M4maths Previous Puzzles

29 Aug 2013 Compound Interest

A sum of Rs.1200 become Rs.1323 in 2 years at compound interest compounded annually. Find the rate of interest.

OPtion

1) 1.5%
2) 2.5%
3) 4%
4) 5%
5) 6%
6) 6.5%
7) 7%
8) 8%
9) 9%
10)None of these

Solution

Let the rate of interest be x%
1200{1+(x/100)}^2=1323
solve
x cannot be -ve, x = 5%
option 4

Correct Option: 4 Best Solution (64)

Md Intshamuddin   11 years AGO

(1323/1200)=(1+r/100)^2
=> 441/400=(1+r/100)^2
=> 21/20=1+r/100
=> r=5%

Like? Yes (18) | No (6)  

Avinash Kumar   11 years AGO

here,
Amt=1323
p=1200
n= 2 years
Now,
(amt/p)=( 1+ r/100)^n
1323/1200 = ( 1+ r/100)^2
solving this eqn we get
r=5

Like? Yes (9) | No (1)  

Abhijeet Kumar   11 years AGO

Formula of a compound interest:
A=P(1+R)^T
1323=1200(1+r%)^2
1323=1200(1+r/100)^2
1323=1200{(100+r)/100}^2
1323=12{(100+r)^2}/100
132300/12=(100+r)^2
11025=(100+r)^2 -- taking sqrt on both side
105=100+r
r=5% Ans

Like? Yes (10) | No (2)  

Alok Kumar   11 years AGO

A = p (1+ r/100)^2
r/100 = 1.05 - 1
r = 5%

Like? Yes (6) | No (2)  

ABHISHEK KUMAR MISHRA   11 years AGO

according to question
1323=1200(1+r/100)^2
1323/1200 = (1+r/100)^2
1.05 = (1+r/100)
105-100 = r
so r = 5%

Like? Yes (5) | No (2)  

Md.Sultan Khichi   11 years AGO

Compound intrst formula is a=p(1+r/100)^n
so 1323=1200(1+r/100)^2
After solving we get r=5
Ans:5%.obtn 4.

Like? Yes (4) | No (1)  

Sourav Maity   11 years AGO

1200*5%=60 for one year.so in two years that is 120. so the ans is option 4

Like? Yes (3) | No   

k.d.n padma rao   11 years AGO

amount=principle(1+rate/100)^time period
principle=1200
amount=1323
time period=2yrs
so, rate=((1323/1200)^(1/2)-1)100=5%

Like? Yes (4) | No (1)  

nishant kumar   11 years AGO

principal=1200;amount=1323;time=2 years
let the rate be R% per annum
then,[1200(1+R/100)^2]=1323
or, (1+R/100)^2=1323/1200
=>(1+R/100)=21/20
=>R/100=1/20
=>R=5%
hence rate is 5%pa

Like? Yes (3) | No   

KUNAL GAURAV   11 years AGO

Let the rate of interest be R.
We have,
P(1+R/100)^n=A

Here in the question we have P=1200
A=1323
n=2
So R can be calculated from the above formula.

1200(1+R/100)^2=1323

On solving we get

R=5%

Like? Yes (9) | No (7)  

manasa epuru   11 years AGO

1323=1200(1+r/100)(1+r/100)
on solving we get 5..

Like? Yes (3) | No (1)  

MOVVA ROHITH   11 years AGO

ans:5%
a=p(1+r/(100*n))^(n*t)
here n=1;t=2;a=1323;p=1200
by solving r=5%

Like? Yes (2) | No   

Harleen Chadha   11 years AGO

A=P(1+(r/100))^n
1323/1200=(1+r/100)^2
r=5%

Like? Yes (3) | No (1)  

Kushal Saraf   11 years AGO

formula for compound interest is

M=P(1+i)^n
here i-rate of interest
solving we get- 5% Ans

Like? Yes (3) | No (1)  

Chandni Maurya   11 years AGO

ans=(4) 5%
P=1200Rs,A=1323Rs,n=2years,r=?
where P=principal ,A=amount/sum, n= year,r=rate
A=P(1+r/100)^n
=>1323=1200(1+r/100)^2
=>sq.rt(1.1025)= 1+r/100
=>1.05 = 1+r/100
=>r= (1.05-1)*100
=>r=5%

Like? Yes (3) | No (1)  

JITENDRA KR.BAROLIYA   11 years AGO

1323=1200(1+r/100)^2
441/400=((100+r)^2)/10000
21/20=(100+r)/100
r=5 ans

Like? Yes (2) | No   

SHIVANI PRAKASH   11 years AGO

1200(1+r/100)^2=1323/1200
r=5%

Like? Yes (3) | No (1)  

FRANKLIN VIJAY   11 years AGO

P = Rs. 1,200, A = Rs. 1,323, N = 2
1,323 = 1,200 (1+ r)2
R = 100 × 0.051 = 5.1%.

Like? Yes (2) | No   

Chandra sekhar reddy   11 years AGO

formula: compound interest sum=p(1+r/100)n
here p is intial amount
r is rate
n is no of years

substitute you will get 1200(1+r/100)^2=1320
now solve for r...
Answer is 5%
Choice is 4...

Like? Yes (2) | No   

LOGAN WOLVERINE   11 years AGO

p(1+r/100)2=1323;
(1+r/100)2=1323/1200;
1.1025;
(100+r)2=1.1025*10000;
(100+r)2=11025;
(100+r)2=105^2;
100+r=105;
r=105-100;
interest=5%
1=

Like? Yes (2) | No   

HARIHARAN S   11 years AGO

By using compound interest formulae

Like? Yes (2) | No   

ch.jyothsna   11 years AGO

P=1200 amount=1323 n=2 for calculating R we have formula amount=P(1+(R/100))^n 1323=1200(1+(R/100))^2
(1+(R/100))^2=1.1025
1+(R/100)=1.05
100+R=105
R=5
so R=5%

Like? Yes (4) | No (2)  

KARTHIGA DEVI.J   11 years AGO

c.i=p((1+r/100)^n -1)
p=1200;n=2
sub,
r=5%.

Like? Yes (2) | No   

KALIDASS   11 years AGO

Amount=P(1+(R/100))^n
1323=1200(1+(R/100))^2
R=5%

Like? Yes (2) | No   

Gokul Krishnan   11 years AGO

P=A/(1+(R/100))^n
R=5%

Like? Yes (3) | No (1)  

Madhu pericherla   11 years AGO

A=P(1+(r/100))^n

1,323=1=200(1+r/100)^2

1+(r/100)^2=(1,323/1,200)

(1+(r/100))^2=441/400

(1+(r/100))^2=(21/20)^2

(1+r/100)=21/20

(r/100)=1/20

r=5
the required rate of compound interest per annum is 5%

Like? Yes (3) | No (2)  

ritesh kumar pandey   11 years AGO

let the rate be x;
so 1323 = 1200(1+x/100)^2
so x = 5;

Like? Yes (2) | No (1)  

vigneshwari   11 years AGO

the formula to find the amount compounded annually, final amount = P[1+ R/100]^n
here, amount = Rs.1323, p = Rs.1200, n= 2 years.
Then, 1323 = 1200x[1+ R/100]^2
= 1200[100+R]^2 / 10000
(100+R)^2 = 11025
R^2 + 200R - 1025 = 0
R = 5.
required rate of interest = 5%.

Like? Yes (4) | No (3)  

Alok kumar   11 years AGO

P=1200
A=1323
N=2
LET RATE OF INTEREST IS R
WE KNOW A=P(1+R/100)^N
USING ABOVE
WE GET R=5%

Like? Yes (4) | No (3)  

rahul jha   11 years AGO

A=P(1+(r/100))^n
n=2 yr,A=1323,P=1200
1323=1200(1+(r/100))^2
1+(r/100)=.05
r=5 %

Like? Yes (5) | No (4)  

nirmal   11 years AGO

1323=1200(1+R/100)2
=>then calculate R from this formula

Like? Yes (2) | No (1)  

Mayuresh Srivastava   11 years AGO

A=P(1+(r/100))^n
Here A=1323,P=1200,n=2
thus r=5

Like? Yes (2) | No (1)  

Abhik Kumar Dey   11 years AGO

A=P(1+r/100)^n
n=2
P=1200
A=1323

on calculating we get r=5%

Like? Yes (2) | No (1)  

ashiwani kumar singh   11 years AGO

ans=4
1323=1200*(1+r/100)^2;
(1323/1200)^1/2=(1+r/100);
(21*100/20)=100+r;
105-100=r;
r=5%

Like? Yes (2) | No (1)  

Aniruddha Bhattacharjee   11 years AGO

x=1200
y=1323
x*5/100+x=1260
1260*5/100=63
NOw 1260+63=1323

Like? Yes (4) | No (3)  

Ankit Shukla   11 years AGO

1323=1200(1+(r/100))*(1+(r/100))
1323/1200=(1+(r/100))*(1+(r/100))
421/400=(1+(r/100))*(1+(r/100))
21=20=(1+(r/100))
r/100=1/20
s0 r=5%

Like? Yes (2) | No (1)  

Bingumalla Abinay   11 years AGO

A=P(1+r/n)2t

1323/1200=[1+r]^2

r=5%

Like? Yes (2) | No (1)  

Jasmine Khan   11 years AGO

A=P(1+r/100n)^nt
1323=1200(1+r/100)^2
1323/1200=(100+r/100)^2
36.37/34.64=100+r/100
r=3637/34.64-100
r=4.99%

Like? Yes (2) | No (1)  

vky   11 years AGO

1323=1200*(1+r/100)square
r=5%

Like? Yes (2) | No (1)  

ravi singh   11 years AGO

answer is 5%

A=P * (1 + R/100)^T

where A=amount
P=principle
R=rate
T=time
from question
1323=1200* (1 + R/100)^2
by solving this we get R=5

hence answer is option 4

Like? Yes (3) | No (2)  

Mariappan   11 years AGO

P=1200, A=1323, n=2yrs

A=P(1+r/100)^n

therfore, r =5%

Like? Yes (2) | No (1)  

kishor kumar   11 years AGO

amt = sum (1+r/100)^t
1323 = 1200(1+r/100)^2
or r= 5%

Like? Yes (2) | No (1)  

Naveen Jayan   11 years AGO

A=P(1+r)^t formula. Substituting the values give us 0.05 which is 5%

Like? Yes (2) | No (1)  

Balakrishnan M   11 years AGO

A=P[1+(r/100)]^n
1323=1200[1+(r/100)]^2
(1323/1200)=[1+(r/100)]^2
(441/400)=[1+(r/100)]^2
(21/20)^2=[1+(r/100)]^2
(21/20)=[1+(r/100)]
(21/20)-1=(r/100)
(1/20)=(r/100)
r=5%

Like? Yes (3) | No (2)  

Shubhankur Manuja   11 years AGO

Amount = P*(1+R/100)^2

1323 = 1200*(1+R/100)^2

1.05 = 1+R/100

R=5%

Like? Yes (2) | No (1)  

Harsh Poddar   11 years AGO

By formula
1200*(1+(r/100))^2=1323
(1+(r/100))^2=441/400
1+(r/100)=21/20
r/100=1/20
r=5
rate =5%
option 4

Like? Yes (2) | No (1)  

aman kumar tiwari   11 years AGO

1323=1200(1+(r/100))(1+(r/100))=r=5% ans

Like? Yes (1) | No   

brijesh   11 years AGO

Amount=P(1+(R/100))^N
1323=1200(1+(R/100))^2

calculating it we get R=5%

Like? Yes (1) | No   

Bhargaviinduri   11 years AGO

we know compound interest=p(1+r/100)^t

here p=sum,t=time,r=rate of interest

1323=1200(1+r/100)^2 by solving this we get
r=5%

Like? Yes (1) | No   

sanjay singh   11 years AGO

principal=1200(P)
amount=1323(A)
A=P(1+r/100)^2
1323=1200(1+r/100)^2
solving it ,r=5

Like? Yes (1) | No   

Jagadeesh Rampati   11 years AGO

Here, P = Rs. 1,200, A = Rs. 1,323, N = 2, i = ? 1,323 = 1,200 (1+ i)^2
(1+i)^2=1323/1200
(1+i)^2=1.1025
take logs on both sides
2 log(1+i)=1og 1.1025
2 log(1+i)=0.0426
log (1 + i) = 0.0213
(1 + i) = Antilog 0.0213 = 1.051
i = 0.051 and R = 100 × 0.051 = 5.1%.

