# CRE 3 - Weighted Average | Arithmetic - Average

**CRE 3 - Weighted Average | Arithmetic - Average**

A class has 20 boys and 30 girls. The average age of boys is 13 years and of girls is 14 years. What is the average age of the whole class?

- (a)
13.6

- (b)
14

- (c)
13

- (d)
13.4

Answer: Option A

**Explanation** :

Class average age = $\frac{20\times 13+30\times 14}{50}$ = 13.6

Hence, option (a).

Workspace:

**CRE 3 - Weighted Average | Arithmetic - Average**

A man purchased 3 blankets at Rs. 50 each, 5 blankets at Rs. 100 each and 2 blankets at a certain rate which he can’t recall. If the average price of all the blankets was Rs. 100, find the rate of the 2 blankets.

- (a)
Rs. 100

- (b)
Rs. 175

- (c)
Rs. 200

- (d)
Rs. 350

Answer: Option B

**Explanation** :

Let the ratio of 2 blankets referred in the problem be x, then (3 × 50 + 5 × 100 + 2x)/10 = 100.

⇒ x = 175.

Hence, option (b).

Workspace:

**CRE 3 - Weighted Average | Arithmetic - Average**

Class 10^{th} in a school has two sections, A and B. The average marks of section A is 50 and that of section B is 80. The combined average of the class is 65. Find the number of students in section A if there are a total of 60 students in the class.

- (a)
20

- (b)
25

- (c)
45

- (d)
30

Answer: Option D

**Explanation** :

Let the number of students in section A be x and in section B = (60 - x)

Then, (50 × x + 80 × (60 - x))/60 = 65

⇒ 50x + 4800 – 80x = 3900

⇒ 900 = 30x

⇒ x = 30

**Alternately,**

Here, the weighted average of the two section is same as the simple average of averages. That is possible only when the number of students (i.e. the weights) are same in both the sections.

Hence, both the section has same number of students, i.e. number of students in section A = 60/2 = 30.

Hence, option (d).

Workspace:

**CRE 3 - Weighted Average | Arithmetic - Average**

The average weight of a group of students is 39 kg. The average weight of the boys is 43 kg and that of the girls is 34 kg. If there are 25 boys, find the number of girls.

- (a)
20

- (b)
25

- (c)
45

- (d)
5

Answer: Option A

**Explanation** :

Let the number of girls in the group be x.

Then, $\frac{25\times 43+34\mathrm{x}}{\left(25+x\right)}$ = 39.

⇒ x = 20

Hence, option (a).

Workspace:

**CRE 3 - Weighted Average | Arithmetic - Average**

In a class, there are three divisions. The number of students and the average marks in Mathematics in three divisions are 30, 40, 30 and 50, 40, 60 respectively. What are the average marks in Mathematics of the class?

- (a)
49

- (b)
40

- (c)
51

- (d)
45

Answer: Option A

**Explanation** :

From the problem, $\frac{30\times 50+40\times 40+30\times 60}{\mathrm{30+40+30}}$ = 49

Hence, option (a).

Workspace:

**CRE 3 - Weighted Average | Arithmetic - Average**

The average score of a cricketer for 10 matches is 48.9 runs. If the average for the first 6 matches is 52. Find the average for last 4 matches.

- (a)
43.25

- (b)
44.25

- (c)
45.25

- (d)
46.25

Answer: Option B

**Explanation** :

Let the average score for the last 4 matches be x. Then, (6 × 52 + 4x)/10 = 48.9.

⇒ x = 44.25.

Hence, option (b).

Workspace:

**CRE 3 - Weighted Average | Arithmetic - Average**

The average of marks obtained by 120 candidates was 45. If the average of marks of candidates who passed was 49 and that of those who failed was 25, find the number of candidates who passed the examination.

- (a)
100

- (b)
110

- (c)
120

- (d)
20

Answer: Option A

**Explanation** :

Let p be the number of passed candidates. Then, number of failed candidates = (120 – p)

By the problem, 49p + 25(120 – p) = 120 × 45

⇒ p = 100

Hence, option (a).

Workspace:

**CRE 3 - Weighted Average | Arithmetic - Average**

A shopkeeper earned Rs. 624 in 12 days. His average for the first four days was Rs. 50 a day. Find his average income for the remaining eight days.

- (a)
Rs. 51

- (b)
Rs. 52

- (c)
Rs. 53

- (d)
Rs. 54

Answer: Option C

**Explanation** :

Let the average income of last 8 days = Rs. x

Then, 4 × 50 + 8x = 624.

⇒ x = Rs. 53

Hence, option (c).

Workspace:

**CRE 3 - Weighted Average | Arithmetic - Average**

In a class, there are three divisions A, B and C. The average marks in Mathematics in three divisions are 55, 40, 60 respectively whereas the weighted average of three division is 55. What is the ratio of number of students in section B and C?

- (a)
1 : 1

- (b)
1 : 3

- (c)
2 : 3

- (d)
Cannot be determined

Answer: Option B

**Explanation** :

Let the number of students in three section be a, b and c.

⇒ $\frac{55a+40b+60c}{a+b+c}$ =55

⇒ 55a + 40b + 60c = 55a + 55b + 55c

⇒ 40b + 60c = 55b + 55c

⇒ 5c = 15b

⇒ b : c = 1 : 3

**Alternately,**

Since, the average of section A is same as the combined average of the three division, even if we remove section A, the combined average of other two will not change.

Hence, weighted average of section B and C is also 55.

⇒ $\frac{40b+60c}{b+c}$ = 55

⇒ 40b + 60c = 55b + 55c

⇒ 5c = 15b

⇒ b : c = 1 : 3

Hence, option (b).

Workspace:

**CRE 3 - Weighted Average | Arithmetic - Average**

The total population of a town consists of 6,000 men and women. The average height of a man is 165 cm and that of a woman is 155 cm. and that of the whole population is 162.5 cm. Find the number of women in the town.

- (a)
1500

- (b)
1375

- (c)
1625

- (d)
None of these

Answer: Option A

**Explanation** :

Let m be the number of men

Then, (6000 – m) = number of women

By the problem, $\frac{m\times 165+\left(6000-m\right)\times 155}{6000}$ = 162.5

⇒ 165m + 930000 – 155m = 975000

⇒ 10m = 45,000 or m = 4500

∴ Number of women = 6,000 – 4,500 = 1,500.

Hence, option (a).

Workspace:

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