# PE 2 - Average | Average, Mixture & Alligation

**PE 2 - Average | Average, Mixture & Alligation**

The average of 11 results is 60 marks. If the average of first six results in 59 marks and that of the last six is 62 marks, then the sixth result is:

- A.
65

- B.
61

- C.
66

- D.
60

Answer: Option C

**Explanation** :

Sum of all 11 results = 60 × 11 = 660

Sum of first 6 results = 6 × 59 = 354

Sum of first 6 results = 6 × 62 = 372

∴ 354 + 372 – 6^{th} result = 660

⇒ 6^{th} result = 726 – 660 = 66

Hence, option (c).

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**PE 2 - Average | Average, Mixture & Alligation**

The average of 3 numbers is 120. The largest number is 165 while the difference between the other two is 15. Find the smallest number:

- A.
135

- B.
150

- C.
155

- D.
140

Answer: Option D

**Explanation** :

Sum of the three numbers = 3 × 120 = 360

Let the other two numbers be ‘x’ and ‘x – 15’.

⇒ 165 + x + x – 15 = 360

⇒ 2x = 210

⇒ x = 155

∴ The smallest number = x – 15 = 140

Hence, option (d).

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**PE 2 - Average | Average, Mixture & Alligation**

The population of a town increased from 25000 to 375000 in a decade. The average percent increase of population per year is:

- A.
50%

- B.
5%

- C.
10%

- D.
None of these

Answer: Option B

**Explanation** :

Total percentage increase in 10 years = $\frac{(375000-250000)}{250000}$× 100% = 50%

Average % change per year = 50/10 = 5%

Hence, option (b).

Workspace:

**PE 2 - Average | Average, Mixture & Alligation**

Three years ago, the average age of a family of 5 members was 21 years. A baby having been born, the average age of the family is again 20.5 years. The present age of the baby is?

- A.
3 years

- B.
2 years

- C.
2.5 years

- D.
1.5 years

Answer: Option A

**Explanation** :

Average age of 5 members 3 years ago = 21 years.

Average age of these 5 members today = 21 + 3 = 24 years.

⇒ Total age of these 5 members (excluding the baby) = 24 × 5 = 120.

⇒ Total age of all 6 members (including the baby) = 20.5 × 6 = 123.

∴ 123 = 120 + baby’s age

⇒ Baby’s age = 3 years.

Hence, option (a).

Workspace:

**PE 2 - Average | Average, Mixture & Alligation**

In a class of 50 students, the mean marks obtained in a subject is 60 and in another class of 25 students the mean marks obtained in the same subject is 90. The mean marks obtained by the students of two classes taken together is:

- A.
68

- B.
72

- C.
70

- D.
65

Answer: Option C

**Explanation** :

The mean marks of the combined class of 75 students = $\frac{50\times 60+25\times 90}{50+25}$ = $\frac{5250}{75}$ = 70.

Hence, option (c).

Workspace:

**PE 2 - Average | Average, Mixture & Alligation**

The mean weight of 34 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.
66

- B.
56

- C.
55

- D.
57

Answer: Option B

**Explanation** :

Let the weight of the teacher be ‘x’ kgs.

⇒ New average = 42.4 = (34 × 42 + x)/35

⇒ 1484 = 1428 + x

⇒ x = 56

Hence, option (b).

Workspace:

**PE 2 - Average | Average, Mixture & Alligation**

Roshan bought 5 pants at Rs. 25 each, 10 shirts at Rs. 50 each and 15 ties at Rs. 35 each. Find the average price of all the articles.

- A.
Rs. 38.33

- B.
Rs. 45

- C.
Rs. 60

- D.
Rs. 45.33

Answer: Option A

**Explanation** :

Total price of 5 pants = 5 × 25 = Rs. 125

Total price of 10 shirts = 10 × 50 = Rs. 500

Total price of 15 ties = 15 × 35 = Rs. 525

Total price of all 30 items = 125 + 500 + 525 = 1150

∴ Average price of all items = 1150/30 = 38.33

Hence, option (a).

Workspace:

**PE 2 - Average | Average, Mixture & Alligation**

The average of the test scores of a class of ‘m’ students is 70 and that of ‘n’ students is 91. When the scores of both the classes are combined, the average is 80. What is n/m?

- A.
11/10

- B.
13/10

- C.
10/13

- D.
10/11

Answer: Option D

**Explanation** :

Combined average = 80 = $\frac{70m+91n}{m+n}$

⇒ 80n + 80m = 70m + 91n

⇒ 10m = 11n

⇒ $\frac{n}{m}$ = $\frac{10}{11}$

Hence, option (d).

Workspace:

**PE 2 - Average | Average, Mixture & Alligation**

10 years ago the average age of all the 25 teachers of the Girls college was 45 years. 4 years ago, the principal has retired from her post at the age of 60 year. So after one year a new principal whose age was 54 years recruited from outside. The present average age of all the teachers is, if principal is also considered as a teacher:

- A.
$54\frac{18}{25}$

- B.
$55\frac{17}{25}$

- C.
49.5

- D.
$49\frac{2}{3}$

Answer: Option A

**Explanation** :

10 years ago average age of 25 teachers = 45 years.

Average age of original 25 teachers (including the retired principal) = 45 + 10 = 55 years.

∴ Total age of original 25 teachers (including the retired principal) = 55 × 25 = 1375.

Present age of the retired principal = 60 + 4 = 64 years.

Present age of the new principal = 54 + 3 = 57 years.

∴ Total age of new set of 25 teachers (excluding the retired principal but including the new principal) = 1375 – 64 + 57 = 1368.

∴ New average age of teachers = $\frac{1368}{25}$ = $54\frac{18}{25}$.

Hence, option (a).

Workspace:

**PE 2 - Average | Average, Mixture & Alligation**

There were 35 students in a hostel. If the number of students increases by 7, the expenses of the mess increase by Rs. 42 per day while the average expenditure per head decreases by Rs. 1. Find the original expenditure of the mess?

- A.
Rs. 420

- B.
Rs. 520

- C.
Rs. 450

- D.
Rs. 550

Answer: Option A

**Explanation** :

Let the average expenditure of the mess initially = Rs. ‘a’ per head per day.

Total expenditure initially = 35a

New total expenditure 35a + 42

New average expenditure = a – 1 = $\frac{35a+42}{35+7}$

⇒ 42a - 42 = 35a + 42

⇒ 7a = 84

⇒ a = 12

∴ Initial expenditure of the mess = 35 × 12 = 420

Hence, option (a).

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