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Math Revolution GMAT Instructor
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07 Sep 2018, 00:20
[ Math Revolution GMAT math practice question] If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, which of the following must be a factor of n? A. 16 B. 32 C. 36 D. 48 E. 64 => Since 4 = 2^2,14 = 2*7, 27 = 3^3 are factors of n and n is a perfect cube, the smallest possible value of n is 2^3*3^3*7^3. When n = 2^3*3^3*7^3, 16 = 2^4, 32 = 2^5, 48=2^4*3 and 64 = 2^6 can’t be factors of n. The only answer choice that is a factor of n is 36 = 2^2*3^2. Therefore, the answer is C. Answer: C
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09 Sep 2018, 17:28
[ Math Revolution GMAT math practice question] What is the sum of roots of the equation x^2 – 40x + 399 = 0? A. 19 B. 20 C. 21 D. 40 E. 399 => Let p and q be the roots of the equation x^2 – 40x + 399 = 0. Then x^2 – 40x + 399 = (xp)(xq) = x^2 – (p+q)x + pq. Equating coefficients gives p + q = 40 from the coefficient of x. Therefore, the answer is D. Answer: D
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09 Sep 2018, 17:29
[ Math Revolution GMAT math practice question] If n is the greatest positive integer for which 5^n is a factor of 50!, what is the value of n? A. 10 B. 11 C. 12 D. 13 E. 14 => 50! = 1*2*…*5*…*10*…*15*…*20*…*25*…*30*…*35*…*40*…*45*…*50 =1*2*…*(5)…*(2*5)*…*(3*5)*…*(4*5)*…*(52)*…*(6*5)*…*(7*5)*…*(8*5)*…*(9*5)*…*(2*5^2) Since 5 is a prime number, no further factors of 5 appear in the prime factorization of 50!. The number of 5s in the above expansion of 50! is 12. Therefore, the answer is C. Answer: C
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12 Sep 2018, 01:43
[ Math Revolution GMAT math practice question] If 5^{m2}2^{m+2} is an ndigit integer, where m is an integer greater than 2, what is the value of n, in terms of m? A. n=m4 B. n=m2 C. n=m1 D. n=m E. n=2m => 5^{m2}2^{m+2} = 5^{m2}2^{m2+4} = 5^{m2}2^{m2}*2^4 = (5*2)^{m2}*2^4= (10)^{m2}*2^4 = 16*(10)^{m2} = 16*(100…0)= 1600…0 with m2 0’s. 10^{m2} has m2 digits that are 0 and 16*(10)^{m2} has digits including 1, 6 and m2 digits that are 0. Thus, 16*(10)^{m2} has m – 2 + m = m digits. Therefore, D is the answer. Answer: D
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13 Sep 2018, 00:37
[ Math Revolution GMAT math practice question] What is the value of 22C 19? A. 770 B. 1540 C. 3080 D. 4620 E. 6160 => Since nCnr = nCr, 22C 19 = 22C 3 = (22*21*20)/(1*2*3) = 11*7*20 = 1540. Therefore, the answer is B. Answer: B
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13 Sep 2018, 23:40
[ Math Revolution GMAT math practice question] If the interior angles of a triangle are in the ratio 3 to 4 to 5, what is the measure of the largest angle? A. 30 B. 45 C. 60 D. 75 E. 90 => Let x, y and z be interior angles of the triangle. Since x:y:z = 3:4:5, we can write x = 3k, y = 4k and z = 5k. Since the interior angles of the triangle add to 180, x + y + z = 3k + 4k + 5k = 12k = 180, and so k = 180/12 = 15. Therefore, the largest angle of the triangle is z = 5k = 5(15) = 75. Therefore, the answer is D. Answer: D
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16 Sep 2018, 17:33
[ Math Revolution GMAT math practice question] (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=? A. 41/3 B. 41/6 C. 41 D. 210 E. 420 => Now, 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6 and 1 + 2 + 3 + … + n = n(n+1)/2. So, (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20) = ( 20*21*41/6 ) / ( 20*21/2 ) = (41/6) / (1/2) = 41/3 Therefore, the answer is A. Answer: A
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16 Sep 2018, 17:35
