1. The average wages of a worker during a fortnight comprising 15 consecutive working days was Rs.90 per day. During the first 7 days, his average wages was Rs.87/day and the average wages during the last 7 days was Rs.92 /day. What was his wage on the 8th day?

a) 83

b) 92

c) 90

d) 97

Explanation: The total wages earned during the 15 days that the worker worked = 15 × 90

= Rs. 1350

The total wages earned during the first 7 days = 7 × 87

= Rs. 609

The total wages earned during the last 7 days = 7 × 92

= Rs. 644

Total wages earned during the 15 days,

= wages during first 7 days + wage on 8

^{th}day + wages during the last 7 days.

1350 = 609 + wage on 8

^{th}day + 644

wage on 8

^{th}day = 1350 - 609 - 644 = Rs. 97

2. The average temperature on Wednesday, Thursday and Friday was 25°. The average temperature on Thursday, Friday and Saturday was 24°. If the temperature on Saturday was 27°, what was the temperature on Wednesday?

a) 24°

b) 21°

c) 27°

d) 30°

Explanation: Total temperature on Wednesday, Thursday and Friday was 25 × 3 = 75°

Total temperature on Thursday, Friday and Saturday was 24 × 3 = 72°

Hence, difference between the temperature on Wednesday and Saturday = 3°

If Saturday temperature =27°, then

Wednesday's temperature = 27 + 3 = 30°

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?

a) 57

b) 56.8

c) 58.2

d) 52.2

Explanation: Let the average weight of the 59 students be X. Therefore, the total weight of the 59 of them will be 59X

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

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

59X + $$\frac{{45}}{{60}}$$ = X - 0.2

59X + 45 = 60X - 12

45 + 12 = 60X - 59X

X = 57

4. The difference between two angles of a triangle is 24°. The average of the same two angles is 54°. Which one of the following is the value of the greatest angle of the triangle?

a) 45°

b) 60°

c) 66°

d) 72°

Explanation: Let a and b be the two angles in the question, with a > b. We are given that the difference between the angles is 24°.

a – b = 24

Since the average of the two angles is 54°, we have $$\frac{{{\text{a}} + {\text{b}}}}{2}$$ = 54

Solving for b in the first equation yields b = a – 24, and substituting this into the second equation yields,

$$ {\frac{{\left\{ {{\text{a}} + \left( {{\text{a}} - 24} \right)} \right\}}}{2}} = 54$$

2a − 24 = 54 × 2

2a − 24 = 108

2a = 108 + 24

2a = 132

a = 66

b = a − 24 = 66 − 24 = 42

Now, let c be the third angle of the triangle. Since the sum of the angles in the triangle is

180°, a + b + c = 180°

Putting the previous results into the equation yields 66 + 42 + c = 180°

Solving for c yields c = 72°

The greatest of the three angles a, b and c is c, which equal.

5. The average age of a family of 5 members is 20 years. If the age of the youngest member be 10 years then what was the average age of the family at the time of the birth of the youngest member?

a) 13.5

b) 14

c) 15

d) 12.5

Explanation: At present the total age of the family = 5 × 20 =100

The total age of the family at the time of the birth of the youngest member,

= 100 - 10 - (10 × 4)

= 50

Average age of the family at the time of birth of the youngest member,

$$ = \frac{{50}}{4} = 12.5$$

6. The average of five different positive numbers is 25. x is the decrease in the average when the smallest number among them is replaced by 0. What can be said about x?

a) x is less than 5

b) x is greater than 5

c) x is equal to 5

d) Cannot be determined

Explanation: Let a, b, c, d, and e be the five positive numbers in the decreasing order of size such that e is the smallest number.

We are given that the average of the five numbers is 25. Hence, we have the equation

$$\frac{{{\text{a}} + {\text{b}} + {\text{c}} + {\text{d}} + {\text{e}}}}{5} = 25$$

a + b + c + d + e = 125 ----- (1) by multiplying by 5.

The smallest number in a set is at least less than the average of the numbers in the set if at least one number is different.

