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Math Revolution Approach (PS)

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New post 15 Feb 2018, 04:40
[GMAT math practice question]

How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’ in a straight line are possible?

A. 60
B. 120
C. 180
D. 240
E. 420

=>

The seven letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’ include three s’s and two c’s.
The number of permutations is 7! / (3!*2!) = 420.

Therefore, the answer is E.

Answer: E
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New post 16 Feb 2018, 05:08
[GMAT math practice question]

A committee has 10 members. How many different selections of a chairman, an editor, and a secretary can be made from the members of the committee?

A. 180
B. 360
C. 540
D. 720
E. 810

=>

There are 10 ways of choosing the chairman. Once the chairman has been chosen, there are 9 remaining possibilities for the editor. Once both the chairman and the editor have been chosen, 8 choices remain for the secretary. So, the total number of choices for these three positions is:
10*9*8 = 720.
This may also be written as
10P3 = 10 * 9 * 8 = 720.

Therefore, the answer is D.

Answer: D
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New post 18 Feb 2018, 19:23
[GMAT math practice question]

The length and width of a certain rectangle are in the ratio 4 to 5. The area of the rectangle is 2,000 square units. If the length and width of the rectangle are both increased by 10 units, what is the increase in the area of the rectangle?

A. 800
B. 900
C. 1,000
D. 1,100
E. 1,250

=>

Suppose the length is L = 4k and the width is W = 5k for some k. Then
L*W = 4k*5k = 20k^2 = 2,000.
k^2 = 100 and k = 10.
Thus, L = 4k = 40 and W = 5k = 50.
The area of the rectangle after increasing both the length and the width is (L+10)(W+10) = (40 + 10)(50 + 10) = 50*60 = 3,000 square units.
Thus, the increase in the area is 3,000 – 2,000 = 1,000 square units.

Therefore, the answer is C.

Answer : C
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New post 18 Feb 2018, 19:25
[GMAT math practice question]

Attachment:
2.16.png
2.16.png [ 4.66 KiB | Viewed 362 times ]


What is the value of x?

A. 1.5
B. 3.0
C. 4.5
D. 5.0
E. 6.0

=>

The two triangles are similar since they have the same angles. Consequently, each pair of corresponding sides is in the same proportion: 39:3 = (18+x):x.
Rearranging this proportion yields
39x = 3(18+x)
39x = 3*18 + 3x
36x = 54
x = 1.5

Therefore, A is the answer.

Answer: A
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New post 21 Feb 2018, 01:06
[GMAT math practice question]

The points A, B, C and D lie on the number line in that order. If AB=BC/3, AC=(4/7)CD, A=-1, and D=10, what is the value of B?

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

=>

Attachment:
2.21.png
2.21.png [ 1015 Bytes | Viewed 348 times ]


Let BC = 3d.
Then AB = BC/3 = d and AC = 4d.
Also, AC = 4d = (4/7)CD and CD = 7d.
Therefore,
AD = AC + CD = 4d + 7d = 11d = 10 – (-1) = 11.
So,
d = 1, and
B = -1 + d = -1 + 1 = 0.

Therefore, the answer is C.
Answer: C
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New post 22 Feb 2018, 02:11
[GMAT math practice question]

Let A=2^{50}, B=3^{30}, and C=5^{20}. Which of the following is true?

A. A<B<C
B. A<C<B
C. C<A<B
D. B<C<A
E. C<B<A

=>

If we wish to compare these numbers, we need to either make their bases the same or make their exponents the same. In this case, it is easiest to make all exponents the same as follows:
A=2^{50} = (2^5)^{10} = 32^{10}
B=3^{30} = (3^3)^{10} = 27^{10}
C=5^{20} = (5^2)^{10} = 25^{10}

Since 32 > 27 > 25, we must have A > B > C.

Therefore, the answer is E.

Answer: E
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New post 25 Feb 2018, 18:40
[GMAT math practice question]

Which of the following is closest to 11^5*9^5–7*10^7?

A. 10^2
B. 10^7
C. 10^8
D. 10^9
E. 10^10


=>

Since 11 is close to 10 and 9 is close to 10, 11^5*9^5 is close to 10^5*10^5 = 10^10.
Since 7*10^7 is a relatively small number compared with 10^10, 11^5*9^5 – 7*10^7 is approximately equal to 10^10.

