Test Yourself
Q1: Multiple Choice Type
i. ab and ba are two 2-digit numbers then ab + ba is equal to:
Let ab = 10a + b and ba = 10b + a
Sum = (10a + b) + (10b + a) = 11a + 11b = 11(a + b)
Answer: c. 11(a + b)
ii. In a 2-digit number ab and a > b then ab − ba is equal to:
ab = 10a + b, ba = 10b + a
Difference = (10a + b) − (10b + a) = 9(a − b)
Answer: b. 9(a − b)
iii. If a = b, then ab ÷ ba is equal to:
Let a = b = 2 → ab = 22, ba = 22
Then ab ÷ ba = 22 ÷ 22 = 1
None of the given options show 1
Answer: d. none of these
iv. If a > b, then for 3-digit numbers abc and cba; we have abc − cba equals to:
Let abc = 100a + 10b + c, cba = 100c + 10b + a
Then abc − cba = (100a − a) + (10b − 10b) + (c − 100c)
= 99a − 99c = 99(a − c)
If a > c, then result is positive.
Try a = 4, b = 2, c = 1 → 421 − 124 = 297
Check: 99(a − c) = 99(3) = 297 ✔
So, abc − cba = 99(a − c)
Options do not show this. None are correct.
Answer: d. none of these
v. Find the value of A for:
3 2 A + 1 A 1 ------- 5 0 9
Let’s add column-wise:
Units: A + 1 = 9 → A = 8
Check: Tens: 2 + 8 = 10 → carry 1
Hundreds: 3 + 1 + carry = 5 ✔
Answer: c. 8
vi. Without actual calculation, the quotient, when the sum of 59 and 95 is divided by 11, is:
Step 1: Observe that 59 and 95 are reverse of each other.
Let the number be of the form: 10a + b
Then its reverse is: 10b + a
Step 2: Sum of number and its reverse =
(10a + b) + (10b + a) = 11(a + b) ✔
Step 3: Therefore, this sum is always divisible by 11.
Quotient = a + b ✔
Now apply to 59 and 95:
→ 59: a = 5, b = 9 ⇒ a + b = 5 + 9 = 14
→ So, quotient = 14 ✔
Answer:
a. 14
vii. Without actual calculation, the quotient when 74 − 47 is divided by 9, is:
Step 1: Recognize the numbers:
74 and 47 are reverse of each other.
Let the original number be of the form: 10a + b
Its reverse = 10b + a
Step 2: Now compute their difference:
(10a + b) − (10b + a) = 9(a − b) ✔
This shows that the difference is always divisible by 9.
Step 3: Hence, the quotient is:
\(\frac{74 – 47}{9} = a – b\)
In 74, a = 7, b = 4 ⇒ a − b = 3
Answer:
c. 3
viii. The smallest two digit number 10a + 4 is divisible by 6 if a + 4 divisible by:
For number to be divisible by 6, it must be divisible by 2 and 3.
10a + 4 ends in 4 → even ✔
Sum = a + 4 must be divisible by 3
So, a = 2, Number = 24
So, a + 4 (2 + 4 = 6) divisible by 6
Answer: b. 6
ix. 3z4 divisible by 9, if the digit z is equal to:
Sum = 3 + z + 4 = z + 7
z + 7 divisible by 9 → z = 2 ✔
Answer: b. 2
x. Statement 1: When the sum of a two-digit number and the number obtained by reversing its digits is divided by 11, the quotient is equal to the sum of the two digits.
Statement 2: When the sum of a two digit number and the number obtained by reversing its digits is divided by sum of the two digits, the quotient is always 11.
Which of the following options is correct?
Let number be ab = 10a + b and reversed = ba = 10b + a
Sum = ab + ba = 11(a + b)
Divide by 11 → (a + b) ✔ Statement 1 is true
Now divide by (a + b): 11(a + b) ÷ (a + b) = 11 ✔
Statement 2 is also true
Answer: a. Both the statements are true.
xi. Assertion (A): \(759=100\times7+10\times5+1\times9\)
Reason (R): In a three-digit number 100p + 10q + 1r, the digit p at hundred’s place is any whole number from 0 to 9, the digit q at ten’s place is any whole number from 0 to 9 and the digit r at unit place is any whole number from 0 to 9.