Like? Yes (1) | No   

praneeth   11 years AGO

A=P(1+r/n)^nt
A=1323 rs. t=2years;n=1(compounded annually);P=1200;
1323=1200(1+r)^2
r=0.05
r(%)=0.05*100=5%

Like? Yes (1) | No   

jeevan reddy mandali   11 years AGO

323 rs increment per 2 years which means 61.5 rs per a year. so its , 1200*rate = 61.5 => 1/rate = 1200/61.5 => rate=0.05 => 5%

Like? Yes (1) | No   

MAYANK AGARWAL   11 years AGO

a=p(1+r/100)^n
given
a=1323,p=1200,n=2
we get r=5%

Like? Yes (1) | No   

richa ahuja   11 years AGO

p=1200
a=1323
n=2
a=p(1+r/100)^n
1323=1200(1+r/100)^2
r=5%

Like? Yes (1) | No   

Jayaraman Baskar   11 years AGO

amount=principal*(1+R/100)^n
R-rate of interest;n=no. of years
1323=1200*(1+R/100)^2
R=5%

Like? Yes (1) | No   

ravi chandra singh   11 years AGO

A=p(1+R/100)^n
put all values then
1323=1200(1+R/100)^2

1323/1200=(1+R/100)
1.05=1+R/100

0.05=R/100

so,R=5%

Like? Yes (1) | No   

sumanth   11 years AGO

5%+5%+((5*5)/100)=123
and 1323-1200=123

Like? Yes (1) | No   

mohini gupta   11 years AGO

A = P ( 1 + r/n )^ nt,
where,
P = principal amount (initial investment)
r = annual interest rate (as a decimal)
n = number of times the interest is compounded per year
t = number of years
A = amount after time t.
by solving, 1323=1200(1+r)^2,
r=.5*100=5%

Like? Yes (1) | No   

roushan   11 years AGO

total amount=principal value(1+r/100)^t
1323=1200(1+r/100)^2
1323/1200=(1+r/100)^2
1.05=1+r/100
.05=r/100
r=5%

Like? Yes (1) | No (1)  

Ch Kumaraswamy   11 years AGO

principal amnt p=1200,
total amnt =1323
years n=2,
let r be the rate of intrst.,
by using compound intrst annually formula

toal amnt=p*(1+r/100)^n

by solving we get r=5%.

Like? Yes (2) | No (2)  

Chithra   11 years AGO

5% compound interest from 1200 is 60+60+3=123, total 1200+123=1323

Like? Yes (2) | No (2)  

Raju Sharma   11 years AGO

bcoz
ci =p(1+R/100)^n
usig this formula we get rate of 5%

Like? Yes (1) | No (1)  

balaji ponnambalam   11 years AGO

C.I=P(1+(x/100))^n
so, ans 4%

Like? Yes (1) | No (1)  

28 Aug 2013 walk

A man walks up a mountain at the rate of 2 km an hour and down by a way 6 km longer at the rate of 7/2 km an hour. He is out eight hours altogether. How far has he walked?

OPtion

1) 43 km
2) 38 km
3) 32 km
4) 29 km
5) 26 km
6) 22 km
7) 20 km
8) 19 km
9) 17 km
10) None of these

Solution

Let the way up be x km
Time taken to climb up = x/2 hours.
The way down = (x+6)km
Time taken to climb down =(x+6)/(7/2)= 2(x+6)/7
Therefore the total time of the journey = (x/2)+{2(x+6)/7}=8
solve x = 8
Total distance = x+x+6 = 22km.
option 6

Correct Option: 6 Best Solution (33)

Vinay Sipani    11 years AGO

Let time taken to walk up the mountain = Tu
Let time taken to travel down the mountain = Td
=> T1 + T2 = 8 hrs
Let distance up the mountain = D kms
=> Distance travelled down the mountain= (D+6) kms

=> Tu = D/2 ,
=> Td = (D+6)/(7/2) = (2D + 12) / 7

=> [ D/2 ] + [(2D + 12) / 7 ] = 8
=> 7D + 4D + 24 = 112
=> D = 8 kms

=> Total distance travelled = D + D + 6 = 22 kms

Like? Yes (12) | No (4)  

Md Shahnawaz Shahab   11 years AGO

let the distance traveled by the man uphill be 'x' km, therefore the distance traveled by him downhill will be (x+6) km.
Now, adding the time taken by both the journeys we get,

x/2 + 2(x+6)/7 = 8

solving for x, we get x = 8 km
therefore total distance traveled by the man will be x + (x+6) = 8+8+6 = 22km.

Like? Yes (6) | No (3)  

KUNAL GAURAV   11 years AGO

Let the distance covered while climbing up the mountain be d.
So the distance covered while coming downwards will me d+6.

Time taken during whole journey = d/2 + (d+6)2/7
Since, total time of the journey is given as 8 hrs
So,
On solving we get d = 8
So total distance covered will be = 8+8+6 = 22

Like? Yes (4) | No (1)  

mohini gupta   11 years AGO

let initialdistance be x
hence,x/2+(x+6)/3.5=8
by solving
x=8
total distance=8+8+6=22

Like? Yes (4) | No (2)  

ShAg   11 years AGO

for upward journey:
distance = x
speed = 2kmph
time = x/2
for downward journey:
distance = x+6
speed = (7/2)kmph
time = (x+6)/(7/2) = (2x+12)/7

total time = 8hrs.
hence (x/2) + ((2x+12)/7) = 8
=> x = 8

hence total distance = x + (x+6) = 22km

Like? Yes (1) | No   

Abhijeet Kumar   11 years AGO

let
x be the distance to walk up at 2 km/h
& (x+6) be the distance to walk down at 7/2=3.5 km/h.

(x/2)+(x+6)/3.5 = 8
after solving this equation x=8 & x+6=14
so, the total walking distance = (x)+(x+6) = 22.

Like? Yes (3) | No (2)  

priyanka kumari   11 years AGO

x/2+(x+6)*2/7=8
x=8
total distance=8+(8+6)=22

Like? Yes (1) | No   

sanjay kumar saha   11 years AGO

x is upward distance

x/2+(x+6)*2/7=8
x=8
total dis=8+(8+6)=22

Like? Yes (2) | No (1)  

richa ahuja   11 years AGO

let distance covered by him up the mountain is x km
therefore distance covered by him down the mountain is x+6 km
distance =speed*time
time=distance/speed
total time=8 hours
x/2+2(x+6)/7=8
on solving this
x=8
total distance=x+x+6=8+8+6=22km

Like? Yes (2) | No (1)  

Alok Kumar   11 years AGO

let time is t hr
2t - 7/2*(8-t) = 6
so t = 4
and 2t + 7/2*(8-t) = 22 km

Like? Yes (2) | No (2)  

Kumar Ankit   11 years AGO

Suppose total distance to walks up a mountain be xkm and down the mountain be (x+6)km.
Time taken to walks up a mountain be x/2 hr(Since given speed to walks up a mountain be 2km/hr).

Time taken to walks down a mountain be 2(x+6)/7 hr(Since given speed to walks down a mountain be 7/2km/hr).

Hence,total time taken=x/2+2(x+6)/7=8
on solving this we get x=8
Hence total distance cover is x+(x+6) i.e
8+(8+6)=22

Like? Yes (2) | No (2)  

rahul jha   11 years AGO

going UP @ 2km/hr
so cover distance:8km need
time:4hr
going down @ 7/2km/hr
6km longer so need to cover 14 km distance..
so required time=(2*14)/7=4 hr
so,total time=8hr
distance=22 KM

Like? Yes (3) | No (3)  

Avinash Kumar   11 years AGO

let the distance covered by the man while going up be as 'x'
so distance covered while coming down will be 'x+6'
So the equation will be
x/2 + 2x+12/7 =8
solving this we get x=8
so total distance covered by the man is x+x+6
or 8+8+6
or 22

Like? Yes (1) | No (1)  

Alok kumar   11 years AGO

let he walks up s distance
so, s=2*t1
now he walks down (s+6)
so, (s+6)=7/2*t2
now t1+t2=8 (given)
solving these eqtns we will get s=8(upward distance)
so, downward distance=(s+6)=14
so,total distance= (8+14)=22

Like? Yes (1) | No (1)  

SATYAM SAURABH   11 years AGO

from above qus:
8 =d/2 + (d+6)/3.5)
d=8
total walking distance 8+8+6=22

Like? Yes (1) | No (1)  

Jacky shevkani   11 years AGO

let t1 and t2 be the time taken when the men walls up and walls down the mountain respectively.

let x km be the distance covered while going up and (x+6) km be the distance covered while coming down the mountain.

therefore
t1=(x/2)
t2=(x+6)/(7/2)

adding both the above equations we get
t1+t2=(x/2)+((x+6)/(7/2))

since total time taken is 8hours
therefore
t1+t2=8

substituting in above we equation and solving it we he
x=8km
and
x+6=14km
therefore total distance covered is 8+14=22km

Like? Yes (1) | No (1)  

Jasmine Khan   11 years AGO

assume that he travelled for xkm while going up and while coming back he has travelled (x+6)km
let us denote as,
d1=x,d2=x+6,s1=2km/hr ands2=7/2km/hr
d1/s1+d2/s2=8
x/2+2(x+6)/7=8
7x+4x+24=8*14
11x=112-24
11x=88
x=8
he has travelled for 8+(8+6)=22km

Like? Yes (1) | No (1)  

Madhu pericherla   11 years AGO

d=xkm,s=2kmph,t1
d=x+6,s=7/2,t2
given
T=t1+t2=8hrs
speed=distance/time
time=distance/speed
8hrs=(x/2)+((x+6)/(7/2))
while solving we will get x=8
total journey=x+(x+6)=8+8+6=22km
(so ans is 22km)

Like? Yes (1) | No (1)  

Devendra Marghade   11 years AGO

If up distance=x & down distance=x+6
[(x/2)+((x+6)2/7)]=8,solving x=8
Total distance walked=x+(x+6)=8+14=22 km

Like? Yes (1) | No (1)  

T Lokesh   11 years AGO

let the distance covered by him while going up =x
then time t1 =x/2
distance covered by him while coming down =x+6
time t2 =(x+6)/(7/2)
total time =8
x/2 + (2x+12)/7 =8
so x=8
so total distance = 8+8+6 =22

