100/105)*42=40. for the earlier structure sold price is=80*70/100=56. So he decided to get actual profit=((56-40)/40)*100%=40%. hence option 7 is correct.

Like? Yes (1) | No   

Nitin Srivastava   10 years AGO

ans 40%

Like? Yes (1) | No   

sayan pal   10 years AGO

actual cost = y
printed price = x
case 1 (20%+30%) discount
price after discount= (0.8*x)*.7=0.56*x
case 2 (30%+40%) discount
price after discount=(0.7*x)*0.6=0.42*x
gain in case 2 is =.05 *y
hence, (0.42*x-y)=0.05*y
x=2.5*y
gain in case 1 is
=(0.56*x-y)*100/y %
=(0.56*2.5*y - y)*100/y
= 40%. (ans)

Like? Yes (1) | No   

Arpit Singh   10 years AGO

1.30+20-(30*20/100)=44%
2.30+40-(30*40/100)=58%
Printed price=100
Case 2 SP=100-58=42
As profit is 5% so CP=40
Actual SP=100-44=56
Profit that he plannes=(56-40)/40*100=40%

Like? Yes (1) | No   

Devendra Marghade   10 years AGO

If Actual cost=x & printed price=y, then with later condition,
y*(70/100)*(60/100)=(105/100)*x
y=2.5x
It shows, Printed price is 2.5 times the cost price
So at gradual discount of 20% & 30%
y*(80/100)*(70/100)=0.56y
=0.56*2.5x=1.4x
Which is 40% above the cost price

Like? Yes (1) | No (1)  

Rahul rahul   10 years AGO

Let mrp =100
30+40 percent= sp=42
Then by cp=100/100+profit ×sp
cp=40
Now sp in earlier structure
20+30 percent=56
gain=56-40/40 ×100= 40 perc.

Like? Yes (1) | No (1)  

Ravi Bhalani   10 years AGO

suppose CP=100
First case: 44% discount
Second case:58% discount
He gets 5% of actual cost so, SP=105
By selling 30% +40% discount he gets only 5% profit.So
Marked price=10500/42=250
Now in 1st case discount was 20% + 30% =44%
Discount=250*0.44=110
So,S,P=140
=> %profit =40%

Like? Yes (1) | No (1)  

naina   10 years AGO

profit to be added = 40%

Like? Yes (1) | No (1)  

HARDIP BHALANI   10 years AGO

suppose CP=100
First case: 44% discount
Second case:58% discount
He gets 5% of actual cost so, SP=105
By selling 30% +40% discount he gets only 5% profit.So
Marked price=10500/42=250
Now in 1st case discount was 20% + 30% =44%
Discount=250*0.44=110
So,S,P=140
=> %profit =40%

Like? Yes (1) | No (1)  

Harshad    10 years AGO

suppose CP=100
First case: 44% discount
Second case:58% discount
He gets 5% of actual cost so, SP=105
By selling 30% +40% discount he gets only 5% profit.So
Marked price=10500/42=250
Now in 1st case discount was 20% + 30% =44%
Discount=250*0.44=110
So,S,P=140
=> %profit =40%

Like? Yes (1) | No (1)  

suman kumar   10 years AGO

let marked price be x.
and let c.p be 100.
so x*0.7*0.6=0.42x=105(5%gain).
since, x=250,
so,x*0.8*0.7=0.56x=140
since 40%gain

Like? Yes (1) | No (1)  

DIPAK TIWARI   10 years AGO

Let printed price be ₹100
From the 1st condition S.P=₹56
From the 2nd condition S.P=₹42
Let cost price be ₹ X
Therefore 105X/100=42
=>X=₹40
% Profit in earlier =(16/40)*100=40%
Option 7

Like? Yes (2) | No (2)  

Aakash Jain   10 years AGO

assume that printed price of watch=100
then 20%+30% discount=44 so selling price=56
and 30%+40% discount=58 so selling price=42
5%profit when it sold at price 42 so cost price=40
when it sold in Rs.56 ,profit % will be 40%

Like? Yes (1) | No (1)  

sanjeev kumar   10 years AGO

ans=40%

Like? Yes (1) | No (1)  

vartika    10 years AGO

Let market price be rs 100.
then after 2 successive discounts of 20 and30 %
the SP will be rs 56
bt with d successive discounts of 30 and 40%
new SP will be rs 42.
according to question.. his gain is 5% on new discounts,
so CP = (100/105)*42 = rs 40
gain with his previous discounts ={( 56-40)/56}*100 = 40%

Like? Yes (1) | No (1)  

Tark Patel   10 years AGO

Let market price is 100 Rs.

In Second case discount 30% + 40% so effective discount is 58% so selling price will be 42 Rs. and he gain 5% profit so cost price will be 40 Rs.

In first case discount 20% + 30% so effective discount is 44% so selling price will be 56 Rs. and cost price is same as above calculated which is 40 Rs.
Hence profit gain by him in first case is 40% (SP = 56 Rs and CP = 40 Rs)

Like? Yes (1) | No (1)  

posirajesh   10 years AGO

cost price= 40rs since 5%profit when he sell for second discount..
hence for 20%+30% discount gain=16rs which is 40%of actual cost

Like? Yes (1) | No (1)  

ashok   10 years AGO

ans==40%

assume the cost price is c

and printed price is 1

now calculate discounted cost-
#30+40@
1-->(30% discount)-->0.7-->(additional 40% discount)-->0.42

ie. give 5 % gain
so 1.05*c=0.42
c=0.42/1.05

#20+30@
1-->(20% discount)-->0.8-->(additional 30% discount)-->0.56

assume selling cost amount earlier is x*c

x*c=0.56
x*0.42/1.05=0.56
x=.56*1.05/.42
x=1.40

so profit=(1.40c-c)/c*100=40%

Like? Yes (1) | No (1)  

Vanika   10 years AGO

Gain= ( (0.7 * 0.8 SP) - CP ) / CP * 100 =5 (GIVEN )
=14 SP= 35 CP
SO FOR DISCOUNT 20% +30%
Gain = ( (0.8 * 0.7 SP -CP)/CP * 100
Solving this
Gain= 40 %

Like? Yes (1) | No (1)  

ANIL KUMAR   10 years AGO

cost price is 40,a/t previous discount selling price will be 56,so

(16/40*100)=40%

Like? Yes (1) | No (1)  

Rajat Goyal   10 years AGO

Using simple C.P and S.P. formulas

Like? Yes (1) | No (1)  

Chiranjit Agarwal   10 years AGO

let the printed price=100
@ 20+30 % discount: S.P= 56
@ 30+40 % discount: S.P=42 (here gain= 5%)
so actual cost= 40
if S,P=56
gain= 40%

Like? Yes (1) | No (1)  

hiren vavadiya   10 years AGO

here 5% for 100% then 120% it is 40%

Like? Yes (1) | No (1)  

shivam tiwari   10 years AGO

Let M.P = 100 Rs.
On giving Successive discount of 20% and 30% the S.P will be 56 Rs.

Also, giving Successive discounts of 30% and 40% the new S.P will be 42 Rs.

According to the Question ,he gets only 4% profit
Therefore C.P will be 40 Rs.

So, the actual Profit = (56-40)*100 / 40 = 16*100/40 = 40 %

Like? Yes (1) | No (1)  

RAVI RANJAN   10 years AGO

30%+40%=58% discount
20%+30%=44%discount
42% price gives 5% profit
then 56% price will give 40% discount

Like? Yes (1) | No (1)  

MRITUNJAY KUMAR   10 years AGO

option 7)

Like? Yes (1) | No (1)  

Gaurav shah   10 years AGO

sp=100rs let
hence mp will be 42 rs and cp will be 40 rs
now for dis of 20% and 30% the sp=56 rs and cp=40rs. so profit=40%

Like? Yes (1) | No (1)  

Sukanya   10 years AGO

30 % + 40%

30%= 30
40% of 70=28
So 42 .
If 42=(100+5)% then 40 is the Cost Price.


20% + 30 %

100-20=80
30% 0f 80=24
80-24=56

56 =140%

Like? Yes (1) | No (1)  

mnsai   10 years AGO

initially ASSUME printed price 100
after giving 20% 80 it becomes then 30% on it is 56 is selling price
but due to to banner mistake after giving 30 it becomes 70 after 40% on it becomes 42
so actual cost is 42-2.1 =39.9 rupees as 5% profit on actual cost
but he sells at 56 means 39.9 he decide to earn 40 % profit

Like? Yes (1) | No (1)  

Md.Sultan Khichi   10 years AGO

Let the printed price is 100.
He wants to give discount of 20%+30%..
So there fore sp=56..
But printed discount is 30%+40%
So the actual dp=42..
Now let the actual cl is x..
So he get 5% profit..
5%*x+x=42
X=40..
Now if he give the discount of 20%+30% then his profit % =(56-40)/40*100=40%
Ans..40%

Like? Yes (1) | No (1)  

Deb    10 years AGO

If the printed price is x
Then selling price would be (70*x/100)*(80/100).
But due to misprint,It is now (60*x/100)*(70/100)=42*x/100.
As he gained 5% of actual cost...
so ,the actual cost is (42*x/100)*(100/105)=2*x/5.
If if he sells it in the old format(20%+30%)
he will gain (56*x/100)-(2x/5)=16*x/100
now the gain is (16*x/100)/(2*x/5)*100 %=40%

Like? Yes (1) | No (1)  

deepesh singh panwar   10 years AGO

p=16/40*100

Like? Yes (1) | No (1)  

Yogesh   10 years AGO

Like? Yes (1) | No (1)  

Shashi nath jha   10 years AGO

Ans= 40

Like? Yes (1) | No (2)  

Kumar Vishal   10 years AGO

Let us assume a total of Rs 100.Then 30%+40% corresponds to Rs 42,and if makes a gain of 5% then c.p=40.Now 20%+30% means 56,i.e a gain of rs 16 which is 40% of c.p Rs40.

Like? Yes (1) | No (2)  

26 Aug 2014 Probability

What will be the probability that in a month of 30 days there are 5 Monday.

OPtion

1) 1/8
2) 2/3
3) 1/4
4) 2/9
5) 1/5
6) 1/6
7) 3/7
8) 2/7
9) 2/5
10)None of these

Solution

30 days = 4 week + 2 days

2 days can be taken as SUN + MON and MON + TUE for getting Monday, and there are total 7 possibilities so
Probability = 2/7

Option 8)

Correct Option: 8 Best Solution (112)

BHANU KUMAR   10 years AGO

in a month of 30 days =30/7=4 weeks and 2 days.
now our sample spaces are (sun mon),(mon tue),(tue wed),(wed thu),(thu fri),(fri sat),(sat sun)
event is of having monday are (sun mon),(mon tue)

probability=n(e)/n(s)=2/7(OPTION 8)

Like? Yes (25) | No (8)  

vivek   10 years AGO

for 28 days there are 4 monday.
so remaing 2 day will be favorable if (sun,mon) ,(mon,tues) occurrs in 2 ways
total day in week=7
so required probability= 2/7

Like? Yes (6) | No (1)  

ATTARU NASIMUNNI   10 years AGO

IF THE MONTH STARTS WITH MONDAY OR SUNDAY THERE WILL BE 5 MONDAYS IN A MONTH OF 30 DAYS . .
SO 1/7 + 1/7 = 2/7

Like? Yes (7) | No (3)  

jagdish sharma   10 years AGO

in 30 days 4 weeks complete and 2 days
7 casas for two days
(s,m),(m,t),(t,w),(w,t),(t,f),(f,s),(s,s)
in two casa monday will come
so probability=2/7

Like? Yes (5) | No (1)  

ashok   10 years AGO

7*4=28days
remaining days=2
probability is 2/7

Like? Yes (4) | No   

kundan kumar   10 years AGO

in a month 28 days of complete week and 2 day extra
so
2/7

Like? Yes (4) | No (1)  

Samrat Adhikari   10 years AGO

To have 5 any days (Sun,Mon,...,Sat) in a month of 30 days, thr probability is 1/7.
As there will be maximum 2 Mondays in that month, the answer is 2/7.

Like? Yes (3) | No   

Jeevanantham.V.M   10 years AGO

Total no of day in a week is 7 and if a 30 days month have 5 days then month's first day should be sunday or monday so 2/7 is answer.

Like? Yes (3) | No   

Rajat sadhu   10 years AGO

out of 30days 28 days contains 4 sun-days and the probability that in the remaining 2 days there is a sunday=2/7

Like? Yes (3) | No (1)  

Tapash Dutta   10 years AGO

we may have two conditions that is Sun+Mon or Mon+Tue
so probability =2/7

Like? Yes (2) | No   

anandhakumar   10 years AGO

to get 5 monday in a month,the month should start with monday .
we can get 5 monday only when first two days are monday,
so probability is 2/7
option 8

Like? Yes (3) | No (1)  

sudhakar   10 years AGO

The any one week having 2 mondays

Then only we get 5 mondays in 30 days

Probability is 2/7

Like? Yes (3) | No (1)  

selvakumar   10 years AGO

The answer is 2/7

Like? Yes (3) | No (1)  

kumar aryan   10 years AGO

in a week of 7 days on 2days it occurs 5 times

Like? Yes (2) | No   

Rahul Sah   10 years AGO

Max no. of weeks in a month=4 (28 days)
Remaining (30-28) 2 days can be either of the following combination :
1)Mon-Tue
2)Tue-Wed
3)Wed-Thurs
4)Thurs-Fri
5)Fri-Sat
6)Sat-Sun
7)Sun-Mon
Out of these 7 combination there are 2 such combination containing a Monday
Therefore the no of ways of selecting these 2 days out of the the 7 possible combination=2/7
Answer-option 8)

Like? Yes (3) | No (1)  

Dharampal   10 years AGO

30/7=2days remaining.....in that 2days...(mon,tue),(tue,wed),(wed,thurs),(thurs,fri),(fri,sat),(sat,sun),(sun,mon)....so in these 7pair there are two mondays ....n(p)=2/7

Like? Yes (4) | No (2)  

shruthi   10 years AGO

30 days has 4 weeks + 2 odd days. So we have 4 mondays and extra one monday can be obtained in 2 ways
S={ (s,m) (m,t) (t,w) (w,th) (th,fr) (fr,sat) (sat,s)}

Like? Yes (2) | No   

sumit   10 years AGO

sample spaces are:
(sun,mon),(mon,tue),(tue,wed),(wed,thur),(thur,fri),(fri,sat),(sat,sun)
so required probability=2/7

Like? Yes (2) | No   

MP    10 years AGO

It is possible only if 1st of the month is falling on Sunday or Monday out of 7 days of week.
so probability = 2/7

Like? Yes (2) | No   

vartika    10 years AGO

there will be 5 mondays if the remaining 2 days be sun mon or mon tue, then probability= 2/7

Like? Yes (2) | No   

GAURAV CHOUDHARY   10 years AGO

P(e) = n(e)/t(e)
= 2/7
Where n(e) = no of event occurred
t(e) = total no of events

Like? Yes (2) | No   

santu ghosh   10 years AGO

30-28=2 days remaining
so the probability that in a month of 30 days there are 5 Monday is P(E)=2/7

Like? Yes (3) | No (1)  

PURIDI MOUNIKA   10 years AGO

in 30 days normally we get minimum 4 mondays, because 30/7 gives coefficient as 4 and remainder=2.
in order to get 5th monday we have remaining two days.
for that probability is =(6+6)/(7*6)=2/7.

