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A
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Difficulty:
75%
(hard)
Question Stats:
69% (03:49) correct
31%
(03:23)
wrong
based on 362
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A, B, C and D were the members of a team. The average runs of the team decreases by 2 if another member E is added. It is known that E scored 45 runs. No player scored less than E or more than 65 runs. If the runs scored by A and B are in the ratio 13:12 and C scored more than A, what will be the the ratio of the runs scored by B to the average runs scored by C & D ? (Assume that the runs scored by all the members is a natural number).
A. 4:5
B. 5:7
C. 7:9
D. 8:9
E. 16:19
(adpted from gmatfree)
A. 4:5
B. 5:7
C. 7:9
D. 8:9
E. 16:19
(adpted from gmatfree)
31 Aug 2022, 23:34
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Suraj0184
Solution:
Let us assume the average of A, B, C and D be M. Thus we can say \(\frac{A+B+C+D}{4}=M⇒A+B+C+D=4M.....(i)\)
We are said that this average decreased when E with a score of 45 joins, thus we can say \(\frac{A+B+C+D+E}{5}=\frac{A+B+C+D+45}{5}=M-2\)
\(⇒A+B+C+D+45=5M-10\)\(\)
\(⇒4M+45=5M-10\)
\(⇒M=55\)
\(⇒A+B+C+D=4M=220.....(ii)\)
We are told that A and B scored in a ratio \(13:12\). So, let us assume \(A=13x\) and \(B=12x\)
We also know that \(45<A<65\) or \(45<13x<65\) or \(3.4<x<5\) which means \(x= 4\)
and \(A=13x=52\) and \(B=12x=48\)
We are asked the ratio of B to average of C and D
Average of C and D \(= \frac{C+D}{2}=\frac{220-A-B}{2}\) (from (ii))
\(=\frac{220-52-48}{2}\)
\(=\frac{120}{2}\)
\(=60\)
Thus, the required ratio \(=\frac{B}{60}=\frac{48}{60}=\frac{4}{5}\)
Hence the right answer is Option A
31 Aug 2022, 16:10
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Let, A,B,C,D total run x.average x/4
According to question,
x/4-x+45/5= 2
X=220
A:B =13:12 =52:48( as every score is above 45 not more than 65, and C score is more Than a)
52+48+C+D=220
C+D=120
average of C+D=60
B=48, C+D average 60
48:60=4:5
Posted from my mobile device
According to question,
x/4-x+45/5= 2
X=220
A:B =13:12 =52:48( as every score is above 45 not more than 65, and C score is more Than a)
52+48+C+D=220
C+D=120
average of C+D=60
B=48, C+D average 60
48:60=4:5
Posted from my mobile device
