M4maths Previous Puzzles

03 Dec 2014 Time and Word

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill it in 7 hours. The number of hours taken by C alone to fill the cistern, is:

OPtion

1) 10
2) 21
3) 12
4) 22
5) 14
6) 20
7) 16
8) 15
9) 18
10)None of these

Solution

Part filled in 2 hours = 2/6 = 1/3. Remaining part = 2/3
(A + B)'s 7 hours' work = 2/3.
So, (A+B)'s 1 hr work = 2/21
Therefore C's 1 hour's work
= [(A+B+C)'s 1 hour]-[(A+B)'s 1 hr work]
=(1/6 - 2/21) = 1/14
C alone can fill the tank in 14 hours

Option 5)

Correct Option: 5 Best Solution (34)

ITHA Naresh Babu   10 years AGO

suppose that the capacity of tank is 42 units(lcm of 6,7)
a,b,c can fill tank in 6 hours=>a,b,c can fill (42/6=)7 units per hour.---------->(1)
after 2 hours 14 units of tank is filled and 28 units is left.Given a,b takes 7 hours to fill the remaining tank(28 units)=>a,b fills 4 units of tank per hour.---------->(2)
From (1) &(2) we get that c fills 3 units of tank per hour.
Hence to fill the full(42 units) tank it takes (42/3=)14 hours

Like? Yes (65) | No (7)  

K.ARun   10 years AGO

Three Pipes A,B and C together filled a tank in 6 hours
so 1hr work of A,B and C is 1/6
2hr working together 2/6
rem part nedd to be filled is (1-2/6)
i.e 2/3

(A+B)'s 7hr work = 2/3
(A+B)'s 1hr work= 2/3*7--> 2/21

C's 1hr work= (A=B+C)'s 1hr work -(A+B)'s 1hr Work
so.. 1/6-2/21
by solving leads to the answer 1/14
so 14hrs :)

Like? Yes (24) | No (7)  

priyanka    10 years AGO

a,b,c one hour work = 6 hour
1 hour work =1/6
2 hour =1/3
left work 2/3 coverd in 7 hours by a nd b
c work 1/6-2/21=1/14
14 hour...

Like? Yes (9) | No (1)  

P.Pandeeswari   10 years AGO

1/a+1/b+1/c=1/6

after working for 2 hrs. the work done is 1/3. so the

remaining work is 2/3.

which is done by a and b in 21 /2 hrs.

1/a+1/b+1/c=1/6

1/a+1/b=2/21

so 1/c-1/6-2/21= 14 hrs.

ans is

option 5) 14

Like? Yes (15) | No (8)  

Vishal Chugh   10 years AGO

a+b+c=6
efficiency=1
a+b effi.12/21
c effi=9/21
6*21/9=14

Like? Yes (24) | No (18)  

Udhaya kumar   10 years AGO

6 (1/a + 1/b +1/c) = 1
2(1/a + 1/b +1/c) + 7(1/a + 1/b) = 1
let (1/a + 1/b) = x and 1/c as y
6x + 6y = 1
9x + 2y = 1
by solving y = 0.07142
c = 1/y = 14..........

Like? Yes (8) | No (3)  

lakshmi   10 years AGO

Part filled in 2 hours = 2/6 = 1/3

Remaining part = (1-1/3) = 2/3 .

(A + B)'s 7 hour's work =2/3

(A + B)'s 1 hour's work =2/21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=1/6 - 2/21 =1/14

C alone can fill the tank in 14 hours.

Like? Yes (6) | No (4)  

Sanu Panji   10 years AGO

A+B+C do whole work in 6 hrs.

A+B+C works in 2 hr 2/6= 1/3 fraction of whole work
rest (1-1/3)= 2/3 fraction A+B do in 7 hr

so, A+B do whole work in 7/(2/3)=7*3//2=10.5 hrs

so, C does whole work in (10.5*6)/(10.5-6)= 14 hrs

Correct option is (5)

Like? Yes (3) | No (1)  

Chirag Ranjan Lahiri   10 years AGO

Part filled in 2 hours = 1/3
Remaining = 2/3
(A + B)'s 7 hour's work = 2/3
(A + B)'s 1 hour's work = 2/21
So, C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work } = 1/6 - 2/21 = 1/14

Hence, C alone can fill the tank in 14 hours.

Like? Yes (5) | No (3)  

Surit Datta   10 years AGO

Part filled in 2 hours = 2/6 = 1/3
Remaining part = ( 1 - 1/3) = 2/3
(A + B)'s 7 hour's work = 2/3
(A + B)'s 1 hour's work = 2/21
C's 1 hour's work =
{ (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }
(1/6 - 2/21) = 1/14
C alone can fill the tank in 14 hours.
That's it.

Like? Yes (5) | No (3)  

A.P.Singh   10 years AGO

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed ( After filling 1/3 tank) and A and B can fill ( balance 2/3 tank ) it in 7 hours. A & B can fill full tank in 7*3/2=21/2 hrs.
A&B can fill 2/21 tank in 1 hr.
C can fill 1/6 -2/21 = 3/42 or 1/14 tank in 1 hr.

so
The number of hours taken by C alone to fill the cistern, is 14

Like? Yes (4) | No (2)  

Ashwin Dinesh   10 years AGO

Let A, B, C can fill a, b, c part of the tank in 6 hrs.
1/a + 1/b + 1/c = 1/6
In 2 hrs: 2(1/a + 1/b + 1/c) = 1/3
A and B can fill the tank in 7 hrs after this...
7(1/a + 1/b) = 7(1/6 - c) = 2/3
Solving the above equation, we get c= 1/14
So pipe C can fill the tank in 14 hrs alone.
Which is option 5)

Like? Yes (6) | No (4)  

Shreya Banerjee   10 years AGO

Part filled in 2 hours = 2/6 = 1/3

Remaining part = (1 - 1/3) = 2/3

Therefore (A + B)'s 7 hour's work = 2/3

(A + B)'s 1 hour's work = 2/21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

= 1/6 - 2/21 = 1/14

Thus, C alone can fill the tank in 14 hours.

Like? Yes (5) | No (3)  

Abhishek singh   10 years AGO

as all can fill in 1 hr=1/6
c closed after 2 hr so total tank filled =1/3
left=2/3
a n b can fill remaining in 7 hr so i.e 2/3 so fully they can fill in 10.5 hr
so 1/10.5+1/c=1/6
solving we get c can fill in 14 hr.

Like? Yes (4) | No (2)  

Sundar Kumar   10 years AGO

Part filled in 2 hours = 2 /6 = 1/3

Remaining part = (1 - 1/3) = 2 /3 .

(A + B)'s 7 hour's work = 2/3

(A + B)'s 1 hour's work = 2/21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

= (1 /6- 2 /21) = 1/14

C alone can fill the tank in 14 hours.
Answer: Option 5) 14

Like? Yes (3) | No (2)  

Devendra Marghade   10 years AGO

(A+B+C)'s 1 hr. filling= 1/6,
In 2 hrs. (A+B+C) will fill 2/6=1/3, & Remaining=2/3
For (A+B) it takes 7 hrs. to fill 2/3
So in 1 hr. (A+B) will fill (2/3)/7=2/21
1 hr. filling of C= 1 hr. filling of[ (A+B+C) - (A+B)]
=(1/6) -(2/21)=3/42=1/14
So 'C' alone can fill the cistern in 14 hours

Like? Yes (3) | No (2)  

Azhar Syed   10 years AGO

1/A+1/B+1/C=1/6
working for 2 hrs A,B,C can do 1/3rd of work and remaining work is done by A,B in 7 hrs if A,B does this total work together it takes 10.5 hrs.
hence work done by C alone is 1/6-1/10.5=1/14
there fore option 5 is correct it takes 14 hours

Like? Yes (3) | No (2)  

Manasa Kotha   10 years AGO

A+B+C 1 hours work is 1/6
in 2 hours they complete 1/3rd work and remaining 2/3rd work is completed by a and b in 7 hours so a+b one hour work is 2/21
so c's 1 hour work=1/6-2/21= 9/126=1/14
no of hour's taken by c alone is 14 hours

Like? Yes (2) | No (1)  

purushothama   10 years AGO

Part filled in 2 hours = 2/6 = 1/3

Remaining part = (1-1/3) = 2/3 .

(A + B)'s 7 hour's work =2/3

(A + B)'s 1 hour's work =2/21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=1/6 - 2/21 =1/14

C alone can fill the tank in 14 hours.

Like? Yes (2) | No (2)  

jayalakshmi   10 years AGO

1/a+1/b+1/c=1/6
after working for 2 hrs. the work done is 1/3. so the remaining work is 2/3.
which is done by a and b in 21 /2 hrs.
1/a+1/b+1/c=1/6
1/a+1/b=2/21
so 1/c-1/6-2/21= 14 hrs.

Like? Yes (2) | No (2)  

A M ANKIT KALLURAYA   10 years AGO

if a,b,c can do the work together in 6 days.....
(1/a+1/b+1/c)=1/6.............................1st eqn
if they work for two days then one third of the work will be done...
remaining two thirds of the work will be done by a and b in 7 days
so a and b will do the 2/21 work in 1 day
substituting this in the 1st equation we get c=14

Like? Yes (1) | No (1)  

ravi prakash   10 years AGO

let complete is x
then 2/3of x=7hours
x=21/2
now 1/a +1/b=2/21
so 1/c=1/6-2/21=1/14
c can fill in 14 hours

Like? Yes (1) | No (1)  

Rahul Rathod   10 years AGO

1/a+1/b+1/c=1/6
after working for 2 hrs. the work done is 1/3. so the remaining work is 2/3.
which is done by a and b in 21 /2 hrs.
1/a+1/b+1/c=1/6
1/a+1/b=2/21
so 1/c-1/6-2/21= 14 hrs.

Like? Yes (2) | No (3)  

Pramod Gaikwad   10 years AGO

1/a+1/b+1/c=1/6
remaining work 2/3 done by a and b in 21 /2 hrs.
1/a+1/b+1/c=1/6
1/a+1/b=2/21
so 1/c-1/6-2/21= 14 hrs.

Like? Yes (1) | No (2)  

bhagya   10 years AGO

Part filled in 2 hours = 2/6 = 1/3
Remaining part = 1 - 1/3 = 2/3
∴ (A + B)'s 7 hour's work = 2/3
⇒ (A + B)'s 1 hour's work = 2/21
∴ C's 1 hour's work = [(A + B +C)'s 1 hour's work - (A + B)'s 1 hour's work] = 1/6 - 2/21 = 1/14
∴ C alone can fill the tank in 14 hours.

Like? Yes (2) | No (3)  

CHANDA VIJAYALAKSHMI   10 years AGO

A+B+C = 1/6
For 2 hours = 2 (1/6) = 1/3
Remaining = 2/3
(A+B)’s 7 hours work = 2/3
(A+B)’s 1 hours work = 2/21
C’s 1 hour work = (A+B+C)’s work – (A+B)’s work
= 1/6 -2/21
= 9/126 = 1/14
C alone can fill the tank in 14 hours

Like? Yes (1) | No (2)  

nitesh gupta   10 years AGO

Part filled in 2 hours =1/3
Remaining =2/3
A+B 7 hour work = 2/3
A+B 1 hour work = 2/21
C one hour work = A+B+C one hour work - A+B one hour work
1/6-2/21=1/14

Like? Yes (2) | No (3)  

arun   10 years AGO

part filled in 2 hrs = 2/6 = 1/3
remaining work = (1-1/3) = 2/3

(A+B) do 2/3 work in 7hrs
1hr (A+B) do ,(1/A+1/B)= 2/21
so we hv 1/A + 1/B + 1/c = 1/6
2/21 + 1/C = 1/6
1/C = 1/6 - 2/21
1/C = 1/14
so C alone can complete in 14 hrs

Like? Yes (1) | No (2)  

Anubhav jain   10 years AGO

a+b+c=6 hour's
a+b =7 hours after c is closed
c alone a and b work completed in 14 hours

Like? Yes (6) | No (7)  

Anand Dwivedi   10 years AGO

total part filled in two hour:- 2/6=1/3

So reamaining= 1-1/2=2/3

so total A and B work in 7h= 2/3

So subtract C s Work -A+B work = 1/6-2/21=1/14

so total fill tank fill in 14 hour

Like? Yes (1) | No (3)  

Arul Shaijan   10 years AGO

A+B+C=1/6 part of tank per hour

i.e, in 2 hours, A+B+C=2/6 part of tank
so, 4/6 part of tank remaining
4/6 part filled in 7 hours by A+B
so, A+B=4/42=2/21 part of tank per hour

(A+B+C)-(A+B)=1/6-2/21.

hence C=1/14 = 14 hours

Like? Yes (1) | No (3)  

sanjeev kumar   10 years AGO

Ans 14

Like? Yes (1) | No (3)  

Mithun Kumar Kesari   10 years AGO

14 h

Like? Yes (4) | No (6)  

Manojkumar.K   10 years AGO

A+B(alone)=7hours
A,B,C each has 2 hours
c alone means=7*2=14 hours

Like? Yes (6) | No (12)  

02 Dec 2014 Probability

There are two bags. First contain 10 white and other 8 red balls. One ball from any bag is shifted to other bag at random. What will be the probability that second ball taken after first event would be red. First event is to drawn first ball.

