Question

Divide 56 into two parts such that three times the first part exceeds one third of the second by 48. the parts are
by 48. The parts are.
a) (20,36) b) (25,31)
c) (24,32)
d) none of these​

Answer 1

$\bf\large\underline\pink{Answer:-}$

a) (20,36)

=============================

let one part be x

so the other part is 56-x

ATQ,

3x-(56-x/3)=48

(9x-56+x)/3=48

10x-56=144

10x=200

x=20

so one part is 20 and the other part is 36.

Answer 2

$\huge\bold\red{ANSWER }$

$\rule{300}{2}$

Let the 1st part be x

then, Other part is (56 - x)

$\rule{300}{2}$

Three times the first part exceeds one third of the second part by 48.

$\rule{300}{2}$

$3x = \frac{1}{3}(56 - x) + 48 \\3x = \frac{56 - x + 144}{3} \\ 9x = 56 - x + 144 \\ 9x - x = 200 \\ 10x = 200 \\ x = 20$

so,$\bold\green{x=20}$

second part = 56 - x

= 56 - 20

= 36

$\boxed{first\:part = 20}$

$\boxed{Second\:part = 36}$