Like? Yes (1) | No (1)  

sivaraman   11 years AGO

consider x is distance
The mountain up time is x/2km/hr
The mountain down time is (x+6)/ 7/2kh/hr
so

x/2 = (x+6)/(7/2)
x/2= 2(x+6)/7
x/2=(2x+12)/7

do cross multiply

7x=2(2x+12)
7x=4x+24
7x-4x=24
3x=24
x=8

so up time is 8, down time is 8+6=14
add up and down time is =8+14
22km

Like? Yes (1) | No (1)  

ravi chandra singh   11 years AGO

t1=1hr
t2=6*2/7=12/7
t=t1+t2=19/7
required distance=19/7*(2+6)=21.71=22km

Like? Yes (1) | No (1)  

Parthiban C   11 years AGO

(x/2)+((x+6)2)/7=8
then x=8
finally ans =8+8+6=22

Like? Yes (1) | No (1)  

Chandni Maurya   11 years AGO

ans=(6). 22km
let x is the distance to walk up the mountain and t1 is the corresponding time,
t2 is the return time.
we know,
distance=velocity*time
according to question,
x=2*t1 ---(1)
x+6=(7/2)*t2 ----(2)
(2)-(1)
-2*t1 + (7/2)*t2 =6 ----(3)
t1+t2 = 8 (given) -----(4)
solving 3 and 4
t1=t2 = 4 hr => x= 8 km
total distance= x+x+6 = 8+8+6 = 22 km

Like? Yes (1) | No (1)  

IFFAT IRAM   11 years AGO

let the distance for the forward journey be x km
distance for backward journey=(x+6) km
time for forward journey=x/2 hr
time taken for backward journey=2(x+6)/7 hr
x/2 + 2(x+6)/7 = 8
(7x+ 4(x+6))/14 = 8
11x + 24 = 8*14
11x=112-24
11x=88
x=8 km
he has walked 8+(8+6)=22km

Like? Yes (1) | No (1)  

mohamed ismail   11 years AGO

let x be distance travelled ,y be. time taken

according to the condition, the eqns will be
2x+7y equal to 44
2y-x equal to 0
solving we get. 22 kms

Like? Yes (1) | No (1)  

sudipta sadhukhan   11 years AGO

LET, THE DISTANCE HE COVERED AT THE TIME OF CLIMBING UP IS D KM.
THE DISTANCE HE COVERED AT THE TIME OF CLIMBING DOWN IS (D+6) KM.
NOW,
ACCORDING TO THE QUESTION,
D/2+((D+6)/3.5)=8
=>D=8
SO HE COVERED A TOTAL DISTANCE OF 8+(8+6)=22 KM.☻

Like? Yes (1) | No (1)  

Bingumalla Abinay   11 years AGO

x=2t
x+6=3.5(8-t)
t=4

so total distance = 2t+3.5t=8+14=22

Like? Yes (1) | No (2)  

Rats   11 years AGO

time to go up(t1)=x/21
time to go down(t2)=((x+6)2)/7
t1+t2=8
On solving all these eq. we get
x=8
So total distance=x+(x+6)=8+14=22

Like? Yes (1) | No (2)  

gaurav   11 years AGO

x/2 + 2(x+6)/7 = 8
on solving we get x= 8
so total distance = 8+8+6 = 22

Like? Yes (1) | No (2)  

Pankaj   11 years AGO

up=2km/hr
d=3.5km/hr
av=2.75
dis=2.75*8=22 ANS.

Like? Yes (2) | No (3)  

Akshay Hanswal   11 years AGO

x/2 +(x+6)/(7/2)=8
On solving we get x=8 so total walk=8+8+6 i.e. 22km

Like? Yes (1) | No (2)  

rahul   11 years AGO

8 km walks up in 4 hr at 2km/h nd 14 km walks down in hr at speed of 7/2 km/h

Like? Yes (1) | No (2)  

27 Aug 2013 average

Average of prime numbers ranging 1 to 50 is

OPtion

1) 15.43
2) 16.54
3) 17.78
4) 17.93
5) 18.72
6) 19.13
7) 19.46
8) 21.33
9) 21.87
10) none of these

Solution

prime number in this range are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
so average = [2+3+5+7+11+13+17+19+23+29+31+37+41+43+47]/15=328/15=21.87

Correct Option: 9 Best Solution (44)

Dr.Dipin Singh   11 years AGO

Average = (2+3+5+7+11+13+17+19+23+29+31+37+41+43+47)/15 = 328/15= 21.8666.. or 21.87 ..option 9

Like? Yes (8) | No   

ShAg   11 years AGO

There are 15 prime numbers from 1 to 50, they are:
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
Their sum = 328
Average = 328/15 = 21.87

Like? Yes (6) | No   

soham kundu   11 years AGO

add of prime numbers are 328
then divide by 15 as there are 15 prime numbers..
result 21.866667

Like? Yes (5) | No   

rahul jha   11 years AGO

answer--->21.87
prime no are those which z only divisble by 1 and itself...so between 1-50 ,
prime no. =2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
sum=328
avg=(sum of prime no/no. of prime no.)=328/15=21.87

Like? Yes (6) | No (3)  

T Lokesh   11 years AGO

sum of prime numbers between 1 and 50 = 328
prime numbers = 2,3, 5 ,7 11,13,17,19,23,29,31,37,41,43,47
so average =328/15
=21.866
=21.87

Like? Yes (4) | No (1)  

Alok Kumar   11 years AGO

as any prime accept 2,3 can be represented by
(6n-1) or (6n+1)
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
avg 21.87

Like? Yes (13) | No (11)  

Madhu pericherla   11 years AGO

prime numbers from 1 to 50 is
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
average=(2+3+5+7+11....+47)/15=328/15=21.87
so ans is 21.87

Like? Yes (3) | No (2)  

devendra kumar gond   11 years AGO

prime nos.are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 and their avg is 21.866

Like? Yes (3) | No (2)  

Chandni Maurya   11 years AGO

ans=(9).21.87
prime no range 1 to 50 are,
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47=>15 no.s
avg=addition of above prime no/15
=328/15=21.8666 = 21.87

Like? Yes (4) | No (3)  

Mayuresh Srivastava   11 years AGO

(2+3+5+7+11+13+17+19+23+29+31+37+41+43+47)/15=21.87

Like? Yes (3) | No (2)  

roushan   11 years AGO

(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47)/15=21.87

Like? Yes (3) | No (2)  

manasa epuru   11 years AGO

prime numbers between 1 and 50
2+3+5+7+11+13+17+19+23+29+31+37+41+43+47=328 divided by 15 = 21.87

Like? Yes (3) | No (2)  

Mithlesh Prasad   11 years AGO

taking all prime no.s sum=328
then divide it by no. of prime no.s.
i.e,328/15=21.8667

Like? Yes (2) | No (1)  

sudipta sadhukhan   11 years AGO

2 3 5 7 11 13 17 19 23 29
31 37 41 43 47

these are the 15 prime nos b/w 1-50.
so avg=328/15=21.87

Like? Yes (3) | No (2)  

SATYAM SAURABH   11 years AGO

(2+3+5+7+11+13+17+19+23+29+31+37+41+43+47)/15=21.87
option is 9
note:- 1 is not a prime number

Like? Yes (2) | No (1)  

rajakumar   11 years AGO

There are 15 prime numbers between 1 and 50.
They are: 2 3 5 7 11 13 17 19 23 29
31 37 41 43 47
sum=2+ 3 +5 +7+ 11 +13+ 17+ 19+ 23+ 29+
31 +37+ 41+ 43+ 47=328
average of no's=(total sum/no of prime no b/w 1 to 50)=328/15=21.866666=21.87

Like? Yes (3) | No (2)  

AISHWARYA   11 years AGO

there are 15 prime numbers between 1 to 50
2+3+5+7+11+13+17+19+23+29+31+37+41+43+47=328
avg =328/15=21.87

Like? Yes (2) | No (1)  

aman kumar tiwari   11 years AGO

prime no are...2'+3+'+5+'+7='+11+'1+3'+17='1+9'+23+'29+'31+'37+'41+'43+'47=328/15=21.666666667

Like? Yes (1) | No   

FAISAL RIZWAN   11 years AGO

sum of prime no between 1 to 50 is
2+3+5+7+11+13+17+19+23+29+31+37+41+43+47=328
total no. is 15 then 328/15=21.87

Like? Yes (7) | No (6)  

kumar    11 years AGO

the total number of prime number's 1 to 50 is 15
sum of prime numbers 328
average = 328/15
=21.8666

Like? Yes (2) | No (1)  

Debopam Barik   11 years AGO

prime numbers between 1 to 50 are :
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47

Like? Yes (2) | No (1)  

Grishma Shah   11 years AGO

there are total of 15 prime no. between 1 to 50
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
total =328
average: 328/15=21.866666
avg.= 21.87 ans

Like? Yes (2) | No (1)  

Bingumalla Abinay   11 years AGO

2+3+5+7+11+13+17+19+23+29+31+37+41+43+47=328

328/15=21.866

Like? Yes (2) | No (1)  

sanjay singh   11 years AGO

the prime nos from 1 to 50 are 2,3,5,7,11,13,17,23,27,29,31,37,41,43,47,(15 nos.)
their total=328
average=328/15=21.866=21.87

Like? Yes (2) | No (1)  

LOGAN WOLVERINE   11 years AGO

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
sum is 328
average is 328/15
is 21.87

Like? Yes (1) | No   

Aniruddha Bhattacharjee   11 years AGO

prime numbers between 1 to 50 are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
There is total 15 number and sum is 328
Average=328/15
=21.87

Like? Yes (2) | No (2)  

Rats   11 years AGO

Prime numbers between 1 and 50 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
So average (2+3+5+...+47)/15=21.87.....Ans

Like? Yes (2) | No (2)  

Devendra Marghade   11 years AGO

Prime numbers from 1 to 50 are total 15 as
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
Sum=328, Average=328/15=21.87

Like? Yes (2) | No (2)  

sridevi   11 years AGO

there are 15 prime numbers between 1 to 50
so 328/15=21.87

Like? Yes (2) | No (2)  

KUNAL GAURAV   11 years AGO

We have total 15 primes between 1 and 50 they are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

Adding them all we get 328.
And then dividing the sum by 15 we get 21.87
So the correct answer is 21.87

Like? Yes (2) | No (2)  

IFFAT IRAM   11 years AGO

prime no.s=> 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
avg=21.87

Like? Yes (2) | No (2)  

Avinash Kumar   11 years AGO

Simply adding the prime nos and dividing..... We get 21.87

Like? Yes (2) | No (2)  

arjun   11 years AGO

total number prime numbers ranging 1 to 50 is
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43

total numbers is 15 and their avg will be 21.87

Like? Yes (2) | No (2)  

nishant kumar   11 years AGO

2+3+5+7+11+13+17+19+23+29+31+37+41+43+47
=328
avg=328/15=21.87

Like? Yes (3) | No (3)  

sivaraman   11 years AGO

number of primenumbers is 15. sum of prime numbers 328
so the average is 21.866666=21.87

Like? Yes (2) | No (2)  

Raju Sharma   11 years AGO

prime no =15
and average=sum/15

Like? Yes (2) | No (2)  

Abhishek misra   11 years AGO

sum of all 15 prime numbers between 1 to 50 =328
avg=328/15=21.87

Like? Yes (2) | No (2)  

ABDUL KHALAQUE   11 years AGO

(1+2+3+5+7+11+13+17+19+23+29+31+37+41+43+47)/15=21.86666666666666666666666666

Like? Yes (1) | No (1)  

shraboni bishwas   11 years AGO

ans is 21.8666= 21.87

Like? Yes (2) | No (2)  

ritesh kumar pandey   11 years AGO

2+3+5+7+11+13+17+19+23+29+31+37+41+43+47=328
328/15 = 21.87

Like? Yes (1) | No (1)  

