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Math Revolution Approach (PS)

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Re: Math Revolution Approach (PS) [#permalink] New post   18 Jan 2018, 01:44
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[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|^2-|x+1|= 0
⇔ |x+1|(|x+1|-1) = 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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Re: Math Revolution Approach (PS) [#permalink] New post   19 Jan 2018, 01:20
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[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
(x-1)(x-5) = 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^2-6x+8 = (x-2)(x-4) = 0. Its roots are 2 and 4.

Therefore, the answer is C.
Answer: B
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Re: Math Revolution Approach (PS) [#permalink] New post   21 Jan 2018, 18:41
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[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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Re: Math Revolution Approach (PS) [#permalink] New post   21 Jan 2018, 18:42
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[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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Re: Math Revolution Approach (PS) [#permalink] New post   24 Jan 2018, 01:33
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[GMAT math practice question]

How many solutions has the equation ||x-3|-2|=1?

A. 0
B. 1
C. 2
D. 3
E. 4

=>

||x-3|-2|=1
⇔|x-3|-2=±1
⇔|x-3|=2±1
⇔|x-3|=1 or |x-3|=3
⇔x-3=±1 or x-3=±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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Re: Math Revolution Approach (PS) [#permalink] New post   25 Jan 2018, 04:06
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[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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Re: Math Revolution Approach (PS) [#permalink] New post   26 Jan 2018, 01:20
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[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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Re: Math Revolution Approach (PS) [#permalink] New post   28 Jan 2018, 18:46
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[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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Re: Math Revolution Approach (PS) [#permalink] New post   28 Jan 2018, 18:47
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[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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Re: Math Revolution Approach (PS) [#permalink] New post   31 Jan 2018, 02:52
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[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+p-30 = 0
⇔ (p-5)(p+6) = 0
⟹ p = 5, since prices cannot be negative.

Therefore, the answer is C.

Answer: C
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Re: Math Revolution Approach (PS) [#permalink] New post   01 Feb 2018, 01:38
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[GMAT math practice question]

If m<n<0, x=m^2+n^2, and y=2mn, what is the value of √x^2-y^2, in terms of m and n?

A. m^2+n^2
B. 2m^2+n^2
C. m^2+2n^2
D. m^2-n^2
E. n^2-m^2

=>

x^2-y^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^2-n^2)^2

Therefore,
√x^2-y^2
=√(m^2-n^2)^2
= |m^2-n^2|
= m^2-n^2, since m^2 > n^2.

Therefore, the answer is D.

Answer: D
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Re: Math Revolution Approach (PS) [#permalink] New post   02 Feb 2018, 02:03
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[GMAT math practice question]

If a⊙b=(a+b)^2-2ab, 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)^2-2ab=a^2+2ab+b^2-2ab = 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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Re: Math Revolution Approach (PS) [#permalink] New post   04 Feb 2018, 18:14
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[GMAT math practice question]

If a≠b, and a^2/(a-b)=b, which of the following could be a?

A. -1
B. 0
C. 1
D. 2
E. a does not exist

=>
a^2/(a-b)=b
⇔ a^2=b(a-b)
⇔ 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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Re: Math Revolution Approach (PS) [#permalink] New post   04 Feb 2018, 18:16
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[GMAT math practice question]

If 2^n+2^{n-2}=5120, then n=?

A. 8
B. 9
C. 10
D. 11
E. 12

=>
Factoring yields
2^n+2^{n-2}=2^22^{n-2}+2^{n-2}=(2^2+1)2^{n-2}=5*2^{n-2}=5120=5*1024.
Therefore,
2^{n-2}=1024=2^{10}
and
n-2 = 10.
It follows that
n = 12.

Therefore, the answer is E.
Answer : E
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Re: Math Revolution Approach (PS) [#permalink] New post   07 Feb 2018, 00:47
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[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 (5-1)! = 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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Re: Math Revolution Approach (PS) [#permalink] New post   08 Feb 2018, 01:12
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[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 L-F. Now, Range / 2 + 1 = (L-F)/2 + 1 = 29. So, Range / 2 = 28, and, hence, Range = 56

Therefore, the answer is A.

Answer: A
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Re: Math Revolution Approach (PS) [#permalink] New post   09 Feb 2018, 01:43
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[GMAT math practice question]

How many values of x satisfy the equation ||x-7|-9|=11?

A. 1
B. 2
C. 3
D. 4
E. 5

=>

||x-7|-9| = 11
⇔ |x-7|-9 = ±11
⇔ |x-7| = 9 ±11
⇔ |x-7| = 20 or |x-7| = -2
⇔ |x-7| = 20 since |x-7| ≥ 0
⇔ x-7 = ±20
⇔ x = 7±20
⇔ x = 27 or x = -13


Therefore, the answer is B.

Answer: B
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Re: Math Revolution Approach (PS) [#permalink] New post   11 Feb 2018, 18:00
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[GMAT math practice question]

Which of the following inequalities holds if 3.24<x<5.36?

A. |x+4.3|<1.06
B. |x-4.3|<1.06
C. |x-2.1|<0.53
D. |x+2.1|<0.53
E. |x-2.1|<1.06

=>

The inequality |x-a|<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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Re: Math Revolution Approach (PS) [#permalink] New post   11 Feb 2018, 18:02
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[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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Re: Math Revolution Approach (PS) [#permalink] New post   14 Feb 2018, 01:24
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[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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Re: Math Revolution Approach (PS) [#permalink]
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