Aptitude Interview Questions (2026): 50 Q&A

The Euclidean algorithm is the fast way to get HCF: divide the larger number by the smaller, then keep dividing the previous divisor by the remainder until the remainder hits zero, the last non-zero divisor is the HCF. Once you have the HCF, LCM falls out of a single relationship: for any two numbers, HCF times LCM equals the product of the two numbers.

HCF(84, 126):
126 = 1×84 + 42
84 = 2×42 + 0  →  HCF = 42

LCM(84, 126) = (84 × 126) / 42 = 252

That shortcut only works for two numbers at a time. For three or more, you either extend the Euclidean method pairwise or fall back to prime factorization, which is slower but never wrong.

For 3 and 9, sum the digits and check if that sum is divisible by 3 or 9 respectively, and you can repeat the digit sum if the first pass still leaves a big number. For 11, take the alternating sum of the digits, add the digits in the odd positions, subtract the sum of the digits in the even positions, and check if the result is divisible by 11, including zero.

Is 8712 divisible by 11?
Digits (right to left): 2,1,7,8
Odd positions: 2 + 7 = 9
Even positions: 1 + 8 = 9
9 - 9 = 0  →  divisible by 11

Split the number into the leading digits and the trailing 5. Take the leading digits as a single number, multiply it by itself plus 1, and append 25 to the end of that result. For 85, the leading digit is 8, so compute 8 times 9, which is 72, then append 25 to get 7225.

95^2:
Leading part = 9
9 × (9+1) = 90
Append 25  →  9025

This only works for numbers ending in exactly 5. It's one of the few aptitude shortcuts that's genuinely faster than a calculator once it's automatic, since it turns a two-digit multiplication into a one-digit one.

You can't just subtract 20 from 40. Work it through on a cost price of 100: the marked price becomes 140, and a 20 percent discount on 140 is 28, leaving a selling price of 112. That's a profit of 12 on a cost of 100, so the profit percentage is 12 percent, not 20.

Cost = 100
Marked price = 100 × 1.40 = 140
Discount = 20% of 140 = 28
Selling price = 140 - 28 = 112
Profit = 112 - 100 = 12  →  12% profit

No, the final value is always lower than the original whenever you increase and then decrease by the same percentage, because the second percentage change is applied to a bigger base than the first one was. Take 100: a 20 percent increase gives 120, then a 20 percent decrease on 120 is 24, leaving 96, not 100.

The general result is that increasing by x percent and then decreasing by x percent changes the original value by minus x-squared over 100 percent. People get it wrong because they mentally treat both percentage changes as if they were happening on the same fixed base, which is only true if you're adding and subtracting a fixed amount, not a percentage of a moving number.

Crossing a signal pole means covering only the train's own length, so that gives you speed directly: 150 meters in 5 seconds is 30 meters per second. Crossing a platform means covering the train's length plus the platform's length, so 30 meters per second times 15 seconds is 450 meters total, and subtracting the train's own 150 meters leaves a 300 meter platform.

Speed = 150 / 5 = 30 m/s
Distance for platform crossing = 30 × 15 = 450 m
Platform length = 450 - 150 = 300 m

Averaging the two times gives the wrong answer because work rates add, not the number of days. A does 1/10 of the job per day, B does 1/15 per day, so together they do 1/10 + 1/15 = 1/6 of the job per day, which means the whole job takes 6 days.

A's rate = 1/10 job/day
B's rate = 1/15 job/day
Combined = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6
Time together = 6 days

Averaging days only happens to look plausible here because 10 and 15 aren't far apart. Try it with 2 days and 100 days and it's obvious that "51 days" (the average) is nowhere close to the real answer, which is under 2 days.

Work from total sums, not averages directly. The original 30 students total 30 times 12, which is 360. After the new student joins, 31 students total 31 times 12.2, which is 378.2. The new student's age is the difference, 378.2 minus 360, which is 18.2.

Original total = 30 × 12 = 360
New total = 31 × 12.2 = 378.2
New student's age = 378.2 - 360 = 18.2

The shared term, b, has to be made equal across both ratios before you can chain them together. Here b is 3 in the first ratio and 4 in the second, so scale the first ratio by 4 and the second by 3 to make b equal to 12 in both, then read off the combined ratio.

a:b = 2:3  →  ×4  →  8:12
b:c = 4:5  →  ×3  →  12:15
Combined: a:b:c = 8:12:15

A permutation counts arrangements where order matters, first place and second place are different outcomes even with the same two people. A combination counts selections where order doesn't matter, choosing the same two people for a committee is one outcome no matter which one you picked first.

