Problems on age are similar to those of Ratio and Proportion. However in certain questions,calculations can be shortened using the concepts of Singapore Maths. For example the age of the father 3 years ago was 7 times that of his son. At present the father's age is 5 times that of his son than to find the ages of father and son we proceed as follows:

F: S

7:1 <-- 3 Years Ago

5:1 <-- Present

Here if the ratios were the actual ages then Present ratio should be more than that of 3 years ago. Also the same years would be added to father's age and son's age. So what we do here is we make the ratio of present age to be more than that of 3 years ago and also make the difference of the ratio to be equal. To do that we find the difference of the ratios and cross multiply as:

F: S

7:1 <-- Difference is 6

5:1<--- Difference is 4 Thus the ratio of difference is 6:4 or 3:2, cross multiplying the upper ratio by 2 and lower by 3, we have:

F: S

14:2 <-- three years ago

15:3 < --Present

-------

1 :1

-----

Here we see that the difference is 1: 1 which is same and also the present age is more than that of 3 years ago.

Now the calculations is very simple :1 year difference in ratio--> actual 3 years difference so Father's age that is 15 present ---> actual will be 3/1 x 15=45 years and son's age will be 3/1 x 3 = 9 years

This concept becomes very handy when we solve similar problems in ratio and proportions.

Similarly if Three years earlier the father's age 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. We go like this:

F: S

7:1 <- Three years earlier

4:1 <-- Three years hence

----

3:0

----

To make the difference same we cross multiply by the difference

F: S

7:1 <--6~2 in ratio

\ /

/ \

4:1 <--3~1 in ratio

F: S

7:1 <-- -3

8:2<-- +3

---

1:1

---

Now if difference in ratio is 1--> Actual difference is 3+3=6 years. If the age 3 years earlier is 7 in ratio--> actual age is 6/1 x 7= 42. So present age is 42+3=45. similarly 1-->6 then 1-->6 so present age of son is 6+3=9 years.

The concept is very simple and gives the answer almost intuitively.

Sometimes the questions are asked like this : The sum of ages of a mother and her daughter is 50 years. Also 5 years ago, the mother's age was 7 times the age of the daughter. What are the present ages of the mother and the daughter. Here 5 years ago, the sum of the ages will be 40 years. Divide 40 in the ratio of 7:1, Thus 8~40 so 7~40/8 x 7 =35 years and 1~ 40/8= 5 years. So mothers age today is 35+5= 40 years and daughter's age today is 5+5=10 years.

Most of the problems on ages fall into one or more of these categories

You can now evaluate yourself with this test.

F: S

7:1 <-- 3 Years Ago

5:1 <-- Present

Here if the ratios were the actual ages then Present ratio should be more than that of 3 years ago. Also the same years would be added to father's age and son's age. So what we do here is we make the ratio of present age to be more than that of 3 years ago and also make the difference of the ratio to be equal. To do that we find the difference of the ratios and cross multiply as:

F: S

7:1 <-- Difference is 6

5:1<--- Difference is 4 Thus the ratio of difference is 6:4 or 3:2, cross multiplying the upper ratio by 2 and lower by 3, we have:

F: S

14:2 <-- three years ago

15:3 < --Present

-------

1 :1

-----

Here we see that the difference is 1: 1 which is same and also the present age is more than that of 3 years ago.

Now the calculations is very simple :1 year difference in ratio--> actual 3 years difference so Father's age that is 15 present ---> actual will be 3/1 x 15=45 years and son's age will be 3/1 x 3 = 9 years

This concept becomes very handy when we solve similar problems in ratio and proportions.

Similarly if Three years earlier the father's age 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. We go like this:

F: S

7:1 <- Three years earlier

4:1 <-- Three years hence

----

3:0

----

To make the difference same we cross multiply by the difference

F: S

7:1 <--6~2 in ratio

\ /

/ \

4:1 <--3~1 in ratio

F: S

7:1 <-- -3

8:2<-- +3

---

1:1

---

Now if difference in ratio is 1--> Actual difference is 3+3=6 years. If the age 3 years earlier is 7 in ratio--> actual age is 6/1 x 7= 42. So present age is 42+3=45. similarly 1-->6 then 1-->6 so present age of son is 6+3=9 years.

The concept is very simple and gives the answer almost intuitively.

Sometimes the questions are asked like this : The sum of ages of a mother and her daughter is 50 years. Also 5 years ago, the mother's age was 7 times the age of the daughter. What are the present ages of the mother and the daughter. Here 5 years ago, the sum of the ages will be 40 years. Divide 40 in the ratio of 7:1, Thus 8~40 so 7~40/8 x 7 =35 years and 1~ 40/8= 5 years. So mothers age today is 35+5= 40 years and daughter's age today is 5+5=10 years.

Most of the problems on ages fall into one or more of these categories

You can now evaluate yourself with this test.

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