Question

Four years ago, a mother was four times as old as her daughter. Six years later, the mother will be two and a half times as old as her daughter at that time. Find the present ages of mother and her
daughter.
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Answer 1

Solution :-

Let the the present ages of mother and her daughter be x years and y years respectively.

Case I : Four years ago, a mother was four times as old as her daughter.

=> x - 4 = 4(y - 4)

=> x - 4 = 4y - 16

=> x = 4y - 12 _______(i)

Case II : Six years later, the mother will be two and a half times as old as her daughter at that time.

=> x + 6 = 5/2(y + 6)

=>2(4y - 12 + 6) = 5y + 30 [from equation (i)]

=> 8y - 12 = 5y + 30

=> 3y = 42

=> y = 14

Putting the value of y in equation (i) we get,

=> x = 4 × 14 - 12

=> x = 56 - 12 = 44

Hence,

The present ages of mother and her daughter are 44 years and 14 years respectively.

Answer 2

Answer:

Step-by-step explanation:

To Find : The present ages of mother and daughter (x and y)

Solution : Let the present age of her mother be x and let the present age of daughter be y

Now in 1st case,

x - 4 = 4 (y-4)

x - 4 = 4y - 16

Therefore,  x - 4y = -12 _______ eq 1

Now in 2nd case,

x + 6 =   2 1/2(y + 6)  now, solve the mixed fraction

x + 6 = 5/ 2 (y + 6)

2x + 12 = 5y + 30

2x - 5y = 30 - 12

Therefore,  2x - 5y = 18 _______ eq 2

Now, from equation 1 we get,

x = -12 + 4y ____________eq 3  

Put the value of x in eq 2

2(-12 + 4y) - 5y = 18

-24 + 8y - 5y = 18

3y = 42

Therefore,  y = 14  

Now, put the value of y in eq 3

x = -12 + 4(14)

x = -12 + 56

x = 44  

Therefore, the present age of mother is 44 and the present age of daughter is 14.

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