Numbers, Divisibility, HCF/LCM and Progressions Previous Year Questions (PYQs) for XAT: 6+ Solved Questions with Step-by-Step Solutions

Solve 6+ Numbers, Divisibility, HCF/LCM and Progressions previous year 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

The Madhura Fruits Company is packing four types of fruits into boxes. There are 126 oranges, 162 apples, 198 guavas and 306 pears. The fruits must be packed in such a way that a given box must have only one type of fruit and must contain the same number of fruit units as any other box.

What is the minimum number of boxes that must be used?

A.
21
B.
18
C.
44
D.
42
E.
36

Correct Answer:

Step-by-Step Solution

Key idea: This is an HCF and distribution problem. The trigger is packing items into boxes of equal size with no leftovers, while minimizing the total number of boxes.

Step 1: Understand the conditions.

Each box contains only one type of fruit.
Every box has the same number of fruits.
We need to minimize the number of boxes.

Step 2: Relate to HCF.

To minimize the number of boxes, we must maximize the number of fruits per box. The number of fruits per box must perfectly divide the total count of each fruit type. Therefore, the capacity of each box is the Highest Common Factor (HCF) of the quantities.

Step 3: Calculate the HCF.

Quantities: 126, 162, 198, 306.

Prime factorizations:

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

$162 = 2 \times 3^{4}$

$198 = 2 \times 3^{2} \times 11$

$306 = 2 \times 3^{2} \times 17$

The common factors are $2$ and $3^{2}$ .

$HCF = 2 \times 3^{2} = 18$ .

So, each box contains 18 fruits.

Step 4: Calculate the number of boxes for each fruit.

Oranges: $126 \div 18 = 7$ boxes.

Apples: $162 \div 18 = 9$ boxes.

Guavas: $198 \div 18 = 11$ boxes.

Pears: $306 \div 18 = 17$ boxes.

Step 5: Find the total number of boxes.

Total boxes = $7 + 9 + 11 + 17 = 44$ .

Answer: 44 (Option C).

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

125 is multiplied by either 10 or 15. The resultant number is again multiplied by either 10 or 15. This process continues.

Which of the following CANNOT be a resultant number at any point in time?

A.
$2^{235} \times 3^{453} \times 5^{691}$
B.
$2^{253} \times 3^{824} \times 5^{1080}$
C.
$2^{689} \times 3^{912} \times 5^{1604}$
D.
$2^{476} \times 3^{455} \times 5^{1034}$
E.
$2^{689} \times 3^{453} \times 5^{1145}$

Correct Answer:

Step-by-Step Solution

Key idea: This is an invariant-tracking question using prime factorization. The number is repeatedly multiplied by 10 or 15, so the exponents of 2, 3 and 5 must satisfy a fixed linear relation.

Step 1: Factor the starting number and the multipliers.

Starting number:

$125 = 5^{3}$ .

Multipliers:

$10 = 2 \times 5$ ,

$15 = 3 \times 5$ .

Step 2: Let the number of multiplications by 10 be $a$ , and by 15 be $b$ .

Each multiplication by 10 adds one factor of 2 and one factor of 5.

Each multiplication by 15 adds one factor of 3 and one factor of 5.

Therefore the final number has the form:

$N = 5^{3} \times (2 \times 5)^{a} \times (3 \times 5)^{b}$ .

Step 3: Simplify and create the invariant.

$N = 2^{a} \times 3^{b} \times 5^{3 + a + b}$ .

Let the final exponents be $E_{2}$ , $E_{3}$ and $E_{5}$ .

Then:

$E_{2} = a$ ,

$E_{3} = b$ ,

$E_{5} = 3 + a + b$ .

Substitute $a = E_{2}$ and $b = E_{3}$ :

$E_{5} = 3 + E_{2} + E_{3}$ .

Therefore every valid resultant number must satisfy:

$E_{5} - E_{2} - E_{3} = 3$ .

Step 4: Test the options using the invariant.

Option A: $E_{2} = 235$ , $E_{3} = 453$ , $E_{5} = 691$ .

$691 - 235 - 453 = 3$ . Valid.

Option B: $E_{2} = 253$ , $E_{3} = 824$ , $E_{5} = 1080$ .

$1080 - 253 - 824 = 3$ . Valid.

Option C: $E_{2} = 689$ , $E_{3} = 912$ , $E_{5} = 1604$ .

$1604 - 689 - 912 = 3$ . Valid.

Option D: $E_{2} = 476$ , $E_{3} = 455$ , $E_{5} = 1034$ .

$1034 - 476 - 455 = 103$ , not 3.

Equivalently, if $E_{2} = 476$ and $E_{3} = 455$ , the required $E_{5}$ would be:

$3 + 476 + 455 = 934$ , not 1034.

