Question

the difference of 2 nos is 80 and the greater no exceeds twice the smaller by 11. Find the no.​

Answer 1

Answer:

Step-by-step explanation:

Hint: Let the greater number be x and the smaller number be y.

Then, form linear equations using the information provided in the question.

Finally, solve the pair of linear equations to get the values of x and y i.e. greater number and smaller number respectively.

Complete step-by-step answer:

Let the greater number be x and the smaller number be y.

It is given that the sum of the numbers is 80.

⇒x+y=80 … (1)

Also, it is given that the greater number exceeds twice the smaller number by 11.

∴x=2y+11 … (2)

Now, substituting the equation (2) in equation (1) to get the value of y.

⇒2y+11+y=80⇒3y=80−11⇒3y=69⇒y=693⇒y=23

Substituting y = 23, in equation (1) gives the value of x.

⇒x+23=80⇒x=80−23⇒x=57

Thus, the greater number is 57 and the smaller number is 23.

Note: Alternate method:

Let the greater number be x and the smaller number be y.

It is given that the sum of the numbers is 80.

⇒x+y=80 … (1)

Also, it is given that the greater number exceeds twice the smaller number by 11.

⇒x=2y+11

⇒x−2y=11 … (2)

Now, we will find the value of y, by multiplying the equation (1) by -1.

⇒−1(x+y)=−80⇒−x−y=−80

So,

−x−y=−80x−2y=11

−3y=−69

⇒y=−69−3⇒y=23

Substituting y = 23, in equation (1) gives the value of x.

 ⇒x+23=80⇒x=80−23⇒x=57  

Thus, the greater number is 57 and the smaller number is 23.