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The inside of a rectangular carton is 48 centimeters long, 3 [#permalink]

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03 Sep 2012, 06:14

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The inside of a rectangular carton is 48 centimeters long, 32 centimeters wide, and 15 centimeters high. The carton is filled to capacity with k identical cylindrical cans of fruit that stand upright in rows and columns, as indicated in the figure above. If the cans are 15 centimeters high, what is the value of k?

(1) Each of the cans has a radius of 4 centimeters. (2) Six of the cans fit exactly along the length of the carton.

Practice Questions Question: 35 Page: 278 Difficulty: 600

Re: The inside of a rectangular carton is 48 centimeters long, 3 [#permalink]

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03 Sep 2012, 06:14

1

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SOLUTION

The inside of a rectangular carton is 48 centimeters long, 32 centimeters wide, and 15 centimeters high. The carton is filled to capacity with k identical cylindrical cans of fruit that stand upright in rows and columns, as indicated in the figure above. If the cans are 15 centimeters high, what is the value of k?

(1) Each of the cans has a radius of 4 centimeters --> radius=4 means that diameter=8, which implies that along the 48 centimeter length of the carton 48/8=6 cans can be placed and along the 32 centimeter width of the carton 32/8=4 cans can be placed. Thus, k=6*4=24. Sufficient.

(2) Six of the cans fit exactly along the length of the carton --> the diameter of the can is 48/6=8 centimeters. So, we have the same info as above. Sufficient.

Re: The inside of a rectangular carton is 48 centimeters long, 3 [#permalink]

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03 Sep 2012, 06:23

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Bunuel wrote:

The inside of a rectangular carton is 48 centimeters long, 32 centimeters wide, and 15 centimeters high. The carton is filled to capacity with k identical cylindrical cans of fruit that stand upright in rows and columns, as indicated in the figure above. If the cans are 15 centimeters high, what is the value of k?

(1) Each of the cans has a radius of 4 centimeters. (2) Six of the cans fit exactly along the length of the carton.

Since Height of Box and Height of Cans are equal so only one stack of cans is there in the box. No need to bother about height. So inside dimensions of Box = 48 X 32 St 1: Sufficient: Each can has radius of 4 = Dia is 8 cms. ie (48/8) 6 nos of cans can be in one row. And since cans are identical (32/8) ie 4 cans can fit in colums. So 24 nos of can can fit in box. St 2: Sufficient: 6 cans can fit along the length. ie dia of each can = 48/6 ie 8 cms. As discussed in St 1.

Hence Answer is D) _________________

Regards SD ----------------------------- Press Kudos if you like my post. Debrief 610-540-580-710(Long Journey): http://gmatclub.com/forum/from-600-540-580-710-finally-achieved-in-4th-attempt-142456.html

Re: The inside of a rectangular carton is 48 centimeters long, 3 [#permalink]

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07 Sep 2012, 06:22

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SOLUTION

The inside of a rectangular carton is 48 centimeters long, 32 centimeters wide, and 15 centimeters high. The carton is filled to capacity with k identical cylindrical cans of fruit that stand upright in rows and columns, as indicated in the figure above. If the cans are 15 centimeters high, what is the value of k?

(1) Each of the cans has a radius of 4 centimeters --> radius=4 means that diameter=8, which implies that along the 48 centimeter length of the carton 48/8=6 cans can be placed and along the 32 centimeter width of the carton 32/8=4 cans can be placed. Thus, k=6*4=24. Sufficient.

(2) Six of the cans fit exactly along the length of the carton --> the diameter of the can is 48/6=8 centimeters. So, we have the same info as above. Sufficient.

Answer: D.

Kudos points given to everyone with correct solution. Let me know if I missed someone. _________________

Re: The inside of a rectangular carton is 48 centimeters long, 3 [#permalink]

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24 Apr 2013, 02:53

Bunuel wrote:

SOLUTION

The inside of a rectangular carton is 48 centimeters long, 32 centimeters wide, and 15 centimeters high. The carton is filled to capacity with k identical cylindrical cans of fruit that stand upright in rows and columns, as indicated in the figure above. If the cans are 15 centimeters high, what is the value of k?