Like? Yes (4) | No (3)  

K Sudhakar   10 years AGO

In month of 30 days, there will be 4 weeks(mon,tue,etc).. and we are left with 2 days which can be {sun,mon},{mon,tue},{tue,wed},{wed,thu},{thu,fri},{fri,sat}, and {sat,sun}..

therefore the probality of 5 mondays in a month will be 2/7

Like? Yes (1) | No   

rohit sakhi   10 years AGO

It is possibal only if either 1st or 2 nd will be sunday so 2/7

Like? Yes (1) | No   

himanshu priyadarshi   10 years AGO

only 2 ways when monday comes 5 times in a month i.e.
1 8 15 22 29
2 9 16 23 30
and total no. of ways of monday in month=7
probability= 2/7

Like? Yes (1) | No   

chanish agarwal   10 years AGO

there are atleast 4 monday in a month 7*4=28
remaining days 30-28=2
so possibility of getting another monday is only 2 times either it could be (sun , mon) or (mon, tuesday)
and total no. of days are 7 in a week
so the probability is 2/7

Like? Yes (1) | No   

prem kumar   10 years AGO

in 30 days to obtain 5 Mondays
the day should start with Monday or Sunday
so probability 2/7

Like? Yes (1) | No   

deepak khurana   10 years AGO

1/7+1/7

Like? Yes (1) | No   

vishal   10 years AGO

7*4=28 remaining 2 days are may be mon tues or sunday monday

Like? Yes (1) | No   

kajol nigam   10 years AGO

30-28=2
possibilities are: SM MT TW WT TF FS SS
P=2/7

Like? Yes (1) | No   

Ravi Bhalani   10 years AGO

30/7=2(remainder)
Remaining two days may be any of the following.
sunday & monday
monday & tuesday
tuesday & wednesday
wednsday & thursday
thursday & friday
friday & saturday
saturday & sunday
For having 5 mondays in a month one of the remaining two days must be monday.
So,2/7

Like? Yes (1) | No   

Ranjan kumar   10 years AGO

for this the remaining 2 days of week(7 days) must have a monday

Like? Yes (1) | No   

HARDIP BHALANI   10 years AGO

30/7=2(remainder)
Remaining two days may be any of the following.
sunday & monday
monday & tuesday
tuesday & wednesday
wednsday & thursday
thursday & friday
friday & saturday
saturday & sunday
For having 5 mondays in a month one of the remaining two days must be monday.
So,2/7

Like? Yes (1) | No   

Harshad    10 years AGO

30/7=2(remainder)
Remaining two days may be any of the following.
sunday & monday
monday & tuesday
tuesday & wednesday
wednsday & thursday
thursday & friday
friday & saturday
saturday & sunday
For having 5 mondays in a month one of the remaining two days must be monday.
So,2/7

Like? Yes (1) | No   

shahsanka   10 years AGO

4weeks =2 days 2*(1/7)=2/7

Like? Yes (1) | No   

GAURAV BOHRA   10 years AGO

THAT WILL BE IN ONLY TWO CASE WHEN 1ST OF THE MONTH IS EITHER ON SUNDAY OR ON MONDAY ,FROM THE TOTAL OF SEVEN DAYS.
TOTAL NO OF FAVOURABLE OUTCOMES=2
TOTAL NO OF POSSIBLE OUTCOMES=7
HENCE THE PROBABILITY=2/7

Like? Yes (1) | No   

PAWAN KUMAR   10 years AGO

In a month of 30 days, 28 days are for 4 weeks, so we are left with 2 extra days.
Also, the total no. of different days of a week is 7.
Therefore, probability = 2/7

Like? Yes (1) | No   

Imran Mohammed   10 years AGO

obviously 28 days contains 4 mondays..
now we'll be left 2 days and the combinations can be (sun,mon),(mon,tue),(tue,wed),(wed,thu),(thu,fri),(fri,sat),(sat,sun)
among the 7 combintaions 2 pairs contain monday. hence the probability is 2/7

Like? Yes (1) | No   

SAYAN CHATTERJEE   10 years AGO

in 28 days each of the 7 days will recur 4 times.
the remaining two days may be any of the 7 days.
so, probability of 5 Mondays in the week is 2/7. (8)

Like? Yes (1) | No   

bathula praveena   10 years AGO

Total 30days lo 4 mndays cnfrm and remaining 2dayslo the possibility of being mnday is2/7

Like? Yes (1) | No   

SUNITEE GARIMA   10 years AGO

Sample Space=(Sunday,monday);(monday,tuesday);(tuesday,wednesday);wednesday,thrusday);(Thrusday,Friday),(Friday,Saturday);(Saturday,Sunday)
n(S)=7
n(E)=2
P=2/7

Like? Yes (1) | No   

sanjeev kumar   10 years AGO

30 day= (7*4)+2 day
the probability in last 2 day is 2/7

Like? Yes (1) | No   

Rashmi bharti   10 years AGO

2/7 is the right answer.

Like? Yes (1) | No   

HERSAL   10 years AGO

sample space=7
no of event=2
p=2/7

Like? Yes (1) | No   

Pushpendra singh chundawat   10 years AGO

after 4 weeks , 2 days remaining in a month
so-2/7

Like? Yes (1) | No   

sandeep singh   10 years AGO

since 30-(4*7)=2
so this 2 will come out of 7 days
so probability of being monday is 2/7

Like? Yes (1) | No   

Deb Das   10 years AGO

Sunday&monday
monday & tuesday

so 2/7

Like? Yes (1) | No   

Saravjeet Singh   10 years AGO

1st is monday or 2nd is monday then there are 5 monday..
thus 2 days out of i week
2/7

Like? Yes (1) | No   

MRITUNJAY KUMAR   10 years AGO

option 8) 2/7.

Like? Yes (1) | No   

ANAND KUMAR   10 years AGO

4 monday are fixed in 1st 28 days and for next tdays probability=2/7

Like? Yes (1) | No   

suman kumar   10 years AGO

30days ....
in a week-7 days
4week -28 days-4 monday
so, in rest 2 days probability of monday =2/7.

Like? Yes (1) | No   

Vinoth Kumar   10 years AGO

If a month of 30 days have 5 mondays then either the first day or the second day of the month should be a monday.Since a week has 7 days and there are two possibilites for the event to occur;the probability is 2/7.

Like? Yes (1) | No   

Vanika   10 years AGO

there will be 4 mondays in 28 days
so probability will be = 2/7

Like? Yes (1) | No   

Jasmita Parmar   10 years AGO

Obviously there will be 4 Mondays in the month of 30 days because 7 * 4 = 28 .
Here 30 - 28 = 2 days remaining

The remaining 2 days can have anyone combination out of this.

sun-mon , mon-tue , tue-wed ,wed-thu , thu-fri , fri-sat , sat-sun

so there are total 7 combinations and out of this 2 combinations have Monday in it.

so probability of having 5 Monday in the month of 30 days is P(E) = 2 / 7

Like? Yes (1) | No   

raj ranjan verma   10 years AGO

last 2 days after 4 weeks can be
sun-mon,
mon-tue,
tue-wed,
wed-thur,
thur-fri,
fri-sat,
sat-sun
out of these mon occurs 2 times out of 7
so, prob. of occuring mon 5 times will be 2/7

Like? Yes (1) | No   

Rahul Rathod   10 years AGO

There are 30 days.
so 4 monday in 28 days.
and remianing days=2
so probability=2/7

Like? Yes (1) | No   

Stephen Reji   10 years AGO

There are two ways in which Monday arrives.
(Sunday,Monday),(Monday,Tuesday)

Like? Yes (1) | No   

Rajat Rokade   10 years AGO

there are 7 possibilities of a perticular day to come 5 times and twice it can contain monday.

Like? Yes (1) | No   

Tark Patel   10 years AGO

In 30 days month 5 monday will occur only on if 1st date of month is starting from sunday or monday.
So these are only 2 possibilities and total 7 days
Hence probability of 5 monday in 30 days month is 2/7 .

Like? Yes (1) | No   

ashok   10 years AGO

30=7*4+2
so at-least 7 times all every days will come
remaining 2 days
there are 7 possible combinations of 2days-
1.sun-MON
2.MON-tue
3.tue-wed
4.wed-thu
5.thu-fri
6.fri-sat
7.sat-sun

so there are 2 cases out of all possible 7 where monday will come

so probability=2/7

Like? Yes (1) | No   

Ajay Kumar Gupta   10 years AGO

any 30 day month have 4 mondays i.e 28 days are full rest 2 days are vacent . Therefore the probability is 2/7.

Like? Yes (1) | No   

swarup kumar prusti   10 years AGO

there are 2 cases when monday start from 1 0r 2.

Like? Yes (1) | No   

Urvashi Jain   10 years AGO

assume total days in a month is 30
4 mondays are sure in 28 days means 4 weeks
30-28=2
so the possible monday are 2/7

Like? Yes (1) | No   

Dipak Singh   10 years AGO

28 days for normal 4 mondays.
rest 2 day (30-28=2) to have a probabilty to be a monday is=2/7
hence, 2/7 is the answer.

Like? Yes (1) | No   

MANU   10 years AGO

in a 30(4 week) days month 2 days are remaining so P=2/7

Like? Yes (1) | No   

Enukonda Ramakrishna   10 years AGO

(1*6c1)/7c2

Like? Yes (1) | No   

Proloy Mukherjee   10 years AGO

30%7=2
2/7

Like? Yes (1) | No   

RANJAN KUMAR PANDIT   10 years AGO

because 30-7*4=2 daye remaining
so probability = 2/7

Like? Yes (1) | No   

Tania Chatterjee   10 years AGO

1-8-15-22-29
or
2-9-16-23-30
So, only 2 times it can happen out of 7 days i.e.P(M)=2/7

Like? Yes (1) | No   

yagnik pandya   10 years AGO

Only in Sunday n monday no. Of monday is 5. Otherwise 4.
2/7

Like? Yes (1) | No   

Harsh Poddar   10 years AGO

30 days
4 mondays confirm of 28 days
next 2 days outcome=
mon-tue
tue-wed
wed-thu
thu-fri
fri-sat
sat-sun
sun-mon
2 mondays possible outcome of 7 total outcome
probablity=2/7
option 8

Like? Yes (1) | No (1)  

RAKESH   10 years AGO

30 = 4*7 + 2

if month starts from Sunday or Monday then there are 5 Monday
p = 2/7

Like? Yes (2) | No (2)  

chandan sharma   10 years AGO

there will be 4 weeks consisting of 28 days and 4 mondays.of the remaining 2 days we can have 2 combination (sun-mon and mon-tue) out of total of seven combination.therefore probability will be 2/7.

Like? Yes (3) | No (3)  

krishna kanth   10 years AGO

for 28 days we get 4 mondays
next 2 days may be (sun,mon) or (mon,tue) among 7 possibilities
so probability is 2/7

Like? Yes (1) | No (1)  

Shashi nath jha   10 years AGO

2|7 ans

Like? Yes (1) | No (1)  

DEEPAK SINGHAL   10 years AGO

BECOZ IN A A MONTH ON 30 DAYS EVERY DAY COMES 4 TIMES NECESSARY BUT THE PROBABLITY OF 5 MAONDAY .. THERE ARE TWO DAYS REMAINING AND TOTAL DAYS 7
SO PROB=2/7

Like? Yes (1) | No (1)  

KAUSHAL KUMAR   10 years AGO

Probability= no of days left/no of days un a week
=2/7

Like? Yes (1) | No (1)  

Md.Sultan Khichi   10 years AGO

1st and 2nd date of the month should v started at Mon..for 5 Mon in the month..there is only 2 possible way..total no. of days are 7..so the probability is
2/7..
Ans..2/7..

Like? Yes (1) | No (1)  

S.SHAFI   10 years AGO

For 4 weeks of 28 days, 4 mondays will come..
Remaining 2 days, 1 monday will have the probability of 2/7.

So, the option is 8..

