Q14. Ranjan is younger than his father by 20 yrs. 5 years ago his father was twice elder than him. Find their present ages.
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$\Rightarrow$ Let Ranjan's present age be = $x$ yrs
$\therefore$ Father's age = $20 + x$
$\Rightarrow$ Father's age five years ago = $(20 + x) - 5 = (15 + x)$
$\Rightarrow$ Ranjan's age five years ago = $3(x - 5) = (3x - 15)$
So, we have the following equation:
$\Rightarrow 3x - 15 = 15 + x$
$\Rightarrow 3x - x = 15 + 15$
$\Rightarrow 2x = 30$
$\Rightarrow x = \dfrac{30}{2} = 15$
$\therefore$ Ranjan's present age = 15 yrs Ans
$\Rightarrow$ So, father's age = $20 + 15 = 35$ yrs Ans
$\therefore$ Father's age = $20 + x$
$\Rightarrow$ Father's age five years ago = $(20 + x) - 5 = (15 + x)$
$\Rightarrow$ Ranjan's age five years ago = $3(x - 5) = (3x - 15)$
So, we have the following equation:
$\Rightarrow 3x - 15 = 15 + x$
$\Rightarrow 3x - x = 15 + 15$
$\Rightarrow 2x = 30$
$\Rightarrow x = \dfrac{30}{2} = 15$
$\therefore$ Ranjan's present age = 15 yrs Ans
$\Rightarrow$ So, father's age = $20 + 15 = 35$ yrs Ans
Q15. It was Binita and Jacob's $12^{th}$ marriage anniversary. At the time of marriage, Binita was $\dfrac{3}{4}$ of Jacob's age, but now she is $\dfrac{5}{6}$ of his age. Find their present ages.
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$\Rightarrow$ Let Jacob's age at the time of marriage be = $x$ yrs
$\therefore$ Jacob's present age = $x + 12$
$\Rightarrow$ Being $\dfrac{3}{4}$ of Jacob's age; Binita's present age:
$\Rightarrow (\dfrac{3}{4} \text{ of } x) + 12 = \dfrac{3x}{4} + 12$
$\Rightarrow$ Being $\dfrac{5}{6}$ of Jacob's age; Binita's present age:
$\Rightarrow \dfrac{5}{6} \text{ of } x + 12 = \dfrac{5x + 60}{6}$
So, we have the equation:
$\Rightarrow \dfrac{3x}{4} + 12 = \dfrac{5x + 60}{6}$
$\Rightarrow \dfrac{3x}{4} + 12 = \dfrac{5x}{6} + \dfrac{60}{6}$
$\Rightarrow \dfrac{3x}{4} + 12 = \dfrac{5x}{6} + 10$
$\Rightarrow 12 - 10 = \dfrac{5x}{6} - \dfrac{3x}{4}$
$\Rightarrow 12 - 10 = \dfrac{10x - 9x}{12}$
$\Rightarrow 2 = \dfrac{x}{12}$
$\Rightarrow x = 2 \times 12 = 24$
So, we have Jacob's age at the time of marriage = 24 yrs
$\therefore$ Jacob's present age = $24 + 12 = 36$ yrs Ans
$\Rightarrow$ And, Binita's present age = $\dfrac{5}{6} \text{ of } 36 = 5 \times 6 = 30$ yrs Ans
$\therefore$ Jacob's present age = $x + 12$
$\Rightarrow$ Being $\dfrac{3}{4}$ of Jacob's age; Binita's present age:
$\Rightarrow (\dfrac{3}{4} \text{ of } x) + 12 = \dfrac{3x}{4} + 12$
$\Rightarrow$ Being $\dfrac{5}{6}$ of Jacob's age; Binita's present age:
$\Rightarrow \dfrac{5}{6} \text{ of } x + 12 = \dfrac{5x + 60}{6}$
So, we have the equation:
$\Rightarrow \dfrac{3x}{4} + 12 = \dfrac{5x + 60}{6}$
$\Rightarrow \dfrac{3x}{4} + 12 = \dfrac{5x}{6} + \dfrac{60}{6}$
$\Rightarrow \dfrac{3x}{4} + 12 = \dfrac{5x}{6} + 10$
$\Rightarrow 12 - 10 = \dfrac{5x}{6} - \dfrac{3x}{4}$
$\Rightarrow 12 - 10 = \dfrac{10x - 9x}{12}$
$\Rightarrow 2 = \dfrac{x}{12}$
$\Rightarrow x = 2 \times 12 = 24$
So, we have Jacob's age at the time of marriage = 24 yrs
$\therefore$ Jacob's present age = $24 + 12 = 36$ yrs Ans
$\Rightarrow$ And, Binita's present age = $\dfrac{5}{6} \text{ of } 36 = 5 \times 6 = 30$ yrs Ans
Q16. David's age is 5 times his son's age. After 3 years, David's age will be 4 times his son's age. Find their present ages and its ratio.
