Exercise: 5-A
Q1: Multiple Choice Type
i. The quotient, when sum of 68 and 86 is divided by 14, is:
Step 1: Add 68 and 86.
68 + 86 = 154
Step 2: Divide the sum by 14.
154 ÷ 14 = 11
Answer: c. 11
ii. The quotient when 75 – 57 is divided by 9 is:
Step 1: Subtract 57 from 75.
75 – 57 = 18
Step 2: Divide the result by 9.
18 ÷ 9 = 2
Answer: b. 2
iii. Find the value of A for:
1 A + A 3 ----- 1 0 1
Step 1: Add the units digit: A + 3 should give 1 in units place.
Only possible if A + 3 = 11, so A = 8 (carry 1)
Step 2: Now tens place: 1 (carry) + 1 + 8 = 10 → tens digit is 0 and carry 1 again.
Step 3: Hundreds place: 1 + 0 = 1
So, A = 8 is correct.
Answer: d. 8
iv. The quotient, when difference of 53 and the number obtained on reversing its digits is divided by 2:
Step 1: Reverse digits of 53 → 35
Step 2: Find the difference: 53 – 35 = 18
Step 3: Divide by 2: 18 ÷ 2 = 9
Answer: b. 9
v. Find the value of A for:
1 2 A
- A 9
-------
4 8
Step 1: Subtract the units place: A – 9 = 8 → not possible unless we borrow.
So, borrow from 2: 12 becomes 11, and A becomes (10 + A)
Try A = 6 → 16 – 9 = 7 ✘
Try A = 7 → 17 – 9 = 8 ✔
Now check tens: 1 (borrowed) → 1 – 1 = 0 (since 2 became 1)
Then hundreds: 1 – A = 4 → 12 – A = 4 → A = 8 ✘
Try again properly:
Let’s test A = 7
12 7
– 7 9
= 4 8
✔ Works correctly
Answer: b. 7
Q2:
i. Find the quotient, when the sum of 94 and 49 is divided by:
Step 1: Observe that 94 and 49 are reverse of each other.
Let the original 2-digit number be ab = 10a + b
Then the reverse = ba = 10b + a
Step 2: Their sum = (10a + b) + (10b + a) = 11(a + b)
This shows that the sum of a two-digit number and the number obtained by reversing its digits is always divisible by 11 ✔
Step 3 (a): So, 94 + 49 = 143
Since 143 = 11 × 13
Quotient when divided by 11: 13 ✔
Step 4 (b): Similarly, 143 ÷ 13 = 11 ✔
So, this sum is also divisible by 13
Answer: a. 13 b. 11
ii. Find the quotient when the difference between 94 and 49 is divided by:
Step 1: Again use the property of reverse digits:
Difference = (10a + b) − (10b + a) = 9(a − b)
So, the difference is always divisible by 9 ✔
Step 2: 94 − 49 = 45
Check if divisible by 9: 9 × 5 = 45 ✔
Check if divisible by 5: 5 × 9 = 45 ✔
Answer:
a. 9 b. 5
Q3: If a = b, show that abc = bac.
Step 1: Given: a = b
So wherever we see ‘a’, we can replace it with ‘b’.
Step 2: Start with Left Hand Side (LHS): abc
Since a = b, replace ‘a’ with ‘b’:
abc = bbc
Step 3: Now check Right Hand Side (RHS): bac
Since b = a, we can also write:
bac = aac
But since a = b, aac = bbc
So, both sides become the same:
abc = bac = bbc
Answer: Hence proved: abc = bac when a = b.
Q4: Write the quotient when the difference between 694 and number obtained on interchanging its ones and hundreds digits is divided by:
Step 1: Original number = 694
Interchanging hundreds and units digit: 4 and 6
New number = 496
Step 2: Find the difference:
694 − 496 = 198
i. 33
Divide 198 by 33:
33 × 6 = 198 ✔
Quotient: 6
Answer: 6
ii. 99
Divide 198 by 99:
99 × 2 = 198 ✔
Quotient: 2
Answer: 2
iii. 2
Divide 198 by 2:
2 × 99 = 198 ✔
Quotient: 99
Answer: 99
Q5: If the sum of the number 842 and two other numbers obtained by arranging the digits in cyclic order is divided by:
Concept:
For any 3-digit number like \(abc\), the cyclic permutations are: \(abc,\ bca,\ cab\)
The sum of these numbers is always divisible by 111 and the sum of the digits.
This is similar to the property for 2-digit numbers, where:
The sum of a number and its reverse is divisible by 11.
Step 1: Given number = 842
Cyclic permutations = 842, 428, and 284
Step 2: Find the sum of digits:
8 + 4 + 2 = 14
i. 111
(842 + 428 + 284) ÷ 111 = 14 (Sum of digits)
Answer: 14
ii. 14
(842 + 428 + 284) ÷ 14 = 111
Answer: 111
iii. 37
(842 + 428 + 284) ÷ 37 = 42
Answer: 42
Q6: If a * b = a + 2b + 5, then find the value of 4 * 5.
