Two students appeared an examination- One of them secured 9 marks than the other and his marks was 56- of the of their marks- The marks obtained by them are-41-3239-3042-3343-34

Question

Toppr Admin · Accepted Answer

Let their marks be -x-9- and xThen- x-9-56100-x-9-x-x21D2- 25-x-9-14-2x-9-x21D2- 3x-99-x21D2- x-33So- their marks are 42 and 33

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:
$41, 32$
$39, 30$
$42, 33$
$43, 34$

Let their marks be $(x + 9)$ and $x$
Then, $x + 9 = \frac{56}{100} (x + 9 + x)$
$\Rightarrow$ $25 (x + 9) = 14 (2 x + 9)$
$\Rightarrow$ $3 x = 99$
$\Rightarrow$ $x = 33$
So, their marks are $42$ and $33$

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:41,3239,3042,3343,34

Let their marks be (x+9) and xThen, x+9=56100(x+9+x)⇒ 25(x+9)=14(2x+9)⇒ 3x=99⇒ x=33So, their marks are 42 and 33

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:
$41, 32$
$39, 30$
$42, 33$
$43, 34$

Let their marks be $(x + 9)$ and $x$
Then, $x + 9 = \frac{56}{100} (x + 9 + x)$
$\Rightarrow$ $25 (x + 9) = 14 (2 x + 9)$
$\Rightarrow$ $3 x = 99$
$\Rightarrow$ $x = 33$
So, their marks are $42$ and $33$