Question

The sum of two numbers is 80 and the greater number exceeds twice the smaller by 11.find the numbers.​

Answer 1

Question:

The sum of two numbers is 80 and the greater number exceeds twice the smaller by 11.Find the numbers?

Answer:

Greater no. = x = 57

Smaller no. = 23

Solution:

It is said that,

The sum of two numbers is 80.

Thus,

Let the greater number be "x" then smaller number will be "(80-x)"

Also,

It is given that,

The greater number exceeds twice the smaller by 11.

Thus,

=> Greater no. = 2•(Smaller no.) + 11

=> x = 2•(80-x) + 11

=> x = 160 - 2x + 11

=> x + 2x = 160 + 11

=> 3x = 171

=> x = 171/3

=> x = 57

Hence,

Greater no. = x = 57

Smaller no. = 80-x = 80-57 = 23

Answer 2

AnswEr :

• Sum of Two Numbers is 80.
• Greater No. is 11 more twice than Smaller No.
• Find the Numbers.

Let's assume the Smaller No. be x

» Greater No. = 11 + Twice Smaller No.

» Greater No. = 11 + 2x

• According to Question Now ;

⇒ Greater No. + Smaller No. = 80

• Plugging the Values

⇒ ( 11 + 2x ) + ( x ) = 80

⇒ 11 + 3x = 80

⇒ 3x = 80 - 11

⇒ 3x = 69

• Dividing Both term by 3

⇒ x = 23

_________________________________

◑ Greater No.

» 11 + 2x

» 11 + 2 × 23

» 11 + 46

» 57

◑ Smaller No.

» x

» 23

჻ Required Numbers are 57 and 23.