Step 1: Given operation:
a * b = a + 2b + 5
Step 2: Substitute a = 4 and b = 5 into the expression:
4 * 5 = 4 + 2×5 + 5
Step 3: Simplify step-by-step:
= 4 + 10 + 5
= 19
Answer: 19
Q7: If a * b = 2a – b – 2, find the value of 8 * 7.
Step 1: Given operation:
a * b = 2a – b – 2
Step 2: Substitute a = 8 and b = 7:
8 * 7 = 2×8 – 7 – 2
Step 3: Simplify step-by-step:
= 16 – 7 – 2
= 9 – 2 = 7
Answer: 7
Q8: In each of the following cases, find the least value/value of letters used in place of digits:
a.
3 A + 2 5 ----- B 2
Step 1: A + 5 = 2 (units place)
But 2 < 5, so we take carry 1 ⇒ A + 5 = 12 ⇒ A = 7
Step 2: 3 + 2 + 1 (carry) = 6 ⇒ B = 6
Answer: A = 7, B = 6
b.
9 8 + 4 A ----- C B 3
Step 1: Units place: 8 + A = 3 → with carry
Take A = 5 ⇒ 8 + 5 = 13 ⇒ units digit = 3, carry = 1
Step 2: Tens: 9 + 4 + 1 (carry) = 14 ⇒ B = 4, carry = 1
Step 3: Hundreds: carry = 1 ⇒ C = 1
Answer: A = 5, B = 4, C = 1
c.
A 1 + 1 B ----- B 0
Step 1: Units: 1 + B = 0 → impossible without carry → 1 + B = 10 ⇒ B = 9
Step 2: Tens: A + 1 + carry = 9 ⇒ A = 7
Answer: A = 7, B = 9
d.
2 A B + A B 1 ------- B 1 8
Step 1: Units: B + 1 = 8 → B = 8 – 1 ⇒ B = 7
Step 2 Tens: A + B = 1 → impossible without carry → A + B = 11 ⇒ A + 7 = 11 ⇒ A = 11- 7 = 4
Verify:
2 4 7 + 4 7 1 ------- 7 1 8
Answer: A = 4, B = 7
e.
1 2 A + 6 A B ------- A 0 9
Try A = 7, B = 7:
1 2 7
+ 6 7 7
= 8 0 4 → ✘
Try A = 8, B = 1:
1 2 8
+ 6 8 1
= 8 0 9 ✔
Answer: A = 8, B = 1
f.
1 A × A ----- 9 A
Try A = 3:
13 × 3 = 39 → ✘
A = 4:
14 × 4 = 56 → ✘
A = 5:
15 × 5 = 75 → ✘
A = 6:
16 × 6 = 96 ✔
9A = 96, so A = 6
Answer: A = 6
g.
A B × 6 ----- B B B
Try B = 0
6 × A cannot be 00
Try B = 2
6 × 2 = 12, So. B = 2 carry 1
Now, 6 × A + 1 = 22, No number is exist for A which gives 22
Try B = 4
6 × 4 = 24, So. B = 4 carry 2
Now, 6 × A + 2 = 44, So A = 7
Answer: A = 7, B = 4
h.
A B × 3 ----- C A B
Try B = 0
3 × A = CA, only 5 satisfies the condition (3 × 5 = 15). So A = 5, C = 1
Answer: A = 5, B = 0, C = 1
i.
A B × 5 ----- C A B
Try B = 0
5 × A = CA, only 5 satisfies the condition (5 × 5 = 25). So A = 5, C = 2
Answer: A = 5, B = 0, C = 2
j.
8 A 5 + 9 4 A --------- 1 A 3 3
Step 1: Units: 5 + A = 3 → impossible without carry → 5 + A = 13 ⇒ A = 8
Verify:
8 8 5 + 9 4 8 --------- 1 8 3 3
Answer: A = 8
k.
6 A B 5 + D 5 8 C --------- 9 3 5 1
Step 1: Units: 5 + C = 1 → impossible without carry → 5 + C = 11 ⇒ C = 6
Step 2: Tens: B + 8 – 1 = 5 → impossible without carry → B + 8 + 1 = 15 ⇒ B = 6
Step 3: Hundreds: A + 5 + 1 = 3 → impossible without carry → A + 5 + 1 = 13 ⇒ A = 7
Step 4: Thousands: 6 + D + 1 = 9 ⇒ D = 2
Answer: A = 7, B = 6, C = 6, D = 2
Q9: For 3-digit number abc, what will be the quotient if abc – cba is divided by 11?
Step 1: Let the 3-digit number be represented as:
abc = 100a + 10b + c
cba = 100c + 10b + a
Step 2: Find the difference:
abc – cba = (100a + 10b + c) – (100c + 10b + a)
= 100a – a + c – 100c + 10b – 10b
= (99a – 99c)
Step 3: Factor the result:
abc – cba = 99(a – c)
Step 4: Divide by 11:
Quotient = [99(a – c)] ÷ 11
= 9(a – c)
Answer: 9(a − c)
