Algebraic Word Problems and Number Properties Previous Year Questions (PYQs) for XAT: 7+ Solved Questions with Step-by-Step Solutions

Solve 7+ Algebraic Word Problems and Number Properties previous year questions for XAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Algebraic Word Problems and Number Properties

Chapter Roadmap

Step 1: Number Properties & Algebraic Expressions

Master parity, divisibility, prime constraints, symmetric identities, and the behavior of exponents in integer domains.

Step 2: Algebraic Word Problems & Data Sufficiency

Translate real-world scenarios into equations, optimize variables, and rigorously test if given statements provide a unique solution.

Topic Hero: Number Properties & Algebraic Expressions

The Core Intuition

Algebra in competitive exams is about understanding the behavior of numbers and manipulating expressions to reveal hidden constraints.

The Golden Rule of Domains

Integers? Look for factor pairs and divisibility.

Primes? Look for restricted factorizations.

Positive? Eliminate negative roots and zero.

First-Principles Clarity:
Constraints are not obstacles; they are the primary filters that narrow down infinite real solutions to a single integer answer.

Algebraic Word Problems and Number Properties: Solved Questions with Step-by-Step Explanations (5 Problems)

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

Nadeem’s age is a two-digit number X, squaring which yields a three-digit number,whose last digit is Y.

Consider the statements below: Statement I: Y is a prime number Statement II: Y is one-third of X To determine Nadeem’s age uniquely:

A.
either of I and II, by itself, is su cient.
B.
only II is su cient, but I is not.
C.
only I is su cient, but II is not.
D.
it is necessary and su cient to take I and II together.
E.
even taking I and II together is not su cient.

Correct Answer:

Step-by-Step Solution

Key idea: Evaluate each statement independently to see if it yields a unique value for Nadeem's age, then check if they are sufficient together.

Step 1: Define constraints. Nadeem's age $X$ is a two-digit number ( $10 \leq X \leq 99$ ). $X^{2}$ is a three-digit number, so $100 \leq X^{2} \leq 999$ , which means $10 \leq X \leq 31$ . $Y$ is the last digit of $X^{2}$ .

Step 2: Analyze Statement I. $Y$ is a prime number. The possible last digits of any perfect square are 0, 1, 4, 5, 6, 9. The only prime among these is 5. So $Y = 5$ . If $X^{2}$ ends in 5, $X$ must end in 5. Within the range $10 \leq X \leq 31$ , $X$ can be 15 or 25. Statement I alone is not sufficient.

Step 3: Analyze Statement II. $Y$ is one-third of $X$ , so $X = 3 Y$ . Since $Y$ is a possible last digit of a square (0, 1, 4, 5, 6, 9), we test these:

If $Y = 4$ , $X = 12$ . $1 2^{2} = 144$ (ends in 4). Valid.
If $Y = 5$ , $X = 15$ . $1 5^{2} = 225$ (ends in 5). Valid.
If $Y = 9$ , $X = 27$ . $2 7^{2} = 729$ (ends in 9). Valid.

Statement II gives $X \in {12, 15, 27}$ , so it is not sufficient alone.

Step 4: Analyze Statements I and II together. From I, $X \in {15, 25}$ . From II, $X \in {12, 15, 27}$ . The only common value is $X = 15$ .

Step 5: Conclusion. Both statements together are necessary and sufficient to uniquely determine Nadeem's age.

Answer: D

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

If a, b, and c are all positive integers, with $4 a > b$ , then which of the following conditions is BOTH NECESSARY AND SUFFICIENT for the expression $3 3^{4 a - b} \cdot 7^{3 a + 3 b + 2 c}$ to be a positive integer?

A.
a- b+ 2c is divisible by 3
B.
None of the other conditions is both necessary and sufficient
C.
a, b, and c are divisible by 3
D.
a- b= c
E.
(a- b) and c are divisible by 3

Correct Answer:

Step-by-Step Solution

Key idea: For a cube root of a product of prime powers to be a positive integer, each exponent must be a non-negative multiple of 3.

