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. Find the marks obtained by them.

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Hint: Here we will solve this question by assuming the marks of two students and one has more 9 marks than the others. Then we will get the equation with one variable and after that we will combine the marks of both of the students and will consider it as 100%.

Step-1
Let us consider the marks of the both of the students as x and y respectively.
According to question one has 9 marks more than the others.
Let’s x= y+9………………(1)
Step-2
Also according to the question x is 56% of the sum of the x and y
i.e. $x = \dfrac{{56}}{{100}}(x + y)$
Step-3
Multiplying 100 on both side we get,
100x=56x+56y
Step-4
Taking 56x to the left hand side we get,
100x-56x=56y
Or, 44x=56y
Or, $x = \dfrac{{56}}{{44}}y$
Step-5
Cancelling numerator and denominator by 4,
Or, $x = \dfrac{{14}}{{11}}y$…………………..(2)
Step-6
Equating equation (1) and (2) we get,
$y + 9 = \dfrac{{14}}{{11}}y$
Step-7
Multiplying 11 on both of the side we get,
11(y+9) = 14y
Step-8
Upon simplifying the equation we get,
11y + 99 =14y
Step-9
Step-10
Putting terms containing y on the same side we get,
14y – 11y = 99
Or, 3y = 99
Or, y = 99/3
Or, y = 33
Step-11
Putting the value of y in equation (1) we get,
x = y + 9
Or, x = 33 + 9
Or, x = 42
Step-12
Hence, two of the students secured 42 and 33 marks in the examination respectively.

Note: We can check our answer with the below process
Checking error-
x = y + 9
42= 33+9 (checked)
$x = \dfrac{{56}}{{100}}(x + y)$
$\dfrac{{56}}{{100}}(33 + 42)$
= 4200/100
= 42 (checked)