The fastest way to tell which one a question wants: ask whether swapping two chosen items around changes the outcome. If yes, it's a permutation question. If the two arrangements describe the same real-world result, it's a combination question.

Five distinct people in a row is a straight factorial, 5 factorial, which is 120 arrangements. When two specific people must sit together, treat that pair as a single block, so you're really arranging 4 units (the block plus the other 3 people), which is 4 factorial, times 2 for the two internal orders of the pair itself.

5 people, no restriction: 5! = 120
2 must sit together: 4! × 2 = 24 × 2 = 48

Check the units and the base value each chart represents before touching a single question, because a pie chart usually shows a percentage breakdown of some total that's given separately, often in the bar graph or in a line of text above both charts. Answering percentage-of-percentage questions without first converting the pie chart's shares back into absolute numbers using that shared base is the most common way to burn time redoing a question.

Take the differences between consecutive terms first: 4, 6, 8, 10. Those differences themselves increase by a constant amount, 2 each time, which means the original series follows a quadratic pattern rather than a simple constant-difference or constant-ratio one. The next difference should be 12, so the next term is 30 + 12, which is 42.

Spotting this fast just means checking the first differences before assuming the series is arithmetic or geometric. If the first differences aren't constant, check whether they themselves form a clean pattern before giving up and guessing.

Compare each letter's position in the alphabet to its coded counterpart. C to D is plus 1, A to B is plus 1, T to U is plus 1, so the rule is simply shift every letter forward by one position in the alphabet. Apply that same shift to DOG: D becomes E, O becomes P, G becomes H, giving EPH.

Every number here except one is a prime number, 3, 5, 7, and 11 are all prime, while 9 factors into 3 times 3 and isn't. The justification has to be a rule that applies cleanly to four of the five and clearly fails for the fifth, not a vague sense that one number "looks different," since in a list of all-odd numbers, oddness itself can't be the classifying rule.

The moment a question mentions a common remainder, subtract that remainder from both numbers first, then find the HCF of the reduced numbers. Here that means finding HCF(289-5, 391-5), which is HCF(284, 386).

386 = 1×284 + 102
284 = 2×102 + 80
102 = 1×80 + 22
80 = 3×22 + 14
22 = 1×14 + 8
14 = 1×8 + 6
8 = 1×6 + 2
6 = 3×2 + 0  →  HCF = 2

Candidates who skip the "subtract the remainder first" step almost always end up computing HCF(289, 391) directly, which gives the wrong answer entirely because the numbers themselves aren't exact multiples of anything meaningful once you leave the remainder in.

Trailing zeros come from factors of 10, and every 10 is a 2 paired with a 5. In a factorial there are always far more factors of 2 than factors of 5, so the count of trailing zeros is just the count of factors of 5 in the number. Divide 100 by 5, then by 25, then by 125 (stopping once the divisor exceeds the number), and add up the quotients.

100 / 5   = 20
100 / 25  = 4
100 / 125 = 0 (stop)
Trailing zeros = 20 + 4 = 24

Unit digits repeat in a cycle of at most 4 as you raise a base to successive powers, so the trick is finding that cycle and then working out where the given exponent falls inside it. For 7, the unit digits of 7^1, 7^2, 7^3, 7^4 are 7, 9, 3, 1, and then the cycle repeats.

Cycle for 7: 7, 9, 3, 1 (length 4)
123 mod 4 = 3  →  matches the 3rd position in the cycle
7^123 has unit digit 3

The one gotcha: if the exponent mod 4 comes out to 0, that maps to the last position in the cycle, not a fresh cycle starting at position 0. People who forget that off-by-one usually get the unit digit for a multiple-of-4 exponent wrong.

A successive change means the second percentage is applied to the already-changed value, not the original one, so you can't just add the two percentages together and call it done except as a rough estimate. The correct formula for two successive changes of a percent and b percent is a + b + (ab/100), and that correction term, ab over 100, is exactly what candidates forget under time pressure.

