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Math Revolution Approach (PS)

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New post 18 Dec 2017, 00:29
[GMAT math practice question]

In the 5-digit integer 54xy2, x and y are single digits. What is the probability that 54xy2 is divisible by 4?

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

=>
The divisibility of 54xy2 by 4 depends on the last two digits only. So, 54xy2 is divisible by 4 exactly when 10y + 2 is divisible by 4.
Out of the possible values for 10y + 2, 12, 32, 52, 72, 92 are divisible by 4, but 02, 22, 42, 62, 82 are not. Therefore, the required probability is 1/2.

Therefore, the answer is C.

Answer : C
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New post 20 Dec 2017, 02:02
[GMAT math practice question]

46, 47, 48, 49, 50, 51, 52, 53, 54
The standard deviation of the 9 numbers in the above list lies between 2 and 3. How many of the 9 numbers are within one standard deviation of the average (arithmetic mean)?

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

=>
The average of the 9 numbers is 50. So, if n lies within one standard deviation of the mean, then
50 – 2.xxx < n < 50 + 2.xxx
47.xxx < n < 52.xxx
and n = 48, 49, 50, 51, or 52.
There are five numbers in this range.

Therefore, the answer is A.
Answer : A
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New post 21 Dec 2017, 00:47
[GMAT math practice question]

Suppose you have a collection of 1 cent, 5 cent, 10 cent, 25 cent, and 50 cent coins. If you make 50 cents with these coins, which of the following could be the total number of coins used?

I. 41
II. 31
III. 26

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

=>

Let a, b, c, d and e be the numbers of 1 cent, 5 cent, 10 cent, 25 cent and 50 cent coins, respectively.
Then a + 5b + 10c + 25d + 50d = 50.
The possible numbers of each coin are as follows:
(a,b,c,d,e) = (50,0,0,0,0) : 50 coins
(a,b,c,d,e) = (45,1,0,0,0) : 46 coins
(a,b,c,d,e) = (40,2,0,0,0) : 42 coins
(a,b,c,d,e) = (40,0,1,0,0) : 41 coins
(a,b,c,d,e) = (35,1,1,0,0) : 37 coins
(a,b,c,d,e) = (30,2,1,0,0) : 33 coins
(a,b,c,d,e) = (30,0,2,0,0) : 32 coins
(a,b,c,d,e) = (25,1,2,0,0) : 28 coins
(a,b,c,d,e) = (25,0,0,1,0) : 26 coins

Only 41 and 26 are in this list.
Therefore, the answer is D.

Answer: D
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New post 22 Dec 2017, 06:47
[GMAT math practice question]

Which of the following is closest to (1+10^26)/(5^4+10^10)?

A. 10^15
B. 10^16
C. 10^17
D. 10^18
E. 10^19

=>

(1+10^26)/(5^4+10^10)
≒ 10^26/10^10 = 10^16

Therefore, the answer is B.

Answer: B
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New post 25 Dec 2017, 18:06
[GMAT math practice question]

Alice and Bob were asked to solve the equation x^2+px+q=0. Alice obtained the solution set x=1 and x=4, which was incorrect. She obtained the solution set of the equation x^2+px+s=0. Bob found the solution set x=-2 and -3, which was also incorrect. Bob’s answer was the solution set of the equation x^2+tx+q=0. What are the solutions of the original equation?

A. 2,-3
B. -2,3
C. 2,3
D. -2,-3
E. -1,-4

=>
Since (x-1)(x-4) = x^2 - 5x + 4 = x^2 + px +s, p = -5 and s = 4.
Since (x+2)(x+3) = x^2 + 5x + 6 = x^2 + ts + q, t = 5 and q = 6.
Thus x2 + px + q = x^2 -5x + 6. This factors as (x-2)(x-3) = 0, and so its solutions are x = 2 and x = 3.
Therefore, the answer is C.

Answer: C
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New post 25 Dec 2017, 18:07
[GMAT math practice question]

Alice drives her car from town M to town N at a constant rate of 35 miles per hour. One hour after Alice sets out, Bob starts to drive from town N to town M along the same road at a constant rate of 45 miles per hour. If the distance between towns M and N is 355 miles, then what distance in miles has Alice traveled when she and Bob pass each other?

A. 70 miles
B. 105 miles
C. 140 miles
D. 175 miles
E. 210 miles

=>
After 1 hour, Alice has traveled 35 miles. So, the distance between Alice and Bob is 320 miles. Alice and Bob approach each other at a speed of 35 + 45 = 80 miles per hour. They pass each other after 320/80 = 4 hours. Thus, when they pass each other, Alice has been driving for 1 + 4 = 5 hours. The total distance she has traveled is 5 x 35 = 175 miles.

Therefore, the answer is D.

Answer : D
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New post 27 Dec 2017, 01:04
[GMAT math practice question]

4 points a, b, c and d are placed on the number line in that order. If abcd>0, which of the following must be positive?

