The sum of the ages of a father and son is \(45\) years. Five years ago, the product of their ages was four times the father's age at that time. The present age of father and son
Answer: C
Let son's age \(=x\) years.
Then father's age\(=(45-x)\) years.
\(\Rightarrow\) \((x-5)(45-x-5) = 4(45- x - 5)\)
\(\Rightarrow\) \((x-5)=4\)
\(\Rightarrow\) \(x=9\)
Their ages are \(36\) years and \(9\) years.
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Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar's age at present?
Answer: A
Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.
then,\(\frac{ (6x + 6) + 4 }{(5x + 6) + 4}= \frac{11}{10}\)
\(\Rightarrow 10(6x + 10) = 11(5x + 10)\)
\(\Rightarrow 5x = 10\)
\(\Rightarrow x = 2\)
\(\therefore\) Sagar's present age \(= (5x + 6) = 16\) years.
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The ratio of the present ages of P and Q is 3 : 4. Five years ago, the ratio of their ages was 5 : 7. Find their present ages.
Answer: B
As the ratio of their present ages is 3 : 4 , let their present ages be \(3x\) and \(4x\) respectively.
So, 5 years ago, as the ratio of their ages was 5 : 7, we can write \((3x-5) : (4x-5) = 5 : 7\)
Solving, we get \(x=10\)
Hence, their present ages are \(3x = 30\) and \(4x=40\)
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The average of the ages of Sumit, Krishna and Rishabh is 43 years and the average of the ages of Sumit, Rishabh and Rohit is 49 years. If Rohit is 54 years old, what is Krishna'a age ?
Answer: B
Let present ages of Sumit = \(w\), Krishna = \(x \), Rishabh = \(y\) and Rohit = \(z\) , then
\(w+x+y = 43*3 = 129 ..(i)\)
\(w+y+z = 49*3 = 147 ..(ii)\)
Subtracting (i) from (ii),
\((w+y+z) - (w+x+y)=147 -129 , z- x =18 ..(iii)\)
Given Rohit's age => z = 54, so from (iii), x = 54-18 = 36
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The present average age of a family of five members is 26 years. If the present age of the youngest member in the family is ten years, then what was the average age of the family at the time of the birth of the youngest member ? (Assume no death occurred in the family since the birth of the youngest
Answer: C
Present total age of the members = 26(5) = 130 years
Present age of the youngest member = 10 years
Present total age of the remaining four members = 130 - 10 = 120 years
Their average age at the time of the birth of the youngest member = \(\frac{120-(4*10)}{4}\)
= 30 - 10 = 20 years
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Rohit is younger than Rahul by 7 years. If the ratio of their ages is 7:9, find the age of Rohit
Answer: D
lets Rahul's age is \(x\) years
Then Rohit's age \(=(x-7)\) years
so, \(\frac{(x-7)}{x}=\frac{7}{9}\)
\(\Rightarrow \) \(9x-63=7x\)
\(\Rightarrow\) \(2x=63\)
\(\Rightarrow\) \(x=31.5\)
\(\therefore \) Rohit's age \(31.5-7=24.5\) years
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Father is aged three times more than his son. After 8 years, father would be two and a half times of son's age. After further 8 years, how many times would father be of son's age?
Answer: A
Assume that son's present age \(= x \)
Then, father's present age = \(3x+x=4x\)
After \(8\) years, father's age \(=2\tfrac{1}{2}\) times of son's age.
\(\Rightarrow\) \((4x+8)=2\tfrac{1}{2}(x+8)\)
\(\Rightarrow\) \(4x+8=\tfrac{5}{2}(x+8)\)
\(\Rightarrow\) \(8x+16=5x+40\)
\(\Rightarrow\) \(3x=40-16=24\)
\(\Rightarrow\) \(x=\frac{24}{3}=8\)
After further \(8\) years,
Son's age \( =x+8+8=8+8+8=24\)
Father's age \(=4x+8+8=4×8+8+8=48\)
Father's age/Son's age \(=\frac{48}{24}=2\)
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Six years ago Anita was P times as old as Ben was. If Anita is now 17 years old, how old is Ben now in terms of P ?
Answer: A
Let Ben’s age now be \(B
\)
Anita's age now be \(A\)
\((A - 6) = P(B - 6)\)
But \(A\) is 17 and therefore, \(11 = P(B - 6)\)
\(\Rightarrow 11/P = B-6\)
\(\Rightarrow (11/P) + 6 = B\)
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Raju got married 8years ago. His present age is 6/5 times his age at the time of his marriage Raju's sister was 10years younger to him at the time of his marriage. The present age of Raju's sister is ?
Answer: B
Let Raju's age at the time of his marriage = \(x\) years
Then from given data,
\(\Rightarrow x +8=\frac{6}{5}x\)
\(x=40\)
Since got married 8 years ago, his present age = 48
His sister is 10 years younger => 48 - 10 = 38 years.
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A man is \(24\) years older than his son. In two years, his age will be twice the age of his son. What is the present age of his son?
Answer: C
Let present age of the son \(= x \) years
Then, present age the man \(=(x+24)\) years
Given that, in \(2\) years, man's age will be twice the age of his son
\(\Rightarrow\) \((x+24)+2=2(x+2)\)
\(\Rightarrow\) \(x=22\)
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