Two successive increases of 10% and 20%:
Naive (wrong): 10 + 20 = 30%
Correct: 10 + 20 + (10×20)/100 = 32%

The sign of that correction term flips when one of the changes is a decrease, so a 10 percent increase followed by a 20 percent decrease works out to 10 - 20 - (10×20)/100 = -12 percent, an overall 12 percent decrease.

For a single year, simple interest and compound interest give exactly the same amount, because there's nothing yet to compound on. The difference only starts showing up from the second year onward, since compound interest earns interest on the previous year's interest and simple interest never does.

Principal 1000, rate 10%, 2 years:
SI = 1000 × 10 × 2 / 100 = 200
CI = 1000 × (1.10^2 - 1) = 1000 × 0.21 = 210
Difference at year 2 = 10 (this equals P×(r/100)^2 for 2 years)

That's also a useful shortcut for a specific question type: the difference between CI and SI over exactly two years always equals the principal times the square of the rate, expressed as a decimal.

This is a two-step trap. The claimed 25 percent markup applies to the price he charges, but he's also handing over less actual product than he's charging for, so the real profit compounds both effects rather than just adding them. The standard formula for profit from a false weight alone is (true weight minus claimed weight) over claimed weight, times 100, which for 1000 grams claimed but 800 grams delivered is 200/800 = 25 percent.

Combine that 25 percent weight-based profit with the 25 percent markup he's already charging using the successive-percentage formula: 25 + 25 + (25×25)/100 = 56.25 percent. That's the number most candidates miss, they compute one of the two 25 percents and stop, instead of combining both sources of profit.

Moving toward each other or apart from each other, relative speed is the sum of the two individual speeds, because the gap between them is closing or opening at the combined rate. Moving in the same direction, relative speed is the difference of the two speeds, because only the speed gap matters for how fast one train pulls away from or catches up to the other.

The trip-up is almost always a sign error under pressure, adding when the question actually calls for subtraction, especially in a "train A catches up with train B" question where both are moving the same way. Reading the direction words carefully before touching the formula saves more time than memorizing the formula itself.

Downstream speed is the boat's own speed plus the current, upstream speed is the boat's own speed minus the current. Once you've computed both of those speeds from distance and time, the boat's still-water speed is their average and the stream's speed is half their difference.

Downstream speed = 20 km/h, Upstream speed = 12 km/h
Boat's speed in still water = (20 + 12) / 2 = 16 km/h
Stream's speed = (20 - 12) / 2 = 4 km/h

Yes, because the filling rate (1/6 per hour) is faster than the emptying rate (1/8 per hour), so the net rate is still positive. Subtract the two rates the same way you'd add rates for two people working together, since an outlet pipe is really just negative work.

Fill rate = 1/6 per hour
Empty rate = 1/8 per hour
Net rate = 1/6 - 1/8 = 4/24 - 3/24 = 1/24 per hour
Time to fill = 24 hours

Add their individual daily rates, 1/a plus 1/b plus 1/c, to get the fraction of the job they finish together in one day, then take the reciprocal of that sum to get the total number of days. The step people skip under time pressure is that final reciprocal, they compute the combined daily fraction correctly and then report that fraction itself as the answer instead of flipping it.

a=8, b=12, c=24 days
Combined rate = 1/8 + 1/12 + 1/24 = 3/24 + 2/24 + 1/24 = 6/24 = 1/4
Total time together = 4 days

In 25 liters at a 4:1 ratio, milk is 20 liters and water is 5 liters. Adding water only increases the water side, milk stays fixed at 20 liters, so set up the new ratio with milk fixed and solve for the new water amount.

Milk = 25 × 4/5 = 20 L, Water = 25 × 1/5 = 5 L
New ratio needed: milk:water = 2:1, milk stays at 20
20 / new_water = 2/1  →  new_water = 10 L
Water to add = 10 - 5 = 5 L

Alligation is a shortcut for finding the ratio in which two ingredients at different values (price, concentration, whatever the quantity being mixed represents) must be combined to produce a mixture at a target value in between. Cross-subtract diagonally: the ratio of the two quantities equals (higher value minus mean value) to (mean value minus lower value).

Mixing 10%-fat milk and 20%-fat cream to get 15%-fat mixture:
Ratio of 10% : 20% = (20-15) : (15-10) = 5 : 5 = 1:1

It's faster than an equation for anything with exactly two ingredients and a target mean, which is the majority of mixture questions on a timed test. Once there are three or more components involved, it stops being a clean shortcut and an equation is usually less error-prone.

