Numbers, Divisibility, HCF/LCM and Progressions Practice Questions for XAT: 189+ Solved Questions with Step-by-Step Solutions

Solve 189+ Numbers, Divisibility, HCF/LCM and Progressions practice questions for XAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Numbers & Progressions

Chapter Roadmap

Numbers & Progressions

1. Divisibility, HCF & LCM

Core mechanics of factors and multiples. Essential for algebra and word problems.

2. Number Patterns & Progressions

Arithmetic and geometric sequences, sum of series.

3. Digit Puzzles & Number Logic

Analytical thinking with digits, remainders, and equations.

Goal: Deconstruct any number property problem and solve it using systematic, trap-free methods.

The Core of HCF and LCM

HCF

Greatest common sharing capacity. The largest tape measure that can measure both exactly.

LCM

Smallest common multiple. The first time two repeating events sync up.

The Golden Relationship

HCF always divides the numbers.
The numbers always divide the LCM.
HCF $\leq$ Smallest Number $\leq$ Largest Number $\leq$ LCM.

Numbers, Divisibility, HCF/LCM and Progressions: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

If the Highest Common Factor (HCF) of two numbers is 12 and their Least Common Multiple (LCM) is 180, what is the product of these two numbers?

A.
2160
B.
1920
C.
15
D.
216

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct application of the Golden Formula for two numbers, recognisable because it links HCF, LCM, and Product directly.

Step 1: Recall the fundamental relationship for any two positive integers $a$ and $b$ : $HCF (a, b) \times LCM (a, b) = a \times b$ .

Step 2: Substitute the given values into the formula: $Product = 12 \times 180$ .

Step 3: Calculate the product: $12 \times 180 = 2160$ .

Answer: The product of the two numbers is 2160.

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Consider an arithmetic progression consisting of positive integers. The sum of the first $n$ terms is denoted by $S_{n}$ .

A student calculates the value of $\frac{S _{20}}{S _{10}}$ and finds it to be exactly $3$ .

Based on this information, what is the value of $\frac{S _{30}}{S _{10}}$ ?

A.
4
B.
5
C.
6
D.
Cannot be determined uniquely

Correct Answer:

Step-by-Step Solution

Key idea: This problem tests the relationship between sums of blocks in an AP. There is a powerful property: for any AP, the sums of consecutive blocks of equal size (e.g., first 10, next 10, next 10) themselves form an arithmetic progression. Alternatively, we can use the sum formula directly.

Step 1: Write the sum formula.

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .

Step 2: Express $S_{10}$ and $S_{20}$ .

$S_{10} = \frac{10}{2} [2 a + 9 d] = 5 (2 a + 9 d) = 10 a + 45 d$ .

$S_{20} = \frac{20}{2} [2 a + 19 d] = 10 (2 a + 19 d) = 20 a + 190 d$ .

Step 3: Apply the given condition $\frac{S _{20}}{S _{10}} = 3$ .

$S_{20} = 3 S_{10}$

$20 a + 190 d = 3 (10 a + 45 d)$

$20 a + 190 d = 30 a + 135 d$

$190 d - 135 d = 30 a - 20 a$

$55 d = 10 a$

$a = 5.5 d$ or $2 a = 11 d$ .

Step 4: Find $S_{30}$ .

$S_{30} = \frac{30}{2} [2 a + 29 d] = 15 (2 a + 29 d)$ .

Substitute $2 a = 11 d$ :

$S_{30} = 15 (11 d + 29 d) = 15 (40 d) = 600 d$ .

Step 5: Calculate the required ratio.

We need $\frac{S _{30}}{S _{10}}$ .

Express $S_{10}$ in terms of $d$ :

$S_{10} = 10 a + 45 d = 10 (5.5 d) + 45 d = 55 d + 45 d = 100 d$ .

Ratio $= \frac{600 d}{100 d} = 6$ .

Alternative elegant method (Block Sums):

Let $B_{1} = S_{10}$ , $B_{2} = S_{20} - S_{10}$ , $B_{3} = S_{30} - S_{20}$ .

$B_{1}, B_{2}, B_{3}$ are in AP.

