Let the present age of father be x years and the present age of his son be y years.
After 10 years, father’s age will be (x+10) years and son’s age will be (y + 10) years. Thus using the given information, we have
`x +10 =2(y+10)`
`⇒ x+ 10 = 2y+ 20`
` ⇒ x - 2y -10 =0`
Before 10 years, the age of father was `(x - 10)` years and the age of son was `(y - 10)` years. Thus using the given information, we have
` x-10 =12 (y -10)`
`⇒ x + 10 = 2y +20`
`⇒ x - 12y +110 =0`
So, we have two equations
`x -2y -10 =0`
` x- 12y +110 =0`
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
` x / ((-2)xx110 -(-12)xx(-10))=(-y)/(1xx110 -1(-10))= 1/(1xx(-12)-1xx(-2))`
`⇒ x / (-220-120)=(-y)/(110+10)+ 1/(-12+2)`
`⇒ x/-340 =(-y)/120 = 1/(-10)`
`⇒ x/(340)= y/120 = 1/10`
`⇒ x = 340/10,y=120/10`
`⇒ x = 34 , y = 12`
Hence, the present age of father is 34 years and the present age of son is 12 years.
Let’s the present ages of the father as a years and that of his son’s age as b years.
According to the question,,
After 10 years, the age of father will be $(a+10)$ years and son’s age will be $(b+10)$ years.
So, the equation is
$a+10=2(b+10)$
$a–10=2b+20$
$a–2b–10=0$……… (i)
Also again from the question it’s given as,
Ago 10 years, the age of father was $(a–10)$ years and the age of son was $(b–10)$ years.
therefore, the relation between their 10 years ago is given below
$a–10=12(b–10)$
$a–10=12b–120$
$a–12b+110=0$……… (ii)
Therefore, by solving (i) and (ii), we get the solution
Using cross-multiplication method , then
$\begin{array}{l}
\frac{a}{{( – 2) \times 110 – ( – 12) \times ( – 10)}} = \frac{b}{{1 \times 110 – 1 \times ( – 10)}} = \frac{1}{{1 \times ( – 12) – 1 \times ( – 12)}}\\
\frac{a}{{ – 220 – 120}} = \frac{{ – b}}{{110 + 10}} = \frac{1}{{ – 12 + 2}}\\
\frac{a}{{ – 340}} = \frac{{ – b}}{{120}} = \frac{1}{{ – 10}}\\
\frac{a}{{340}} = \frac{b}{{120}} = \frac{1}{{10}}
\end{array}$
⇒ $a=34$, $b=12$
So, the present age of father is $34$ years and the present age of the son is $12$ years.
Solution:
The present age of father = 42 years
The present age of son = 14 years
(a) Ratio of the present age of father to the present age of son = Present age of father / Present age of the son
= 42 / 14
= 3 : 1
Therefore, 3:1 is the ratio of the present age of the father to the present age of the son.
(b) The son's age will be 12 two years back (14 - 2 = 12) , so we will calculate the father's age two years ago.
Hence, two years ago father's age will be = 42 – 2 = 40 years
Hence, required ratio = 40 / 12
= (4 × 10) / (4 × 3)
= 10 / 3
Therefore, 10:3 is the ratio of the father's age to the son's age, when the son was 12 years old.
(c) Father will be 52 years old after 10 years ( 42 + 10 = 52 years)
Son's age after 10 years = 14 + 10 = 24 years
Hence, required ratio = 52 / 24
= (4 × 13) / (4 × 6)
= 13 / 6
Therefore, the ratio of the father's age after 10 years to the son's age after 10 years is 13 : 6.
(d) 12 years ago father’s age was 30 years (42 - 12 = 30)
Age of son at that time = 14 – 12 = 2 years
Required ratio = 30 / 2
= (2 × 15) / 2
= 15 / 1
Therefore, the ratio of father's age to the son's age when father was 30 years old is 15:1
NCERT Solutions for Class 6 Maths Chapter 12 Exercise 12.1 Question 16
Summary:
It is given that the present age of the father is 42 years and that of his son is 14 years. The ratio of (a) the Present age of the father to the present age of the son is 3:1. (b) Age of the father to the age of the son, when the son was 12 years old is 10:3.c(c) Age of father after 10 years to the age of son after 10 years is 13:6 (d) Age of father to the age of son when father was 30 years old is 15:1
☛ Related Questions:
Answer
Hint: Assume the age of father to be y, the age of son to be x and let after z years father’s age will be twice his son’s age. From the question above create two linear equations in x,y and z. Solve for z by eliminating the variables x and y. If the elimination is not possible, then there is no definite answer to the question. Otherwise, the value of z which is found by eliminating x and y is the answer.
Complete step-by-step solution -
Let the age of son be x, the age of father be y and let after z years fathers age become twice that of his son.Since 10 years ago father’s age was 35 years more than twice the age of his son we havey-10 = 2(x-10)+35Applying distributive law in RHS, we get y-10 = 2x -20 + 35i.e. y-10 = 2x+15Subtracting 15 from both sides, we gety-10-15 = 2x+15-15i.e. y-25 = 2x (i)Also, since z years from now father’s age is twice that of his son, we havey+z = 2(x+z)Applying distributive law in RHS, we gety+z = 2x+2zSubtracting z from both sides, we gety+z-z= 2x+2z-zi.e. y = 2x+zSubstituting the value of 2x from equation (i) we gety= y-25+zAdding 25-y on both sides, we gety+25-y = y -25+z+25-yi.e. z = 25.Hence 25 years from now father’s age will be twice that of his son’s age.Note: [1] Although the given system of equations has infinitely many solutions, the value of z = 25, for the system to have a solution. Hence in the given system x and y take infinitely many values, but z is fixed. The given system can be visualised as a line in a plane parallel to the x-y plane at a distance of z = 25 from it.[2] Age of an individual is an integer. Hence x, y and z only take integral values. Hence if the given system of equations had solutions at non-integral values of z, e.g. 25.5, then the age was not deterministic, and in fact, no such situation is possible.
Let father's and son's age, 10 years ago, be f and s respectively.
From the given information,
f = 2s + 35 ... (i)
Present age of father = f + 10
Present age of son = s + 10
Let after x number of years, father be twice his son's age. Then, we have:
f + 10 + x = 2(s + 10 + x)
On simplifying, we get, x = f - 2s - 10
Using (i), we get,
x = 35 - 10 = 25
Thus, after 25 years father will be twice his son's age.
Answered by Expert 23rd May 2012, 10:51 PM
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