The product of two successive integral multiples of 5 is 300. Determin

1.	The product of two successive integral multiples of 5 is 300. Determine the multiples.
Answer» Given that the product of two successive integral multiples of 5 is 300 Let’s assume the integers be 5x and 5(x+1), where x and x+1 are two consecutive multiples Then, according to the question, we have 5x[5(x + 1)] = 300 ⇒ 25x(x + 1) = 300 ⇒ x²+ x = 12 ⇒ x²+ x – 12 = 0 Solving for x by factorization method, we have ⇒ x²+ 4x – 3x – 12 = 0 ⇒ x(x + 4) – 3(x + 4) = 0 ⇒ (x + 4)(x – 3) = 0 Now, either x + 4 = 0 ⇒ x = -4 Or, x – 3 = 0 ⇒ x = 3 For, x = – 4 5x = – 20 and 5(x + 1) = -15 And, for x = 3 5x = 15 and 5(x + 1) = 20 Thus, the two successive integral multiples can be 15, 20 or -15 and -20 respectively.

The product of two successive integral multiples of 5 is 300. Determine the multiples.

Answer»

Given that the product of two successive integral multiples of 5 is 300

Let’s assume the integers be 5x and 5(x+1), where x and x+1 are two consecutive multiples

Then, according to the question, we have

5x[5(x + 1)] = 300

⇒ 25x(x + 1) = 300

⇒ x²+ x = 12

⇒ x²+ x – 12 = 0

Solving for x by factorization method, we have

⇒ x²+ 4x – 3x – 12 = 0

⇒ x(x + 4) – 3(x + 4) = 0

⇒ (x + 4)(x – 3) = 0

Now, either x + 4 = 0 ⇒ x = -4

Or, x – 3 = 0 ⇒ x = 3

For, x = – 4

5x = – 20 and 5(x + 1) = -15

And, for x = 3

5x = 15 and 5(x + 1) = 20

Thus, the two successive integral multiples can be 15, 20 or -15 and -20 respectively.