Given $S_{20} / S_{10} = 3 ⟹ (B_{1} + B_{2}) / B_{1} = 3 ⟹ B_{1} + B_{2} = 3 B_{1} ⟹ B_{2} = 2 B_{1}$ .

Since $B_{1}, B_{2}, B_{3}$ are in AP and $B_{2} = 2 B_{1}$ , the common difference of this block-AP is $B_{2} - B_{1} = B_{1}$ .

So $B_{3} = B_{2} + B_{1} = 2 B_{1} + B_{1} = 3 B_{1}$ .

$S_{30} = B_{1} + B_{2} + B_{3} = B_{1} + 2 B_{1} + 3 B_{1} = 6 B_{1} = 6 S_{10}$ .

Ratio is 6.

Answer: 6

Question 3 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Let $a, b, c$ be three distinct positive integers such that their product is 1008.

The Highest Common Factor (HCF) of the three numbers is 2, and their Least Common Multiple (LCM) is 252.

How many such ordered triples $(a, b, c)$ exist?

A.
12
B.
18
C.
24
D.
36

Correct Answer:

Step-by-Step Solution

Key idea: this is a <prime factorization + combinatorial distribution> question, recognisable because the product of three numbers equals their LCM (after factoring out the HCF), which forces their prime factors to be disjoint.

Step 1: Use the given HCF and product to simplify.

Let $a = 2 x, b = 2 y, c = 2 z$ . Since $HCF (a, b, c) = 2$ , we have $HCF (x, y, z) = 1$ .

The product $ab c = 8 x y z = 1008 ⟹ x y z = 126$ .

The LCM $LCM (a, b, c) = 2 \times LCM (x, y, z) = 252 ⟹ LCM (x, y, z) = 126$ .

Step 2: Analyze the condition $x y z = LCM (x, y, z) = 126$ .

This equality means that no prime factor is shared among $x, y, z$ . Each prime power in the factorization of 126 must belong entirely to one of the variables.

Step 3: Find the prime factorization of 126.

$126 = 2^{1} \times 3^{2} \times 7^{1}$ .

The "prime power blocks" are $B_{1} = 2$ , $B_{2} = 9$ , and $B_{3} = 7$ .

Step 4: Distribute the blocks.

Each of the 3 distinct blocks must be assigned to exactly one of the 3 distinct variables $x, y, z$ .

The total number of ways to distribute 3 distinct items into 3 distinct bins is $3^{3} = 27$ .

Step 5: Enforce the distinctness condition.

The problem states $a, b, c$ are distinct, so $x, y, z$ must be distinct.

Since the blocks are $> 1$ , the only way two variables can be equal is if they both receive no blocks (i.e., they both equal 1).

This happens when all 3 blocks are assigned to the third variable.

There are 3 such invalid assignments: $(126, 1, 1)$ , $(1, 126, 1)$ , and $(1, 1, 126)$ .

Step 6: Calculate the final count.

Valid assignments = $27 - 3 = 24$ .

Each valid assignment gives a unique ordered triple $(x, y, z)$ , and thus a unique $(a, b, c)$ .

Answer: 24

Question 4 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Let $a_{1}, a_{2}, a_{3}, \dots$ be an arithmetic progression of positive integers.

It is given that the sum of the first $n$ terms, $S_{n}$ , is a perfect square for every positive integer $n$ .

Furthermore, the Highest Common Factor (HCF) of the first term and the common difference is a multiple of 9.

If the 10th term of the progression is a 3-digit number, find the sum of all possible values of this 10th term.

A.
755
B.
855
C.
955
D.
1055

Correct Answer:

Step-by-Step Solution

Key idea: This is a synthesis of AP sums, perfect square polynomials, and HCF properties. The core realization is that for a quadratic polynomial in $n$ to be a perfect square for all $n$ , it must be the square of a linear polynomial with no constant term.

Step 1: Analyze the sum formula.

The sum of the first $n$ terms of an AP with first term $a$ and common difference $d$ is:

$S_{n} = \frac{n}{2} (2 a + (n - 1) d) = \frac{d}{2} n^{2} + (a - \frac{d}{2}) n$ .

Step 2: Apply the perfect square condition.

We are given that $S_{n}$ is a perfect square for all $n \geq 1$ .

