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In an examination where full marks were 800, A gets 20% more than B, B gets 20% more than C and C gets 15% less than D. If A got 576, what percentage of full marks did D get approximately?
$
  (a){\text{ 45}}{\text{.7% }} \\
  (b){\text{ 51}}{\text{.2% }} \\
  (c){\text{ 58}}{\text{.8% }} \\
  (d){\text{ 61}}{\text{.7% }} \\
 $

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Last updated date: 29th Mar 2024
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Views today: 8.14k
MVSAT 2024
Answer
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Hint: The relation between the marks of A, B, C and D are given in the question. Start with formulating equations to get numerical relation between them as the overall marks of A is given. Use this concept to get D’s percentage.

Complete Step-by-Step solution:
Total marks were = 800.
Now it is given that A got 576 marks.
Now again it is given that A gets 20% more than B.
So construct the linear equation according to this information we have,
Therefore marks of A = B + 20%B
$ \Rightarrow A = B + \dfrac{{20}}{{100}}B = 1.2B$
$ \Rightarrow B = \dfrac{A}{{1.2}} = \dfrac{{576}}{{1.2}} = 480$
Now again it is given that B gets 20% more than C.
So construct the linear equation according to this information we have,
Therefore marks of B = C + 20%C
$ \Rightarrow B = C + \dfrac{{20}}{{100}}C = 1.2C$
$ \Rightarrow C = \dfrac{B}{{1.2}} = \dfrac{{480}}{{1.2}} = 400$
Now again it is given that C gets 15% less than B.
So construct the linear equation according to this information we have,
Therefore marks of C = D - 15%B
$ \Rightarrow C = D - \dfrac{{15}}{{100}}D = 0.85D$
$ \Rightarrow D = \dfrac{C}{{0.85}} = \dfrac{{400}}{{0.85}}$
So the percentage of marks D gets is the ratio of marks D got to the total marks multiplied by 100.
$ \Rightarrow D = \dfrac{{\dfrac{{400}}{{0.85}}}}{{800}} \times 100 = \dfrac{{100}}{{2 \times 0.85}} = \dfrac{{100}}{{1.7}} = 58.8$ %
So this is the required percentage of full marks D get approximately.
Hence option (C) is correct.

Note: Whenever we face such types of problems the key point is that although marks of a candidate are given but however the data asked is in terms of percentage. So use the basic concept of percentage to get everything in percentage during formulation of equations.
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