Therefore, the answer is E.

Answer: E
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New post 25 Feb 2018, 18:42
[GMAT math practice question]

What is the units digit of 2^105+3^105+4^105+5^105?

A. 3
B. 4
C. 5
D. 6
E. 7

=>

The units digit of any integer raised to the exponent 5 is the same as the units digit of the integer:
0^1 and 0^5 have remainder 0
1^1 and 1^5 have remainder 1
2^1 and 2^5 have remainder 2

9^1 and 9^5 have remainder 9
when divided by 10.

Therefore, the units digit of 2^105+3^105+4^105+5^105 is same as the units digit of 2 + 3 + 4 + 5 = 14.
It is 4

Therefore, B is the answer.

Answer: B
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New post 26 Feb 2018, 02:35
[GMAT math practice question]

When both positive integers a and b are divided by 9, their remainders are 7. What is the reminder when a2b is divided by 9?

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

=>
The best way to approach remainder questions is to plug in a number. We shall plug in 7 for both a and b because its remainder, when divided by 9, is 7, as required by the question. Then a^2b=7^27=343, and the remainder of 343 when it is divided by 9 is 1.

Therefore, the answer is D.

Answer : D
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New post 28 Feb 2018, 01:29
[GMAT math practice question]

The 20% acid liquid solution is produced by adding a gallons of 10% acid liquid solution to b gallons of 50% acid liquid solution. To get 10 gallons of 20% acid liquid solution, how many gallons of 50% acid liquid solution are needed?

A. 1
B. 1.5
C. 2
D. 2.5
E. 3

=>

We have a + b = 10.

( 0.1a + 0.5b ) / 10 = 0.2
⇔ 0.1a + 0.5b = 2
⇔ 0.1(10-b) + 0.5b = 2
⇔ 1 – 0.1b + 0.5b = 2
⇔ 0.4b = 1
⇔ b = 2.5

Therefore, D is the answer.

Answer: D
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New post 01 Mar 2018, 03:49
[GMAT math practice question]

Which of the following is 6/7 times as far from 7/3 as is 17/6 from 2/3?

A. 88/21
B. 21/88
C. 77/20
D. 20/77
E. 78/21

=>

We have |x – 7/3| = (6/7)|17/6 – 2/3|.
|x - 7/3| = (6/7)(13/6)
x – 7/3 = ±(6/7)(13/6) = ±13/7
x = ±13/7 + 7/3
x = (39 + 49)/21 = 88/21 or x = (-39 + 49)/21=10/21

Therefore, the answer is A.

Answer: A
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New post 02 Mar 2018, 01:21
[GMAT math practice question]

When x/y = 2.6, (x-y)/(x+y)=?

A. 2/7
B. 3/8
C. 4/9
D. 5/9
E. 7/10


=>

( x – y ) / ( x + y ) = ( x/y – 1 ) / ( x/y + 1 ) = ( 2.6 – 1 ) / ( 2.6 + 1 ) = 1.6 / 3.6 = 16 / 36 = 4 / 9

Therefore, the answer is C.

Answer: C
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New post 04 Mar 2018, 19:25
[GMAT math practice question]

There are 4 machines with the same work rate. If it took k hours for 3 machines to work and did k-2 hours for 4 machines when working together, what is the value of k?

A. 4
B. 5
C. 6
D. 7
E. 8

=>

Work Amount = Number of machines * Work Rate * Time
If r is the machine’s work rate, we have 3rk = 4r(k-2) or 3k = 4k -8.
Thus k = 8.

Therefore, E is the answer.

Answer: E
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New post 04 Mar 2018, 19:28
[GMAT math practice question]

John traveled the entire 50 km trip. If he traveled the first 20 km of the trip at a constant rate 40 km per hour and the remaining trip at a constant rate 20 km per hour, what is his average speed, in km per hour?

A. 20 km/h
B. 23 km/h
C. 25 km/h
D. 26 km/h
E. 27 km/h

=>
Ave Speed = ( Total Distance ) / ( Total Time )
The total distance = 50 km.
Total time is 20 / 40 + 30 / 20 = 0.5 + 1.5 = 2
The average speed is 50 / 2 = 25.
Therefore, the answer is C.

