Copy<\/button><\/div>\n\nHCF(84, 126):\n\n126 = 1\u00d784 + 42\n\n84 = 2\u00d742 + 0 \u2192 HCF = 42\nLCM(84, 126) = (84 \u00d7 126) \/ 42 = 252\n<\/code><\/pre><\/div><\/p>\nThat 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.<\/p>\n
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How do you check if a large number is divisible by 3, 9, or 11 without doing the full division?<\/span>Divisibility<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
text<\/span>Copy<\/button><\/div>\n\nIs 8712 divisible by 11?\n\nDigits (right to left): 2,1,7,8\n\nOdd positions: 2 + 7 = 9\n\nEven positions: 1 + 8 = 9\n\n9 \u2013 9 = 0 \u2192 divisible by 11\n<\/code><\/pre><\/div>What's the shortcut for squaring a number that ends in 5, like 85, without long multiplication?<\/span>Squaring Shortcuts<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
text<\/span>Copy<\/button><\/div>\n\n95^2:\n\nLeading part = 9\n\n9 \u00d7 (9+1) = 90\n\nAppend 25 \u2192 9025\n<\/code><\/pre><\/div><\/p>\nThis 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.<\/p>\n
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If a shopkeeper marks up goods 40 percent above cost and gives a 20 percent discount, what's the actual profit percentage?<\/span>Profit and Loss<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
text<\/span>Copy<\/button><\/div>\n\nCost = 100\n\nMarked price = 100 \u00d7 1.40 = 140\n\nDiscount = 20% of 140 = 28\n\nSelling price = 140 \u2013 28 = 112\n\nProfit = 112 \u2013 100 = 12 \u2192 12% profit\n<\/code><\/pre><\/div>A number is increased by 20 percent and then decreased by 20 percent. Is the final value equal to the original? Why do people get this wrong?<\/span>Percentages<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
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.<\/p>\n
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A train 150 meters long crosses a platform in 15 seconds and a signal pole in 5 seconds. What's the length of the platform?<\/span>Trains<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
text<\/span>Copy<\/button><\/div>\n\nSpeed = 150 \/ 5 = 30 m\/s\n\nDistance for platform crossing = 30 \u00d7 15 = 450 m\n\nPlatform length = 450 \u2013 150 = 300 m\n<\/code><\/pre><\/div>If A can do a job in 10 days and B in 15 days, how long do they take working together, and why isn't it just 12.5 days?<\/span>Time and Work<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
text<\/span>Copy<\/button><\/div>\n\nA\u2019s rate = 1\/10 job\/day\n\nB\u2019s rate = 1\/15 job\/day\n\nCombined = 1\/10 + 1\/15 = 3\/30 + 2\/30 = 5\/30 = 1\/6\n\nTime together = 6 days\n<\/code><\/pre><\/div><\/p>\nAveraging 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.<\/p>\n
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The average age of a class of 30 students is 12. A new student joins and the average becomes 12.2. What's the new student's age?<\/span>Averages<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
text<\/span>Copy<\/button><\/div>\n\nOriginal total = 30 \u00d7 12 = 360\n\nNew total = 31 \u00d7 12.2 = 378.2\n\nNew student\u2019s age = 378.2 \u2013 360 = 18.2\n<\/code><\/pre><\/div>If a to b is 2 to 3 and b to c is 4 to 5, what's a to b to c, and what's the general method for combining two ratios that share a common term?<\/span>Ratios<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
text<\/span>Copy<\/button><\/div>\n\na:b = 2:3 \u2192 \u00d74 \u2192 8:12\n\nb:c = 4:5 \u2192 \u00d73 \u2192 12:15\n\nCombined: a:b:c = 8:12:15\n<\/code><\/pre><\/div>What's the actual difference between a permutation and a combination, in your own words, not the formula?<\/span>Permutations and Combinations<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
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.<\/p>\n
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In how many ways can 5 people be seated in a row, and how does that change if 2 specific people must sit together?<\/span>Permutations<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
text<\/span>Copy<\/button><\/div>\n\n5 people, no restriction: 5! = 120\n\n2 must sit together: 4! \u00d7 2 = 24 \u00d7 2 = 48\n<\/code><\/pre><\/div>In a DI set with a bar graph and a pie chart shown together, what's the first thing you should check before answering any question?<\/span>Data Interpretation<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
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What's the pattern in the number series 2, 6, 12, 20, 30, and how do you spot second-difference patterns fast?<\/span>Number Series<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
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.<\/p>\n
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In coding-decoding, if CAT is coded as DBU, how is DOG coded, and what's the underlying rule you're looking for?<\/span>Coding-Decoding<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
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Find the odd one out: 3, 5, 7, 9, 11. What logic actually justifies calling one of these the odd one, versus just picking the one that looks like an outlier?<\/span>Classification<\/span>Easy<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
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<\/span>Medium questions<\/h2>23<\/span><\/div><\/div><\/div>\n
A trick question: what's the largest number that divides both 289 and 391, leaving a remainder of 5 in each case?<\/span>HCF<\/span>Medium<\/span><\/span><\/span><\/button><\/p>\n
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).<\/p>\n
text<\/span>Copy<\/button><\/div>\n\n386 = 1\u00d7284 + 102\n\n284 = 2\u00d7102 + 80\n\n102 = 1\u00d780 + 22\n\n80 = 3\u00d722 + 14\n\n22 = 1\u00d714 + 8\n\n14 = 1\u00d78 + 6\n\n8 = 1\u00d76 + 2\n\n6 = 3\u00d72 + 0 \u2192 HCF = 2\n<\/code><\/pre><\/div><\/p>\nCandidates 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.<\/p>\n
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How many trailing zeros does 100! have, and what's the shortcut?<\/span>Factorials<\/span>Medium<\/span><\/span><\/span><\/button><\/p>\n
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.<\/p>\n
text<\/span>Copy<\/button><\/div>\n\n100 \/ 5 = 20\n\n100 \/ 25 = 4\n\n100 \/ 125 = 0 (stop)\n\nTrailing zeros = 20 + 4 = 24\n<\/code><\/pre><\/div>How do you find the unit digit of a number raised to a large power, like 7 to the power 123?<\/span>Unit Digit<\/span>Medium<\/span><\/span><\/span><\/button>