OPtion

1) 7/13
2) 1/33
3) 33/49
4) 1/49
5) 13/33
6) 1/99
7) 50/101
8) 49/99
9) 47/97
10)None of these

Solution

Initial condition 10W 8R
After first event two possibilities arise
9W
8R;1W
10W, 1R
7R
Choice of any has probability 1/2 so overall probability for getting red ball
= 1/2(1/2*0 + 1/2*8/9) + 1/2(1/2*1/11 + 1/2*1)
solve
49/99
Option 8)

Correct Option: 8 Best Solution (3)

sravani   10 years AGO

for the first event the ball can be a white or red...
if condition is for the 1st event the answer would be
8/9(red balls bag)+1/11(white balls bag)
97/99
for 2nd event it would be
98/99 *1/2 =49/99...

Like? Yes (3) | No (5)  

Rahul Rathod   10 years AGO

2/9+1/4+1/44=49/99

Like? Yes (2) | No (6)  

Ashwin Dinesh   10 years AGO

48.5/99 = 49/99 (approx)

Like? Yes (2) | No (7)  

01 Dec 2014 Sequence

What is the next no of the following sequence
10, 40, 270, 2560,.........

OPtion

1) 1250
2) 2250
3) 6250
4) 12500
5) 31250
6) 62500
7) 125000
8) 225000
9) 625000
10) None of these

Solution

10 = 1^1*10
40 = 2^2*10
270 = 3^3*10
2560 = 4^4*10
5^5*10 = 31250

Option 5)

Correct Option: 5 Best Solution (20)

K.ARun   10 years AGO

1^1*10=10
2^2*10=40
3^3*10=270
4^4*10=2560
So, 5^5*10=31250

Like? Yes (8) | No (4)  

RAKESH   10 years AGO

n^n * 10

n = 1,2,3,4,5...

Like? Yes (6) | No (4)  

Shekhar Suman   10 years AGO

Explanation
1^1*10=10
2^2*10=40
3^3*10=270
4^4*10=2560
5^5*10=31250

Like? Yes (3) | No (2)  

Surit Datta   10 years AGO

According to the question the series is
10,40,270,2560....
1^1 = 1
we write it as 10 (adding a zero at last)
2^2 = 4
we write it as 40 (adding a zero at last)
3^3 = 27
we write it as 270 (adding a zero at last)
4^4 = 256
we write it as 2560 (adding a zero at last)
similarly
5^5 = 3125
we write it as 31250 (adding a zero at last)
so the answer is 31250
Option 5
That's it..

Like? Yes (2) | No (3)  

Devendra Marghade   10 years AGO

10*1^1=10
10*2^2=40
10*3^3=270
10*4^4=2560
10*5^5=31250

Like? Yes (1) | No (2)  

sravani   10 years AGO

1^1 *10 = 10
2^2 * 10 = 40
3^3 * 10 = 270
4^4 * 10 = 2560
5^5 * 10 = 31250....

Like? Yes (2) | No (3)  

Md.Sultan Khichi   10 years AGO

option 5-31250
1^1=1
2^2=4
3^3=27
4^4=256
5^5=3125
And 1 zero hi added with every numbers.

Like? Yes (1) | No (2)  

susmit   10 years AGO

1(0)
2^2(0)
3^3(0)
4^4(0)
5^5(0)= 3125(0)=31250

Like? Yes (1) | No (2)  

Ashwin Dinesh   10 years AGO

1 ^ 1 x 10 = 10
2 ^ 2 x 10 = 40
3 ^ 3 x 10 = 270
4 ^ 4 x 10 = 2560
5 ^ 5 x 10 = 31250

Like? Yes (2) | No (3)  

Tark Patel   10 years AGO

1^1 put 0 after it = 10
2^2 put 0 after it = 40
3^3 put 0 after it = 270
4^4 put 0 after it = 2560
5^5 put 0 after it = 31250

Like? Yes (1) | No (2)  

Azhar Syed   10 years AGO

1*10=10
2*2*10=40
3*3*3*10=270
4*4*4*4*10=2560
5*5*5*5*5*5*10=31250
so answer is option 5) 31250

Like? Yes (1) | No (3)  

tirupathi reddy   10 years AGO

Ans: 5^5*10= 31250

1^1*10=10
2^2*10=40
3^3*10=270
4^4*10=2560
5^5*10=31250

Like? Yes (1) | No (3)  

PAWAN KUMAR   10 years AGO

10 = 1*10
40 = 2^2*10
270 = 3^3*10
2560 = 4^4*10
5^5*10 = 3125*10 = 31250 Ans.

Like? Yes (1) | No (3)  

Rahul Rathod   10 years AGO

1^3=1*10=10
2^2=4*10=40
3^3=27*10=270
4^4=256*10=2560
similarily
5^5=3125*10=31250

Like? Yes (1) | No (3)  

Akshay   10 years AGO

1=1
2*2=4
3*3*3=27
4^4=256
5^5=3125
and after that there is a zero
so,31250

Like? Yes (1) | No (3)  

Rishikesh Ranjan   10 years AGO

1^1*10
2^2*10
3^ 3*10
4^4*10
5^5*10=31250

Like? Yes (1) | No (3)  

sunil bajpai   10 years AGO

10*1^1=10
10*2^2=40
10*3^3=270
10*4^4=2560
10*5^5=31250

Like? Yes (1) | No (3)  

Anubhav jain   10 years AGO

1^1=10
2^2=40
3^3=270
4^4=2560
5^5=31250

Like? Yes (1) | No (3)  

arun   10 years AGO

(1^1)*10,(2^2)*10,(3^3)*10,(4^4)*10
same way next term will be (5^5)*10 = 31250

Like? Yes (1) | No (3)  

Rats   10 years AGO

(1^1)*10=10
(2^2)*10=40
(3^3)*10=270
(4^4)*10=2560
So, the next term is...
(5^5)*10=31250

Like? Yes (1) | No (4)  

28 Nov 2014 Possible Combination

If x^2 - y^3 is a positive non zero integer for all possible integral values of x, y satisfying 1 =< x =< 9 and 1 =< y =< 9 then number of possible ordered combinations of (x, y) are.

OPtion

1) 8
2) 10
3) 12
4) 16
5) 18
6) 20
7) 21
8) 24
9) 25
10)27

Solution

If y = 1, then x can have values ranging 2 to 9 so 8 possibilities.
If x = 2, then x can have values ranging 3 to 9 so 7 possibilities.
If y = 3, then x can have values ranging 6 to 9 so 4 possibilities.
If y = 4, then x can have values only 9 so 1 is possibilities.

Total cases = 8+7+4+1 = 20

Option 6)

Correct Option: 6 Best Solution (28)

BHASKAR MANDAL   10 years AGO

possibilities are
1) y=1 , then x will be from 2 to 9 ---> 8
2)y=2 , then x will be from 3 to 9 -----> 7
3) y= 3, then x will be from 6 to 9------>4
4) y= 4, then x will be from 9 ------------> 1
so total = 8+7+4+1 = 20

Like? Yes (11) | No (2)  

kamal   10 years AGO

1,2,3,4,5,6,7,8,9(possible value of x,y)
1,4,,9,16,25,36,49,64,81 (x^2 possible)
1,8,27,64,125,216,343,512,729 (all possible y^3)
now take x^2=1 and y^3 from 1 to 729 there is no pair such possible (x,y) such that x^2-y^3=non zero positve number,,
solve further and u'll get 20 such pairs---(2,1) (3,1) (3,2) (4,1) (4,2) (5.1)----------------(9,4).

Like? Yes (4) | No   

Aditya Verma   10 years AGO

according to question...
x^2 - y^3 > 0
where...
1 =< x =< 9 and
1 =< y =< 9
so all possible combinations of (x,y) are...
(2,1),(3,1),(3,2),(4,1),(4,2),(5,1),(5,2),(6,1),(6,2),(6,3),(7,1),(7,2),(7,3),(8,1),(8,2),(8,3),(9,1),(9,2),(9,3),(9,4)
there are total 20 possible combinations of x and y
so answer is 20

Like? Yes (2) | No   

Jayanta Dutta   10 years AGO

First consider y = 1 so x will be (2,3,4,5,6,7,8,9)
Next y=2
So x will have to be(3,4,5,6,7,8,9)
Next y=3
X will hve to be (6,7,8,9)
Next y=4
X will hav to be (9).
So total combo =[8+7+4+1] optn 6

Like? Yes (2) | No   

RAKESH   10 years AGO

(2,1)
(3,1),(3,2)

------------------
(8,1),(8,2),(8,3)
(9,1),(9,2),(9,3),(9,4)

total = 1+2+2+2+3+3+3+4 = 20

Like? Yes (1) | No   

ANIL KUMAR   10 years AGO

by jst chkng all the possible combination of x^2 - y^3
got it as 20.

Like? Yes (2) | No (1)  

Devendra Marghade   10 years AGO

For x^2 - y^3 to be positive,
possible ordered combinations of (x, y) are
(2,1), (3,1), (3,2), (4,1), (4,2), (5,1), (5,2), (6,1), (6,2), (6,3), (7,1), (7,2), (7,3), (8,1), (8,2), (8,3), (9,1), (9,2), (9,3), (9,4)

Total 20

Like? Yes (1) | No   

PAWAN KUMAR   10 years AGO

For
x=9, y=1,2,3,4
x=8, y=1,2,3
x=7, y=1,2,3
x=6, y=1,2,3
x=5, y=1,2
x=4, y=1,2
x=3, y=1,2
x=2, y=1

Like? Yes (1) | No   

Narasapu Hari prasad   10 years AGO

ans is option6 20:
the possible x,y combinations are
(2,1)(3,1)(4,1)(5,1)(6,1)(7,1)(8,1)(9,1)
(3,2)(4,2)(5,2)(6,2)(7,2)(8,2)(9,2)
(6,3)(7,3)(8,3)(9,3)
(9,4)

Like? Yes (1) | No   

Rajat Rokade   10 years AGO

check for all combinations of x and y
The answer is 20

Like? Yes (1) | No   

SUBHASH TULSYAN   10 years AGO

x^2-y^3>0
1

Like? Yes (1) | No   

Azhar Syed   10 years AGO

possible combinations if (x,y) are (2,1) (3,1) (3,2) (4,1) (4,2) (5,1) (5,2) (6,1) (6,2) (6,3) (7,1) (7,2) (7,3) (8,1) (8,2) (8,3) (9,1) (9,2) (9,3) (9,4) total of 20 combinations

Like? Yes (1) | No   

CHANDAN KUMAR   10 years AGO

x^2-y^3=+ve>0
xy=21,31,32,41,42,51,52,61,62,63,71,72,73,1,82,83,91,92,93,94
total=20

Like? Yes (1) | No   

sravani   10 years AGO

x can have values from 2 to 9
but to satisfy the condition y takes only two values 1 for x=2.
values 1,2 for x=3,4,5
values 1,2,3 for x=6,7,8.
values 1,2,3,4 for x=9.......
so total (x,y) pairs are 20.......

Like? Yes (1) | No   

Pramod Gaikwad   10 years AGO

put the number between 1 to 9,
number of solution =20.
so
option is 6)20

Like? Yes (1) | No   

prem   10 years AGO

(2,1)...........................(9,1) total 8 combination
(3,2).................(9,2) total 7 combination
(6,3).....................(9,3) total 4 combination
(9,4) 1 combination hence total 20 combination
hence 6 option

Like? Yes (1) | No   

Rishikesh Ranjan   10 years AGO

8+7+4+1

Like? Yes (1) | No   

poornima    10 years AGO

if k is the positive int, then assume X =6,Y=3
it satisfy x^2-Y^3=k
then ordered combination of (6,3) is 20
ans is 20

Like? Yes (1) | No   

khushboo goenka   10 years AGO

when y=1
8 combinations
y=2, 7 combn, y=3,4 cominatons
y=4, one comination total= 8+7+4+1=20

Like? Yes (1) | No   

MRITUNJAY   10 years AGO

X^2-y^3=non zero
by heat and trial
x=9&y=1,2,3,4
x=8&y=1,2,3
x=7&y=1,2,3
x=6&y=1,2,3
x=5&y=1,2
x=4&y=1,2
x=3&y=1,2
x=2&y=1
total 20.

