Math, asked by amartyaas, 1 year ago

Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their age
in years was 48.
enel of nerimatar Romandarea
22​

Answers

Answered by Steph0303
14

Answer:

Let the age of friends be 'x' and 'y'.

⇒ x + y = 20

⇒ x = 20 - y ... ( 1 )

Also, it is given that 4 years ago, the product of their ages was 48 years.

⇒ ( x - 4 ) ( y - 4 ) = 20

Substituting the value of x from ( 1 ) we get,

⇒ ( 20 - y - 4 ) ( y -  4 ) = 48

⇒ ( 16 - y ) ( y -  4 ) = 48

⇒ 16y - y² - 64 + 4y = 48

⇒ -y² + 20y - 64 = 48

⇒ y² - 20y + 64 + 48 = 0

⇒ y² - 20y = 112 = 0

Since we cannot factorise the above condition, we can say that the given situation is not possible.

Answered by Anonymous
27

Let present age of one friend be (M) years and present age of another friend be (N) years.

» The sum of the ages of two friends is 20 years.

According to question,

→ M + N = 20

→ M = 20 - N ________ (eq 1)

» Four years ago, the product of their age in years was 48.

According to question

→ (M - 4) (N - 4) = 48

→ M(N - 4) -4(N - 4) = 48

→ MN - 4M - 4N + 16 = 48

→ MN - 4M - 4N = 48 - 16

→ MN - 4M - 4N = 32

→ (20 - N)N - 4(20 - N) - 4N = 32 [From (eq 1)]

→ 20N - N² - 80 + 4N - 4N = 32

→ - N² + 20N - 80 - 32 = 0

→ - N² + 20N - 112 = 0

→ N² - 20N + 112 = 0

Here.. a = 1, b = - 20 and c = 112

Now.. solve it by quadratic formula

D = √[(b)² - 4ac]

D = √[(-20)² - 4(1)(112)]

D = √(400 - 448)

D = √(-48)

From above calculations it is clear that.. we cannot solve the above equation or we cannot factorise the above equation.

So, the given situation is not possible.

____________________________

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