Sum of ages = 40 x 10 = 400

At the time of birth, i.e., 9 years back,

\({400-(10×9) \over {9}}\) = 34.4. Hence, (a).

Sum of age of children = 72

72 + M+ F = 24 x 8 = 192.

M + F = 120 and F = M + 4

On solving both the equations we get, M = 58. Hence, (c).

\({(20×60)-55+35 \over {20}}\) = 59. Hence, (b).

10B + (14 × 20) = 12 × (

Sum of ages of P + N + M 2 yrs ago = 32 × 2 = 64

So after 2 yrs, i.e. at present their total = 64 + (2 × 2) = 68

Similarly, sum of present age of N and M = (26 × 2) + (4 × 2) = 60

Therefore, present age of Pritham = 68 – 60 = 8. Hence, (c).

(50 × 8) – [(30 × 4) + (40 × 2)] = 200. Hence, (b).

Assume there are 100 students in the class, then in category A = 25 students, in B = 20 students, remaining = 55 students. Let x be the average of these 55 students.

Therefore, (20 × 36) + (25 × 60) + (55 × x) = 45 × 100 = 41.45. Hence, (a).

Average age of D and M age 2 yrs ago = 48

Present average age of D and M = 48 + 2 = 50. Hence, (c).

20 men age = 20 × 50 =1000

10 men age = 10 × 40 = 400

New average age = \({1000+400 \over {30}}\) = 46.67. Hence, (b).

90 = 1:2 = 30:60 = Officer : Clerk

90 × 20000 = (30 × x) + (60 × \({x \over {2}}\))

18,00,000 = 60x

x = 30000. Hence, (a).

Initial number of persons = x

= 24x + (20 × 15) – 18(x + 20)

6x = 60

x = 10. Hence, (d).

(40 × 3) + (36 × 3) – (30 × 6) = 120 + 108 – 180 = 48. Hence, (b).

Total weight increased = (6 x 2) kg = 12 kg

Weight of new person = (48 + 12) kg = 60 kg. Hence, (d).

Weight of new person = (48 + 12) kg = 60 kg. Hence, (d).