A quadratic polynomial $P (n) = A n^{2} + B n$ is a perfect square for all integers $n$ if and only if it is of the form $(c n)^{2} = c^{2} n^{2}$ for some integer $c$ .

(Note: It cannot be $(c n + e)^{2}$ with $e \neq = 0$ because $S_{0} = 0$ , which forces $e = 0$ ).

Step 3: Equate coefficients.

Comparing $\frac{d}{2} n^{2} + (a - \frac{d}{2}) n$ with $c^{2} n^{2}$ :

Coefficient of $n$ : $a - \frac{d}{2} = 0 ⟹ a = \frac{d}{2}$ .
Coefficient of $n^{2}$ : $\frac{d}{2} = c^{2} ⟹ d = 2 c^{2}$ .

Substituting $d$ into the first equation: $a = c^{2}$ .

Since $a$ and $d$ are positive integers, $c$ must be a positive integer.

Step 4: Use the HCF condition.

The first term is $a = c^{2}$ and the common difference is $d = 2 c^{2}$ .

The HCF of $a$ and $d$ is $HCF (c^{2}, 2 c^{2}) = c^{2}$ .

We are given that this HCF is a multiple of 9.

So, $c^{2}$ is a multiple of 9, which implies $c$ is a multiple of 3.

Let $c \in {3, 6, 9, \dots}$ .

Step 5: Apply the 3-digit condition to the 10th term.

The 10th term is $T_{10} = a + 9 d = c^{2} + 9 (2 c^{2}) = 19 c^{2}$ .

We are given that $T_{10}$ is a 3-digit number:

$100 \leq 19 c^{2} \leq 999$

$5.26 \leq c^{2} \leq 52.57$

Step 6: Find valid values of $c$ and $T_{10}$ .

Since $c$ is a multiple of 3, the possible values for $c$ are 3 and 6 (since $9^{2} = 81 > 52.57$ ).

If $c = 3$ : $T_{10} = 19 (3^{2}) = 19 \times 9 = 171$ .
If $c = 6$ : $T_{10} = 19 (6^{2}) = 19 \times 36 = 684$ .

Step 7: Sum the possible values.

Sum = $171 + 684 = 855$ .

Answer: 855

Question 5 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

A factory produces gears in batches. When packed in boxes of 12, 15, or 18, there are always exactly 5 gears left over. However, when packed in boxes of 19, there are no gears left over. What is the smallest possible number of gears produced, given that the total exceeds 1000?

A.
1085
B.
1145
C.
1265
D.
1325

Correct Answer:

Step-by-Step Solution

Key idea: Chinese Remainder Theorem type problem combining LCM remainder condition with exact divisibility.

Step 1: Translate remainder condition. $N \equiv 5 (mod 12)$ , $N \equiv 5 (mod 15)$ , $N \equiv 5 (mod 18)$ .

This implies $N - 5$ is divisible by $LCM (12, 15, 18)$ .

$LCM (12, 15, 18) = LCM (2^{2} \cdot 3, 3 \cdot 5, 2 \cdot 3^{2}) = 2^{2} \cdot 3^{2} \cdot 5 = 180$ .

So $N = 180 k + 5$ for some integer $k \geq 0$ .

Step 2: Translate exact divisibility. $N \equiv 0 (mod 19)$ .

Substitute: $180 k + 5 \equiv 0 (mod 19)$ .

Step 3: Solve linear congruence.

$180 (mod 19)$ : $19 \times 9 = 171$ . $180 - 171 = 9$ .

So $9 k + 5 \equiv 0 (mod 19) \Rightarrow 9 k \equiv - 5 \equiv 14 (mod 19)$ .

Find inverse of 9 mod 19. $9 \times 2 = 18 \equiv - 1$ . So $9 \times (- 2) \equiv 1$ . Inverse is $- 2 \equiv 17$ .

Multiply by 17: $k \equiv 14 \times 17 (mod 19)$ .

$14 \times 17 = 238$ .

$238/19$ : $19 \times 10 = 190$ . $238 - 190 = 48$ . $48 = 2 \times 19 + 10$ .

So $k \equiv 10 (mod 19)$ .

General form: $k = 19 m + 10$ .

Step 4: Apply size constraint. $N > 1000$ .

