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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8471
GMAT 1: 760 Q51 V42
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[GMAT math practice question]

(probability) How many arrangements of the letters A, C, C, E, N, T include at least one letter between the two C’s?

A. 60
B. 120
C. 240
D. 360
E. 720

=>

The easiest way to solve this problem is to find the number of arrangements satisfying the complementary condition that there is no letter between the two C’s and subtract this from the total number of arrangements of the letters. The best approach to solving probability or counting problems, including the words, ‘at least’, is to use complementary counting.

The total number of arrangements of the letters A, C, C, E, N and T is (6!)/(2!) = 360.
To count the number of arrangements with no letter between the two C’s, we consider CC to be one letter. Thus, the number of arrangements satisfying the complementary condition is 5! = 120.

Thus, the number of arrangements of the letters A, C, C, E, N, T with at least one letter between the two C’s is 360 – 120 = 240.

Attachments 6.17.png [ 3.26 KiB | Viewed 581 times ]

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GMAT 1: 760 Q51 V42
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[GMAT math practice question]

(arithmetic, equation) What is the value of 1 / (√2+√1) + 1 / (√3+√2) + 1/(√4+√3) + … + 1/(√25+√24)?

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

=>

We rationalize the denominator of each fraction to give
1 / (√2+ √1) + 1 / (√3+ √2) + 1/( √4+ √3) + … + 1/( √25+ √24)
= (√2-√1)/[(√2+√1)(√2-√1)] + (√3-√2)/[(√3+√2) (√3-√2)] + (√4-√3)/[(√4+√3) (√4-√3)] + … + (√25-√24)/[(√25+√24)(√25-√24)]
= (√2-√1)/[2-1] + (√3-√2)/[3-2] + (√4-√3)/[4-3] + … + (√25-√24)/[25 - 24]
= (√2-√1) + (√3-√2) + (√4-√3) + … + (√25-√24)
= √25 - √1 = 5 - 1 = 4

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Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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[GMAT math practice question]

(number properties) m is an odd integer. If m^3n^4=432, what is the value of n?

1) n is positive.
2) n is an integer.

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since m is an odd integer, n is a positive integer and 432 = 33*24, the unique solution pair is m = 3 and n = 2.
Thus, both conditions are sufficient, when applied together.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
m = 3 and n = 2 is a pair of solutions.
m = 1 and n = 4√432 is another pair of solutions.
Since condition 1) doesn’t yield a unique solution, it is not sufficient.

Condition 2)
m = 3 and n = 2 is a pair of solutions.
m = 3 and n = -2 is another pair of solutions.
Since condition 2) doesn’t yield a unique solution, it is not sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
_________________
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[GMAT math practice question]

(work rate) A factory has two machines. Machine X can produce 200 pins per hour and machine Y can produce 500 pins per hour. At least one machine is in operation throughout the 8-hour work day. The factory must produce 5,000 pins every day. What is the least possible number of hours that machines X and Y must work together on each day?

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

=>

In order to minimize the number of hours that the two machines work together, the more efficient machine Y should work for the entire day.
Thus, Y works for 8 hours and produces 8 * 500 = 4,000 pins.
There are 1,000 pins remaining for machine X to produce. This takes 1,000/200 = 5 hours.

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[GMAT math practice question]

(inequality) -2 ≤ x ≤ -1 and 1 ≤ y ≤ 2. What is the greatest possible value of ( x + y ) / x?

A. -1
B. –1/2
C. 0
D. 1/2
E. 1

=>

(x+y)/x = x/x + y/x = 1 + y/x
Since -2 ≤ x ≤ -1 and 1 ≤ y ≤ 2, we have -2 ≤ y/x ≤ -1/2 and -1 ≤ 1 + y/x ≤ 1/2.
Thus, the greatest possible value of (x+y)/x is ½.

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[GMAT math practice question]

(number properties) p and q are different positive integers. What is the remainder when p^2 + q^2 is divided by 4?

1) p and q are prime numbers.
2) p and q are not consecutive integers.

