CRE 2 - Numbers Replaced/Added/Removed | Average, Mixture & Alligation

Answer: Option B

Explanation :

Mean of 100 observations = 50

Correction to be made = 40 – 50 = -10

New average, $\frac{\mathrm{50\; \times \; 100\; -\; 10}}{100}$ = 49.9

Alternately,

Since the overall sum of 100 numbers is reduced by 10 (= 50 - 40), the average reduction of each number will be 10/100 = 0.1

⇒ New average = old average – reduction = 50 – 0.1 = 49.9

Hence, option (b).

Answer: 150

Explanation :

Mean of 100 observations = 60

Correction to be made = x – 50

New average, $\frac{\mathrm{60\times 100\; +\; (x\; -\; 50)}}{100}$ = 61

⇒ 6000 + x – 50 = 6100

⇒ x = 150.

Alternately,

Since the average of 100 numbers increases by 1, the sum increases by 100 × 1 = 100.

Hence the number added is 100 more than the number removed

⇒ 100 = number added – number removed = x - 50

⇒ x = 150.

Hence, 150.

Answer: Option C

Explanation :

Mean of 100 observations = 60

Correction to be made = x – 80

New average, $\frac{\mathrm{60\times 100\; +\; (x\; -\; 80)}}{100}$ = 59.5

⇒ 6000 + x – 80 = 5950

⇒ x = 30.

Alternately,

Since the average of 100 numbers decreases by 0.5, the sum decreases by 100 × 0.5 = 50.

Hence the number added is 50 less than the number removed

⇒ 50 = number removed – number added = 80 – x.

⇒ x = 30.

Hence, option (c).

Answer: Option A

Explanation :

No. of runs scored in 7 innings is 7 × 12 = 84

If he scored x runs in the next innings, then, new average will be, (84 + x)/8 = 14 (given) 

or, 84 + x = 112

⇒ x = 28 runs.

Alternately,

Average after 7 innings is 12. If the batsman scores 12 runs in 8th innings, his final average will not change.

But, since his final average score increases by 2 runs after 8th innings, this means that he scores 8 × 2 = 16 runs more than earlier average in the 8th innings.

⇒ Score in 8th innings = earlier average + extra runs scored = 12 + 16 = 28 runs.

Hence, option (a).

Answer: Option A

Explanation :

Let x be the value of number removed.

Given, (6 × 12 - x)/5 = 13

⇒ 72 - x = 65

⇒ x = 7.

Alternately,

Had the number removed been same as average (i.e. 12), the average of remaining 5 numbers would not have changed.

Since average of now 5 numbers increases by 1 ⇒ the total increase for remaining numbers = 5 × 1 = 5.

This means that the number removed was 5 less than the old average.

⇒ x = 12 – 5 = 7

Hence, option (a).

Answer: Option D

Explanation :

Let x be the value of number removed.

Given, (6 × 12 - x)/5 = 11

⇒ 72 - x = 55

⇒ x = 17.

Alternately,

Had the number removed been same as average (i.e. 12), the average of remaining 5 numbers would not have changed.

Since average of now 5 numbers decreases by 1 ⇒ the total decrease for remaining numbers = 5 × 1 = 5.

This means that the number removed was 5 more than the old average.

⇒ x = 12 + 5 = 17.

Hence, option (d).

Answer: 19

Explanation :

Let x be the value of number added.

Given, (6 × 12 + x)/7 = 13

⇒ 72 + x = 91

⇒ x = 19.

Alternately,

Had the number added been same as average (i.e. 12), the average of now 7 numbers would not have changed.

Since average of now 7 numbers increases by 1 ⇒ the total increase in all numbers = 7 × 1 = 7.

This means that the number added was 7 more than the old average.

⇒ x = 12 + 7 = 19.

Hence, 19.

Answer: 5

Explanation :

Let x be the value of number added.

Given, (6 × 12 + x)/7 = 11

⇒ 72 + x = 77

⇒ x = 5.

Alternately,

Had the number added been same as average (i.e. 12), the average of now 7 numbers would not have changed.

Since average of now 7 numbers decreases by 1 ⇒ the total decrease of all numbers = 7 × 1 = 7.

This means that the number added was 7 less than the old average.

⇒ x = 12 - 7 = 5.

Hence, 5.

Answer: Option D

Explanation :

Total wages earned by 6 employees = 6 × 310

Let I be the increment awarded. Then, (6 × 310 + i)/6 = 315

⇒ i = 30

Alternately,

Since the average of same 6 people increase by (315 – 310 =) 5, the total increases by 5 × 6 = 30.

∴ increment received = Rs. 30.

Hence, option (d).

Answer: Option D

Explanation :

Aggregate marks (%) of 7 students = 7 × 42 = 294

With a new member in the group, whose marks is x%,

we have, (294 + x)/8 = 48

⇒ 294 + x = 384.

⇒ x = 90

Hence, option (d).

Answer: 93

Explanation :

Let the batting average for first 20 innings be a.

Total runs scored in first 20 innings = 20 × a.

∴ Total score after 21 innings = 20a + 135

⇒ Average after 21 innings = a + 2 = (20a + 135)/21

⇒ 21a + 42 = 20a + 135

⇒ a = 93

Alternately,
Since the average of now 21 innings increases by 2 runs hence he scored 21 × 2 = 42 runs more than old average.

∴ 135 - a = 42

⇒ a = 93

Hence, 93.