I. ab
II. bc
III cd

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

=>

There are three cases we need to consider. Note that if a < 0 < b < c < d or a < b < c < 0 < d, then abcd < 0.

Case 1) 0 < a < b < c < d:
ab > 0, bc > 0 and cd > 0

Case 2) a < b < 0 < c < d:
ab > 0, bc < 0, cd > 0

Case 3) a < b < c < d < 0:
ab > 0, bc > 0, cd > 0

Only statements I and III are always true.

Therefore, the answer is D.

Answer : D
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New post 28 Dec 2017, 00:33
[GMAT math practice question]

The manager of a certain company gives its employees one rostered day off between Monday and Friday each week. In how many ways can four employees take their leave, if they cannot all be rostered off together?

A. 120
B. 125
C. 620
D. 625
E. 3125

=>

Each employee may be rostered off on 5 different days. This gives 5^4 = 625 ways in which they can take their rostered days off. We need to exclude the cases where are all employees take their leave together on Monday, Tuesday, Wednesday, Thursday or Friday. There are 5 such cases.

So, the number of ways in which the four employees can take their leave is 625 – 5 = 620.

Therefore, the answer is C.

Answer: C
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New post 28 Dec 2017, 22:57
[GMAT math practice question]

2 teachers and 3 students line up in a row. How many different arrangements are possible if the teachers cannot be adjacent to each other?

A. 60
B. 72
C. 81
D. 90
E. 100

=>

Since at least one student should be between the two teachers, we should consider complementary cases. This means we should calculate the difference between the total number of arrangements and the number of arrangements in which the teachers are adjacent to each other.


The total number of ways in which 5 people can stand in a row is 5!= 120.
The total number of arrangements with the teachers adjacent to each other is 4! * 2!.
Thus, the total number of permitted arrangements is 5! – 4! * 2! = 120 – 48 = 72.

Therefore, the answer is B.

Answer: B
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New post 28 Dec 2017, 23:34
Possibilities of difference 3 are - ( 6,3),(3,6); (1,4)(4,1);(2,5)(5,2)..total cases area 6..so probability is 6/36= 1/6

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New post 01 Jan 2018, 02:24
[GMAT math practice question] 12.2

How many different sums can be formed by adding 2 different numbers from the set {1, 2, 4, 8, 16, 32, 33}?

A. 16
B. 17
C. 18
D. 19
E. 20

=>
All choices of two different numbers from this set give rise to different sums apart from 1 + 33 and 2 + 32.
The number of ways to choose 2 numbers from the set {1, 2, 4, 8, 16, 32, 33} of 7 numbers is 7C2 = (7*6) / (1*2) = 21.
Since two of these choices, 1, 33 and 2, 32, have the same sum, we need to subtract 1 from 21.
Then we have 21 – 1 = 20 possible choices.

Therefore, the answer is E.
Answer: E
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New post 03 Jan 2018, 01:42
[GMAT math practice question]

For positive integer n, with distinct prime factors p1, p2,…,pn, the function f(n) = n(1-1p1)(1-1p2)(1-1p3)….(1- 1pk) gives the number of positive integers less than n which have no common factor with n except 1. What is the value of f(30) ?

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

=>


Since 30 = 2*3*5, f(30) = 30*(1-1/2)(1-1/3)(1-1/5) = 30*(1/2)(2/3)(4/5)=8.

Therefore, the answer is D.

Answer : D
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New post 04 Jan 2018, 01:28
[GMAT math practice question]

The probability that an afternoon storm will occur on Monday is 40%, the probability that an afternoon storm will occur on Tuesday is 50%, and the probability that an afternoon storm will occur on Wednesday is 40%. What is the probability that afternoon storms will occur on exactly two of these three days?

A. 0.28
B. 0.30
C. 0.32
D. 0.34
E. 0.36

=>

There are three cases to consider.

Case 1: Afternoon storms occur on Monday and Tuesday, but not on Wednesday.
Probability = 0.4 * 0.5 * 0.6 = 0.12

Case 2: Afternoon storms occur on Monday and Wednesday, but not on Tuesday.
Probability = 0.4 * 0.5 * 0.4 = 0.08

Case 3: Afternoon storms occur on Tuesday and Wednesday, but not on Monday.
Probability = 0.6 * 0.5 * 0.4 = 0.12

The probability that one of these three cases will occur is 0.12 + 0.08 + 0.12 = 0.32

Therefore, the answer is C.

Answer: C
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New post 05 Jan 2018, 00:58
[GMAT math practice question]

In how many ways can two integers x and y (with x>y) be selected from -10 to 10 (inclusive)?

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


=>

This is the number of ways of selecting two different numbers from a set of 21 numbers. The order of choosing the numbers does not matter: we simply assign the larger number to x once the choice has been made. So, the number of ways of choosing the two numbers is

21C2 = 21*20 / (1*2) = 21 * 10 = 210

Therefore, the answer is D.

