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GMAT Perp Questions

by raj.gmat.2011 » Thu Feb 17, 2011 3:09 am
Can someone please provide explanations for below questions?

1. For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is
A. between 2 and 10
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40

2. The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged:
A. 100% decrease
B. 50% decrease
C. 40% decrease
D. 40% increase
E. 50% increase

3. If X4 + Y4 = 100, then the greatest possible value of x is between
A. 0 & 3
B. 3 & 6
C. 6 & 9
D. 9 & 12
E. 12 & 15

4. For which of the following functions is f(a+b) = f(a) + f(b) for all positive numbers a and b
A. f(x) = x2
B. f(x) = x+ 1
C. f(x) = √x
D. f(x) = 2/x
E. f(x) = -3x
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by Anurag@Gurome » Thu Feb 17, 2011 4:41 am
h(100) = 2 * 4 * 6 * 8 * ... * 100
= (2^50)(1 * 2 * 3 * 4 * ... * 50)
All integers till 50 are factors of h(100), which includes the prime numbers till 50 also.
So, h(100) + 1 cannot have any prime factor below 50, it will always give a remainder of 1.
Hence, from the given answer choices the correct answer should be E.
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by Anurag@Gurome » Thu Feb 17, 2011 5:59 am
Q2.

Let the rate of a chemical reaction = R
Then R = k(A^2)/B ...Equation 1
Let the new concentration of chemical A and B be a and b respectively. B is increased by 100% means that B is doubled, which means b = 2B.

Then R = k(a^2)/2B ...Equation 2
From equation 1 and 2, k(a^2)/2B = k(A^2)/B
a^2 = 2.A^2
a = A√2
a = 1.4A, which implies there is 40% increase.

The correct answer is D.
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by Anurag@Gurome » Thu Feb 17, 2011 6:12 am
Q3.

x can be maximum if y is the least and the least possible value of y = 0.
If x = 3 then 3^4 = 81, which implies x should be greater than 3.

The correct answer is B.
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by Anurag@Gurome » Thu Feb 17, 2011 6:23 am
Q4.
Let a = 2 and b = 3.
A. f(2 + 3) = f(5) = 5^2 = 25
f(2) = 2^2 = 4 and f(3) = 3^2 = 9, then f(2) + f(3) = 4 + 9 = 13; not true
B. f(2 + 3) = f(5) = 5 + 1 = 6
f(2) = 3, f(3) = 4, then f(2) + f(3) = 7; not true
C. f(2 + 3) = f(5) = √5
f(2) + f(3) = √2 + √3; not true
D. f(2 + 3) = f(5) = 2/5
f(2) + f(3) = 2/2 + 2/3; not true
E. f(5) = -3 * 5 = -15
f(2) + f(3) = -3 * 2 + -3 * 3 = -6 - 9 = -15; True

The correct answer is E.
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by raj.gmat.2011 » Fri Feb 18, 2011 10:41 pm
Thanks Anurag!
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by sanju09 » Sat Feb 19, 2011 1:08 am
raj.gmat.2011 wrote:Can someone please provide explanations for below questions?

1. For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is
A. between 2 and 10
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40
Please post one question per thread.


In case n is even, h (n) = n (n - 2) (n - 4)...4 × 2 and

h (100) + 1 = 100 × 98 × 96...4 × 2 + 1 = 2^50 (50 × 49 × 48...2 × 1) + 1

Key fact, 1 added to a number divisible by a prime say q makes it indivisible by q. Here h (100) is already divisible by all primes from 2 to 47, hence if 1 is added to h (100), it's no more divisible by any prime from 2 to 47. Therefore, if h (100) + 1 is still divisible by some prime p, then p must be [spoiler]more than 47.

E is the best choice[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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by amma4u » Sun Feb 20, 2011 6:38 am
Hi Sanju09,

h(100) is not divisible by 3, 7, 11, right?

This is what I did product of all evens between 2-100 is 2550. Taking primes I get 51*5^2*2, here primes are 51,5 and 2.

Please could you explain how you say all primes between 2-47 divide h(100) ?

Thanks,
S
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by sanju09 » Sun Feb 20, 2011 11:02 pm
amma4u wrote:Hi Sanju09,

h(100) is not divisible by 3, 7, 11, right?

This is what I did product of all evens between 2-100 is 2550. Taking primes I get 51*5^2*2, here primes are 51,5 and 2.

Please could you explain how you say all primes between 2-47 divide h(100) ?

Thanks,
S
No, in fact h (100) is divisible by 3, 7, 11 too. See, h (100) = 100 X 98 X 96...4 X 2 = 2^50 (50 X 49 X 48...2 X 1), in which the bracket (50 X 49 X 48...2 X 1) includes all primes 2 through 47.

"product of all evens between 2-100 is 2550"

This again is not true, in fact this product is too big to express :)
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