Like? Yes (1) | No (1)  

Aakash Jain   10 years AGO

only two possibility if starting day of month is sunday or monday so ans is 2/7

Like? Yes (1) | No (1)  

Manikumar R   10 years AGO

the month will have 5 sundays if it started on saturday or sunday so the probability is 2/7

Like? Yes (1) | No (1)  

Mrityunjay   10 years AGO

we can have 5 monday in a month only if the month starts with monday or sunday, as in 4 weeks 4 mondays are for sure
so the probability becomes =no. of events/no. of days
=2/7

Like? Yes (1) | No (1)  

NAVEEN KUMAR   10 years AGO

30 days=4weeks+2days
so, in 4 week 4 monday come .
In last two days .
{sun,mon}or(mon,tue}or{tue,wed}or,{wed,thu}or,{thu.fri},or{fri,sat},or{sat,sun}
so, there are two possiblities to come monday in seven possiblities so probability=2/7

Like? Yes (3) | No (3)  

T.VIGNESWARA REDDY   10 years AGO

4 weeks completed in 28 days so 4 mondays is completed
So remaining 2 days have a chance of monday is 2 ways i.e sun,mon or mon tue

Like? Yes (1) | No (1)  

abhik k. dey   10 years AGO

as in a week there are 7 days.
so in 30 days of a month there are 4 full week+2 extra day.
the 2 extra day will be(sunday, monday),(monday,tuesday),(tuesday,wednesday),..... and so on..
so there are total 7 chance to come 2 day consecutive. but we have to find five monday. so 4 wk= 4 mon day and extra 1 will come from these seven probable pair.
so probability of 5 monday will be 2/7............................ans

Like? Yes (1) | No (1)  

Tapobrata Banerjee   10 years AGO

there will be 4 mondays in 28 days...so in remaining 2 days there can be following 7 combination of days,i.e. sun-mon,mon-tue,tue-wed,wed-thurs,thurs-fri,fri-sat,sat-sun.so,there are 7 combinations out of dem 2 can be monday.
so, the probability is 2/7
ANS 8

Like? Yes (1) | No (1)  

DIPAK TIWARI   10 years AGO

For 5th Monday the last two days(29th,30th) will have combinations of sautrday-sunday & sunday-monday from sautrday-sunday, sunday-monday, monday-tuesday,Tuesday- Wednesday,Wednesday- thrusday,thrusday-friday,friday-sautrday.
Hence probability=2/7
Option 8

Like? Yes (1) | No (1)  

Dark Knight   10 years AGO

1/7+1/7=2/14

Like? Yes (1) | No (1)  

Arijit Acharyya   10 years AGO

n(E)=2 {(sun,mon),(mon,tue)}
n(S)={(sun,mon),(mon,tue),(tue,wed),(wed,thu),(thu,fri),(fri,sat),(sat,sun)}
P(E)=n(E)/n(S)=2/7
option 8)

Like? Yes (1) | No (1)  

Anurag   10 years AGO

since there are 2 odd days in 30 days
the 2 days can be Sun Mon or Mon Tue or Tue Wed or Wed Thu or Thu Fri or Fri Sat or Sat Sun

thus the probability of occurrence of mon will be 2/7

Like? Yes (1) | No (1)  

Pranoy Bej   10 years AGO

for 4 mondays in a month of 30day probability will be 1
for the remaining 2days combination will be
sum-mon
mon-tue
tue-wed
wed-fri
fri-sat
sat-sun

so probability will be = 2/7

Like? Yes (1) | No (1)  

LOKESH SAMANT   10 years AGO

26 AUG 2014 : Probability

Like? Yes (1) | No (1)  

pushpa   10 years AGO

30 days means 4 weeks n 2 days
4 weeks means we hv 4 mondy already

n rest two days can b
1(sundy,mondy)
2(mondy,tusdy)
3(tusdy,wednsdy)
4(wednsdy,thursdy)
5(thursdy,fridy)
6(fridy,satrdy)
7(satrdy,sundy)

so for d 5th mondy
we hv 2 probablity out of 7
n answr is 2/7


so for 5 mondy in this week we

Like? Yes (1) | No (1)  

Shikha   10 years AGO

A month of 30 days contain 4 full weeks (7*4=28) and 2 extra days which may be(sun,mon) (mon,tue) (tue,wed).(wed,thu) (thu,fri) (fri,sat)(sat ,sun)
so,P(getting 5th monday)=2/7

Like? Yes (1) | No (1)  

teja srinivas   10 years AGO

there are 4 weeks 4 mondays will come there after there are 2 days left therefore probabailty is 2 in 7 days
i..e 2/7

Like? Yes (1) | No (1)  

sadaq vali   10 years AGO

30 days = 4 weeks 2 days

For 2 days chances to become Monday n(m)= 2

Probability p(m)= 2/7

Like? Yes (1) | No (1)  

Venkateswarlu   10 years AGO

If the month either start with monday , or sunday in the week,there will be 5 mondays.

====>>2/7

Like? Yes (1) | No (1)  

ADARSH GAURAV   10 years AGO

2/7 is the ans

Like? Yes (1) | No (1)  

Nitin Srivastava   10 years AGO

as only last 2 days ie 29 and 30 is responsilble for 5 monday
ie sunday and monday
monday and tuesday
and total number of combination of 2 days ar 7
so probability is 2/7

Like? Yes (2) | No (2)  

vikash kumar   10 years AGO

2/7 is the ans

Like? Yes (2) | No (2)  

MUKESH YADAV   10 years AGO

there will be surely 4 monday,4tue,4wed,4thu,4fri,4sat,4sun in 28 days...remaining 2 days and total days in a week is 7 then probability will be 2/7..

Like? Yes (1) | No (1)  

gurneet singh   10 years AGO

there are $weeks which complete one month+2 additional days...
these days can be sunday& Monday OR monday& tusesday..

Like? Yes (1) | No (1)  

Saroj kumar   10 years AGO

P=no of Mon/no of total days
P=2/7

Like? Yes (1) | No (1)  

Varunraj Reddy   10 years AGO

2 odd days in a month of 30 days.
So the probability getting 5 Monday is 2/7.

Option 8)

Like? Yes (1) | No (2)  

Utkarsh    10 years AGO

out of 30 days there are 4 complete weeks (28 days) having 4 mondays. So now left with 2 days
favourable outcomes are
- Sunday and Monday
- Monday and Tuesday

probability= fav. events/total event

here fav. event=2
total event=7
therefore, 2/7 must be the ans.

Like? Yes (2) | No (3)  

nisha kumari singh   10 years AGO

ina a month of 30 daysthere are total of 4 weeks except remaining 2 days.
(sunday ,monday),(monday,tuesday),(tuesday,wednesday),(wednesday,friday),(friday,saturday),(saturday,sunday).
for the month to be of 5 mondays ,these 2 days can be any of the first two combination.
therefore probability=2/7

Like? Yes (1) | No (2)  

suman kumar   10 years AGO

In a month of 30 days....4 mon is sure in 4 weeks....so we have 2 more days
if the month start with sunday, then it will contain 5 mon
and if month start with monday then also it will contain 5 mon,
so the probability becomes =no. of events/no. of days
=2/7

Like? Yes (1) | No (3)  

ARKAJIT NANDY   10 years AGO

30=7*4+2
so 4 complete weeks i.e. 4 Mondays and an incomplete week of 2 days.
in 7 days week the probability of the day will be Monday in those 2 days will be=2/7

Like? Yes (1) | No (3)  

tendulkar   10 years AGO

for having five mondays the moth should end up with either monday or tuesday
there are 7 days a week and the probabikity of getting two days out of 7 days is 2/7
hence the probability of getting 5 mondays a month is equal to probability of getting two days out of 7 days
therefore the answer is 2/7

Like? Yes (1) | No (3)  

posirajesh   10 years AGO

28 days 4 Monday
Remaining either Mon,Tue or sun,Mon
Hence 2/7 is probability

Like? Yes (1) | No (3)  

yatish kumar sharma   10 years AGO

problity will be 2/7

Like? Yes (1) | No (4)  

25 Aug 2014 Sequence

What is the next number of the following sequence
3, 14, 39, 84, ?

OPtion

OPTION
1) 98
2) 125
3) 150
4) 115
5) 190
6) 175
7) 200
8) 110
9) 155
10)None of these

Solution

1^3+1^2+1,
2^3+2^2+2,
3^3+3^2+3,
4^3+4^2+4,
5^3+5^2+5 = 155 Correct

Correct Option: 9 Best Solution (141)

Kulbeer Singh Sidhu   10 years AGO

3=1*3
14=2*7
39=3*13
84=4*21
?=5*(n)
7-3=4, 13-7=6, 21-13=8, n-21=10 =>n=31
5*31=155

Like? Yes (21) | No (10)  

prem kumar   10 years AGO

1*3=3
2*7=14
3*13=39
4*21=84
5*31=155

Like? Yes (9) | No   

nisha sharma   10 years AGO

as difference between multiple is in series 4,6,8
3*1=3
7*2=14
13*3=39
21*4=84
31*5=155

Like? Yes (10) | No (2)  

Niveditha Ravichandran   10 years AGO

n^3 + n^2 + n

n=1, 1+1+1=3
n=2, 8+4+2 =14
n=3, 27+9+3 =39
n=4, 64+16+4 =84
n=5, 125+25+5 =155

Like? Yes (6) | No   

sanjeev kumar   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84= 4 * 21

3,7,13,21...
7-3=4
13-7=6
21-13=8
_ -21=10
so next number 21+10=31
and 5*31=155

Like? Yes (6) | No   

suraj tiwary   10 years AGO

the general term of this series is n^3+n^2+n..so for n=5...the term will be..125+25+5=155

Like? Yes (4) | No   

Amit Kumar Pandit   10 years AGO

3*1=3
7*2=14
13*3=39
21*4=84
31*5=155

Like? Yes (4) | No   

posirajesh   10 years AGO

use nth number =n×[n*n+(n+1)]

Like? Yes (4) | No   

Venkateswarlu   10 years AGO

14-3=11
39-14=25 25-11=14
84=39=45 45-25=20 14,20,26,-,-,--
39=11+14+14=39
84=25+39+20
// 45+84+26=155

Like? Yes (3) | No   

ESHA CHATTERJEE   10 years AGO

1st no.=(1^2+2)*1=3
2nd no.=(2^2+3)*2=14
3rd no.=(3^2+4)*3=39
4th no=(4^2+5)*4=84
5th no=(5^2+6)*5=155

Like? Yes (3) | No   

Neel Patel   10 years AGO

3=1+1^2+1^3
14=2+2^2+2^3
36=3+3^2+3^3
84=4+4^2+4^3
so
5+5^2+5^3=155

Like? Yes (3) | No   

ashok   10 years AGO

s= 3, 14, 39, 84, ?
-s= 3, 14, 39, 84, A
-------------------------------------
s1=3, 11 25 45 B
-s1= 3, 11 25 45
---------------------------------------
s2=3, 8, 14, 20 C
-s2= 3, 8, 14 20
------------------------------------
s4=3, 6, 6, 6 D

So D=6
C=D+20=26
B=C+45=71
A=B+84=155

Ans=155

Like? Yes (4) | No (1)  

lebaka pavan   10 years AGO

Given numbers 3,14,39,84,?
Difference between given numbers are 11,25,45,?
Then the difference between above numbers are 14,20,?
Therefore,the difference is 6
So the next number for the above difference is 26
So the next number is 26+45+84=155

Like? Yes (3) | No   

Tanya Kumar   10 years AGO

basic series

Like? Yes (3) | No   

ravikumar   10 years AGO

let us take the series in quadratic form
an^2 + bn +c
replace a=1 n=1 1(1)+1(1)+1=3
a=2 n=2 2(4)+2(2)+2=14
similarly
a=4 n=4 4(16)+4(4)+4=84
when a=5 n=5 5(25)+5(5)+5= 155

hence 155 is correct

Like? Yes (3) | No   

Sourojit Dhua   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

answer = option 9, 155

Like? Yes (8) | No (6)  

sumit   10 years AGO

series is n^3+n*(n+1)
1^3+1*(1+1)=3
2^3+2*(2+1)=14
so the 5th term will be
5^3+5*(5+1)=155 opt(9) ans

Like? Yes (7) | No (5)  

SAYAN CHATTERJEE   10 years AGO

3+11=14
14+25=39
39+45=84
differences are 11, (11+14=25), (25+20=45)
so, next difference is (45+26=71).
so, next no. is (45+71=155) (9)

Like? Yes (2) | No   

urvish   10 years AGO

here 3 = 1 * 3,14 = 2 * 7,39 = 3 * 13,84 = 4 * 21.....
T(n) = n[(n+1)^2 - n]
so next n=5
=5*((5+1)^2-5)
=5*(36-5)
=5*31
=155

Like? Yes (2) | No   

mohana lakshmi v   10 years AGO

by adding sequence no

Like? Yes (2) | No   

palasiva   10 years AGO

t(n)=n(n^2+n+1)
t(5)=5(5^2+5+1)=5(25+5+1)=155

Like? Yes (2) | No   

akash kumar   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (2) | No   

jitendra varshney   10 years AGO

t(n)=n^3 + n^2 + n
we have to find 5th term so put the value n=5
t(5)=5^3+5^2+5
=125+50+5
=155 ans..

Like? Yes (2) | No   

NAVEEN KUMAR   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n
Put n=5
T(5)=125+25+5

Like? Yes (3) | No (1)  

chandra mondal   10 years AGO

3*1=3
(3+4)*2=14
13*3=39
21*4=84
31*5=155

Like? Yes (2) | No   

Soham Chakraborty   10 years AGO

(3*1),(7*2), (13*3),(21*4),(31*5)

Like? Yes (2) | No   

chandanailupuru   10 years AGO

3*1,7*2,13*3,21*4=3,14,39,84
7-3=4,13-7=6,21-13=8
Next number is 21+10=31
31*5=155
Ans 155

Like? Yes (2) | No   

Partha Pratim Mahata   10 years AGO

3=1*[(1+1)^2-1]
14=2*[(2+1)^2-2]
39=3*[(3+1)^2-3]
...
..
155=5*[(5+1)^2-5]=5*31

Like? Yes (2) | No   

santu ghosh   10 years AGO

u(n)=n^3+n^2+n,n=1,2,3,4,5.....
u(1)=1^3+1^2+1=3
u(2)=2^3+2^2+2=14
.........................................
.....................................
u(5)=5^3+5^2+5=155

Like? Yes (2) | No   

madhurima   10 years AGO

1st no is:(1^2+2)*1
2nd no is:(2^2+3)*2
3rd no is:(3^2+4)*3
4th no is:(4^2+5)*4
so the 5th no will be:(5^2+6)*5=155
so,option 9 is correct

Like? Yes (2) | No   

D.HARINATH REDDY   10 years AGO

1^3+2=3
2^3+6=14
3^3+12=39
4^3+20=84
5^3+30=155

Like? Yes (2) | No   

G.Jaya Krishna   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
155=5*31

Like? Yes (2) | No   

Harsh Poddar   10 years AGO

1^3+1^2+1=3
2^3+2^2+2=14
3^3+3^2+3=39
4^3+4^2+4=84
next number=
5^3+5^2+5=155

Like? Yes (2) | No   

Chirag Ranjan Lahiri   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (2) | No   

deekshith   10 years AGO

it is the answer coz check out solution contact me @deekshith.nani@gmail.com

Like? Yes (2) | No   

Chiranjit Agarwal   10 years AGO

(3*1) (7*2) (13*3) (21*4) (31*5)
the difference between 3,7,13,21,31 is of the order 4,6,8,10 respectively

Like? Yes (3) | No (1)  

suman kumar   10 years AGO

differences are 11,25,45;
differences b\w them are 14,20 so next time it will be26.....
so adding it to 45 we get 71
since 71 is difference thus adding it to 84 we get 155.

