**Answer:** Option D

**Explanation:**

Let the number of women workers be x

According to the question 22*x+16*8 = 15(16+x)

=> 22x+128 =240 +15x

=> 22x-15x = 240 -128

=> 7x = 112

Therefore, x = \(\frac{112}{7}\) = 16

Therefore, Unmarried women workers =(16-10) = 6

**2. The mean marks of 30 students in a class is 58.5. Later on it was found that 75 was wrongly recorded as 57. Find the correct them.
**

**Answer:** Option D

**Explanation:** Correct mean = 30x 58.5 – 57 + \(\frac{75}{30}\)

\(\frac{1755 + 18}{30}\) = \(\frac{1773 }{30}\) = 59.1

**3. A person travels from x to y at a speed of 40Km/h and returns by increasing his speed 50%. What is his average speed for both the trips?
**

**Answer:** Option C

**Explanation:**

Speed of person from x to y =40 km/h

Speed of person from y to x = \(\frac{(40 × 150) }{100 }\) = 60km/h

Since the distance travelled is same

Therefore, Average Speed = \(\frac{(2 × 40 × 60) }{40 + 60}\) = 48km/h

**4. The average of 18 observations was calculated and it was 124. Later on it was discovered that two observations 46 and 82 were incorrect. The correct values are 64 and 28. The correct average of 18 observations is
**

**Answer:** Option D

**Explanation:**

Sum of 18 observations = 18×124 =2232

Correct sum of 18 observations = 2232-46-82+64+28

= 2196

Therefore, Correct average = \(\frac{2196 }{18}\) = 122.

**5. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:
**

**Answer:** Option D

**Explanation:**

Let A, B, C represent their respective weights. Then, we have: A + B + C = (45 x 3) = 135 …. (i) A + B = (40 x 2) = 80 …. (ii) B + C = (43 x 2) = 86 ….(iii) Adding (ii) and (iii), we get: A + 2B + C = 166 …. (iv) Subtracting (i) from (iv), we get : B = 31. B’s weight = 31 kg.

**Answer:** Option B

**Explanation:**

Sum of the present ages of husband, wife and child = (27 * 3 + 3 * 3) = 90 years.

Sum of the present age of wife and child = (20 * 2 + 5 * 2) = 50 years.

Husband’s present age = (90 – 50) = 40 years.

**2. The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is:**

**Answer:** Option B

**Explanation:**

Let the total number of workers be x. Then,

8000x = (12000 * 7) + 6000(x – 7)

= 2000x = 42000

= x = 21.

**3. In an examination a pupil’s average marks were 63 per paper. If he had obtained 20 more marks for his Geography paper and 2 more marks for his History paper, his average per paper would have been 65. How many papers were there in the examination?
**

**Answer:** Option D

**Explanation:**

Let the number of papers be x. Then,

63x + 20 + 2 = 65x

= 2x = 22

= x = 11.

**4. The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ration of the number of boys to the number of girls in the class is:
**

**Answer:** Option B

**Explanation:**

Let the ratio be k : 1. Then,

k * 16.4 + 1 * 15.4 = (k + 1) * 15.8

= (16.4 – 15.8)k = (15.8 – 15.4)

= k = \(\frac{0.4}{0.6}\) = \(\frac{2}{3}\)

Required ratio = \(\frac{2}{3}\) : 1 = 2:3.

**5. The average monthly salary of 20 employees in an organisation is Rs. 1500. If the manager’s salary is added, then the average salary increases by Rs. 100. What is the manager’s monthly salary?
**

**Answer:** Option C

**Explanation:**

Manager’s monthly salary

= Rs. (1600 * 21 – 1500 * 20) = Rs. 3600

**Answer:** Option A

**Explanation:**

Let the average of the whole team be x years.

11x – (26 + 29) = 9(x – 1)

= 11x – 9x = 46

= 2x = 46 => x = 23

So, average age of the team is 23 years.

**2. A cricketer has a certain average for 10 innings. In the eleventh inning, he scored 108 runs, there by increasing his average by 6 runs. His new average is:**

**C**. 55 runs

**D**. 60 runs

**Answer:** Option A

**Explanation:**

Let average for 10 innings be x. Then,

\(\frac{(10x + 108)}{11}\) = x + 6

= 11x + 66 = 10x + 108

= x = 42.

New average = (x + 6) = 48 runs.

**3. A cricketer whose bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and there by decreases his average by 0.4. The number age of the family now is:**

**D**. 85

**Answer:** Option D

**Explanation:**

Let the number of wickets taken till the last match be x. Then,

\(\frac{(12.4x + 26)}{(x + 5)}\) = 12

= 12.4x + 26 = 12x + 60

= 0.4x = 34

= x = \(\frac{340}{4}\) = 85.

**4. The average age of a husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:**

**B**. 23 years

**C**. 28.5 years

**D**. 29.3 years

**Answer:** Option A

**Explanation:**

Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years.

Required average = \(\frac{57}{3}\) = 19 years.

**5. The mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The corrected new mean is:**

**B**. 36.1

**C**. 36.5

**D**. 39.1

**Answer:** Option C

**Explanation:**

Correct sum = (36 * 50 + 48 – 23) = 1825.

Correct mean = \(\frac{1825}{50}\) = 36.5