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# Math Revolution Approach (PS)

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15 Mar 2018, 02:24
[GMAT math practice question]

X is proportional to Y. If Y is increased by 30%, by approximately what percent is X^2 increased?

A. 30%
B. 40%
C. 50%
D. 60%
E. 70%

=>

X = kY for some k.
Since (1.3kY)^2 = 1.69k^2Y^2 = 1.69(kY)^2 = 1.69X^2, we have (1.69X^2 – X^2) / X^2 = 0.69, which is approximately 70%.

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16 Mar 2018, 00:55
[GMAT math practice question]

Out of 75 students, 17 students enrolled in a Physics class, 28 students enrolled in a Chemistry class, and 39 students enrolled in a Biology class. 5 students enrolled in both the Physics and Chemistry classes, 7 students enrolled in both the Chemistry and Biology classes, and 6 students enrolled in both the Biology and Physics classes. If 4 students enrolled in Physics, Chemistry, and Biology, how many students did not enroll in any of the three science classes?

A. 2
B. 3
C. 4
D. 5
E. 6

=>

Attachment:

777.png [ 10.71 KiB | Viewed 562 times ]

The total number of students is given by
a + b + c + d + e + f + g + h = 75.
Of these,
a + d + f + g = 17 are enrolled in Physics,
b + d + e + g = 28 are enrolled in Chemistry, and
c + e + f + g = 39 are enrolled in Biology.
Adding these three equations gives ( a + b + c ) + 2( d + e + f ) + 3g = 84.
We also know that
d + g = 5 study both Physics and Chemistry,
e + g = 7 study both Chemistry and Biology, and
f + g = 6 study both Physics and Biology.
Adding these three questions yields ( d + e + f ) + 3g = 18.
Since 4 students are enrolled in all three classes, we have g = 4,

Plugging this value for g into the above equation yields
(d + e + f) + 3g = 18
(d + e + f) + 12 = 18
d + e + f = 6.

So,
( a + b + c ) + 2( d + e + f ) + 3g = 84 yields
(a + b + c) + 2(6) + 3(4) = 84
(a + b + c) + 12 + 12 = 84
a + b + c = 60.

Finally, using the equation, a + b + c + d + e + f + g + h = 75, we see that
(a + b + c) + (d + e + f) + g + h = 75
60 + 6 + 4 + h = 75
h = 5.

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18 Mar 2018, 23:20
[GMAT math practice question]

What is the minimum distance between the point (3,4) and points on the circle x2+y2=1?

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

Attachment:

19.png [ 6.84 KiB | Viewed 540 times ]

=>

Since the distance between (0,0) the center of the circle x^2+y^2=1 and (3,4) is 5, and the radius of the circle is 1, the minimum distance between a point on the circle and (3,4) is 5 – 1 = 4.

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18 Mar 2018, 23:22
[GMAT math practice question]

Suppose y=2^{x^2}+2^{x+3}. Which of these x-values gives the smallest value of y?

A. -3
B. -2
C. -1
D. 1
E. 2

=>

Since the exponent function y=2^x is an increasing function, the smallest value of x^2+2x+3 gives the smallest value of y=2^{x^2}+2^{x+3}.
Since x^2+2x+3 = (x+1)^2 + 2, the exponent (x+1)^2 + 2 has a minimum value when x = -1.

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21 Mar 2018, 02:51
[GMAT math practice question]

In how many ways can two integers m and n, with m > n, be selected from the whole numbers from 12 to 32, inclusive?

A. 150
B. 180
C. 190
D. 210
E. 240

=>

Since the order of m and n is fixed, we only need to count the number of ways to choose 2 numbers from 12, 13, …, 32.

We have 21 numbers to choose from since 32 – 12 + 1 = 21.

The number of selections is
21C2 = (21*20) / (1*2) = 210.

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22 Mar 2018, 17:55
[GMAT math practice question]

Machine A can produce balls at a constant rate of 2 balls per hour, and machine B can produce balls at a constant rate of 3 balls per hour. If at least one of machine A and machine B produces balls at any time, what is the smallest possible number of hours that machine A and machine B must work together at their constant rates to produce 70 balls in 20 hours?

A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs

=>

Since machine B has the faster working rate, it must work for the entire period to minimize the number of hours that machine A has to work.
When machine B works for 20 hours, it produces 60 balls. This means that machine A must produce 70 – 60 = 10 balls, and it must work for 5 hours, since the work rate of machine A is 2 balls per hour.

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23 Mar 2018, 00:22
[GMAT math practice question]

Alice and Bob traveled in the same direction along the same route at their respective constant speeds of 12 km per hour and 6 km per hour. They each started to travel from their own houses. Bob’s house is halfway between Alice’s house and the destination. After passing Bob, Alice took 10 minutes to reach the destination. How many minutes did it take Bob to reach the destination after Alice passed him?

A. 5 min
B. 6 min
C. 8 min
D. 10 min
E. 20 min

=>

After Alice passed Bob, she traveled for 10 more minutes.
Since Bob’s speed is half of Alice’s speed, he took a further 20 minutes to reach the destination.

