{"id":62370,"date":"2024-04-22T09:40:17","date_gmt":"2024-04-22T16:40:17","guid":{"rendered":"https:\/\/gmatclub.com\/blog\/gmat-sample-data-sufficiency-practice-questions\/"},"modified":"2026-04-02T02:48:47","modified_gmt":"2026-04-02T09:48:47","slug":"gmat-sample-data-sufficiency-practice-questions","status":"publish","type":"post","link":"https:\/\/gmatclub.com\/blog\/gmat-sample-data-sufficiency-practice-questions\/","title":{"rendered":"GMAT Sample Data Sufficiency Practice Questions"},"content":{"rendered":"

Unlike Problem Solving<\/a>, Data Sufficiency<\/a> is a question-type unique to the GMAT.\u00a0 Data Sufficiency is part of the Data Insights section on the GMAT.<\/p>\n

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GMAT Data Sufficiency Practice Questions<\/h2>\n

1. Maggie is 15 years older than Bobby.\u00a0 How old is Bobby?<\/p>\n

Statement #1<\/span>: In 3 years, Maggie’s age will be 50% larger than Bobby’s age.<\/p>\n

Statement #2<\/span>: Years ago, when Maggie was 25 years old, Bobby was 10 years old.<\/p>\n

2. By abc we denote a three digit number with digits a, b, and c.\u00a0 Is abc divisible by 3?<\/p>\n

Statement #1<\/span>: The product of (a) times (b) is a number divisible by 3<\/p>\n

Statement #2<\/span>: c = 3<\/p>\n

3. The ratio of underclassmen to upperclassmen in a particular school is 4:5.\u00a0 How many underclassmen are there?<\/p>\n

Statement #1<\/span>:\u00a0 If the number of upperclassmen increased by 20%, then underclassmen would constitute 40% of this larger student body.<\/p>\n

Statement #2<\/span>:\u00a0 If 10 more underclassmen joined the school, the new ratio of underclassmen to upperclassmen would be 9:10.<\/p>\n

4. An oddly shaped die has N sides, numbered from 1 to N, that are equally likely to appear on a throw.\u00a0 Let P be the probability that at least one “1” appears in two throws.\u00a0 Is P < (1\/5)?<\/p>\n

Statement #1<\/span>: N is a prime number<\/p>\n

Statement #2<\/span>: N > 5<\/p>\n

5. Set S consists of all the positive multiples of 5 that are less than K, and K is a positive integer not divisible by 5.\u00a0 The mean of Set S is not divisible by 5.\u00a0 Let N be the number of members of the set.\u00a0 N is not divisible by 5.\u00a0 What does N equal?<\/p>\n

Statement #1<\/span>: N < 52<\/p>\n

Statement #2<\/span>: K\/5 > 48<\/p>\n

6. The numbers D, N, and P are positive integers, such that D < N, and N is not a power of D.\u00a0 Is D a prime number?<\/p>\n

Statement #1<\/span>:\u00a0 N has exactly four factors, and D is a factor of N<\/p>\n

Statement #2<\/span>:\u00a0 D = (3^P) + 2<\/p>\n

Answers and Solutions<\/h2>\n

1. Let M = Maggie’s age right now, and let B = Bobby’s age right now.\u00a0 The prompt gives us the equation M = B + 15.<\/p>\n

Statement #1 gives us a second equation, (M + 3) = 1.5*(B + 3).\u00a0 Notice that I used a multiplier for the percent increase<\/a>.\u00a0\u00a0 With this equation and the prompt equation, we have two equations with two unknowns, so we can solve for the values of M & B.\u00a0 We don’t need to perform those calculations: it is enough to know that we could.\u00a0 This statement, alone and by itself, is sufficient<\/strong>.<\/p>\n