Like? Yes (1) | No   

ARPAN SHAH   10 years AGO

for y only 4 possibilities 1,2,3,4 ...now for y=1 x can 8 values 2 to 9,for y=2 ,x=3 to 9,for y=3,x=6 to 9,for y=4,x=9.so total 20

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sanjeev kumar   10 years AGO

ans 20

Like? Yes (1) | No   

abeer naskar   10 years AGO

as the whole equation is positive then the only options for y are 1,2,3 and 4
for y=5, y^3 will be 125 but max value of x^2 will be 81
so for y=1 we get x from 2 to 9 that is 8 no. of x
for y=2 we get x from 3 to 9 that is 7 choice of x
for y=3 we get x from 6 to 9 that is 4
for y=4 we get x from 9 to 9 that is 1
so answer that is total choice is
8+7+4+1=20

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DEERGHA AGGARWAL   10 years AGO

solving through the equation

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Manju Prakash R   10 years AGO

Condition is that y^3 must be greater than x^2
There are 20 possible ordered pairs
y=9 -> 4 pairs
y=8 -> 3 pairs
y=7 ->3 pairs
y=6 -> 3 pairs
y=5 => 2 pairs
y=4=>2 pairs
y=3 =>2 pairs
y=2 =>1 pair

Ans 6) 20

Like? Yes (1) | No   

Sourabh Prasad   10 years AGO

x 2 - y 3 =
Satisfying values are

2 1
3 1
3 2
4 1
4 2
5 1
5 2
6 1
62
6 3
7 1
7 2
7 3
8 1
8 2
8 3
9 1
9 2
9 3
9 4

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Md.Sultan Khichi   10 years AGO

x^2-y^3= non zero +ve value.
By it and trial method
The combinations are
(2,1),(3,1)(3,2),(4,1),(4,2),(5,1),(5,2),(6,1),6,2),(6,3),(7,1),(7,2),(7,3),(8,1),(8,2),(8,3),(9,1),(9,2),(9,3)
Total=20.
Ans ob. 6

Like? Yes (1) | No (1)  

Rahul Rathod   10 years AGO

check all combinations of all (x,y) pair
we get 20 solutions.

Like? Yes (1) | No (1)  

27 Nov 2014 Number

The sum of a number and its reciprocal is equal to 1/10 multiple of the number of playing card in a deck. x is?

OPtion

1) 10
2) 6
3) 3
4) 4
5) 12
6) 8
7) 9
8) 7
9) 11
10)None of these

Solution

According to the given condition
x + (1/x) = 52/10
or x + (1/x) = 26/5
On solving, we get x = 5, 1/5

Option: 10)

Correct Option: 10 Best Solution (53)

P.Pandeeswari   10 years AGO

let the number be x.
x + (1/x) = 52/10

x + (1/x) = 26/5

5(x^2)-26x+5=0

On solving, we get x = 5, 1/5

Like? Yes (7) | No (3)  

Rajat Rokade   10 years AGO

Suppose a number is x.
so x + 1/x = 52/10

By solving this equation, we get x=5.

Like? Yes (3) | No (2)  

Chirag Ranjan Lahiri   10 years AGO

x+(1/x)=(1/10)*52
5x^2-26x+5=0
(5x-1)(x-5)
So x=5, 1/5

Like? Yes (3) | No (2)  

RAKESH   10 years AGO

x + 1/x = 52/10 = 26/5 = 5 + 1/5

=> x = 5 or 1/5

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khushboo goenka   10 years AGO

x+1/x=52/10
x=5.,1/5

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MINTU KUMAR GUPTA   10 years AGO

x+1/x=52/10
x=5,1/5

Like? Yes (2) | No (1)  

K.ARun   10 years AGO

x+1/x=1/10(52)
10x^2-52x+10=0
on solving,
X=5 or X=1/5
So, Option 10) None of these

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Udhaya kumar   10 years AGO

x + (1/x) = (1/10) 52
x = 5......

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santosh kumar patel   10 years AGO

x + 1/x=52/10
5 + 1/5=26/5
so
5 or 1/5

Like? Yes (2) | No (2)  

ravi prakash   10 years AGO

5 or 1/5

x + 1/x=52/10
5 + 1/5=26/5
so
5 or 1/5

Like? Yes (1) | No (1)  

Nitish Kumar Dubey   10 years AGO

5x^2-26x+5=0

After sollving we are getting X=1/5 , 5

So ans is 10- none of these.

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sanjeev kumar   10 years AGO

x+(1/x)=52/10
By solving the equation we get x= 5, 1/5.

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lavanya   10 years AGO

5 or 1/5

x + 1/x=52/10
5 + 1/5=26/5
so
5 or 1/5

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Manju Prakash R   10 years AGO

Ans will be 5.

Like? Yes (1) | No (1)  

Shreya Banerjee   10 years AGO

let the no be x
x + 1/x = 1/10 * 52
x + 1/x = 26/5
by hit and trial
x=5
or solving eq. 5 x^2 - 26 x + 5 = 0
x=5

Like? Yes (1) | No (1)  

MAHARAJAN.G   10 years AGO

x+1/x=52/10

x^2+1/x=52/10

(x^2+1)10=52x

10x^2+10-52x

divide by 2

5x^2-26x+5=0

(x-5)(x-1/5)=0

x=5 x=1/5

answer x= 5, 1/5

none of option has this answer so

ans) 10 option

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DEERGHA AGGARWAL   10 years AGO

x + 1/x = 52/10

Like? Yes (1) | No (1)  

Sourabh Prasad   10 years AGO

x + 1/x = 52/10

Solving
x= 5 and 1/5

Like? Yes (1) | No (1)  

ravi   10 years AGO

5,0.2

Like? Yes (1) | No (1)  

Jayanta Dutta   10 years AGO

ans 5

Like? Yes (1) | No (1)  

Sonam Gupta   10 years AGO

x+1/x =(1/10)*52
x^x+1=5.2*x
x^x-5.2*x+1=0

Like? Yes (1) | No (1)  

r   10 years AGO

if we solve it answer will be 5

Like? Yes (1) | No (1)  

Sanjay Bhatt   10 years AGO

x+1/x=5.2
So,
x=5

Like? Yes (1) | No (1)  

MRITUNJAY   10 years AGO

X+1?X=52/10
solve this then we get x=5.
so 5 is the answer

Like? Yes (1) | No (1)  

Tark Patel   10 years AGO

According to the given condition
x + (1/x) = 52/10
or x + (1/x) = 26/5
On solving, we get x = 5, 1/5

Like? Yes (1) | No (1)  

Sanu Panji   10 years AGO

ans is 5 or 1/5

Like? Yes (1) | No (1)  

sailash prasad chowdhury   10 years AGO

x+1/x=56/10by solving the equation we get x=5....so ans is 10-none of these

Like? Yes (1) | No (1)  

shrihari   10 years AGO

ans, 5

Like? Yes (1) | No (1)  

padmakar kumar   10 years AGO

no of cards = 52
1/10 *52= 5.2
x+1/x= 5.2
x=5 1/x= .2
x+1/x= 5.2

Like? Yes (1) | No (1)  

A SIRISHA    10 years AGO

LET NUMBER = X
THEN ITS RECIPROCAL = 1/X
NO.OF CARDS IN A PACK =52
THN A/C TO QTN
X+1/X=(1/10 )* (52)
X^2+1=5.2X
BY SOLVING THIS X= 0.2 OR 5 BOTH VALUES ARE POSSIBLE

Like? Yes (2) | No (2)  

Sandeep Kumar   10 years AGO

x+1/x = 52/10 = 26/5.
on solving this equation we will get x = 5;

Like? Yes (1) | No (1)  

Sanchit Kapoor   10 years AGO

x+1/x=1/10*52 [Since a deck consists of 52 playing cards]
Solving, x=5
So option 10 is correct.

Like? Yes (3) | No (4)  

Anand   10 years AGO

x+1/x = (1/10*52)
solving above equation we get
x=5 and its reciprocal 1/5

Like? Yes (1) | No (2)  

Rahul Rathod   10 years AGO

x+(1/x)=52*(1/10)
x=5 or 1/5

Like? Yes (2) | No (3)  

Pramod Gaikwad   10 years AGO

X+(1/X)=(1/10)*52;
X=5
Therefore Answer=10

Like? Yes (2) | No (3)  

sravani   10 years AGO

Let a be the number
a + 1/a =1/10 *52
a=5.
but no option so, none of these....

Like? Yes (1) | No (2)  

BHASKAR MANDAL   10 years AGO

x = 5
5+1/5 = 5.2
1/10*no. of playing cards = 5.2

Like? Yes (1) | No (2)  

Md.Sultan Khichi   10 years AGO

1/10*52=5.2
Let x be the no.
So (x+1/x)=5.2
After solving we get x=5.
Ans 5

Like? Yes (1) | No (2)  

Anand Dwivedi   10 years AGO

x + 1/x = 52/10

=> x + 1/x = 5.2

=> x=5

Like? Yes (1) | No (2)  

purushothama   10 years AGO

+ 1/x = 52/10

by solving this quadratic of form -
5x^2-26x+5=0
we get, x=5,

Like? Yes (1) | No (2)  

jayalakshmi   10 years AGO

x+(1/x)=(1/10)*52
x^2-5.2x+1=0
by solving we get x= 5,0.2(1/5)

Like? Yes (1) | No (2)  

Azhar Syed   10 years AGO

Answer is 5,
x+1/x=1/10*52 => x² +1=5.2x;
solving we get x=5 or 1/5

Like? Yes (1) | No (2)  

lakshmi   10 years AGO

x+(1/x)=(1/10)*52
x^2-5.2x+1=0
by solving we get x= 5,0.2(1/5)

Like? Yes (1) | No (2)  

Gaurav Sharan   10 years AGO

n+1/n =(1/10)*52
5n*n-26*n+5=0
n=5

Like? Yes (1) | No (2)  

SUBHASH TULSYAN   10 years AGO

a/q
x+1/x=5.2
on solving
x=5

Like? Yes (1) | No (2)  

Abhishek Porwal   10 years AGO

x + 1/x=52/10
5 + 1/5=26/5
LHS=RHS
so
5 or 1/5

Like? Yes (1) | No (2)  

bhagya   10 years AGO

x+(1/x)=(1/10)*52
x^2-5.2x+1=0
by solving we get x= 5,0.2(1/5)

Like? Yes (1) | No (2)  

Rajnikant   10 years AGO

let's assume no. be x
A/c to ques,
x+1/x=52/10
=>(x^2+1)/x=26/5
=>5(x^2+1)=26x
=>5(x^2)-26x+5=0
So, x=5 and 1/5

Like? Yes (1) | No (2)  

CHANDA VIJAYALAKSHMI   10 years AGO

x+(1/x)=(1/10)*52
x^2-5.2x+1=0
by solving we get x= 5,0.2(1/5)

Like? Yes (1) | No (2)  

poornima    10 years AGO

the value of deck is 52
if X is d unknown number..
then X+1/X=52/10
ans is 5 or 1/5

Like? Yes (1) | No (2)  

RAMAKRISHNA   10 years AGO

x+1/x=5.2
solving this we get 5
so none of this

Like? Yes (1) | No (2)  

Vineedh CV   10 years AGO

x is 5.2
because 5.2+1/5.2=5.4=1/10 multiple of number of cards in a deck=5.4

Like? Yes (1) | No (2)  

Siddharth Kaushik   10 years AGO

No. of playing cards in a deck is 52
then equation will form
x+1/x=52/10
10x^2-52x+10=0
x= 5 , 1/5
So sol is non of these

Like? Yes (1) | No (2)  

26 Nov 2014 Speed and Time

Ramesh hears a sound of whistle after 2 sec. of whistling by Shyam and Shyam hears a sound of whistle after 3 sec. of whistling by Ramesh. If Ramesh and Shyam are in the line of flow of wind, then velocity of wind in m/s is
(Given that the speed of sound is 300 m/s)

OPtion

1) 80
2) 50
3) 40
4) 30
5) 70
6) 90
7) Wind is not blowing
8) Data are insufficient
9) 60
10)20

Solution

Ramesh hears a sound of whistle after 2 sec. of whistling by Shyam and Shyam hears a sound of whistle after 3 sec.
(greater than 2 sec.) of whistling by Ramesh. These data indicate that wind is blowing from Shyam to Ramesh
According to first condition l = (300 + x) 2,
l represents distance between them
According to second condition l = (300 - x) 3
On solving these equations, we get x = 60

Option 9)

Correct Option: 9 Best Solution (29)

RAKESH   10 years AGO

velocity of wind = x m/s then
equate distance b/w Ramesh & Shyam

(300+x)*2 = (300-x)*3
=> x = 60

Like? Yes (6) | No (4)  

Pramod Gaikwad   10 years AGO

consider speed of wind x m/s
distance between shyam and ramesh is same,
so v=distance /time.
distance=(300+x)*2
and
distance =(300-X)*3
by solving this we get
x=60
i.e. option 9)60.