Like? Yes (2) | No   

ravi shah   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No   

vidhijain   10 years AGO

Differnce of differnce is in a.p.

Like? Yes (1) | No   

vishva shikha   10 years AGO

3*1=3
7*2=14 (7-3)=4
13*3=39 (13-7)=6
21*4=84 (21-13)=8
21+10=31
31*5=155
hence answer=155

Like? Yes (1) | No   

Janaga mageshkumar.v   10 years AGO

diff bt two values 14-3,39-14,84-39 are in the order of 11 25 45 and diff bt these numbers are 14 20 so then diff as 26. so add 26+45=71 then 71+84=155

Like? Yes (1) | No   

prabhat kumar   10 years AGO

1*3, 2*7, 3*13, 4*21, 5*31

Like? Yes (1) | No   

potani manikumar   10 years AGO

3= 1*(1+1)^2 -1
.
.
.
5*(5+1)^2 -5

Like? Yes (1) | No   

Tark Patel   10 years AGO

1 ( 1^3 + 1^2 + 1) ,
14 ( 2^3 + 2^2 + 2) ,
39 ( 3^3 + 3^2 + 3) ,
84 ( 4^3 + 4^2 + 4) ,

so next number is
5^3 + 5^2 + 5 = 155

Like? Yes (1) | No   

vikash kumar   10 years AGO

3=1^3+1^2+1
14=2^3+2^2+2
39=3^3+3^2+3
84=4^3+4^2+4
155=5^3+5^2+5

Like? Yes (1) | No   

ADARSH GAURAV   10 years AGO

3=1^3+1^2+1^1
14=2^3+2^2+2^1
39=3^3+3^2+3^1
84=4^3+4^2+4^1
155=5^3+5^2+5^1

Like? Yes (1) | No   

Abhishek Mitra   10 years AGO

I think. It’s a sequence, the formula of which is for any number ‘n’, (n-1)3+(n-1)2+(n-1).

Like? Yes (1) | No   

Gaurav shah   10 years AGO

3*1,7*2,13*3,21*4,31*5

Like? Yes (1) | No   

SUNITEE GARIMA   10 years AGO

1^3+2=3
2^3+6=14
3^+12=39
4^3+20=84
5^3+10=155
so the next number is 155

Like? Yes (1) | No   

akshay   10 years AGO

5*31

Like? Yes (1) | No   

abhishek   10 years AGO

(1^3)+(1*2)=3
(2^3)+(2*3)=14
(3^3)+(3*4)=39
(4^3)+(4*5)=84
then
(5^3)+(5*6)=155

Like? Yes (1) | No   

Aravinth kumar   10 years AGO

a(n) = n^3 + n^2 + n
a(1)=1^3 + 1^2 + 1 = 3
... a(5)=5^3 + 5^2 +5 = 155

Like? Yes (1) | No   

sruthi   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No   

NIRUPOM GHOSH   10 years AGO

next no 155

Like? Yes (1) | No   

Enukonda Ramakrishna   10 years AGO

3, 14, 39, 84, x=155
11 25 45 71
14 20 26
6 6

Like? Yes (1) | No   

ABHIJIT DUTTA   10 years AGO

1*3 = 3
2*7 = 14
3 *13 = 39
4*21 = 84
similarly, 5*31 = 155

option 9)

Like? Yes (1) | No   

Tapobrata Banerjee   10 years AGO

we can get it using back triangle method.i.e. after finding the differences of given numbers we can get the series as 11,25,45 and the unknown no...then again 14,20 and the no..like this....finally 6 will be the common difference and by back calculation we can get the no. 155

Like? Yes (1) | No   

Ravi Bhalani   10 years AGO

5^3 + 5^2 +5=155

Like? Yes (1) | No   

HARDIP BHALANI   10 years AGO

5^3 + 5^2 +5=155

Like? Yes (1) | No   

Harshad    10 years AGO

5^3 + 5^2 +5=155

Like? Yes (1) | No   

Varunraj Reddy   10 years AGO

n^3+n^2+n

Like? Yes (1) | No   

Ankit Raj   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No   

shaurya neel   10 years AGO

n

Like? Yes (1) | No   

Sachin Bhakt   10 years AGO

(n-1)3+(n-1)2+(n-1).
n= nth number of term

Like? Yes (1) | No   

Sai Indu   10 years AGO

3*1=3
7*2=14
13*3=39
21*4=84
31*5=155
because the difference between 3 and 7 is 4,7 and 13 is 6,13 and 21 is 8,21 and 31 is 10...
as the difference order is 4,6,8,10.....so the answer is 155

Like? Yes (1) | No   

Rahul shaw   10 years AGO

By finding common differences...

Like? Yes (1) | No   

jagriti goomer   10 years AGO

general term of series:
(n)^3 + (n)^2 + (n)^1

putting n=5 for 5th term v get:
125+25+5=155

Like? Yes (1) | No   

latika   10 years AGO

3*1+0, 3*3+5,3*7+18,3*13+45,3*21+92

Like? Yes (1) | No   

shruthi   10 years AGO

(3+4)*2=14
(7+6)*3=39
(13+8)*4=84
(21+10)*5=155

Like? Yes (1) | No   

surya   10 years AGO

n^3+n^2+n

Like? Yes (1) | No   

gouthamdaneti   10 years AGO

n((n+1)^2-n) n denotes the position of number

Like? Yes (1) | No   

Hari Nani   10 years AGO

n((n+1)^2-n) n denotes the position of number

Like? Yes (1) | No   

Jasmita Parmar   10 years AGO

3 = 3 * 1
14 = 7 * 2
39= 13 * 3
84= 21 * 4
155= 31 * 5

Like? Yes (1) | No   

RAKESH KUMAR RANJAN   10 years AGO

1^3+2=3
2^3+6=14
3^3+12=39
4^3+20=84
5^3+30=155

Like? Yes (1) | No   

Niraj   10 years AGO

T(n) = n^3 + n^2 + n
So T(5) = 5(25+5+1) = 5*31 = 155

Like? Yes (1) | No   

ravi chandra singh   10 years AGO

3=1*3
14=2*7
39=3*13
84=4*21
155=5*31

Like? Yes (1) | No   

Shubhangi   10 years AGO

number-2*2+11

Like? Yes (1) | No   

IMRAN KHAN   10 years AGO

3*1+0
3*3+2+3
3*7+4+14

Like? Yes (1) | No   

Ajay Kumar Gupta   10 years AGO

(2-1)^3+(2-1^)2+(2-1)= 1 + 1 + 1 = 3
(3-1)^3+(3-1)^2+(3-1) = 8 + 4 + 2 = 14
continuing
(5-1)^3+(5-1)^2+(5-1)=64+16+4=84
then-(6-1)^3+(6-1)^2+(6-1)=125+25+5=155 ans.

Like? Yes (1) | No   

pavankumar   10 years AGO

choose ‘n’= (n-1)3+(n-1)2+(n-1).
choose n as 1
choose n as 2
choose n as 7

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RAJEEV KUMAR   10 years AGO

5*31=155

Like? Yes (1) | No   

Satya Karteek   10 years AGO

3= 1^3+1^2+1
14=2^3+2^2+2
similarly,
5^3+5^2+5=125+25+5=155

Like? Yes (1) | No   

vandna   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
155=5*31

Like? Yes (1) | No   

Ramu Giri   10 years AGO

n^3+n^2+n

where 0

Like? Yes (1) | No   

shree   10 years AGO

155.......

Like? Yes (1) | No   

shiv pujan   10 years AGO

sequence is n^3+n^2+n
putting n=5
ans is 155

Like? Yes (1) | No   

prakash kumar   10 years AGO

1^3+2=3,2^3+6=14,3^3+12=39,4^3+20=84,
5^3+30=155

Like? Yes (1) | No   

KRISHAN MURARI   10 years AGO

use n^3+n^2+n
put n=1,2,3........

Like? Yes (1) | No   

Mahak   10 years AGO

2*1+1 ,3*4+2 ,4*9+3, 5*16+4,
i.e.
2*(1)power2+1
3*(2)power2+2
4*(3)power2+3
5*(4)power2+4
so next number will be
6*(5)power2+5=155

Like? Yes (1) | No   

sudipta maji   10 years AGO

n3+n2+n=no
5^3+5^2+5=155

Like? Yes (1) | No (1)  

mantena phanindra v varma   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
155=5*31

Like? Yes (1) | No (1)  

Pushpendra singh chundawat   10 years AGO

(1^3+1^2+1)=3
(2^3+2^2+2)=14
(3^3+3^2+3)=39
(4^3+4^2+4)=84
(5^3+5^2+5)=155

opt-9

Like? Yes (1) | No (1)  

chandan jha   10 years AGO

(3+4)*2=14
(7+6)*3 =39
(13+8)*4=84
(21+10)*5=155..

Like? Yes (1) | No (1)  

ANKITA prasad   10 years AGO

1^3+2=3
2^3+6=14
3^3+12=39
4^3+20=84
5^3+30=155.
the number being added are increasing in multiples of 2.

Like? Yes (1) | No (1)  

Inline Geek   10 years AGO

Look at the series:

(1+1) * (1+1) - 1 = 3
(2+1) * (2+3) - 1 =14
(3+1) * (5+5) - 1 =39
(4+1) * (10+7) - 1 =84

so

(5+1) * (17+9) - 1 =155
is the answer

Like? Yes (1) | No (1)  

KRISHNA CHANDRA UPADHYAY   10 years AGO

Use expression n^3+n^2+n
and put n=1,2,3,4 and 5
after putting n=5 in above expression we get 155
HENCE.OPTION (9) IS CORRECT.

Like? Yes (1) | No (1)  

Dipak Singh   10 years AGO

webbed A.P
by caculating it comes 155

Like? Yes (1) | No (1)  

manish kumar singh   10 years AGO

1*3 2*7 3*13 4*21 5*31 6*43........

Like? Yes (1) | No (1)  

yatish kumar sharma   10 years AGO

12*12+11

Like? Yes (1) | No (1)  

Siva Prakash   10 years AGO

T(n) = n * (n^2 + n + 1)

1.) T(1) = 1 * (1^2 + 1 + 1) = 1 * 3 = 3

2.) T(2) = 2 * (2^2 + 2 + 1) = 2 * 7 = 14

3.) T(3) = 3 * (3^2 + 3 + 1) = 3 * 13 = 39

4.) T(4) = 4 * (4^2 + 4 + 1) = 4 * 21 = 84

5.) T(5) = 5 * (5^2 + 5 + 1) = 5 * 31 =155

Like? Yes (1) | No (1)  

Yogesh   10 years AGO

3*1=3
7*2=14
13*3=39
21*4=84
31*5=155

Like? Yes (1) | No (1)  

vishal   10 years AGO

1^3 +2 , 2^3 +6 , 3^3+12, 4^3 + 20 so ans is 5^3 +30

Like? Yes (1) | No (1)  

kajol nigam   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

13=(3^2+3+1)
21= (4^2 + 4 + 1)

for 5th = (25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (3) | No (3)  

Mogilipuri.Harish   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
here n=5 then
T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (1)  

rahul anand   10 years AGO

5*31

Like? Yes (1) | No (1)  

vinusha reddy   10 years AGO

1+1^2+1^3=3
2+2^2+2^3=14
i.e, t(n)=n+n^2+n^3
5+5^2+5^3=5+25+125=155

Like? Yes (1) | No (1)  

suman kumar   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (1)  

Mrityunjay   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (1)  

Nitin Srivastava   10 years AGO

let x be next number
take the consecutive difference
step 1: 11 25 45 x-84
step 2: 14 20 x-129
step 3: 6 x-149
let next step difference be 0
x-149-6=0
x=155

Like? Yes (1) | No (1)  

Amritha   10 years AGO

3=1*3
14=2*7
34=3*13
84=4*21
the difference between 3 and 7 is 4, 7 and 13 is 6, 13 and 21 is 8 hence next difference must be 10.
So 5 * (21+10)= 155

Like? Yes (1) | No (1)  

raju manpreet singh   10 years AGO

2*1+1=3
4*3+2=14
6*6+3=39

8*10+4=84
10*15+5=155

Like? Yes (1) | No (1)  

Sneha   10 years AGO

3=1*3
14=2*7
39=3*13
84=4*21
155=5*31
So option 9

Like? Yes (1) | No (1)  

Dr.Dipin Singh   10 years AGO

nth term = n^3 + n^2 + n
5th term = 5^3+5^2+5= 125+25+5=155

Like? Yes (1) | No (1)  

TINKU MITTAL   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (1)  

Manikumar R   10 years AGO

n^3 + n^2 + n
1^3 + 1^2 + 1=3
2^3+2^2+2=14
.....5^3+5^2+5=155

Like? Yes (1) | No (1)  

selvakumar   10 years AGO

answer option 9)155

Like? Yes (1) | No (1)  

chirag shrimali   10 years AGO

N^3+N^2+N

Like? Yes (1) | No (1)  

Ramya   10 years AGO

1^3+1^2+1=3
2^3+2^2+2=14
.
.
5^3+5^2+5=155

Like? Yes (1) | No (1)  

vaishnavi   10 years AGO

3=1*3
14=2*7
39=3*13
84=4*21
so, next number will be 5*31=155

Like? Yes (4) | No (4)  

S.SHAFI   10 years AGO

Formula = n^3 + n^2 + n

n=1 --> 1 + 1 + 1 = 3
n=2 --> 8 + 4 + 2 = 14
n=3--> 27+9+ 3 = 39
n=4--> 64 + 16 + 4 = 84
n=5 --> 125 + 25 + 5 = 155..

so ans is 155..

