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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
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A, B, C and D were the members of a team. The average runs of the team...
Let the average runs of the team before adding E be x.
Total runs scored by A and B = 13x + 12x = 25x
Total runs scored by C and D = 2x (since average runs decreases by 2 when E is added)
Total runs scored by the team before adding E = 25x + 2x = 27x
After adding E, the average runs of the team decreases by 2. So, the new average runs = x - 2.
Total runs scored by the team after adding E = (x - 2) * 5
Since E scored 45 runs, we can write the equation:
(x - 2) * 5 = 45
x - 2 = 9
x = 11
So, the average runs before adding E = 11.
Total runs scored by the team before adding E = 27x = 27 * 11 = 297
Now, let's find the individual runs scored by each player.
Since no player scored less than E or more than 65 runs, we can conclude that A scored 65 runs and B scored 65 - 13 = 52 runs.
Let's assume that C scored y runs. Since C scored more than A, y > 65. Also, since y is a natural number, the minimum value of y is 66.
Total runs scored by the team = 65 + 52 + y + 2x = 297
117 + y + 22 = 297
y = 158
So, C scored 158 runs and D scored 297 - 65 - 52 - 158 = 22 runs.
The ratio of the runs scored by B to the average runs scored by C & D = 52 : ((158 + 22)/2) = 52 : 90 = 4 : 5
Therefore, the correct answer is A: 4:5.
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Most Upvoted Answer
A, B, C and D were the members of a team. The average runs of the team...
Understanding the Problem
To solve this problem, we need to analyze the impact of adding member E on the team's average runs and the relationships between the scores of A, B, C, and D.
Initial Conditions
- Let the total runs of A, B, C, and D be T.
- The average runs of the team is given by \( \text{Average} = \frac{T}{4} \).
- When E is added, the new average becomes \( \frac{T + 45}{5} \).
- Given that the new average is 2 runs less, we have:
\[
\frac{T + 45}{5} = \frac{T}{4} - 2
\]
Solving the Equation
1. Clear the fractions by multiplying through by 20:
\[
4(T + 45) = 5T - 40
\]
2. Expanding and simplifying gives:
\[
4T + 180 = 5T - 40 \implies T = 220
\]
Average Runs Calculation
- The average runs of A, B, C, and D is:
\[
\text{Average} = \frac{220}{4} = 55
\]
Scores of A, B, C, and D
- Let \( A = 13x \) and \( B = 12x \) (from the ratio).
- Then, \( C + D = 220 - (13x + 12x) = 220 - 25x \).
- Since \( C > A \) implies \( C > 13x \).
Constraints on Scores
- Each player scores between E (45) and 65.
- Thus, \( 13x < c="" />< 65="" />
Finding Values of x
- If \( 13x < 65="" \),="" then="" \(="" x="" />< 5="" />
- Possible values of \( x \) are 1, 2, 3, 4.
- Testing \( x = 4 \):
- \( A = 52 \), \( B = 48 \), \( C \) can be 53.
- Then \( D = 220 - 52 - 48 - 53 = 67 \) (not valid).
- Testing \( x = 3 \):
- \( A = 39 \), \( B = 36 \), \( C \) can be 54.
- Then \( D = 220 - 39 - 36 - 54 = 91 \) (not valid).
- Testing \( x = 2 \):
- \( A = 26 \), \( B = 24 \) and \( C \) can be 55.
- Then \( D = 220 - 26 - 24 - 55 = 115 \) (not valid).
Final Ratio Calculation
- Valid scores found with \( x = 1 \):
- \( A = 13 \), \( B = 12 \), \( C = 54 \), \( D = 141 \).
- Average of \( C \) and \( D
Free Test
Community Answer
A, B, C and D were the members of a team. The average runs of the team...
Let the average runs of the team before adding E be x.
Total runs scored by A and B = 13x + 12x = 25x
Total runs scored by C and D = 2x (since average runs decreases by 2 when E is added)
Total runs scored by the team before adding E = 25x + 2x = 27x
After adding E, the average runs of the team decreases by 2. So, the new average runs = x - 2.
Total runs scored by the team after adding E = (x - 2) * 5
Since E scored 45 runs, we can write the equation:
(x - 2) * 5 = 45
x - 2 = 9
x = 11
So, the average runs before adding E = 11.
Total runs scored by the team before adding E = 27x = 27 * 11 = 297
Now, let's find the individual runs scored by each player.
Since no player scored less than E or more than 65 runs, we can conclude that A scored 65 runs and B scored 65 - 13 = 52 runs.
Let's assume that C scored y runs. Since C scored more than A, y > 65. Also, since y is a natural number, the minimum value of y is 66.
Total runs scored by the team = 65 + 52 + y + 2x = 297
117 + y + 22 = 297
y = 158
So, C scored 158 runs and D scored 297 - 65 - 52 - 158 = 22 runs.
The ratio of the runs scored by B to the average runs scored by C & D = 52 : ((158 + 22)/2) = 52 : 90 = 4 : 5
Therefore, the correct answer is A: 4:5.
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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)4:5b)5:7c)7:9d)8:9e)16:19Correct answer is option 'A'. Can you explain this answer?
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