Step 1: 759 = 100×7 + 10×5 + 1×9 = 700 + 50 + 9 = 759 ✔
Step 2: Given: Digit p at hundred’s place is any whole number from 0 to 9.
If p = 0, then it will become 2 digit number.
So reason is False.
Answer: c. Statement 1 is true, and statement 2 is false.
xii. Assertion (A): If 36p52q9 is divisible by 9, then p + q = 2
Reason (R): A number is divisible by 3 if the sum of its digits is divisible by 3.
Step 1: Divisibility rule for 9: sum of digits must be divisible by 9 ✔
Sum = 3 + 6 + p + 5 + 2 + q + 9 = (25 + p + q)
Let (25 + p + q) be divisible by 9
Try p + q = 2 → 25 + 2 = 27, which is divisible by 9 ✔
So Assertion is correct.
Step 2: Reason (R) talks about divisibility rule of 3, which is correct.
Answer: b. Both A and R are correct, and R is not the correct explanation for A.
xiii. Assertion (A): Factors of the sum of a three-digit number 542 and the numbers obtained by changing the order of the digits cyclically are 1, 11, 111, 5 + 4 + 2.
Reason (R): The sum of a three-digit number and the two numbers obtained by changing the digits cyclically is completely divisible by (i) 11, (ii) 111 and (iii) the sum of the digits.
Cyclic numbers: 542, 425, 254
Sum = 542 + 425 + 254 = 1221
Now, 1221 ÷ 11 = 111 ✔
1221 ÷ 111 = 11 ✔
Sum of digits = 5 + 4 + 2 = 11 ✔ → 1221 ÷ 11 = 111 ✔
So, A is true and R clearly explains it.
Answer: a. Both A and R are correct, and R is the correct explanation for A.
xiv. Assertion (A): 2574 is divisible by 11 but 7083 is not divisible by 11.
Reason (R): A number is divisible by 11 if the difference between the sum of its digits in even places and the sum of its digits in odd places is either 0 or divisible by 11.
2574: (2 + 7) − (5 + 4) = 9 − 9 = 0 ✔ → divisible by 11
7083: (7 + 8) − (0 + 3) = 15 − 3 = 12 ❌ not divisible by 11
So A is correct. Now verify Reason:
R gives correct rule of 11 (alternate sum) ✔
Answer: a. Both A and R are correct, and R is the correct explanation for A.
Q2: For three 3-digit numbers abc, cab and bca, show that abc + cab + bca is divisible by 37.
Step 1: Let the digits a, b, and c be at hundreds, tens, and units place respectively in number abc.
Then:
abc = 100a + 10b + c
cab = 100c + 10a + b
bca = 100b + 10c + a
Step 2: Add all three numbers:
abc + cab + bca = (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)
Step 3: Group like terms:
= (100a + 10a + a) + (10b + b + 100b) + (c + 100c + 10c)
= a(100 + 10 + 1) + b(10 + 1 + 100) + c(1 + 100 + 10)
= a×111 + b×111 + c×111
= 111(a + b + c)
Step 4: Now check divisibility:
111 = 3 × 37 ✔
So, 111(a + b + c) is divisible by 37 ✔
Answer: abc + cab + bca is divisible by 37.
Q3: If the three-digit number 24x is divisible by 9, find the value of digit x.
Step 1: Use the divisibility rule of 9:
A number is divisible by 9 if the **sum of its digits** is divisible by 9.
Step 2: Given number = 24x
Sum of digits = 2 + 4 + x = 6 + x
Step 3: We want 6 + x to be divisible by 9.
Try values of x from 0 to 9:
6 + x = 9 → x = 3 ✔
This is the smallest value that makes the sum divisible by 9.
