Question

Two numbers are such that their sum is 19 and their product is 8 times the greater number.The numbers are

Answer 1

Given :-

The sum of two numbers is 19.

Let the two numbers be x and y respectively.

➡ x + y = 19

➡ x = 19 - y ------(i)

ATQ,

The product of the two numbers is 8 times the greater number.

Let the greater number among the two numbers be x.

Therefore xy = 8x

➡ (19 - y)y = 8(19 - y) [from equation (i)]

➡ 19y - y² = 152 - 8y

➡ -y² + 19y + 8y - 152 = 0

➡ y² - 27y + 152 = 0

By factorisation, we get

➡ y² - (19y + 8y) + 152 = 0

➡ y² - 19y - 8y + 152 = 0

➡ y(y - 19) - 8(y - 19) = 0

➡ (y - 19) (y - 8) = 0

➡ y = 19 or y = 8

If y = 19, then x = 19 - 19 = 0 (which is not possible cause xy will be = 0)

If y = 8, then x = 19 - 8 = 11

Hence the two numbers are 8 and 11.

Answer 2

Let two numbers be M and N.

Sum of two numbers is 19.

=> M + N = 19

=> M = 19 - N _____ (eq 1)

The product of numbers is 8 times the greater number.

Let the greater number be M and smaller number be N.

According to question,

=> MN = 8M

=> (19 - N)N = 8(19 - N) [From (eq 1)]

=> 19N - N² = 152 - 8N

=> - N² + 19N + 8N - 152 = 0

=> N² - 27N + 152 = 0

=> N² - 19N - 8N + 152 = 0

=> N(N - 19) -8(N - 19) = 0

=> (N - 8) (N - 19) = 0

=> N = 8, 19

Put value of N in (eq 1)

If N = 8

=> M = 19 - 8

=> M = 11

If N = 19

=> M = 19 - 19

=> M = 0

(Neglected)

∴ Numbers are 8 and 11

______________________________

☆ Verification :

From above calculations we have M = 11 and N = 8

Put value of M and N in (eq 1)

=> 11 = 19 - 8

=> 11 = 11