Answer: D
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New post 07 Jan 2018, 18:17
[GMAT math practice question

Attachment:
20180104_173047.png
20180104_173047.png [ 1.22 KiB | Viewed 307 times ]


The areas of faces A, B, and C of the rectangular solid shown in the above figure are 12, 15, and 20, respectively. What is the volume of the solid?

A. 36
B. 48
C. 60
D. 64
E. 72

=>
Let a and b be the edges of face A, and let b and c be the edges of face B. Then the edges of face C are a and c. Moreover, the volume of the solid is abc. Now,
ab = 12, bc = 15 and ca = 20.

Multiplying these together yields
(ab)(bc)(ca) = a^2b^2c^2 = (abc)2 = 12*15*20 = 3600 = 60^2
So abc = 60.

Therefore, the answer is C.
Answer: C
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New post 10 Jan 2018, 01:09
[GMAT math practice question]

If m, n and p are positive integers such that (p^m)^n=256, which of the following could be the value of n?

I. 1
II. 2
III. 4

A. I and II
B. I and III
C. II and III
D. II only
E. I, II and III

=>
Since 256 = 2^8, (p^m)^n = p^{mn} = 2^8.
So, p = 2 and mn = 8.

Since n is a factor of 8, n could be any of the values 1, 2, 4, and 8.

Therefore, the answer is E.
Answer: E
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New post 11 Jan 2018, 01:46
[GMAT math practice question]

Suppose x=-1 and k is a positive number. If n is a positive root of the equation n4 – 24k = 0, what is the value of x+x^n+x^{n+1}+x^{n+2}?

A. -2
B. -1
C. 0
D. 1
E. 2

=>
n^4 – 2^{4k} = 0
⇔ (n^2-2^{2k})(n^2+2^{2k}) = 0
⇔ (n^2-2^{2k}) = 0, since n^2+2^{2k} ≠ 0
⇔ (n+2^k)(n-2^k) = 0
⇔ n = 2^k
Since k > 0, n must be even.

Thus n = 2m for some integer m. It follows that
x+x^n+x^{n+1}+x^{n+2}
= (-1)+(-1)^n+(-1)^{n+1}+(-1)^{n+2}
= (-1)+(-1)^{2m}+(-1)^{2m +1}+(-1)^{2m+2}
= (-1) + 1 + (-1) + 1 = 0

Therefore, the answer is C.
Answer: C
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New post 12 Jan 2018, 00:53
[GMAT math practice question]

A palindromic number is a number that remains the same when its digits are reversed. For example, 16461 is a palindromic number. If a 4 digit integer is selected randomly from the set of all 4 digit integers, what is the probability that it is palindromic?

A. 1/20
B. 1/50
C. 1/60
D. 1/90
E. 1/100

=>
4-digit palindromic numbers have the form ‘xyyx’, where
x is one of values 1,2,…,9 and y is one of values 0,1,2,…,9.
So, there are 9 x 10 = 90 four-digit palindromic numbers.
The total number of 4-digit numbers between 1000 and 9999, inclusive, is 9000 ( = 9999 – 1000 + 1 ).

Therefore, the probability that the selected 4-digit is palindromic is 90 / 9000 = 1/100.
Therefore, the answer is E.
Answer: E
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Re: Math Revolution Approach (PS) [#permalink]

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New post 14 Jan 2018, 18:34
[GMAT math practice question]

Let f(x)= (x-p)(x-q). If f(11)=f(20)=0, then f(10)=?

A. 1
B. -1
C. 5
D. -5
E. 10

=>

Since 11 and 20 are roots of f(x) = 0, we must have f(x) = (x-11)(x-20).
Thus, f(10) = (10-11)(10-20) = (-1)(-10) = 10.

Therefore, the answer is E.
Answer : E
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New post 17 Jan 2018, 01:06
[GMAT math practice question]

In the 6-digit integer 543,2xy, x and y are chosen from the digits 0, 2, 4, 6 and 8. What is the probability that 543,2xy is divisible by 8?

A. 1/5
B. 6/25
C. 7/25
D. 8/25
E. 9/25

=>

An integer with three or more digits is divisible by 8 if and only if its last three digits from a number that is divisible by 8.

The values of 2xy that are divisible by 8 are 200, 208, 224, 240, 248, 264, 280, and 288.
The total number of 6-digit integers of the form 543,2xy is equal to the number of ways of choosing two digits from 0, 2, 4, 6 and 8, which is 5*5 = 25.
Thus, the probability that 543,2xy is divisible by 8 is 8/25

Therefore, the answer is D.
Answer: D
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Re: Math Revolution Approach (PS)   [#permalink] 17 Jan 2018, 01:06

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Math Revolution Approach (PS)

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