GMAT Perp Questions

Question

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

Anurag@Gurome · Answer

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.

Anurag@Gurome · Answer

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.

Anurag@Gurome · Answer

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.

Anurag@Gurome · Answer

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.

raj.gmat.2011 · Answer

Thanks Anurag!

sanju09 · Answer

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

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

amma4u · Answer

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

sanju09 · Answer

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

The mind is everything. What you think you become. –Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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