Like? Yes (3) | No (5)  

sudhakar   10 years AGO

The answer is 155

Like? Yes (2) | No (4)  

anandhakumar   10 years AGO

answer is 155
option 9

Like? Yes (2) | No (4)  

Kushagra Tiwari   10 years AGO

we can see that
3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

we know that

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

so

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

answer = option 9, 155

Like? Yes (2) | No (5)  

Samrat Adhikari   10 years AGO

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (4)  

Amrita priyadarshi   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (4)  

Jeevanantham.V.M   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (4)  

Surit Datta   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155
thats it...

Like? Yes (1) | No (4)  

abhik k. dey   10 years AGO

sequence is 3, 14, 39, 84
now
3=1^3+2=1+2
14=2^3+6=8+6
39=3^3+12=27+12
84=4^3+20=64+20
now next number will be 5^3+30
as 2,6,12,20 in a common series where
6=2+4
12=6+6
20=12+8
so next will be 20+10=30
so for the above series as 3, 14, 39, 84 the next number will be 5^3+30=125+30=155

Like? Yes (1) | No (5)  

Abhishek Vishwakarma   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

ans=155

Like? Yes (1) | No (5)  

RAKESH   10 years AGO

n^3+n^2+n

Like? Yes (1) | No (5)  

Sarabeshwar   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
155=5*31 Here consider 155 as term

while, Term(n) = n [ ( (n+1)to the power of 2 )- n ]
so Term (5) = 5[ (6)square - 5 ]
= 5 * 31 =155

Like? Yes (1) | No (5)  

Devendra Marghade   10 years AGO

n+n^2+n^3
1+1^2+1^3= 1+1+1= 3
2+2^2+2^3=2+4+8= 14
3+3^2+3^3=3+9+27=39
4+4^2+4^3=4+16+64=84
5+5^2+5^3=5+25+125=155

Like? Yes (1) | No (5)  

KAUSHAL KUMAR   10 years AGO

14-3=11 25-11=14 20-14=6
39-14=25 45-25=20 next no=20+6=26
84-39=45 next no.=45+26=71
next no=84+71=155. option(9).

Like? Yes (1) | No (5)  

kaushal kishor   10 years AGO

difference between 3,14,39,84 is 11,25 and 45 respectively.
and the difference between their diferences are 14,20,
which means 6, so next will be 26 and so 45+26=71
and so the next sequence will be 84+71=155

Like? Yes (2) | No (6)  

Arijit Acharyya   10 years AGO

1*3=3
2*7=14
3*13=39
4*21=84
5*31=155
next number=n*x
where n=1,2,3,4,5....
x=3,7,13,21,31,43... (difference is increasing by 2 starting from 4 )

Like? Yes (1) | No (5)  

sabir ahmd   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
hence next no will be =5*31=155

Like? Yes (1) | No (5)  

Aakash Jain   10 years AGO

1*3=3 leave next odd no-5
2*7=14 leave next 2 odd no's- 9,11
3*13=39 leave next 3 odd no's- 15,17,19
4*21=84 leave next 4 odd no's- 23,25,27,29
5*31=155

Like? Yes (1) | No (5)  

Md.Sultan Khichi   10 years AGO

The series is like that
3*1=3
After this 5 is prime no. but leave it
7*2=14
Agn 11 is the next prime no. So leave it too.
13*3=39
But from now two prime no. Will be leave.
21*4=84.
Next two prime no. 23,29 will be leave so next is
31*5=155.
Ans. 155

Like? Yes (1) | No (5)  

Rahul Sah   10 years AGO

3 14 39 84
11 25 45
14 20
6

anwer will be 6+20+45+84=155

Like? Yes (1) | No (5)  

Rashmi bharti   10 years AGO

31*5=155

Like? Yes (1) | No (5)  

AMIT SINGH   10 years AGO

difference between two nos. are=11,25,45
difference between the diff. are=14,20
so, next no.=20+6=26
so, next diff.=45+26=71
therefore, next sequence no.=84+71=155.

Like? Yes (2) | No (7)  

22 Aug 2014 time interval

From a town, cars start at regular intervals of 30 sec and run towards another town with constant speed of 60 km/hr. At some point of time all of the cars simultaneously have to reduce their speed to 40 km/hr due to bad weather conditions what will become the time interval between arrivals of the cars at the second town during the bad weather?

OPtion

1) 30 sec
2) 35 sec
3) 38 sec
4) 40 sec
5) 45 sec
6) 48 sec
7) 50 sec
8) 60 sec
9) 80 sec
10) none of these

Solution

In normal weather, distance between two cars = 30*60 unit
in bad weather distance between two car = t*40 unit
on comparing 30*60 = t*40
t = 45 sec

Correct Option: 5 Best Solution (61)

Devendra Marghade   10 years AGO

As the speed reduced from 60 km/hr to 40 km/hr i.e. 2/3 rd of the original speed.
So the time of arrival will increase from 30 sec. to 30*3/2= 45 sec.

Like? Yes (38) | No (7)  

Himanshu Garg   10 years AGO

First we calculate the distance between the consecutive cars
Distance(x) = Speed * Velocity

x= 60 * 30* 5/18= 500 m

Initial speed is 60 km/hr
time interval 30 sec
We have to multiply by 5/18 to convert km/hr to m/sec
1000/3600= 5/18

Now it remains same after decreasing the speed because all cars change their speed at same time so

500= 40 * y * 5/18
y=45

final speed is 40 km/hr
y is the time interval
5/18 is the changing factor

Like? Yes (13) | No   

ashok   10 years AGO

(60/40)*30=45

Like? Yes (7) | No (1)  

Dr.Dipin Singh   10 years AGO

At any time , distance between 2 cars travelling at 60km/hr starting at a time gap of 30 secs will be 1/2 kms.
This gap of 1/2 kms will be covered by cars running @ 40 km/hr in 3/4 mins or 45 secs.

Like? Yes (7) | No (2)  

sruthi   10 years AGO

given time=30 sec and speed =60km/hr
so, distance =speed*time=60*30=1800km
now, speed is reduced to 40km/hr
therefore time taken=distance/speed
===>1800/40=45sec

Like? Yes (10) | No (5)  

SUNITEE GARIMA   10 years AGO

The gap between any two cars at any time is the distance they can travel in 30 seconds(because they start 30 secs apart).
now at some time they all reduce their speeds to 40,so that gap(extra distance) will be covered in 3/2*30 seconds.
so they arrive at an interval of 45 secs.

Like? Yes (3) | No   

BHANU KUMAR   10 years AGO

distance between the cars when speed is 60km/h=60*(30/60*60)=1/2km
hence when speed is 40 km/h time taken to complete this distance =(0.5/40)*60*60=45sec (OPTION 5)

Like? Yes (3) | No (1)  

Amit Toor   10 years AGO

When Travelling at 60 kmph interval is of 30 sec. so distance bw cars is speed * time=500 meter.

Since speed is reduced to 40 kmph, the distance will remian same i.e 500 meter.

time taken to cover distance 500 meter at speed 40 kmph is 45 seconds.

So answer is 45 seconds.

Like? Yes (2) | No   

kajol nigam   10 years AGO

x=60*30/40
=45

Like? Yes (2) | No   

Aman   10 years AGO

Distance between two cars starting at a time gap of 30 secs & running @ 60 kmph will be covered in 30*60/40= 45 secs at a speed of 40 kmph.

Like? Yes (2) | No   

Jasmita Parmar   10 years AGO

The gap between any two cars at any time is the distance they can travel in 30 seconds . now at same time they all reduce their speeds to 40,so that gap will be covered in = 3/2*30 s. =45 s

Like? Yes (3) | No (1)  

Neel Patel   10 years AGO

distance between cars =60*30/3600=0.5 km
after one car reach to second town at speed of 40km/hr
time taken by second car =(0.5/40)*3600second
=45 second

Like? Yes (3) | No (2)  

Rahul shaw   10 years AGO

he was running @1m/min. i.e. diffrenece in distances initially would be 1/2 m. and den it will increase and diffrnce in tym = 45 sec.

Like? Yes (1) | No   

palasiva   10 years AGO

5/18*60*30=40*5/18*x
x=45sec

Like? Yes (1) | No   

T.VIGNESWARA REDDY   10 years AGO

in 30sec lack of distance with speed of 60kmph is 500meters so we have to cover these 0.5km with a speed of 40kmph. so time required is 0.5/40 hours i.e 45minutes

Like? Yes (2) | No (1)  

suman kumar   10 years AGO

in the initial time difference was 30sec.
so, distance travelled during that time=30*60*5/18=500m.
b\w journey it started raining so new speed=40km\h.
distance gap=500m.
since arrival time=500/(40*5/18)=45sec.

Like? Yes (1) | No   

Parshant garg   10 years AGO

when at any point first change its speed then the 2nd car is behind at
distance=(60*30)/3600=0.5 km
time travel by 2nd car is=(0.5*3600)/40=45 sec

Like? Yes (1) | No   

sabir ahmd   10 years AGO

at speed of 60 km/h time interval is =30 sec

hence, at speed of 40 km/h time interval will be=(30*60)/40=45 sec

Like? Yes (2) | No (1)  

Aravinth kumar   10 years AGO

The gap between any two cars at
any time is the distance they can
travel in 30 seconds(cos they start
30 secs apart).now at some time
they all reduce their speeds to 40,so
that gap(extra distance) will be
covered in 3/2*30 seconds.so they
arrive at an interval of 45 secs.

Like? Yes (1) | No   

Kushagra Tiwari   10 years AGO

The gap between any two cars at any time is the distance they can travel in 30 seconds(cos they start 30 secs apart).now at some time they all reduce their speeds to 40,so that gap(extra distance) will be covered in 3/2*30 seconds.so they arrive at an interval of 45 secs.

Like? Yes (3) | No (2)  

Abhishek Vishwakarma   10 years AGO

The gap between any two cars at any time is the distance they can travel in 30 seconds(cos they start 30 secs apart).now at some time they all reduce their speeds to 40,so that gap(extra distance) will be covered in 3/2*30 seconds.so they arrive at an interval of 45 secs.

Like? Yes (1) | No   

Pushpendra singh chundawat   10 years AGO

(500*9)/100=45 min

Like? Yes (1) | No   

Nitin Srivastava   10 years AGO

converting 60km/hr and 40km/hr to m/s
60*5/18=16.777
distance between cars = 16.777*30=500m
40*5/18=11.1111
time interval to maintain 500m distance is = 500/11.11=45sec

Like? Yes (2) | No (1)  

Ajay Kumar Gupta   10 years AGO

The gap between any two cars at any time is the distance they can travel in 30 seconds(cos they start 30 secs apart).now at some time they all reduce their speeds to 40,so that gap(extra distance) will be covered in 3/2*30 seconds.so they arrive at an interval of 45 secs

Like? Yes (2) | No (1)  

Chiranjit Agarwal   10 years AGO

distance cover by car 1 in 30 sec @ 60kmph=500m
time taken by car2 to cover 500m @ 40 kmph= 45sec

Like? Yes (1) | No   

ANKITA prasad   10 years AGO

suppose d distance b/w towns is covered in x mins.
d/40-d/50=x/4 mins=x mins 15 secs.
so 30+15=45sec.

Like? Yes (1) | No   

Mahak   10 years AGO

Since,
speed=distance/time
Initially at 60 km/hr,cars reaches at 30 sec
60=distance/30 - (eq. 1)
after decreasing speed to 40 km/hr,Let time be t
40=distance/t. -(eq. 2)
Dividing eq.1 by eq. 2,we get

t=45 seconds.(Answer -option 5).

Like? Yes (2) | No (1)  

G.Jaya Krishna   10 years AGO

(60/40)*30

Like? Yes (2) | No (1)  

Alok Mandal   10 years AGO

time taken by car to reach second town from first town is =30 sec
speed with which it running=(60*1000)/3600 meters/sec.
= 16.66 meter/sec.
therefore distance between two town = velocity * time
= 16.66 * 30
=500 approx.
when wether is bad,
speed reduced to 40 km/h, i.e. 11.11 meters/sec.
distance remains same between two town.

hence, velocity=distance/time
time=distance/velocity

time=500/11.11
=45.00 seconds

Like? Yes (1) | No   

vandna   10 years AGO

since the cars are starting after 30 secs
then there will be 30 secs always between two consecutive cars.
but as due to bad weather,
for 30 secs only, their speeds will be different for the journey as first car will travel with speed 60 km/hr for 30 secs more than the second car.
if distance is D then,
first car will travel D/2 distance in 30 secs with 60 km/hr speed. whereas,
second car will travel only D/3 distance for 30 secs with speed 40 km/hr
so difference in their distance travelled during that 30 secs is D/6
so second car will travel this distance with 40 km/hr speed so it will take time= D/6*3600/40= 15 secs
so the second car was supposed to come after 30 secs but due to bad weather it will come after 45 secs(30+15).
car

Like? Yes (1) | No   

abhishek   10 years AGO

suppose 1st car arrives at destination
then 2nd will be 30 sec behind
now if the speed is (60*5)18 m/s then is takes 30 sec
so if the speed will be (40*5)18 m/s
it will take 45 sec

Like? Yes (1) | No   

posirajesh   10 years AGO

t1 be time of first car bad weather starting
then t1-30 will be time for second car duration with 60kmph velocity
extra distance by 2nd car with 40 kmph distance is 500m
hence time interval=45sec

Like? Yes (2) | No (1)  

Alok Maurya   10 years AGO

The gap between any two cars at any time is the distance they can travel in 30 seconds(cos they start 30 secs apart).now at some time they all reduce their speeds to 40,so that gap(extra distance) will be covered in 3/2*30 seconds.so they arrive at an interval of 45 secs.

Like? Yes (1) | No   

rakesh yadav   10 years AGO

gap between simultanious two car =60*5/18*30=500m
again time interwal=500*18/200=45sec

Like? Yes (1) | No   

ARKAJIT NANDY   10 years AGO

distance between two cars=(60/3600)*30=1/2km
as all the cars slow down simultaneously to 40km/hr, their distance gap will be same.
when one car rech the destination the second car will be 1/2 km away from the destination.
so time taken by second car is=(3600/40)*(1/2)=45sec
so the time interval at arrival will be 45 sec.