Total for the original 5 numbers is 5 times 20, which is 100. Total after the replacement is 5 times 22, which is 110. Since only one number changed and the other four stayed exactly the same, the entire 10-point jump in the total came from that one replaced number, so it increased by 10.

"At least 1" is almost always faster to solve by subtracting the case you don't want from the total, rather than adding up every valid case separately. Total ways to choose any 3 people from 9 is C(9,3) = 84. Subtract the one case that violates the condition, all-men committees, C(5,3) = 10, leaving 74.

Total = C(9,3) = 84
All-men (invalid) = C(5,3) = 10
At least 1 woman = 84 - 10 = 74

"At least one" has many ways to succeed, at least one head could mean 1, 2, or 3 heads, but exactly one way to fail entirely, zero heads. It's almost always faster to compute the probability of the one failing case and subtract from 1 than to add up every succeeding case.

P(no heads in 3 tosses) = (1/2)^3 = 1/8
P(at least one head) = 1 - 1/8 = 7/8

If every letter were distinct, an 11-letter word would have 11 factorial arrangements. But MISSISSIPPI repeats letters, 4 I's, 4 S's, and 2 P's, and swapping two identical letters produces an arrangement that looks exactly the same on paper, so those swaps have already been overcounted and need to be divided back out.

Total letters = 11 (M-1, I-4, S-4, P-2)
Arrangements = 11! / (4! × 4! × 2!) = 39,916,800 / 1,152 = 34,650

A full pie chart is 360 degrees representing 100 percent, so each degree is exactly 100/360, which is 5/18 percent. Instead of dividing by 360 directly, multiply the given degree value by 5/18, which is usually faster to do in your head, especially when the degree value is a multiple of 18 or 36.

A sector spans 72 degrees.
Percentage = 72 × 5/18 = 20%

CAGR, compound annual growth rate, is the single constant annual rate that would take the starting value to the ending value over the given number of years if it compounded every year, and it's found by taking the nth root of the ratio of ending to starting value, then subtracting 1. A simple average of the year-by-year percentage changes ignores compounding entirely and treats each year's growth as if it were happening off the original base every time, which overstates or understates the real trend depending on the pattern of the numbers.

Sales grow from 100 to 200 over 5 years
CAGR = (200/100)^(1/5) - 1 = 2^0.2 - 1 ≈ 0.1487  →  ~14.87%

Cross-multiply instead of converting both to decimals. Compare 47 times 71 against 52 times 63, whichever product is bigger tells you which fraction is bigger, since cross-multiplication preserves the inequality direction as long as both denominators are positive.

47 × 71 = 3337
52 × 63 = 3276
3337 > 3276  →  47/63 > 52/71

Work outward from the fixed point, not from the pronoun. The grandfather's only son is the man's father, since the man himself is a descendant of that grandfather and the son being referred to has to be the direct link in that generation. The daughter of the man's father is either the man's sister, or the man himself if he were female, but since the question already establishes "she" as a separate person being pointed at, the answer is his sister.

Syllogisms get evaluated by drawing every possible Venn diagram consistent with the given statements and checking whether the proposed conclusion holds in all of them, not just the one diagram that comes to mind first. "All A are B" only guarantees A is fully inside B, it says nothing about whether B has anything outside A. "Some A are not B" is genuinely weaker than it sounds, it's consistent with anywhere from almost-no-overlap to A and B being completely separate categories.

The most common mistake is treating "some A are B" as if it implies "some A are not B," which isn't guaranteed, it's entirely possible that all A are B and the statement "some A are B" is still technically true. Syllogism questions exploit exactly that gap between what a statement logically guarantees and what it merely leaves open as a possibility.

Track each leg on a mental (or scratch-paper) coordinate grid rather than trying to hold the whole path in your head. Starting at (0,0): 5km north puts him at (0,5). Turning right (now facing east) and walking 3km puts him at (3,5). Turning right again (now facing south) and walking 5km puts him at (3,0).

Start: (0,0)
After 5km north: (0,5)
After 3km east: (3,5)
After 5km south: (3,0)
Distance from start = sqrt(3^2 + 0^2) = 3 km, due east

The two 5km legs, one north and one south, cancel out completely, which is exactly the kind of detail the question is testing for, whether you actually plot the path or just add up all the distances walked regardless of direction.