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$\Rightarrow$ Let son's age be = $x$ yrs
$\therefore$ David's age = $5x$ yrs
$\Rightarrow$ David's age 3 years hence = $5x + 3$
$\Rightarrow$ Given, in 3 yrs David's age will be = $4(x + 3) = 4x + 12$
So, we have the equation:
$\Rightarrow 5x + 3 = 4x + 12$
$\Rightarrow 5x - 4x = 12 - 3$
$\Rightarrow x = 9$
$\therefore$ Son's present age = 9 yrs
$\Rightarrow$ And, Father's age = $5 \times 9 = 45$ yrs Ans
$\Rightarrow$ Required ratio = $\dfrac{45}{9} = \dfrac{5}{1} = 5:1$ Ans
$\therefore$ David's age = $5x$ yrs
$\Rightarrow$ David's age 3 years hence = $5x + 3$
$\Rightarrow$ Given, in 3 yrs David's age will be = $4(x + 3) = 4x + 12$
So, we have the equation:
$\Rightarrow 5x + 3 = 4x + 12$
$\Rightarrow 5x - 4x = 12 - 3$
$\Rightarrow x = 9$
$\therefore$ Son's present age = 9 yrs
$\Rightarrow$ And, Father's age = $5 \times 9 = 45$ yrs Ans
$\Rightarrow$ Required ratio = $\dfrac{45}{9} = \dfrac{5}{1} = 5:1$ Ans
Q17. The ratio between the present ages of Ratan and Sameer is 5:3. The ratio between Ratan's age 4 years ago and Sameer's age 4 years hence is 1:1. Find their present ages and the ratio between Ratan's age 4 years hence and Sameer's age 4 years ago?
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$\Rightarrow$ Let present ages of Ratan and Sameer be = $x$ yrs
$\therefore \dfrac{5x - 4}{3x + 4} = \dfrac{1}{1}$
$\Rightarrow 5x - 4 = 3x + 4$
$\Rightarrow 5x - 3x = 4 + 4$
$\Rightarrow 2x = 8$
$\Rightarrow x = \dfrac{8}{2} = 4$
As such, Ratan's age = $5 \times 4 = 20$ yrs Ans
And, Sameer's age = $3 \times 4 = 12$ yrs Ans
Given, Ratan's age 4 years hence and Sameer's age 4 years ago:
$\Rightarrow$ Required ratio = $\dfrac{20 + 4}{12 - 4} = \dfrac{16}{8} = \dfrac{2}{1} = 2:1$ Ans
$\therefore \dfrac{5x - 4}{3x + 4} = \dfrac{1}{1}$
$\Rightarrow 5x - 4 = 3x + 4$
$\Rightarrow 5x - 3x = 4 + 4$
$\Rightarrow 2x = 8$
$\Rightarrow x = \dfrac{8}{2} = 4$
As such, Ratan's age = $5 \times 4 = 20$ yrs Ans
And, Sameer's age = $3 \times 4 = 12$ yrs Ans
Given, Ratan's age 4 years hence and Sameer's age 4 years ago:
$\Rightarrow$ Required ratio = $\dfrac{20 + 4}{12 - 4} = \dfrac{16}{8} = \dfrac{2}{1} = 2:1$ Ans
Q18. Average age of 19 students in a class is 21 yrs. If a teacher's age is included, then the average increases to 22 yrs. Find teacher's age.