For example, the average of 1, 2, and 3 is 2, and the smallest number in the set 1 is less than the average 2. Hence, we have the inequality

0 < e < 25

0 > -e > -25 by multiplying both sides of the inequality by -1 and flipping the directions of the inequalities. Adding this inequality to equation (1) yields

0 + 125 > (a + b + c + d + e) + (-e) > 125 - 25

125 > (a + b + c + d) > 100

125 > (a + b + c + d + 0) > 100 by adding by 0

25 > $$\frac{{{\text{a}} + {\text{b}} + {\text{c}} + {\text{d}} + 0}}{5}$$ ⇒ 20 by dividing the inequality by 5

25 > The average of numbers a, b, c, d and 0 > 20

Hence, x equals

(Average of the numbers a, b, c, d and e) – (Average of the numbers a, b, c, and d)

= 25 − (A number between 20 and 25)

⇒ A number less than 5

Hence, x is less than 5

7. In 2011, the arithmetic mean of the annual incomes of Ramesh and Suresh was Rs. 3800. The arithmetic mean of the annual incomes of Suresh and Pratap was Rs. 4800, and the arithmetic mean of the annual incomes of Pratap and Ramesh was Rs. 5800. What is the arithmetic mean of the incomes of the three?

a) Rs. 4000

b) Rs. 4200

c) Rs. 4400

d) Rs. 4800

Explanation: Let a, b, and c be the annual incomes of Ramesh, Suresh, and Pratap, respectively.

The arithmetic mean of the annual incomes of Ramesh and Suresh was Rs. 3800.

$$\frac{{{\text{a}} + {\text{b}}}}{2}$$ = 3800

a + b = 2 × 3800 = 7600

The arithmetic mean of the annual incomes of Suresh and Pratap was Rs. 4800.

$$\frac{{{\text{b}} + {\text{c}}}}{2}$$ = 4800

b + c = 2 × 4800 = 9600

The arithmetic mean of the annual incomes of Pratap and Ramesh was Rs. 5800.

$$\frac{{{\text{c}} + {\text{a}}}}{2}$$ = 5800

c + a = 2 × 5800 = 11,600

Adding these three equations yields:

(a + b) + (b + c) + (c + a) = 7600 + 9600 + 11,600

2a + 2b + 2c = 28,800

a + b + c = 14,400

The average of the incomes of the three equals the sum of the incomes divided by 3,

$$\eqalign{ & \frac{{{\text{a}} + {\text{b}} + {\text{c}}}}{3} \cr & = \frac{{14,400}}{3} \cr & = {\text{Rs}}{\text{.}}\,4800 \cr} $$

8. In Arun's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?

a) 67 kg

b) 68 kg

c) 69 kg

d) Data inadequate

Explanation: Let Arun's weight by X kg.

According to Arun,

65 < X < 72

According to Arun's brother,

60 < X < 70

According to Arun's mother,

X $$ \leqslant $$ 68

The values satisfying all the above conditions are 66, 67 and 68.

Required average,

$$\eqalign{ & = \frac{{66 + 67 + 68}}{3} = \frac{{201}}{3} = 67\,{\text{kg}}{\text{.}} \cr} $$

9. A student finds the average of 10 positive integers. Each integer contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this, the average becomes 1.8 less than the previous one. What was the difference of the two digits a and b?

a) 8

b) 6

c) 2

d) 4

Explanation: Let the original number be (10a + b)

After interchanging the digits, the new number becomes (10b + a)

The question states that the average of 10 numbers has become 1.8 less than the original average. Therefore, the sum of the original 10 numbers will be (10 × 1.8 = 18 more than the sum of the 10 numbers with the digits interchanged.

10a + b = 10b + a + 18

9a - 9b = 18

a - b = 2

10. The average of first five multiples of 3 is:

a) 9

b) 10

c) 8

d) 11

Explanation: First five multiples of three are:

3, 6, 9, 12, 15.

$$\eqalign{ & {\text{Average}} = \frac{{ {3 + 6 + 9 + 12 + 15} }}{5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{45}}{5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 9 \cr} $$