Question
In a cricket match, five batsmen A, B, C, D and E scored an average of 36 runs. D Scored 5 more than E; E scored 8 fewer than A; B scored as many as D and E combined; and B and C scored 107 between them. How many runs did E score ?
Answer: Option D
Total runs scored = (36 x 5) = 180.
Let the runs scored by E be x.
Then, runs scored by D = x + 5; runs scored by A = x + 8;
runs scored by B = x + x + 5 = 2x + 5;
runs scored by C = (107 - B) = 107 - (2x + 5) = 102 - 2x.
So, total runs = (x + 8) + (2x + 5) + (102 - 2x) + (x + 5) + x = 3x + 120.
Therefore 3x + 120 =180 ⇔ 3X = 60 ⇔ x = 20.
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Total runs scored = (36 x 5) = 180.
Let the runs scored by E be x.
Then, runs scored by D = x + 5; runs scored by A = x + 8;
runs scored by B = x + x + 5 = 2x + 5;
runs scored by C = (107 - B) = 107 - (2x + 5) = 102 - 2x.
So, total runs = (x + 8) + (2x + 5) + (102 - 2x) + (x + 5) + x = 3x + 120.
Therefore 3x + 120 =180 ⇔ 3X = 60 ⇔ x = 20.
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