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**AGE PROBLEMS WITH SOLUTIONS**QUANTITATIVE APTITUDE STUDY NOTES FOR BANK AND SSC EXAM
You know that quantitative
aptitude section is most important in

**bank exams in PO and****Clerk**and for other competitive exams because if you want good score in bank exam then you have to score good in maths. In competitive exams the most important thing is time management, if you know how to manage your time then you can do well in**Bank Exams.**That’s where maths shortcut tricks and formula are comes into action. So continuously we are providing shortcut tricks on different maths topics.
Today’s topic is

**AGE PROBLEMS.**This is the one of the most important topic in quantitative aptitude section in bank and ssc exam. You should know how to calculate Age problems questions and answers in very short time for bank exam. From this chapter around 1-2 questions are given in the**SBI and IBPS exams.**For this here we are providing**maths shortcut tricks**and quicker method to calculate quantitative apptitude questions in very short time.
To solve the problems based on
ages, students require the knowledge of linear equations. This method needs
some basic concepts as well as some more time than it deserves. Sometimes it is
easier to solve the problems by taking the given choices in account. But this
hit-and-trial method proves costly sometimes, when we reach our solution much
later. We have tried to evaluate some easier as well as quicker methods to
solve this type of questions on

**.**Although, we are not able to cover each type of questions in this section, our attempt is to minimise your difficulties.**Have a look at the following questions on****age problems:****Ex. 1:**The age of the father 3 years ago was 7 times the age of his son. At present, the father’s age is five times that of his son. What are the present ages of the father and the son?

**Ex. 2:**At present, the age of the father is five times the age of his son. Three years hence, the father’s age would be four times that of his son. Find the present ages of the father and the son.

**Ex. 3:**Three years earlier, the father was 7 times as old as his son. Three years hence, the father’s age would be four times that of his son. What are the present ages of the father and the son?

**Age problems solutions by the conventional method:**

**Solution 1:**let the present age of son = x yrs

Then, the present age of father =
5x yrs

3 years ago,

7 (x – 3) = 5x – 3

Or, 7x – 21 = 5x – 3

Or, 2x = 18

Or, x = 9

Therefore, son’s age = 9 yrs

Father’s age = 45 yrs

**Solution 2:**let the present age of son = x yrs

Then, the present age of son = x
yrs

Then, the present age of father =
5 yrs

3 yrs hence,

4(x + 3) = 5x + 3

Or, x + 12 = 5x + 3

X = 9 yrs. Therefore, son’s age =
9 yrs

And father’s age = 45 yrs

**Solution 3:**let the present age of son = x yrs

And the present age of father = y
yrs

3 yrs earlier, 7 (x – 3) = y – 3

Or, 7x – y = 18 ........... (1)

3 yrs hence, 4 (x + 3) = y + 3

Or, 4x + 12 = y + 3

Solving (1) & (2) we get, x =
9 yrs & y = 45 yrs

by maths shortcut trick:

Solution: son’s age = 3*(7-1) /
7-5

= 9 yrs

And father’s age = 9*5 = 45 yrs.

Undoubtedly you get confused with
the about method, but it is very easy to understand and remember. see the
following form of question:

**Age Problems**Maths short tricks and quicker method:**Q: t1 yrs earlier the father’s age was x times that of his son. At present, the father’s age is y times that of his son. What are the present ages of the son and the father?**

Solution: son’s age = (4-1)*3 /
5-4

= 9 yrs

And father’s age = 9*5 = 45 yrs.

**Q: The present age of the father is y times the age of his son. t2 yrs hence, the father’s age become z times the age of his son. What are the present ages of the father and his son?**

Solution:
3(4-1) + 3(7-1) / 7-4

=
9+18 / 3

=
9 yrs

**Q: t1 yrs earlier the age of the father was x time the age of his son t2 yrs hence, the age the father becomes z times the age of his son. what are the present ages of the son and the father?**

**Some other examples on age problems with solutions:**

**Ex:**Ten years ago, A was half of B in age. If the ratio of their present age is 3 : 4, what will be the total of their present ages?

**Solution:**10 yrs ago, A was ½ of B’s age.

At present, A is ¾ of B’s age.

B’s age

use formula (1)

10 ( ½ – 1) / ½ - ¾

= 20 yrs

A’s age = ¾ of 20 = 15 yrs

**Ex:**After 5 yrs, the age of a father will be thrice the age of his son, whereas five years ago, he was 7 times as old as his son was. What are their present ages?

**Solution:**formula (3) will be used in this case. So

Son’s age = 5(7-1)+5(3-1) / 7-3

= 10 yrs

From the first relationship of
ages, if F is the age of the father then

F + 5 = 3(10 + 5)

F = 40 yrs.

**Ex:**10 yrs ago, Sita’s mother was 4 times older than her daughter. After 10 yrs, the mother will be two times older than the daughter. What is the present age of Sita?

**Solution:**In this case also, formula (3) will be used.

Daughter’s age = 10(4-1) +
10(2-1) / 4-2

= 20 yrs

**To view quantitative aptitude study notes on other topics CLICK HERE**

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