Ages - quantitative aptitude questions for problems on ages in different shortcut keys

In many interview problem on age is important one to face here we list out more than short cut key to understood clearly with question and answer

short cut key to understood problems on age

1. Odd Days:

   We are supposed to find the day of the week on a given date.

   For this, we use the concept of 'odd days'.

   In a given period, the number of days more than the complete weeks are called odd days.

2. Leap Year:

   (i). Every year divisible by 4 is a leap year, if it is not a century.

   (ii). Every 4^th century is a leap year and no other century is a leap year.

   Note: A leap year has 366 days.

   Examples:

   1. Each of the years 1948, 2004, 1676 etc. is a leap year.

   2. Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.

   3. None of the years 2001, 2002, 2003, 2005, 1800, 2100 is a leap year.

3. Ordinary Year:

   The year which is not a leap year is called an ordinary years. An ordinary year has 365 days.

4. Counting of Odd Days:

   1. 1 ordinary year = 365 days = (52 weeks + 1 day.)

      1 ordinary year has 1 odd day.

   2. 1 leap year = 366 days = (52 weeks + 2 days)

      1 leap year has 2 odd days.

   3. 100 years = 76 ordinary years + 24 leap years

        = (76 x 1 + 24 x 2) odd days = 124 odd days.

        = (17 weeks + days) 5 odd days.

      Number of odd days in 100 years = 5.

      Number of odd days in 200 years = (5 x 2) 3 odd days.

      Number of odd days in 300 years = (5 x 3) 1 odd day.

      Number of odd days in 400 years = (5 x 4 + 1) 0 odd day.

      Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days

question and answer

1) A man divides Rs.8600 among 5 sons,4 daughters and 2 nephews. If each daughter receives four times as much as each nephew, and each son receives five times as much as each nephew, how much does each daughter receive?

  Sol : Let the share of each nephew be Rs.x

  Then, share of each daughter = Rs.(4x); share of each son = Rs.(5x).

  So, 5 * 5x + 4 * 4x + 2 * x = 8600

  => 25x + 16x + 2x = 8600 => 43x = 8600

=> x = 200. Therefore, Share of each daughter = Rs. (4 * 200) = Rs.800

2) Kumar spends 75% of his income. His income is increased by 20% and he increased his expenditure by 10%. Find the percentage increase in his savings.

  Sol : Let original income = Rs. 100. Then, expenditure = Rs.75 and savings = Rs. 25.

  New income = Rs.120 & New expenditure = Rs.[110/100 * 75] = Rs.165/2

  New savings = Rs.[120-165/2] = Rs.75/2

  Increase in savings = Rs.[75/2 - 25] = Rs. 25/2

  Therefore, Increase% = [25/2 * 1/25 * 100]% => 50%

3) A man sells an article at a profit of 25%. If he had bought it at 20% less and sold it for Rs.10.50 less, he would have gained 30%. Find the cost of the article.

  Sol: Let the C.P. be Rs,x.

  First S.P. = 125% of x = 125/100x = 5x/4;

  2nd S.P. = 80% of x = 80/100x = 4x/5

  2nd S.P. = 130% of 4x/5 = [130/100 * 4x/5] = 26x/25

  => 5x/4 - 26x/25 = 10.50

  => 21x/100 = 10.50 =>x=[10.50 * 100 / 21] => 50

  Hence, C.P. = Rs. 50

4) A is twice as good as workman as B and together they finish a piece of work in 18 days. In how many days will A alone finish the work?

  Sol: (A's 1 day's work) : (B's 1 day's work) = 2:1

  (A+B)'s 1 day's work = 1/18.

  Therefore A's 1 day's work = [1/18 * 2/3] = 1/27.

  Hence, A alone can finish the work in 27 days.

5) Two pipes A and B can fill a tank in 36 min. and 45 min. respectively. A water pipe C can empty the tank in 30 min. First A and B are opened. After 7 minutes, C is also opened. In how much time, the tank is full?

  Sol: Part filled in min. = 7[1/36 + 1/45] = 7/20

  Remaining part = [1-7/20] = 13/20

  Net part filled in 1 min. when A,B and C are opened = [1/36 + 1/45 - 1/30] = 1/60

  Now, 1/60 part is filled in 1 min.

  13/20 part is filled in [60 * 13/20] = 39 min.

  Therefore, Total time taken to fill the tank = (39+7)min. => 46 min.

6) From a group of boys and girls 15 girls leave. There are then left 2 boys for each girl. After this, 45 boys leave. There are then 5 girls for each boy. Find the number of girls in the beginning.

  Sol: Let at present there be x boys. Then number of girls at present = 5x.

  Before the boys had left : Number of boys = x+45 and number of girls = 5x.

  => x+45 = 2*5x =>9x = 45 => x=5.

  Hence, number of girls in the beginning = 5x+15 = 25+15 = 40.

7) A cricketer has a certain average for 10 innings. In the eleventh inning, he scored 108 runs, thereby increasing his average by 6 runs. His new average is :

  Sol: Let average for 10 innings be x. Then,

  10x + 108/11 = x+6

  => 11x+66 = 10x+108 => x=42.

  Therefore, New Average = (x+6) = 48 runs.

8) Present ages of Abi and Suji are in the ratio of 5:4 respectively. Three years hence, the ratio of their ages will become 11:9 respectively. What is Suji's present age in years?

  Sol: Let the present ages Abi and Suji be 5x years and 4x years respectively.

  Then, 5x+3 / 4x+3 = 11/9

  => 9(5x+3) = 11(4x+3) => x=6.

  Therefore, Suji's present age = 4x = 24 years.

9) The difference between the ages of two person is 10 years. Fifteen years ago, the elder one was twice as old as the younger one. The present age of the elder person is

  Sol: Let the ages be x years and (x+10) years respectively.

  Then, (x+10) - 15 = 2(x-15)

  => x-5 = 2x-30 => x=25.

  Therefore, present age of the elder person = (x+10) = 35 years.

10) Pranav spends 30% of his monthly income on food articles, 40% of the remaining on conveyance and clothes and saves 50% of the remaining. If his monthly salary is Rs.18400, how much money does he save every month?

  Sol: Saving = 50% of (100-40)% of (100-30)% of Rs.18400

  => Rs. 50/100 * 60/100 * 70/100 * 18400 => 3864

  Therefore, he saves Rs.3864.