Answer :
Sure, let's break down the solution step by step for the given problem:
### (i) Write Sunayana's present age in terms of Bishwant's age (x)
Given that Sunayana is 7 years younger than Bishwant, if Bishwant's present age is [tex]\( x \)[/tex] years, then Sunayana's present age can be written as:
[tex]\[ \text{Sunayana's age} = x - 7 \][/tex]
### (ii) Find their present ages
We know the product of their ages is 540. Therefore, we can write this equation:
[tex]\[ x \times (x - 7) = 540 \][/tex]
This is a quadratic equation, which simplifies to:
[tex]\[ x^2 - 7x - 540 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = -540 \)[/tex], we get:
[tex]\[ x = \frac{7 \pm \sqrt{49 + 2160}}{2} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{2209}}{2} \][/tex]
[tex]\[ x = \frac{7 \pm 47}{2} \][/tex]
So, the two solutions are:
[tex]\[ x = \frac{7 + 47}{2} = 27 \][/tex]
[tex]\[ x = \frac{7 - 47}{2} = -20 \][/tex]
Since age cannot be negative, Bishwant's present age must be:
[tex]\[ x = 27 \][/tex]
Therefore, Sunayana's present age is:
[tex]\[ 27 - 7 = 20 \][/tex]
So, Bishwant is 27 years old and Sunayana is 20 years old.
### (iii) Find how many years ago the sum of the squares of their ages was 569
Let [tex]\( y \)[/tex] be the number of years ago that we are looking for.
The ages of Bishwant and Sunayana [tex]\( y \)[/tex] years ago would be:
- Bishwant's age [tex]\( y \)[/tex] years ago: [tex]\( 27 - y \)[/tex]
- Sunayana's age [tex]\( y \)[/tex] years ago: [tex]\( 20 - y \)[/tex]
The sum of the squares of their ages [tex]\( y \)[/tex] years ago is given as 569. Thus, we can write:
[tex]\[ (27 - y)^2 + (20 - y)^2 = 569 \][/tex]
Expanding the squares, we get:
[tex]\[ (27 - y)^2 = 729 - 54y + y^2 \][/tex]
[tex]\[ (20 - y)^2 = 400 - 40y + y^2 \][/tex]
Adding these, we get:
[tex]\[ 729 - 54y + y^2 + 400 - 40y + y^2 = 569 \][/tex]
[tex]\[ 1129 - 94y + 2y^2 = 569 \][/tex]
Simplifying, we get a quadratic equation in terms of [tex]\( y \)[/tex]:
[tex]\[ 2y^2 - 94y + 1129 - 569 = 0 \][/tex]
[tex]\[ 2y^2 - 94y + 560 = 0 \][/tex]
Dividing the whole equation by 2, we get:
[tex]\[ y^2 - 47y + 280 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -47 \)[/tex], and [tex]\( c = 280 \)[/tex]:
[tex]\[ y = \frac{47 \pm \sqrt{2209 - 1120}}{2} \][/tex]
[tex]\[ y = \frac{47 \pm \sqrt{1089}}{2} \][/tex]
[tex]\[ y = \frac{47 \pm 33}{2} \][/tex]
So, the two solutions are:
[tex]\[ y = \frac{47 + 33}{2} = 40 \][/tex]
[tex]\[ y = \frac{47 - 33}{2} = 7 \][/tex]
Thus, the sum of the squares of their ages was 569:
- 7 years ago, when Bishwant was [tex]\( 27 - 7 = 20 \)[/tex] and Sunayana was [tex]\( 20 - 7 = 13 \)[/tex].
- 40 years ago, when Bishwant was [tex]\(-13\)[/tex] and Sunayana was [tex]\(-20\)[/tex], but these are not realistic (negative ages).
Hence, the valid answer is:
[tex]\[ \boxed{7 \text{ years ago}} \][/tex]
### (i) Write Sunayana's present age in terms of Bishwant's age (x)
Given that Sunayana is 7 years younger than Bishwant, if Bishwant's present age is [tex]\( x \)[/tex] years, then Sunayana's present age can be written as:
[tex]\[ \text{Sunayana's age} = x - 7 \][/tex]
### (ii) Find their present ages
We know the product of their ages is 540. Therefore, we can write this equation:
[tex]\[ x \times (x - 7) = 540 \][/tex]
This is a quadratic equation, which simplifies to:
[tex]\[ x^2 - 7x - 540 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = -540 \)[/tex], we get:
[tex]\[ x = \frac{7 \pm \sqrt{49 + 2160}}{2} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{2209}}{2} \][/tex]
[tex]\[ x = \frac{7 \pm 47}{2} \][/tex]
So, the two solutions are:
[tex]\[ x = \frac{7 + 47}{2} = 27 \][/tex]
[tex]\[ x = \frac{7 - 47}{2} = -20 \][/tex]
Since age cannot be negative, Bishwant's present age must be:
[tex]\[ x = 27 \][/tex]
Therefore, Sunayana's present age is:
[tex]\[ 27 - 7 = 20 \][/tex]
So, Bishwant is 27 years old and Sunayana is 20 years old.
### (iii) Find how many years ago the sum of the squares of their ages was 569
Let [tex]\( y \)[/tex] be the number of years ago that we are looking for.
The ages of Bishwant and Sunayana [tex]\( y \)[/tex] years ago would be:
- Bishwant's age [tex]\( y \)[/tex] years ago: [tex]\( 27 - y \)[/tex]
- Sunayana's age [tex]\( y \)[/tex] years ago: [tex]\( 20 - y \)[/tex]
The sum of the squares of their ages [tex]\( y \)[/tex] years ago is given as 569. Thus, we can write:
[tex]\[ (27 - y)^2 + (20 - y)^2 = 569 \][/tex]
Expanding the squares, we get:
[tex]\[ (27 - y)^2 = 729 - 54y + y^2 \][/tex]
[tex]\[ (20 - y)^2 = 400 - 40y + y^2 \][/tex]
Adding these, we get:
[tex]\[ 729 - 54y + y^2 + 400 - 40y + y^2 = 569 \][/tex]
[tex]\[ 1129 - 94y + 2y^2 = 569 \][/tex]
Simplifying, we get a quadratic equation in terms of [tex]\( y \)[/tex]:
[tex]\[ 2y^2 - 94y + 1129 - 569 = 0 \][/tex]
[tex]\[ 2y^2 - 94y + 560 = 0 \][/tex]
Dividing the whole equation by 2, we get:
[tex]\[ y^2 - 47y + 280 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -47 \)[/tex], and [tex]\( c = 280 \)[/tex]:
[tex]\[ y = \frac{47 \pm \sqrt{2209 - 1120}}{2} \][/tex]
[tex]\[ y = \frac{47 \pm \sqrt{1089}}{2} \][/tex]
[tex]\[ y = \frac{47 \pm 33}{2} \][/tex]
So, the two solutions are:
[tex]\[ y = \frac{47 + 33}{2} = 40 \][/tex]
[tex]\[ y = \frac{47 - 33}{2} = 7 \][/tex]
Thus, the sum of the squares of their ages was 569:
- 7 years ago, when Bishwant was [tex]\( 27 - 7 = 20 \)[/tex] and Sunayana was [tex]\( 20 - 7 = 13 \)[/tex].
- 40 years ago, when Bishwant was [tex]\(-13\)[/tex] and Sunayana was [tex]\(-20\)[/tex], but these are not realistic (negative ages).
Hence, the valid answer is:
[tex]\[ \boxed{7 \text{ years ago}} \][/tex]