Like? Yes (3) | No (2)  

khushboo goenka   10 years AGO

dis between ramesh and shyam is 300-w/2=150
simi;arly, 300+w/3
both are eql
so w=60

Like? Yes (1) | No   

Deepak prajapati   10 years AGO

(300+v)*2=(300-v)*3
v=60

Like? Yes (2) | No (1)  

K.ARun   10 years AGO

Ramesh hears sound after 2secs. of whistling by shyam so 2(300+x)
Shyam hears after 3secs. of whistling by Ramesh
so 3(300-x)
Note: (-x) due to line of flow of wind
so 2(300+x)=3(300-x)
600+2x=900-3x
5x=300;
x=60

Like? Yes (3) | No (3)  

ANIL KUMAR   10 years AGO

let the speed of air =x m/s
so , the equation formed is:

(300+x)2=(300-x)3
600+2x=900-3x
5x=300
x=60

Like? Yes (1) | No (1)  

PAWAN KUMAR   10 years AGO

Let the speed of wind be t m/s and distance between them be x mtr.
Then, x/(300+t) = 2 ------------(i)

and x/(300-t) = 3------------(ii)
on solving (i) and (ii), we get;
600+2t = 900-3t
=> t = 60 m/s Ans.

Like? Yes (1) | No (1)  

Pinaki Ghosh   10 years AGO

((300 * 3) - (300*2)) / (2+3) = 60

Like? Yes (1) | No (1)  

Ranjit Somisetty   10 years AGO

v=d/t => d=v*t Here d is constant.
(300+v)2=(300-v)3
v=60m/s, velocity of wind.

Like? Yes (1) | No (1)  

sravani   10 years AGO

the distance between them is same..so v1t1=v2t2
v1 be the velocity of sound+wind at ramesh side
v2 at shyam side
as shyam takes more time to listen he is opposite to wind blow...
so
2*(300+x)=3*(300-x) (let X is the speed of wind)
x=60..

Like? Yes (1) | No (1)  

Sagar Kharab   10 years AGO

2(300+v)=3(300-v)
600+2v=900-3v
5v=300
V=60

Like? Yes (1) | No (1)  

Swagatam Kumar Das   10 years AGO

Let the speed of wind is u
So 2(300+u)=3(300-u)
So u=60

Like? Yes (1) | No (1)  

kanchapu ramanjan   10 years AGO

(300-60)*3=(300+60)*2

Like? Yes (1) | No (1)  

prem kumar chaudhary   10 years AGO

v(w)=velocity of wind
2(300+v(w))=(300-v(w))3
v(w)=60
ans=60

Like? Yes (1) | No (1)  

vaitesh mudaliar   10 years AGO

let the direction of wind be from shyam to ramesh and has speed =s.
when shyam blows whistle, speed of sound traveled will be = wind speed + actual speed of sound
dist traveled (d) = speed of sound travelled * time
d= (s+300)*2
when ramesh blows whistle,
d= (300-s) *3
equating both eqn.
we get s= 60 m/s

Like? Yes (1) | No (1)  

sanjeev kumar   10 years AGO

ans 60

Like? Yes (1) | No (1)  

Divyen Jain   10 years AGO

(300+Vs)2 = d = (300-Vs)3
Vs= 60

Like? Yes (1) | No (1)  

CHANDAN KUMAR   10 years AGO

speed of sound=s
speed of wind=w
2(s+w)=3(s-w)
2(300+w)=3(300-w)
5w=300
w=300/5=60m/s

Like? Yes (1) | No (1)  

Dipu Kumar   10 years AGO

since ramesh takes less time to hear whistal it means direction of blow of wind is shayam to ramesh.
Let x is velocity of wind.
distace cover by whistal from shyam to ramesh=2*(300+x)
''' '' '' '' '' ' '''' ''' '' ramesh to shyam=3*(300-x)
both distance are equal
2*(300+x)=3*(300-x)
x=60 m/s

Like? Yes (1) | No (1)  

RAJESH   10 years AGO

60

Like? Yes (1) | No (1)  

SUBHASH TULSYAN   10 years AGO

let the distance b/w them is k
let the speed of wind is x
a/q
we hv the following eqn
k/(300+v)=2............(1)
k/(300-v)=3...........(2)
from (1) & (2)
2(300+v)=3(300-v)
on solving
x=60

Like? Yes (2) | No (2)  

akhil sai   10 years AGO

from v=d/t
both are separated by same distance but wind is in one direction so (v+x)=d/2
(v-x)=d/3 where v=300m/s velocity of sound x=wind velocity
same direction less time opposite direction more time

Like? Yes (2) | No (2)  

Ramesh sai   10 years AGO

300/2+3=60m/s

Like? Yes (1) | No (1)  

Sourabh Prasad   10 years AGO

60 m/s

Like? Yes (1) | No (1)  

sriharidatta   10 years AGO

300/ (3+2)

Like? Yes (1) | No (1)  

ravi prakash   10 years AGO

let speed of wind=x m/sec
then (300+x)2=(300-x)3
x=300/5
60

Like? Yes (1) | No (2)  

Md.Sultan Khichi   10 years AGO

Let the velocity of wind be x m/s.
So (x+300)/2=(300-x)/3
After solving x=-60
-ve sign indicate that wind blowing in opposite direction..
So speed of wind is 60 m/s.
Ans. Ob 9

Like? Yes (1) | No (2)  

Rajat Rokade   10 years AGO

Suppose, the speed of wind is X m/s.
Now the distance between Ramesh and Shyam is same.
so,
the ratio of the time must be equal to the relative speed of wind.

(300 + X)/(300 - X) = 3/2.
Solving this equation, we get X=60.
Speed of wind is 60 m/s.

Like? Yes (3) | No (4)  

Rahul Rathod   10 years AGO

let velocity of wind=x;
dist(Ramesh)=dist(Shyam)
so
(300+x)/(300-x)=3/2
so x=60
answer is 60

Like? Yes (2) | No (3)  

25 Nov 2014 Age Problem

The age of a man is equal the sum of ages of his 3 sons at present. What was the ratio of his age & his eldest son's age 3 year before. Given that the difference of age of his sons is 3 & age of father is 48 at present.

OPtion

1) 25/16
2) 4/5
3) 41/10
4) 45/16
5) 11/9
6) 25/9
7) 36/25
8) 49/25
9) 45/26
10)None of these

Solution

According to given condition
f = a+b+c
f = c-6+c-3+c (c is the age of eldest son)
f = 3c - 9
f/3 = c-3
So ratio
(f-3)/(c-3) = (f-3)/(f/3)
45/16
Option 4)

Correct Option: 4 Best Solution (67)

K.ARun   10 years AGO

x+x+3+x+6=48
3x=39
x=13
so 1st 3sons 13,16,19 respectively
eldest son=19 father=48..
3years before
eldest son 16 and father=45
so.. 45/16

Like? Yes (12) | No (3)  

Sanjay Bhatt   10 years AGO

Let the present age of father be X.
Let the present age of eldest son be s1.
Let the present age of second son be s2.
Let the age of youngest son be s3.

X=s1+s2+s3.
Also, s1=s2+3
s2=s3+3
then,
X=3*s3+9
s3=13
&s1=19

Before 3 years,
Age of Father=45
Age of Eldest Son=16(s1-3)

So Ratio,
45/16

Like? Yes (2) | No   

sailash prasad chowdhury   10 years AGO

x+x+3+x+6=48,so x=13
now the age of the elder son is x+6=19,3 years ago his age is 16. fathers age 3yrs ago is 45....so the answer is 45:16

Like? Yes (2) | No   

Sanu Panji   10 years AGO

Say three sons age is (x-3), x and (x+3)
so, (x-3)+x+(x+3)=48
or, x=16
elder son's age = 16+3= 19

3 yrs before elder son's age = 19-16
3 yrs before father's age = 48-3 =45

so ratio is 45/16

so, correct option is (4)

Like? Yes (3) | No (2)  

Pramod Gaikwad   10 years AGO

suppose present age of man=48
suppose middle son age is x
48=x+(x+3)+(x+3)
x=16
so age of sons 13,16,19
age of eldest son 3 year before 48-3=45
so ratio is 45/16
answer is 4)45/16

Like? Yes (4) | No (3)  

veeranjaneyulu   10 years AGO

48=19+16+13
m/e=48-3/19-3
m/e=45/16

Like? Yes (1) | No   

Chirag Ranjan Lahiri   10 years AGO

Let, Father's age = a
and his son's age are x, y, z, where x is the eldest one.
Now, a/q, a = x + y + z
so, 48 = (x+3) + x + (x-3)
so x = 16
so, eldest son's age = 16 + 3 = 19 years.

3 years ago, a = 45 and x = 16
so ratio = 45/16.........

Like? Yes (3) | No (2)  

CHANDAN KUMAR   10 years AGO

m=x+y+z
48=x+x-3+x-6
x=19
m-3/x-3=48-3/19-3=45/16

Like? Yes (1) | No   

shireesha chinthagunta   10 years AGO

19+16+13=48
3 yrs ago eldest son age=16
father's age 45
ratio of father and eldest son=45/16

Like? Yes (1) | No   

vibhor   10 years AGO

Let x be the age of his youngest son,
then the age of other son's as x+3,x+6
now,
(x)+ (x+3)+(x+6)=48
3x+9=48
3x=39
x=13

so the present age of his sons
13,16,19

now we have to find the ratio of age af father and eldest son at 3 years before,

father's age at 3 years before=48-3=>45
eldest son's age at 3 years before=19-3=>16

so the ratio is 45/16
option 4

Like? Yes (2) | No (1)  

santosh kumar patel   10 years AGO

man age (48) = his son age()
diff of his son = 3
so son age is - 13 , 16, 19
father and eldest son ratio before 3 year=45/16
ans: 45/16

Like? Yes (2) | No (1)  

nikhil pogale   10 years AGO

m=s1+s2+s3
s1=s3+6
S2=s3+3
3*s3+9=48
s3=13
s1=19
ratio befor 3 yr
f-3/s1-3=45/16

Like? Yes (1) | No   

Shravya   10 years AGO

let agesbe x,x+3,x+6
3x+9=48
48:16(x+6+3)

Like? Yes (1) | No   

Anoop Tiwari   10 years AGO

let son's ages =x+3,x,x-3
so sum=x+3+x+x-3=48
3x=48
x=16
ages=19,16,13
elder son's age =19
3 years before= father's age=45,eldest son's age=16
ratio=45/16
hence 4) is correct

Like? Yes (1) | No   

prem   10 years AGO

if current age of eldest son is x+9
then other two son's age is x+6 and x+3
and current age of father is 48
3year before father age is 45 and their son's age is x,x+3,x+6
since current age of father =sum of son's age
48=x+3+x+6+x+9
x=10
hence ratio of 3year before father age and eldest son age is 45/16
hence option 4 is currect

Like? Yes (1) | No   

pushpa   10 years AGO

sum of ages of his 3 sons= his present age=48
n difference of age of his sons is 3
so ages of his 3 sons can be
(19+16+13)=48
so the age of his elder son is 19 at present
therefore the ratio of his age & his eldest son's age 3 year before,
(48-3)/(19-3)=45/16
ans- 4

Like? Yes (1) | No   

sanjeev kumar   10 years AGO

ans 4

Like? Yes (1) | No   

VIVEK KUMAR BARANWAL   10 years AGO

LET AGES OF SONS ARE X-3, X, X+3
BY PROBLEM
(X-3)+X+X+3=48
X=16
THEREFORE ELDEST SONS PRESENT AGE=16+3=19
THEREFORE REQUIRED RATIO=(48-3):(19-3)=45:16

Like? Yes (1) | No   

MRITUNJAY   10 years AGO

ages of the suns is 13 16 19.
ratio of father and his eldest son three years back is 45/16

Like? Yes (1) | No   

Deepak prajapati   10 years AGO

let age of eldest son=x+6
then age of second son=x+3
age of younger son=x
so
ATQ
3x+9=48
x=13
now
(48-3)/(x+6-3)=45/16
option 4 is correct

Like? Yes (2) | No (1)  

abhishek narayan   10 years AGO

f=e1+e2+e3
=>48 =3e1-9
=>e1=19
f-3/e1-3=45/16

Like? Yes (1) | No   

PRANAV PRAKASH   10 years AGO

Suppose younger son age is A...
Then A+(A+3)+(A+6)=48....SO A=13...
Eldest son age is 13+6=19...A/q ratio of father to the eldest son age 3 years before=(48-3)/(19-3)=45/16

Like? Yes (1) | No   

Deepak Kumar   10 years AGO

let x be the age of the eldest son
then
48= x + (x - 3) + (x - 6)
then
x = 19
father age is 48
three years before
eldest son's age was 16
and father's age was 45
then the ratio is
45/16

Like? Yes (1) | No   

AMREEN   10 years AGO

Let the age of eldest son is A
the age of second son =A-3
age of 3rd son=A-6
Then A+A-3+A-6=48
A=19
Then ratio of eldest son and fathe before 3 years is=(48-3)/(19-3)
=45/16 ans.

Like? Yes (1) | No   

vaitesh mudaliar   10 years AGO

sum of ages of 3 sons =48
x1+x2+x3= 48
x1-x2=3
x2-x3=3
solving these 3 equn v get values of x1=19, x2=16, x3=13
ratio =(X-3)/(x1-3)=48-3/19-3 =45/16

Like? Yes (1) | No   

shubham   10 years AGO

43434

Like? Yes (1) | No   

Jayanta Dutta   10 years AGO

x+x+3+x+6=48

so x+6=19;

now 3 years before

dad was 48-3=45
eldest son was=19-3=16
so ratio is: 45/16

Like? Yes (1) | No   

Sourabh Prasad   10 years AGO

Let son's age be x, y and z.
We have considered x is the youngest son.