Vinay Sipani    11 years AGO

sum of (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47)/15
= 328/15
= 21.87

Like? Yes (1) | No (2)  

Sourav Maity   11 years AGO

the rime numbers is-2,3,5,7,11,13,17,19,23,29,31,37,41,43,47.
so the avg. is 21.87

Like? Yes (1) | No (2)  

gaurav newatia   11 years AGO

prime no. =2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
sum=328
avg=328/15=21.87

Like? Yes (2) | No (3)  

Alok kumar   11 years AGO

first 50 prime nos are = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

add them n divide by 15

Like? Yes (2) | No (3)  

26 Aug 2013 Sequence

What is the missing number.
8,22,44,.....,112

OPtion

1) 52
2) 59
3) 68
4) 74
5) 81
6) 89
7) 94
8) 99
9) 103
10)None of these

Solution

1*3+5 = 8
3*5+7 = 22
5*7+9 = 44
7*9+11 = 74 (option 4)
9*11+13 = 112

Correct Option: 4 Best Solution (68)

ritesh kumar pandey   11 years AGO

8 = 3^2 -1
22 = 5^2-3
44 = 7^2 - 5
74 = 9^2 - 7
112 = 11^2 - 9
so 74 is the answer;

Like? Yes (10) | No (3)  

Abhijeet Kumar   11 years AGO

(2*2)+4=8
(4*4)+6=22
(6*6)+8=44
(8*8)+10=74 ---> Ans
(10*10)+12=112

Like? Yes (5) | No (1)  

kishor kumar   11 years AGO

8*1-0= 8
8*3-2= 22
8*6-4= 44
8*10-6=74
8*15-8=112

Like? Yes (4) | No (2)  

lakshmi chaitanya   11 years AGO

difference between digits increasing by 8
8 22 44 74 112
22-8=14 44-22=22 74-44=30 112-74=38

Like? Yes (3) | No (1)  

Dr.Dipin Singh   11 years AGO

Every term is given by
Tn= 2*(2n^2+n+1)
T4= 2*(2*4^2 +4+1)=74

Like? Yes (2) | No   

Vinay Sipani    11 years AGO

74

22 - 8 = 14
44 - 22 = 22
x - 44 = y
112 - x = z

To form a sequence,assume 14 , 22 ,x , z are in AP
=> y = 30 , z = 38

=> x = 44 + 30 = 112 - 38 = 74

Like? Yes (6) | No (5)  

Rahul Singh Tomar   11 years AGO

8+14=22
22+(14+8)=44
44+(14+8+8)=74

Like? Yes (1) | No   

SOMASHEKAR MACHANI   11 years AGO

the given series can b written as
3^2 - 1 = 8
5^2 - 3 = 22
7^2 - 5 = 44
9^2 - 7 = 74
11^2 - 9 = 112
ie.,series is (odd number square - previous odd number)

Like? Yes (2) | No (1)  

Namratha   11 years AGO

difference between 8 and 22 is 14..
and the difference between 22 and 44 is 22..
now the difference between 14 and 22 is 8..
so,now add 22+8=30..
now add 44+30=74..so,the next numberis 74..

Like? Yes (1) | No   

Grishma Shah   11 years AGO

8= 8*1+0
22=8*2+6*1+0
44=8*3+6*2+(6*1+2) previous addition no.+2
74=8*4+6*3+(6*2+(6*1+2)+4)
112=8*5+(6*4+(6*3+(6*2+(6*1+2)+4)+6)

Like? Yes (3) | No (2)  

brijesh   11 years AGO

22-8=14
44-22=22
74-44=30
112-74=38...

and 14,22,30,38 all are in difference of 8...

Like? Yes (1) | No   

LOGAN WOLVERINE   11 years AGO

2^2+4=8;
4^2+6=22;
6^2+8=44;
..so

8^2+10=74;
ans 74 . option4.

Like? Yes (1) | No   

Alok Kumar   11 years AGO

ANS : 74
the difference in digit is 14 , 22 ..that is 8
(22-8) =14
(44-22)=22
added in the difference so the 74
so hence the difference 14 , 22 , 30 , 36
(22-8) =14
(44-22)=22
(74-44)=30
(112-74)=36

Like? Yes (2) | No (1)  

Alok kumar   11 years AGO

find 1st difference
then 2nd difference which is constant equal to 8
so missing no is 74

Like? Yes (2) | No (1)  

saurabh verma   11 years AGO

22-8=14
44-22=22
74-44=30
112-74=38

so difference is +8 in each case......

Like? Yes (2) | No (1)  

Raju Sharma   11 years AGO

8+7*2+0=22
22+7*3+1=44
44+7*4+2=74
74+7*5+3=112

Like? Yes (2) | No (1)  

Girjesh Chandravanshi   11 years AGO

3^2-1=8
5^2-3=22
7^2-5=44
9^2-7=74 Ans
11^2-9=112

Like? Yes (1) | No   

avinash   11 years AGO

22-8=14
44-22=22
so difference is increamenting by 8
thus
22+8=30
=> 44+30=74
and 74+38=112

Like? Yes (1) | No   

jatin patware   11 years AGO

2^2+4=8
4^2+6=22
6^2+8=44
8^2+10=74
10^2+12=112
SO OPTION 4 IS CORRECT

Like? Yes (1) | No   

varun kumar .r   11 years AGO

8/2=4
(4+7)*2=22
(11+(7+4))*2=44
(22+(11+4))*2=74
(37+(15+4))*2=112
hence series is 8,22,44,74,112.
observe that a number is obtained by adding the previous number's half and adding number 4 simultaneously to 7

Like? Yes (1) | No   

Madhu pericherla   11 years AGO

2^2+4=8 4^2+6=22 6^2+8=44 8^2+10=74 10^2+12=112 so ans is 74

Like? Yes (1) | No   

Jagadeesh Rampati   11 years AGO

8,22,44,...,112
first difference:22-8=14,44-22=22
after first differences series is 14,22
now 22-14=8
back track 22+8=30
44+30=74
next number in the series is 74

Like? Yes (1) | No   

Udhaya kumar   11 years AGO

8+14=22
22+22=44
44+30=74
74+38=112
so ans is 74

Like? Yes (1) | No   

Sudharsan   11 years AGO

the difference between the numbers is an AP series
14,22,30,38
so the given sequence is 8,22,44,74,112

Like? Yes (1) | No   

Anilkumar   11 years AGO

8
8+14=22
22+14+8=44
44+14+8+8=74
74+14+8+8+8=112

Like? Yes (1) | No   

Ashutosh Anshu   11 years AGO

8+6 = 14
14+8 = 22
22+10 = 32
32+12 = 44
44+14 = 58
58+16 = 74
74+18 = 92
92+20 = 112

Like? Yes (1) | No   

deepalakshmi   11 years AGO

22-8=14
44-22=22
(7*2=14)
(11*2=22)
(15*2=30)
so,44+30=74

Like? Yes (1) | No   

vignesh   11 years AGO

it is sequence
subtracting first two numbers we get
22-8=14
then next two numbers is
44-22=22
and the difference of difference is 22-14=8

by adding this 8 with previous difference is
22+8=30
now add it with 44, it gives the next number
44+30=74

thus the option is 4

Like? Yes (1) | No   

karthik chodagiri   11 years AGO

14,14+8=22,22+8=30,30+8=38 are added consecutively

Like? Yes (1) | No   

Lakshmi Narayanan   11 years AGO

8+14=22
22+22=44
44+30=74
74+38=112.......14,22,30,38 are 8 added to previous..

Like? Yes (1) | No   

ASHA.S   11 years AGO

8,22,44
22-8=14
44-22=22 22-14=8
22+8=30
44+30=74
30+8=38
74+38=112
so the missing no is 74

Like? Yes (1) | No   

Md Intshamuddin   11 years AGO

8---0+8
22--8+14
44--22+22
.....--44+30==74
112--74+38=112

Like? Yes (2) | No (2)  

Hasina Abdulhakkim   11 years AGO

8
22==8+14
44==22+22
.....==44+30=74
112== 74+38

Like? Yes (1) | No (1)  

Rats   11 years AGO

3^2-1=8
5^2-3=22
7^2-5=44
9^2-7=74(......Ans)
11^2-9=112

Like? Yes (1) | No (1)  

susmit   11 years AGO

22-8=14
44-22=22
74-44=30
112-74=38
14,22,30,38 are in AP WITH DIFFERENCE=8
HENCE THE ANSWER IS 74 OPTION NO.4

Like? Yes (1) | No (1)  

ABHISHEK KUMAR MISHRA   11 years AGO

22-8=14
44-22=22
74-44=30
112-74=38
so 74 is the right answer

Like? Yes (1) | No (1)  

Aniruddha Bhattacharjee   11 years AGO

simply 22=8+14
then 44=22+22
74=44+30
112=74+38
AS
22-8=14
44-22=22 and 22-14=8
next no will be 44+22+8=74
Next 74+30+8=112

Like? Yes (1) | No (1)  

praveen sb   11 years AGO

3^2-1=8
5^2-3=22
7^2-5=44
9^2-7=74
11^2-9=112

Like? Yes (1) | No (1)  

Devendra Marghade   11 years AGO

(3^2-1=8)(5^2-3=22)(7^2-5=44)(9^2-7=74)(11^2-9=112)
OR
(2^2+4=8)(4^2+6=22)(6^2+8=44)(8^2+10=74)(10^2+12=112)

Like? Yes (1) | No (1)  

Deepak Sharma   11 years AGO

3^2-1=8
5^2-3=22
7^2-5=44
9^2-7=74
11^2-9=112
so the option 4 is correct

Like? Yes (1) | No (1)  

Mayuresh Srivastava   11 years AGO

8 22 44 74 112
4*2 11*2 22*2 37*2 56*2
7 11 15 19

Like? Yes (1) | No (1)  

Jacky shevkani   11 years AGO

the difference between the first and the second number is
(22-8)=14

the difference between the second and the third number is
(44-22)=22

the height between the differences is
(22-14)=8

series of difference is as follows
14,22,30,38...

so the difference between the third and fourth number should be 30
let the fourth number be x
so
x-44=30
x=74

Like? Yes (1) | No (1)  

richa ahuja   11 years AGO

difference is in AP with common difference 8

Like? Yes (1) | No (1)  

Balakumaran N.A   11 years AGO

sequence increasing by 14,22,30,38...

Like? Yes (1) | No (1)  

Sourav Maity   11 years AGO

22-8=14
44-22=14+8=22
74-44=22+8=30
112-74=30+8=38
so the ans.=74

Like? Yes (1) | No (1)  

SARASVATHI   11 years AGO

22-8=14;
44-22=22;22-14=8
so every segment we add 8.so
22+8=30;44+30=74
30+8=38;74+38=112

Like? Yes (1) | No (1)  

Chandni Maurya   11 years AGO

ans=(4) 74
8+14=22
22+(14+8)=44
44+(22+8)=74
74+(30+8)=112

Like? Yes (1) | No (1)  

Md.Sultan Khichi   11 years AGO

22-8=14
44-22=14+8=22
74-44=30=22+8
112-74=38=30+8
so diference between the numbers and adding 8 we will get the next no.
Ans:74.obtion.4.

