| A certain sum is divided among P, Q and R in such a way that P gets Rs. 40 more than the\[\frac{1}{2}\]of the sum. Q gets Rs. 120 less than \[\frac{3}{8}\text{th}\] of the sum and R get Rs. 200. What is the total sum? [IBPS (PO/MT) Pre 2015] |
A) Rs. 960
B) Rs. 850
C) Rs. 906
D) Rs. 805
E) None of these
Correct Answer: A
Solution :
| Let the sum be x. |
| P gets \[=\frac{x}{2}+40\] |
| Q gets \[=\frac{3}{8}x-120\] |
| R gets \[=200\] |
| According to the question, |
| \[\frac{x}{2}+40+\frac{3}{8}x-120+200=x\] |
| \[\Rightarrow \] \[\frac{x}{2}+\frac{3}{8}x-x=-\,200+120-40\] |
| \[\Rightarrow \]\[\frac{4x+3x-8x}{8}=-\,240+120\] |
| \[\Rightarrow \]\[\frac{-x}{8}=-120\]\[\Rightarrow \]\[\frac{x}{8}=120\] |
| \[\Rightarrow \]\[x=8\times 120=960\] |
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