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Math Revolution GMAT Instructor
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18 Jan 2018, 01:44
[GMAT math practice question] What is the product of all roots of the equation (x+1)^2=x+1? A. 2 B. 1 C. 0 D. 1 E. 2 => Now, (x+1)^2=x+1 ⇔ x+1^2=x+1 ⇔ x+1^2x+1= 0 ⇔ x+1(x+11) = 0 ⇔ x+1 = 0 or x+1 = 1 ⇔ x = 1 or x+1 = ±1 ⇔ x = 1 or x = 1 ±1 ⇔ x = 1, x= 2 or x = 0 The product of these solutions is (1)*(2)*0 = 0. Therefore, the answer is C. Answer: D
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Math Revolution GMAT Instructor
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19 Jan 2018, 01:20
[GMAT math practice question] Two students were asked to solve the equation x^2+px+q=0. Alice’s answer was x=1 and 5. This answer is incorrect for the original equation, but correct for the equation x^2+px+r=0. Bob’s answer was x=2 and 4. This answer is incorrect for the original equation, but correct for x^2+sx+q=0. What are the solutions of the original equation? A. 1, 8 B. 1,8 C. 2,4 D. 1,5 E. 1,5 => Using the factor theorem with Alice’s solution yields (x1)(x5) = x^2 – 6x + 5 = x^2+px+r. So, p = 6. Using the factor theorem with Bob’s solution yields (x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q. So, q = 8. Thus, the original equation is x^26x+8 = (x2)(x4) = 0. Its roots are 2 and 4. Therefore, the answer is C. Answer: B
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Math Revolution GMAT Instructor
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21 Jan 2018, 18:41
[GMAT math practice question] If 0<2x+3y<10 and 10<3x+2y<0, then which of the following must be true? I. x<0 II. y<0 III. x<y A. I only B . II only C. I & II D.I & III E. I, II, &III => Label the inequalities as follows: 0<2x+3y<10  (1) 10<3x+2y<0  (2) We consider each statement individually. Statement I: Multiplying (1) by 2 yields 20 < 4x – 6y < 0, and multiplying (2) by 3 yields 30 < 9x + 6y < 0. Adding these inequalities gives 50 < 5x < 0 or 10 < x < 0. This statement is true. Statement II: Multiplying (1) by 3 yields 0 < 6x + 9y < 30, and multiplying (2) by 3 yields 0 < 6x – 4y < 20. Adding these inequalities gives 0 < 5y < 50 or 0 < y < 10. This statement is false. Statement III: Multiplying (1) by – 1 yields 10 < 2x – 3y < 0. Adding this to inequality (2) yields 20 < x – y < 0. This implies that x < y, and statement III is true. Therefore, the answer is D. Answer: B
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Math Revolution GMAT Instructor
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21 Jan 2018, 18:42
[GMAT math practice question] If 8＜x＜6 and 2＜y＜3, which of the following could be the value of √(xy)? A. 3 B. 4 C. 5 D. 6 E. 7 => Multiplying 8 < x < 6 by 1 yields 6 < x < 8. Since 2 < y < 3, it follows (by multiplication) that 12 < (x)y < 24. Taking square roots (since all numbers are positive) yields √12 < √(xy) < √24. Since 3 < √12 and 5 > √24, 4 is one possible value of √(xy). Therefore, the answer is B. Answer : C
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Math Revolution GMAT Instructor
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24 Jan 2018, 01:33
[GMAT math practice question] How many solutions has the equation x32=1? A. 0 B. 1 C. 2 D. 3 E. 4 => x32=1 ⇔x32=±1 ⇔x3=2±1 ⇔x3=1 or x3=3 ⇔x3=±1 or x3=±3 ⇔x=3±1 or x=3±3 ⇔x=2, x = 4, x = 0 or x = 6 Thus, the equation has four solutions. Therefore, the answer is E. Answer: E
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25 Jan 2018, 04:06
[GMAT math practice question] If x=343y, where y is a positive integer, and x/196 is a terminating decimal, what is the smallest possible value of y? A. 1 B. 3 C. 5 D. 7 E. 9 => x/196 = (343*y)/196 = (7^3*y)/(14^2) = (7^3*y)/(2^2)(7^2) = (7y)/4 As the denominator has only 2 as a prime factor, it is a terminating decimal, regardless of the value of y. Thus, the smallest possible value of y is 1. Therefore, the answer is A. Answer: A
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26 Jan 2018, 01:20