Like? Yes (1) | No (1)  

Bingumalla Abinay   11 years AGO

3^2 9-1= 8
5^2 25-3= 22
7^2 49-5= 44
9^2 81-7= 74
11^2 121-9= 112

Like? Yes (1) | No (1)  

rahul jha   11 years AGO

answer:74
22-8=14
44-22=22
74-44=30
112-74=38
we can see--:14+8=22
22+8=30
30+8=38

Like? Yes (1) | No (1)  

siddanth   11 years AGO

8
(14) + 8 = 22
(14+8) + 22 = 44
(14+8+8) + 44 = 74
(14+8+8+8) + 74 = 112

Like? Yes (1) | No (1)  

sudipta sadhukhan   11 years AGO

2*3+2=8
4*5+2=22
6*7+2=44
8*9+2=74
10*11+2=112

Like? Yes (1) | No (1)  

Sunil Kumar   11 years AGO

divide by 2 to each no.
4,11,22,...,56
so
4+7=11
11+11=22
22+15=37
37+19=56

so 37*2=74

Like? Yes (2) | No (2)  

Abhishek misra   11 years AGO

difference between the no is 14,22,30,38
fourth no will be 44+30=74

Like? Yes (1) | No (1)  

rohit punekar   11 years AGO

14
22
30
48

Like? Yes (1) | No (1)  

arjun   11 years AGO

4*2,11*2,22*2,37*2,56*2...
so that 74 is the answer
its like even*2,odd*2,even*2,odd*2,even*2

Like? Yes (1) | No (1)  

sivaraman   11 years AGO

8,22,44,x,122
difference
14,22 difference 4
so
14,22,30,38

so 44+30=74

Like? Yes (1) | No (1)  

ponjothinathan   11 years AGO

The difference between first and second value is 14.and then difference between second and third is 22.
Therefore the difference between 14 and 22 is 8.so,8 will be increase the value in series. add 8+22=30.and then add 30+44=74.
The answer is 74.

Like? Yes (1) | No (1)  

JITENDRA KR.BAROLIYA   11 years AGO

8+14=22,
22+14+8=44,
44+14+8+8=74,
74+14+8+8+8=112
ans=74

Like? Yes (1) | No (1)  

T Lokesh   11 years AGO

given series :3^2-1,5^2-3,7^2-5,x,11^2-9
so x=9^2-7
=74

Like? Yes (1) | No (1)  

sanjay singh   11 years AGO

8=2^2+4
22=(2+2)^2+(4+2)
44=(2+2+2)^2+(4+2+2)
so the next term=(2+2+2+2)^2+(4+2+2+2)
=8^2+10=74

Like? Yes (1) | No (1)  

Aditya Baijal   11 years AGO

The difference increases by 8.
22-8=14
44-22=22....&, 14+8=22
74-44=30......&, 22+8=30
112-74=38.....&, 30+8=38

Like? Yes (1) | No (1)  

ravi singh   11 years AGO

answer is 74

8 + 2*7 = 22

22 + 2*(7 + 4) = 44

44 + 2*(11 + 4) = 74

74 + 2*(15 + 4) = 112

it should also be check from the option as see in series aa numbers are multiple of 2 and in option only 3 option has a multiple of 2 and as see in series that the difference between all the no. is in increasing order and in option only 74 verify this condition so answer is 74

Like? Yes (1) | No (1)  

Tapesh mishra   11 years AGO

4,22,44,?,112 can be shown by
(3)^2-(1)=8
(5)^2-(3)=22
(7)^2-(5)=44
(9)^2-(7)=74
(11)^2-(9)=112

so if u agree please like this

Like? Yes (2) | No (3)  

DNYANESHWAR PURUSHOTTAM AGARKAR   11 years AGO

22-8=14
44-22=22
therefore44+30=74

Like? Yes (1) | No (2)  

Akshay Hanswal   11 years AGO

22-8=14
44-22=22
Nd 22-14=8
Since 74-44=30 which is 22+8
Or 30-22=8
Nd also 112-74=38
Which is 30+8. . .hence 74 is answer

Like? Yes (1) | No (2)  

Gokul Krishnan   11 years AGO

the series have difference of
8,22,44,74,112
difference is 14,22,30,38
hence the difference is 8 in the difference.

Like? Yes (1) | No (2)  

kamlesh kumar vishwakarma   11 years AGO

8-------22-------44-------x-------112
--7*2+0----7*3+1---7*4+2----7*5+3----
x=44+(7*4+2)=44+30=74....
hence missing number is 74.

Like? Yes (1) | No (2)  

24 Aug 2013 In a group

In a group of children each child gives a gift to every other. If the number of gifts is 132, find the number of children.

OPtion

1) 11
2) 12
3) 14
4) 16
5) 18
6) 22
7) 24
8) 28
9) 32
10) None of these

Solution

Let there are n children, then each child will have n-1 gifts.
Total number of gift=total number of children*number of gifts per child
n(n-1)=132
solve
n can not be -ve
n=12

option 2

Correct Option: 2 Best Solution (20)

devendra kumar gond   11 years AGO

Let there are N children,then each of them will give (N-1) gifts,(each children will not gift himself)
Total gifts =N*(N-1) =N^2 -N
N^2 -N =132
N^2 -N -132 =0
(N-12)(N+11) =0
N = 12 ,-11
as number of children can not be negative,
so number of children =12

Like? Yes (22) | No (6)  

Dr.Dipin Singh   11 years AGO

If there are n chidren, then each gave (n-1)gifts.
so n(n-1)=132=12*11
so n=12

Like? Yes (20) | No (4)  

Anu   11 years AGO

assume that there are n children and the number of gifts will be defined as n-1 as each child gives a gift to every other.
clearly n(n-1)=132
n^2-n-132=0
(n-12)(n+11)=0
n=12 is taken as it is positive!!!
therefore the answer is 12.

Like? Yes (20) | No (6)  

Chandni Maurya   11 years AGO

ans=(2). 12
let x be the no of children,
x(x-1)=132
=>x^2-x-132=0
=>(x-12)(x+11)=0
=> x=12,-11 => x=12

Like? Yes (12) | No (1)  

SOMASHEKAR MACHANI   11 years AGO

x(x-1) = 132
solve for x ....
we get x = 12

Like? Yes (15) | No (4)  

sudipta sadhukhan   11 years AGO

say their are n members in the group.
so, their are (n-1)options to each of them,so that the child can give gifts to other (n-1)members.
So the result will be n*(n-1).
here n*(n-1)=132 =>n=12

Like? Yes (9) | No   

Mehak Koul   11 years AGO

each child will give 1 gift less than the total no of children in a group. so if there are 12 children in a group each will give 11 gifts. toatl becomes 12*11=132 gifts.

Like? Yes (7) | No (2)  

Devendra Marghade   11 years AGO

If n=number of children &
if each child gives a gift to every other,then total number of gifts=n*(n-1)=132 or
n^2-n-132=0 ,(n-12)(n+11)=0, n=12

Like? Yes (7) | No (3)  

T Lokesh   11 years AGO

each children gives a gift to other children
hence no of gits when n children are present =n*(n-1)
so n*(n-1)=132
by solving we get n=12 and n=-11
n cant be a negative number
so no of students =12

Like? Yes (6) | No (3)  

praveen sb   11 years AGO

let there be X students .. if each student gives a gift to everyone it means one student gives gift to X-1 students ... therefore X(X-1)=132
by solving we get X=12

Like? Yes (5) | No (2)  

krishnapriya   11 years AGO

Let the no. of children be 'n'
Each child gives a gift to others.
n(n-1)=132
n^2-n-132=0
so, n=12

Like? Yes (5) | No (3)  

Abhishek   11 years AGO

Suppose we have n children in the group, so each child will give gifts to every other child and so the required gifts will be n*(n-1).
n*(n-1) = 132
As we have total 132 gifts available.
After solving we get 12.
So there must be 12 children in the group.

Like? Yes (4) | No (2)  

Neeraj Kumar   11 years AGO

Let the number of the children be x.
Since every children gives gift to all remaining children therefore every children will produce x-1 gifts.
Hence total number of gifts = x(x-1)
that is x(x-1) = 132=12*11
gives x= 12
Hence number of children is 12.
Option 2 is correct.

Like? Yes (4) | No (2)  

Shikha Sinha   11 years AGO

let no of children be n.... now if we find permutation i.e nP2=132... on finding n... it gives no of children.

Like? Yes (4) | No (4)  

manasa epuru   11 years AGO

n(n-1)=132=12*11
no of students=12

Like? Yes (3) | No (3)  

Ananya Gupta   11 years AGO

let the no. of children be n.
each child gives (n-1) gifts to everyone except himself.
then
n*(n-1)=132
we get n=-11,12
so 12 is the correct answer.

Like? Yes (2) | No (3)  

Vinay Sipani    11 years AGO

nP2 = 132
n(n-1) = 132
n = 12

Like? Yes (4) | No (6)  

Susmit Bansod   11 years AGO

if there are x students in a group, each student will give (x-1) gifts. so, total gifts are x(x-1). therefore,x(x-1)=132. so, x=12

Like? Yes (1) | No (3)  

rohit punekar   11 years AGO

if the nomber of children are 12 then,
each child gives one gift to the other 11 childrens.
therefore,12*11=132
hence,the answer is 12.

Like? Yes (2) | No (4)  

vky   11 years AGO

{(n-1)*n}/2=132
n=12
since one will gift to (n-1) children and then (n-2) and so on.
so (n-1)+(n-2)+......+3+2+1={(n-1)*n}/2

Like? Yes (2) | No (5)  

23 Aug 2013 Words

How many words can be formed by using 4 letters at a time of word "SURPRISE" (words may be meaningful or meaningless according to dictionary)

OPtion

1) 498
2) 510
3) 544
4) 586
5) 606
6) 642
7) 698
8) 720
9) 780
10) None of these

Solution

(1) No. of words having all letters different = 6*5*4*3 = 360
(2) No. of words having any one letter repeated
=(2*1)*(5*4*3*2*1/3*2*1*2*1)*(4*3*2*1/2*1) = 240
(3) No. of words having two letter repeated = (1)*(4*3*2*1/2*1*2*1) = 6
Total words = 360+240+6 = 606
ans 606
option 5

Correct Option: 5 Best Solution (2)

Vinay Sipani    11 years AGO

606 ways

SSRRUPIE
1st case i.e. no repetition of letters
SRUPIE => 6C2 * 4! = 360

2nd case => repetition of S
=> selection of two letters from other five letters
=> 5C2 * 4! / 2! = 120

3rd case => repetition of R
=> selection of two letters from other five letters
=> 5C2 * 4! / 2! = 120

4th case => repetition of bot S and R
arrangement of letters SSRR => 4! / (2! * 2!) = 6

=> Total ways = 360 + 120 + 120 + 6 = 606 ways

Like? Yes (16) | No (1)  

KUNAL GAURAV   11 years AGO

We have 6 distinct letters so 6P4 words can be formed from them.
We have 2 groups of 2 letters S and R. Any one of them can be selected by 2C1 and rest 2 letters can be selected from the remaining 5 letters and no. of words can be fromed is 4!/2!.
And if we select both the groups of S and R letters then we can form 4!/(2!*2!)
Finally adding all we get
6P4+(2C1*5C2 4!/2!)+4!/(2!*2!) = 606

Like? Yes (3) | No   

21 Aug 2013 A teacher

A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left over. When he increased the size of the square by one student he found that he was short of 25 students. Find the number of students.