Statement #2 is already obvious from the prompt.\u00a0 If Maggie is 15 years older than Bobby, then when Maggie was 25, Bobby would have had to have been 10.\u00a0 There’s no surprise there.\u00a0 We are given absolutely no indication when these people were these ages, so this doesn’t help us at all.\u00a0 This statement, alone and by itself, is not sufficient<\/strong>.<\/p>\n

Answer = (A)<\/strong><\/p>\n

2. Remember, the divisibility rule<\/a>: if the sum of all the digits in a large number is divisible by 3, then the number itself is divisible by three.<\/p>\n

The problem with statement #1, for any given values of a & b that work, the one’s digit, c, could be anything.\u00a0 For example, a = 1, and b = 3 satisfy this condition, so that would be all the numbers in the 130’s: {130, 131, 13, 133, 134, 135, 136, 137, 138, 139}.\u00a0 Clearly, some of these are divisible by 3, namely {132, 135, 138} and the others aren\u2019t.\u00a0 This statement allows for multiple answers, so this statement, alone and by itself, is not sufficient<\/strong>.<\/p>\n

The problem with statement #2 is that we now know the one’s digit, but the other two digits could be anything.\u00a0 Think about the first few possible numbers of this form: {103, 113, 123, 133, 143, 153, 163, 173, 183, 193, 203}.\u00a0 Of those, only (123, 153, 183) are divisible by three, and the others aren\u2019t.\u00a0 In fact, you would be expected to know this, but as it happens, {103, 163, 173, 193} are prime numbers, not divisible by anything!\u00a0 This statement also allows for multiple answers, so this statement, alone and by itself, is not sufficient<\/strong>.<\/p>\n

Combined statements: now, we have to obey both constraints.\u00a0 One number that obeys both constraints is 133, which is not<\/strong> divisible by 3.\u00a0 Another number that obeys both constraints is 333, which is<\/strong> divisible by 3.\u00a0 Even with both statements in place, we can find numbers that produce either a “yes” or a “no” to the prompt.\u00a0 Even together, the statements are not sufficient<\/strong>.<\/p>\n

Answer = (E)<\/strong><\/p>\n

3. We are given “ratio information” and we want actual numbers.\u00a0 To get counts, actual numbers of people, we need something besides more ratio information.<\/p>\n

Statement #1 just gives us more “ratio information,” so we can’t solve for counts.\u00a0 This statement, alone and by itself, is not sufficient<\/strong>.<\/p>\n

Statement #2 gives us some actual counts.\u00a0 Let N and P represent the current underclassmen and upperclassmen, respectively, in the school.\u00a0 We know N\/P = 4\/5 from the prompt.\u00a0 From this statement, we know (N + 10)\/P = 9\/10.\u00a0 Together, we have two equations for two unknowns, so we could solve for the values of N & P.\u00a0 This statement, alone and by itself, is sufficient<\/strong>.<\/p>\n

Answer = (B)<\/strong><\/p>\n

4. First of all, if the N sides are equally likely, the probability of getting “1” on a single throw, or any particular side on a single throw, is 1\/N.\u00a0 The probability of not getting “1” on a single throw, or not getting any particular side, is (1 \u2013 1\/N) = (N \u2013 1)\/N.<\/p>\n

For relatively low values of N, then getting “1” is relatively more likely, and getting something else is relatively less likely.\u00a0 As N gets bigger, the probability of getting “1” on a single throw decreases, and the probability of getting something else increase.<\/p>\n

What really makes the difference in this probability is the size of N.\u00a0 This is an at least probability<\/a>, so we calculate the probability of getting something other than one on two consecutive throws, and then subtract this from 1.<\/p>\n

Statement #1<\/span>: this statement is not very helpful.\u00a0 If N = 2, then essentially, the “two-sided die” is a coin, and if H = 1, then the probability of not getting a “1” is (1\/2).\u00a0 The probability of not getting a “1” on two consecutive throws is (1\/4).\u00a0 That’s the probability of getting no “1” on two throws.\u00a0 We subtract this from one to get the probability of at least one “1”: P = (3\/4), which greater than 1\/5.\u00a0 Answer to the prompt = “no.”<\/p>\n

