A, B, C and D were the members of a team. The average runs

Question

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)

SaquibHGMATWhiz · Accepted Answer

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

Ahamed Imtiyaz · Answer

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

Shwetabh54 · Answer

Lets the average of A,B,C & D is x. Then, sum of runs (A,B,C & D) = 4x

After joining E, average drops by 2. So, the new average of (A,B,C,D & E) = (x-2)
Sum of runs (A,B,C,D & E) = 5*(x-2)

E scored 45 runs.
=> Sum of runs (A,B,C,D & E) - sum of runs (A,B,C & D) = 5*(x-2) - 4x = 45
=> x = 55

Now, it is mentioned that A,B,C & D runs are not more than 65 and not less than 45 (E's score)
=> 45 < A, B, C, D < 65

A:B = 13:12, what are the possible pairs for A & B - 13:12, 26:24, 39:36, 52:48, 65:60 and so on

We know that A & B should be in the range of 45 to 60. Hence only possible value of A & B is 52 & 48 respectively.

We know, A+B+C+D = 4x = 220
=> C+D = 220 - (A+B) = 220 - (52 + 48) = 120
Average of C & D = (120/2) = 60

B : Average of C & D = 48 : 60 = 4:5

pappal · Answer

A+B+C+D=4x---------i let average of these 4 natural numbers is x
A+B+C+D+E=5(x-2)=5x-10---------ii
from i and ii E=5x-10=45 given
so x=55 or A+B+C+D=220
all the values A,B,C,D are greater than E but less than 65.
also given A:B=13:12 which come out to be 52 and 48; or A+B=100
so C+D=220-100=120
now we have to find the value of B/{(C+D)/2}=2B/C+D=2*48/120=4:5
so A

A, B, C and D were the members of a team. The average runs

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