[GMAT math practice question] What is the least common multiple of 6, 12, 20, 30 and 42? A. 120 B. 180 C. 240 D. 360 E. 420 => These numbers have the following prime factorizations: 6 = 2^1*3^1, 12 = 2^2*3^1, 20 = 2^2*5^1, 30 = 2^1*3^1*5^1 and 42 = 2^13^17^1 In order to find their LCM, we multiply together each of the primes, raised to their maximum exponents found in the above prime factorizations. The LCM of the numbers is 2^2*3^1*5^1*7^1 = 420 since the maximum exponent of 2 is 2, the maximum exponent of 3 is 1, the maximum exponent of 5 is 1 and the maximum exponent of 7 is 1. Therefore, the answer is E. Answer: E
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Math Revolution GMAT Instructor
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28 Jan 2018, 18:46
[GMAT math practice question] If x and y are integers, which of the following CANNOT be the value of x^2+y^2? A. 121 B. 122 C. 123 D. 125 E. 130 => Squares of even integers, (2k)^2 = 4k^2 have the remainder 0, when they are divided by 4. Squares of odd integers, (2k+1)^2 = 4k^2 + 4k + 1 have the remainder 1, when they are divided by 4. Hence, squares of integers can have remainders of 0 or 1 only, when they are divided by 4. So, the sum of two squares of integers cannot have the remainder of 3 when it is divided by 4. Thus, 123 cannot be the value of x^2+y^2. Therefore, the answer is C. Answer: C
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Math Revolution GMAT Instructor
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28 Jan 2018, 18:47
[GMAT math practice question] Given that 1+2+.......+n=n(n+1)/2 and 1^2+2^2+…..+n^2=n(n+1)(2n+1)/6, what is the sum of the integers between 1 and 100 (inclusive) that are not squares of integers? A. 4050 B. 4665 C. 4775 D. 5000 E. 5050 => Since there are 10 squares of integers between 1 and 100, inclusive, we need to find 1+2+…+100 – (1^2+2^2+…+10^2) = (100*101)/2 – (10*11*21)/6 = 5050 – 385 = 4665 Therefore, the answer is B. Answer : B
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31 Jan 2018, 02:52
[GMAT math practice question] A grocery store sells apples by the pound. If the price per pound is increased by $1, $12 will buy 0.4 pounds less of apples than if the price remains at the current level. What is the current price per pound of apples at the grocery store? A. $4 B. $4.5 C. $5 D. $5.5 E. $6 => Suppose p is the current price of apples at the grocery store. Then 12/p – 12/(p+1) = 0.4 ⇔ 12(p+1) – 12p = 0.4p(p+1) if we multiply both sides by p(p+1) ⇔ 12 = 0.4(p^2+p) ⇔ 30 = p^2+p after multiplying by 2.5 ⇔ p^2+p30 = 0 ⇔ (p5)(p+6) = 0 ⟹ p = 5, since prices cannot be negative. Therefore, the answer is C. Answer: C
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01 Feb 2018, 01:38
[GMAT math practice question] If m<n<0, x=m^2+n^2, and y=2mn, what is the value of √x^2y^2, in terms of m and n? A. m^2+n^2 B. 2m^2+n^2 C. m^2+2n^2 D. m^2n^2 E. n^2m^2 => x^2y^2 = (m^2+n^2)^2 – (2mn)^2 = (m^2+n^2)^2 – 4m^2n^2 = m^4 + 2m^2n^2 + n^4 – 4m^2n^2 = m^4 – 2m^2n^2 + n^4 = (m^2n^2)^2 Therefore, √x^2y^2 =√(m^2n^2)^2 = m^2n^2 = m^2n^2, since m^2 > n^2. Therefore, the answer is D. Answer: D
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Math Revolution GMAT Instructor
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02 Feb 2018, 02:03
[GMAT math practice question] If a⊙b=(a+b)^22ab, which of the following is(are) true? Ⅰ. a⊙b=b⊙a Ⅱ. (a⊙b)⊙c=a⊙(b⊙c) Ⅲ. a⊙1=a^2+1 A. Ⅰ only B. Ⅱ only C. Ⅲ only D. Ⅰ and Ⅲ E. Ⅰ,Ⅱ and Ⅲ => a⊙b=(a+b)^22ab=a^2+2ab+b^22ab = a^2 + b^2 Statement I b⊙a = b^2+a^2 = a^2 + b^2 = a⊙b Therefore, statement I is true. Statement II (a⊙b)⊙c = (a^2+b^2) ⊙c = (a^2+b^2)^2 +c^2 = a^4 + 2a^2b^2 + b^4 + c^2 a⊙(b⊙c) = a⊙ (b^2+c2) = a^2+ (b^2+c^2)^2 = a^2 + b^4 + 2b^2c^2 + c^4 We can easily find a counterexample. If a = 1, b = 2 and c = 3, then (a⊙b)⊙c = (1⊙2)⊙3 = (1^2+2^2) ⊙3 = 5⊙3 = 5^2 + 3^2 = 25 + 9 = 34 and a⊙(b⊙c) = 1⊙(2⊙3) = 1⊙(2^2+3^2) = 1⊙13 = 1^2+13^2 = 1 + 169 = 170. Thus, statement II is false. Statement III a⊙1 = a^2 + 1^2 = a^2 + 1 Therefore, statement III is true. Therefore, the answer is D. Answer: D
