Question

If the product of two numbers is 21 and their difference is 4, then the ratio of the sum of the cubes of the numbers to the difference of the cubes of the numbers is
[a] 185:165
[b] 165:158
[c] 185:158
[d] 158:145

Answer 1

Hint: Assume that the numbers are x and y. Using the fact that the product of numbers is 21 and the difference is 4, form two equations in two variables x and y. Eliminate y from the equations and hence form a quadratic equation in x. Solve for x and hence find the numbers. Hence find the ratio of the sum of the cubes of the numbers to the difference of the cubes of the numbers.

Complete step-by-step answer:
Let the two numbers be x and y, with x>y.
Since the product of the two number is 21, we have
$xy=21\text{ }\left( i \right)$
Since the difference of the number is 4, we have
$x-y=4\text{ }\left( ii \right)$
From equation (i), we have
$xy=21$
Dividing both sides by x, we get
$y=\dfrac{21}{x}$
Substituting the value of y in equation (ii), we get
$x-\dfrac{21}{x}=4$
Multiplying both sides of the equation by x, we get
${{x}^{2}}-21=4x$
Subtracting 4x from both sides, we get
${{x}^{2}}-4x-21=0$
We will use splitting the middle term method to find the roots of the equation
We have $-7x+3x=-4x$ and $\left( -7 \right)\times \left( 3 \right)=-21$
Hence, we have
${{x}^{2}}-4x-21={{x}^{2}}-7x+3x+\left( -7 \right)\times 3=0$
From the first two terms taking x as common and from the last two terms taking 3 as common, we get
$x\left( x-7 \right)+3\left( x-7 \right)=0$
Taking x-7 common from LHS, we get
$\left( x-7 \right)\left( x+3 \right)=0$
Using zero product property, i.e. if ab = 0, the a =0 or b= 0, we get
$x-7=0\text{ or }x+3=0$
Hence, we have
x=7 or x = -3
When x = 7, we have from equation (ii)
$7-y=4$
Adding y on both sides, we get
$4+y=7$
Subtracting 4 from both sides, we get
$y=3$
When x = -3, we have from equation (ii)
$-3-y=4$
Adding y on both sides, we get
$4+y=-3$
Subtracting 4 from both sides, we get
y =-7
Hence the numbers are
(7,3) or (-3,-7)
For the pair 3,7, we have
Sum of cubes $={{7}^{3}}+{{3}^{3}}=343+27=370$
Difference of cubes $={{7}^{3}}-{{3}^{3}}=316$
Hence, we have
$\dfrac{\text{Sum of cubes}}{\text{Difference of cubes}}=\dfrac{370}{316}=\dfrac{185}{158}$
Hence the ratio of the sum of the cubes to the difference of the cubes is 185:158
For the pair -3,-7, we have
Sum of cubes $=-{{7}^{3}}-{{3}^{3}}=-343-27=-370$
Difference of cubes $=-{{7}^{3}}+{{3}^{3}}=-316$
Hence, we have
$\dfrac{\text{Sum of cubes}}{\text{Difference of cubes}}=\dfrac{-370}{-316}=\dfrac{185}{158}$
Hence the ratio of the sum of the cubes to the difference of the cubes is 185:158
Hence we conclude that option [c] is correct.

Note: [1] A student can make a mistake in reporting the answer. Instead of reporting the ratio of the sum of cubes to the difference of cubes, he may return the ratio of the cubes and end up having incorrect results.
[2] Solutions involving solving of equations should often be verified by substituting the values.
Here we have product of numbers =(7)(3) = (-7)(-3) = 21
Difference of the numbers = 7-3 = -3-(-7) = 4
Hence the solution is verified to be correct.