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (p and q) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since p and q are prime numbers, which are not consecutive integers, p and q are odd integers.
So, both p^2 and q^2 have remainder 1 when they are divided by 4.
Thus, p^2 + q^2 has remainder 2 when it is divided by 4.
Since conditions 1) & 2) yield a unique solution, when they are applied together, they are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If p = 2 and q = 3, then p^2 + q^2 = 4 + 9 = 13, which has remainder 1 when it is divided by 4.
If p = 3 and q = 5, then p^2 + q^2 = 9 + 25 = 34, which has remainder 2 when it is divided by 4.

Condition 1) is not sufficient since it doesn’t yield a unique solution.

Condition 2)
If p = 3 and q = 5, then p^2 + q^2 = 9 + 25 = 34, which has remainder 2 when it is divided by 4.
If p = 3 and q = 6, then p^2 + q^2 = 9 + 36 = 45, which has remainder 1 when it is divided by 4.

Condition 2) is not sufficient since it doesn’t yield a unique solution.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Joined: 16 Aug 2015
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[GMAT math practice question]

(number properties) m and n are positive integers. Is m^2 + n^2 is divisible by 3?

1) m = 1234
2) n = 4321

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify the conditions if necessary.

Recall that the remainder when an integer is divided by 3 is the same as the remainder when the sum of all of its digits is divided by 3.

The square of an integer k has remainder 0 or 1 when it is divided by 3.
If k is a multiple of 3, then k = 3a for some integer a, and k^2 = (3a)^2 = 3(3a^2) has remainder 0 when it is divided by 3.
If k has remainder 1 when it is divided by 3, then k = 3a + 1 for some integer a, and k^2 = (3a+1)^2 = 9a^2 +6a + 1 = 3(3a^2 +2a) + 1 has remainder 1 when it is divided by 3.
If k has remainder 2 when it is divided by 3, then k = 3a + 2 for some integer a, and k^2 = (3a+2)^2 = 9a^2 +12a + 4 = 3(3a^2 +4a+1) + 1 has remainder 1 when it is divided by 3.

Since m=1234 has remainder 1 when it is divided by 3, m^2 has remainder 1 when it is divided by 3. Since n^2 could have remainder 0 or 1 when it is divided by 3, m^2 + n^2 is never divisible by 3, regardless of the value of n. Condition 1) is sufficient, since it yields the unique answer, ‘no’.

Since n=4321 has remainder 1 when it is divided by 3, n^2 has remainder 1 when it is divided by 3. Since m^2 could have remainder 0 or 1 when it is divided by 3, m^2 + n^2 is never divisible by 3, regardless of the value of m. Condition 2) is sufficient, since it yields the unique answer, ‘no’.

Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, both conditions are sufficient.
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[GMAT math practice question]

(arithmetic) A palindrome, such as 12321, is a number that remains the same when its digits are reversed. The numbers x and x+32 are three-digit and four-digit palindromes, respectively. What is the sum of the digits of x?

A. 18
B. 24
C. 28
D. 32
E. 36

=>

Let x = ABA and x + 32 = CDDC.
Then CDDC = ABA + 32
Since CDDC has a thousands digit, we must have A = 9.
Then the units digit of CDDC is equal to the units digit of A + 2 = 11.
And, looking at the tens digits, we must have B + 3 + 1 ≥ 10.
Therefore, B ≥ 6.
The possible values for B are:
B = 6, 7, 8, 9.
Let’s check which value gives a palindrome for ABA + 32:
969 + 32 = 1001
979 + 32 = 1011
989 + 32 = 1021
999 + 32 = 1031
The only palindrome is 1001, so x = 969.
Thus, the sum of the digits of x is 9 + 6 + 9 = 24.

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[GMAT math practice question]

(probability) A palindrome, such as 12321, is a number that remains the same when its digits are reversed. How many 4-digit palindromes are divisible by 4?

A. 10
B. 20
C. 25
D. 28
E. 30

=>

Palindromes between 1000 and 10,000 have the form ABBA.
When we check divisibility by 4, we need only check the divisibility of BA by 4.
The values of BA which are divisible by 4 are 04, 08, 12, 16, 24, 28, 32, 36, 44, 48, 52, 56, 64, 68, 72, 76, 84, 88, 92, and 96 (note that 20, 40, 60, 80 and 00 are not possible values of BA as they do not give rise to 4-digit numbers ABBA). The possible values for ABBA are 4004, 8008, 2112, 6116, 4224, 8228, 2332, 6336, 4444, 8448, 2552, 6556, 4664, 8668, 2772, 6776, 4884, 8888, 2992 and 6996.
Thus, there are 20 4-digit palindromes.