But, suppose N is larger.\u00a0 For easy of calculation, I will pick N = 15, even though 15 is not prime.\u00a0 If N = 15 produces a probability less than 1\/5, then so will any prime number greater than 15.<\/p>\n

If N = 15, then the probability of not getting a “1” on a single throw is 14\/15.\u00a0 The probability of not getting a “1” on two consecutive throws is this squared, 196\/225.\u00a0 If this is more than 4\/5, then subtracting it from one will be less than 1\/5.\u00a0 Well, 225\/5 = 45, and 45 times 4 is 180.\u00a0 Thus, (180\/225) = (4\/5), and (196\/225) > (4\/5).\u00a0 Subtracting this from one will produce a probability less than 1\/5.\u00a0 Answer to the prompt = “yes.”<\/p>\n

This statement is consistent with different answers to the prompt question, so it does not lead to a definitive answer.\u00a0 This statement, alone and by itself, is not sufficient<\/strong>.<\/p>\n

Statement #2<\/span>: We have already seen that N = 15, or any number larger than that, produces a “yes” answer to the prompt.\u00a0 Is it possible to produce a “no” answer to with a value of N greater than 5?<\/p>\n

I will use N = 7.\u00a0 The probability of not getting a “1” on a single throw is 6\/7.\u00a0 The probability of not getting a “1” on two consecutive throws is this squared, 36\/49.\u00a0 If this is less than 4\/5, then subtracting it from one will be more than 1\/5.<\/p>\n

Well, clearly, 36\/45 = 4\/5, and making the denominator larger makes a fraction smaller, so<\/p>\n

4\/5 = 36\/45 > 36\/49.\u00a0 So, this fraction is less than 4\/5, so subtracting it from one will result in a probability that is greater than 1\/5.\u00a0 This gives a “no” answer to the prompt.<\/p>\n

This statement is consistent with different answers to the prompt question, so it does not lead to a definitive answer.\u00a0 This statement, alone and by itself, is not sufficient<\/strong>.<\/p>\n

Combined statements<\/span>: As we saw in the previous statements, the prime number N = 7 gives a “no” answer, and the non-prime N = 15 gives a “yes” answer, which implies that any prime number greater than 15 (such as 17, 19, 23, etc.) would also give a “yes” answer.\u00a0 Different answers are possible even with the constraints of both statements.\u00a0 Together, both statements are still not sufficient<\/strong>.<\/p>\n

Answer = (E)<\/strong><\/p>\n

5. This is a tricky problem, because there are several constraints in the problem.\u00a0 first of all, the number of members is N, so the numbers in the set go from 5 to 5N.\u00a0 The integer K has the quality that 5N < K < (5N + 5), because K is bigger than the biggest member of the set, but 5N has to be the largest multiple of 5 less than K.<\/p>\n

We know that the mean of the set is not divisible by 5, and this is important.\u00a0 For an evenly spaced set, the mean & median are identical.\u00a0 If there were an odd number of members, the mean & median would simply be the middle number on the list, and this of course would be some multiple of 5.\u00a0 The fact that mean & median is not divisible by 5 necessarily means that there must be an even number of members on the list. This way, the mean & median would be the average of the middle two numbers, which would be a non-integer, certainly not an integer divisible by 5.\u00a0 Therefore, N is an even number.<\/p>\n

Finally, we are also told that N itself is not divisible by 5\u2014an odd constraint that may be relevant.<\/p>\n

Statement #1<\/span>: Here, N could be any even number less than 52, as long as it’s not divisible by 5.\u00a0 It could be {48, 46, 44, 42, 38, \u2026.}\u00a0 Multiple possibilities.\u00a0 This statement, alone and by itself, is not sufficient<\/strong>.<\/p>\n