(1) Each of the cans has a radius of 4 centimeters --> radius=4 means that diameter=8, which implies that along the 48 centimeter length of the carton 48/8=6 cans can be placed and along the 32 centimeter width of the carton 32/8=4 cans can be placed. Thus, k=6*4=24. Sufficient.

(2) Six of the cans fit exactly along the length of the carton --> the diameter of the can is 48/6=8 centimeters. So, we have the same info as above. Sufficient.

Answer: D.

Kudos points given to everyone with correct solution. Let me know if I missed someone.

I know the measurement of the carton, which is 48*32. Now the can's radius is 4 cm. Although the height of the carton is the same as the height of the cans i.e. 15, which I am ignoring as it will eventually cancel out in the calculation. My question is around the solution that is provided in the O.G. They have simply divided the length of the carton by diameter and width by diameter and then further multiplied the result.

(48/8)*(32/8) =6*4 =24.

Btw I got the right answer since its a DS problem. I am worried coz had this been a p.s problem, I might have got this one wrong. Now the way I would have solved this is

1st find the circumference of the circle (eliminating the height as the it is same). 2IIr=2*22/7*4 and then divided it by (48*32)/(176/7). My answer in this case is different from the OA =61.09

Why II (pie) was not considered. Why circumference was not considered instead of diameter. I failed to understand this. Can you please explain this.

Further when I use a formula for rectangle's area =L*B, where both lenght and breadth is present. But in case of circle I don't have a II(pie) anywhere. Why do we use this. What is the significance of II (pie).

Re: The inside of a rectangular carton is 48 centimeters long, 3 [#permalink]

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24 Apr 2013, 05:38

Expert's post

davidfrank wrote:

Bunuel wrote:

SOLUTION

The inside of a rectangular carton is 48 centimeters long, 32 centimeters wide, and 15 centimeters high. The carton is filled to capacity with k identical cylindrical cans of fruit that stand upright in rows and columns, as indicated in the figure above. If the cans are 15 centimeters high, what is the value of k?

(1) Each of the cans has a radius of 4 centimeters --> radius=4 means that diameter=8, which implies that along the 48 centimeter length of the carton 48/8=6 cans can be placed and along the 32 centimeter width of the carton 32/8=4 cans can be placed. Thus, k=6*4=24. Sufficient.

(2) Six of the cans fit exactly along the length of the carton --> the diameter of the can is 48/6=8 centimeters. So, we have the same info as above. Sufficient.

Answer: D.

Kudos points given to everyone with correct solution. Let me know if I missed someone.

I know the measurement of the carton, which is 48*32. Now the can's radius is 4 cm. Although the height of the carton is the same as the height of the cans i.e. 15, which I am ignoring as it will eventually cancel out in the calculation. My question is around the solution that is provided in the O.G. They have simply divided the length of the carton by diameter and width by diameter and then further multiplied the result.

(48/8)*(32/8) =6*4 =24.

Btw I got the right answer since its a DS problem. I am worried coz had this been a p.s problem, I might have got this one wrong. Now the way I would have solved this is

1st find the circumference of the circle (eliminating the height as the it is same). 2IIr=2*22/7*4 and then divided it by (48*32)/(176/7). My answer in this case is different from the OA =61.09

Why II (pie) was not considered. Why circumference was not considered instead of diameter. I failed to understand this. Can you please explain this.

Further when I use a formula for rectangle's area =L*B, where both lenght and breadth is present. But in case of circle I don't have a II(pie) anywhere. Why do we use this. What is the significance of II (pie).

Why are you calculating the circumference? _________________

Re: The inside of a rectangular carton is 48 centimeters long, 3 [#permalink]

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26 Apr 2013, 22:38

Bunuel wrote:

davidfrank wrote:

Bunuel wrote:

SOLUTION

The inside of a rectangular carton is 48 centimeters long, 32 centimeters wide, and 15 centimeters high. The carton is filled to capacity with k identical cylindrical cans of fruit that stand upright in rows and columns, as indicated in the figure above. If the cans are 15 centimeters high, what is the value of k?