[ Math Revolution GMAT math practice question] If 1/n(n+1) = 1/n – 1/(n+1), then what is the value of 1/(1*2) + 1/(2*3) + 1/(3*4) + … + 1/(99*100) A. 1/100 B. 1/50 C. 49/50 D. 99/100 E. 1/2 => 1/(1*2) + 1/(2*3) + 1/(3*4) + … + 1/(99*100) = (1/1 – 1/2) + (1/2 – 1/3) + (1/3 – 1/4) + … + (1/99 – 1/100) = 1/1 – 1/100 = 1 – 1/100 = 99/100 after cancellation of the inner terms. Therefore, the answer is D. Answer: D
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19 Sep 2018, 01:13
[ Math Revolution GMAT math practice question] x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear polynomials with integer coefficients. Which of following cannot be a factor of x^4+12x^3+49x^2+78x+40? A. x+1 B. x+2 C. x+3 D. x+4 E. x+5 => Assume x^4+px^3+qx^2+rx+s = (x+a)(x+b)(x+c)(x+d). Then, s = abcd. The constant terms a, b, c and d of the linear factors are factors of the constant term s of the original polynomial. Since 3 is not a factor of 40, x + 3 cannot be a factor of the original polynomial x^4+12x^3+49x^2+78x+40. Therefore, C is the answer. Answer: C
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20 Sep 2018, 01:02
[ Math Revolution GMAT math practice question] Which of the following functions satisfies f(a+b)=f(a)f(b) for all positive numbers a, b ? A. f(x)=x+1 B. f(x)=x^2+1 C. f(x)=√x D. f(x)=1/x E. f(x)=2^x => A. f(1) = 2, f(2) = 3, f(1)f(2) = 6, but f(1+2) = f(3) = 4. Choice A is incorrect. B. f(2) = 5, f(3) = 10, f(2)f(3) = 50, but f(2+3) = f(5) = 26. Choice B is incorrect. C. f(9) = 3, f(16) = 4, f(9)f(16) = 12, but f(9+16) = f(25) = 5. Choice C is incorrect. D. f(1) = 1, f(2) = 1/2, f(1)f(2) = 1/2, but f(1+2) = f(3) = 1/3. Choice D is incorrect. E. Let a,b > 0. Then f(a+b) = 2^{a+b} = 2^a2^b = f(a)f(b). Choice E is correct. Therefore, the answer is E. Answer: E
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21 Sep 2018, 00:25
[ Math Revolution GMAT math practice question] √32+3√2+5+√3+1√2=? A. 0 B. 2√2 C. 2√3 D. 9 E. 11 => A=A when A>0, 0=0, and A=A when A<0 Since √32 < 0, we have √32 = (√32). Since 3√2 > 0, we have 3√2 = 3√2. Since 5+√3 > 0, we have 5+√3 = 5+√3 Since 1√2 < 0, we have 1√2 = (1√2) So, √32+3√2+5+√3+1√2= (√32) +(3√2) + (5+√3) (1√2) = √3+2 + 3√2 + 5+√3 – 1 +√2 = 9. Therefore, the answer is D. Answer: D
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24 Sep 2018, 05:13
[ Math Revolution GMAT math practice question] How many multiples of 33 lie between 101 and 1,000, inclusive? A. 24 B. 27 C. 33 D. 36 E. 48 => Consider the arithmetic sequence 132, 165, …, 990 of multiples of 33. The number of terms in this sequence is (990132)/33 + 1 = 858 / 33 + 1 = 26 + 1 = 27. Therefore, the answer is B. Answer: B
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24 Sep 2018, 05:17
[ Math Revolution GMAT math practice question] 1/(2√3)^2=？ A. 2+√3 B. 2√3 C. 7+4√3 D. 74√3 E. 4+7√3 => 1/(2√3)^2= [1/(2√3)]^2= {(2+√3)/{(2√3)(2+√3)}^2={(2+√3)/(43)}^2=(2+√3)^2= 4+4√3+3 = 7+4√3. Therefore, the answer is C. Answer: C
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26 Sep 2018, 04:05
[ Math Revolution GMAT math practice question] In the coordinate plane, what is the area of the region enclosed by x≥1, y≥2 and x + y ≤ 5? A. 1 B. 2 C. 3 D. 4 E. 5 => The region can be sketched in the xyplane as shown below. Attachment:
a.png [ 9.94 KiB  Viewed 180 times ]
The area of the shaded triangle above is (1/2)(31)(42) = (1/2)*2*2 = 2 Therefore, B is the answer. Answer: B
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26 Sep 2018, 23:58
