Problems on Ages Quiz Set 004 | Maths Aptitude Mock Test on Problems of Ages Set 4 | mock tests for Bank PO, SSB/SSC Exams, SO Clerical Exam, GATE Exam, LIC AO, GIC AO, etc.

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Question 1

Each year the ages of three friends are in an AP(arithmetic progression). The age of middle friend today is 19 years. What would be the sum of their ages 10 years from now?

87 years.

88 years.

86 years.

89 years.

Soln.
Ans: a

Let the present ages of the three friends be a - d, a and a + d. As of today a = 19. Ten years later their ages would be a + 10 - d, a + 10, a + 10 + d. Adding these we get 3a + 30 which equals 3 × 19 + 30 = 87 years.

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Question 2

The sum of ages of two friends is 15, whereas the product of their ages is 54. What is the sum of squares of their ages?

117 years.

118 years.

116 years.

119 years.

Soln.
Ans: a

Let the ages be x and y. We are given x + y = 15, and xy = 54. Substituting in the identity $x^2 + y^2 = (x + y)^2 - 2 × xy$, we get $x^2 + y^2 = 15^2 - 2 × 54$ = 117.

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Question 3

After 5 years from today the ages of three friends will be in an AP(arithmetic progression), and their sum would be 120. What is the age of the middle friend today?

35 years.

36 years.

34 years.

37 years.

Soln.
Ans: a

Let the ages after 5 years be a - d, a and a + d. The sum is given to us. So (a - d) + a + (a + d) = 3a = 120. We get a = 120/3 = 40. So, the age of the middle friend today is a - 5 = 35 years.

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Question 4

The sum of ages of two friends is 23, whereas the product of their ages is 126. What is the sum of squares of their ages?

277 years.

278 years.

276 years.

279 years.

Soln.
Ans: a

Let the ages be x and y. We are given x + y = 23, and xy = 126. Substituting in the identity $x^2 + y^2 = (x + y)^2 - 2 × xy$, we get $x^2 + y^2 = 23^2 - 2 × 126$ = 277.

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Question 5

The ages of two friends are in the ratio 11:19. What is the age of the younger friend if the sum of their ages is 90 years?

33 years.

34 years.

32 years.

35 years.

Soln.
Ans: a

Let the ages be 11r and 19r. The younger is 11r. We have been given their sum. So (11 + 19)r = 90. Solving, we get r = 3. The younger is 11 × 3 = 33 years.

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