Answer
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Hint: Assume that the total marks were x. Calculate what $5\%$ of x is and equate it to $35\%$ of x-216 since Rishu failed by $5\%$ marks. This will give a linear equation in x. Solve for x. This will provide you with the aggregate marks of the examination.
Complete step-by-step answer:
Let us assume that the total marks were x.
Total passing marks = $35\%$ of x = 0.35x
Marks obtained by Rishu = 216.
Hence Rishu failed the examination by 0.35x-216 marks.
Hence Rishu failed the examination by $\dfrac{0.35x-216}{x}\times 100\%$ marks.
But given that Rishu failed the examination by $5\%$ marks.
So we have
$\begin{align}
& \dfrac{0.35x-216}{x}\times 100\%=5\% \\
& \Rightarrow \dfrac{0.35x-216}{x}\times 100=5 \\
\end{align}$
Multiplying both sides by x we get
$\left( 0.35x-216 \right)\times 100=5x$
Using distributive property of multiplication over addition and subtraction, i.e. a(b+c) = ab+ac and a(b-c) = ab -ac , we get
$35x-21600=5x$
Subtracting 5x from both sides we get
35x-5x-21600=5x-5x
i.e. 30x-21600 =0
Adding 21600 to both sides we get
30x-21600 = 21600+0
i.e. 30x=21600
Dividing both sides by 30, we get
x= 720
Hence the total marks of the examination were 720.
Hence option [b] is correct.
Note:Alternatively we have
Since Rishu failed by $5\%$marks
216 = $\left( 35-5 \right)\%$ of marks
Hence Total marks $=\dfrac{216\times 100}{30}=720$
which is the same as obtained above.
Complete step-by-step answer:
Let us assume that the total marks were x.
Total passing marks = $35\%$ of x = 0.35x
Marks obtained by Rishu = 216.
Hence Rishu failed the examination by 0.35x-216 marks.
Hence Rishu failed the examination by $\dfrac{0.35x-216}{x}\times 100\%$ marks.
But given that Rishu failed the examination by $5\%$ marks.
So we have
$\begin{align}
& \dfrac{0.35x-216}{x}\times 100\%=5\% \\
& \Rightarrow \dfrac{0.35x-216}{x}\times 100=5 \\
\end{align}$
Multiplying both sides by x we get
$\left( 0.35x-216 \right)\times 100=5x$
Using distributive property of multiplication over addition and subtraction, i.e. a(b+c) = ab+ac and a(b-c) = ab -ac , we get
$35x-21600=5x$
Subtracting 5x from both sides we get
35x-5x-21600=5x-5x
i.e. 30x-21600 =0
Adding 21600 to both sides we get
30x-21600 = 21600+0
i.e. 30x=21600
Dividing both sides by 30, we get
x= 720
Hence the total marks of the examination were 720.
Hence option [b] is correct.
Note:Alternatively we have
Since Rishu failed by $5\%$marks
216 = $\left( 35-5 \right)\%$ of marks
Hence Total marks $=\dfrac{216\times 100}{30}=720$
which is the same as obtained above.
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