GMAT Quiz: Systems Of Equations

What this quiz covers

This quiz focuses on Systems Of Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 60 (correct answer)
2. 70
3. 80
4. 90

Explanation: From the given equations: $x + y = 10$ and $x - y = 6$. Adding these equations: $2x = 16$, so $x = 8$. Subtracting the second from the first: $2y = 4$, so $y = 2$. Now, $x^2 - y^2 = (x+y)(x-y) = (10)(6) = 60$. Alternatively, we could calculate $x^2 - y^2 = 8^2 - 2^2 = 64 - 4 = 60$. Choice B might result from incorrectly calculating $(x+y)(x-y)$ as $10 \times 7 = 70$. Choice C might come from $8 \times 10 = 80$. Choice D might come from $9 \times 10 = 90$.

Question 2

1. $k = 1$
2. $k = 2$ (correct answer)
3. $k = 3$
4. $k = 4$

Explanation: For infinitely many solutions, one equation must be a scalar multiple of the other. Notice that the second equation $4x + 6y = 12$ can be simplified by dividing by 2: $2x + 3y = 6$. For the first equation $kx + 3y = 6$ to be identical to this, we need $k = 2$. Let's verify: when $k = 2$, the system becomes $2x + 3y = 6$ and $4x + 6y = 12$. The second equation is exactly twice the first equation, so they represent the same line, giving infinitely many solutions. Choice A gives different coefficients for $x$. Choice C makes the coefficient of $x$ too large. Choice D also makes the coefficient of $x$ too large.

Question 3

1. All values of $k$ (correct answer)
2. $k = 0$ only
3. $k = 3$ only
4. No value of $k$ (system always has solutions)

Explanation: The second equation can be rewritten as $2(3x + 2y) = 2k + 3$, or $3x + 2y = \frac{2k + 3}{2}$. For the system to be consistent, we need $k = \frac{2k + 3}{2}$. Solving: $2k = 2k + 3$, which simplifies to $0 = 3$, a contradiction. Since the coefficients of the second equation are exactly twice those of the first equation, but the constant terms are never proportional (regardless of the value of $k$), the system has no solution for any value of $k$.

Question 4

1. The system has exactly one solution since the coefficients are proportional
2. The system has no solution since $12 \neq 2 \times 18$ (correct answer)
3. The system has infinitely many solutions since the first equation equals twice the second
4. The system has no solution since $12 \neq 2 \times 9$

Explanation: Given $a = 2c$ and $b = 2d$, the first equation becomes $2cx + 2dy = 12$, which simplifies to $2(cx + dy) = 12$, or $cx + dy = 6$. But the second equation states $cx + dy = 18$. Since $6 ≠ 18$, we have a contradiction: the same expression ($cx + dy$) cannot equal both 6 and 18. Therefore, the system has no solution. Choice A is wrong because proportional coefficients with non-proportional constants lead to no solution, not one solution. Choice C is wrong because the equations are inconsistent. Choice D has the wrong calculation ($2 \times 9 = 18$, but we need $2 \times 18 = 36$ to match 12, which it doesn't).