Step 1: Analyze the exponent of 3. We are given $4 a - b$ . For this to be a multiple of 3, we can write $4 a - b = 3 a + (a - b)$ . Since $3 a$ is always divisible by 3, $(a - b)$ must be divisible by 3.

Step 2: Analyze the exponent of 7. We are given $3 a + 3 b + 2 c$ . This can be written as $3 (a + b) + 2 c$ . Since $3 (a + b)$ is always divisible by 3, $2 c$ must be divisible by 3. Since 2 and 3 are coprime, $c$ must be divisible by 3.

Step 3: Check non-negativity. The problem states $a, b, c$ are positive integers and $4 a > b$ , so $4 a - b > 0$ . Also $3 a + 3 b + 2 c > 0$ . Thus, the exponents are strictly positive multiples of 3, ensuring the cube root is a positive integer.

Step 4: Combine conditions. Both $(a - b)$ and $c$ must be divisible by 3. This matches the last option.

Answer: E

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

Let x and y be two positive integers and p be a prime number. If x(x- p)- y(y+ p)= 7p, what will be the minimum value of x- y?

A.
1
B.
3
C.
5
D.
7
E.
None of the above

Correct Answer:

Step-by-Step Solution

Key idea: This is a Diophantine equation involving a prime number. We factor the expression and use parity analysis to check for integer solutions.

Step 1: Expand and group terms: $x^{2} - p x - y^{2} - p y = 7 p ⟹ (x^{2} - y^{2}) - p (x + y) = 7 p$ .

Step 2: Factor the difference of squares: $(x + y) (x - y) - p (x + y) = 7 p ⟹ (x + y) (x - y - p) = 7 p$ .

Step 3: Let $A = x + y$ and $B = x - y - p$ . Then $A \cdot B = 7 p$ .

Step 4: Analyze parity. For any integers $x, y$ , the sum $x + y$ and difference $x - y$ must have the same parity (both even or both odd).

Thus, $A$ and $B + p$ must have the same parity.

Step 5: Test possible values for $A$ . Since $x, y \geq 1$ , $A = x + y \geq 2$ . The factors of $7 p$ are $1, 7, p, 7 p$ .

If $p = 2$ , $7 p = 14$ . Possible $A \in {2, 7, 14}$ .

$A = 2, B = 7$ : $A$ is even, $B + p = 9$ is odd. Mismatch.
$A = 7, B = 2$ : $A$ is odd, $B + p = 4$ is even. Mismatch.
$A = 14, B = 1$ : $A$ is even, $B + p = 3$ is odd. Mismatch.

If $p$ is an odd prime, $7 p$ is odd, so $A$ and $B$ are both odd.

Then $B + p = odd + odd = even$ .

But $A$ is odd. Mismatch.

Step 6: Since parity mismatches occur for all possible factor pairs and all primes $p$ , there are no positive integer solutions for $x$ and $y$ .

Answer: None of the above

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

Wilma, Xavier, Yaska and Zakir are four young friends, who have a passion for integers. One day, each of them selects one integer and writes it on a wall. The writing on the wall shows that Xavier and Zakir picked positive integers, Yaska picked a negative one, while Wilma’s integer is either negative, zero or positive. If their integers are denoted by the rst letters of their respective names, the following is true:

$W + 4 \leq X + 3$

$Y + 2 \leq Z \leq 4$

$X + 3 \geq Z \geq 2$

$W + 4 \leq Y \leq 2$

$2 Y + 2 Z \geq 3$

$W + 2 X + 2 Y + 2 Z \geq 2$

Given the above, which of these can possibly evaluate to?

A.
9
B.
0
C.
4
D.
6
E.
1

Correct Answer:

Step-by-Step Solution

Key idea: This is a system of linear inequalities with integer constraints. We determine the bounds for each variable by cascading the inequalities.

