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

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

The mean of 100 observations is 50. If one observation 50 is replaced by 40, find the resulting mean?

- A.
50

- B.
49.90

- C.
70

- D.
40

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).

Workspace:

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

The average of 100 numbers is 60. One observation 50 is removed and another number x is added such that the average increases by 1. Find x?

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.

Workspace:

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

The average of 100 numbers is 60. One observation 80 is removed and another number x is added such that the average decreases by 0.5. Find x?

- A.
50

- B.
40

- C.
30

- D.
20

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).

Workspace:

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

A batsman scored an average of 12 runs in 7 innings. He played one more innings and the average then became 14. How many runs did he score in the last innings?

- A.
28

- B.
14

- C.
12

- D.
7

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).

Workspace:

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

The average of 6 values is 12. A number is removed and the average increases by 1. Find the number removed.

- A.
7

- B.
8

- C.
5

- D.
11

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).

Workspace:

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

The average of 6 values is 12. A number is removed and the average decreases by 1. Find the number removed.

- A.
7

- B.
18

- C.
13

- D.
17

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).

Workspace:

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

The average of 6 values is 12. A number is added and the average increases by 1. Find the number added.

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.

Workspace:

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

The average of 6 values is 12. A number is added and the average decreases by 1. Find the number added.

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.

Workspace:

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

The average monthly income of 6 employees is Rs. 310. After one of them receives an increment the average rises to Rs. 315. Find the amount of increment.

- A.
Rs. 2

- B.
Rs. 25

- C.
Rs. 20

- D.
Rs. 30

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).

Workspace:

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

The average marks of 7 students is 42%. When a student joined this group, the average increases to 48%. The marks of the student who joined is

- A.
85

- B.
78

- C.
94

- D.
90

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).

Workspace:

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

A batsman scored 135 runs in the 21^{st} innings and thus his batting average increases by 2. Find the batting average for first 20 innings.

[Batting Average = Total Runs scored / Number of innings]

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.

Workspace:

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

The mean weight of 24 students of a school is 42 kg. If the weight of the teacher be included, the mean rises by 400 g. Find the weight (in kg) of the teacher?

- A.
52 kgs

- B.
45 kgs

- C.
51.6 kgs

- D.
51.5 kgs

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).

Workspace:

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