A conclusion "follows" only if it's true in every single Venn diagram consistent with the given statements, no exceptions. A conclusion is merely "possible" if there's at least one valid diagram where it's true and at least one other valid diagram where it's false, meaning the given statements don't guarantee it either way.

Test-writers exploit this because a possible conclusion often reads as intuitively likely, the kind of thing a reasonable person would guess is true, which is exactly why candidates mark it as following when it's actually still open. The discipline required here is mechanical: draw the most extreme diagram that still satisfies every given statement, and if the conclusion breaks in that extreme case, it doesn't follow, no matter how reasonable it sounds.

A ratio only fixes the relative size of two numbers, not their actual values, so you need a second piece of information, here the LCM, to pin down a common multiplier. If the numbers are 3k and 4k, their LCM (since 3 and 4 share no common factor) is 12k. Setting 12k equal to 84 gives k = 7, so the numbers are 21 and 28.

Numbers: 3k and 4k
gcd(3, 4) = 1, so LCM(3k, 4k) = 12k
12k = 84  →  k = 7
Numbers: 21 and 28

This only works cleanly because 3 and 4 are coprime. If the ratio terms shared a common factor, you'd have to strip that out first before multiplying to get the LCM formula right.

It's always a loss, and the size of that loss follows a clean formula: when the same selling price gives a profit of x percent on one item and a loss of x percent on the other, the overall result is a loss of (x/10) squared percent. For x = 20, that's a 4 percent loss overall, not breakeven.

Item A: sold at 120% of its cost (20% profit)
Item B: sold at 80% of its cost (20% loss)
Both sold at the same price P, so:
Cost A = P / 1.20, Cost B = P / 0.80
Total cost = P/1.2 + P/0.8, Total SP = 2P
Loss % = (x/10)^2 = (20/10)^2 = 4%

The reason it's never breakeven: equal percentage profit and loss aren't symmetric in absolute terms, because they're each measured against a different cost price, and the item sold at a loss had a higher original cost than the one sold at a profit.

Average speed for equal distances at two different speeds is the harmonic mean, 2xy over x plus y, not the arithmetic mean, because he spends more time traveling at the slower speed. Average speed is always total distance over total time, and since the distances are equal but the times aren't, the arithmetic mean overweights the faster leg.

Distance D each way, speeds x and y
Time out = D/x, Time back = D/y
Total distance = 2D, Total time = D/x + D/y = D(x+y)/(xy)
Average speed = 2D / [D(x+y)/(xy)] = 2xy / (x+y)

For x = 40 and y = 60, the harmonic mean gives 48, while the naive arithmetic average would wrongly give 50. The gap between the two grows as x and y get further apart.

Each time you remove a fraction of the mixture and replace it with pure water, the remaining original substance gets multiplied by the same fraction again, so after n repetitions the fraction of the original substance left is (1 minus removed-fraction) raised to the power n, not multiplied by n.

Container: 40 L, remove 4 L and replace with water, 3 times
Fraction remaining each time = (40-4)/40 = 0.9
After 3 replacements: original substance left = 40 × 0.9^3 = 29.16 L

The mistake people make is treating this as a linear reduction, subtracting the same 4 liters worth of concentration three separate times, which massively understates how much of the original substance is actually gone, because the mixture being removed each round is progressively more diluted.

The full sample space for two dice is 36 outcomes, not 11 (the count of possible sums), because each die is distinct and (3,4) is a different outcome from (4,3) even though both sum to 7. Miscounting the sample space as 11 possible sums, each treated as equally likely, is the single most common error in this topic.

The prime sums possible with two dice are 2, 3, 5, 7, and 11. Counting the actual dice-pair outcomes that produce each: sum 2 has 1 way, sum 3 has 2, sum 5 has 4, sum 7 has 6, sum 11 has 2, for a total of 15 favorable outcomes out of 36, giving a probability of 15/36, which simplifies to 5/12.

Build a small table or grid as you read the paragraph the first time, rather than reading it once for comprehension and once again to extract numbers. Most caselets describe relationships between a handful of categories, people, products, years, that map cleanly onto rows and columns, and writing the numbers into that grid as you go means you only read the paragraph once.

The other habit worth building: note explicitly which values the caselet gives you directly versus which ones you're expected to derive. Test-writers often bury one condition that has to be worked out from two others rather than read off directly, and candidates who treat every number in the paragraph as a given tend to miss that derived value entirely.