[ Math Revolution GMAT math practice question] The sequence S n has the terms S 1=1, S 2=2, S 3=6, S 4=15, S 5=31, ……. If n is a positive integer, what could be the value of S nS n1 in terms of n? A. n B. n^2 C. –n D. n^2 E. (n1)^2 => We begin by writing out the terms of the sequence: S 2S 1 = 2 1 = 1=1^2 S 3S 2 = 6 – 2 = 4 = 2^2 S 4S 3 = 15 – 6 = 9 = 3^2 S 5S 4 = 3115 = 16 = 4^2 … Following the above pattern gives S nS n1 = (n1)^2 Therefore, the answer is E. Answer: E
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27 Sep 2018, 23:41
[ Math Revolution GMAT math practice question] What is the sum of the solutions of the equation (x1)^2=x1? A. 1 B. 0 C. 1 D. 2 E. 3 => (x1)^2=x1 => x1^2=x1 => x1^2x1=0 => x1 (x11)=0 => x1 = 0 or x11=0 => x1 = 0 or x1=1 => x1 = 0 or x1=±1 => x=1 = 0 or x=1±1 => x=1, x=0 or x=2 The sum of the solutions is 0 + 1 + 2 = 3. Therefore, the answer is E. Answer: E
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30 Sep 2018, 21:56
[ Math Revolution GMAT math practice question] The 2 lines x+2y=3, 2x+py=q have infinitely many points of intersection in the xyplane. Which of the following could be the value of p? A. 0 B. 1 C. 2 D. 3 E. 4 => If the 2 lines have infinitely many points of intersection, their equations must specify the same straight line. The equation x+2y=3 is equivalent to 2x+4y=6. So, p = 4 and q = 6. Therefore, the answer is E. Answer: E
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30 Sep 2018, 21:59
[ Math Revolution GMAT math practice question] When 2 people are selected at random from a group of 4 females and 4 males, what is the probability that at least one female is selected? A. 5/14 B. 7/14 C. 9/14 D. 11/14 E. 13/14 => The probability that at least one female is selected is equal to 1 minus the probability that two males are selected. There are 4C 2 ways of selecting 2 males from the four males and 8C 2 ways of selecting 2 people from the group of 8. Therefore, the probability of selecting two males from the group is 4C 2/ 8C 2, and the probability of selecting at least one female is 1 – 4C 2/ 8C 2 = 1 – { (4*3)/(1*2) } / { (8*7)/(1*2) } = 1 – (4*3)/(8*7) = 1 – 3/14 = 11/14 Therefore, the answer is D. Answer: D
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03 Oct 2018, 01:28
[ Math Revolution GMAT math practice question] Five letters A, P, P, L and E are listed in a row. How many arrangements have at least one letter between the two Ps? A. 24 B. 30 C. 36 D. 42 E. 48 => When we encounter “at least” in counting questions or probability questions, we should consider complementary counting. The total number of arrangements of the 5 letters is 5!/2! (5! Counts each arrangement of the two Ps 2! times). The number of arrangements with no letter between the two Ps is 4!. Thus, the number of arrangements in which at least one letter lies between the two Ps is 5!/2! – 4! = 60 – 24 = 36. Therefore, C is the answer. Answer: C
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03 Oct 2018, 23:57
[ Math Revolution GMAT math practice question] If 3 different numbers are selected from the first 8 prime numbers, what is the probability that the sum of the 3 numbers selected is even? A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 3/8 => Suppose p, q and r are prime numbers. In order for p + q + r to be even, one of them must equal 2, since 2 is the only even prime number. Once 2 has been selected, there are 7 prime numbers remaining from which to select 2 numbers. Thus, the number of selections in which the sum of the 3 numbers is even is 7C 2 = 21. The total number of selections is 8C 3 = 56. Thus, the probability that the sum of the three numbers is even is 21/56 = 3/8. Therefore, the answer is E. Answer: E
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