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Math questions

by [email protected] » Tue Oct 28, 2008 7:04 pm
If abc = b^3 , which of the following must be true?

ac = b^2

b = 0

ac = 1

None
I only
II only
I and III
II and III

If (3 4 )(5 6 )(7 3 ) = (35 n )( x ), where x and n are both positive integers, how many different possible values of n are there?

1
2
3
4
6


The integers a , b , and c are positive, a/b=5/2 , and a/c=7/5 . What is the smallest possible value of 2 a + b ?

63
70
84
95
105


A gambler bought $3,000 worth of chips at a casino in denominations of $20 and $100. That evening, the gambler lost 16 chips, and then cashed in the remainder. If the number of $20 chips lost was 2 more or 2 less than the number of $100 chips lost, what is the largest amount of money that the gambler could have received back?

$2,040
$2,120
$1,960
$1,920
$1,400



A woman is planning a trip that involves 3 connecting trains that depart from Stations X , Y , and Z , respectively. The first train leaves Station X every hour, beginning at 6 a.m., and arrives at Station Y hours later. The second train leaves Station Y every half hour, beginning at 9 a.m., and arrives at Station Z hours later. The third train leaves Station Z every 45 minutes, beginning at 8 a.m. What is the least total amount of time the woman must spend waiting between trains if all trains depart and arrive on schedule, and if she arrives at Station Z no later than 3:30 p.m.?

15 minutes
25 minutes
1 hour 15 minutes
1 hour 40 minutes
4 hours 30 minutes

The weight of every type A widget is the same, the weight of every type B widget is the same, and the weight of every type C widget is the same. If the weight of 8 type A widgets is equal to the weight of 3 type B widgets, and the weight of 5 type B widgets is equal to the weight of 7 type C widgets. What is the ratio of the total weight of 1 type A widget and 1 type B widget, to the total weight of 1 type B widget and 1 type C widget?

a. 12/23
b. 21/40
c. 2/3
d. 77/96
e. 10/7
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by cramya » Tue Oct 28, 2008 7:42 pm
The weight of every type A widget is the same, the weight of every type B widget is the same, and the weight of every type C widget is the same. If the weight of 8 type A widgets is equal to the weight of 3 type B widgets, and the weight of 5 type B widgets is equal to the weight of 7 type C widgets. What is the ratio of the total weight of 1 type A widget and 1 type B widget, to the total weight of 1 type B widget and 1 type C widget?

a. 12/23
b. 21/40
c. 2/3
d. 77/96
e. 10/7


Soln:

8A = 3B 5B=7C

Ratio =
total weight of 1 type A widget +total weight of 1 type B widget / total weight of 1 type B widget + 1 type C widget

= 3/8B+7C/5 / 7C/5+5B/7

= 7(15B+56C) / 8(49C+25B)

We know 56C = 40B and 49C = 35B since 5B=7C

= 7(55B) / 8(60B) = 77/96

OA.please?
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by logitech » Tue Oct 28, 2008 7:44 pm
Dear Sophia,

One question at a time please.

Cheers!
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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by cramya » Tue Oct 28, 2008 7:46 pm
The integers a , b , and c are positive, a/b=5/2 , and a/c=7/5 . What is the smallest possible value of 2 a + b ?

63
70
84
95
105



Is it 84? OA? I will then post the solution if its right.
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by cramya » Tue Oct 28, 2008 7:49 pm
Sophia,

Friendly request: Please post one question per thread. This way its easier to track the question and the methodlogy to solve by all users.

Also kindly post the official answers for these with the SPOILER function(on the message body line).

For example [spoiler]OA:YYYYY[/spoiler]

Good luck!
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by cramya » Tue Oct 28, 2008 7:50 pm
If abc = b^3 , which of the following must be true

Problem already solved in this forum. Answer is none

If u cant find it let me know.
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by cramya » Tue Oct 28, 2008 7:55 pm
A gambler bought $3,000 worth of chips at a casino in denominations of $20 and $100. That evening, the gambler lost 16 chips, and then cashed in the remainder. If the number of $20 chips lost was 2 more or 2 less than the number of $100 chips lost, what is the largest amount of money that the gambler could have received back?


I am getting 2120 B)

OA?
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by cramya » Tue Oct 28, 2008 8:00 pm
If (3 4 )(5 6 )(7 3 ) = (35 n )( x ), where x and n are both positive integers, how many different possible values of n are there?

