MCA Practice Questions (2026) — Free Sample Questions with Answers

Answer: B

1. Divide the coefficients: $3.6÷9=0.4$.
2. Subtract the exponents: $10^{15}÷10^3=10^{12}$.
3. $0.4\times 10^{12}=4\times 10^{11}$ (coefficient now between 1 and 10).
4. Check: $(4\times 10^{11})(9\times 10^3)=36\times 10^{14}=3.6\times 10^{15}$.

$4\times 10^5$: Divides the exponents ($15÷3$) instead of subtracting them.

$4\times 10^{11}$: Correct. $\frac{3.6}{9}=0.4$ and $10^{15-3}=10^{12}$; $0.4\times 10^{12}=4\times 10^{11}$.

$4\times 10^{12}$: Rewrites 0.4 as 4 without lowering the exponent by 1.

$4\times 10^{18}$: Adds the exponents, which is the rule for multiplying, not dividing.

When dividing, subtract exponents; if the coefficient falls below 1, multiply it by 10 and lower the exponent by 1.

Answer: A

1. Write each number as a decimal to four places: $3.\overline{1}=3.1111$, $\pi \approx 3.1416$, $\sqrt{10}\approx 3.1623$, $\frac{19}{6}\approx 3.1667$.
2. Compare place by place: 3.1111 < 3.1416 < 3.1623 < 3.1667.

$3.\overline{1},\pi ,\sqrt{10},\frac{19}{6}$: Correct. $3.\overline{1}\approx 3.1111$, $\pi \approx 3.1416$, $\sqrt{10}\approx 3.1623$ and $\frac{19}{6}\approx 3.1667$.

$\pi ,3.\overline{1},\sqrt{10},\frac{19}{6}$: Treats the repeating decimal as larger than $\pi$ because its digits go on forever; $3.111\dots <3.141\dots$.

$3.\overline{1},\pi ,\frac{19}{6},\sqrt{10}$: Compares only to two decimal places, where $\sqrt{10}$ and $\frac{19}{6}$ both look like 3.16; the third decimal place shows $\sqrt{10}$ is smaller.

$3.\overline{1},\sqrt{10},\pi ,\frac{19}{6}$: Uses the rough estimate $\sqrt{10}\approx 3.1$; in fact $\sqrt{10}\approx 3.162$, which is greater than $\pi$.

To order close real numbers, convert each to a decimal with enough places to tell them apart.

Answer: A

1. 1. Let $a$ = adult tickets and $s$ = student tickets: $a+s=180$ and $11a+6s=1320$.
2. 2. Substitute $s=180-a$: $11a+6(180-a)=1320$.
3. 3. Simplify: $5a+1080=1320$, so $5a=240$ and $a=48$.
4. 4. Then $s=132$. Check: $11(48)+6(132)=528+792=1320$.

48: Correct. $a+s=180$ and $11a+6s=1320$. Substituting $s=180-a$ gives $5a+1080=1320$, so $a=48$.

90: Splits the 180 tickets evenly; 90 adult and 90 student tickets would bring in $1,530.

120: Divides the total by the adult price, $1320÷11$, as if every ticket were an adult ticket.

132: This is the number of student tickets, $180-48$, not the number of adult tickets.

Write one equation for the count and one for the value, substitute, and answer the variable the question asks for.

Answer: D

1. 1. $|x+1|\le 3$ means $-3\le x+1\le 3$.
2. 2. Subtract 1 from all three parts: $-4\le x\le 2$.
3. 3. Graph: closed dots at $-4$ and 2, shaded between.
4. 4. Check $x=0$: $|1|=1\le 3$ is true; check $x=3$: $|4|=4\le 3$ is false.

Graph A: Shows $-2\le x\le 4$, the solution of $|x-1|\le 3$; the sign inside was changed.

Graph B: Shows $x\le -4$ or $x\ge 2$, the solution of $|x+1|\ge 3$.

Graph C: Uses open dots, which leave out $-4$ and 2; $\le$ includes them because $|-4+1|=3$.

Graph D: Correct. $-3\le x+1\le 3$ gives $-4\le x\le 2$, closed dots with the segment between them shaded.

|x − a| ≤ c is the "and" case: all points within c units of a; |x − a| ≥ c is the "or" case.

Answer: B

1. 1. Write the inequality: $85+12s\ge 650$.
2. 2. Subtract 85 from both sides: $12s\ge 565$.
3. 3. Divide by 12: $s\ge 47.08\dots$
4. 4. The number of shirts is a whole number, so the least value is 48.
5. 5. Check: $85+12(48)=661\ge 650$, but $85+12(47)=649<650$.