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Case 1:
$\Rightarrow$ Let sum of age of 19 students be = $x$ yrs
$\therefore$ Average age of 19 students will be = $\dfrac{x}{19}$
$\Rightarrow$ Given average = 21
$\therefore$ We have = $\dfrac{x}{19} = 21$
$\Rightarrow x = 21 \times 19 = 399$ yrs
Case 2:
$\Rightarrow$ Let sum of age of 19 students and a teacher be = $x$ yrs
$\therefore$ Average age of 20 people will be = $\dfrac{x}{20}$
$\Rightarrow$ Given average = 22
$\therefore$ We have = $\dfrac{x}{20} = 22$
$\Rightarrow x = 22 \times 20 = 440$ yrs
$\therefore$ Required teacher's age = New sum of age $-$ Old sum of age
$\Rightarrow 440 - 339 = 41$ yrs Ans
$\Rightarrow$ Let sum of age of 19 students be = $x$ yrs
$\therefore$ Average age of 19 students will be = $\dfrac{x}{19}$
$\Rightarrow$ Given average = 21
$\therefore$ We have = $\dfrac{x}{19} = 21$
$\Rightarrow x = 21 \times 19 = 399$ yrs
Case 2:
$\Rightarrow$ Let sum of age of 19 students and a teacher be = $x$ yrs
$\therefore$ Average age of 20 people will be = $\dfrac{x}{20}$
$\Rightarrow$ Given average = 22
$\therefore$ We have = $\dfrac{x}{20} = 22$
$\Rightarrow x = 22 \times 20 = 440$ yrs
$\therefore$ Required teacher's age = New sum of age $-$ Old sum of age
$\Rightarrow 440 - 339 = 41$ yrs Ans
Q19. The sum of ages of 4 children born at the interval of 2 yrs each is 50. Find the age of youngest child.
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$\Rightarrow$ Let age of youngest child be = $x$ yrs
$\therefore$ Ages of rest 3 children = $(x + 2) + (x + 4) + (x + 8)$
$\Rightarrow$ Given sum of age = 50 yrs
As such, we have the following equation:
$\Rightarrow x + (x + 2) + (x + 4) + (x + 8) = 50$
$\Rightarrow 4x + 14 = 50$
$\Rightarrow 4x = 50 - 14$
$\Rightarrow 4x = 36$
$\Rightarrow x = \dfrac{36}{4} = 9$ yrs Ans
$\therefore$ Ages of rest 3 children = $(x + 2) + (x + 4) + (x + 8)$
$\Rightarrow$ Given sum of age = 50 yrs
As such, we have the following equation:
$\Rightarrow x + (x + 2) + (x + 4) + (x + 8) = 50$
$\Rightarrow 4x + 14 = 50$
$\Rightarrow 4x = 50 - 14$
$\Rightarrow 4x = 36$
$\Rightarrow x = \dfrac{36}{4} = 9$ yrs Ans
Q20. The ratio of age between 2 girls is 3:2. The product of their ages is 27. Find their age.
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$\Rightarrow$ Let ages of two girls be = $x$ yrs
$\therefore 2x \times 1x = 27$
$\Rightarrow 3x^2 = 27$
$\Rightarrow x^2 = \dfrac{27}{3} = 9$
$\Rightarrow x^2 = 3^2$
$\Rightarrow x = 3$
Now, age of $1^{st}$ girl = $2 \times 3 = 6$ yrs Ans
And, age of $2^{nd}$ girl = $1 \times 3 = 3$ yrs Ans
$\therefore 2x \times 1x = 27$
$\Rightarrow 3x^2 = 27$
$\Rightarrow x^2 = \dfrac{27}{3} = 9$
$\Rightarrow x^2 = 3^2$
$\Rightarrow x = 3$
Now, age of $1^{st}$ girl = $2 \times 3 = 6$ yrs Ans
And, age of $2^{nd}$ girl = $1 \times 3 = 3$ yrs Ans
Q21. Present ratio of a Father and his daughter's age is 4:1. The product of their ages is 196. Find ratio of their age 4 yrs ago and 5 yrs hence.
