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PS Problem Set 01

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by gdk800 » Wed Dec 08, 2010 7:56 am
Q1) When a tree was planted it was 4 feet. Its height increased by a constant amount each year for 6 years. After 6 years the tree was 1�5 taller than at the end of 4th year. By how many feet did the height of the tree increase each year?
a. 3�10
b. 2�5
c. 1�2
d. 2�3
e. 6�5

OA is D. Don't know where I am missing the point but just unable to reach option D.


Q2) For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n inclusive. If p is the smallest prime factor of h(100) + 1, then p is between
a. 2 and 10
b. 10 and 20
c. 20 and 30
d. 30 and 40
e. > 40

OA is E. Can someone come up with a fast approach of solving this question? Thanks


Q3) Image attached. Correct answer is marked in Red.

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by Rahul@gurome » Wed Dec 08, 2010 8:03 am
gdk800 wrote:Q1) When a tree was planted it was 4 feet. Its height increased by a constant amount each year for 6 years. After 6 years the tree was 1�5 taller than at the end of 4th year. By how many feet did the height of the tree increase each year?
a. 3�10
b. 2�5
c. 1�2
d. 2�3
e. 6�5
Say each year the height is increased by x feet.
Height at the end of 4th year = (4 + 4x) feet
Height at the end of 6th year = (4 + 6x) feet

According to question,
  • Height after 6th year = Height after 4th year + (1/5)*(Height after 4th year)
Thus, (6x + 2) = (4x + 4) + (1/5)*(4x + 4)
=> (4x + 4)/5 = 2x
=> (4x + 4) = 10x
=> 6x = 4
=> x = 2/3

The correct answer is D.
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by Rahul@gurome » Wed Dec 08, 2010 8:10 am
gdk800 wrote:Q2) For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n inclusive. If p is the smallest prime factor of h(100) + 1, then p is between
a. 2 and 10
b. 10 and 20
c. 20 and 30
d. 30 and 40
e. > 40

OA is E. Can someone come up with a fast approach of solving this question? Thanks
Discussed many times in the forums.
May be this is the fastest approach.

h(100) + 1 = (2*4*6*8*10*...*100) + 1 = 2*(1*2*3*4*5*...*50) + 1

Thus when [h(100) + 1] is divided by any integer (including all the primes) less than or equal to 50, it'll leave a remainder of 1. Thus p must be greater than 50 in turn p > 40.

The correct answer is E.
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by Rahul@gurome » Wed Dec 08, 2010 8:32 am
Q3. A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?
  • (A) 16
    (B) 24
    (C) 26
    (D) 30
    (E) 32
Total number of ways to select 3 people from 8 = 8C3 = 56

Now these possible 56 committees will contain some committees in which there will be two people who are married to each other. We have to discard them.

Total number of ways such that the committee will contain two people who are married to each other = (Number of ways to select a couple from 4)*(Number of ways to select third person from 6) = (4C1)*(6C1) = 4*6 = 24

Number of favorable committees = (56 - 24) = 32

The correct answer is E.
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by prajanya » Wed Dec 08, 2010 9:41 am
Discussed many times in the forums.
May be this is the fastest approach.

h(100) + 1 = (2*4*6*8*10*...*100) + 1 = 2*(1*2*3*4*5*...*50) + 1

Thus when [h(100) + 1] is divided by any integer (including all the primes) less than or equal to 50, it'll leave a remainder of 1. Thus p must be greater than 50 in turn p > 40.

The correct answer is E.

Hi Rahul,

Please give an elaborate explanation so as to gain a better understanding of the concept.

thanks,
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by prajanya » Wed Dec 08, 2010 9:42 am
Discussed many times in the forums.
May be this is the fastest approach.

h(100) + 1 = (2*4*6*8*10*...*100) + 1 = 2*(1*2*3*4*5*...*50) + 1

Thus when [h(100) + 1] is divided by any integer (including all the primes) less than or equal to 50, it'll leave a remainder of 1. Thus p must be greater than 50 in turn p > 40.

The correct answer is E.

Hi Rahul,

Please give an elaborate explanation so as to gain a better understanding of the concept.

thanks,
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by Rahul@gurome » Wed Dec 08, 2010 8:24 pm
prajanya wrote:Hi Rahul,

Please give an elaborate explanation so as to gain a better understanding of the concept.

thanks,
h(100) = (2*4*6*8*10*...*100) = 2*(1*2*3*4*5*...*50) => This means h(100) is a product of first 50 natural numbers multiplied with 2. Thus h(100) is divisible by all the first 50 natural numbers (including all the primes like 2, 3, 5, 7..., 29, 31... 43, 47).

Now if we add 1 to h(100), the quantity [h(100) + 1] will always give a remainder of 1 when divided by any natural number from the first 50 natural numbers. This is because h(100) is divisible by them.
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by gdk800 » Thu Dec 09, 2010 7:20 am
Thanks Rahul for the explanations :)
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by goalevan » Sat Jul 16, 2011 4:20 pm
Here's a correction to the previous posts that doesn't change the answer:

h(100) = 2 * 4 * 6 * ... * 99 * 100
= 2 * (1 * 4 * 6 * ... * 99 * 100)
= 2^2 * (1 * 2 * 6 * ... * 99 * 100)
= 2^3 * (1 * 2 * 3 * ... * 99 * 100)
= 2^50 * (1 * 2 * 3 * ... * 49 * 50)
= 2^50 * 50!

50! is divisible by all integers from 1 to 50. Thus when we add 1 to this quantity and divide by any integer from 1 to 50, we will have a remainder of 1.
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