Aptitude Discussion

Q. |
Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are: |

✖ A. |
39, 30 |

✖ B. |
41, 32 |

✔ C. |
42, 33 |

✖ D. |
43, 34 |

**Solution:**

Option(**C**) is correct

Let their marks be $(x + 9)$ and $x$.

Then,

$x + 9 = \left(\dfrac{56}{100}\right) \times (x + 9 + x)$

$\Rightarrow 25(x + 9) = 14(2x + 9)$

$\Rightarrow 25x + 25 \times 9 = 28x + 14 \times 9$

$\Rightarrow 9 \times (25-14) = 28x -25x$

$\Rightarrow 9 \times 11 = 3x$

$\Rightarrow 3x = 99$

$\Rightarrow x = 33$

So, their marks are $\textbf{42 and 33}$.

**PULOK KHAN**

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**Syed**

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one more solution there could be

one number is 9 more than other so 56%-44%=12% which show 9

so we can move to 100% by =100/12*9 = 75 now you can calculate the 56% of 75 that is 42 and 44% that is 33.

**Anu**

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There is no need to write equations for such sums . Since 56% is given , we just to calculate total of each option and multiply by 56.

Only a total of 75 will give 56% as an integer hence directly option 3.

**Jagdeep Singh**

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Let total marks 100x

One student got - 56x............ (i)

secondt got - 44x..............(ii)

Now,

56x-44x =9 (given)

x = 9/12

put the value of x in eq. i and ii

we get 42 and 33

Let,

total mark=x

ATQ,

56x/100-44x/100=9

=> x=75

as their marks difference =9

so, one secured 42 and other secured 33.