Question

Dvide 180 into two parts such that first part is 12 less than twice the second part

Answer 1

$\mathfrak{\large{\underline{\underline{Answer :-}}}}$

The numbers are 64 and 116.

$\mathfrak{\large{\underline{\underline{Explanation :-}}}}$

Given :

The number = 180

First part should be 12 less than twice the second part

To find :

The numbers

Solution :

Consider the second part as x

Consider the first part as 2x - 12

$\tt{\implies} \: (2x - 12) + x = 180$

$\tt{\implies} \: 2x + x =180 + 12$

$\tt{\implies} \: 3x = 192$

$\tt{\implies} \: x = \dfrac{192}{3}$

$\tt{\implies} \: x = 64$

Second part is 64.

Value of 2x - 12

$\tt{\implies} \: 2x - 12$

$\tt{\implies} \: 64 \times 2 - 12$

$\tt{\implies} \: 128 - 12$

$\tt{\implies} \: 116$

One number = 64

Second Number = 116

$\therefore$ The numbers are 64 and 116.

$\mathfrak{\large{\underline{\underline{Verification :- }}}}$

$\tt{\implies} \: (128 - 12) + 64 = 180$

$\tt{\implies} \: 116 + 64 = 180$

$\tt{\implies} \: 180 = 180$

$\therefore$ The numbers are 64 and 116.

Answer 2

$\huge\boxed{\fcolorbox{orange}{red}{Answer}}$

• The first part =2x-12
• The second part=x

All the parts =180

$2x - 12 + x = 180$

$3x = 180 + 12$

$3x = 192$

$\times = 192 \div 3$

$x = 64$

Hence,

Second part=x=64

First part=2x-12=2 .

(64-12=128)-12)

(128−12)+64=180

$\tt{\implies}\:116+64=180$

**116+64=180**

Answer =116