(1) Each of the cans has a radius of 4 centimeters --> radius=4 means that diameter=8, which implies that along the 48 centimeter length of the carton 48/8=6 cans can be placed and along the 32 centimeter width of the carton 32/8=4 cans can be placed. Thus, k=6*4=24. Sufficient.

(2) Six of the cans fit exactly along the length of the carton --> the diameter of the can is 48/6=8 centimeters. So, we have the same info as above. Sufficient.

Answer: D.

Kudos points given to everyone with correct solution. Let me know if I missed someone.

I know the measurement of the carton, which is 48*32. Now the can's radius is 4 cm. Although the height of the carton is the same as the height of the cans i.e. 15, which I am ignoring as it will eventually cancel out in the calculation. My question is around the solution that is provided in the O.G. They have simply divided the length of the carton by diameter and width by diameter and then further multiplied the result.

(48/8)*(32/8) =6*4 =24.

Btw I got the right answer since its a DS problem. I am worried coz had this been a p.s problem, I might have got this one wrong. Now the way I would have solved this is

1st find the circumference of the circle (eliminating the height as the it is same). 2IIr=2*22/7*4 and then divided it by (48*32)/(176/7). My answer in this case is different from the OA =61.09

Why II (pie) was not considered. Why circumference was not considered instead of diameter. I failed to understand this. Can you please explain this.

Further when I use a formula for rectangle's area =L*B, where both lenght and breadth is present. But in case of circle I don't have a II(pie) anywhere. Why do we use this. What is the significance of II (pie).

Why are you calculating the circumference?

Hi Bunuel,

I am calculating the circumference because I know the area of the rectangular (ignoring the height 15 cm as it is common to both can and carton). Once I know the area, I can divide the area by circumference of the circle to know the no of cans.

Re: The inside of a rectangular carton is 48 centimeters long, 3 [#permalink]

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27 Apr 2013, 05:32

Expert's post

davidfrank wrote:

Bunuel wrote:

davidfrank wrote:

I know the measurement of the carton, which is 48*32. Now the can's radius is 4 cm. Although the height of the carton is the same as the height of the cans i.e. 15, which I am ignoring as it will eventually cancel out in the calculation. My question is around the solution that is provided in the O.G. They have simply divided the length of the carton by diameter and width by diameter and then further multiplied the result.

(48/8)*(32/8) =6*4 =24.

Btw I got the right answer since its a DS problem. I am worried coz had this been a p.s problem, I might have got this one wrong. Now the way I would have solved this is

1st find the circumference of the circle (eliminating the height as the it is same). 2IIr=2*22/7*4 and then divided it by (48*32)/(176/7). My answer in this case is different from the OA =61.09

Why II (pie) was not considered. Why circumference was not considered instead of diameter. I failed to understand this. Can you please explain this.

Further when I use a formula for rectangle's area =L*B, where both lenght and breadth is present. But in case of circle I don't have a II(pie) anywhere. Why do we use this. What is the significance of II (pie).

Why are you calculating the circumference?

Hi Bunuel,

I am calculating the circumference because I know the area of the rectangular (ignoring the height 15 cm as it is common to both can and carton). Once I know the area, I can divide the area by circumference of the circle to know the no of cans.

It seems that you don't understand the question. You need neither circumference of the cans nor the area.

Simpler example might help:

Attachment:

Untitled.png [ 10.24 KiB | Viewed 6124 times ]

The carton is 16 centimeters long and 16 centimeters wide. If the diameter of the cans is 8 centimeters, how many cans can be placed in the carton? _________________

Re: The inside of a rectangular carton is 48 centimeters long, 3 [#permalink]

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11 Sep 2014, 12:53

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The inside of a rectangular carton is 48 centimeters long, 3 [#permalink]

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05 Oct 2014, 19:40

When I did this problem, I inadvertinly read over the specification desigmating, width, length and height. In this case 15 was the height, allowing us to cram the most amount of cyclinder cans.

Here is my question, if height was not deginated as 15 in the case, would the solution be E? ie, we have the 15×32×48 dimensions but we don't know what is length, width, height. E is what I came up with.

Re: The inside of a rectangular carton is 48 centimeters long, 3 [#permalink]

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27 Feb 2016, 07:16

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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