CRE 2 - Numbers 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.

Answer: Option A

Explanation :

If teacher's weight was 42 kgs, the new average would not have changed.

New average of 25 people (including teacher) increases by 400 gms because teacher's weight is more than the original average weight.

Extra weight brought by the teacher = 25 × 400 = 10000 gms = 10 kgs.

∴ Weight of the teacher = 42 + 10 = 52 kgs.

Hence, option (a).