Step 1: Identify domain constraints. $X, Z \geq 1$ ; $Y \leq - 1$ ; $W \in Z$ .

Step 2: From $Y + 2 \leq Z \leq 4$ and $Z \geq 2$ , we get $Z \in {2, 3, 4}$ .

Step 3: From $W + 4 \leq Y \leq 2$ and $Y \leq - 1$ , we get $Y \in {- 1, - 2, \dots}$ .

Step 4: From $2 Y + 2 Z \geq 3 ⟹ Y + Z \geq 2$ . Since $Z \leq 4$ , $Y \geq 2 - 4 = - 2$ . Thus $Y \in {- 1, - 2}$ .

Step 5: Test $Y = - 1$ . Then $Z \geq 3 ⟹ Z \in {3, 4}$ .

From $W + 4 \leq - 1 ⟹ W \leq - 5$ .

From $W + 2 X + 2 Y + 2 Z \geq 2 ⟹ W + 2 X + 2 Z \geq 4$ .

If $W = - 5, Z = 3$ , then $- 5 + 2 X + 6 \geq 4 ⟹ 2 X \geq 3 ⟹ X \geq 2$ .

Check other constraints for $W = - 5, X = 2, Y = - 1, Z = 3$ : all hold.

Step 6: Test $Y = - 2$ . Then $Z \geq 4 ⟹ Z = 4$ .

From $W + 4 \leq - 2 ⟹ W \leq - 6$ .

From $W + 2 X - 4 + 8 \geq 2 ⟹ W + 2 X \geq - 2$ .

If $W = - 6$ , then $2 X \geq 4 ⟹ X \geq 2$ .

Check constraints for $W = - 6, X = 2, Y = - 2, Z = 4$ : all hold.

Step 7: Evaluate possible expressions. For the valid solutions, $X + Z$ can be $2 + 3 = 5$ or $2 + 4 = 6$ . Among the given options, 6 is a possible value (for $X + Z$ ).

Answer: 6

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

The sum of the cubes of two numbers is 128, while the sum of the reciprocals of their cubes is 2.

What is the product of the squares of the numbers?

A.
64
B.
256
C.
16
D.
48
E.
32

Correct Answer:

Step-by-Step Solution

Key idea: Use algebraic manipulation of symmetric expressions involving reciprocals to find the product of the variables.

Step 1: Define variables. Let the two numbers be $x$ and $y$ .

Step 2: Write the given equations. We have $x^{3} + y^{3} = 128$ and $\frac{1}{x ^{3}} + \frac{1}{y ^{3}} = 2$ .

Step 3: Simplify the reciprocal equation. Find a common denominator: $\frac{x ^{3} + y ^{3}}{x ^{3} y ^{3}} = 2$ .

Step 4: Substitute the known value. Replace $x^{3} + y^{3}$ with 128: $\frac{128}{x ^{3} y ^{3}} = 2$ .

Step 5: Solve for the product of cubes. $x^{3} y^{3} = \frac{128}{2} = 64$ . This means $(x y)^{3} = 64$ .

Step 6: Solve for $x y$ . Taking the cube root of both sides gives $x y = 4$ .

Step 7: Find the target value. The question asks for the product of the squares of the numbers, which is $x^{2} y^{2} = (x y)^{2}$ . Substituting $x y = 4$ gives $4^{2} = 16$ .

Answer: C

Algebraic Word Problems and Number Properties Previous Year Questions (PYQs) for XAT: 7+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Algebraic Word Problems and Number Properties

Chapter Roadmap

Step 1: Number Properties & Algebraic Expressions

Step 2: Algebraic Word Problems & Data Sufficiency

Topic Hero: Number Properties & Algebraic Expressions

The Core Intuition

The Golden Rule of Domains

Algebraic Word Problems and Number Properties: 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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