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Cricket/Marks Average Problems

Introduction

In aptitude tests, cricket/marks average problems are very common. These problems usually involve finding a missing score, calculating total runs/marks, or determining a player’s or student’s performance over several matches/tests.

The trick is to apply the average formula correctly: Average = Total ÷ Number of terms.

Pattern: Cricket/Marks Average Problems

Pattern

The key idea is to calculate the total from the average or vice versa, then use it to find the missing data.

Steps:
1. Write the formula: Total = Average × Number of terms.
2. Subtract known scores/runs from total to find the missing one.
3. Or add new scores to update the average.

Step-by-Step Example

Question

A cricketer has scored an average of 50 runs in 9 innings. How many runs must he score in the 10th inning to raise his average to 55?

Options:
A. 80 runs
B. 90 runs
C. 100 runs
D. 110 runs

Solution

  1. Step 1: Compute current total runs

    Runs in 9 innings = 9 × 50 = 450.

  2. Step 2: Compute required total for new average

    For 10 innings with average 55 → total = 10 × 55 = 550.

  3. Step 3: Find runs required in the 10th inning

    Runs required in 10th inning = 550 - 450 = 100.

  4. Final Answer:

    100 runs → Option C

  5. Quick Check:

    (450 + 100) ÷ 10 = 550 ÷ 10 = 55 ✅

Quick Variations

Marks Problems: - If the average of 5 subjects is 60, total = 60 × 5 = 300. - If marks in 4 subjects are known, subtract from total to get the 5th.

Updating Averages: - If a new match/subject is added, recalculate the total and divide by the new count.

Trick to Always Use

  • Step 1: Multiply average by count to get total.
  • Step 2: Subtract/add known values.
  • Step 3: Divide to get updated average.

Summary

Summary

In Cricket/Marks Average Problems, always start with Total = Average × Number of terms. Use this to calculate missing runs, marks, or new averages quickly.

  • Convert average → total when number of terms is known.
  • Find missing score by subtracting known totals.
  • Update average when new entry is added.

Practice

(1/5)
1. A student scores an average of 60 marks in 4 subjects. If his marks in three subjects are 58, 62, and 64, find his marks in the fourth subject.
easy
A. 56
B. 60
C. 64
D. 68

Solution

  1. Step 1: Compute total marks

    Total marks = 60 × 4 = 240.

  2. Step 2: Compute known marks total

    Sum of three subjects = 58 + 62 + 64 = 184.

  3. Step 3: Find missing marks

    Fourth subject = 240 - 184 = 56.

  4. Final Answer:

    56 → Option A

  5. Quick Check:

    (58 + 62 + 64 + 56) ÷ 4 = 240 ÷ 4 = 60 ✅
Hint: Multiply average by count to get total, then subtract known marks.
Common Mistakes: Forgetting to multiply average by the number of subjects before subtracting.
2. A batsman has scored 360 runs in 9 innings. What must he score in the 10th inning to increase his average by 5 runs?
medium
A. 85
B. 90
C. 95
D. 100

Solution

  1. Step 1: Compute current average

    Current average = 360 ÷ 9 = 40.

  2. Step 2: Compute new average

    Required average = 40 + 5 = 45.

  3. Step 3: Compute required total

    Total runs needed for 10 innings = 10 × 45 = 450.

  4. Step 4: Compute required 10th inning score

    Runs required in 10th inning = 450 - 360 = 90.

  5. Final Answer:

    90 → Option B

  6. Quick Check:

    (360 + 90) ÷ 10 = 450 ÷ 10 = 45 ✅
Hint: Compute required total from new average and subtract existing total.
Common Mistakes: Adding the increase directly to the last score instead of adjusting the total.
3. The average score of 8 students in an exam is 72. If one student’s score of 64 is replaced by 80, what is the new average?
medium
A. 73
B. 74
C. 75
D. 76

Solution

  1. Step 1: Compute original total

    Original total = 8 × 72 = 576.

  2. Step 2: Compute change in total

    Replacement increases total by 80 - 64 = 16.

  3. Step 3: Compute new total

    New total = 576 + 16 = 592.

  4. Step 4: Compute new average

    New average = 592 ÷ 8 = 74.

  5. Final Answer:

    74 → Option B

  6. Quick Check:

    Increase per student = 16 ÷ 8 = 2 → 72 + 2 = 74 ✅
Hint: Adjust the old total by the change, then divide by the same count.
Common Mistakes: Recomputing all scores instead of applying the net change method.
4. The average score of 10 innings of a batsman is 32. How many runs must he score in the next inning to raise the average to 35?
medium
A. 60
B. 62
C. 65
D. 68

Solution

  1. Step 1: Compute current total

    Current total = 10 × 32 = 320.

  2. Step 2: Compute required total

    Required total for 11 innings = 11 × 35 = 385.

  3. Step 3: Compute required runs

    Runs needed = 385 - 320 = 65.

  4. Final Answer:

    65 → Option C

  5. Quick Check:

    (320 + 65) ÷ 11 = 385 ÷ 11 = 35 ✅
Hint: Required runs = (new average × new count) - old total.
Common Mistakes: Forgetting to include the new inning in the count when calculating the new total.
5. A student has an average of 70 marks in 6 subjects. If he scores 88 in the seventh subject, what is his new average (to two decimal places)?
medium
A. 72
B. 72.5
C. 72.57
D. 73

Solution

  1. Step 1: Compute original total

    Original total = 6 × 70 = 420.

  2. Step 2: Compute new total

    New total = 420 + 88 = 508.

  3. Step 3: Compute new average

    New average = 508 ÷ 7 = 72.571428.... To two decimal places this is 72.57.

  4. Final Answer:

    72.571428... → Option C (72.57 to two decimal places).

  5. Quick Check:

    Increase = 88 - 70 = 18 → Increase per subject = 18 ÷ 7 ≈ 2.5714 → 70 + 2.5714 ≈ 72.5714 ✅
Hint: Add the new mark to the old total and divide by the new count; express to required decimal places.
Common Mistakes: Rounding too early or misplacing the divisor when computing the new average.

Mock Test

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