Start with whichever clue fixes a relationship the most tightly, usually one naming two specific people as immediate neighbors or directly opposite each other, and place those two first before touching any other clue. Placing the loosest clue first, something like "A is not next to B," gives you almost no information to build on and tends to produce false starts that need backtracking.

After placing the tightest clue, work through the remaining clues in order of how many people they involve, fewest first, and eliminate positions rather than trying to solve the whole arrangement in one pass. Most six or eight person circular arrangements resolve fully after two or three clues are placed correctly, the remaining clues usually just confirm the arrangement or fix the last one or two open seats.

The real method counts "odd days," the remainder when the total number of days since a known reference point is divided by 7, since the day of the week just cycles every 7 days regardless of the actual date. A normal year contributes 1 odd day, a leap year contributes 2, because 365 mod 7 is 1 and 366 mod 7 is 2.

January 1, 1900 was a Monday (a commonly memorized reference point). 1900 was not a leap year despite being divisible by 4, because century years are only leap years if divisible by 400, and 1900 isn't. So 1900 contributes exactly 1 odd day, moving January 1, 1901 forward by one day from Monday to Tuesday. The century-year leap rule is the exact detail candidates forget, treating 1900 as a leap year purely because it's divisible by 4 gives the wrong day entirely.

The minute hand moves 6 degrees per minute and the hour hand moves 0.5 degrees per minute, so the minute hand gains 5.5 degrees on the hour hand every minute. At 3:00 exactly, the hour hand is at 90 degrees and the minute hand is at 0, a gap of 90 degrees. For the hands to be exactly opposite, that gap needs to become 180 degrees.

At 3:00, hour hand is ahead by 90 degrees.
For hands to be opposite, minute hand must gain 90+180 = 270 degrees on hour hand
(since minute hand needs to go from behind to 180 degrees past)
Time needed = 270 / 5.5 = 49 1/11 minutes
Answer: 3:49 1/11

The step people skip is realizing the minute hand starts behind the hour hand at 3:00, so it first has to close the 90 degree gap and then open up an additional 180 degrees on the other side, not just close 90 degrees of gap to reach 180.

Speed and time are inversely proportional for a fixed distance, so a 25 percent reduction in speed means the new speed is 3/4 of the original, which means the new time is 4/3 of the original time, since time is the reciprocal relationship. The time increased by 1/3 of the original time, and that 1/3 corresponds to the 20 minute delay given in the question.

New speed = 3/4 of original  →  New time = 4/3 of original time
Extra time = (4/3 - 1) × original time = (1/3) × original time = 20 minutes
Original time = 60 minutes

Setting up two separate distance equations with an unknown distance and unknown original speed works too, but it's slower and gives more places to make an arithmetic slip. Going straight from the percentage change in speed to the reciprocal percentage change in time skips the unknowns entirely.

The formula itself, subtracting rates instead of adding them, isn't usually the problem. The actual failure point is candidates forgetting that the emptying rate needs to be subtracted from the correct combined filling rate first, before comparing it against the intended time, rather than being applied at the very end as an adjustment to a final answer that's already been computed without it.

A typical setup: two pipes together fill a tank in 6 hours, but with a leak present it actually takes 8 hours. The mistake is computing "6 hours normally, 8 hours with the leak, so the leak alone would empty it in 2 hours," which looks reasonable but is wrong, because rates don't subtract like that. The leak's own rate has to be found by comparing the combined-with-leak rate (1/8) against the combined-without-leak rate (1/6), and the leak's rate is the difference of those two rates, 1/6 minus 1/8, which is 1/24, meaning the leak alone would empty the tank in 24 hours, not 2.

Misreading which quantity the question is actually asking for, after correctly solving for a different, related quantity. A time-work question asks for the number of days remaining after some work is already done, and the candidate correctly computes total days for the job, then submits that number instead of subtracting the days already elapsed. A percentage question asks for the new value after a change, and the candidate correctly computes the percentage change itself, then submits that percentage instead of applying it to get the final value.

The pattern across almost every version of this mistake: the math itself is done correctly, and the error happens in the last 5 seconds, translating a correctly-computed intermediate result back into the specific answer the question actually asked for. Reading the question a second time after finishing the calculation, specifically to check what's being asked rather than to redo the math, catches most of these before they cost a mark.