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by phoenix9801 » Sat May 19, 2012 1:14 am
Not So such how to solve them. thanks
1-
Company A has half as many employees as Company B, and Company C has three times as many employees as Company B. Which of the following could be the number of employees at Company C?

A) 39
B) 45
C) 50
D) 57
E) 66


2-
On Monday night, 2/5 of the attendees at a certain film were female, and 1/3 of the females and 1/6 of the males received a student discount. If an additional 1/3 of the males had received a student discount, then a total of 26 of all attendees would have received a student discount. How many people attended the film on Monday night?


A) 45
B) 60
C) 75
D) 90
E) 120

3-
A certain pet store currently carries three times as many dogs as cats, and twice as many cats as rabbits. However, if the store carried 10 fewer dogs than it currently does, then it would have an equal number of dogs and rabbits. How many cats does the store currently carry?


A) 2
B) 4
C) 8
D) 12
E) 15
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Source: — Problem Solving |

by aneesh.kg » Sat May 19, 2012 2:54 am
phoenix9801 wrote:Not So such how to solve them. thanks
1-
Company A has half as many employees as Company B, and Company C has three times as many employees as Company B. Which of the following could be the number of employees at Company C?

A) 39
B) 45
C) 50
D) 57
E) 66
A = B/2 or B = 2A -- (1)
C = 3B = 6A -- (2)
A + B + C
(Substituting B and C in terms of A from (1) and (2))
= A + 2A + 6A = 9A
Since A is an integer,
A + B + C has to be a multiple of 9.
Only option [spoiler](B)[/spoiler] fits the criteria.

2-
On Monday night, 2/5 of the attendees at a certain film were female, and 1/3 of the females and 1/6 of the males received a student discount. If an additional 1/3 of the males had received a student discount, then a total of 26 of all attendees would have received a student discount. How many people attended the film on Monday night?

A) 45
B) 60
C) 75
D) 90
E) 120
Let the number of people be N.
2N/5 = F and 3N/5 = M -- (1)
Number of people who received a discount = (1/3)F + (1/6)M
(substituting F and M in terms of N from (1))
= (1/3)(2N/5) + (1/6)(3N/5)
= 2N/15 + N/10
But an additional of M/3 or (1/3)*(3N/5) = N/5 people received a discount.

total number of people who received a discount = 2N/15 + N/10 + N/5 = 26 (given)
or (4N + 3N + 6N)/30 = 26
N = 60

[spoiler](B)[/spoiler] is correct.
3-
A certain pet store currently carries three times as many dogs as cats, and twice as many cats as rabbits. However, if the store carried 10 fewer dogs than it currently does, then it would have an equal number of dogs and rabbits. How many cats does the store currently carry?


A) 2
B) 4
C) 8
D) 12
E) 15
D = 3C -- (1)
C = 2R -- (2)
(D - 10) = R -- (3)

Substituting D and R in terms of C from (1) and (2) in (3), we have
3C - 10 = C/2
C = 10/2.5
C = 4
[spoiler](B)[/spoiler] is correct
Aneesh Bangia
GMAT Math Coach
[email protected]

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by GmatKiss » Sat May 19, 2012 9:44 am
phoenix9801 wrote:Not So such how to solve them. thanks
1-
Company A has half as many employees as Company B, and Company C has three times as many employees as Company B. Which of the following could be the number of employees at Company C?

A) 39
B) 45
C) 50
D) 57
E) 66


2-
On Monday night, 2/5 of the attendees at a certain film were female, and 1/3 of the females and 1/6 of the males received a student discount. If an additional 1/3 of the males had received a student discount, then a total of 26 of all attendees would have received a student discount. How many people attended the film on Monday night?


A) 45
B) 60
C) 75
D) 90
E) 120

3-
A certain pet store currently carries three times as many dogs as cats, and twice as many cats as rabbits. However, if the store carried 10 fewer dogs than it currently does, then it would have an equal number of dogs and rabbits. How many cats does the store currently carry?


A) 2
B) 4
C) 8
D) 12
E) 15
Join the discussion