Statement #2<\/span>: The statement tells us that K > 5*48, so it could be that N = 48, so that K would be between 5(48) and 5*(49).\u00a0 That’s a possibility, or N could be any larger even number that is not divisible by 5.\u00a0 N could be {48, 52, 54, 56, 68, 62, \u2026}\u00a0 Again, multiple possibilities.\u00a0 This statement, alone and by itself, is not sufficient<\/strong>.<\/p>\n

Combined statements<\/span>: with both constraints, the only possible value if N = 48.\u00a0 The statements together are sufficient<\/strong>.<\/p>\n

Answer = (C) <\/strong><\/p>\n

6. This is a tricky one.<\/p>\n

Statement #1<\/span>: two kinds of numbers have exactly four factors: (a) products of two distinct prime numbers, and (b) cubes of prime numbers.<\/p>\n

The product of two distinct prime numbers S and T would have factors {1, S, T, ST}.\u00a0 For example, the factors of 10 are (1, 2, 5, 10), and the factors of 21 are {1, 3, 7, 21}.<\/p>\n

The cube of a prime number S would have as factors 1, S, S squared, and S cubed.\u00a0 For example, 8 has factors {1, 2, 4, 8} and 27 has factors {1, 3, 9, 27}.<\/p>\n

We know N is not a power of D, so the second case is excluded.\u00a0 N must be the product of two distinct prime numbers.\u00a0 We know D < N, so of the four factors, D can’t be the product of the two prime numbers.\u00a0 D could be either of the prime number factors, or D could be 1, which is not a prime number<\/a>.\u00a0\u00a0 Because D could either be a prime number or 1, we cannot give a definitive answer to the question.\u00a0 This statement, alone and by itself, is not sufficient<\/strong>.<\/p>\n

Statement #2<\/span>: this is tricky.\u00a0 The first few plug-ins seem to reveal a pattern.<\/p>\n

\"ghdmpp_img25\"<\/p>\n

Even if you sense a pattern, it’s important to remember that plugging in numbers alone is never enough to establish that a DS statement is sufficient.\u00a0 Here, if we persevered to one more plug-in, we would find the one that breaks the pattern.<\/p>\n

\"ghdmpp_img26\"<\/p>\n

That gives another answer to the prompt, so we know this statement is not sufficient<\/strong>.<\/p>\n

To avoid a lot of plugging in, it’s also very good to know that in mathematics, prime numbers are notorious for not following any easy pattern. \u00a0It is impossible to produce an algebraic formula that will always produce prime numbers. <\/strong>\u00a0In fact, this is more than you need to know, but the hardest unsolved question in higher mathematics, the Riemann Hypothesis<\/a>, concerns the pattern of prime numbers; mathematicians have been working on this since 1859, and no one has proven it yet.\u00a0 Suffice to say that no one-line algebraic formula is going to unlock the mystery of prime numbers!<\/p>\n

Combined statements<\/span>: according to the information in statement #1, either D = 1 or D is a prime number.\u00a0 Well, statement #2 excludes the possibility that D = 1, because that number cannot be written as two more than a power of 3.\u00a0 Therefore, D must be a prime number.\u00a0 We have a definitive answer to the prompt question.\u00a0 Combined, the statements are sufficient<\/strong>.<\/p>\n

Answer = (C)<\/strong><\/p>\n

DS Probability Questions<\/h2>\n

1. A class contains boys and girls.\u00a0 What is the probability of selecting a boy from a class?<\/p>\n

Statement #1<\/span>: there are 35 students in the class<\/p>\n

Statement #2<\/span>: the ratio of boys to girls is 3:4<\/p>\n

2. Bert has $1.37 of loose change in his pocket\u00a0 —- pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25).\u00a0 He reaches into his pocket and pulls out one coin at random.\u00a0 What is the probability that the coin is a nickel?<\/p>\n