Answer: x = 3
Q4: If 31y5 is divisible by 3, find the value(s) of digit y.
Step 1: Use the divisibility rule of 3:
A number is divisible by 3 if the **sum of its digits** is divisible by 3.
Step 2: Given number = 31y5
Sum of digits = 3 + 1 + y + 5 = 9 + y
Step 3: We want (9 + y) to be divisible by 3.
So, check values of y from 0 to 9 such that (9 + y) is divisible by 3:
Values that satisfy:
9 + 0 = 9 ✔
9 + 3 = 12 ✔
9 + 6 = 15 ✔
9 + 9 = 18 ✔
Answer: y = 0, 3, 6, or 9
Q5: In a 3-digit number, the ten’s digit is thrice the unit digit and hundred’s digit is twice the unit digit. If the sum of all the three digits is 12, find the number.
Step 1: Let the unit digit = x
Then ten’s digit = 3x
Hundred’s digit = 2x
Step 2: According to the question:
Sum of digits = x + 3x + 2x = 6x = 12
Step 3: Solve for x:
6x = 12 ⇒ x = 2
Step 4: Now plug values:
Unit digit = 2
Ten’s digit = 3×2 = 6
Hundred’s digit = 2×2 = 4
So, the number = 462
Answer: 462
Q6: Find the digits A, B and C, if
7 3 A - B C 9 --------- 3 4 8
Step 1: Start from the units place:
A − 9 should give 8, but 8 is less than 9, so borrowing is needed.
So A + 10 − 9 = 8 ⇒ A + 1 = 8 ⇒ A = 7 ✔
Step 2: Now move to the tens place.
We had borrowed 1 from 3 (tens digit), so it becomes 2.
2 − C = 4 is not possible (since 2 − C = 4 ⇒ C = -2 ❌)
So again we borrow 1 from 7 (hundreds), making 2 into 12.
Now: 12 − C = 4 ⇒ C = 8 ✔
Step 3: Finally, the hundreds place:
We borrowed 1 from 7, so it becomes 6.
6 − B = 3 ⇒ B = 3 ✔
Step 4: Verify all digits:
7 3 7 - 3 8 9 --------- 3 4 8 ✔
Answer: A = 7, B = 3, C = 8
Q7: Find the digits A and B
i.
2 A B + A B 1 --------- B 1 8
Step 1: Units place: B + 1 = 8 → B = 7 ✔
Step 2: Tens place: A + 7 = 1 (with carry 1)
So A + 7 + 1 = 11 → A = 3 ✔
Step 3: Hundreds place: 2 + A = B (with carry 1)
2 + 3 + 1 = 6 → B = 6 ❌, but we earlier found B = 7
Try again: 2 + A = B (carry 1 already used)
2 + 3 = 5 → not 7 ❌
Try A = 4 → 2 + 4 = 6, still not 7 ❌
But A = 4, B = 7 satisfies tens and units:
Check full addition:
2 4 7 + 4 7 1 ------- 7 1 8 ✔
Answer: A = 4, B = 7
ii.
A B + 3 7 ------ 9 A
Step 1: Units: B + 7 = A → Cannot find the value of 2 variables simmultaneously
Step 2: Tens: A + 3 = 9 → impossible without carry → A + 3 + 1 = 9 ⇒ A = 5
Step 3: Units: B + 7 = A → B + 7 = 15 ⇒ B = 8
Verify:
5 8 + 3 7 ----- 9 5
Answer: A = 5, B = 8
iii.
1 B × B ------ 9 B
Try B = 3 → 1B = 13, 13 × 3 = 39 → not in form 9B ❌
Try B = 6 → 1B = 16, 16 × 6 = 96 ✔
So B = 6
Answer: B = 6
iv.
8 B 4 - 5 1 B --------- B 2 1
Step 1: Units: 4 − B = 1 → B = 3 ✔
Step 2: Tens: B − 1 = 1 → 3 − 1 = 2 ✔
Step 3: Hundreds: 8 − 5 = B → 8 − 5 = 3✔
Answer: B = 3
