Questions based on Ages

Example 01: At the time of marriage of A and B, B's age was three fourth of A's age, but now B's age is only five sixth of A's age; If combined age of A and B right now is 77 years. Find A's age at the time of marriage (in years).

Solution: Suppose at the time of marriage, the ages of A and B are \(4x\) and \(3x\), and their present ages are \(6y\) and \(5y\), then

\(6y + 5y = 77 \Rightarrow y = 7\)

So, their present ages are 42 and 35, that means difference of their ages at present is 7 years. As the difference of the ages does not change, so at the time of their marriage, the difference would have been the same. Therefore \(4x - 3x = 7 \Rightarrow x = 7\) and age of A at the time of marriage = 28 years.

Example 02: \(n\) years ago, Ravi was one fifth as old as Tanmay. In \(n\) years, Tanmay will be twice as old as Ravi. What is the ratio of Ravi's current age to Tanmay's current age?

Solution: Suppose ages of Ravi and Tanmay are \(r\) and \(t\), then \((t - x) = 5(r - x)\) or \(t = 5r - 4x\) 

Also \((t + x) = 2(r + x) \Rightarrow t = 2r + x\)

Solving the above equations \(5t = 13r\)

Hence the required ratio is \(5 : 13\).

Questions based on Numbers

Example: Find the biggest two digit number such that the number is four times of the sum of the digits of the number.

Solution: Suppose the two digit number is \(xy\) which can be written as \(10x + y\), now

(x + y) =\(\frac{1}{4}(10x + y)\) or

\(4x + 4y =10x + y\)  or \(y = 2x\)

To maximize \(y, x\) has to be maximum and maximum value of \(x\) can be 4, hence \(y = 8\) or the number is 48.

Example: A is a three digit number; B is formed by reversing the digits of \(A\). If \(A - B = 495\), find maximum value of \(A\).

Solution: Suppose \(A = xyz\), which can be written as 

\(100x + 10y + z\)

\(A =100x + 10y + z\)

\(B =100z + 10y + x\), hence \(A - B = 99 (x - z)\)

\(99 (x - z) = 495\) \( \Rightarrow x{\rm{ }}-{\rm{ }}z = {\rm{ }}5\) or \(x = z + 5\) 

To maximize \(A\), \(x\) must be maximum. If \(x = 9\) then \(z = 4; y\) can take any value from 0 to 9. To maximize the number \(y = 9\). Thus maximum value of \(A\) is 994

Example: \(A\) is greater than \(B\) by one fourth of the sum of \(A\) and \(B\). If \(B\) is increased by 45, it becomes greater than twice of A by 10. Find \(2A + B\).

Solution: It is given that, \(A = B + \frac{1}{4}(A + B)\) and 

\(B + 45 = 2A + 10\) 

Solving we get \(\frac{3}{4}A = \frac{5}{4}B\)  and \(2A - B = 35\)

\[ \Rightarrow B = {\rm{ }}{\bf{15}},A = {\rm{ }}{\bf{25}},\] hence \(2A + B =  65\).

Example: Mohan had certain number of apples, half of which he gave to his brother. The latter gave 75% of the apples that he had gotten to be equally shared among his three cousins: Anil, Jitendra and Mandeep. Anil bought 7 apples more and gave 50% of all his apples to his brother Mandeep. Mandeep’s apples then totaled 17. How many apples did Jitendra get? 

Solution: apple with Mohan = \(a\)

Given to brother = \(\frac{a}{2}\)

Given to each cousin = \(\frac{a}{2} \times \frac{3}{4} \times \frac{1}{3} = \frac{3}{{24}}a\)\(\)

Anil bought 7 more apple, hence total number of apples = \(\left( {\frac{3}{{24}}a + 7} \right)\), he gave half of this to Mandeep, and total number of apples with Mandeep = 17, hence  \(\frac{3}{{24}}a + \left( {\frac{3}{{24}}a + 7} \right) \times \frac{1}{2} = 17\)

Solving this equation;

\(6a + {\rm{ }}3a + {\rm{ }}24 \times 7{\rm{ }} = {\rm{ }}17 \times 48\Rightarrow a = 72\)

Jitendra got \(\frac{3}{{24}}a\) apples =  9 apples.

Alternate method

Working by choice we can easily prove that Jitendra got 27 Apples, because answer can be only an odd number so that after adding 7 more apples it can be divided into 2 parts.