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Math Revolution Approach (PS)

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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (PS)  [#permalink]

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New post 07 Sep 2018, 01:20
[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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Math Revolution GMAT Instructor
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Joined: 16 Aug 2015
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Re: Math Revolution Approach (PS)  [#permalink]

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New post 09 Sep 2018, 18:28
[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 = (x-p)(x-q) = 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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New post 09 Sep 2018, 18:29
[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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Re: Math Revolution Approach (PS)  [#permalink]

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New post 12 Sep 2018, 02:43
[Math Revolution GMAT math practice question]

If 5^{m-2}2^{m+2} is an n-digit integer, where m is an integer greater than 2, what is the value of n, in terms of m?

A. n=m-4
B. n=m-2
C. n=m-1
D. n=m
E. n=2m

=>

5^{m-2}2^{m+2} = 5^{m-2}2^{m-2+4} = 5^{m-2}2^{m-2}*2^4 = (5*2)^{m-2}*2^4= (10)^{m-2}*2^4
= 16*(10)^{m-2} = 16*(100…0)= 1600…0 with m-2 0’s.
10^{m-2} has m-2 digits that are 0 and 16*(10)^{m-2} has digits including 1, 6 and m-2 digits that are 0.
Thus, 16*(10)^{m-2} has m – 2 + m = m digits.

Therefore, D is the answer.
Answer: D
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New post 13 Sep 2018, 01:37
[Math Revolution GMAT math practice question]

What is the value of 22C19?

A. 770
B. 1540
C. 3080
D. 4620
E. 6160

=>

Since nCn-r = nCr,
22C19 = 22C3 = (22*21*20)/(1*2*3) = 11*7*20 = 1540.

Therefore, the answer is B.
Answer: B
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New post 14 Sep 2018, 00:40
[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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New post 16 Sep 2018, 18:33
[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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New post 16 Sep 2018, 18:35
[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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New post 19 Sep 2018, 02:13
[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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Re: Math Revolution Approach (PS)  [#permalink]

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New post 20 Sep 2018, 02:02
[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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Math Revolution GMAT Instructor
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Joined: 16 Aug 2015
Posts: 6227
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Re: Math Revolution Approach (PS)  [#permalink]

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New post 21 Sep 2018, 01:25
[Math Revolution GMAT math practice question]

|√3-2|+|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 √3-2 < 0, we have |√3-2| = -(√3-2).
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, |√3-2|+|3-√2|+|5+√3|+|1-√2|= -(√3-2) +(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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Re: Math Revolution Approach (PS) &nbs [#permalink] 21 Sep 2018, 01:25

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