Like? Yes (1) | No   

ANKUR ANAND   10 years AGO

constant distance maintained b/w two cars is 60*30/3600=.5km
so time interval is the time required to cover this distance
i.e. 0.5*3600/40=45 sec

so, answer is 5)

Like? Yes (1) | No   

Urvashi Jain   10 years AGO

firstly convert 30 sec into hr =1/120
distance=speed*time
d=60*1/120=1/2
then speed reduced by 40km/hr
t=distance/speed
t=1/2*40
t=1/80
again convert into sec
so 45 sec
option (5)

Like? Yes (1) | No   

vikash kumar   10 years AGO

let the total distance covered by cars is 1000
the average speed will be 2*60*40/100=48km/h
then 1000*18/5*48 -500*18/60*5=75-30=45sec

Like? Yes (1) | No   

sudipta maji   10 years AGO

The gap between any two cars at any time is the distance they can travel in 30 seconds(cos they start 30 secs apart).now at some time they all reduce their speeds to 40,so that gap(extra distance) will be covered in 3/2*30 seconds.so they arrive at an interval of 45 secs.

Like? Yes (1) | No   

iman   10 years AGO

Dist. is 2L
assume in half way they reduce speed
L/60 -> 30
L/40 -> 45
Ans. 45

Like? Yes (1) | No   

Dikesh Kumar Gumansingh   10 years AGO

The gap between any two cars at any time is the distance they can travel in 30 seconds(cos they start 30 secs apart).now at some time they all reduce their speeds to 40,so that gap(extra distance) will be covered in 3/2*30 seconds.so they arrive at an interval of 45 secs.

Like? Yes (1) | No   

Kapil Shirsath   10 years AGO

The distance covered by car in 30 seconds at the speed of 60km/hr is 0.5km
Each car is lagging by 0.5km from another car.
When the speed of car becomes 40km/hr, the time required to cover 0.5km is 45 seconds. So the time interval betwen arrival of cars at the town is 45 seconds.

Like? Yes (1) | No   

Guru Prasath   10 years AGO

The gap between any two cars at any time is the distance they can travel in 30 seconds(cos they start 30 secs apart).now at some time they all reduce their speeds to 40,so that gap(extra distance) will be covered in 3/2*30 seconds.so they arrive at an interval of 45 secs.

Like? Yes (2) | No (1)  

santosh kumar   10 years AGO

60*30=45*x,so x=45sec

Like? Yes (1) | No   

NAVEEN   10 years AGO

distance between first and second car when first car just reaches destination = 500m so, time taken to cover that with speed = 100/9= 5*9 = 45 secs

Like? Yes (1) | No   

Ramu Giri   10 years AGO

They have gap of 30 sec intervals.
Now they reduce thier speed due to bad weather .
so remaining distance will cover in=(3/2)*30=45sec

Like? Yes (1) | No   

Avinash   10 years AGO

Total distance gap b/w two car=speed*time=60*30*5/18=500mtrs
New time taken to cover this distance coz of bad wether=500*18/(40*5)=45sec

Like? Yes (1) | No   

Surendra   10 years AGO

when car reduce their speed to 40km/h then second car far from first car is 1/2 km and second car cover this distance by 40km/h so this time is between two cars is=
1/2*(60*60)/40=45sec

Like? Yes (1) | No   

kundan kumar   10 years AGO

distance travelled in 30 sec at 60km/h=distance travelled in x sec at 40km/h
so
30*(60*5/18)=x*(40*5/18)
so x=45

Like? Yes (1) | No   

avishek   10 years AGO

speed of car=60km/hr
interval time=30 sec

dist traveled=60*5/18/*30=500m
reduced speed=40 km/hr
time to reach =500/(100/9)=45sec

Like? Yes (1) | No   

KAUSHAL KUMAR   10 years AGO

const. speed is 60kmph i.e=50/3 m/s.
distance between 2 cars= (50/3)*30= 500m.

new speed=40kmph =100/9 m/s
so, new time gap= (500)/(100/9)= 45 sec.

Like? Yes (1) | No (1)  

Ashok Reddy Kalam   10 years AGO

FOR 30SEC WITH SPEED 60KM/HR SO DISTANCE IS 500M SO FOR
TO FIND TIME FOR SPEED 40KM/HR CAR NEEDS 45 SEC TO COVER THE SAME DISTANCE

Like? Yes (1) | No (1)  

krishna kanth   10 years AGO

20*30/40=15 sec
ans is 30+15=45 sec

Like? Yes (1) | No (1)  

Abhinav khare   10 years AGO

initially the car are at distance of 1/2 km from each other
so at the speed of 40 they will take 45 seconds to cover that distance

Like? Yes (1) | No (1)  

AMIT SINGH   10 years AGO

const. speed is 60kmph i.e=50/3 m/s.
distance between 2 cars= (50/3)*30= 500m.

new speed=40kmph =100/9 m/s
so, new time gap= (500)/(100/9)= 45 sec.

Like? Yes (1) | No (1)  

kaushal kishor   10 years AGO

const. speed is 60kmph i.e=50/3 m/s.
distance between 2 cars= (50/3)*30= 500m.

new speed=40kmph =100/9 m/s
so, new time gap= (500)/(100/9)= 45 sec.

Like? Yes (1) | No (1)  

Md.Sultan Khichi   10 years AGO

First car covers the distance in 30 sec=60*30*1000÷3600=500 m
All cars are moving with the constant speed so the distance between the cars are always 500m.
Whenever they get the signal to slow down the speed to 40km/hr.
So after this they should cover only 500m with the speed of 40km/hr..
So the time between them is=500*3600/1000/40=45 sec.
Ans. 45 secs.

Like? Yes (1) | No (1)  

ajit   10 years AGO

the lead taken by the first car will be 60*[(30)/60*60]=0.5km.
now the distance between two cars remain 0.5km even when they decrease their speeds. the difference in time is x+1/40-x/40=(1/80)*60*60 sec=45 sec

Like? Yes (1) | No (1)  

Tapobrata Banerjee   10 years AGO

difference between each car before reducing speed is=500m
As,at a speed of 60km/hr,car can cover 500 m in 30 sec
so at the time of reducing speed each preceding car is 500m ahead.
so the next car will reach after the time taken to cover 500m at 40km/hr.
so the answer will be (3600/40000)*500=45sec

Like? Yes (1) | No (1)  

Pranoy Bej   10 years AGO

Let the distance = 60km
so the time taken to cover this distance @60kmph is 1
and time time taken to coveraa this distance @ 40kmph is 1.5
so time gap is also 1.5*30 =45sec

Like? Yes (1) | No (1)  

manideep   10 years AGO

speed is inversly proportional to time, by taking that relation we can get the answer.

Like? Yes (1) | No (2)  

21 Aug 2014 bike race

Two boys on their bikes start a race around a triangular track ABC. Starting from the point A simultaneously, one of them traverse the track in clockwise sense A-> B -> C & the other in anticlockwise sense A-> C-> B with their respective constant speed. After 4 min from the start, they first meet at B and then continue the race. After what minimum time they meet first at B, will they again meet at B? Distance AB=500m, BC=400m, CA=300m

OPtion

1) 12 min
2) 16 min
3) 24 min
4) 30 min
5) 36 min
6) 42 min
7) 48 min
8) 64 min
9) 74 min
10) none of these

Solution

Speed of boy moving clockwise =500/4 m/min
Speed of boy moving anticlockwise =700/4 m/min
Let us suppose they they will meet again after t min of first meeting then according to given condition

700*t/4 - 500 *t/4 = 1200n where n is minimum possible integer

So 200*t/4 = 1200 n
=>t = 24 n
on putting t on 700*t/4 & 500*t/4

we get these two as 700*(24n)/4; 500*(24n)/4
= 4200n; 3000n

we should choose n in such a way that 4200n & 300n would be multiple of 1200 & this is possible only for n=2 (min value)

so t = 48 min;

Correct Option: 7 Best Solution (54)

suman kumar   10 years AGO

Speed of First person =500/4=125m/min

Speed of second person=700/4=175m/min

speed of first person:speed of second person=125:175
=5/7 i.e. they will meet at B if A completes 5 revolution or b completes 7 revolution beacuse time taken by both of them are same. So, required time =5*(1200/125)=48 min

Like? Yes (25) | No (7)  

SUNITEE GARIMA   10 years AGO

speed of C=700/4=175m/min
speed of B=500/4=125m/min
time taken by C to complete one round acroos the traingular field=4+(500/175)=48/7min
time taken by B to complete one revolution=4+(700/125)=48/5min
time when they will meet again =LCM of 48/5 and 48/7=48

Like? Yes (13) | No (4)  

sruthi   10 years AGO

as time is same so speed ratio is similar to distance ratio= 700:500= 7:5
so totally they will meet at 12 points before they reach at B (since,7+5=12)
so at 13 th time they will be meet at B
since after 4 min they start, therefore 12*4=48 min they will meet again

Like? Yes (8) | No (2)  

NAVEEN KUMAR   10 years AGO

both meet at B
distance travell by first and second boy is 500&700
speed of 1 boy=500/4=125m/s
speed of 2nd boy=700/4=175m/s
if both meet at b then next dist. travell byboth in multiple of 1200.
so,for first boy (1200*(X1))/125=I(integer)
=(48*X1)/5= min val of x1=5
total time taken by first boy=(1200*5)/125=48
second boy(1200(X2))/175=I2(integer)
similarly ,X2=7
tatal time taken=(1200*7)/175=48
so ,we can say that minimum time they meet first at B is 48min

Like? Yes (5) | No (1)  

PURIDI MOUNIKA   10 years AGO

let's say the time to meet at point B =t
two persons are P & Q.
suppose P travels through A->B->C
Q travels through A->C->B
they both meet At B in 4 mins
So the distance travelled by P in 4 min=500 m
the speed of P =500/4=125 m/min
the distance travelled by Q in 4 min=700 m
the speed of Q =700/4=175 m/min
then the time for meeting At B for both of them =t (say)
for P, t=1200*x/175
for Q, t=1200*y/125
->1200*x/175=1200*y/125
->x/y=7/5
then x=7,y=5
t=1200*x/175=1200*7/175=48
so time=48 mins

Like? Yes (5) | No (1)  

sabir ahmd   10 years AGO

A finished one round in (4/500)*1200=48/5 min
B finished one round in (4/700)*1200=48/7 min

they will meet together after 48 mins.

Like? Yes (3) | No   

RAKESH   10 years AGO

their speeds = 500/4 , 700/4 = 125,175 m/min
time taken to complete track = 1200/125 , 1200/175
lcm(1200/125 , 1200/175 ) = 1200/25 = 48 min

Like? Yes (3) | No (1)  

Md.Sultan Khichi   10 years AGO

First man covers 500m in 4 mins so his speed is 500/4=125m/m.
2nd person covers 700m to reach B in 4 mins so his speed is 700/4=175 m/m.
Acorting to the ques they have to meet at B again..so if they wants to meet at B they should have to cover 1200m..now we have to check when they reached B together.when their speed totally divide the distance of 1200m or 1200m's multiple together at the same tym..
First person=1200/125=9.6(not totally divide).
2400/125=19.2
3600/125=28.8
4800/125=38.4
6000/125=48.
After 48 mins first person agn come at B.
Now at this tym where is the 2nd person..
2nd person=48*175=7200.
7200 is multiple of 1200..so 2nd person is also at B.
Ans..48 mins..

Like? Yes (2) | No (1)  

suman kumar   10 years AGO

let two boy be x and y.
taking x rotating clockwise from a to b (500m in 4min.)
then y rotating anticlockwise from a-c- b (700m in 4min.).
starting from b if both have to meet next time at b
total triange length=1200.
taking revolving time for x=1200/(500/4)min
taking revolving time for y=1200/(700/4)min
after taking the ratio since both will not rotate equal=5:7.
since after 5 rotation of x and 7 of y they will meet together.
thus, time=5*1200/(500/4)
ans :-48 min

Like? Yes (1) | No   

Malisetti Mounika   10 years AGO

they will meet at B on 13th time so
after 12 meetings which is 12*4=48

Like? Yes (2) | No (1)  

Himanshu Garg   10 years AGO

They meet first at B so one travels 500m and another travels 700m so they are meeting again & again by the ratio of 5:7
So, we calculate the distance by adding 5 and evaluate when the again 5 comes in our series

Hence 0 5 10 3 8 1 6 11 4 9 2 7 12 5
1 2 3 4 5 6 7 8 9 10 11 12

Now, 12* 4= 48min

Like? Yes (1) | No   

Abhishek Jain   10 years AGO

Time taken by first boy for 500 m and by second boy for 700 m is same. And they meet every time they covers the above distance. So They will meet at point B again on 12th time. Which will take 4*12 mins = 48 mins.

Like? Yes (2) | No (1)  

HARDIP BHALANI   10 years AGO

This is similar to a circular motion except for its triangular with different dimensions ! so once we have reached B so lets designate this as our starting point because time has to be calculated from here !!

So LCM (1200/125 , 1200/175) = 48 min

Like? Yes (2) | No (1)  

Ajay Kumar Gupta   10 years AGO

As time is same, so speed ratio = dist ratio = 7 : 5
so they will meet each other at 12 distinct positions in the track... hence at B they will meet again the 13th time
meeting time = 4 min
so after 12 such meetings i.e 12 * 4 = 48 mins they will again meet at B

Like? Yes (1) | No   

anandhakumar   10 years AGO

In 48 min both are travelling in 8400,6000 respectively

8400 and 6000 are divide by 1200

Answer is option 7) 48 min

Like? Yes (2) | No (1)  

selvakumar   10 years AGO

In 48 min both are travelling in 8400,6000 respectively

8400 and 6000 are divide by 1200

Answer is option 7) 48 min

Like? Yes (2) | No (1)  

ANKUR ANAND   10 years AGO

700/4x=1200*7
=>x=48min
answer 7)

Like? Yes (1) | No   

abhishek   10 years AGO

52 min.

