Practice Ages weights - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) The average age of Ram and his two children is 17 years and the average age of Ram’s wife and the same children is 16 years. If the age of Ram is 33 years, the age of his wife is (in years):
(a)
(b)
(c)
(d)
Ram + two children = 51 years
His wife + two children = 48 years
∴ Ram – wife = 3 years
⇒ 33 – wife = 3 years
∴ Wife = 33 – 3 = 30 years
Q-2) Average age of 6 sons of a family is 8 years. Average age of sons together with their parents is 22 years. If the father is older than the mother by 8 years, the age of mother (in years) is :
(a)
(b)
(c)
(d)
Let the mother’s age = x years
∴ Father’s age = (x + 8) years
Sum of age of 6 sons = 8 × 6 = 48 years
Sum of age of 6 sons and parents = 22 × 8 = 176 years.
∴ Age of Parents = 176 – 48 = 128 years
⇒ x + x + 8 = 128
⇒ 2x = 120
⇒ x = 60
Hence, mother’s age = 60 years
Q-3) The average age of 30 students is 9 years. If the age of their teacher is included, the average age becomes 10 years. The age of the teacher (in years) is
(a)
(b)
(c)
(d)
Total age of 30 students = 9 × 30 = 270 years
Total age of 30 students and a teacher = 31×10 = 310 years
∴ Age of the teacher = 310 – 270 = 40 years
Q-4) The average age of four boys A, B, C and D is 5 years and the average age of A, B, D, E is 6 years. C is 8 years old. The age of E is (in years)
(a)
(b)
(c)
(d)
A + B + C + D = 20 years
⇒ A + B + D = 20 – 8 = 12 years
Now, A + B + D + E = 24 years
∴ E = 24 – 12 = 12 years
Q-5) In a school with 600 students, the average age of the boys is 12 years and that of the girls is 11 years. If the average age of the school is 11 years and 9 months, then the number of girls in the school is
(a)
(b)
(c)
(d)
Number of girls = x
Number of boys = 600 – x
∴ (600 – x ) × 12 + 11x
= 11$3/4$×600 = $47/4$ × 600
⇒ 7200 – 12x + 11x = 7050
⇒ x = 7200 – 7050 = 150
Q-6) In a school, the average age of students is 6 years, and the average age of 12 teachers is 40 years. If the average age of the combined group of all the teachers and students is 7 years, then the number of students is :
(a)
(b)
(c)
(d)
Let the number of students be n. Then,
7 = ${n × 6 + 12 × 40}/{n + 12}$
⇒ 7n + 84 = 6n + 480
⇒ n = 480 – 84 = 396
Q-7) The average age of a husband and a wife was 27 years when they married 4 years ago. The average age of the husband, the wife and a new-born child is 21 years now. The present age of the child is
(a)
(b)
(c)
(d)
Sum of the present age of husband and wife
= 2 × 27 + 8 = 62 years
Sum of the present age of husband, wife and child
= 21 × 3 = 63 years
∴ present age of the child
= 63 – 62 = 1 year
Q-8) The average age of a husband and wife, who were married 4 years ago, was 25 years at the time of their marriage. The average age of the family consisting of husband, wife and a child, born during the interval is 20 years today. The age of the child is
(a)
(b)
(c)
(d)
Sum of the present age of husband and wife
= 2 × 25 + 2 × 4 = 58 years
Sum of the present age of husband, wife and child
= 3 × 20 = 60 years
&there; Child’s present age
= 60 – 58 = 2 years
Q-9) The average age of 12 players of a team is 25 years. If the captain’s age is included, the average age increases by 1 year. The age of the captain is :
(a)
(b)
(c)
(d)
Total age of 12 players = 12 × 25 = 300
Total age including captain = 13 × 26 = 338
∴ Age of the captain = 338 – 300 = 38 years
Aliter : Using Rule 23,
( If in a group, one member is replaced by a new member, then,
Age of new member = ( age of replaced member) ± xn
where, x = increase (+) or decrease (–) in average n = Number of members. )
Here, x = 1, n = 12
Age of the captain = Average + x (n + 1)
= 25 + 1 (12 + 1) = 38 years
Q-10) 3 years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby is :
(a)
(b)
(c)
(d)
Total age of 5 members, 3 years ago = 17 × 5 = 85 years
Total age of 5 members, now
= (85 + 3 × 5) = 100 years
Total age of 6 members, now
= 17 × 6 = 102 years
∴ Age of the baby
= 102 – 100 = 2 years
Aliter : Using Rule 17,
( 't’ years before, the average age of N members of a family was 'T’ years. If during this period 'n’ children increased in the family but average age (present) remains same, then,
Present age of n children = n.T – N.t )
Here, t = 3, N = 5, T = 17, n = 1
Present age of baby = nT – Nt
= 1 × 17 – 5 × 3
= 17 – 15 = 2 years