Q. 6

# Solved each

For unit’s place,

We have two conditions, when 7 + B ≤ 9 and 7 + B > 9

For 7 + B ≤ 9

7 + B = A

A B = 7 (1)

In ten’ place,

B + A = 8 …(2)

Solving 1 and 2 simultaneously,

2A = 15 which means A = 7.5 which is not possible

Hence, our condition 7 + B ≤ 9 is wrong.

7 + B > 9 is correct condition

Hence, carrying one in ten’s place and subtracting 10 from unit’s place,

7 + B – 10 = A

B A = 3 (3)

For ten’s place,

B + A + 1 = 8

B + A =7 (4)

Solving 3 and 4 simultaneously,

2B = 10

B = 5 and A = 2

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses RELATED QUESTIONS :

Solved eachRD Sharma - Mathematics

If the 4-digit nuRS Aggarwal - Mathematics

If 1A2B5 is exactRS Aggarwal - Mathematics

Find all possibleRS Aggarwal - Mathematics

If 37y4 is exactlRS Aggarwal - Mathematics

Replace A, RS Aggarwal - Mathematics

If 486*7 isRS Aggarwal - Mathematics

Find the vaRS Aggarwal - Mathematics

Solved eachRD Sharma - Mathematics

Solved eachRD Sharma - Mathematics