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25 Mar 2018, 18:21
[GMAT math practice question]

When you buy a wrap from a certain store, you must select at least one of the three kinds of meat (beef, pork and fish), and one of lettuce, spinach and cucumber. You can also add one of oriental sauce, spicy mayo and sesame soy sauce. How many kinds of wraps can you buy?

A. 6
B. 9
C. 12
D. 18
E. 27

=>
There are three ways to choose one meat (beef, pork or fish), three ways to choose one vegetable (lettuce, spinach or cucumber), and 3 ways to choose one sauce (oriental sauce, spicy mayo or sesame soy sauce).
So, the total number of different wraps that can be bought is

3*3*3 = 27.

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25 Mar 2018, 18:22
[GMAT math practice question]

2018 has 365 days and the 294th day, Oct 31. 2018 is on a Wednesday. On which day will the last day of 2018 fall?

A. Mon
B. Tue
C. Wed
D. Thu
E. Fri

=>

Since we have 30 days in November and 31 days in December, we have 61 days remain in 2018 after Oct. 31, 2018.

Now,
61 = 7*8 + 5.

Since 61 has a remainder of 5 when it is divided by 7, the last day of the year 2018 must be a Monday (count 5 days on from Wednesday).

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28 Mar 2018, 02:02
[GMAT math practice question]

Which of the following is the greatest?

A. (1+2)^5
B. 10^3
C. (1^2+2^2)^4
D. (2^2+2^2+2^2)^3
E. (2^2+2^2+2^2+2^2)^2

=>

A. (1+2)^5 = 3^5 = 243
B. 10^3 = 1000
C. (1^2+2^2)^4 = 5^4 = 625
D. (2^2+2^2+2^2)^3 = 12^3 = 1728
E. (2^2+2^2+2^2+2^2)^2 = 16^2 = 256

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30 Mar 2018, 02:07
[GMAT math practice question]

Which of the following inequalities is equivalent to |2x-|x||<3?

A. 0<x<2
B. 0<x<3
C. -1<x<3
D. 0<x<1
E. -3<x<1

=>

|2x-|x||<3
=> -3 < 2x – |x| < 3

Case 1: If x ≥ 0, then |x| = x, and so
-3 < 2x – |x| < 3
=> -3 < x < 3
=> 0 ≤ x < 3, since x ≥ 0.

Case 2: If x < 0, then |x| = - x, and so
-3 < 2x – |x| < 3
=> -3 < 3x < 3
=> - 1< x < 1
=> - 1< x < 0, since x < 0.

Thus, -1 < x < 3.

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01 Apr 2018, 18:46
[GMAT math practice question]

How many different three-digit numbers can be formed which contain two digits that are the same, and a third digit that is different from the other two?

A. 196
B. 216
C. 243
D. 256
E. 316

=>

These three-digit numbers can have one of the forms XXY, XYX and YXX.
Note that 0 cannot be the hundreds digit.

Case 1): XXY
There are 9 possibilities for X (X is not 0), and 9 possibilities for Y (Y≠X).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.

Case 2): XYX
There are 9 possibilities for X (X is not 0), and 9 possibilities for Y (Y≠X).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.

Case 3): YXX
There are 9 possibilities for Y (Y is not 0), and 9 possibilities for X (X≠Y).
This gives a total of 9*9 = 81 possible three-digit numbers of this form.

Thus, the total number of possible three-digit numbers is 81 + 81 + 81 = 243.

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01 Apr 2018, 18:48
[GMAT math practice question]

An organization constructs a committee of 3 people from A, B, C, D, E, F and G. A and B are relatives, so they cannot both be committee members at the same time. How many different committees can be formed?

A. 25
B. 30
C. 35
D. 40
E. 45

=>
The total number of possible committees of three 3 people chosen from 7 people is 7C3 = 35.
However, we need to exclude the committees containing both A and B.
The number of committees containing A, B and one other person is equal to the number of ways of choosing 1 person from the C, D, E, F, and G, which is 5C1 = 5.
Therefore, the total number of committees 35 – 5 = 30.

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04 Apr 2018, 02:52
[GMAT math practice question]

The number 123,k50 is a 6-digit integer, and k is a positive 1 digit integer. Which of the following cannot be a factor of 123,k50?

A. 2
B. 3
C. 4
D. 5
E. 6

=>

The last two digits tell us whether the number is divisible by 4.
Since 50 is not a multiple of 4, the number cannot be a multiple of 4.

Let’s see why the number could be divisible by each of the other options:

A: Since the units digit is an even number, the whole number is a multiple of 2.
B: A number is divisible by 3 if the sum of its digits is divisible by 3. If k = 4, then the sum of the digits is 1 + 2 + 3 + 4 + 5 + 0 = 15, which is a multiple of 3, and so the number is a multiple of 3.
D: Since the units digit is a multiple of 5, the number is a multiple of 5.
E: If k = 4, the number is divisible by 3 as seen in part B. Since it is also divisible by 2 (see part A), the number is divisible by 6.
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06 Apr 2018, 00:52
[GMAT math practice question]

Car A drives from P to Q at a constant rate of 100 km per hour. After car A has driven for 1 hour, train B begins traveling from Q to P at a constant rate of 150 km per hour. If the distance between P and Q is 600 km, then what distance has car A traveled when it meets train B?

