3 Problems from the GMAT Prep Program

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ocean: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Fri Jan 02, 2009 11:27 am

3 Problems from the GMAT Prep Program

by ocean » Sat Jan 03, 2009 3:35 am

hi guys,
another kind request for explanation:

"If n is a positive integer less than 200 and 14n/60 is an integer, then n has how many different positive prime factors?"

a) two
b) three
c) five
d) six
e) eight

and another one:

"A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the area, in square meters, of the circular and the square regions in terms of r?"

a) Pi*r^2
b) Pi*r^2+10
c) Pi*r^2+1/4*Pi^2*r^2
d) Pi*r^2+(40-2*Pi*r)^2
e) Pi*r^2+(10-1/2*Pi*r)^2

and another one:

"If 2^x - 2^(x-2) = 3(2^13), what is the value of x?"

a) 9
b) 11
c) 13
d) 15
e) 17

All questions are from the GMAT Prep Program.

thx again,
ocean

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Source: Beat The GMAT — Problem Solving | Sat Jan 03, 2009 3:35 am

jimmylow: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Jan 03, 2009 5:50 am; Thanked: 1 times

by jimmylow » Sat Jan 03, 2009 7:13 am

Let me attempt this one:

"If 2^x - 2^(x-2) = 3(2^13), what is the value of x?"
a) 9
b) 11
c) 13
d) 15
e) 17

[Working]

2^x - 2^(x-2) = 3 (2^13)
= [(2^2)(2^x) - (2^x)] / (2^2) = 3 (2^13)
= 2^(2+x) - 2^x = 3 (2^13) (2^2) <--- this is done by multiplying 2^2 both sides

3 = 2^2 - 2^0 <---- anything to power of 0 (zero) is 1.

Subtitute 3 with the above

= 2^(2+x) - 2^x = (2^2 - 2^0) (2^13) (2^2)
= 2^(2+x) - 2^x = (2^2 - 2^0) (2^15)
= 2^(2+x) - 2^x = 2^(2+15) - 2^(0+15)
Thus,
(2 + x) - x = (2 +15) - (15)
x is 15

Ans: D. Sorry for the long-winded working. I want to make sure you can follow every steps.

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Zipper: Master | Next Rank: 500 Posts; Posts: 119; Joined: Sun Nov 30, 2008 2:47 am; Thanked: 11 times

by Zipper » Sat Jan 03, 2009 7:35 am

I have no idea what you just did to solve that but I think there is a much easier way.

You can easily forget about A, B and C.

Which leaves only two possible answers. We have to solve for only one of them.... if we get the right one then everything is OK. If we don't it's the other one.

1st we "simplify" a bit.

3(2^13)= 2^13 + 2^13 + 2^13 = 2^14 + 2^13

seeing 2^13 (this is a giveaway what we should try 1st) we should try D since 15-2=13

so

2^15 - 2^13 = 2^14 + 2^13
2^15 = 2^14 + 2^13 + 2^13
2^15 = 2^15

Hence, D
Simple math. Under one minute.

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jimmylow: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Jan 03, 2009 5:50 am; Thanked: 1 times

by jimmylow » Sat Jan 03, 2009 7:59 am

This question appeared in Official Guide for GMAT Review 11th Ed (Orange) pg 186 Q249

"If n is a positive integer less than 200 and 14n/60 is an integer, then n has how many different positive prime factors?"

a) two
b) three
c) five
d) six
e) eight

Ans: B

14 * n / 60 = (7 * 2 * n) / (3 * 5 * 2 * 2) <--- Breakdown into prime factors
= 7n / (2 * 3 * 5)

For 7n / (2 * 3 * 5) to remain an integer, n must be a product of 2 , 3 AND 5. Thus, there are 3 different positive prime factors.

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DanaJ: Site Admin; Posts: 2567; Joined: Thu Jan 01, 2009 10:05 am; Thanked: 712 times; Followed by:550 members; GMAT Score:770

by DanaJ » Sat Jan 03, 2009 8:02 am

First problem:

It says there that 14n/60 is an integer, which is the equivalent of 7n/30 being an integer. That means that n is divisible by 30 = 2*3*5. So there are at least 3 different positive prime factors in n: 2, 3 and 5. There could be more, but watch out: the next positive prime factor after 5 is 7 and 30*7 = 210, which is greater than 200. Therefore there are exactly 3 positive prime factors in n.

Answer: B

Second one:

You have a circle with r = radius and a square with n = side (unknown). You used all the wire, so:
40 = 2*Pi*r + 4n
20 = Pi*r + 2n, which means that n = (10 - Pi*r/2)
That means that the area you are looking for is:
Pi*r^2 + (10 - Pi*r/2)^2 (area of circle + area of square)

Answer: E

Third one:

You have: 2^x - 2^(x-2) = 2^(x-2)[2^2 - 1] = [2^(x-2)]*3. That means that your x-2 = 13 so x= 15.

