GMAT Perp Questions

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raj.gmat.2011: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Thu Feb 17, 2011 3:07 am

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) = âˆšx
D. f(x) = 2/x
E. f(x) = -3x

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Source: Beat The GMAT — Problem Solving | Thu Feb 17, 2011 3:09 am

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

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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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

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 = Aâˆš2
a = 1.4A, which implies there is 40% increase.

The correct answer is D.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

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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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

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) = âˆš5
f(2) + f(3) = âˆš2 + âˆš3; 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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raj.gmat.2011: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Thu Feb 17, 2011 3:07 am

by raj.gmat.2011 » Fri Feb 18, 2011 10:41 pm

Thanks Anurag!

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

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 Ã— 2 and

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

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amma4u: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Mon Feb 22, 2010 9:54 am

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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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

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

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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