A. 200 km
B. 220 km
C. 250 km
D. 270 km
E. 300 km

=>

After car A has driven for 1 hour, the distance between car A and train B is 500 km.

Car A and train B approach each other at a speed of 250 km/hr. This means that they will take 2 hours to meet each other.
When they meet, car A will have traveled for 3 hours, and have covered a distance of 3 * 100 = 300 km.

It is important that both vehicles will have traveled for the same amount of time after train B has started moving. This gives the equation 100 + 100t + 150t = 600, from which we may deduce that t = 2. Since 100 + 100 *2 = 300, E is the answer.
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08 Apr 2018, 18:15
[GMAT math practice question]

If x > y >0, a < 0 and b > 0, then which of the following is (are) true?

Ⅰ. ax+by
Ⅱ.ax-by
Ⅲ. by-ax

A. Ⅰonly
B. Ⅱ only
C. Ⅲ only
D.Ⅰ& Ⅱ only
E. Ⅱ & Ⅲ only

=>

Statement I.
x = 2, y = 1, a = -1, b = 1: False

Statement II.
x = 2, y = 1, a = -1, b = 1: False

Statement III.
Since a < 0 and x > 0, -ax > 0. And by > 0 since b > 0 and y > 0.
by – ax = by + (-ax) > 0. True

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08 Apr 2018, 18:18
[GMAT math practice question]

If Rn+1-Rn=(-1/2)^n for positive integers n, which of the following is true?

A. R1>R3>R2
B. R1>R2>R3
C. R3>R1>R2
D. R2>R3>R1
E. R3>R2>R1

=>
If n = 1, then R2 – R1 = -(1/2) < 0, and we have R1 > R2.
If n = 2, then R3 – R2 = 1/4 > 0, and we have R3 > R2.
Furthermore, R3 – R1 = ( R3 – R2 ) + ( R2 – R1 ) = 1/4 + (-1/2) = -1/4 < 0, and we have R3 < R1.
Thus R1 > R3 > R2.

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11 Apr 2018, 05:57
[GMAT math practice question]

A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from the jar, and its number is recorded before it is returned to the jar. If a second ball is then selected from the jar, what is the probability that the difference between the numbers on the two balls selected is 1?

A. 1/3
B. 3/4
C. 1/2
D. 5/8
E. 3/8

=>

The total number of ways the two balls may be selected is 4C2 = ( 4 * 3 ) / ( 1 * 2 ) = 6.
There are three ways in which the numbers on the two balls selected can have a difference of 1: ( 1, 2 ), ( 2, 3 ), and ( 3, 4 ).

Thus, the probability is 3/6 or 1/2

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12 Apr 2018, 01:59
[GMAT math practice question]

In the x-y plane, what is the slope of the line segment joining the y-intercept and the positive x-intercept of the curve y=x^2-5x-6?

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

=>

First, we find the x-intercepts of the curve. These occur when y = 0. Now,
y=x^2-5x-6
=> y = (x+1)(x-6).

Thus, the x-intercepts are the points (-1,0) and (6,0), and the positive x-intercept is (6,0).
As the y-intercept occurs when x = 0, it is the point (0,-6).

Thus, the required slope is (6 – 0) / ( 0 – (-6) ) = 1.

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13 Apr 2018, 02:44
[GMAT math practice question]

Each letter of a 5-letter password must be a different letter of the alphabet. Each password must contain 2 vowels and 3 consonants, and the vowels must not be adjacent to each other. How many different passwords of this form can be made?

A. 95760
B. 128000
C. 159600
D. 256000
E. 720000

=>

The total number of passwords with 2 vowels and 3 consonants is determined as follows:
1) Choose the 2 different vowels in 5C2 ways
2) Choose the 3 consonants in 21C3 ways
3) Arrange the 5 letters in 5! ways
So, the number of passwords is
21C3 * 5C2 * 5! = ( 21*20*19 ) / (1*2*3) * (5*4)/(1*2) * 120 = 133 * 10 * 120 = 159600.

The number of passwords with 2 adjacent vowels and 3 consonants is
determined as follows:
1) Choose the 2 different vowels in 5C2 ways
2) Choose the 3 consonants in 21C3 ways
3) Choose a position for the two adjacent vowels in 4! ways
4) Rearrange the adjacent vowels in 2! ways

21C3 * 5C2 * 4!*2! = ( 21*20*19 ) / (1*2*3) * (5*4)/(1*2) * 48 = 133 * 10 * 48 = 63840.

159600 – 63840 = 95760.

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# Math Revolution Approach (PS)

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