Answer: D and I think it takes less time than the previous methods....

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rajataga: Senior | Next Rank: 100 Posts; Posts: 87; Joined: Tue Dec 23, 2008 11:06 pm; Thanked: 2 times

by rajataga » Sat Jan 03, 2009 8:13 am

Infact, i also have a very easy solution for the third problem....

just removing 2^(x-2) common from the LHS,

we get, 2^(x-2) [2^2 - 1] = 3(2^13)

therefore, 2^(x-2).3 = 3(2^13)

hence, x-2 = 13

therefore, x = 15.

Therefore, answer for third problem, D

For the first problem, i think the answer will be 3....B

For the second problem,

since the total wire is 40 meters,

length used for circle = circumference of the circle = 2pi.r

area of circle = pi(r^2)

therefore, length used by square,

40 - 2pi.r

side of square = 10 - pi.r/2

area of square = (10 - pi.r/2)^2

add both the areas and you get your answer, which is E

Ocean, please confirm these answers.

Thanks

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jimmylow: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Jan 03, 2009 5:50 am; Thanked: 1 times

by jimmylow » Sat Jan 03, 2009 8:22 am

DanaJ wrote: There could be more, but watch out: the next positive prime factor after 5 is 7 and 30*7 = 210, which is greater than 200. Therefore there are exactly 3 positive prime factors in n.

Yes, DanaJ is right about this one. Although I got B as the answer, I missed out the constraint that n < 200.

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jimmylow: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Jan 03, 2009 5:50 am; Thanked: 1 times

by jimmylow » Sat Jan 03, 2009 8:27 am

Thks Zipper, Dana and Raja for simplifying third problem. Always good to see alternative (quicker) solutions

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AWR83: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Wed Oct 22, 2008 10:17 pm

by AWR83 » Sun Jan 04, 2009 12:26 am

just removing 2^(x-2) common from the LHS,

we get, 2^(x-2) [2^2 - 1] = 3(2^13)

Thanks[/quote]

would someone plz explain how does takin common 2^(x-2) from LHS gives 2^(x-2) [2^2 - 1] ?

Thanks

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jimmylow: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Jan 03, 2009 5:50 am; Thanked: 1 times

by jimmylow » Sun Jan 04, 2009 1:30 am

2^x - 2^(x - 2)
= 2^x - (2^x / 2^2)
= (2^2 / 2^2) * 2^x - (2^x / 2^2)
= 2^2 * (2 ^ (x - 2)) - 2^(x - 2)
= 2^(x - 2) (2^2 - 1)

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udayasurya v: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Wed Jan 27, 2010 9:10 pm; Location: India, Chennai

by udayasurya v » Mon May 03, 2010 9:18 am

First problem:

It says there that 14n/60 is an integer, which is the equivalent of 7n/30 being an integer. That means that n is divisible by 30 = 2*3*5. So there are at least 3 different positive prime factors in n: 2, 3 and 5. There could be more, but watch out: the next positive prime factor after 5 is 7 and 30*7 = 210, which is greater than 200. Therefore there are exactly 3 positive prime factors in n.

In the explaination given above I am stumped at the constraint given ... could somebody please help me out with it ??

Thanks !!

Udaya

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harshavardhanc: Legendary Member; Posts: 526; Joined: Sat Feb 21, 2009 11:47 pm; Location: India; Thanked: 68 times; GMAT Score:680

by harshavardhanc » Mon May 03, 2010 11:00 am

udayasurya v wrote:
First problem:

It says there that 14n/60 is an integer, which is the equivalent of 7n/30 being an integer. That means that n is divisible by 30 = 2*3*5. So there are at least 3 different positive prime factors in n: 2, 3 and 5. There could be more, but watch out: the next positive prime factor after 5 is 7 and 30*7 = 210, which is greater than 200. Therefore there are exactly 3 positive prime factors in n.

In the explaination given above I am stumped at the constraint given ... could somebody please help me out with it ??

you first need to prime-factorize all the numbers : 14 and 60.

14 = 7*2
60 = 2^2 * 3 * 5

observe that the denominator (60) has a 2, a 3 and a 5 more than what 14 has . So, as the product of n and 14 is divisible by 60, these additional numbers will be supplied by n.

hence, n will definitely have 2,3 and 5.

Now, if you go further and ask yourself "can n have a 7 as well, the next prime number? "

the answer will be No, it can't. Because, 2 * 3 * 5 * 7 will be 210, which violates the condition given in the question :
(n<200). So, n can only have three different prime factors.

Hope this is clear now.

Regards,
Harsha