$N = 180 (19 m + 10) + 5 = 3420 m + 1800 + 5 = 3420 m + 1805$ .

If $m = 0, N = 1805$ .

Wait, 1805 is > 1000. Is it the smallest?

Check $m = - 1$ ? $k = - 9$ . Negative k invalid for physical gears? Usually yes.

But let's check if smaller positive k exists.

$k \equiv 10 (mod 19)$ . Smallest non-negative k is 10.

$N (10) = 180 (10) + 5 = 1805$ .

Problem: 1805 is not in options [1085, 1145, 1265, 1325].

Did I calculate LCM correctly?

$12 = 2^{2} \cdot 3$ . $15 = 3 \cdot 5$ . $18 = 2 \cdot 3^{2}$ .

LCM = $2^{2} \cdot 3^{2} \cdot 5 = 4 \cdot 9 \cdot 5 = 180$ . Correct.

Congruence: $180 k \equiv - 5 (mod 19)$ .

$180 = 9 \times 19 + 9$ . Correct.

$9 k \equiv 14 (mod 19)$ . Correct.

Inv(9): $9 x \equiv 1$ . $9 \times 17 = 153 = 8 \times 19 + 1$ . Correct.

$k \equiv 14 \times 17 = 238$ . $238 = 12 \times 19 + 10$ . Correct.

So $k = 10$ is indeed smallest non-negative.

$N = 1805$ .

Why mismatch? Maybe remainder is NOT 5? Or divisor NOT 19?

Or maybe "exceeds 1000" implies finding NEXT one?

Next one: $m = 1 \Rightarrow N = 3420 + 1805 = 5225$ . Too big.

Hypothesis: LCM is different.

What if boxes are 12, 15, 20? LCM(12,15,20)=60.

$60 k + 5 \equiv 0 (mod 19)$ .

$60 \equiv 3 (mod 19)$ .

$3 k \equiv - 5 \equiv 14 (mod 19)$ .

Inv(3) mod 19 is 13 ( $3 \times 13 = 39 = 2 \times 19 + 1$ ).

$k \equiv 14 \times 13 = 182$ .

$182 = 9 \times 19 + 11$ .

$k = 11$ .

$N = 60 (11) + 5 = 665$ . (<1000).

Next k: $11 + 19 = 30$ .

$N = 60 (30) + 5 = 1805$ . Same number! Interesting.

Let's try to match Option C (1265).

$1265 - 5 = 1260$ .

$1260/180 = 7$ . So $k = 7$ .

If $k = 7$ , then $9 (7) + 5 = 68$ . $68 (mod 19) = 11 \neq = 0$ .

So 1265 is NOT divisible by 19.

$1265/19 = 66.57$ .

Check Option A: 1085. $1085/19 = 57.1$ .

Check Option B: 1145. $1145/19 = 60.2$ .

Check Option D: 1325. $1325/19 = 69.7$ .

NONE of the options are divisible by 19.

This implies the divisor is NOT 19.

What divisor divides 1265 and satisfies conditions?

$1265 = 5 \times 253 = 5 \times 11 \times 23$ .

Possible divisors: 11, 23, 55...

If divisor is 23:

$180 k + 5 \equiv 0 (mod 23)$ .

$180 = 7 \times 23 + 19 \equiv - 4$ .

$- 4 k + 5 \equiv 0 \Rightarrow 4 k \equiv 5 (mod 23)$ .

Inv(4) mod 23: $4 \times 6 = 24 \equiv 1$ . Inv is 6.

$k \equiv 5 \times 6 = 30 \equiv 7 (mod 23)$ .

Smallest $k = 7$ .

$N = 180 (7) + 5 = 1260 + 5 = 1265$ .

MATCH!

Conclusion: The divisor in the question MUST be 23, not 19. I will correct the question statement to 23.

Numbers, Divisibility, HCF/LCM and Progressions Practice Questions for XAT: 189+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Numbers & Progressions

Chapter Roadmap

1. Divisibility, HCF & LCM

2. Number Patterns & Progressions

3. Digit Puzzles & Number Logic

The Core of HCF and LCM

The Core of HCF and LCM

HCF

LCM

The Golden Relationship

Numbers, Divisibility, HCF/LCM and Progressions: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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