Question

The product of two consecutive multiples of three is four hundred sixty-eight
more than twenty times their sum. Find the second multiple.

Answer 1

Given - The product of two consecutive multiples of three is four hundred sixty-eight more than twenty times their sum.

Find - Second multiple.

Solution - Let the two consecutive multiples of three be X and X+3.

The product of these multiples will be - X(X+3).

The sum will be - X+X+3 - 2x+3.

Twenty times of their sum will be represented as - 20(2x+3).

Since it is 468 more then twenty time of sum, the final equation will be represented as -

x(X+3) = 468 + 20(2x+3)

x²+3x = 468 + 40x + 60

x² - 37x - 528 = 0

37 ± ✓(37² + 4*528*1)/2*1

37 ± ✓(1369+2112)/2

37 ± 59/2

Considering only addition, 96/2

48

First multiple - X - 48

Second multiple - X+3 - 48+3 - 51

Thus, second multiple is 51.

Answer 2

SOLUTION

GIVEN

The product of two consecutive multiples of three is four hundred sixty-eight more than twenty times their sum.

TO DETERMINE

The second multiple.

EVALUATION

Let the two consecutive multiples of three are 3x and 3x + 3

So by the given condition

$\sf{3x(3x + 3) = 20(3x +3x + 3) + 468 }$

$\sf{ \implies \: 9x(x + 1) = 60(2x + 1) + 468 }$

$\sf{ \implies \: 3x(x + 1) = 20(2x + 1) + 156 }$

$\sf{ \implies \: 3 {x}^{2} + 3x = 40x + 20+ 156 }$

$\sf{ \implies \: 3 {x}^{2} - 37x - 176 = 0}$

$\sf{ \implies \: 3 {x}^{2} - 48x + 11x - 176 = 0}$

$\sf{ \implies \: 3x(x - 16) + 11(x - 16) = 0}$

$\sf{ \implies \: (x - 16)(3x + 11) = 0}$

$\displaystyle \sf{ \implies \:x = 16 \: , \: - \frac{11}{3} }$

∴ x = 16

So the required consecutive multiples of three are 48 and 51

Hence the required second multiple = 51

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