Like? Yes (1) | No   

ravi shah   10 years AGO

speed of 1st bike covering distance AB=500/4=125m/min
speed of 2nd bike covering distance ACB=700/4=175m/min
Relative speed=175-125=25
so,time after which they meet=(500+300+400)/25=48 min

Like? Yes (1) | No   

KAPIL   10 years AGO

Speed of biker who traverse in clockwise sense=125 m/min
Speed of biker who traverse in anticlockwise sense=175 m/min
Perimeter of triangle= 1200 m
LCM of 125 and 1200 = 6000
So, Ist biker will reach at B after 6000/125 = 48 min
Similarly LCM of 175 and1200 =8400
So 2nd biker will reach at B after 8400/175 = 48 min

Like? Yes (1) | No   

Kushagra Tiwari   10 years AGO

This is similar to a circular motion except for its triangular with diferent dimensions ! so once we have reached B so lets designate this as our starting point because time has to be calculated from here !!

So LCM (1200/125 , 1200/175) = 48min !

Like? Yes (2) | No (1)  

Swechchha Agarwal   10 years AGO

As time is same, so speed ratio = dist ratio = 7 : 5

so they will meet each other at 12 distinct positions in the track... hence at B they will meet again the 13th time

meeting time = 4 min

so after 12 such meetings i.e 12 * 4 = 48 mins they will again meet at B

Like? Yes (1) | No   

yatish kumar sharma   10 years AGO

time for

Like? Yes (1) | No   

abhishek   10 years AGO

speed of 1st boy(b1) = 500/4
speed of 2nd boy(b2)=700/4
time taken by b1 to complete the whole round distance i.e 1200m,from pt b =1200/500*4=48/5
time taken by b2 to complete the whole round distance i.e 1200m, from pt b=1200/700*4=48/7
now they will meet second time @ pt b in (L.C.M of 48/5,48/7)= 48 min

Like? Yes (2) | No (1)  

TUSHAR    10 years AGO

8400

Like? Yes (1) | No   

mukesh   10 years AGO

speed of a and b is 48/5 and 48/7
so they will meet after 48 min

Like? Yes (1) | No   

Suresh Kashyap   10 years AGO

let clockwise biker x1 and anticlockwise biker x2
x1 travel 500m in 4 min
x2 travel 700m in 4 min
now they are at point B
they start again
after 12 comlete round they meet again at point B
so after 12x4=48min later ther meet again at B

Like? Yes (1) | No   

lalit   10 years AGO

after every 4 minutes both boy meet at a point.
the next time when boy who reach point b from a in clockwise direction will be on spot b when he has covered 6000m distance.
time reqired to cover 6000meter is (6000/500)*4=48minute

Like? Yes (1) | No   

sudipta maji   10 years AGO

THE RATIO OF THE DISTANCE COVER BYTHE TWO BOY A->B->C : A->C->B =5:7 AND THEIR SPEED WILL BE 125M/mand 175M/m
so the time is1200*5/125=48min

Like? Yes (1) | No   

Pushpendra singh chundawat   10 years AGO

let x and y be the no of rounds to be completed by the boys respectively.
hence speed will be 125 and 175 metr/min
so x=5 and y=7
now
time taken by both boys to meet-(1200*5)/125=48min
or
(1200*7)/175=48 min

Like? Yes (1) | No   

chanda.ajaykishore babu   10 years AGO

to reach b the they has to travel 1200 km with speeds 125 &161.5 after 12 loops that is 48 min only the slower reach point B at that time only the faster one reach the point B

Like? Yes (1) | No   

sankalp yadav   10 years AGO

its question similar to hc verma relative velocity topic.

Like? Yes (1) | No   

shiv pujan   10 years AGO

time to cover =48 min

Like? Yes (1) | No   

Dikesh Kumar Gumansingh   10 years AGO

total distance=1200m
time for 500 m=4
12*4=48

Like? Yes (1) | No (1)  

KAUSHAL KUMAR   10 years AGO

after min 48min they will again meet at B

Like? Yes (1) | No (1)  

AMIT SINGH   10 years AGO

after 48 mint they will meet at B point again

Like? Yes (1) | No (1)  

arpit farkya   10 years AGO

Lcm of(48/5,48/7)=48

Like? Yes (1) | No (1)  

Rajat sadhu   10 years AGO

ans.-48min
let x and y moving clockwise and anti clockwise around the triangular track ABC ,
speed of x=500/4=125m/min
speed of y=700/4=175m/min
they again meet at B when x completes n rounds and y completes m round of the track,then
1200*n/125=1200*m/175 (1200 is total distance of the track)
n/m=5/7
hence x completes 5 rounds and y completes 7 rounds after which they meet at B
required time=5*1200/125=48min

Like? Yes (1) | No (1)  

S.SHAFI   10 years AGO

Check option Substitution..
For p1, 4min ---> 500m
48min--> 6000m
For p2 4min-->700m
48min -->8400m
Total distance --> 1200m

6000m, 8400m are divisible by 1200m ... so after 48min they will meet at B.

Like? Yes (2) | No (2)  

Ravi Bhalani   10 years AGO

This is similar to a circular motion except for its triangular with different dimensions ! so once we have reached B so lets designate this as our starting point because time has to be calculated from here !!

So LCM (1200/125 , 1200/175) = 48 min

Like? Yes (1) | No (1)  

Harshad    10 years AGO

This is similar to a circular motion except for its triangular with different dimensions ! so once we have reached B so lets designate this as our starting point because time has to be calculated from here !!

So LCM (1200/125 , 1200/175) = 48 min

Like? Yes (1) | No (1)  

Chiranjit Agarwal   10 years AGO

determine the hcf of the individual speed of the boys and divide it with the total length of the track will give you the time after which they will meet again at B

Like? Yes (1) | No (1)  

RAJEEV KUMAR   10 years AGO

LCM (1200/125 , 1200/175) = 48 min

Like? Yes (1) | No (1)  

kajol nigam   10 years AGO

v1=700/4=175
v2=500/4=125
lcm of 1200/175, 1200/125 = 48min

Like? Yes (1) | No (1)  

jagdish sharma   10 years AGO

respective speed of both bikers =50m/min
total dist. b/w BB=1200
time=1200/50=24
but in 24 min bikers r not meet at B
multiple of 24 is 48 is right ans.

Like? Yes (1) | No (1)  

Parshant garg   10 years AGO

ist 5 10 3 8 1 6 11 4 9 2 7 12 5
2nd 7 2 9 4 11 6 1 8 3 10 5 12 7
after 48 min...

Like? Yes (1) | No (1)  

Urvashi Jain   10 years AGO

according to the question both boy1 and boy 2 meet at 4 min so
we know that speed=distance /time
boy1 speed=500/4=125
boy 2 speed=700/4=175
they again meet so total distance=1200
time =1200/125=48/5
time=1200/175=48/7
take lcm and hcf of 48/7 and 48/5
lcm/hcf=48/1=48 min
option (7)

Like? Yes (1) | No (1)  

sudhakar   10 years AGO

In 48 min both are travelling in 8400,6000 respectively

8400 and 6000 are divide by 1200

Answer is option 7) 48 min

Like? Yes (2) | No (2)  

Abhishek Vishwakarma   10 years AGO

This is similar to a circular motion except for its triangular with diferent dimensions ! so once we have reached B so lets designate this as our starting point because time has to be calculated from here !!

So LCM (1200/125 , 1200/175) = 48min !

Like? Yes (1) | No (1)  

naina   10 years AGO

This is similar to a circular motion except
for its triangular with diferent dimensions !
so once we have reached B so lets
designate this as our starting point because
time has to be calculated from here !!
So LCM (1200/125 , 1200/175) = 48min !

Like? Yes (1) | No (1)  

Jasmita Parmar   10 years AGO

This is similar to a circular motion except for its triangular with different dimensions . so once we have reached B so lets designate this as our starting point because time has to be calculated from here

So LCM (1200/125 , 1200/175) = 48min

Like? Yes (1) | No (1)  

prem kumar   10 years AGO

1200 divide into 12 parts so every 4 mins their meet one part
so 12*4=48mins

Like? Yes (1) | No (1)  

gouthamdaneti   10 years AGO

one of the boy cover 500m in 4 min and other cover 700m
so we get each of there speed and ratio between there speed is 7:5;there fore there distance covered ratio is 7:5;
that is they meet at B after one of them cover 7*1200 divide this distance by the highest speed among them

Like? Yes (1) | No (1)  

krishna kanth   10 years AGO

speed of 1st is 125m/min
2nd person is 175m/min
time taken by first person to complete 1 cycle is 48/5
second person is 48/7
lcm of 48/5,48/7=48
ans is 48 min

Like? Yes (1) | No (2)  

20 Aug 2014 find the capital share

A B C start a business with capitals in the ratio 1/3:1/4:1/5. A withdraws half his capitals at the end of 4 months Out of a total annual profit of Rs 847, A's share is?

OPtion

1) Rs 152
2) Rs 222
3) Rs 252
4) Rs 280
5) Rs 292
6) Rs 320
7) Rs 380
8) Rs 392
9) Rs 410
10) None of These

Solution

Ratio of capital in the beginning 1/3:1/4:1/5 = 20:15:12
suppose the originally invest Rs 20, Rs 15, Rs 12

ratio of the investment of 12 months = (20*4+10*8):15*12:12*12
=40:45:36
Therefore A's share = Rs (847*40/121) = Rs 280

Correct Option: 4 Best Solution (83)

Ravi Bhalani   10 years AGO

Investment ratio will be 20:15:12
So, A's investment=20x
B's investment=15x
C's investment=12x
A withdraws half his capitals at the end of 4 months.
So,A=20x*4+10x*8=160x
B=15x*12=180x
C=12x*12=144x
Total=160x+180x+144x=484x
A's share=160x*847/484x=280

Like? Yes (22) | No (8)  

HARDIP BHALANI   10 years AGO

Investment ratio will be 20:15:12
So, A's investment=20x
B's investment=15x
C's investment=12x
A withdraws half his capitals at the end of 4 months.
So,A=20x*4+10x*8=160x
B=15x*12=180x
C=12x*12=144x
Total=160x+180x+144x=484x
A's share=160x*847/484x=280

Like? Yes (8) | No (2)  

abhik k. dey   10 years AGO

capital ratio of A,B and c is 1/3:1/4:1/5
so A:B:C=20:15:12
A withdraws his half of the capital after 4 months;
so effective capital ratio for A,B,C=(20*4+10*8):15*12:12*12
so, A:B:C=160:180:144

so in annual profit of 847 A\'s profit is 847*160/(160+180+144) =280 ................................. (answer)

Like? Yes (5) | No (1)  

Devendra Marghade   10 years AGO

Initial capital share of A:B:C = 1/3:1/4:1/5 = 20:15:12
For 1 year, share of A:B:C= [(20*4)+(10*8)]:(15*12):(12*12)
=160 : 180 : 144
A's share in annual profit of Rs 847
= 160*847/(160+180+144)= 280

Like? Yes (4) | No   

anandhakumar   10 years AGO

(1/3)*4+(1/6)*8:(1/4)*12:(1/5)*12
2.66:3:2.4
847/(2.66+3+2.4)=105
105*2.666=280

Like? Yes (4) | No   

vikash kumar   10 years AGO

1/3x*4+1/6x*8:1/4x*12:1/5x*12
8x/3:3x:12x/5
total share =8x/3+3x+12x/5=121x/15
A's share=(847*(8x/3))/(121x/15)=280

Like? Yes (4) | No   

DIPAK TIWARI   10 years AGO

A 's investment for annual=4*(1/3)+8*(1/6)=8/3
B 's investment for annual=12*(1/4)=3
C 's investment for annual=12*(1/5)=12/5
Hence A+B+C=(8/3)+(3)+(12/5)=121/15
Since total profit=₹847
Hence A's profit=(8/3)*847/(121/15)
=(8*847*15)/(3*121)
=₹280
Option 4

Like? Yes (3) | No   

sudhakar   10 years AGO

(1/3):(1/4):(1/5)
'A' withdraw half of the capital after 4 months (1/3)*4 +(1/6)*8=>2.67

Total annual profit of Rs 847

Total share is= 2.67+3+2.4=>8.07

A share is (847/8.07)*2.67=280.2

Answer is 280

Like? Yes (3) | No   

shiv pujan   10 years AGO

Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12

Profit ratio (20*4+10*8) : 15*12 : 12*12 = 40:45:36 A's share (40/121)*847 = 280

Like? Yes (3) | No   

Sneha   10 years AGO

1/3:1/4:1/5=20:15:12
Let their initial investments be 20x, 15x and 12x
A:B:C=(20x*4+10x*8):(15x*12):(12x*12)
=160x:180x:144x
=40:45:36
A's share=847*(40/121)=280
Option 4

Like? Yes (2) | No   

Harshad    10 years AGO

Investment ratio will be 20:15:12
So, A's investment=20x
B's investment=15x
C's investment=12x
A withdraws half his capitals at the end of 4 months.
So,A=20x*4+10x*8=160x
B=15x*12=180x
C=12x*12=144x
Total=160x+180x+144x=484x
A's share=160x*847/484x=280

Like? Yes (2) | No   

Shashi nath jha   10 years AGO

Let the capital be 60x.
A:B:C=20:15:12
a ns=280

Like? Yes (4) | No (2)  

Saraswathy   10 years AGO

A:B:C => (4*1/3 + 8*1/6) : 12/4 : 12/5
A:B:C => 8/3 : 3 : 12/5
A+B+C = 121/15
A's Share => 847 * (8/3)/(121/15) = 280

Ans : 280 (Option 4)

Like? Yes (2) | No   

mrunali s gaonkar   10 years AGO

Ratio of capitals in the beginning
=
1/3:1/4:1/5
: : 20:15:12
Ratio of investment for the whole year
= (20x4+10x8) : (15x12) : (12x12)
= 40 : 45 : 36
A's share = 847 x 40121 = Rs. 280

Like? Yes (2) | No   

naina   10 years AGO

let money invested by a = x /3
money invested by b = x/4
money invested by c = x/5
total profit = x/3*4+x/6*8+x/4*12+x/5*12=847
=>x=105
profit of a= 105/3*4+105/3*8=280

Like? Yes (2) | No   

Md.Sultan Khichi   10 years AGO

Ratio of their shares=1/3:1/4:1/5=20:15:12
Let total capital is x..
So A's share=20/47*x
4 months share of A=4*(20/47*x)=80/47*x.
For next 8 months share of A =8*(20/47/2*x)=80/47*x.
B's and c's shares for the 12 month=12*15/47 and 12*15/47 respectively.
So new ratio becomes=160/47:180/47:144:47=160:180:144
Profit=847
Ba's profit=160*847/484=280.
Ans ob. 4.