47: Rounds $47.08\dots$ down. Selling 47 shirts gives $85+12(47)=\$649$, which is less than $650.

48: Correct. $85+12s\ge 650$ gives $s\ge 47.08\dots$; the least whole number of shirts is 48, which raises $661.

55: Ignores the $85 already raised: $650÷12\approx 54.2$, rounded up to 55.

62: Adds the $85 to the goal instead of subtracting it: $735÷12\approx 61.3$, rounded up to 62.

For "at least" problems with whole-number answers, round a decimal solution up, then check the value just below it.

Answer: A

1. Distribute: $y+2=-3x+3$.
2. Subtract 2: $y=-3x+1$.
3. Add $3x$: $3x+y=1$.
4. Check the point $(1,-2)$ from the original form: $3(1)+(-2)=1$. ✓

$3x+y=1$: Correct. $y+2=-3x+3$, so $y=-3x+1$; adding $3x$ gives $3x+y=1$.

$3x+y=5$: Adds 2 to both sides instead of subtracting it, giving $y=-3x+5$.

$y=-3x-5$: Multiplies $-3$ by $-1$ incorrectly, getting $-3x-3$, then subtracts 2.

$y=-3x+3$: Distributes correctly but forgets to subtract 2 from both sides.

Point-slope, slope-intercept and standard form can describe the same line; convert step by step and check with the point $(x_1,y_1)$.

Answer: D

1. Rate: $\frac{114-186}{10-4}=\frac{-72}{6}=-12$ ft/s.
2. Starting height: $186=-12(4)+b$, so $b=186+48=234$.
3. $h=-12t+234$. Check: $-12(10)+234=114$. ✓

$h=-12t+186$: Uses the height at 4 seconds as the starting height; the starting height is the height at $t=0$.

$h=12t+138$: Uses a positive rate, so the elevator would be rising, and subtracts 48 instead of adding it to find the start.

$h=-72t+474$: Uses the change in height, $-72$ feet, as the rate without dividing by the 6 seconds.

$h=-12t+234$: Correct. The rate is $\frac{114-186}{10-4}=-12$ feet per second, and $186+12(4)=234$ feet is the height at $t=0$.

The rate is change in output over change in input; the constant term is the value at time 0, which may not be one of the given data points.

Answer: B

Associative property of addition: Step 3 does not regroup the same terms; it rewrites $6x-6x$ as one product, $(6-6)x$.

Distributive property: Correct. $6x-6x=(6-6)x$ is the distributive property used in reverse to combine like terms.

Commutative property of addition: This justifies Step 2, where $3$ and $-6x$ change places. Step 3 combines like terms.

Identity property of multiplication: No factor of 1 is used in Step 3. The step factors out the common $x$.

Combining like terms, $ax+bx=(a+b)x$, is the distributive property; moving terms past each other is the commutative property.

Answer: A

1. 1. The rule is $f(n)=200\cdot 3^n$ for $n=0,1,2,\dots$.
2. 2. List values: 200, 600, 1,800, 5,400, 16,200, 48,600.
3. 3. The first value above 40,000 is $f(5)=48,600$, so day 5. Check: $f(4)=16,200<40,000$.

Day 5: Correct. Day 4 has $200\cdot 3^4=16,200$ cells, which is too few, and day 5 has $200\cdot 3^5=48,600$, which exceeds 40,000.

Day 6: Counts day 0 as the first day, so the day number is one too high.

Day 10: Leaves out the starting 200 cells and finds when $3^n$ first exceeds 40,000 ($3^{10}=59,049$).

Day 11: Leaves out the starting 200 cells and also counts day 0 as the first day.

Write the geometric rule with the starting amount as the coefficient and the number of steps as the exponent; with a day-0 start, the exponent equals the day number.

Answer: C

1. 1. Write the rule: $y=15x+25$.
2. 2. $x=1$: 40. $x=2$: 55. $x=4$: $60+25=85$.
3. 3. Only the third set matches every pair.

$(1,40),(2,80),(3,120)$: Treats $40 as the cost of every lesson, so the fee is charged again each time.

$(1,15),(2,30),(3,45)$: Leaves out the $25 registration fee.

$(1,40),(2,55),(4,85)$: Correct. $y=15x+25$: $15+25=40$, $30+25=55$, $60+25=85$.

$(1,40),(2,55),(4,70)$: Adds only $15 from $x=2$ to $x=4$, but that is 2 more lessons, so the cost rises $30.

Write the rule from the description, then check every pair; when inputs skip a value, the output changes by the rate times the change in input.