Given,
Present age of father is 48.
x+y+z=48 -----------------(1)

Also Given, Age difference of his sons is 3 years.
x=y+3 -----------------(2)
y=z+3 -----------------(3)


So we have 3 equation & 3 variable,

Soving we get x=19

Therefore, Ratio before 3 years is (48-3)/(19-3)
=45/16

Like? Yes (1) | No   

Yekollu Muneeswar   10 years AGO

given age of father at present =48 & sons age difference is 3 yrs
let age of young son be X ,middle X+3 , and Elder be X+6
then given that at present, sum of ages of sons=father age
---> X+X+3+X+6=48
3X+9=48
x=13.

before 3 yrs Father age=45 and Elder Son Age=16
so Ratio is 45/16

Like? Yes (2) | No (1)  

ARJUN PRASAD   10 years AGO

let present age of three sons are x,x+3,x+6 ;
atp....
x+(x+3)+(x+6)=48
so x=13.
present age of elder son=19
before 3 ys
age of father=45 and age of elder son=16
so ratio=45/16

Like? Yes (1) | No   

Praneet   10 years AGO

At Present
let smallest son age be x the
x+x+3+x+6=48
so, x=13
therefore age of son is 13,16,19


Now 3 years ago age of eldest son is 19-3=16
so, ratio of their ages is 45/16

Like? Yes (1) | No   

Hemadri   10 years AGO

present age of man=48
His 3 sons ages having difference of 3
So the ages of his sons are 19 16 13(19+16+13)=48
Ratio of his eldest son to his age before 3 years is
48-3=45 and 19-3=16
Ratio =45/16

Like? Yes (1) | No   

nitin Maheshwari   10 years AGO

Let the age of children are x,x-3,x-6
given,
x+x-3+x-6=48,
x=19


ratio 45/16 ans.

Like? Yes (1) | No   

A M ANKIT KALLURAYA   10 years AGO

45/16
x+x-3+x-6=48
x=19
three years before
x=x-3=16
fathers present age=48
three yrs ago father was 45
=>45/16

Like? Yes (1) | No   

Pankaj Malik   10 years AGO

Present age of father=sum of ages of his sons…n difference b/w age of sons is 3...
So 48=13+16+19
At present the age of eldest son is 19.
Before 3 years...eldest son's age=16 & father's age=45
Ratio=45/16

Like? Yes (1) | No   

vootukoori laxman   10 years AGO

let the ages are x, x+3,x+6 from young to elder
now x+x+3+x+6=48
then x=13
x+3=16
x+6=19
then ratio before 3 years is 48-3 /19-3 =45/16

Like? Yes (1) | No   

MAHARAJAN.G   10 years AGO

now father age is 48
no. of sons=3
they have difference of 3 in their ages
so,
first son age is 19
second son age is 16
third son age is 13
first son age before 3 years is 16
father age before three years is 45
ratio is=45/16
ans ) 4

Like? Yes (1) | No   

rajkaran singh   10 years AGO

we are given : present age of father=48
differences b/w sons ages=3
therefore sons ages must be = 13,16, 19(as 13+16+19=48 and differences b/w their ages=3)
now
3 years back father's age=45
and eldest son = 19-3=16
therefore ratio of their ages=45/16

Like? Yes (1) | No   

Biswajit Saha   10 years AGO

let the age of youngest son is x.
differences of ages of his son is 3.
man's present age is 48.
x+(x+3)+(x+6)=48
x=13
hence age of eldest son is (x+6)=13+6=19
3 years before the age of father and eldest son was (48-3)=45 and (19-3)=16.
the ratio of father's age & his eldest son's age 3 year before is = 45/16.

Like? Yes (1) | No   

Gagan   10 years AGO

before 3 years the age of father is 45 and the of eldest son is 16 because the age of eldest son is after 3 years 19 and age of other son's are 13 & 16 so difference between ages are 3

Like? Yes (1) | No   

sravani rachakonda   10 years AGO

M=s1+s2+s3
Given s1-s2=3
S2-s3=3
48=s1+s1-3+s1-6------->>s1=19
M-3:s1-3=45:16

Like? Yes (1) | No   

ayushi   10 years AGO

Let a be the age of eldest son :
48=a+(a-3)+(a-6)
=>a=19;
=>age of father/age of eldest son is(3 yrs ago )=(48-3)/(19-3)
=>45/16

Like? Yes (1) | No   

Sonam Gupta   10 years AGO

given x+y+z = 48
let x be the eldest son , y mid son and z youngest son
hence , x= y+3 or y = x-3
y = z+3 or z = y-3 or z= x-3 -3 = x-6

hence , x + x-3 +x - 6 = 48
3x -9 =48
3x = 57
x = 19

3 years before , x = 16
and father was = 45
hence ratio was 45 / 16

Like? Yes (1) | No   

niraj   10 years AGO

48-3/19-3=45/16

Like? Yes (1) | No   

Y V S Chandrakanth   10 years AGO

x,x-3,x-6
given x+x-3+x-6=48
x=19
ratio=(48-3)/(19-3)=45/16

Like? Yes (1) | No   

Tark Patel   10 years AGO

Let age of man = M
age of his 3 sons like S1, S2. S3 (S3 is eldest)

Given M = S1 + S2 + S3
Sons' age difference is 3
so M = (S1) + (S1+3) + (S1+6)

48 = 3S1 + 9
S1 = 13

->S3 = 19
Hence 3 years before ratio of man's age and his eldest son (S3) is,

48 - 3 / 19 - 3 = 45 / 16

Like? Yes (1) | No   

khushboo goenka   10 years AGO

let son ages are x,x-3,x-6
thn father age =x+x-3+x-6=48
so x=19
3 year before father age=45
and eldest son age= 16,13,10

Like? Yes (1) | No   

ANKIT GABA   10 years AGO

48=(x-6)+(x-3)+x
x=19
3year ago
48-3=45 father age
19-3=16 elder son age
45/16

Like? Yes (1) | No   

Manoj Chouhan   10 years AGO

a+b+c= 48
a+(a-3)+(a-6)=48
3a-9 =48
a=19.
so b=16; c=13
now,

48 = 13+16+19 //*

48-3/19-3 ratio

45/16 ans (4)

Like? Yes (1) | No   

vikaskumar   10 years AGO

father age=48;elder son age=x
As there is a difference of 3 between the brother's ages
48=x+(x-3)+(x-6)
48=3x-9
x=57/3=19;
As in que it is asked before 3 years father's age=45 and eldest son=16
::Answer is 45/16

Like? Yes (1) | No   

SUMAN SAURAV   10 years AGO

48 = s + (s-3) + (s-6)
s = 19

age of man 3 yrs ago = 45
age of elder son 3 yrs ago = 16
ratio = 45/16

Like? Yes (1) | No   

Udhaya kumar   10 years AGO

48 = (x + x-3 + x-6)
x = 19
3 years before 16 , 45
45 / 16......

Like? Yes (2) | No (1)  

Amit Kumar   10 years AGO

age of son's are x, x+3,x+6
so
x+x+3+x+6=48
x=13
age of father before 3 years=45
age of the eldest son is=19-3=16

Like? Yes (1) | No   

Rajat Rokade   10 years AGO

Suppose Present age of eldest son = x
Difference between age=3
So,
(x-6) + (x-3) + x=48
(age of eldest son) x=19
Before 3 years,
father's age=45 and eldest son's age=16
So ratio = 45/16.
Option 4 is correct.

Like? Yes (3) | No (3)  

Rahul Rathod   10 years AGO

let ages of sons are=x-3 , x , x+3
present age of man =48
so 48=(x-3)+(x)+(x+3)
x=16
so ages of sons are: 13,16,19
age of eldest son 3 years before=19-3=16
age of man before 3 years=48-3=45
so
ratio=(45/16)
answer is option 4 =45/16

Like? Yes (3) | No (3)  

sravani   10 years AGO

Let the age of elder son be X
given diff between the sons is 3 years each and mans age is 48 at present .
so the eq is x+x-3+x-3-3 =48
3x-9=48
x=19
3 years before
father's age is 45
elder sons age is 16
ratio is 45:16.

Like? Yes (2) | No (2)  

BHASKAR MANDAL   10 years AGO

let father present age be p
age of older = x
p = x + x-3 + x-6
= 3x-9
48 = 3x-9
=> x =19
ratio of father age to son 3 years before = 45/16
option 4

Like? Yes (2) | No (2)  

Md.Sultan Khichi   10 years AGO

Present age of father is 48.
So sum of ages of sons is 48.
Diff of their ages is 3. So eldest son's age is 19.
3 years before fathers age was 45 and eldest sons was 16.
Ratio=45/16
Ans..ob 4

Like? Yes (1) | No (1)  

Azhar Syed   10 years AGO

Because x+x+3+x+6=48;
solving we get x=13 i.e present age of younger son & present age of eldest son is 19.
Before 3 years fathers age was 45 and eldest son's age was 16

Like? Yes (1) | No (1)  

PAWAN KUMAR   10 years AGO

Age of father = 48
let ages of sons be x, x+3 and x+6 respectively.
Then, sum = 3x+9 = 48 =>x= 13
Therefore, ratio = 45/16 Ans.

Like? Yes (1) | No (1)  

RAKESH   10 years AGO

13+16+19 = 48

req. ratio = 48-3 / 19-3 = 45/16

Like? Yes (1) | No (1)  

ganesh babu   10 years AGO

F+S+T=man
F is elder son , S second son, T third son
F+(F-3)+(F-6)=48 Given that the difference of age of his sons is 3
3F=48+9
F=19
What was the ratio of his age & his eldest son's age 3 year before
45/16

Like? Yes (1) | No (1)  

Sukanya   10 years AGO

Age of man= sum of his sons ages= 48
Dif bw the sons ages is 3
x-3+x+x+3=48
x=16
sons ages=13,16,19.

3 years before
father's age =45
elder son age= 16
so the ratio =45/16.

Like? Yes (1) | No (1)  

susheela   10 years AGO

father=sum of age of 3 sons
at present father's age is =48
48=19+16+13
48=19 (age of eldest son)
three year before
45=16
45/16

Like? Yes (1) | No (1)  

Ranjit Somisetty   10 years AGO

48=(x+x+3+x+6) x is age of youngest
x=13
3 years ago fathers age is 45 and eldest sons age is x+6-3=16
Hence ratio becomes 45/16

Like? Yes (1) | No (1)  

Manju Prakash R   10 years AGO

Let ages of sons be x-3,x,x+3 in the order.
So, x-3 + x+ x+3 =48

So age of eldest =19
Ratio 45 :16

Like? Yes (1) | No (1)  

Jeyadev    10 years AGO

given 48=x+(x=3)=((x+3)+3) ==> 48=3x+9 ==> x=13
before 3 yrs their ratio will be ((48-3)/(x+6-3)) ==> 45/(13+6-3)
==> 45/16

Like? Yes (1) | No (1)  

24 Nov 2014 Sequence

What is the next number of the following sequence
199, 296, 391, 484, ... ?

OPtion

1) 581
2) 576
3) 525
4) 567
5) 564
6) 575
7) 665
8) 595
9) 645
10) 543

Solution

199 => 1+99 = 100
296 => 2^2+96 = 100
391=> 3^2+91 = 100
484 => 4^2+84 = 100
Similarly

575 => 5^2+75 = 100

Option 6)

Correct Option: 6 Best Solution (108)

P.Pandeeswari   10 years AGO

199 => 1+99 = 100
296 => 2^2+96 = 100
391=> 3^2+91 = 100
575 => 5^2+75 = 100

Like? Yes (22) | No (5)  

Kushagra Tiwari   10 years AGO

T(1)=199
T(2)=199+97= 296
T(3)=296+95= 391
T(4)=391+93= 484
So, the difference between terms is reducing by 2 each time.
Therefore, T(5)=484+91= 575..... Option 5)

Like? Yes (17) | No (6)  

Udhaya kumar   10 years AGO

sequence of (n*100) - (n-1) ^ 2
from n = 2,3,4,5,6 respectively.......
n = 2 .... 199
3.........296
4.........391
5.........484
6.........575
or else
simply add 97,95,93,91 respectively..........

Like? Yes (8) | No (1)  

Shreya Tiwari   10 years AGO

T(1)=199
T(2)=199+97= 296
T(3)=296+95= 391
T(4)=391+93= 484
So, the difference between terms is reducing by 2 each time.
Therefore, T(5)=484+91= 575..... Option 5)

Like? Yes (13) | No (6)  

Chirag Ranjan Lahiri   10 years AGO

1 + 1 = 2, 2 * 100 = 200, 200 - (1*1) = 199
2 + 1 = 3, 3 * 100 = 300, 300 - (2*2) = 296
3 + 1 = 4, 4 * 100 = 400, 400 - (3*3) = 391
.
.
.
hence, 5 + 1 = 6, 6 * 100 = 600, 600 - (5*5) = 575, answer...

