Having revised the concepts covered in the Class 8 Maths Chapter 16, let us look at the solutions for the exercise questions.
Exercise 16.1
1.
Solution: Adding the unit’s digits we see A + 5 = 2 in the unit’s digit. The only possible solution for this is 7 + 5 = 12 since 2 is smaller than 5. Thus, A = 7. Carrying over the 1 from 12, we get the next step, i.e., the addition of ten’s digits as (1) + 3 + 2 = B. So, B = 6.
2.
Solution: Adding the unit’s digits we see A + 8 = 3 in the unit’s digit. The possible solution for this is 5 + 8 = 13. Thus, A = 5. Carrying over the 1 from 13, we get the next step, i.e., the addition of ten’s digits as (1) + 4 + 9 = B. So, B = 4, with 1 being carried over. Thus, C = 1.
3.
Solution: Checking the multiplication of the unit’s digits we find that A x A= A in the unit’s digit. The possible values for A, in this case, can be 0, 1, 5, and 6.



We know that 0 x 0 = 0. Since the final product is not 0, this is not the correct answer.

Next, if A = 1, the equation becomes 11 x 1 = 91, which is not correct.

For A = 5, we see that 15 x 5 = 75, so our value does not satisfy the equation.

Checking for A = 6, we find that 16 x 6 = 96. Thus, the correct solution is A = 6.
4.
Solution: Adding the unit’s digits we see B + 7 = A in the unit’s digit. Also, adding the ten’s digits gives us the equation A + 3 = 6.
Let us call these equations as
i. B + 7 = A
ii. A + 3 = 6

Let us see what happens when we take A = 3. We know that 3 + 3 = 6. So, equation ii. becomes correct. But, B + 7 = 3 means that B will be 6, in which case we get a carryover as 1 and equation ii. becomes incorrect. Thus, A cannot be 3.

Taking A = 2, gives us the following solution.

B + 7 = 2 => B = 5. So, 5 + 7 = 2 with 1 carried over.

(1) + 2 + 3 = 6. This equation is also correct. Hence, our solution becomes A = 2, B = 5.
5.
Solution: Checking the multiplication of the unit’s digits we find that B x 3= B in the unit’s digit. There are 2 possible solutions for this B = 5 or 0. Taking the value of B as 5, we get (A x 3) + 1 = A with C as carryover. No value of A can solve this equation. Now, consider B=0. This gives us the value of C=1 and A=5.
6.
Solution: Checking the multiplication of the unit’s digits we find that B x 5= B in the unit’s digit. There are 2 possible solutions for this B = 5 or 0. Taking B=0, we get (A x 5) = A with C as carryover. So, we get the following solution A =5, B = 0, C = 2. Now, taking B=5, we get the next equation as (Ax5)+2=A. There are 2 possible values of A that can satisfy this equation, A = 2 or 7. Considering both we get the values as A=2,B=5,C=1 and A=7,B=5,C=3.
7.
Solution: Checking the multiplication of the unit’s digits we find that B x 6= B in the unit’s digit. There are 2 possible solutions for this B = 4 or 6. Taking B=4, we get (A x6)+2 = 4 with 4 as carryover. A=7 gives us the equation (7x6)+2 = 44. Thus, a possible solution is A=7 and B=4. Now checking with B=6, we get (Ax6)+3=66. There is no value of A that solves this equation.
8.
Solution: Adding the unit’s digits we see 1+B=0 in the unit’s digit. The only possible solution for this is B=9. Thus, addition of tens digits gives us, (1)+A+1=9. So, A=7.
9.
Solution: Adding the unit’s digits we see B+1=8 in the unit’s digit. The only possible solution for this is B=7. Thus, addition of tens digits gives us, A+7=1. Solving this, we get A=4.
10.
Solution: Adding the unit’s digits we see A+B=9 in the unit’s digit. Next, on adding tens digits we get 2+A=0. The possible value of A=8. This satisfies the addition of hundreds digits (1)+1+6=8. Taking A=8, we get B=1.