**Answer:** Option D

**Explanation:**

Prime numbers between 30 and 50 are:

31, 37, 41, 43, 47

Average of prime numbers between 30 to 50 will be

Average = \(\frac{31+37+41+43+47}{5}\) = \(\frac{199}{5}\) = 39.8

**2. Reeya obtained 65, 67, 76, 82 and 85 out of 100 in different subjects, What will be the average.**

**Answer:** Option B

**Explanation:**

\(\frac{65+67+76+82+85}{5}\) = 75

**3. Find the sum of first 30 natural numbers**

**Answer:** Option C

**Explanation:**

Sum of n natural numbers

Average = \(\frac{158 }{10}\) = 15.8

**4. The average of first 10 natural numbers is? **

**Answer:** Option B

**Explanation:**

Sum of 10 natural no. = \(\frac{n(n+1) }{2}\) = 55

= \(\frac{30(30+1) }{2}\) = \(\frac{30(31) }{2}\) = 465

**5. Find the average of first 10 multiples of 7**

**Answer:** Option C

**Explanation:**

Sum of 10 odd no. = 100

Average = \(\frac{100 }{10}\) = 10

**Answer:** Option C

**Explanation:**

= \(\frac{7(1+2+3+…+10) }{10}\) = \(\frac{7(10(10+1)) }{20}\)

\(\frac{7(110) }{20}\) = 38.5

**2. The average of four consecutive odd numbers is 24. Find the largest number. **

**Answer:** Option B

**Explanation:**

Let the numbers are x, x+2, x+4, x+6, then

\(\frac{4x+12)}{4}\) = 24

[x + 3 = 24 =>x = 21

**3. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 students**

**Answer:** Option C

**Explanation:**

Let the average weight of the 59 students be A.

So the total weight of the 59 of them will be 59*A.

The questions states that when the weight of this student who left is added, the total weight of the class = 59A + 45

When this student is also included, the average weight decreases by 0.2 kgs

\(\frac{59A+45}{60}\) = a – 0.2

59 A + 45 = 60 A – 12

45 + 12 = 60 A -59 A

A = 57

**4. A motorist travel to a place 150 km away at an avearge speed of 50 km/hr and returns ar 30 km/hr.His average speed for the whole jouney in km/hr is **

**Answer:** Option B

**Explanation:**

Average speed will be 2xy(x+y) km/hr

= {2(50)(30)}{50+30} km/hr

= 37.5 km/hr

**5. The average of six numbers is X and the average of three of these is Y.If the average of the remaining three is z, then**

**Answer:** Option B

X =((3y+3z)/6)

or

2X= y + z

**Answer:** Option A

**Explanation:**

23 + Sum of the present ages of husband, wife and child

= (27 * 3 + 3 * 3) years

= 90 years.

Sum of the present ages of wife and child

= (20 * 2 + 5 * 2) years

= 50 years.

Husband’s present age

= (90 – 50) years

= 40 years

**2. Average of 10 matches is 32, How many runs one should should score to increase his average by 4 runs.**

**C**. 78

**D**. 80

**Answer:** Option B

**Explanation:**

Average after 11 innings should be 36

So, Required score = (11 * 36) – (10 * 32)

= 396 – 320 = 76

**3. Average of five numbers is 27. If one number is excluded the average becomes 25. The excluded number is **

**D**. 65

**Answer:** Option A

**Explanation:**

Number is (5*27) – (4*25) = 135-100 = 35

**4. Average weight of 10 people increased by 1.5 kg when one person of 45 kg is replaced by a new man. Then weight of the new man is**

**B**. 55

**C**. 60

**D**. 65

**Answer:** Option C

**Explanation:**

Total weight increased is 1.5 * 10 = 15.

So weight of new person is 45+15 = 60

**5. A batsman makes a score of 87 runs in the 17th match and thus increases his average by 3. Find his average after 17th match**

**Answer:** Option D

**Explanation:**

Let the average after 17th match is x

then the average before 17th match is x-3

so, 16(x-3) + 87 = 17x

=> x = 87 – 48 = 39