Like? Yes (2) | No   

krishna kanth   10 years AGO

ratio is 20;15:12
a's share will be 20*4+10*8=160
similarly b'share is 180
and c'share is 144
then 160*847/484=280

Like? Yes (2) | No   

Aakash Jain   10 years AGO

profit ratio will be 40:45:36, add all these and divide 847 with this result will 7 after that A's share 40*7=280.

Like? Yes (2) | No   

Pushpendra singh chundawat   10 years AGO

there contribution will be-(8/3:3:12/5)
so A's share=(8/(3*8.06)*847)
=280

Like? Yes (2) | No   

RAKESH   10 years AGO

A:B:C = 160:180:144
A's share = 160/484 *847 = 280

Like? Yes (2) | No   

ganesh babu   10 years AGO

ratio 20:15:12 (4 month)
ratio 10:15:12(next 8 month)
ratio 160:180:144 = 40:45:36 (of 12 month)
A shares 40/121*847=Rs 280

Like? Yes (2) | No   

Rahul shaw   10 years AGO

1/3:1/4:1/5 = 20:15:12
Then find the ratio of their shares and find A's share.

Like? Yes (2) | No   

Pranay Mukhopadhyay   10 years AGO

ratio of final investment=40:45:36
A's share=(40/121)*847=280rs

Like? Yes (2) | No   

Surit Datta   10 years AGO

according to the ratio 1/3:1/4:1/5
if we simplify it we get the ratio 20:15:12
assuming that A's capital is Rs 20
B's capital is Rs 15
C's capital is Rs 12
so A's capital Rs 20 is rolled for 4 month & Rs 10(half) is rolled for next 8 months.
for B,Rs 15 is rolled for 12 months.
for C,Rs 12 is rolled for 12 months.
so the ratio is now 160:180:144
according to this ratio the profit is distributed.
so A's capital is Rs 280
thats it...

Like? Yes (2) | No   

Pranoy Bej   10 years AGO

A:B:C = 1/3 : 1/4 : 1/5 = 20:15:12
A withdraw half of his capital at the end of 4 months
then A:B:C = (20*4 +10*8) :15*12:12*12 = 40:45:36
A's share = (40/121)*847 = 280

Like? Yes (2) | No   

Urvashi Jain   10 years AGO

A B AND C invest in the ratio 20:15:12
according to the question a shares
20x4+10x8:15x12:12x12
160:180:144
so A shares (160/484)x847
280
option(4)

Like? Yes (2) | No   

S.SHAFI   10 years AGO

A's share is 40/121 of 847 =280

Like? Yes (2) | No   

purushottam sah   10 years AGO

1/3:1/4:15
=20:15:12
=20x4+10x8:15x12:12x12
=40:45:36
A's share =847x40/121=280

Like? Yes (2) | No   

deepesh singh panwar   10 years AGO

a= 40/121*847
a=280

Like? Yes (2) | No   

suman kumar   10 years AGO

ratio of their investments are (1/3):(1/4):(1/5)
=>20:15:12
so their investments for 4 months is
20x:15x:12x &
for 8 months are 10x:15x:12x
so,
interest gain by A=(20x*(4/12)*r)/100)+(10x*(8/12)*r/100);
interest gain by B=15x*1*r/100;
interest gain by C=12x*1*r/100;
ratio of their gain=A:B:C
=40:45:36
THEREFORE A's share=847*40/(40+45+36)
=280rs

Like? Yes (2) | No   

Rashmi bharti   10 years AGO

(40/121)*847=280

Like? Yes (2) | No   

T.VIGNESWARA REDDY   10 years AGO

a investment is 20*4+10*8
b 15*12
c 12*12
a profit is (847/121) *40

Like? Yes (2) | No   

RITESH SINGH   10 years AGO

Ratio of their total investment=(20*4+10*8):15*12:12*12
=160:180:144
=40:45:36
therefore A's share=(40/(40+45+46))*847
=280

Like? Yes (2) | No   

NAVEEN KUMAR   10 years AGO

Ratio of capitals in the beginning
=1/3:1/4:1/5=20:15:12
Ratio of investment for the whole year
= (20x4+10x8) : (15x12) : (12x12)
= 40 : 45 : 36
A\'s share = 847 x(40/121)= Rs. 280

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Ramesh Sriram   10 years AGO

their annual investments in the ratio
40:45:36
total ratio=40+45+36=121
A's share=847*(40/121)=280

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AMRENDRA KUMAR   10 years AGO

4*20+8*10:15*12:12*12
a share=280rs

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aditya wargis   10 years AGO

A:B:C=1/3:1/41:1/5=20:15:12
(20x*4/12+10X*8/12):15X:12X
=40:45:36
A's share =847/(40+45+36)*40=280

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sindhu    10 years AGO

ratio of investments for 12 months is 8/3:3:12/5
annual profit is 847
A's share is 280

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Amit Kumar Pandit   10 years AGO

a:b:c=(20*4+8*10):15*12:12*12=40:45:36
a's share=847*40/121

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Malisetti Mounika   10 years AGO

A,B,C capital ratios are 1/3:1/4:1/5
Take LCM of denominators which is 60,
Now the new ratio is 20/60:15/60:12/60
which is 20:15:12

A invested his total capital only four months after that for remaining time he invested half of his capital so
20x*4+10x*8:15x*12:12x*12
-->160x:180x:144x
-->A\'s share is (160/484)*847
which is Rs 280.

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sompon temtheerakij   10 years AGO

ratio of A : B : C = 1/3 : 1/4 : 1/5 = 33.3 : 25 : 20
average ratio of A = ((33.3*4)+(16.65*8))/12 = 22.2
profit sharing of A = 22.2/(22.2+25+20)*847 = 280

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RAJEEV KUMAR   10 years AGO

(20*4+10*8):(15*12):(12*12)
=>160:180:144
160*847/484=280

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sabir ahmd   10 years AGO

ration of their capital =20:15:12

after 12 months their capital ration would be
=(20*4+10*8) : (15*12) : (12*12)
=40 : 45 : 36
therefore,
A's share is=(40/121)*847=280

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Tark Patel   10 years AGO

Profit ratio = ( 20 * 4 + 10 * 8 ) : 15*12 : 12*12
= 160 : 180 : 144
= 40 : 45 : 36
A\'s share = (40/121) * 847 = 280 Rs.

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NAGENDRA P   10 years AGO

Ratio of capitals of A:B:C=20:15:12
20*4+10*8:15*12:12*12=160:180:144=80:90:72=40:45:36
40/(121)*847=280..
So option (4)..

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Aravinth kumar   10 years AGO

Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12

Profit ratio (20*4+10*8) : 15*12 : 12*12 = 40:45:36 A's share (40/121)*847 = 280

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Jasmita Parmar   10 years AGO

Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12

Profit ratio (20*4+10*8) : 15*12 : 12*12 = 40:45:36 A's share (40/121)*847 = 280

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Kushagra Tiwari   10 years AGO

280.... Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12

Profit ratio (20*4+10*8) : 15*12 : 12*12 = 40:45:36 A's share (40/121)*847 = 280

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Abhishek Vishwakarma   10 years AGO

280.... Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12

Profit ratio (20*4+10*8) : 15*12 : 12*12 =40:45:36
A's share (40/121)*847 = 280

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ravi shah   10 years AGO

profit ratio of A:B:C=[x/3*4+(x/3-1/2*x/3)*8]:x/4*12:x/5*12
=8x/3:3x:12x/5
a's share=(8/3)/(8/3+3+12/5)*847=280

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TUSHAR    10 years AGO

280

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manishjohar17@gmail.com   10 years AGO

280.... Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12

Profit ratio (20*4+10*8) : 15*12 : 12*12 = 40:45:36 A's share (40/121)*847 = 280

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nagineni sukumar   10 years AGO

20:15:12
20*8:15*12:12*12
40:45:36
A's share=(40/121)*847=280

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Chirag Ranjan Lahiri   10 years AGO

280.... Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12

Profit ratio (20*4+10*8) : 15*12 : 12*12 =40:45:36 A's share (40/121)*847 = 280

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mithun mondal   10 years AGO

ratio 1/3:1/4:1/5=20:15:12
so according to question ratio will be
20*4+10*8:15*12:12*12=40:45:36
so A's share will be 847*(40/121)=280 Ans

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sruthi   10 years AGO

given:- A:B:C investment ratio's are 1/3:1/4:1/5
step1: now convert all the fractions by taking lcm
lcm is 60.
step2: the ratio becomes as 20:15:12 when we calculate lcm
step3: now it is given in the question as A withdraws the money after 4 months...so for 4 months he used 20...
then again the remaining 8 months he used half of the amount so it becomes 10..
so we can write A:B:C as : (20*4+10*8):15:12
step4: computing will get A:B:C=40:45:36
so A's share out of annual profit of Rs.847 is ,
step5: therefore A's share=( 40/(40+45+36))*847
==>(40/121)*847=280

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amar   10 years AGO

partnerships

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Bhagwan sahay   10 years AGO

There shares in the last of year will be 40:45:36
so A's share will be. (40/119)*847

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Guru Prasath   10 years AGO

280.... Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12
Profit ratio (20*4+10*8) : 15*12 : 12*12 = 40:45:36
(40/121)*847 = 280

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Naresh   10 years AGO

1/3:1/4:1/5 =>LCM of 3 4 5 is 60 which gives the ratio 20:15:12
(20*4+10*8):15*12:12*12=40:45:36
(40/121)*847=280

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pavankumar   10 years AGO

Ratio of capitals in the beginning
=
1
3
1
4
1
5
: : 20:15:12
Ratio of investment for the whole year
= (20x4+10x8) : (15x12) : (12x12)
= 40 : 45 : 36
A\'s share = 847 x
40
121 = Rs. 280

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mukesh   10 years AGO

A:B:C
1/3*4+1/6*8 : 1/4*12 : 1/5*12
8/3 : 3 : 12/5
so 8/3+3+12/5=121/15
So A\'s share is= (8*15)/(3*121)*847 = 280
ans is 4

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gagan   10 years AGO

approximatly

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yatish kumar sharma   10 years AGO

ratio of a is 8%3 n total ratio is 121%15
so 8*15*847%3*121

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G.Jaya Krishna   10 years AGO

1/3 of 847 is approximately 280

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prashant shroff   10 years AGO

280.... Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12

Profit ratio (20*4+10*8) : 15*12 : 12*12 = 40:45:36 A\'s share (40/121)*847 = 280

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prem kumar   10 years AGO

(8/3*60:12/4*60:12/5*60)=40:45:36
40*847/121=280

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gouthamdaneti   10 years AGO

the ratio can be converted to 20:15:12 for first 4 months and 10:15:12 for rest 8 months combined ratio for 12 months is 40:45:36 therefore profit share of A is 847*40/121

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Hari Nani   10 years AGO

the ratio can be converted to 20:15:12 for first 4 months and 10:15:12 for rest 8 months combined ratio for 12 months is 40:45:36 therefore profit share of A is 847*40/121

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Ashish Shende   10 years AGO

A,B,C Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12
Profit ratio
A IS WITHDRAW HALF IN LAST 4 MONTHS 8*10
AND THE WHOLE 20*4
(10*8+20*4) : 15*12 : 12*12 = 40:45:36
A's share (40/121)*847 = 280

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DIVANSHU   10 years AGO

20:15:12
profit sharing=160:180:144
160/484*847
=280

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santu ghosh   10 years AGO

280.... Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12

Profit ratio (20*4+10*8) : 15*12 : 12*12 = 40:45:36 A's share (40/121)*847 = 280

Like? Yes (1) | No   

sudipta maji   10 years AGO

Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12

Profit ratio (20*4+10*8) : 15*12 : 12*12 = 40:45:36
A's share (40/121)*847 = 280

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Rayapureddy V Sai Kiran   10 years AGO

ratios can be transformed as 20:15:12
ratios according to investments=(20*4+10*8):15*12:12*12
=160:180:144
a's share=(160*847)/(160+180+144)
=280

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avinash   10 years AGO

lcm of 1/3,1/4,1/5 is 20:15:12
20*4+10*8 : 15*12 : 12*12 = 160:180:144
160+180+144=484
a share is 160*847/484=280

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chanda.ajaykishore babu   10 years AGO

lcm becomes 60 for 1/3,1/5,1/4
the rsatio becomes 20:15:12
(20*4*x)+(10*8*x)+(15*12*x)+(12*12*x)=847
then a\\\\\\\'sinvestment is (80*847/484)+(80*847/484)=280

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Avinash   10 years AGO

ratio (20*4+10*8):15*12:21*12
40:45:36
121x=847
x=7
40x=280
A's share on profit is 280

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gangulasiva   10 years AGO

answer is 280.
the ratio of investments 20
:15:12
for20*4+10*8:15*12:12*12 ratio the net profit 847
for a\'s share 847*160/484 equals to the 280...

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Ramu Giri   10 years AGO

Investment ratio (1/3:1/4:1/5) * 60 = 20:15:12

Profit ratio (20*4+10*8) : 15*12 : 12*12 = 40:45:36 A's share (40/121)*847 = 280

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Gourab Dutta   10 years AGO

Ratio =20:15:12
A 's total investment = (20k*4+10k*8)=160k
B's total investment= (15k*12), same for c=(12k*12)
At the end of the year ratio becomes=40:45:36
A's share= 847*(40/121)=280

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selvakumar   10 years AGO

ans is 4

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Vanika   10 years AGO

Annual share of A: B : C
= ( (1/3 *4) + (1/6*8) ) : (1/4 *12) : ( 1/5 * 12)
=8/3 : 3 : 12/5
So A's share in annual profit = 8/3 * ( 847 / ( 8/3 + 3 + 12/5 ) )
= Rs. 280

Like? Yes (1) | No (1)  

Ajay Kumar Gupta   10 years AGO

A:B:C=20:15:12
A=20X,B=15X,C=12X
A:B:C=(20X*4+50/100*20X*8):(15X*12):(12X*12)=160X:180X*144X=40X:45X:36X
A'S SHARE=847*40/121=280

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