# Thread: A Couple of Hard Questions from a CAT Exam

1. ## A Couple of Hard Questions from a CAT Exam

Hello,

I recently did a CAT exam. I missed a few questions on Math and, unfortunately, there is no explanation for the solution. I hoped the wizards here could please help. I would love to hear how you would solve any of these problems

1. 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

A. b/n 2 and 10
B. b/n 10 and 20
C. b/n 20 and 30
D. b/n 30 and 40
E. Greater than 40

2. If x <> 0, then square root (x^2) / x =

A. -1
B. 0
C. 1
D. x
E. abs(x)/x

I thought it was c, but no dice

3. For which of the following functions is f(a+b) = f(a) + f(b) for all positive numbers a and b
A. f(x) = x^2
B. f(x) = x +1
C. f(x) = sqr (x)
D. f(x) = 2/x
E. f(x) = -3x

Thank you kindly  Reply With Quote

2. 3.Ans (E)

f(x) =-3x

f(a)= -3a--(1)
f(b)= -3b---(2)

f(a+b)= -3 (a+b)

(1) + (2)

f(a) + f (b) = -3a + (-3b) = -3 (a+b)

f (a+b) =f (a) + f (b)

2. ans (E)

(sqrt x^2)/ x = (+/- x)/x  Reply With Quote

3. What does b/n mean in question n&#176;1?  Reply With Quote

4. dissa,
sqrt(x^2)= x
not +x and -x

sqrt(9) = 3

if x^2 = 9
then x can be +3 and -3..

I also got C..

jsloan,
are you sure that C is wrong.. and what is b/n in a.. I do not see b in the question.  Reply With Quote

5. Originally Posted by abhishek_mumbai dissa,
sqrt(x^2)= x
not +x and -x

sqrt(9) = 3

if x^2 = 9
then x can be +3 and -3..

I also got C..

jsloan,
are you sure that C is wrong.. and what is b/n in a.. I do not see b in the question.
@ Abhishek,

x^2 = 9==> x = +/- sqrt (9) = +/- 3

y^2 = (x^2)==> y = sqrt(x^2) =+/-x sqrt (x^2) = +/- x

PS: keep in mind; when we take sqrt of anything there are two solutions(+ or -)!!  Reply With Quote

6. 1. 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

A. b/n 2 and 10
B. b/n 10 and 20
C. b/n 20 and 30
D. b/n 30 and 40
E. Greater than 40

I think in this question b/n means between.  Reply With Quote

7. Originally Posted by Md. Minuddin 1. 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

A. b/n 2 and 10
B. b/n 10 and 20
C. b/n 20 and 30
D. b/n 30 and 40
E. Greater than 40

I think in this question b/n means between.

And the answer is A. Because the smallest prime factor is 7

h(100) +1=100* 98* 96* ....*2+1

Now, 2*4*6*8*10+1=3840+1=3841 Smallest prime factor is 7
2*4*6*8*10* 12+1=46080+1=46081 Smallest prime factor is 7

And so on.................................

Here unit digit will always be 1. So is not divisible 2 or, 5. So we should check by 3 and then by 7 and if it is not factor then option is 11, 13 etc. But here it is divisible by 7. So answer is A
Minuddin and everyone else,

Thanks for the responses. I understand 2 and 3, but question 1 is a mystery. The answer is E. The question is from the GMATPrep software.

Is there a way to view an explanation to the solution? The software tells you the answer, but does not explain the answer.

Hopefully someone can figure out why it is E  Reply With Quote

8. hmm.. b/n between? I didn't get it. h(n) = 2 x 4 x .....n
h(100) = 2 x 4 x ...100
h(100) + 1 = (2 x 4 x ...100 + 1) = 2^50 (1 x 2 x...50) +1

if we divide this any number below 50;
1 would be the remainder. So that the smallest prime factor would be greater than 50.

Ans (E)  Reply With Quote

9. take simple eg,,, h(100) = 2 *4*8*10......100 so ,= 2^50(1.2.3......25)= 2^50*2*2^2*2^3*2^4( 1.3.5.6.7.9.10.11.12.13.14.15.17.18.19.20.21.22.23 .24.25) and so on h(100)+1 will have a smallest prime no greater than 50 hence E  Reply With Quote

10. Originally Posted by dissa hmm.. b/n between? I didn't get it. h(n) = 2 x 4 x .....n
h(100) = 2 x 4 x ...100
h(100) + 1 = (2 x 4 x ...100 + 1) = 2 (1 x 2 x...50) +1

if we divide this any number below 50;
1 would be the remainder. So that the smallest prime factor would be greater than 50.

Ans (E)
Just a small correction.
It (red 2) will be 250  Reply With Quote