Like? Yes (4) | No (2)  

RAKESH   10 years AGO

1---99,
2---96,
3 ---91
4--- 84,
5----75

Like? Yes (3) | No (1)  

Robeda   10 years AGO

199, 296, 391, 484, ... ?
97 95 93 next will be 91
so 484+91=575

Like? Yes (2) | No   

A SIRISHA    10 years AGO

GIVEN SERIES

LET ' X' BE THE NEXT NUMBER

199 296 391 484 X

(296-199) ( 391-296) (484-391) ( X-484)

= 97 = 95 = 93 = X-484

(97-95) (95-93) ( 93-X+484)

2 2 2

SO

93-X+484=2

X= 575

Like? Yes (4) | No (2)  

CHANDAN KUMAR   10 years AGO

296-199=97
391-296=95
484-391=93
484+91=575

Like? Yes (1) | No   

NEHA KUMARI   10 years AGO

199,296,391,484,575
296-199=97
391-296=95
484-391=93
575-484=91
so, ans=575

Like? Yes (1) | No   

khushboo goenka   10 years AGO

diff between frst 2 is 100-3, thn 100-5,100-7,100-9

Like? Yes (1) | No   

Deepak Kumar   10 years AGO

200- 1*1 = 199
300- 2*2 = 296
400- 3*3 = 391
500- 4*4 = 484
therefore
600- 5*5 = 575

Like? Yes (2) | No (1)  

esther   10 years AGO

199, 296, 391, 484,....?
296-199=97
391-296=95
484-391=93
575-484=91

Like? Yes (2) | No (1)  

lavanya   10 years AGO

199+97=199,
296+95=391,
391+93= 484,
484+91=575

Like? Yes (1) | No   

karthik sundaresan.S   10 years AGO

296-199=97
391-296=95
484-391=93
575-484=91

Like? Yes (1) | No   

Ravi Ranjan Ojha   10 years AGO

296-199=97
391-296=95
484-391=93
hence next difference will be 91
so
484+91=575

Like? Yes (1) | No   

Raj   10 years AGO

484-391=93; 391-296=95; 296-199=97 , result is decrease so ans is 484+91=575: (6)

Like? Yes (1) | No   

veeranjaneyulu   10 years AGO

add 97 ,95,93,91......

Like? Yes (1) | No   

ISWARYA   10 years AGO

199+97=296
296+95=391
391+93=484
484+91=575

Like? Yes (2) | No (1)  

santosh kumar patel   10 years AGO

To get the value for n=5:

Add one, giving =6.
Multiply by one hundred, giving 600.
Subtract the original number squared, giving =600-25.
So the result is =575.

Like? Yes (2) | No (1)  

swasthika   10 years AGO

296-199 =97
391-296 =95
484-391 =93
x-484=91
x=575

Like? Yes (2) | No (1)  

Rajat Rokade   10 years AGO

199=(2)100 - (1)^2
296=(3)100 - (2)^2
391=(4)100 - (3)^2
484=(5)100 - (4)^2
so,
575=(6)100 - (5)^2

Like? Yes (2) | No (2)  

Pramod Gaikwad   10 years AGO

199=1^2+99=100
296=2^2+96=100
391=3^2+91=100
484=4^2+84=100
575=5^2+75=100

Like? Yes (2) | No (2)  

Rats   10 years AGO

199+97=296
296+95=391
391+93=484
so next term is...
484+91=575

Like? Yes (1) | No (1)  

Tark Patel   10 years AGO

by subtracting two numbers in sequence like,

296 - 199 = 97

391 - 296 = 95

484 - 391 = 93

Difference is between 97 , 95 , 93 is 2 and in decreasing order so next difference will be 91

answer is 484 + 91 = 575

Like? Yes (1) | No (1)  

Anand   10 years AGO

296-199=97
391-296=95
484-391=93
484-575=91
so ans will be 575

Like? Yes (2) | No (2)  

MD SOHAIL AKHTER   10 years AGO

199+97=296
296+95=391
391+93=484
484+91=575

ans option 6

Like? Yes (1) | No (1)  

ravi prakash   10 years AGO

484+91=575

Like? Yes (1) | No (1)  

K.ARun   10 years AGO

199+97
296+95
391+93
484+91
so.. 575

Like? Yes (2) | No (2)  

NAVEEN KUMAR   10 years AGO

199---(1+1,99-3)=296
296-----(2+1,96-5)=391
391------(3+1,91-7)=484
484-------(4+1,84-9)=575 option(6)

Like? Yes (1) | No (1)  

DIPAK TIWARI   10 years AGO

LOGIC-Difference of the next number to its previous number is striclty decreasing by 2
296-199=97
391-296=95
484-391=93
So,next no=484+91=575
Option 6

Like? Yes (1) | No (1)  

Supriya Mondal   10 years AGO

100+99=199
199+97=296
296+95=391
391+93=484
484+91=575
so answer will be 6)575

Like? Yes (1) | No (1)  

sanjeev kumar   10 years AGO

199+97=296,
296+95=391,
391+93= 484,
484+91=575

Like? Yes (1) | No (1)  

ANAND KUMAR   10 years AGO

199+(100-3), 296+(100-5), 391+(100-7), 484+(100-9)=575

Like? Yes (1) | No (1)  

Md.Sultan Khichi   10 years AGO

Ans 575
199+100-3=296
296+100-5=391
391+100-7=484
484+100-9=575

Like? Yes (2) | No (2)  

naresh potturi   10 years AGO

296-199=97;
391-296=95;
484-391=93;
575-484=91

Like? Yes (1) | No (1)  

Dinesh thatti   10 years AGO

differnece b/w numbers are as 97,95,93,91

Like? Yes (1) | No (1)  

lakshmi   10 years AGO

sequence : 199 296 391 484, ......
doing subtraction: 97 95 93 91
by adding : 91 + 484 = 575

Like? Yes (1) | No (1)  

purushothama   10 years AGO

sequence : 199 296 391 484, ......
doing subtraction: 97 95 93 91
by adding : 91 + 484 = 575

Like? Yes (1) | No (1)  

jayalakshmi   10 years AGO

sequence : 199 296 391 484, ......
doing subtraction: 97 95 93 91
by adding : 91 + 484 = 575

Like? Yes (1) | No (1)  

Abhishek Vishwakarma   10 years AGO

T(1)=199
T(2)=199+97= 296
T(3)=296+95= 391
T(4)=391+93= 484
So, the difference between terms is reducing by 2 each time.
Therefore, T(5)=484+91= 575..... Option 5)

Like? Yes (1) | No (1)  

bhagya   10 years AGO

sequence : 199 296 391 484, ......
doing subtraction: 97 95 93 91
by adding : 91 + 484 = 575

Like? Yes (1) | No (1)  

CHANDA VIJAYALAKSHMI   10 years AGO

sequence : 199 296 391 484, ......
doing subtraction: 97 95 93 91
by adding : 91 + 484 = 575

Like? Yes (1) | No (1)  

sravani   10 years AGO

199+97=296
296+95=391
391+93=484
484+91=575

Like? Yes (1) | No (1)  

A M ANKIT KALLURAYA   10 years AGO

199
296=199+97
391=296+95
484=391+93
575=484+91

Like? Yes (1) | No (1)  

Ranjit Somisetty   10 years AGO

common difference is decreasing by 2 starting from 97
484+91=575

Like? Yes (1) | No (1)  

prem   10 years AGO

296-199=97
391-296=95
484-391=93
hence next difference between no is 91
hence no is 575

Like? Yes (1) | No (1)  

kuldeep singh   10 years AGO

200 - 1^2 = 199
300 - 2^2 = 296
400 - 3^2 = 391
500 - 4^2 = 484
600 - 5^2 = 575
600 - 5^2 = 575

Like? Yes (1) | No (1)  

Ramesh Sriram   10 years AGO

+100-3
+100-5
+100-7
+100-9

Like? Yes (1) | No (1)  

sandineni kiran   10 years AGO

difference b/w numbers is 97,95,93,91
result=484+91=575

Like? Yes (1) | No (1)  

PAWAN KUMAR   10 years AGO

In the given series, every term is being added by
100- (2n+1).

199+100-(2*1+1) = 296
296+100-(2*2+1) = 391
391+100-(2*3+1)= 484
484+100-(2*4+1) = 575

Like? Yes (1) | No (1)  

Prabhat   10 years AGO

The given series is in the form - +97,+95,+93 ...
So the next number is = 484+91=575

Like? Yes (1) | No (1)  

akash kumar   10 years AGO

difference between consecutive terms decreasing in order 97,95,93,91...

Like? Yes (1) | No (1)  

Anoop Tiwari   10 years AGO

296-199=97
391-296=95
484-391=93
so next term will be 484+91=575
correct option is 6)

Like? Yes (1) | No (1)  

Shiva Guntuku   10 years AGO

above series increasing order of no.s 97,95,93,91 so on.then next no. is 575

Like? Yes (1) | No (1)  

BHASKAR MANDAL   10 years AGO

if we take the difference from each 100
200 - 199 = 1
300-196 = 4 = 2^2
400-391 = 9 = 3^2
500 - 484 = 16 = 4^2
so the next number will be
600 - 5^2 = 600 - 25 = 575
option 6

Like? Yes (2) | No (2)  

LOKESH KUMAR   10 years AGO

199+97=296
296+95=391
391+93=484
484+91=575

Like? Yes (1) | No (1)  

MRITUNJAY   10 years AGO

296-199=97
391-296=95
484-391=93
next is 484+91=575

Like? Yes (1) | No (1)  

Swagatam Kumar Das   10 years AGO

The difference of 1 & 2 is 97
The difference of 2&3 is 95
The difference of 3&4 is 93
So,The difference of 4&5 is 91
So next is 484+91=575

Like? Yes (1) | No (1)  

Azhar Syed   10 years AGO

199, 199+97=296, 296+95=391, 391+93=484, 484+91=575.

Like? Yes (1) | No (1)  

Manas   10 years AGO

there is adding 97 95 93 odd series so next will add 91

Like? Yes (1) | No (1)  

sravani rachakonda   10 years AGO

The difference follows 97,95,93,91....
484+91=575

Like? Yes (1) | No (1)  

Dowrla Sachin Sanjeevan   10 years AGO

first digit is 5 since it has 1 , 2 ,3 ,4 in their first place. after this the digits other than the hundreds place are in series like they are decreasing by odd number 3, 5 , 7 respectively. so i subtracted 9 from last term 84 i.e 75. so my number is 575

Like? Yes (1) | No (1)  

Anshul Shrivastava   10 years AGO

199
199+97=296
296+95=391
391+93=484
484+91=575

Like? Yes (1) | No (1)  

Manish Singh   10 years AGO

Ans is (6)
296-199=97
391-296=95
484-391=93
NOW THERE SHOULD BE NEXT DIFF WOULD B 91
SO 484+91=575
ANS 6

Like? Yes (2) | No (2)  

susheela   10 years AGO

199-296=97
296-391=95
391-484=93
484-575=91

Like? Yes (1) | No (1)  

Manju Prakash R   10 years AGO

1st digit is 1,2,3,4..5
99-96=3
96-91=5
91-84=7
84-9=75

So Ans=575

Like? Yes (1) | No (1)  

Devendra Marghade   10 years AGO

Difference of 97, 95, 93, 91, .....
199+97 =296
296+95= 391
391+93= 484
484+91= 575

Like? Yes (1) | No (1)  

SRICHANDER P   10 years AGO

296-199=97
391-296=95
484-391=93
575-484=91

Like? Yes (1) | No (1)  

Pravin Raj   10 years AGO

296-199=97
391-296=95
484-391=93
next will be 91
484+91=575

Like? Yes (1) | No (1)  

Tania Chatterjee   10 years AGO

200 - 1^2 = 199
300 - 2^2 = 296
400 - 3^2 = 391
500 - 4^2 = 484
600 - 5^2 = 575

Like? Yes (1) | No (1)  

vivek   10 years AGO

296-199=97
391-296=95
484-391=93

hence next difference should be 91
next digit 575-484=91

Like? Yes (1) | No (1)  

Rishikesh Ranjan   10 years AGO

99-96 =3
96-91=5
91-84=7
next difference should be 9
ans. 576

Like? Yes (1) | No (1)  

Chandra chur anand   10 years AGO

296-199=97
391-296=95
484-391=93
x-484=91 so x=575 ans....