Answer: D

1. Distribute: $-3(x-2)=-3x+6$.
2. Combine: $f(x)=4-3x+6=-3x+10$.
3. Check: $f(0)=4-3(-2)=10$. ✓

It is not linear, because it is not written in the form $f(x)=mx+b$.: A linear function does not have to be written in $mx+b$ form; this one can be rewritten that way.

It is linear, with slope $-3$ and $y$-intercept 2.: Reads the 2 from $(x-2)$ as the $y$-intercept without simplifying.

It is linear, with slope $-3$ and $y$-intercept 4.: Takes the first number, 4, as the $y$-intercept without distributing $-3$ to $-2$.

It is linear, with slope $-3$ and $y$-intercept 10.: Correct. $4-3(x-2)=4-3x+6=-3x+10$, so $m=-3$ and $b=10$.

Simplify a rule to $y=mx+b$ before reading the slope and $y$-intercept; distribute signs carefully.

Answer: C

1. 1. $m_{KL}=\frac{-1-(-3)}{4-(-2)}=\frac{2}{6}=\frac{1}{3}$.
2. 2. $m_{JK}=\frac{-3-2}{-2-(-4)}=-\frac{5}{2}$; $m_{LM}=\frac{3-(-1)}{-1-4}=-\frac{4}{5}$; $m_{MJ}=\frac{2-3}{-4-(-1)}=\frac{1}{3}$.
3. 3. Only $MJ$ has slope $\frac{1}{3}$, so $MJ\parallel KL$.

Side JK: $JK$ is adjacent to $KL$ and has slope $-\frac{5}{2}$, not $\frac{1}{3}$.

Side LM: $LM$ has slope $-\frac{4}{5}$; it can look parallel to $JK$, but neither matches $KL$'s slope $\frac{1}{3}$.

Side MJ: Correct. $MJ$ has slope $\frac{2-3}{-4-(-1)}=\frac{1}{3}$, the same as $KL$.

Compare slopes, not appearance: sides are parallel exactly when their slopes are equal.

Answer: C

1. 1. $P(-4,-1)$ and $Q(2,6)$.
2. 2. Horizontal leg: $|2-(-4)|=6$. Vertical leg: $|6-(-1)|=7$.
3. 3. $d^2=6^2+7^2=85$, so $d=\sqrt{85}\approx 9.22$.
4. 4. Rounded to the nearest tenth: 9.2.

3.6: Subtracts the squares: $\sqrt{7^2-6^2}=\sqrt{13}$. The legs' squares are added.

6.3: Finds the distance from the origin to $Q$, $\sqrt{2^2+6^2}=\sqrt{40}$, instead of from $P$.

9.2: Correct. The horizontal leg is $2-(-4)=6$ and the vertical leg is $6-(-1)=7$, so $d=\sqrt{36+49}=\sqrt{85}\approx 9.2$.

13.0: Adds the horizontal and vertical distances, $6+7$; the straight path is the hypotenuse.

Draw the right triangle formed by the horizontal and vertical changes; the distance is its hypotenuse.

Answer: D

The coach should have used the closest data point instead of the line.: Predictions should come from the line; a data point would not help at 20 hours.

The y-intercept of 35 means every player makes at least 35%.: The intercept is a prediction for 0 hours of practice, not a minimum for every player, and it does not explain this error.

The coach made an error; the line predicts 115% for 20 hours.: The calculation is right: $4.5(20)+35=125$.

Twenty hours is far outside the data, and no player can make over 100%.: Correct. The data go only to 10 hours, and a percent of free throws made cannot exceed 100, so the trend cannot continue that far.

Check a prediction against the context: values outside the data range can give impossible results such as percents above 100.

Answer: B

1. 1. Slope: $\frac{35-75}{10-2}=\frac{-40}{8}=-5$ dollars per year.
2. 2. Substitute $(2,75)$: $75=-5(2)+b$, so $b=85$.
3. 3. $y=-5x+85$. Check $(10,35)$: $-50+85=35$.

$y=-5x+75$: Uses the y-value of $(2,75)$ as the y-intercept; the line crosses $x=0$ at 85, not 75.

$y=-5x+85$: Correct. Slope $=\frac{35-75}{10-2}=-5$; then $75=-5(2)+b$ gives $b=85$.

$y=-0.2x+75.4$: Uses run over rise, $\frac{8}{-40}=-0.2$, for the slope.

$y=5x+65$: Drops the negative sign of the slope; the prices fall as age increases.

With two points on the line, find the slope, then substitute one point to solve for b; do not assume either point is the y-intercept.