So this violates the invariant.

Option E: $E_{2} = 689$ , $E_{3} = 453$ , $E_{5} = 1145$ .

$1145 - 689 - 453 = 3$ . Valid.

Therefore the only option that cannot be a resultant number is option D.

Answer: D.

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

Separately, Jack and Sristi invested the same amount of money in a stock market. Jack’s invested amount kept getting reduced by 50% every month. Sristi’s investment also reduced every month, but in an arithmetic progression with a common difference of Rs. 15000. They both withdrew their respective amounts at the end of the sixth month. They observed that if they had withdrawn their respective amounts at the end of the fourth month, the ratio of their amounts would have been the same as the ratio after the sixth month.
What amount of money was invested by Jack in the stock market?

A.
Rs. 100000
B.
Rs. 120000
C.
Rs. 150000
D.
Rs. 180000
E.
None of the above

Correct Answer:

Step-by-Step Solution

Key idea: This is a mixed progression question. Jack’s amount decays geometrically because it reduces by 50% every month. Sristi’s amount decays arithmetically because it reduces by a fixed rupee amount every month. The equal-ratio condition at two different months gives an equation for the initial investment.

Step 1: Define the initial investment.

Let both Jack and Sristi initially invest $P$ .

Step 2: Model Jack’s amount.

Jack’s amount reduces by 50% every month, so each month it is multiplied by $\frac{1}{2}$ .

After $t$ months:

$J_{t} = P (\frac{1}{2})^{t}$ .

Therefore:

$J_{4} = \frac{P}{16}$ ,

$J_{6} = \frac{P}{64}$ .

Step 3: Model Sristi’s amount.

Sristi’s amount reduces by Rs. 15000 every month, which is an arithmetic decrease.

After $t$ months:

$S_{t} = P - 15000 t$ .

Therefore:

$S_{4} = P - 60000$ ,

$S_{6} = P - 90000$ .

Step 4: Use the ratio condition.

The ratio of Jack’s amount to Sristi’s amount is the same at month 4 and month 6:

$\frac{J _{4}}{S _{4}} = \frac{J _{6}}{S _{6}}$ .

Rearrange by dividing both sides appropriately:

$\frac{J _{6}}{J _{4}} = \frac{S _{6}}{S _{4}}$ .

Step 5: Find Jack’s scaling factor from month 4 to month 6.

$\frac{J _{6}}{J _{4}} = \frac{P /64}{P /16} = \frac{16}{64} = \frac{1}{4}$ .

Since the ratios are equal, Sristi’s amount must also scale by $\frac{1}{4}$ :

$\frac{S _{6}}{S _{4}} = \frac{1}{4}$ .

Step 6: Solve for $P$ .

$\frac{P - 90000}{P - 60000} = \frac{1}{4}$ .

Cross-multiply:

$4 (P - 90000) = P - 60000$ .

Expand:

$4 P - 360000 = P - 60000$ .

Move terms:

$4 P - P = 360000 - 60000$ .

$3 P = 300000$ .

Hence:

$P = 100000$ .

Step 7: Check validity.

$S_{4} = 100000 - 60000 = 40000$ .

$S_{6} = 100000 - 90000 = 10000$ .

Both amounts are positive, and $\frac{10000}{40000} = \frac{1}{4}$ , matching Jack’s scaling.

Answer: A.

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

Ronny uses a 5-digit key for a combination lock, where 5 digits need to be entered in a fixed sequence.

While he remembers that the 5 digits are 9, 8, 7, 5 and 4, he has forgotten the sequence he uses. He also remembers that the sum of the first three digits is a multiple of 3, and so is the sum of the last three digits.

Further, the sum of the last four digits is a multiple of 4.

Which of the following is DEFINITELY FALSE?

A.
4 is the fourth digit of the key
B.
9 is the third digit of the key
C.
8 is the fourth digit of the key
D.
4 is the second digit of the key
E.
8 is the second digit of the key

Correct Answer:

Step-by-Step Solution

Key idea: This is a digit-arrangement puzzle using modular arithmetic. The clues involve sums of selected positions being multiples of 3 or 4, so we work with congruences and the total digit sum.

Step 1: Name the positions and total sum.

Let the key be $d_{1} d_{2} d_{3} d_{4} d_{5}$ .

The digits are 9, 8, 7, 5, 4.

Their total sum is:

$9 + 8 + 7 + 5 + 4 = 33$ .

Step 2: Use the two multiple-of-3 conditions.

We are given:

$d_{1} + d_{2} + d_{3}$ is a multiple of 3,

and:

$d_{3} + d_{4} + d_{5}$ is a multiple of 3.

Add these two sums:

$(d_{1} + d_{2} + d_{3}) + (d_{3} + d_{4} + d_{5}) = 33 + d_{3}$ .