Like? Yes (1) | No (1)  

Sandeep Kumar   10 years AGO

+97, +95, +93, +91

Like? Yes (1) | No (1)  

Amit Roshan   10 years AGO

296-199=97
391-296=95
484-391=93

the difference decreses by 2 hence the next number would be 484+(93-2)=575

Like? Yes (1) | No (1)  

Farid Ahmad   10 years AGO

296-199=97
391-296=95
484-391=93

Here we can see that difference start with 97 and decreasing by 2 after 1 term,so that for the last term difference would be 91,now we add 91 to 484 then we get 575.
so answer would be 575

Like? Yes (2) | No (2)  

Vidyasagar   10 years AGO

296-199= 97
391-296= 95
484-391= 93
***-484= 91
***= 484+91= 575

Like? Yes (1) | No (1)  

Biepashaa Dutta   10 years AGO

number+100-odd number ie 3,5,7,9

Like? Yes (1) | No (1)  

Mayank kumar modi   10 years AGO

DIIF IS 91

Like? Yes (1) | No (1)  

Dipu Kumar   10 years AGO

199-296=-97
296-391=-95
391-484=-93
484-***=-91

484+91=575

Like? Yes (1) | No (1)  

Rahul Kumar   10 years AGO

Clear Question

Like? Yes (1) | No (1)  

Preeti Sagar   10 years AGO

Answer

Like? Yes (1) | No (1)  

MAHARAJAN.G   10 years AGO

296-199=97
391-296=95
484-391=93 difference between sequence is odd number in decreasing order so next number difference is 91, so
x-484=91
484+91=x
575=x
ans ) 6

Like? Yes (1) | No (1)  

Deepak prajapati   10 years AGO

100 place digit increasing by one and last two digit no is decreasing by 3,5,7,9..... so ans will be 575
so option 6 is correct

Like? Yes (1) | No (1)  

Sukanya   10 years AGO

100 +100-1^2=199
100 +200-2^2=296
100 +300-3^2=391
100 +400-4^2=484
100 +500-5^2=575

Like? Yes (1) | No (1)  

harshal pawar   10 years AGO

add 91

Like? Yes (1) | No (1)  

nitin Maheshwari   10 years AGO

series odd nos

Like? Yes (1) | No (1)  

Aravinth kumar   10 years AGO

199+97= 296+95= 391+93= 484+91= 575

Like? Yes (1) | No (1)  

rajesh kumar gupta   10 years AGO

difference of increased by 2 each number

Like? Yes (1) | No (1)  

mohit maddhesiya   10 years AGO

diffrence of 97,95,93,91

Like? Yes (1) | No (1)  

Anees Fathima   10 years AGO

296-199=97
391-296=95
484-391=93
so next number to be added is 91 i.e alternatenumbers are getting added(97,95,93,91)
so 484+91=575

Like? Yes (1) | No (1)  

anupriya sinha   10 years AGO

in this series 296 -199 =97
391-296 =95
484-391 = 93
484+91 =575

Like? Yes (1) | No (1)  

Ashish Shende   10 years AGO

199+(100-3)=296
296+(100-5)=391
391+(100-7)=484
484+(100-9)=575
Ans = option 6

Like? Yes (1) | No (1)  

Devendra Kumar   10 years AGO

200-1^1=199
300-2^2=296
400-3^3=391
500-4^4=484
in the same way
600-5^5=575

Like? Yes (1) | No (1)  

sachin verma   10 years AGO

296-199=97
391-296=95
484-391=93
So,
x( ? ) -484=91
> x=575

Like? Yes (1) | No (1)  

ANJU MURALY   10 years AGO

199+97=296
296+(97-2)=296+95=391
391+(95-2)=391+93=484
484+(93-2)=484+91=575

Like? Yes (1) | No (1)  

vaitesh mudaliar   10 years AGO

the digits in the hundreds place are increasing by one so the next no. vl b having no. 5 at the hundreds place. in the sequence last two digits are being decreased by odd no. 3,5,7 an next no. vl b dec by 9. hence next no. vl be 575

Like? Yes (1) | No (1)  

Satyanarayana   10 years AGO

199+97 = 296
296+95 = 391
391+93 = 484
484+91 = 575

Like? Yes (1) | No (1)  

manya   10 years AGO

it add in 1st no. is 97 then 92 then 85 then 76 then the answer is 575

Like? Yes (1) | No (1)  

Vaibhav Shah   10 years AGO

we just add (100-3) to 199 to get 296, (100-5) to 296 to get 391, (100-7) to 391 to get 484 , (100-9) to 484 to get '575'.

Like? Yes (1) | No (1)  

pettugani tejaswini   10 years AGO

the difference is in these sequence..97,95,93,91

Like? Yes (1) | No (1)  

abhishek agrawal   10 years AGO

1 and 99-0 = 199
2 and 99-3 = 296
3 and 96-5 = 391
4 and 91-7 = 484
5 and 84-9 = 575

Like? Yes (1) | No (1)  

Jigyasa Kumar   10 years AGO

difference of 95 then 93 then 91

Like? Yes (1) | No (1)  

ANIL KUMAR   10 years AGO

199(+97)=296(+95)=391(+93)=484(+91)=575

always adding the no less the previous one.

Like? Yes (1) | No (1)  

abhilasha   10 years AGO

order followed is of the difference 95 then 93 and so next will be at a difference 91

Like? Yes (1) | No (1)  

Rahul Rathod   10 years AGO

199 = 1+99 = 100
296 = 2^2+96 = 100
391= 3^2+91 = 100
484=4^2+84=100
575 = 5^2+75 = 100

Like? Yes (2) | No (3)  

Sonam Gupta   10 years AGO

296 - 199 = 97
391 - 296 = 95
484 - 391 = 93

hence , 484 +91 =575

Like? Yes (1) | No (2)  

21 Nov 2014 Quadratic Equation

The sum of a non zero number, its square, its cube and its four power is 7380. If five power of the number is added to 7380 and divided by the number, then quotient will be?

OPtion

1) 2880
2) 960
3) 1555
4) 7381
5) 1024
6) 363
7) 1364
8) 10918
9) 3904
10) 4679

Solution

x + x^2 + x^3 + x^4 = 7380
Add x^5 on both the side
We get
x + x^2 + x^3 + x^4 + x^5 = 7380 + x^5
Now divide by x on both the side
(x + x^2 + x^3 + x^4 + x^5)/x = (7380 + x^5)/x
1 + x + x^2 + x^3 + x^4 = (7380 + x^5)/x
1+7380 =
7381
Option 4)

Correct Option: 4 Best Solution (58)

Shreya Tiwari   10 years AGO

LET NON ZERO NUMBER =X

X+X^2+X^3+X^4=7380

BY SIMPLFY

X=9

SO A/C TO GIVEN QUESTION "If five power of the number is added to 7380 and divided by the number, then quotient will be?"

SO

(X^5+7380)/X = ((9^5+7380)/9)
=7381
SO QUOTIENT = 7381

Like? Yes (57) | No (7)  

Kushagra Tiwari   10 years AGO

LET NON ZERO NUMBER =X

X+X^2+X^3+X^4=7380

BY SIMPLFY

X=9

SO A/C TO GIVEN QUESTION "If five power of the number is added to 7380 and divided by the number, then quotient will be?"

SO

(X^5+7380)/X = ((9^5+7380)/9)
=7381
SO QUOTIENT = 7381

Like? Yes (45) | No (7)  

P.Pandeeswari   10 years AGO

the non zero number is 9
9+9^2+9^3+9^4=7380
9^5+7380=66429
66429/9=7381.
option:4)7381

Like? Yes (11) | No (4)  

Sanu Panji   10 years AGO

(x^5+x^4+x^3+x^2+x)/x
= x^4+x^3+x^2+x+1
7380+1
=7381

So correct option 4)

Like? Yes (4) | No (3)  

khushboo goenka   10 years AGO

let x+x^2+x^3+x^4=7380
7380+x^5/x=1+x+x^1+x^2+x^3+x^4=7381
so remainder=1
quotient=7381

Like? Yes (1) | No   

CHANDAN KUMAR   10 years AGO

x+x^2+x^3+x^4=7380
x^5+7380/x=x^4+7380/x=7380+1=7381

Like? Yes (1) | No   

ADARSH GAURAV   10 years AGO

7381 is the answer.

Like? Yes (1) | No   

Mohan   10 years AGO

number is 9 ,and ans is 7381,
9+9^2+9^3+9^4=7380 and 9^5+7380=66429
so 66429/9 =7381

Like? Yes (1) | No   

kushagra pareek   10 years AGO

x+x^2+x^3+x^4 = 7380

find (x^5+7380)/x

x^5+x^4+x^3+x^2+x divided by

gives quotient 1+x+x^2+x^3+x^4 which is 7381

Like? Yes (3) | No (2)  

Udhaya kumar   10 years AGO

x+x^2+x^3+x^4 = 7380
x^5+7380 = x^5+x^4 + x^3 + x^2 + x / x
x^4 + x^3 + x^2 + x + 1 = 7381.......

Like? Yes (4) | No (4)  

Chirag Ranjan Lahiri   10 years AGO

Number is 9 and ans is 7381,
9 + 9^2 + 9^3 + 9^4 =7380
and 9^5 +7380 = 66429 so 66429/9 =7381.........

Like? Yes (4) | No (4)  

PAWAN KUMAR   10 years AGO

Let the no be x.
Then, x + x^2 + x^3 + x^4 = 7380
On solving, we get: x=9
Now, 9^5 + 7380 = 59049 +7380 = 66429
66429/9 = 7381 Ans.

Like? Yes (2) | No (3)  

sravani   10 years AGO

x+x^2+x^3+x^4=7380
(x^5+7380)/x=(x^5+x+x^2+x^3+x^4)/x
x(1+x+x^2+x^3+x^4)/x=1+x+x^2+x^3+x^4=1+7380=7381.....

Like? Yes (3) | No (4)  

SUBHASH TULSYAN   10 years AGO

let the no. be x
A/Q
x+x^2+x^3+x^4=7380....(1)
now we have to find
(7380+x^5)/x
=(x+x^2+x^3+x^4+x^5)/x
=x(1+x+x^2+x^3+x^4)/x
=1+x+x^2+x^3+x^4
=1+7380............from(1)
=7381
option(4)

Like? Yes (2) | No (3)  

Rahul Singh   10 years AGO

X+X^2+X^3+X^4 = 7380
X(1+X+X^2+X^3) = 7380
1+X+X^2+X^3 = 7380/X

Now (7380 + X^5)/X = 7380/X + X^5/X
From above = 1+X+X^2+X^3+X^4
= 1+ 7380
= 7381

Like? Yes (2) | No (3)  

Shreya Banerjee   10 years AGO

n+n2+n3+n4 = 7380 (given)
now add n5 to it
and divide that by n,
we wil hv
(n+n2+n3+n4+n5)/n = (1+n+n2+n3+n4) = 7380+1 =
7381

So option 4.

Like? Yes (2) | No (3)  

santosh kumar patel   10 years AGO

x + x^2 + x^3 + x^4 = 7380
then x=9
x^5=66429
then, 66429+7380?9
7381

ans: 7381

Like? Yes (2) | No (3)  

MAHARAJAN.G   10 years AGO

number is 9 ,
sum of its square , cube, four power is 7381, 9+9^2+9^3+9^4=7380
and 9^5+7380=66429 so 66429/9 =7381

Like? Yes (1) | No (2)  

jitu   10 years AGO

9^5=59049+7380

Like? Yes (1) | No (2)  

rajesh reddy   10 years AGO

x+x2+x3+x4=7380,

x+x2+x3+x4+x5= x(1+x+x2+x3+x4)= x(1+7380)

quotient=x(7381)/x

Like? Yes (1) | No (2)  

Anand Dwivedi   10 years AGO

7381

let : X+X^2+X^3+X^4=7380

(X^5+7380)/X = ((9^5+7380)/9) =7381

SO QUOTIENT WILL BE 7381

Like? Yes (1) | No (2)  

niraj   10 years AGO

9*9*9*9*9+7380/9=7381

Like? Yes (1) | No (2)  

Abhishek Vishwakarma   10 years AGO

LET NON ZERO NUMBER =X

X+X^2+X^3+X^4=7380

BY SIMPLFY

X=9

SO A/C TO GIVEN QUESTION "If five power of the number is added to 7380 and divided by the number, then quotient will be?"

