Practice Questions – Average – 1

1. The average of 5 numbers is 306.4. The average of first two numbers is 431 and the average of last two numbers is 214.5. What is the third number?

a) 108

b) 52

c) 321

d) Cannot be determined

e) None of these

Click here to view Answer
e) None of these

Explanation:

Total of five numbers – 306.4 × 5 = 1532

∴ Required third number = 1532 – [(431 × 2) + (214.5 × 2)]

= 1532 – [862 + 429] = 241

2. The sum of five numbers is 555. The average of first two numbers is 75 and the third number is 115. What is the average of last two numbers?

a) 145

b) 290

c) 265

d) 150

e) None of these

Click here to view Answer
a) 145

Explanation:

Sum of first two numbers = 75 × 2 =150

Sum of last two numbers = 555 – 150 – 115 = 290

∴ Average = 290/2 = 145

3. The average of nine numbers is 50. The average of the first five numbers is 54 and that of the last three numbers is 52. Then the sixth number is

a) 30

b) 34

c) 24

d) 44

e) None of these

Click here to view Answer
c) 24

Explanation:

The sixth number

= (9 × 50) – (5 × 54) – (3 × 52)

= 450 – 270 – 156 = 24

4. The average of four consecutive even numbers P, Q, R and S respectively (increasing order) is 51. What is the product of P and R?

a) 2592

b) 2400

c) 2600

d) 2808

e) None of these

Click here to view Answer
e) None of these

Explanation:

(P+Q+R+S)/4 = 51

P + Q + R + S = 204

Let the four consecutive even numbers be

2n-2, 2n, 2n+2, 2n+4

2n-2 + 2n + 2n+2 + 2n+4 = 204

8n + 4 = 204

n = 200/8 = 25

P × R = (2 × 25 – 2) (2 × 25 + 2)

= 48 × 52 = 2496

5. The average height of 27 persons was recorded as 162 cm. If the height of Sheryas was deleted from the observation, the average height reduced by 1 cm. What was Shreyas, height?

a) 184 cm

b) 226 cm

c) 179 cm

d) 186 cm

e) None of these

Click here to view Answer
e) None of these

Explanation:

Method 1: Height of Shreyas =162 + 26 = 188 cm

Method 2: Let the height of Sheryas be ‘x’

Total height height of 27 persons = 27 × 162 = 4374

(4374-x)/(27-1) = 161

4374 – (161 × 26) = x

X = 188 cm

6. Find the average of the following set of scores :

432, 623, 209, 378, 908, 168

a) 456

b) 455

c) 453

d) 458

e) None of these

Click here to view Answer
c) 453

Explanation:

Required average

= (432 + 623+209+ 378+908+168)/6  =  2718/6  = 453

7. The average of five numbers is 34.4. The average of the first and second number is 46.5. The average of fourth and fifth number is 18. What is the third number?

a) 45

b) 46

c) 42

d) 49

e) None of these

Click here to view Answer
e) None of these

Explanation:

Third number

= (5 × 34.4) – (2 × 46.5) – (2 × 18)

= 172 – 93 – 36 = 43

8. Farah was married 8 years ago. Today her age is times to that at the time of marriage. At present her daughter’s age is of her age. What was her daughter’s age 3 years ago?

a) 6 years

b) 7 years

c) 3 years

d) Cannot be determined

e) None of these

Click here to view Answer
a) 6 years

Explanation:

Let Farah’s age 8 years ago be x years

Farah’s present age = (x + 8) years

∴ x + 8 = 9x/7

7x + 56 = 9x

X = 28

Farah’s present age = 28 + 8 = 36 years

Her daughter’s age 3 years ago

= 36 × 1/6 = 6 years

9. The present age of Romila is one fourth of that of her father. After 6 years the father’s age will be twice the age of Kapil. If Kapil celebrated fifth birthday 8 years ago, What is Romila’s present age?

a) 7 years

b) 7.5 years

c) 8 years

d) 8.5 years

e) None of these

Click here to view Answer
c) 8 years

Explanation:

Kapil’s present age = 5 + 8 = 13 years

After 6 years, Kapil’s age = 19 years

∴ Father’s present age = (19 × 2) – 6 = 32 years

∴ Romila’s present age = 1/4 × 32 = 8 years