Question

Two numbers are such that whose sum and difference is
11 and 7 respectively, then the
difference between their respective cubes is ?
(A) 721
(B) 729
(C)737
(D)321​

Answer 1

Answer:

Let us assume those two numbers be x and y respectively whereas x>y.

According to your question:

• x+y=11--------i
• x-y=7--------ii

Adding and Subtracting equation i and ii we get:

=>x=9

&

=>y=2

We have to find difference b/w their respective cubes.

cube of x(x³) is 729 and cube of y(y³) is 343

Therefore its difference will be.

=> x³-y³=729-343=386(your answer)

Answer 2

Given :

• Two numbers are such that whose sum and difference is  11 and 7 respectively.

To find :

• Difference b/w their respective cubes

Solution :

Let the two numbers be a and b respectively.  [a > b]

ATQ,

$\longrightarrow\sf a + b = 11 \dots \dots \dots (\sf i)$

$\longrightarrow\sf a - b = 7 \dots \dots \dots (\sf ii)$

Now adding (i) and (ii) we get,

$\longrightarrow\sf a + b + a - b = 11 + 7 \\\\ \\ \longrightarrow\sf 2a = 18 \\\\ \\ \longrightarrow\sf a = \dfrac{18}{2} \\\\ \\ \longrightarrow\displaystyle\underline{\boxed{\bold{a = 9 }}}$

Now putting value of a in (i) we get,

$\longrightarrow\sf 9 + b = 11 \\\\ \\ \longrightarrow\sf b = 11 - 9 \\\\ \\ \longrightarrow\displaystyle\underline{\boxed{\bold{b = 2 }}}$

$\therefore\bold{Two \ numbers \ are \ 11 \ and \ 2 \ respectively.}$

Now difference b/w their respective cubes,

$\longrightarrow\sf Difference = a^3 - b^3 \\\\ \\ \longrightarrow\sf Difference = 9^3 - 2^3 \\\\ \\ \longrightarrow\sf Difference = 729 - 8 \\\\ \\ \longrightarrow\large\underline{\boxed{\bold{Difference = 721 }}} \ \bigstar$

$\therefore\underline{\underline{\textbf{Difference between their respective cubes = 721 }}}$