SO

(X^5+7380)/X = ((9^5+7380)/9)
=7381
SO QUOTIENT = 7381

Like? Yes (1) | No (2)  

Sanjay Bhatt   10 years AGO

Let the Number be X.
X+X2+X3+X4=7380
(X+X2+X3+X4+X5)/X= 1+X+X2+X3+X4=7381

Like? Yes (1) | No (2)  

Sagar Kharab   10 years AGO

x+x^2+x^3+x^4=7380
now x+x^2+x^3+x^4+x^5=7380 + x^5
x(1+x+x^2+x^3+x^4) / x = 1+x+x^2+x^3+x^4= 1 + 7380

Like? Yes (1) | No (2)  

Y V S Chandrakanth   10 years AGO

x+x2+x3+x4=7380
(7380+x5)/x=(x+x2+x3+x4+x5)/x=1+x+x2+x3+x4=1+7380=7381

Like? Yes (1) | No (2)  

Gagan Goel   10 years AGO

9 + 81 +729+6561=7380
9^5 +7380=66429
66429 / 9=7381

Like? Yes (1) | No (2)  

Rajnikant   10 years AGO

let's assume no.=x
So from ques,
x+x²+x³+x⁴=7380 .......eq. 1

Adding x⁵ with 7380 and dividing it by the no.
= (7380+x⁵)/x
= (x+x²+x³+x⁴+x⁵)/x ......from eq. 1
= x(1+x+x²+x³+x⁴)/x
= 1+x+x²+x³+x⁴
= 1+7380 ......from eq. 1
= 7381

Like? Yes (1) | No (2)  

jagadeesh   10 years AGO

x+X^2+x^3+X^4 = 7380

(X^5+ 7380)/X = X(1+7380)/X = 7381

Like? Yes (1) | No (2)  

Abhishek mani   10 years AGO

9+81+729+6561=7380 (so x=9)
again, (7380+59049)/9=7381
so ans(4)

Like? Yes (2) | No (3)  

NAVEEN KUMAR   10 years AGO

9+9^2+9^3+9^4=7380
9^5+7380=66429
66429/9=7381

Like? Yes (1) | No (2)  

Sahil Nagpal   10 years AGO

let the no be x
so by statement of question x+x^2+x^3+x^4=7380

next statement is (x^5+7380)/x=we have to find

so put the value of 7380 from first in second
therefore, (x^5+x+x^2+x^3+x^4)/x=x^4+1+x+x^2+x^3

1+7380=7381, answer

Like? Yes (1) | No (2)  

ayushi   10 years AGO

By inspection the no. shud be

Like? Yes (1) | No (2)  

pushpa   10 years AGO

x+ x^2 +x^3+ x^4 = 7380

(7380+ x^5)/ x
= (x+ x^2 +x^3+ x^4 + x^5)/ x
=(1+x+ x^2 +x^3+ x^4)
=7381

Like? Yes (1) | No (2)  

prem   10 years AGO

let the no =a
then
a+a2+a3+a4=7381
from GP cf=a
hence by sum of gp
a=9
hence (7380+(9)pow5)/9=7381

Like? Yes (2) | No (3)  

BHASKAR MANDAL   10 years AGO

x+x^2 + x^3 +x^4 +X^5 / x = 7380 + 1 = 7381
option 4

Like? Yes (2) | No (4)  

Sanchit Kapoor   10 years AGO

x+x^2+x^3+x^4=7380
Solving x=9
Now,(9^5+7380)/9=7381

Like? Yes (2) | No (4)  

RAKESH   10 years AGO

x + x^2 + x^3 + x^4 + x^5 / x

= 1 + (x + x^2 + x^3 + x^4)
= 1 + 7380
= 7381

Like? Yes (2) | No (4)  

Surit Datta   10 years AGO

9 square + 9 cube + 9 four power + 9 = 7380
9 five power + 7380 = 66429
66429 / 9 = 7381
That's it.

Like? Yes (1) | No (3)  

Md.Sultan Khichi   10 years AGO

Let a is the no.
a+a^2+a^3+a^4=7380
Now (a+a^2+a^3+a^4+a^5)/a
=1+a+a^2+a^3+a^4=1+7380=7381
Ans ob 4

Like? Yes (2) | No (4)  

Supriya Mondal   10 years AGO

Let the non zero number is a.
So according to the question a+a^2+a^3+a^4=7380
And we have to divide a^5+7380 by a.
So, (a^5+7380)/a
=(a^5+a+a^2+a^3+a^4)/a
=(1+a+a^2+a^3+a^4)
=(1+7380)
=7381
So answer will be 4)7381

Like? Yes (1) | No (3)  

Abhishek Porwal   10 years AGO

Consider the non zero digits i.e. from 9 to 1
Consider 9 and here is your answer 7381

Like? Yes (1) | No (3)  

DIPAK TIWARI   10 years AGO

(9)+ (9)^2 +(9)^3 +(9)^4 =7380
Hence ,{(9)^5 + 7380 }/9 =7381
Quotient=7381
Option 4

Like? Yes (1) | No (3)  

ravi prakash   10 years AGO

9+9^2+9^3+9^4+9^5=7380+59049
=66429/9
=7381

Like? Yes (1) | No (3)  

Swagatam Kumar Das   10 years AGO

Let the no is x
So x(1+x+x^2+x^3)=7380
so (1+x+x^2+x^3)=7380/x
and the quient will be (7380+x^5)/x
=x^4+7380/x
=x^4+(1+x+x^2+x^3)
=1+7380 =7381

Like? Yes (1) | No (3)  

Tark Patel   10 years AGO

Let number be x

by solving x + x^2 + x^3 + x^4 = 7380 we get x = 9

Then ( 9^5 + 7380 ) / 9 we get 7381.

Like? Yes (1) | No (3)  

VIKRAM KUMAR MAHTO   10 years AGO

7380 a bit greater than 9^4 and almost double of 8^4 . So the single digit number is 9.
Now, 9+91+81+729+6561=7380
Now dividing 9+91+81+729+6561+59049 will give quotient as 9^0+9^1+9^2+9^3+9^4=7380 + 1 =7381

Like? Yes (1) | No (3)  

pragya   10 years AGO

Ans is 7381

x+ x^2 + x^3 + x^4 = 7380 ------(1)

((x^5)+7380)/x=quotient ---------(2)

i tried hit & trial method fr this ques

putting x=10, we get

x^4=10000 which is greater than 7380. So, x must be less than 10.

put x=9,

(9^4)=6561 which is less than 7380
9+(9^2)+(9^3)+(9^4)=7380
hence, x=9

now, put x=9 in eqn (2), we get

quotient=[ (9^5) + 7380 ] / 9 = 7381

Like? Yes (1) | No (3)  

Anoop Tiwari   10 years AGO

Number is 9, because
9+9^2+9^3+9^4=7380
and 9^5=59049
So 9^5+7380=66429
66429/9=7381

Like? Yes (1) | No (3)  

Anshul Shrivastava   10 years AGO

Given x+x^2+x^3+x^4=7380,
now,
(x+x^2+x^3+x^4+x^5)/x
=>1+x+x^2+x^3+x^4....... (eqn 1)
now from the que.
x+x^2+x^3+x^4=7380 putting in eqn 1
we get,
=1+7380
=7381 ans.

Like? Yes (1) | No (3)  

A SIRISHA    10 years AGO

LET NON ZERO NUMBER =X

X+X^2+X^3+X^4=7380

BY SIMPLFY

X=9

SO A/C TO GIVEN QUESTION "If five power of the number is added to 7380 and divided by the number, then quotient will be?"

SO

(X^5+7380)/X = ((9^5+7380)/9)
=7381
SO QUOTIENT = 7381

Like? Yes (1) | No (3)  

Ajinkya Kulkarni   10 years AGO

(x+x2+x3+x4+x5)/x=1+x+x2+x3+x4=1+7380=7381

Like? Yes (1) | No (3)  

shubham   10 years AGO

n+n2+n3+n4 = 7380 (given) nw add n5 to it and divide that by n, we wil hv (n+n2+n3+n4+n5)/n = (1+n+n2+n3+n4) = 7380+1 = 7381

Like? Yes (1) | No (3)  

RAMAKRISHNA   10 years AGO

the number is 9
9+9^2+9^3+9^4+9^5=66429
66429/9=7381

Like? Yes (1) | No (3)  

Parag   10 years AGO

i just found out the closest x^4 to 7380
i tried x=10 so x^4=10000 so that means x

Like? Yes (1) | No (3)  

bhavna singh   10 years AGO

when the number is divided, one time power gets cancelled and the number is incremented once

Like? Yes (1) | No (4)  

Rahul   10 years AGO

Suppose the number is x

Then x = 1 + 7380 = 7381

Like? Yes (1) | No (5)  

Ranjit Somisetty   10 years AGO

x*(1+x+x2+x3+x4)/x= 1+7380

Like? Yes (1) | No (6)  

20 Nov 2014 Number of Balls

Balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second of two balls, the third of 3 balls and so on. If 456 more balls are added then all the balls can be arranged in the shape of a square and each of its sides then contains 8 balls less than the balls in each side of the triangle. Determine the initial number of balls.

OPtion

1) 913
2) 1144
3) 1065
4) 988
5) 1225
6) 840
7) 769
8) 700
9) 633
10) 568

Solution

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) + (n-1) + n (1)
or
x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)
adding (1) and (2)
2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)
or 2x = n.(n+1)
or x =n(n+1)/2
According to given condition n(n+1)/2 + 456 = (n-8)2
On solving or checking the options we get n = 49
Thus x = 49*50/2 = 1225

Option 5)

Correct Option: 5 Best Solution (26)

Shreya Tiwari   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (25) | No (8)  

Kushagra Tiwari   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (19) | No (7)  

Dr.Dipin Singh   10 years AGO

If there are n rows in triangle of balls, then
n(n+1)/2 + 456 = (n-8)^2
solving it, we get n=49
No of initial balls = 49*50/2=1225

Like? Yes (6) | No (3)  

CHANDA VIJAYALAKSHMI   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (5) | No (3)  

CHANDAN KUMAR   10 years AGO

let side=n
no of balls=n/2(n+1)
n/2(n+1)+456=(n-8)^2
n=49
no of ballls=n/2(n+1)=49x25=1225

Like? Yes (1) | No   

purushothama   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (5) | No (4)  

Jayanta Dutta   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (7) | No (7)  

MAHARAJAN.G   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (4) | No (4)  

santosh kumar patel   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (5) | No (5)  

Ayushi Gupta   10 years AGO

as 1st row has 1 ball, second has 2 so, nth row will have n balls
total initial ball = (1+2+3+4..n) = n(n+1)/2
side of square = n-8 (n = side of a triangle)
area of square = (n-8)^2
so, n(n+1)/2 + 456 = (n-8)^2
n = 49
total initial balls = (49)(50)/2 = 1225

Like? Yes (3) | No (3)  

lakshmi   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (4) | No (4)  

jayalakshmi   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (4) | No (4)  

Anoop Tiwari   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (3) | No (4)  

P.Pandeeswari   10 years AGO

initial number of balls= (49-8)^2-456 = 41^2-456 = 1225.

Like? Yes (3) | No (4)  

Anand   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (4) | No (5)  

ravi prakash   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (4) | No (5)  

bhagya   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (3) | No (4)  

RAKESH   10 years AGO

equate no. of balls
n(n+1)/2 + 456 = (n-8)^2

=> n^2 -33n -784 = 0
=> (n-49)(n+16) = 0
=> n = 49

initially n*(n+1)/2 = 49*25 = 1225 balls

Like? Yes (4) | No (6)  

BHASKAR MANDAL   10 years AGO

let there are n balls on each side of the triangle then we can write
n(n+1)/2 + 456 = (n-8)^2
n^2 + n + 912 = 2n^2 - 32 n + 128
=> n^2 -33 n - 784 = 0
=>n^2 - 49n + 16 n -784 = 0
=> n = 49.....we will take the positive value
to total number of balls = 49*50 / 2 = 49 * 25 = 1225
so the option is 5

Like? Yes (4) | No (6)  

Sundar Kumar   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (2) | No (4)  

Tark Patel   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (2) | No (4)  

Chirag Ranjan Lahiri   10 years AGO

1+2+3+...+49=1225
1225+456=1681
sqrt of 1681 is 41
41 is 8 less than 49

so initial number of balls are 1225.

Like? Yes (2) | No (4)  

Abhishek Vishwakarma   10 years AGO

Let number of balls in a side of the triangle are n
Thus total number of balls are
x = 1 + 2 + 3 +........+(n-2) +(n-1) + n (1)

or x = n + (n-1) + (n-2) +.......+ 3 + 2 + 1 (2)

adding (1) and (2)

2x =(n+1) +(n+1) + (n+1)+.......+(n+1) +(n+1)+(n+1)

or 2x = n.(n+1)

or x =n(n+1)/2

According to given condition n(n+1)/2 + 456 = (n-8)2

On solving or checking the options we get n = 49

Thus x = 49*50/2 = 1225

Like? Yes (2) | No (4)  

Sourabh Prasad   10 years AGO

n(n+1)/2 + 456 = (n-8)2

Like? Yes (2) | No (4)  

Parag   10 years AGO

Number of balls in a triangle of length n = n(n+1)/2
and number of balls in square of length n = (n)*(n)
now from the given information we can say
number of balls in triangle of (n-1) length + 456 = number of balls in square of length [n-8]

that is n(n+1)/2 + 456 = (n-8)*(n-8)

u get a final quadratic eqn n^2 -33n - 784=0
finally n=49
this the length of one side of triangle.
total number of balls = 49*50/2=1225

Like? Yes (2) | No (4)  

Surit Datta   10 years AGO

let each side of the equilateral triangle be n

n(n+1)/2+456 = (n-8)2

by solving this

n = 49

n(n+1)/2 = 49*50/2

=49*25

=1225

hence initial number of balls = 1225

Like? Yes (1) | No (5)