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$\Rightarrow$ Let their present ages be = $x$ yrs
$\therefore 4x \times 1x = 196$
$\Rightarrow 4x^2 = 196$
$\Rightarrow x^2 = \dfrac{196}{4} = 49$
$\Rightarrow x^2 = 7^2$
$\Rightarrow x = 7$
As such, Father's age = $4 \times 7 = 28$ yrs
And, daughter's age = $1 \times 7 = 7$ yrs
$\Rightarrow$ Required ratio 4 yrs ago = $\dfrac{28 - 4}{7 - 4} = \dfrac{24}{3} = \dfrac{8}{1} = 8:1$ Ans
$\Rightarrow$ Required ratio 5 yrs hence = $\dfrac{28 + 5}{7 + 5} = \dfrac{33}{12} = \dfrac{11}{4} = 11:4$ Ans
$\therefore 4x \times 1x = 196$
$\Rightarrow 4x^2 = 196$
$\Rightarrow x^2 = \dfrac{196}{4} = 49$
$\Rightarrow x^2 = 7^2$
$\Rightarrow x = 7$
As such, Father's age = $4 \times 7 = 28$ yrs
And, daughter's age = $1 \times 7 = 7$ yrs
$\Rightarrow$ Required ratio 4 yrs ago = $\dfrac{28 - 4}{7 - 4} = \dfrac{24}{3} = \dfrac{8}{1} = 8:1$ Ans
$\Rightarrow$ Required ratio 5 yrs hence = $\dfrac{28 + 5}{7 + 5} = \dfrac{33}{12} = \dfrac{11}{4} = 11:4$ Ans
Q22. Manoj is 16 yrs elder than Abhay. 6 yrs ago Manoj's age was thrice as that of Abhay. Find their present ages.
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$\Rightarrow$ Let Abhay's age be = $x$ yrs
$\therefore$ Manoj's age = $x + 16$ yrs
$\Rightarrow$ Abhay's age 6 yrs ago = $(x - 6)$ yrs
$\Rightarrow$ Manoj's age 6 yrs ago = $(x + 16) - 6 = (x + 10)$
Given, 3 yrs ago Manoj's age was thrice that of Abhay's:
$\Rightarrow 3(x - 6) = $
$\Rightarrow 3x - 18 = x + 10$
$\Rightarrow 3x - x = 10 + 18$
$\Rightarrow 2x = 28$
$\Rightarrow x = \dfrac{28}{2} = 14$
$\therefore$ Abhay's present age = 14 yrs Ans
$\Rightarrow$ Ans, Manoj's present age = $14 + 16 = 30$ yrs Ans
$\therefore$ Manoj's age = $x + 16$ yrs
$\Rightarrow$ Abhay's age 6 yrs ago = $(x - 6)$ yrs
$\Rightarrow$ Manoj's age 6 yrs ago = $(x + 16) - 6 = (x + 10)$
Given, 3 yrs ago Manoj's age was thrice that of Abhay's:
$\Rightarrow 3(x - 6) = $
$\Rightarrow 3x - 18 = x + 10$
$\Rightarrow 3x - x = 10 + 18$
$\Rightarrow 2x = 28$
$\Rightarrow x = \dfrac{28}{2} = 14$
$\therefore$ Abhay's present age = 14 yrs Ans
$\Rightarrow$ Ans, Manoj's present age = $14 + 16 = 30$ yrs Ans
Q23. The ratio between the age of two boys is 3:5 and sum of their age is 80 yrs. Find their present age and ratio between their age 10 yrs hence.