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[GMAT math practice question]

(probability) The vertices of a regular pentagon are to be colored using five different colors. In how many ways can the pentagon’s vertices be colored if the 5 colors are to be chosen from a palette of 6 different colors?

A. 64
B. 96
C. 108
D. 144
E. 192

=>

The number of ways to choose 5 colors out of 6 colors is 6C5 = 6.
The number of circular permutations of the 5 colors is (5-1)! = 4! = 24.
Thus, 6*24 = 144 different colorings of the pentagon are possible.

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[GMAT math practice question]

(algebra) x, y, z are 3 consecutive positive integers. What is the value of x?

1) x<y<z
2) 12+13+14+15=x+y+z

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Consecutive integers have two variables for the first number and the number of integers. Since the number of integers is 3, we need one more equation and D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
If x = 1, y = 2 and z = 3, then we have x = 1.
If x = 2, y = 3 and z = 4, then we have x = 2.
Since condition 1) doesn’t yield a unique solution, it is not sufficient.

Condition 2)
Since 12 + 13 + 14 + 15 = x + y + z, we have x + y + z = 54.
If x = 17, y = 18, z = 19, then we have x = 17.
If x = 19, y = 18, z = 17, then we have x = 19.
Since condition 2) doesn’t yield a unique solution, it is not sufficient.

Conditions 1) & 2)
Since y = x + 1 and z = x + 2 from condition 1), we have 12 + 13 + 14 + 15 = 54 = x + y + z = x + x + 1 + x + 2 = 3x + 3 or 3x = 51.
Thus we have x = 17.
Since both conditions together yield a unique solution, they are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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[GMAT math practice question]

(Statistics) A class consists of 30 students. Among them a students have 90 books, b students have 80 books, c students have 70 books and all the remaining students have 60 books. What is the average number of books the students in the class have?

1) a= b+c
2) a is twice b

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (a, b and c) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

The average number of books is
( 90a + 80b + 70c +60(30 – a – b – c) ) / 30
= ( 30a + 20b + 10c + 1800) /30

If a = 2, b = 1 and c = 1, then the average is (60 + 20 + 10 + 1800)/30 = 1890/30 = 63.
If a = 4, b = 2 and c = 2, then the average is (120 + 40 + 20 + 1800)/30 = 1980/30 = 66.

Since both conditions together don’t yield a unique solution, they are not sufficient.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
_________________
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Joined: 16 Aug 2015
Posts: 8471
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(arithmetic) What is 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56?

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

=>

1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56
= 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + 1/(5*6) + 1/(6*7) + 1/(7*8)
= (1/1 – 1/2) + (1/2 – 1/3) + (1/3 – 1/4) + (1/4 – 1/5) + (1/5 – 1/6) + (1/6 – 1/7) + (1/7 – 1/8)
= 1/1 – 1/8 = 7/8

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[GMAT math practice question]

(ALGEBRA) As the figure shows, A = -5 and B = 4. If AC:CD:DB=1:2:3, what are the values of C and D?

A. -3 and -1
B. -3.5 and -0.5
C. -2.5 and 0.5
D. 1 and 3
E. 0 and 1

Attachment: 7.12.png [ 1.64 KiB | Viewed 282 times ]

=>

Since AC:CD:DB = 1:2:3, we can put AC = k, CD = 2k and DB = 3k.
Then k + 2k + 3k = 6k = 4 – (-5) = 9 and k = 3/2.
Thus, C = A + k = -5 + (3/2) = -(7/2) = -3.5 and D = B – 3k = 4 – 9/2 = -(1/2) = -0.5.

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Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(algebra) What is a?

