Question

Five of the batsmen of a team (A, B,
C, D and E) scored at an average of
36 runs in a cricket match. While B
and C together scored 107 runs, B's
score was equal to the scores of D
and E combined. If D's score was 5
more than E and E's score is 8 less
than A, then how much did C score?​

Answer 1

Step-by-step explanation:

Total runs scored =(36×5)=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.

Answer 2

Given:-

A+B+C+D+E /5 = 36.

A+B+C+D+E =180 ——-> 1

D = 5+E ——-> 2

A = 8+E. ——->3

B = D+E.-——- >4

B+C = 107.——->5

Solution:-

Substituting 5 in 1 we get, A+B+E =73. ——>6

Substitute 2 in 6 we get, A + 5 + 2E = 73 ——> 7

Substitute 3 in 7 and solving it, we get 3E = 60

Solving 3E = 60. we get

E = 20.