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$\Rightarrow$ Let age of two boys be = $x$ yrs
$\therefore 3x + 5x = 80$
$\Rightarrow 8x = 80$
$\Rightarrow x = \dfrac{80}{8} = 10$
$\Rightarrow$ Age of $1^{st}$ boy = $3 \times 10 = 30$ yrs Ans
$\Rightarrow$ Age of $2^{nd}$ boy = $5 \times 10 = 50$ yrs Ans
$\Rightarrow$ Age 10 yrs hence = $\dfrac{30 + 10}{50 + 10} = \dfrac{40}{60}$
$\therefore$ Required ratio = $\dfrac{40}{60} = \dfrac{2}{3} = 2:3$ Ans
$\therefore 3x + 5x = 80$
$\Rightarrow 8x = 80$
$\Rightarrow x = \dfrac{80}{8} = 10$
$\Rightarrow$ Age of $1^{st}$ boy = $3 \times 10 = 30$ yrs Ans
$\Rightarrow$ Age of $2^{nd}$ boy = $5 \times 10 = 50$ yrs Ans
$\Rightarrow$ Age 10 yrs hence = $\dfrac{30 + 10}{50 + 10} = \dfrac{40}{60}$
$\therefore$ Required ratio = $\dfrac{40}{60} = \dfrac{2}{3} = 2:3$ Ans
Q24. The sum of a father and his son presently is 70. After 10 yrs son's age turns half of his father. Find their present ages.
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$\Rightarrow$ Let age of father be = $x$ yrs
$\Rightarrow$ Let age of son be = $y$ yrs
$\Rightarrow$ Given, sum = $x + y = 70$....(i)
$\Rightarrow$ Given, after 10 yrs son's age becomes half of his father's age:
$\therefore \dfrac{1}{2}(x + 10) = y + 10$
$\Rightarrow x + 10 = 2(y + 10)$
$\Rightarrow x + 10 = 2y + 20$
$\Rightarrow x - 2y = 20 - 10$
$\Rightarrow x - 2y = 10$
$\Rightarrow x = 10 + 2y$...(ii)
$\Rightarrow$ Including equation (ii) in equation (i):
$\Rightarrow 10 + 2y + y = 70$
$\Rightarrow 3y = 70 - 10$
$\Rightarrow 3y = 60$
$\Rightarrow y = \dfrac{60}{3} = 20$
$\therefore$ Son's age = 20 yrs Ans
$\Rightarrow$ Now, Father's age = $x + 20 = 70$
$\Rightarrow x = 70 - 20 = 50$ yrs Ans
$\Rightarrow$ Let age of son be = $y$ yrs
$\Rightarrow$ Given, sum = $x + y = 70$....(i)
$\Rightarrow$ Given, after 10 yrs son's age becomes half of his father's age:
$\therefore \dfrac{1}{2}(x + 10) = y + 10$
$\Rightarrow x + 10 = 2(y + 10)$
$\Rightarrow x + 10 = 2y + 20$
$\Rightarrow x - 2y = 20 - 10$
$\Rightarrow x - 2y = 10$
$\Rightarrow x = 10 + 2y$...(ii)
$\Rightarrow$ Including equation (ii) in equation (i):
$\Rightarrow 10 + 2y + y = 70$
$\Rightarrow 3y = 70 - 10$
$\Rightarrow 3y = 60$
$\Rightarrow y = \dfrac{60}{3} = 20$
$\therefore$ Son's age = 20 yrs Ans
$\Rightarrow$ Now, Father's age = $x + 20 = 70$
$\Rightarrow x = 70 - 20 = 50$ yrs Ans
Q25. The ratio between the age of two brothers is 5:2. If product of their ages is 1000, find their present ages and ratio of their age 10 hence.