Statement #1<\/span>: There are exactly seven pennies in his pocket<\/p>\n

Statement #2<\/span>: There are exactly three quarters in his pocket<\/p>\n

Answers:<\/strong><\/p>\n

1. Statement #1<\/span>: from this, we know there are 15 employees selected.\u00a0 We don’t know the value of M, which is clearly greater than or equal to 15.\u00a0 If M is very large, it may be highly unlikely that either Andrew or Georgia is seated.\u00a0 If M is closer to 15, then that would change the probability. Without knowing the value of M, we can’t complete the calculation.\u00a0\u00a0 This statement, alone and by itself, is insufficient<\/b>.<\/p>\n

Statement #2<\/span>: Ignore the information in the first statement.\u00a0 If M = N, then all employees, everyone in the pool of selection, takes a seat.\u00a0 It is simply a matter of seating a group of M employees randomly in M seats.<\/p>\n

Think about this.\u00a0 For every seating arrangement with Andrew to the right of Georgia, there will be exactly one seating arrangement, the mirror image, that has Andrew to the left of Georgia.\u00a0\u00a0 Since there’s a one-to-one correspondence between these two, there must be just as many arrangements with Andrew on one side of Georgia as on the other side.\u00a0\u00a0 Both are equally likely, so the probability must be 1\/2.<\/p>\n

This information was enough to calculate a numerical answer to the prompt question.\u00a0\u00a0 This statement, alone and by itself, is sufficient<\/b>.<\/p>\n

First not sufficient, second sufficient.\u00a0 Answer = B<\/b>.<\/p>\n

2. A reminder for non-American students — $1.00 = 100\u00a2.\u00a0 Thus<\/p>\n

penny = $0.01 = 1\u00a2<\/p>\n

nickel = $0.05 = 5\u00a2<\/p>\n

dime = $0.10 = 10\u00a2<\/p>\n

quarter = $0.25 = 25\u00a2<\/p>\n

Statement #1<\/span>: Bert has seven pennies, amounting to 7\u00a2.\u00a0 The other coins total $1.30.\u00a0 This could be all nickels — 26 nickels — so that of the 7 + 26 = 33 coins in the pocket, 26 are nickel, and the probability of picking a nickel would be 26\/33.\u00a0 Or, that $1.30 could be 4 quarters and 3 dimes, so that there were no<\/span> nickels, and the probability of picking a nickel would be zero!\u00a0 Different possible choices lead to different answer to the prompt question, so this statement, alone and by itself, is insufficient<\/b>.<\/p>\n

Statement #2<\/span>: Bert has three quarters, amounting to 75\u00a2.\u00a0 The other coins total 62\u00a2 — this could be two pennies and twelve nickels, that of the 3 + 2 + 12 = 17 coins in the pocket, 12 are nickels, and the probability is 12\/17. Or, that 62\u00a2 could be entirely in pennies, without any nickels or dimes at all: then the probability of picking a nickel would be zero.\u00a0\u00a0 Different possible choices lead to different answer to the prompt question, so this statement, alone and by itself, is insufficient<\/b>.<\/p>\n

Combined statements<\/span>: Now, we know Bert has seven pennies and three quarters, and these ten coins together account for 82\u00a2.\u00a0 The remaining 55\u00a2 must be composed of dimes and nickels.\u00a0 There could be eleven nickels and no dimes, so that of the 10 + 11 = 21 coins in the pocket, 11 are nickels, and P = 11\/21.\u00a0 OR, there could be five dimes and one nickel in the pocket, so of the 10 + 5 + 1 = 16 coins in the pocket, only one is a nickel, and P = 1\/16.\u00a0 Different possible choices lead to different answer to the prompt question, so both statements combined are insufficient<\/b>.<\/p>\n

Nothing is sufficient.\u00a0 Answer = E<\/b><\/p>\n

Try This Tricky Data Sufficiency Question!<\/h2>\n

On each lab that Ren\u00e9 completed he received either 100 points or 85 points. On how many labs did he score 100 points?<\/p>\n