Answer : C
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New post 07 Mar 2018, 03:40
[GMAT math practice question]

If n!/(n-2)!<100, what is the greatest possible value of n?

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

=>

We must have
n! / (n-2)! = n(n-1) < 100.
If n = 10, n(n-1) = 10*9 = 90 < 100.
If n = 11, n(n-1) = 11*10 = 110 > 100.

10 is the greatest value of n for which n!/(n-2)! < 100.

Therefore, C is the answer.

Answer: C
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New post 08 Mar 2018, 02:26
[GMAT math practice question]

10 members of a society, including members A and B, attended a meeting. 2 of these members were selected to form a committee. What is the probability that the 2 members selected were members A and B?

A. 1/10
B. 1/15
C. 1/20
D. 1/30
E. 1/45

=>

The number of committees of 2 people that include members A and B is equal to the number of ways that 2 people can be selected from 2 people, which is 2C2 = 1.
The number different possible committees of 2 people, is equal to the number of ways that 2 people can be selected from 10 people, which is 10C2 = (10*9) / (1*2) = 45.
The probability that the committee contains members A and B is 2C2 / 10C2 = 1/45.

Therefore, the answer is E.

Answer: E
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New post 09 Mar 2018, 00:22
[GMAT math practice question]

The terms of the sequence {An}, where n is a positive integer, satisfy A1=81, A2=82, A3=83, and An+3=An+4. Which of the following cannot be a value of An?

A. 801
B. 802
C. 803
D. 804
E. 805


=>

The terms of the sequence can be divided into three groups:
A1 = 81, A4 = 85, A7 = 89, … : These have a remainder of 1 when they are divided by 4.
A2 = 82, A5 = 86, A8 = 90, … : These have a remainder of 2 when they are divided by 4.
A3 = 83, A6 = 87, A9 = 91, … : These have a remainder of 3 when they are divided by 4.
No term of the sequence is a multiple of 4.

Since 804 is a multiple of 4, it cannot be a term of the sequence.

Therefore, the answer is D.

Answer: D
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New post 11 Mar 2018, 18:33
[GMAT math practice question]

If neither x nor y is divisible by 3, which of the following could be the value of x^2+y^2?

A. 333
B. 334
C. 335
D. 336
E. 337

=>
Consider the squares of the integers that are not divisible by 3:
1^2 = 1, 2^2 = 4, 4^2 = 16, 5^2 = 25, 7^2 = 49, 8^2=64, ….
They all have a remainder of 1 when they are divided by 3.
Thus, the sum of the squares of two integers which are not divisible by 3 must have a remainder of 2 when it is divided by 3.

The only answer choice having this property is 335.

Therefore, C is the answer.

Answer: C
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New post 11 Mar 2018, 18:35
[GMAT math practice question]

Reading from the right, how many consecutive zeros, starting with the units digit, appear in 29!?

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

=>

To find the number of 0’s ending 29!, we need to count the numbers of 2’s and 5’s in the prime factorization of 29!.
Since the number 2’s is greater than the number of 5’s, we only need to count the number of 5’s in the prime factorization of 29!.
The factors of 5 are contributed by 5, 10, 15, 20 and 25. Each of 5, 10, 15 and 20 contributes one 5, while 25 contributes two 5s to the prime factorization.
Thus, there are 6 copies of 5 in the prime factorization of 29!, giving rise to 6 consecutive 0’s at the end of 29!.

Therefore, the answer is B.
Answer : B

Note: The actual value of 29! is 8841761993739701954543616000000.
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New post 14 Mar 2018, 00:24
[GMAT math practice question]

If (n+2)!= n!(an^2+bn+c), then abc=?

A. 2
B. 3
C. 4
D. 6
E. 8

=>

(n+2)! = (n+2)(n+1)n! = (n^2 + 3n + 2)n!
So,
a = 1, b = 3, and c = 2.
Thus, abc = 6.

Therefore, D is the answer.

Answer: D
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Re: Math Revolution Approach (PS) &nbs [#permalink] 14 Mar 2018, 00:24

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