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$\Rightarrow$ Let their ages be = $x$ yrs
$\therefore 5x \times 2x = 1000$
$\Rightarrow 10x^2 = 1000$
$\Rightarrow x^2 = \dfrac{1000}{10} = 100$
$\Rightarrow x^2 = 10^2$
$\Rightarrow x = 10$
$\Rightarrow$ Now, $1^{st}$ brother's age = $5 \times 10 = 50$ yrs Ans
$\Rightarrow$ And, $2^{nd}$ brother's age = $2 \times 10 = 20$ yrs Ans
$\Rightarrow$ Age 10 yrs hence = $\dfrac{50 + 10}{20 + 10} = \dfrac{60}{30}$
$\therefore \dfrac{60}{30} = \dfrac{2}{1} = 2:1$ Ans
$\therefore 5x \times 2x = 1000$
$\Rightarrow 10x^2 = 1000$
$\Rightarrow x^2 = \dfrac{1000}{10} = 100$
$\Rightarrow x^2 = 10^2$
$\Rightarrow x = 10$
$\Rightarrow$ Now, $1^{st}$ brother's age = $5 \times 10 = 50$ yrs Ans
$\Rightarrow$ And, $2^{nd}$ brother's age = $2 \times 10 = 20$ yrs Ans
$\Rightarrow$ Age 10 yrs hence = $\dfrac{50 + 10}{20 + 10} = \dfrac{60}{30}$
$\therefore \dfrac{60}{30} = \dfrac{2}{1} = 2:1$ Ans
Q26. Two years ago, a Father was 4 times as old as his son. 6 yrs later, Father's age will become more than double his son's age by 10 yrs. Find their present age and the ratio between them.
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$\Rightarrow$ Let Father's age be = $x$ yrs
$\Rightarrow$ Let Son's age be = $y$ yrs
Given, 2 yrs ago age of father was 4 times that of son:
$\therefore x - 2 = 4(y - 2)$
$\Rightarrow x - 2 = 4y - 8$
$\Rightarrow 8 - 2 = 4y - x$
$\Rightarrow 4y - x = 6$...(i)
Also, 6 yrs hence Father's age becomes 10 of his son's age:
$\therefore x + 6 = 2(y + 6) + 10$
$\Rightarrow x + 6 = 2y + 12 + 10$
$\Rightarrow x + 6 = 2y + 22$
$\Rightarrow x - 2y = 22 - 6$
$\Rightarrow x - 2y = 16$
$\Rightarrow x = 16 + 2y$...(ii)
$\Rightarrow$ Including equation (ii) in equation (i):
$\Rightarrow 4y - (16 + 2y) = 6$
$\Rightarrow 4y - 16 - 2y = 6$
$\Rightarrow 2y = 6 + 16$
$\Rightarrow 2y = 22$
$\Rightarrow y = \dfrac{22}{2} = 11$
$\therefore$ Son's present age = 11 yrs Ans
$\Rightarrow$ Father's present age = $x - 2(11) = 16$
$\Rightarrow x - 22 = 16$
$\Rightarrow x = 16 + 22 = 38$ yrs Ans
$\therefore$ Required ratio = $\dfrac{38}{11} = 38:11$ Ans
$\Rightarrow$ Let Son's age be = $y$ yrs
Given, 2 yrs ago age of father was 4 times that of son:
$\therefore x - 2 = 4(y - 2)$
$\Rightarrow x - 2 = 4y - 8$
$\Rightarrow 8 - 2 = 4y - x$
$\Rightarrow 4y - x = 6$...(i)
Also, 6 yrs hence Father's age becomes 10 of his son's age:
$\therefore x + 6 = 2(y + 6) + 10$
$\Rightarrow x + 6 = 2y + 12 + 10$
$\Rightarrow x + 6 = 2y + 22$
$\Rightarrow x - 2y = 22 - 6$
$\Rightarrow x - 2y = 16$
$\Rightarrow x = 16 + 2y$...(ii)
$\Rightarrow$ Including equation (ii) in equation (i):
$\Rightarrow 4y - (16 + 2y) = 6$
$\Rightarrow 4y - 16 - 2y = 6$
$\Rightarrow 2y = 6 + 16$
$\Rightarrow 2y = 22$
$\Rightarrow y = \dfrac{22}{2} = 11$
$\therefore$ Son's present age = 11 yrs Ans
$\Rightarrow$ Father's present age = $x - 2(11) = 16$
$\Rightarrow x - 22 = 16$
$\Rightarrow x = 16 + 22 = 38$ yrs Ans
$\therefore$ Required ratio = $\dfrac{38}{11} = 38:11$ Ans