Since both original sums are multiples of 3, their sum is also a multiple of 3.

Also, 33 is a multiple of 3.

Therefore $d_{3}$ must be a multiple of 3.

Among 9, 8, 7, 5, 4, only 9 is divisible by 3.

So:

$d_{3} = 9$ .

Step 3: Use the multiple-of-4 condition.

The sum of the last four digits is:

$d_{2} + d_{3} + d_{4} + d_{5} = 33 - d_{1}$ .

This must be divisible by 4:

$33 - d_{1} \equiv 0 (mod 4)$ .

Since $33 \equiv 1 (mod 4)$ , we need:

$d_{1} \equiv 1 (mod 4)$ .

After fixing $d_{3} = 9$ , the remaining digits are 8, 7, 5, 4.

Their remainders modulo 4 are:

$8 \equiv 0$ , $7 \equiv 3$ , $5 \equiv 1$ , $4 \equiv 0$ .

Only 5 works, so:

$d_{1} = 5$ .

Step 4: Determine possible values of $d_{2}$ .

Used digits are 5 and 9. Remaining digits are 8, 7, 4.

Use the first multiple-of-3 condition:

$d_{1} + d_{2} + d_{3} = 5 + d_{2} + 9 = 14 + d_{2}$ .

This must be divisible by 3.

Since $14 \equiv 2 (mod 3)$ , we need $d_{2} \equiv 1 (mod 3)$ .

Check the remaining digits:

$8 \equiv 2 (mod 3)$ ,

$7 \equiv 1 (mod 3)$ ,

$4 \equiv 1 (mod 3)$ .

Therefore $d_{2}$ can be 7 or 4, but cannot be 8.

Step 5: Identify the definitely false statement.

Option E says “8 is the second digit”.

Since $d_{2}$ can only be 7 or 4, this statement is impossible in every valid arrangement.

So it is definitely false.

Step 6: Check that the other options are not definitely false.

If $d_{2} = 7$ , the remaining digits 8 and 4 can be placed as:

$5, 7, 9, 8, 4$ or $5, 7, 9, 4, 8$ .

These make “8 is fourth” possible and “4 is fourth” possible.

If $d_{2} = 4$ , valid keys include:

$5, 4, 9, 8, 7$ or $5, 4, 9, 7, 8$ .

This makes “4 is second” possible.

Also, $d_{3} = 9$ is forced, so “9 is the third digit” is actually definitely true, not definitely false.

Answer: E.

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

A supplier receives orders from 5 different buyers. Each buyer places their order only on a Monday. The rst buyer places the order after every 2 weeks, the second buyer, after every 6 weeks, the third buyer, after every 8 weeks, the fourth buyer, every 4 weeks, and the fth buyer, after every 3 weeks. It is known that on January 1st, which was a Monday, each of these ve buyers placed an order with the supplier.

On how many occasions, in the same year, will these buyers place their orders together excluding the order placed on January 1st?

A.
1
B.
5
C.
2
D.
4
E.
3

Correct Answer:

Step-by-Step Solution

Key idea: This is an LCM synchronization question, recognisable because several buyers repeat orders at fixed weekly intervals and we need the occasions when all of them coincide again.

Step 1: List the ordering intervals.

Buyer 1: every 2 weeks.

Buyer 2: every 6 weeks.

Buyer 3: every 8 weeks.

Buyer 4: every 4 weeks.

Buyer 5: every 3 weeks.

Step 2: Find the common synchronization interval.

They all order together after the LCM of the intervals.

Prime factorizations:

$2 = 2^{1}$ ,

$6 = 2^{1} \times 3^{1}$ ,

$8 = 2^{3}$ ,

$4 = 2^{2}$ ,

$3 = 3^{1}$ .

Take the highest power of each prime:

$LCM = 2^{3} \times 3^{1} = 8 \times 3 = 24$ .

So all five buyers order together every 24 weeks.

Step 3: Count common occasions in the same year.

January 1st is the starting occasion, so treat it as week 0.

Later common occasions occur at weeks:

$24, 48, 72, \dots$

A year has about 52 weeks.

$24 \leq 52$ , so week 24 counts.

$48 \leq 52$ , so week 48 counts.

$72 > 52$ , so week 72 is outside the year.

Step 4: Exclude January 1st.

The question explicitly says “excluding the order placed on January 1st”.

Therefore only week 24 and week 48 are counted.

Number of occasions = 2.

Common trap: including January 1st, which would incorrectly give 3. Another common trap is computing the LCM as 12 instead of 24 by not accounting for the factor $8 = 2^{3}$ .

Answer: C.

Numbers, Divisibility, HCF/LCM and Progressions Previous Year Questions (PYQs) for XAT: 6+ 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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