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04 Feb 2018, 18:14
[GMAT math practice question] If a≠b, and a^2/(ab)=b, which of the following could be a? A. 1 B. 0 C. 1 D. 2 E. a does not exist => a^2/(ab)=b ⇔ a^2=b(ab) ⇔ a^2=ab – b^2 ⇔ a^2  ab + b^2 = 0 Since a≠b, one of a and b is not zero. If a ≠ 0 or b ≠ 0, a^2  ab + b^2 cannot be zero for the following reason: a^2  ab + b^2 = a^2 – 2a(b/2) + b^2 = a^2 – 2a(b/2) + b^2/3 + (3/4)b^2 = (a – b/2)^2 + (3/4)b^2 > 0 Thus, there is no pair of real numbers (a,b) satisfying the equation a^2  ab + b^2 = 0. Therefore, a cannot exist, and the answer is E. Answer: E
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04 Feb 2018, 18:16
[GMAT math practice question] If 2^n+2^{n2}=5120, then n=? A. 8 B. 9 C. 10 D. 11 E. 12 => Factoring yields 2^n+2^{n2}=2^22^{n2}+2^{n2}=(2^2+1)2^{n2}=5*2^{n2}=5120=5*1024. Therefore, 2^{n2}=1024=2^{10} and n2 = 10. It follows that n = 12. Therefore, the answer is E. Answer : E
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07 Feb 2018, 00:47
[GMAT math practice question] 5 couples sit on 10 chairs around a round table. If each couple must be seated together, how many possible arrangements are there? A. 256 B. 512 C. 768 D. 1,024 E. 1,080 => The number of arrangements of the 5 couples in a circle is (51)! = 4!. The members of each couple can be arranged in 2! ways. Thus, the total number of arrangements is 4! * 2! * 2! * 2! * 2! * 2! = 24*2*2*2*2*2 = 768. Therefore, the answer is C. Answer: C
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08 Feb 2018, 01:12
[GMAT math practice question] What is the range of 29 consecutive even numbers? A. 56 B. 58 C. 60 D. 62 E. 64 => The number of consecutive even integers is (the last term(L) – the first term(F))/2 + 1 and its range is LF. Now, Range / 2 + 1 = (LF)/2 + 1 = 29. So, Range / 2 = 28, and, hence, Range = 56 Therefore, the answer is A. Answer: A
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09 Feb 2018, 01:43
[GMAT math practice question] How many values of x satisfy the equation x79=11? A. 1 B. 2 C. 3 D. 4 E. 5 => x79 = 11 ⇔ x79 = ±11 ⇔ x7 = 9 ±11 ⇔ x7 = 20 or x7 = 2 ⇔ x7 = 20 since x7 ≥ 0 ⇔ x7 = ±20 ⇔ x = 7±20 ⇔ x = 27 or x = 13 Therefore, the answer is B. Answer: B
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11 Feb 2018, 18:00
[GMAT math practice question] Which of the following inequalities holds if 3.24<x<5.36? A. x+4.3<1.06 B. x4.3<1.06 C. x2.1<0.53 D. x+2.1<0.53 E. x2.1<1.06 => The inequality xa<b is equivalent to the inequality a  b < x < a + b. Note that a is the average of a – b and a + b and b is half of the difference between a – b and a + b, that is, b = {( a + b ) – ( a – b )} / 2. Thus, the inequality 3.24 < x < 5.36 is equivalent to the inequality  x – 4.3  < 1.06 since a = ( 3.24 + 5.36 ) / 2 = 8.6 / 2 = 4.3 and b = ( 5.36 – 3.24 ) / 2 = 2.12 / 2 = 1.06. Therefore, B is the answer. Answer: B
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11 Feb 2018, 18:02
[GMAT math practice question] The probability that event A occurs is 1/3, and the probability that event B occurs is 1/12. If events A and B are independent, what is the probability that both events A and B occur? A. 1/6 B. 1/12 C. 1/15 D. 1/18 E. 1/36 => If events A and B are independent, then P(A∩B) = P(A)P(B). So, P(A∩B) = P(A)P(B) = (1/3)*(1/12) = 1/36. Therefore, the answer is E. Answer : E
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14 Feb 2018, 01:24
[GMAT math practice question] What is the least common multiple of 39, 91, 93 and 217? A. 8461 B. 8462 C. 8463 D. 8464 E. 8465 => Since 39 = 3 * 13, the least common multiple must be a multiple of 3. The sum of the digits of a multiple of 3 is a multiple of 3. 8643 is the only answer choice for which the sum of the digits is a multiple of 3: 8 + 4 + 6 + 3 = 21 = 3 * 7. Therefore, C is the answer. Alternatively, since 39 = 3*13, 91 = 7*13, 93 = 3*31, and 217=7*31, the least common multiple of the four numbers is 3*7*13*31 = 8463. Therefore, the answer is C. Answer: C
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