If we take x = 1 to get the maximum possible value of n we end up with just 1 (3*4*6*3) = 216 FOR lhs to equal rhs

OA?
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Re: Math questions

by parallel_chase » Wed Oct 29, 2008 4:39 am
[email protected] wrote:If abc = b^3 , which of the following must be true?

ac = b^2

b = 0

ac = 1

None
I only
II only
I and III
II and III
refer this
https://www.beatthegmat.com/hard-ps-mult ... 18383.html
If (3 4 )(5 6 )(7 3 ) = (35 n )( x ), where x and n are both positive integers, how many different possible values of n are there?

1
2
3
4
6
Is it multiplication or something else.

if it is multiplication

3*4*5*6*7*3 = 35*n*x

3^3 * 2^3 * 5 * 7 = 5*7*n*x

n can have more than 6 values,
n=2 if x=2^2*3^3
n=2^2 if x=2*3^3
n=2^3 if x= 3^3
n=2*3 if x=3^2*2^2
n=2^3*3^2 if x= 3
n=2^2 * 3^2 if x = 2*3
Similarly you can have many possible values of n. Hence either the question is wrongly posted or I am definitely missing something.
The integers a , b , and c are positive, a/b=5/2 , and a/c=7/5 . What is the smallest possible value of 2 a + b ?

63
70
84
95
105
since, a/b = 5/2 , 2a = 5b

2a + b = 5b + b = 6b.

In the answer options there is only multiple of 6 i.e. 84.

I dont know how we could use a/c =7/5 information.

OA?

A gambler bought $3,000 worth of chips at a casino in denominations of $20 and $100. That evening, the gambler lost 16 chips, and then cashed in the remainder. If the number of $20 chips lost was 2 more or 2 less than the number of $100 chips lost, what is the largest amount of money that the gambler could have received back?

$2,040
$2,120
$1,960
$1,920
$1,400
let x be $20 chips
let y be $100 chips

20x + 100y = 3000

lost chips = 16
remaining = x+y-16
lost chips, x = y+20 or x = y-20
The gambler can get the maximum when he/she lost more of $20 chips than $100 chips

therefore it should x=y+2

16 = x+x-2
18 = 2x
x = 9
y = 7
total money lost, 20*9 + 7*100 = 180 +700 = 880

maximum money recieved back = 3000 -880 = 2120.

OA?
A woman is planning a trip that involves 3 connecting trains that depart from Stations X , Y , and Z , respectively. The first train leaves Station X every hour, beginning at 6 a.m., and arrives at Station Y hours later. The second train leaves Station Y every half hour, beginning at 9 a.m., and arrives at Station Z hours later. The third train leaves Station Z every 45 minutes, beginning at 8 a.m. What is the least total amount of time the woman must spend waiting between trains if all trains depart and arrive on schedule, and if she arrives at Station Z no later than 3:30 p.m.?

15 minutes
25 minutes
1 hour 15 minutes
1 hour 40 minutes
4 hours 30 minutes
The question is missing information about hours. Kindly check the question while posting.

The weight of every type A widget is the same, the weight of every type B widget is the same, and the weight of every type C widget is the same. If the weight of 8 type A widgets is equal to the weight of 3 type B widgets, and the weight of 5 type B widgets is equal to the weight of 7 type C widgets. What is the ratio of the total weight of 1 type A widget and 1 type B widget, to the total weight of 1 type B widget and 1 type C widget?

a. 12/23
b. 21/40
c. 2/3
d. 77/96
e. 10/7
Cramya's explanation above is perfect. I am also getting 77/96, OA?
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by vittalgmat » Wed Oct 29, 2008 9:17 am
cramya wrote:A gambler bought $3,000 worth of chips at a casino in denominations of $20 and $100. That evening, the gambler lost 16 chips, and then cashed in the remainder. If the number of $20 chips lost was 2 more or 2 less than the number of $100 chips lost, what is the largest amount of money that the gambler could have received back?


I am getting 2120 B)

OA?
I got the same answer. I tried to solve using equations but could not. So I tried the following to arrive at the answer.

We know that the diff between # of $20 and $100 chips is 2. Total chips lost
is 16. If both were equal then he would have lost 8 and 8. Here diff should be 2, we have 2 combinations. 9 $20 chips and 7 $100 chips OR
7 $20 chips and 9 $100 chips. The gambler got the largest money back. So, the combo should be 9 $20 chips and 7 $100 chips.
ie. he lost 9*20 + 7*100 = $880. 3000 -880 = 2120 is his return.

Pls let me know if there is a better way to solve this.

thanks
-V
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by bhanu75 » Wed Oct 29, 2008 10:50 am
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