Aptitude

## Problems on Age

Hello guys as we all know that word problem consisting of linear equation are hard to solve when we were in schools and especially problems on ages. At some point of our schooling we felt that why these chapter is in syllabus. But at later stage of life we find out that these is the most important topic because every competitive exam has quantitative aptitude section and it consists of age problems. Hence we are here to share some basic concepts that will help in solving age problems.

## Basic Concepts

1. If the present age of ‘P’ is ‘x’ years, then ‘n’ years ago = Age of ‘P’ was (x-n) years.
2. If the present age of ‘P’ is ‘x’ years, then ‘n’ years after = Age of ‘P’ was (x+n) years.
3. In general we have to assume the present age of one person. it is good to assume present age as x years.
4. If the present age of you and your friend is ‘x’ and ‘y’ years respectively then ‘n’ years ago age of you and your friend were (x-n) and (y-n) years respectively or ‘n’ years after from the present age will be (x+n) and (y+n) years respectively.
5. The age difference between you and your friend will remain same throughout your lifetime because clock will run equally for both of you.

Some Examples :-

Example :- Pradeep’s age after 15 years will be 5 times his age 5 years back. What is the present age of Pradeep ?

Suppose Pradeep’s present age be x years. Then

Pradeep’s age after 15 years will be (x+15) years.

Pradeep’s age 5 years ago was (x-5) years.

According to question x+15 = 5(x-5)

x+15 = 5x-25

4x = 40

x = 10

Hence Pradeep’s present age = 10 years.

Example :- One year ago, the ratio of Dhoni’s and Virat’s age was 6:7 respectively.Four years hence, this ratio would become 7:8. How old is Virat ?

Suppose Dhoni’s and Virat ages one year ago be 6x and 7x years respectively. Then,

Dhoni’s age 4 years hence = (6x+1)+4 = 6x+5 years

Virat’s age 4 years hence = (7x+1)+4 = 7x+5 years

(Note :- We have added one because the age we had assumed is for 1 year ago)

(6x+5)/(7x+5) = 7/8

8(6x+5) = 7(7x+5)

48x+40 = 49x+35

x = 5

Hence Virat’s present age is (7x+1) = 7*5 + 1 = 36 years.

Example :- The ages of two persons differ by 16 years. If 6 years ago, the elder one be 3 times as old as the younger one, find their present ages.

Let the age of the younger one be x years.

Then, age of elder person = x+16 (According to question)

3(x-6) = (x+16-6)

3x-18 = x+10

2x = 28

x = 14

Hence there present ages are 14 years and (14+16) = 30 years.

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This topic has been asked by our reader Pradeep singh tomar thats why his name is used in first example. Any doubts then write it down in the comment box below.

Apptitude

## Calendar (How to Find day on Given Date)

Here we will deal with finding the day of the week on a particular date. The process lies on obtaining the no. of odd days.

Odd Days :- In a given period, number of days more than a complete weeks are called odd days.

Leap Year :- Every year which is divisible by 4 is called a leap year. Thus 1992,1996,2004 etc. are leap years. Every 4th century is a leap year like 400,800,1200 etc. but no other century is a leap year like 1500,1900,2100 etc.

An year which is not a leap year is called an ordinary year.

1. An ordinary year has 365 days.
2. A leap year has 366 days.

Counting of odd days :

1 ordinary year = 365 days

= 52 weeks + 1 day

= 1 odd day.

1 leap year = 366 days

= 52 weeks + 2 day

= 2 odd days.

100 years = 76 ordinary years + 24 leap years

= (76 x 52) + 76 days + (24 x 52) + 48 days

= 124 days = 124/7 = 5 odd days

100 years = 5 odd days

200 years = (5+5)/7 = 3 odd days

300 years = (5+5+5)/7 = 1 odd day

400 years = 21 = 0 odd days

(7n+m) odd days, where m<7 is equivalent to m odd days. Thus 8 odd days = 1 odd day.

Lets understand the concept of finding odd days and the day on a particular week by some examples.

Example :- What was the day on 15 August 1947 ?

Solution :- 15 august 1947 = (1946 years + Period from 1 january 1947 to 15 august 1947)

Counting of odd days: 1600 years + 300 years + 47 years

1600 years = 0 odd days.

300 years = 1 odd day.

47 years = 11 leap year + 36 ordinary year

= 11 x 2 + 36 = 58 = 58/7 = 2 odd days.

Jan.     Feb.      March        April       May       June     July        Aug.

31     +   28   +    31     +       30   +     31   +      30     +   31      +    15

= 227 days = 32 weeks + 3 days = 3 odd days

Total No. of odd days = 0+1+2+3 = 6 odd days

Hence , the required day was Saturday.

Example :- What was the day on 16 April 2000 ?

Solution = 16 April 2000 = (1999 year + Period from 1 january 2000 to 16 april 2000)

Counting of odd days:

1600 years + 300 years + 99 years

1600 years = 0 odd day.

300 year = 1 odd day

99 year = 24 leap year + 75 ordinary year

=(24 x 1 + 75 x 2)  = 123 odd days = 123/7 = 4 odd days

Jan.     Feb.      March        April

31     +   29  +    31     +       16

= 107 days = 15 weeks + 2 days = 2 odd days.

Total no. of odd days = 0+1+4+2 = 7 odd days = 0 odd days.

Hence, the required day was Monday.

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