Question

I)the difference between the two natural numbers is 12 and the difference between their squares is 600.find the sum of these two numbers. also find the numbers.​

Answer 1

Answer:

$\sf{The \ numbers \ are \ 31 \ and \ 19 \ respectively, \ their}$

$\sf{sum \ is \ 50.}$

Given:

• The difference between the two natural numbers is 12.

• The difference between their squares is 600.

To find:

• The numbers and the sum of the numbers.

Solution:

$\sf{Let \ the \ two \ numbers \ be \ a \ and \ b \ respectively. }$

$\sf{According \ to \ the \ first \ condition. }$

$\sf{a-b=12...(1)}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{a^{2}-b^{2}=600...(2)}$

$\sf{By \ identity}$

$\sf{a^{2}-b^{2}=(a+b)(a-b)}$

$\sf{...from \ (1) \ and \ (2), \ we \ get}$

$\sf{600=12(a+b)}$

$\sf{\therefore{a+b=\dfrac{600}{12}}}$

$\sf{\therefore{a+b=50...(3)}}$

$\sf{Add \ equations \ (1) \ and \ (3) \ we \ get,}$

$\sf{a-b=12}$

$\sf{+}$

$\sf{a+b=50}$

__________________

$\sf{2a=62}$

$\sf{\therefore{a=\dfrac{62}{2}}}$

$\boxed{\sf{\therefore{a=31}}}$

$\sf{Substitute \ a=31 \ in \ equation(3), \ we \ get}$

$\sf{31+b=50}$

$\sf{\therefore{b=50-31}}$

$\boxed{\sf{\therefore{b=19}}}$

$\sf\purple{\tt{\therefore{The \ numbers \ are \ 31 \ and \ 19 \ respectively,}}}$

$\sf\purple{\tt{their \ sum \ is \ 50.}}$

Answer 2

Let ,

The two numbers are x and y

Given ,

The difference between the two natural numbers is 12

Thus ,

x - y = 12 --- (i)

The difference between their squares is 600

Thus ,

(x)² - (y)² = 600 --- (II)

We know that ,

$\rm \large \fbox{ {(a)}^{2} - {(b)}^{2} = (a + b)(a - b) }$

Thus ,

600 = (x + y) × 12

(x + y) = 50 --- (iii)

Subtract eq (i) from (iii) , we get

2y = 38

y = 19

Put the value y = 19 in eq (i) , we get

x - 19 = 12

x = 31

$\sf \therefore \underline{x + y = 50 \: and \: x = 31 \:, \: y = 19 }$