Question

divide 56 into two parts such that three times the first part exceeds one third of the second by 48 how to solve this in easy way​

Answer 1

Answer:

First part=20 and second part=36

Given:

Two parts such that three times the first part exceeds

One third of second by 48

To find:

The numbers

Solution:-

Let one part be x

other part be 56-x

Now

According to the question

$3(x) - \frac{1}{3}(56 - x) = 48$

$3x - \frac{56}{3} - \frac{x}{3} = 48$

$9x - 56 - x = 144$

$10x = 200$

$x = 20$

__________

One part=20

other part=56-x

=56-20

=36

_____________________

☆Verification☆

Total number=56

One part + other part = 56

20+36 = 56

56 = 56

L.HS=R.HS

Answer 2

Answer:

${\pink{\green{\sf{\therefore First\:number=20}}}}$

${\pink{\green{\sf{\therefore Second\:number=36}}}}$

Step-by-step explanation:

$\huge{\pink{\green{\underline{\red{\sf{SOLUTION-}}}}}}$

• In the given question information given about a nunber which is divided into two parts such that three times the first part exceeds one third of the second part by 48.

• So, we have to find both the numbers.

$\underline \bold{Given : } \\ \implies Let \:first\:number = x \\ \implies Second \: number = 56- x \\ \\ \underline \bold{To \: Find : } \\ \implies First \: number = ? \\ \implies Second \: number = ?$

• According to given question :

$\implies 3x - \frac{56 - x}{3} = 48 \\ \implies \frac{9x -( 56 - x)}{3 } = 48 \\ \implies 10x - 56 = 144 \\ \implies 10x = 144 + 56 \\ \implies x = \frac{200}{10} \\ \bold{\implies x = 20} \\ \\ \bold{ \therefore First \:number = x = 20} \\ \bold{ \therefore Second \: second= 56 - x = 56 - 20 = 36}$