OPtion

1) 450
2) 470
3) 495
4) 525
5) 550
6) 600
7) 700
8) 1200
9) 2400
10) None of these

Solution

Let the number of students in one side = x
Total number of students = x^2+24 (because 24 students were left)
When size of the square in increased by 1
then total number of students = (x+1)^2 - 25 (25 students were short)
Now the number of students in same,
x^2 + 24 = (x+1)^2 - 25
solve
x = 24
Total number of students = 24^2 + 24 = 600
option 6

Correct Option: 6 Best Solution (22)

FRANKLIN VIJAY   11 years AGO

no of students be x
side of square be y
y*y=x-24 => y^2=x-24 =>1st equation
(y+1)(y+1) = x + 25 =>2nd equation
y^2+2y+1=x+25
on substituting 1st equation we get
x-24+2y+1=x+25
we get 2y=48
so=>y = 24
on subst y=24 in 2nd equation
25*25=x+25
x=625-25
x = 600
no of studnts= 600

Like? Yes (20) | No (6)  

Muhammad Nadeem Haider Naseem   11 years AGO

Let the side of the square be x.
No. of students = x
2
+ 24
New side = x + 1
No. of students = (x + 1)2
– 25
APQ ⇒ x
2
+ 24 = (x + 1)2 – 25
⇒ x
2
+ 24 = x
2
+ 2 x + 1 - 25
⇒ 2x = 48
⇒ x = 24
∴ side of square = 24
No. of students = 576 + 24
= 600

Like? Yes (10) | No (4)  

nishant kumar   11 years AGO

let the no. of students be ---- x
the side of the square be ----- y
y*y = x - 24
(y+1)(y+1) = x + 25
on solving them simultaneously.....we have y = 24
x = 600
Total students = 600
ANS: 600 students.

Like? Yes (7) | No (3)  

Mayuresh Srivastava   11 years AGO

Let the no. of students be ---- a
the side of the square be ---- x
x*x=a-24
(x+1)(x+1)=a+25
on solving we have x = 24 a = 600
Total students = 600.

Like? Yes (7) | No (3)  

kishor kumar   11 years AGO

let side of square = x
so no. of st. = x^2+24
now if side of squar increased by 1
than no of student = (x+1)^2-25
becoz no of students are equal in both cases so
x^2+24 = (x+1)^2-25
or x = 24
so no of st. = (24)^2+24 =600

Like? Yes (6) | No (2)  

krishnapriya   11 years AGO

let the no of students be 'x'
the side of square be 'y'
given, y*y=x-24--------(1)
(y+1)(y+1)=x+25--------(2)
solving the equations (1) and (2) we get
x=600 and y=24

Like? Yes (5) | No (2)  

mohan lal garg   11 years AGO

Let no of students be x

& side of square be y

then y *y = x - 24

according to other condition ( y + 1 ) ( y + 1 ) = x + 25

on solving them we get y = 24 and the value of x = 600

therefore no of students are 600 and value of each side of square is 24

Like? Yes (5) | No (2)  

Vinay Sipani    11 years AGO

Let No. of students on each side = x
Arranging students in form of solid square
No of students in square = area = x^2
=> Total no. of students = x^2 + 24

Increasing 1 student on each side
=> No of students in square = area = (x + 1)^2
But 25 students were less
=> Total no of students = (x + 1)^2 - 25

=> x^2 + 24 = (x + 1)^2 - 25
=> 49 = 2x + 1
=> x = 24
=> No. of students = x^2 + 24 = 600

Like? Yes (5) | No (2)  

Anil Kumar Bharti   11 years AGO

Exp:
let the no. of students be ---- x
the side of the square be ----- y
y*y = x - 24
(y+1)(y+1) = x + 25
on solving them simultaneously.....we have y = 24
x = 600
Tot stu = 600 Ans

Like? Yes (6) | No (3)  

Amit kumar   11 years AGO

let the no. of students is x
the side of the square is y
according to ques
y*y = x - 24
(y+1)(y+1) = x + 25
on solving the equations we have y = 24
x = 600
Total students = 600
ANS: 600 students.

Like? Yes (4) | No (2)  

P.L.ANANTHI   11 years AGO

let the side of the square be x m
students be y
1 case
y = x2 + 24---1
2nd case
y = (x + 1) square - 25-------------2
from eq 1 and eq 2
x2 +24 = x + 1 square -25
x = 24
puttin in eq 1
y = 600

Like? Yes (2) | No   

Niraj   11 years AGO

Let no of students be x & side of square be y
then according to given conditions
y*y=x-24 ----------- (1)
(y+1)(y+1)=x+25 ----------- (2)
on solving these 2 equations,
x=600 and y=24
So no of students are 600 and each side of square is 24

Like? Yes (4) | No (2)  

MOVVA ROHITH   11 years AGO

let there are total y students
and in square there is x rows and x columns
so,x^2=y-24 from the 1st condition
and (x+1)^2=y+25 from 2nd condition.
by solving these two we get y=600.

Like? Yes (3) | No (2)  

Md.Sultan Khichi   11 years AGO

let the total no. of students be y and side of the square be x.
according to the ques
x^2=y-24
(x+1)^2=y+25
after solving this x=24.
y=24^2+24=600.
total no of students are 600.
Ans:6.

Like? Yes (3) | No (2)  

deepak kumar   11 years AGO

total no. of students => x^2+24=(x+1)^2-25
solving this we get x=24
so total no. of students =24^2+24=(25)^2-25=600

Like? Yes (1) | No   

SHIVANI PRAKASH   11 years AGO

let initially the square formed was with x students on each side and 24 were left.
thus total students=x^2+24
the side of square was increased by 1 and 25 students were short.
thus total students=(x+1)^2-25
x^2+24=(x+1)^2-25
thus x=24
and x^2+24=600 is the required answer...

Like? Yes (3) | No (2)  

aman kumar tiwari   11 years AGO

let the no. of students be ---- x
the side of the square be ----- y
y*y = x - 24
(y+1)(y+1) = x + 25
on solving them simultaneously.....we have y = 24
x = 600
Total students = 600
ANS: 600 students.

Like? Yes (1) | No   

Devendra Marghade   11 years AGO

If number of student=x and
student to be arranged in a line=n
Ist case:n^2+24=x or n^2=x-24 ---(i) &
IInd case:(n+1)^2=x+25 or n^2+2n+1=x+25 ---(ii)
Substituting value of n^2 in (ii),
(x-24)+2n+1=x+25, n=24,so x=n^2+24=24^2+24=600

Like? Yes (2) | No (2)  

Bingumalla Abinay   11 years AGO

x^2+24=(X+1)^2-25
solving we get x=24
total no students = X^2+24=600

Like? Yes (2) | No (2)  

IFFAT IRAM   11 years AGO

let the no. of students in a row be x
total no. of students forming a square= x^2
overall students=x^2+24
new no. of students forming a square= (x+1)^2
overall student=(x+1)^2-25

x^2+24=(x+1)^2-25
x^2+24=x^2 +1+2x-25
24=2x-24
2x=48
x=24

no. of students x^2+24
=> 24^2+24
600

Like? Yes (2) | No (2)  

siddanth   11 years AGO

x^2 + 24 = (x+1)^2-25

=> x=24
no of students x^2+24 = 600

Like? Yes (2) | No (2)  

Eliza    11 years AGO

let x be the total number of students. Let y be the number of students arranged in each row and column of the square. Hence there will be y^2 students arranged in the form of square.
also x-(y^2)=24 --> x= 24+y^2
(y+1)^2-x=25
(y^2)+2y+1-(y^2)-24=25
2y=48
y=24
hence x=576+24=600

Like? Yes (2) | No (2)  

20 Aug 2013 Triangle

The area of an isosceles triangle is 60 cm^2 and the length of each one of its equal sides is 13 cm. Find its base.

OPtion

1) 25, 12
2) 24, 10
3) 20, 24
4) 19, 20
5) 15, 12
6) 12, 15
7) 10, 24
8) 9, 12
9) 8, 24
10)None of these

Solution

Let base x and altitude p
Area=1/2(x*p)=60
p=120/x
Now applying pythagoras theorem
(x/2)^2+(120/x)^2=13^2
solve
x=24, x=10
option 7

Correct Option: 7 Best Solution (9)

Vinay Sipani    11 years AGO

Area of triangle = sqrt [ s(s-a)(s-b)(s-c) ]
a = c = 13 , s = (a+b+c)/2 = (26 + b)/2
(s - a) = (s - c) = (b/2) , (s - b) = (26-b)/2

=> 60^2 = (26 + b)(26 - b)b^2 / 16
=> 16*60^2 = b^2*(26^2 - b^2)
=> b = 10,24

Like? Yes (32) | No (19)  

Abhishek misra   11 years AGO

Having drawn your isosceles triangle with the length of the equal sides equal to 13 cm, the width of the base and the altitude equal to 2.x and h respectively, then we may write for the area A of the triangle

A = h.x = 60 cm²

so that h = 60/x

and from Pythagoras applied to either of the right-angled triangles

x² + h² = 13²

Substituting for h from above gives

x² + (60/x)² - 13² = 0

Multiplying through by x² and rearranging

x^4 - 169.x² + 3600 = 0

which is a quadratic in z = x² with solutions

z = [169 ±√(169² - 4*3600)]/2

giving z = [169 ± 119]/2 = 25 and 144

so that the solutions for x are 5 and 12, both of which are valid,

The base of the triangle may therefore be either 2.x = 10 or 24 cm in length.

Like? Yes (19) | No (6)  

~Sandeep~   11 years AGO

Divide the isosceles triangle in two halves by dividing the base side from middle .
Now we have 2 right angle triangles and the area of each of two resultant triangles is 30cm^2.
Using prognathous theorem where hypotenuse is fixed as 13 and also 1/2*base*height= 30
We get two equations:
b^2 + h^2 = 13^2
1/2 *b*h= 30
=> b*h = 60

Solving we get 5, 12. As we have divided the base in two equal parts we need to multiply by 2
Hence answer is 5*2, 12*2 = 24, 10

Like? Yes (25) | No (13)  

Jigyasa Kumar   11 years AGO

Let x represent the measure of segement AD. Let y represent segment BD. Since we know from a common geometric proof that the perpendicular from the apex to the base of an isosceles triangle bisects the base, the apex angle, and the triangle itself. Therefore the area of triangle ABD is equal to the area of triangle BCD and the area of each is one-half of the area of the entire triangle ABC.
2x=24,10
Area =xy
x^2+y^2=13
x^2+3600/x^2=169
we get x=12,5
so,
segment BD bisects segment AC, and by construction, AD plus DC is equal to AC, we can deduce that AC is equal to two times AD. Therefore the base measurement that we seek as an answer to this problem is two times the value of
or
2x=24,10

Like? Yes (8) | No (3)  

Aarush Singh   11 years AGO

Let x represent the measure of segement AD. Let y represent segment BD. Since we know from a common geometric proof that the perpendicular from the apex to the base of an isosceles triangle bisects the base, the apex angle, and the triangle itself. Therefore the area of triangle ABD is equal to the area of triangle BCD and the area of each is one-half of the area of the entire triangle ABC.
2x=24,10
Area =xy
x^2+y^2=13
x^2+3600/x^2=169
we get x=12,5
so,
segment BD bisects segment AC, and by construction, AD plus DC is equal to AC, we can deduce that AC is equal to two times AD. Therefore the base measurement that we seek as an answer to this problem is two times the value of
or
2x=24,10

Like? Yes (6) | No (3)  

Mayuresh Srivastava   11 years AGO

Area of an Isosceles triangle=(b/4)sqrt(4a^2-b^2)
where a=Each one of equal sides
b=Base
Equation will be b^4-676b^2+240^2=0
so b^2=576,100
Thus b=24,10.