1) 3x-[7x-{2x-(5-6x)}] = -10x+4
2) –a+5 = 11x

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (a and x) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) and 2)
3x-[7x-{2x-(5-6x)}] = 3x-[7x-{2x-5+6x}] = 3x-[7x-{8x-5}] = 3x-[7x-8x+5] = 3x-[-x+5] = 3x+x-5 = 4x – 5 = -10x + 4
Then, by condition 1), we must have 14x = 9 and x = 9/14.
Since –a + 5 = 11x, we have a = 5 -11x = 5 -11(9/14) = -(29/14)
Thus, both conditions together are sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
_________________
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Joined: 16 Aug 2015
Posts: 8471
GMAT 1: 760 Q51 V42
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[GMAT math practice question]

(algebra) Let x be a real number. (a, b) denotes ax+b. What is (1, 0)?

1) 3*(2,0)=(-1, 4) – (-2, -6)
2) (1, 0)^2 +4 = 4(1, 0)

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since (1, 0)=1*x+0=x, the question asks for the value of x.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
The left hand side is 3*(2,0) = 3*(2x + 0) = 6x and the right hand side evaluates to (-1,4) – (-2,-6) = -x + 4 – (-2x – 6) = x + 10. Equating both sides yields 6x = x + 10 and x = 2.
Thus, condition 1) is sufficient.

Condition 2)
(1,0)^2 + 4 = 4(1,0)
=> (x)^2 + 4 =4(x)
=> x^2 -4x + 4 = 0
=> (x-2)^2 = 0
=> x = 2.
Thus, condition 2) is also sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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[GMAT math practice question]

(ratio) Tom painted 1/3 of a wall red, 1/5 of the wall blue and the remaining 238 m^2 black. What is the area of the wall in m2?

A. 490
B. 500
C. 510
D. 520
E.530

=>

Using the Ivy Approach, we assume 15w is the area of the wall. We have chosen 15 since it is the lcm of the denominators, 3 and 5. So, 15w/3 = 5w is the area painted red, and 15w/5 = 3w is the area painted blue.
The area painted black is given by the equation
15w-(15w·1/3)-(15w·1/5)=15w-5w-3w=7w=238.

Solving for w yields w=34. So, the area of the whole wall is 15w=15·34=510 m^2.

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[GMAT math practice question]

(speed) Tom travels from A to B at a constant speed of 4 miles per hour and returns from B to A at a constant speed of 8 miles per hour. What is Tom’s average speed for the entire journey?

A. 14/3
B. 5
C. 16/3
D. 17/3
E. 18/3

=>

The average speed is calculated by dividing the total distance by the total time. We often encounter this type of question on the GMAT exam.

Let d be the distance between A and B.
The time that Tom takes to travel from A to B is d/4 and the time that he takes to travel from B to A is d/8.
The total time taken is (d/4) + (d/8) = 3d/8.
The average speed is 2d / (3d/8) = 16/3.

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Posts: 19
Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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Example 9:
Stations M and N are connected by two separate, straight and parallel rail lines that are 500 miles long. Freight train A and freight train B simultaneously left Station M and Station N, respectively, and each freight train traveled to the other’s point of departure. The two freight trains passed each other after traveling for 4 hours. When the two freight trains passed, which train was nearer to its destination?

(1) At the time when the two freight trains passed, freight train A had averaged a speed of 60 miles per hour.
(2) Freight train B averaged a speed of 130 miles per hour for the entire trip.

How come the answer to this question is not A?
It is clear that A traveled 4 hours with 60 kmph speed so it covered a distance of 240 miles. that means, 500-240=260.
A is 260 miles away from its destination.

Also if B passed or met A exactly after 4 hours, B must have traveled at least 260 kms which means it is just
(500-260=240)
240 kms away from its destination.
Doesn't this answer our question that Train B is nearer to its destination???

Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8471
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: Overview of GMAT Math Question Types and Patterns on the GMAT  [#permalink]

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[GMAT math practice question]

(geometry) In the figure, lines AB, CD and EF are parallel, and lines BC and DE are parallel. What is the length of CD?

Attachment: 7.23.png [ 14.87 KiB | Viewed 146 times ]

1) AB=3
2) EF=12

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Triangles ABC and CDE are similar, and triangles BDC and DFE are similar. So, we can set up the proportion AB:CD = CD:EF.
Since CD^2 = AB*EF, conditions 1) and 2) together give us sufficient information to calculate the value of CD.

_________________ Re: Overview of GMAT Math Question Types and Patterns on the GMAT   [#permalink] 28 Jul 2019, 18:16

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# Overview of GMAT Math Question Types and Patterns on the GMAT   