Aptitude - Simplification - Discussion
Discussion Forum : Simplification - General Questions (Q.No. 6)
6.
A sum of Rs. 1360 has been divided among A, B and C such that A gets 
of what B gets and B gets 
of what C gets. B's share is:
of what B gets and B gets of what C gets. B's share is:
Answer: Option
Explanation:
Let C's share = Rs. x
| Then, B's share = Rs. | x | , A's share = Rs. |  |
2 | x | x |  |
= Rs. | x |
| 4 | 3 | 4 | 6 |
 |
x | + | x | + x = 1360 |
| 6 | 4 |
 |
17x | = 1360 |
| 12 |
| x = |
1360 x 12 | = Rs. 960 |
| 17 |
| Hence, B's share = Rs. | |
960 | |
= Rs. 240. |
| 4 |
Discussion:
34 comments Page 4 of 4.
MANOJ KUMAR said:
5 years ago
2 + 3 + 12 = 17,
Now we need value of B,
So, 3/17 * 1360=240.
Now we need value of B,
So, 3/17 * 1360=240.
(1)
Kaushlendra said:
5 years ago
a + b + c = 1360 ---> (1)
a=2/3(b), b=(1/4)(c)
Therefore, c = 4b,
using 1,
b=240 ans.
a=2/3(b), b=(1/4)(c)
Therefore, c = 4b,
using 1,
b=240 ans.
(3)
K.sumana said:
4 years ago
Given x + y + z = 1360.
x = 2/3(y)],
y = 1/4(z),
z = 4y.
Substitute x value and z value in eqn;
we get,
2/3(y) + y + 4y = 1360,
2/3(y) + 5y = 1360,
17/3(y) = 1360,
y = 240.
x = 2/3(y)],
y = 1/4(z),
z = 4y.
Substitute x value and z value in eqn;
we get,
2/3(y) + y + 4y = 1360,
2/3(y) + 5y = 1360,
17/3(y) = 1360,
y = 240.
(8)
Chandrika said:
4 years ago
A+B+C=1360-->1.
We don't know the c value.
Let us assume C=x.
A=2/3*B.
B=(1/4) *C.
=(1/4)*x=>x/4.
A=2/3*x/4=>2x/12=>x/6.
From 1,A+B+C=1360.
x/6+x/4+x=1360=>(2x+3x+12x)/12=1360.
17x=1360*12=>x=(1360*12)/17=>960,
B=x/4=>960/4=>Rs.240.
We don't know the c value.
Let us assume C=x.
A=2/3*B.
B=(1/4) *C.
=(1/4)*x=>x/4.
A=2/3*x/4=>2x/12=>x/6.
From 1,A+B+C=1360.
x/6+x/4+x=1360=>(2x+3x+12x)/12=1360.
17x=1360*12=>x=(1360*12)/17=>960,
B=x/4=>960/4=>Rs.240.
(4)
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