Like? Yes (5) | No (2)  

ritesh kumar pandey   11 years AGO

its equal sides are 13 cm..
by Pythagoras theorem, in right triangle with median and base, the median length is of 12 or 5 cm . similarly the half of base is 5 or 12 cm. so total of its base is 12*2 = 24; or 5*2 = 10;
so option 2 & 7 both are correct.

Like? Yes (7) | No (5)  

garvit sharma   11 years AGO

becoz hight is x and base is y then 1/2xy=60
and in right angle tringle (x/2)^2+(y/2)^2=13^2 so by eq. 1 and 2 base may be 24 or 10

Like? Yes (7) | No (6)  

susmit   11 years AGO

Let x be the base and y be the height.
So 1/2(xy)=60.....(1)
and by Pythagoras (x/2)^2 + y^2= 13^2 = 169 (2)
Solving (1)&(2)
We get x= 24 or x=10.
Hence option no. 2.

Like? Yes (9) | No (9)  

19 Aug 2013 sequence

next number of the sequence 18, 46, 94, 63, 52 ...... will be

OPtion

1) 16
2) 36
3) 41
4) 45
5) 56
6) 61
7) 64
8) 78
9) 87
10) none of these

Solution

18----------81
46----------64
94----------49
63----------36
52----------25
so next will be 16 so in reverse it will be 61

Correct Option: 6 Best Solution (59)

Md Intshamuddin   11 years AGO

18, 46, 94, 63, 52 ......
1st reverse the digits in the following way
81,64,49,36,25..
so next term will be 16
and after reversing it 61 will be the answer

Like? Yes (20) | No (3)  

Pargat Pal Singh   11 years AGO

18, 46, 94, 63, 52 .

perfect squares with reversed digits.
next number will be 61 .. reverse of 16=4^2

Like? Yes (19) | No (6)  

~Sandeep~   11 years AGO

Reverse multiples of squares of integers.
9x9 = 81 => 18
8x8 = 64 => 46
7x7 = 49 => 94
6x6 = 36 => 63
5x5 = 25 => 52
4x4 = 16 => 61

Like? Yes (10) | No (4)  

susmit   11 years AGO

Reversing the digits of the numbers we get,
81, 64, 49, 36, 25.
These are the squares of numbers,
9, 8, 7, 6, 5.
Hence the next number will be the
square of 4. which is 16.
Reversing digits again the answer is 61.
Hence option no. 6.

Like? Yes (3) | No (1)  

Vinay Sipani    11 years AGO

reversing all the digits,

81,64,49,36,25
=> next term will be reverse of 16 = 61

Like? Yes (3) | No (1)  

Ankit Raj   11 years AGO

Each one is the square root backwards.

9x9=81
8x8=64
7x7=49
6x6=36
5x5=25
4x4=16
so the next no. is 61

Like? Yes (4) | No (2)  

aman kumar tiwari   11 years AGO

81=9*9......18
64=8*8.....46
.
.
.
4*4=16.....61.........anawer

Like? Yes (2) | No   

richa ahuja   11 years AGO

reverse of square..
81,64,49,36,25,16
therfore nxt no is 61
reverse of 16

Like? Yes (2) | No   

malarvezhi.m   11 years AGO

9*9=81 its reverse is 18
8*8=64 its reverse is 46
7*7=49 its reverse is 94
6*6=36 its reverse is 63
5*5=25 its reverse is 52
so
4*4=16 and its reverse is 61

Like? Yes (2) | No   

SHIVANI PRAKASH   11 years AGO

add to the terms nos. 28,48,68 and so on...
18+28=46
46+48=94
94+68=162(in case of a 3 digit no.:make the 1st digit of the next term in series as the middle digit of sum obtained(here 6) nd 2nd digit as addition of the 1st nd 3rd digit obtained in the sum(here 1+2=3)
therefore the no. becomes 63
now,63+88=151(its a 3 digit no.) thus it becomes 52
now,52+108=160(3 digit no.)thus it becomes 61

Like? Yes (3) | No (1)  

Hasina Abdulhakkim   11 years AGO

18, 46, 94, 63, 52 ......
81,64,49,36,25..
so next term will be 16
and hence required number is 61

Like? Yes (6) | No (5)  

Jigyasa Kumar   11 years AGO

They are backward multiples of square root integers.

9x9 = 81
8x8 = 64
7x7 = 49
6x6 = 36
5x5 = 25
4x4 = 16

Like? Yes (3) | No (2)  

LOGAN WOLVERINE   11 years AGO

18,46,94,63,52,.....
this series is reverse order of square value
81,64,49,36,25,.......
9^2,8^2,7^2,6^2,5^2,
next number is 4^2=16
reverse of 16 is 61.
answer is 61. option 6.

Like? Yes (2) | No (1)  

Niraj   11 years AGO

Reverse the digits of square of integers.
9x9=81, Reverse digits = 18
8x8=64, Reverse digits = 46
7x7=49, Reverse digits = 94
6x6=36, Reverse digits = 63
5x5=25, Reverse digits = 52
4x4=16, Reverse digits = 61

Like? Yes (2) | No (2)  

Anil Kumar Bharti   11 years AGO

Exp:
18, 46, 94, 63, 52 ......
Interchange Unit & tenth place no is sqrt.

Like? Yes (1) | No (1)  

Sagar Jinachandran   11 years AGO

9^2=81 reversing the digits 18
8^2=64 reversing the digits 46
7^2=49 reversing the digits 94
6^2=36 reversing the digits 63
5^2=25 reversing the digits 52
so the next number
4
4^2=16 reversing the digits 61
option 6

Like? Yes (1) | No (1)  

IFFAT IRAM   11 years AGO

9*9=81
8*8=64
7*7=49
6*6=36
5*5=25
4*4=16
reverse the digits
so 61 is the next no.

Like? Yes (1) | No (1)  

Siddhant Sambit   11 years AGO

9*9=81;18
8*8=64;46
7*7=49;94
6*6=36;63
5*5=25;52
4*4=16;61

Like? Yes (1) | No (1)  

Tapesh mishra   11 years AGO

46-18=28
94-46=48
63-94=-31
52-63=-11
next difference will -11+20=9
52+9=61

Like? Yes (1) | No (1)  

SOMASHEKAR MACHANI   11 years AGO

all the numbers are reverse of squares .....
that is 9^2=81....18
8^2=64....46
7^2=49....94
6^2=36....63
5^2=25....52
similarly4^2=16....61

Like? Yes (1) | No (1)  

Chandni Maurya   11 years AGO

ans=(6). 61
18,46,94,63,52
reversing the digits,
81,64,49,36,25=>which are the suares of 9,8,7,6,5 resp.
so next will be 4^2=16=>61

Like? Yes (1) | No (1)  

vky   11 years AGO

swap the no. in 18, it will become 81
similarly we get 64, 49, 36, 25..... these are perfect square of natural numbers in decreasing order so next will be 16. so swap the no.'s and we will get our final answer that is 61...
Hence option 6 is correct.

Like? Yes (1) | No (1)  

Aarush Singh   11 years AGO

They are backward multiples of square root integers.

9x9 = 81
8x8 = 64
7x7 = 49
6x6 = 36
5x5 = 25
4x4 = 16

Like? Yes (2) | No (2)  

Bingumalla Abinay   11 years AGO

9x9 = 81
8x8 = 64
7x7 = 49
6x6 = 36
5x5 = 25
4x4 = 16

Like? Yes (1) | No (1)  

Ayush Jain   11 years AGO

Answer 61

They are backward multiples of square root integers.

9x9 = 81
8x8 = 64
7x7 = 49
6x6 = 36
5x5 = 25
4x4 = 16
--------

which means your answer is 61

Like? Yes (1) | No (1)  

garvit sharma   11 years AGO

interchange the unit's and ten's digit.
u will obtain a series with (9*9), (8*8), (7*7), and so on. so the last no. will be 4*4 which is 16 and on interchanging the values we get 61 which is the answer

Like? Yes (1) | No (1)  

mohamed ismail   11 years AGO

The nos are squared and written backwards from 9 onwards.Thus the inverse of 4 squared 16 is written as
61

Like? Yes (1) | No (1)  

MUKESH KUMAR   11 years AGO

number one is squaring 9 and then flipping the numbers. 9x9=81=18. then 8x8=64=46.the missing number is 61.

Like? Yes (1) | No (1)  

NIDHI   11 years AGO

number one is squaring 9 and then flipping the numbers. 9x9=81=18. then 8x8=64=46.the missing number is 61.

Like? Yes (1) | No (1)  

SWAPNIL PATIL   11 years AGO

9*9=81 Reverse of 81 is 18
8*8=64 Reverse of 64 is 46
7*7=49 Reverse of 49 is 94
6*6=36 Reverse of 36 is 63
5*5=25 Reverse of 25 is 52
SO,NEXT NUMBER WILL BE
4*4=16 So its reverse will be the answer which is "61".

Like? Yes (1) | No (1)  

Abhijeet Kumar   11 years AGO

18,46,94,63,52...
Square the number from 9 to 4 then exchange the digits.
9*9=81=18(exchange digit)
8*8=64=46
7*7=49=94
6*6=36=63
5*5=25=52
then
4*4=16=61 Ans

18,46,94,63,52,61.

Like? Yes (1) | No (1)  

NILOTPAL MRINAL   11 years AGO

because
9*9=81 =>18
8*8=64 =>46
7*7=49 =>94
6*6=36 =>63
5*5=25 =>52
so
4*4=16 =>61

Like? Yes (1) | No (1)  

Mayuresh Srivastava   11 years AGO

18, 46, 94, 63, 52 ......
Do Square and Write in Reverse Order.
i.e. 9*9=81(18)
8*8=64(46)
7*7=49(94)
6*6=36(63)
5*5=25(52)
4*4=16(61) so next term is 61.

Like? Yes (1) | No (1)  

FAISAL RIZWAN   11 years AGO

after reversing the series:-81,64,49,36,25
now 81=9^2
64=8^2
49=7^2
36=6^2
25=5^2
16=4^2
so 61 is reversed as 16

Like? Yes (2) | No (2)  

Grishma Shah   11 years AGO

its backward multiplication
9*9=81==18
8*8=64==46
7*7=49==94
6*6=36==63
5*5=25==52
4*4=16==61
3*3=09==90 e.t.c

Like? Yes (1) | No (1)  

ravi singh   11 years AGO

The first digit of the no. Is last term of the square like 1^2=1,2^2=4,3^2=9,4^2=16=6(last digit),5^2=25=5,hence first digit of required no. is 6^2=36=6,and the second digit is going to decrease like 8,6,4,3,2,hence second digit of the required no. is 1 Hence no. is 61

Like? Yes (1) | No (1)  

Abhishek misra   11 years AGO

They are backward multiples of square root integers.
9x9 = 81
8x8 = 64
7x7 = 49
6x6 = 36
5x5 = 25
4x4 = 16
hence the no.is 61

Like? Yes (1) | No (1)  

Nikhilesh Jhade   11 years AGO

9x9 = 81
8x8 = 64
7x7 = 49
6x6 = 36
5x5 = 25
4x4 = 16
i.e 61

Like? Yes (1) | No (1)  

utkarsh singh   11 years AGO

as given
18 46 94 63 52 here change position of numbers 81(9^2) 64(8^2) 49(7^2) 36(6^2) 25(5^2) so as 16(4^2) so answer will be 61

Like? Yes (1) | No (1)  

kumar    11 years AGO

we shall given number reverse order in each number is
81,64,49,36,25...then
9*9=81
8*8=64
7*7=49
6*6=36
5*5=25
4*4=16
so reverse in each number 18,46,94,63,52,61
finally next number is 61....

