Percentages
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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

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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 $\textbf{42 and 33}$.


(3) Comment(s)


Syed
 ()

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
 ()

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
 ()

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