Like? Yes (1) | No (1)  

preeti sinha   11 years AGO

each number is reverse of the square value as:
9*9= 81
8*8=64
7*7=49
6*6=36
5*5=25
4*4=16
thus the next number is 61

Like? Yes (1) | No (1)  

vinyosna   11 years AGO

18 rev=81=9^2
46 rev=64=8^2
94 rev=49=7^2
63 rev=36=6^2
52 rev=25=5^2
so next no. is rev(4^2)=rev(16)=61

i.e option 6

Like? Yes (1) | No (1)  

Nayan Agrawal   11 years AGO

They are backward multiples of square root integers.

9x9 = 81
8x8 = 64
7x7 = 49
6x6 = 36
5x5 = 25
4x4 = 16
--------
the next number in the sequence would be:
61

Like? Yes (1) | No (1)  

M Rakesh   11 years AGO

9^2=81=reverse it=18
8^2=64=reverse it=46
7^2=49=reverse it=94
6^2=36=reverse it=63
5^2=25=reverse it=52
next will be 4^2=16=reverse it=61

Like? Yes (1) | No (1)  

Jagadeesh Rampati   11 years AGO

9x9 = 81 reverse=18
8x8 = 64 reverse=46
7x7 = 49 reverse=94
6x6 = 36 reverse=63
5x5 = 25 reverse=52
4x4 = 16 reverse=61
the answer is 61
option 6

Like? Yes (1) | No (1)  

ankit chauhan   11 years AGO

answer is 61 because...notice the first digit of every number and observe the sequence. 1+3=4 4+5=9 9+7=16 means 6 6+9=15 means 5 similarly 5+11=16 means 6 ..so 6 at tens place . and for units place it has a decreasing sequence of numbers so the guess is 1 .SO the number is 61

Like? Yes (1) | No (1)  

Nikunj Sharma   11 years AGO

Just reverse all the numbers like.
18 =81 =>9*9
46 =64 =>8*8
94= 49 =>7*7
63= 36 =>6*6
52= 25 =>5*5

So
next digit will be
61 as 4*4=16

Like? Yes (1) | No (1)  

Alok kumar   11 years AGO

1st invert the place of digits den we hv :
81,64,49,36,25
these r the sq of :9,8,7,6,5,4,3,........
so the next no is 4
(4)^2=16
by inverting the place of 16= 61
so ans is 61

Like? Yes (1) | No (1)  

vinod   11 years AGO

9^2=81 write the digits in reverse order-->18
8^2=64---?46
7^2=49--->94
answer:: 4^2=16---> 61

Like? Yes (1) | No (1)  

Praveen   11 years AGO

61! the first number in every digit is units place of perfect squares,if the number lies within 1 to 10,or 10 to 20 subtract 2(eg:18 i.e 8-2=6i.e 46)or else subtract 1(eg:63i.e 6-1=5 i.e 52

Like? Yes (1) | No (1)  

Raju Sharma   11 years AGO

9x9 = 81
8x8 = 64
7x7 = 49
6x6 = 36
5x5 = 25
4x4 = 16

Like? Yes (1) | No (1)  

Mariappan   11 years AGO

9^2 = 81 -> 18
8^2 = 64 -> 46
7^2 = 49 -> 94
6^2 = 36 -> 63
5^2 = 25 -> 52
Therefore,
4^2 = 16 -> 61 (Ans)

Like? Yes (1) | No (1)  

svsrpavankumar vadrevu   11 years AGO

9x9 = 81 8x8 = 64 7x7 = 49 6x6 = 36 5x5 = 25 4x4 = 16

Like? Yes (1) | No (1)  

vignesh   11 years AGO

hai friends
to begin with we have to reverse all the numbers then we get

81,64,49,36,25....
which can be also written as
9^2,8^2,7^2,6^2,5^2.......

hence the next no must be 4^2=16

then reversing the terms we get 61 therefore the answer is option 6

Like? Yes (1) | No (1)  

Pradyumna K S   11 years AGO

They are backward multiples of square root integers.

9x9 = 81
8x8 = 64
7x7 = 49
6x6 = 36
5x5 = 25
4x4 = 16

Like? Yes (1) | No (1)  

ritesh kumar pandey   11 years AGO

series is the reverse of digits of
81, 64, 49, 36,25 next is 16 so reverse of 16 is 61. that is the answer..

Like? Yes (1) | No (2)  

Deepak Sharma   11 years AGO

9X9=81.......18
8X8=64.......46
7X7=49.......94
6X6=36.......63
5X5=25.......52
SO NEXT NUMBER WILL BE
4X4=16.......61
SO OPTION 6 IS CORRECT.

Like? Yes (1) | No (2)  

nishant kumar   11 years AGO

9*9=81=>18
8*8=64=>46
7*7=49=>94
6*6=36=>63
5*5=25=>52
4*4=16=>61

Like? Yes (1) | No (2)  

ASHA.S   11 years AGO

the sequence is 18,46,94,63,52
taking units place
1+3 4 8-2 6
4+5 9 6-2 4
9+7 16 put 6 extra 1 4-2+1(extra) 3
6+9 5 1 3-2+1(extra) 2
5+11 6 1 2-2+1 1
so 61

Like? Yes (1) | No (2)  

16 Aug 2013 A person

A person borrows Rs.4500 and promises to pay back (without any interest) in 30 installment, each of value Rs.10 more than the last (preceding one). Find the last installment.

OPtion

1) 215
2) 225
3) 230
4) 245
5) 265
6) 295
7) 300
8) 310
9) 340
10) None of these

Solution

Let a be the first installment
Successive installment are
a, a+10, a+20
here d=10, n=number of installment =30
Solve by using A.P. Series
S=n/2(2a+(n-1)d)=4500
solve
first intallment 5, last intallment 295
option 6

Correct Option: 6 Best Solution (16)

Devendra Marghade   11 years AGO

From arithmetic sum series formula,
if a=first installment,n=30, d=10 then
n/2[2a+(n-1)10]=4500 or 15[2a+290]=300 ,a=5
and 30th term=a+29d=5+29*10=295

Like? Yes (27) | No (4)  

Tapesh mishra   11 years AGO

S=4500
n=30
d=10
s=n/2[2a+(n-1)d]
a=5
LAST TERM=5+290=295

Like? Yes (7) | No (1)  

mandeep   11 years AGO

let x be the amount of first installment
as every installment is 10 more than the last one so the 2nd installment will be (x+10)
third will be (x+20)
it forms an A.P.
their sum is
(x)+(x+10)+(x+20)+(x+30)..........
this sum is equal to 4500.
as their are 30 installments to be paid
applying the sum formula to this A.P.
S=n/2(2x+(n-1)10)
S is 4500 and n is 30
on substituting values, 'x' will be obtained as 5.
i.e. first installment is of Rs. 5
for the 30th installment(last installment),the amount he pays is determined by using the formula
x+(n-1)10 (x=50)
on putting values it gives the answer 295
so the last instmnt is Rs. 295

Like? Yes (10) | No (4)  

SHIVANI PRAKASH   11 years AGO

let 1st inst. be x
next inst.=x+10
similarly,nxt inst are:x+20,x+30......x+290
the person pays back without any interest,thus sum of all the inst.=4500
thus sum of all the inst.(in A.P.):
(30/2)*(2x+290)=4500
x=5,thus last installment=x+290=295

Like? Yes (3) | No   

Balakrishnan M   11 years AGO

Sn= (n/2)[2a+(n-1)d]
4500=(30/2)[2a+(30-1)10]
Solve this we get,
a=5
Last term=a+(n-1)d
=5+(30-1)10
=5+(29)10
=5+290
=295

Like? Yes (3) | No   

Bingumalla Abinay   11 years AGO

4500=30/2(2a+290)
a=5
therefore Snth term =a+(n-1)d=5+290=295

Like? Yes (3) | No (1)  

siddhartha   11 years AGO

let the first installment is X

then next 2nd installment is X+10
and then next 3rd installment X+20

and so on... last 30th installment is X+290

so according to prob..

4500 = X+(X+10)+(X+20)+(X+30)......+(X+290)
= 30X + 10*(1+2+3+..+29)
= 30X + 4350

so X=5 rs which is first installment

last installment = X+290 = 295 RS

Like? Yes (2) | No   

richa ahuja   11 years AGO

sn=4500
n=30
d=10
sn=n/2[2a+(n-1*d)]
putting values of sn,d,n we get
a=5
l=a+(n-1)*d
l=5+(29)*10=295

Like? Yes (2) | No   

Aarush Singh   11 years AGO

starting from Rs.5 and adding Rs.10 we get the last installment Rs.295 which will payback rs4500 (without any interest)

Like? Yes (2) | No   

Suman Saurabh   11 years AGO

Let first installment=a
Total no. of installment=30
Last installment=a+(30-1)*10=a+290
4500=15(2*a+290)
a=5
so Last installment=295

Like? Yes (3) | No (1)  

vishnu vardhan   11 years AGO

given,4500 to be payed in 30 instalments with out any interest andfor every next instalment there is increase in price of 10.
therefore,let first installment price be 'x'
then total instalments will be in the order,
=>x+(x+10)+(x+10+10)+(x+10+10+10)+............
=>x+(x+10)+(x+20)+.........+(x+290)=4500.
=>30x+(10+20+30+.......+290)=4500.
=>3x+(1+2+3+......+29)=450.
=>3x+435=450.
=>3x=15.
=>x=5.
therefore,last installment=290+x,=>290+5=295 rupees.

Like? Yes (2) | No (1)  

sudipta sadhukhan   11 years AGO

lets assume first installment Rs. x.
it follows a AP series with c.d. 10.
so,
30/2[2*x+(30-1)*10]=4500
solving we get x=5
so ans is 5+(30-1)*10=295.

Like? Yes (1) | No   

ShAg   11 years AGO

this is in Arithmetic Progression.
let first installment is of Rs.x, 30 installments are to be paid with increment of Rs.10 in the previous installment.
So, in our formula L = a + (n-1)d, to find the last installment, a=x, n=30, and d=10
hence our eqn comes out to be:
L = x + (30-1)10
=> L = x + 290
for finding the sum of an Arithmetic Progression, formula is: S = n(a+L)/2
so our eqn comes out to be:
4500 = 30(x+x+290)/2
=> x=5
Putting x in L = x + 290, we get the answer:
L = 5 + 290 = Rs.295

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Chandni Maurya   11 years AGO

ans=(6). 295
let x be the first installment,
x+x+10+x+20+....+x+290=4500
=> 30x + 29/2(2*10+(29-1)10))=4500
=> 30x + 4350 =4500
=> 30x = 150
=> x = 5 (1st installment)
=> 5 + 290 = 295 (last installment)

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paramesh   11 years AGO

let us assume that

in first installmenthe pays - x
in second - x+10
in third - x+20
similarly..
in 30th instlmnt -x+290

By adding all..
we will get 30x+4350=4500
from above equation x=5 ruppes
we know that in 30th installment he should pay X+290 ruppes..here x=5..So answer is 290+5=295..

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GnanaMeena   11 years AGO

295+285+275+265+255+245+235+225+215+205+195+185+175+165+145+135+125+115+105+95+85+75+65+45+35+25+15+5=4500

Like? Yes (2) | No (11)