A Couple of Hard Questions from a CAT Exam

[dissa]

Originally Posted by Md. Minuddin

Just a small correction.
It (red 2) will be 2⁵⁰
your explanation is good

yeah!! Thanks for the correction.I corrected it.

[doctortt]

Hi, here is my explanation to question 2:

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

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

Since x <> 0, it means x is a positive or a negative number. A negative number to the power of 2 is always positive, so in this question, we just need to worry about the x in the denominator. Remember, after the square root, you can't have a negative integer. For example sqrt(4) is 2 but sqrt(-4) is *not* good. Remember, 2^2 is 4 and (-2)^2 is also 4. Therefore, in this question, since the numerator is always postive, and the denominator *could* be negative, you need to absolute value it to make sure the answer is positive. Remember, in absolute value, |2| is 2 and |-2| is 2.

[mightyfellow]

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.

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The answer is certainly E
Please, mind that X is defined as a number different from zero.
Then, it might be positive or negative.
Furthermore, the author of the problem must deffine x <>0, because X is placed as a divisor in the expression.
Square Root (x^2)=|x|
Should the problem stated that X is positive then C would have been the answer.
However, this question is testing the concept of absolute value.
Thus E is the Answer
You can find more challenging GMAT questions for free in the following site
GMAT Online Questions

[aimster7888]

Originally Posted by dissa

hmm.. b/n between? I didn't get it.

My answer is here,
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.

answer (E)

i

Can you please explain how 2^50 was factored out. there's something really simple i'm missing here...thanks.

[india_chintan]

Originally Posted by dissa

hmm.. b/n between? I didn't get it.

My answer is here,
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.

answer (E)

I will elaborate on this more to make it simple.

h(n) = 2 x 4 x .....n
h(100) = 2 x 4 x ...100

Since there are 50 even nos. in 1-100 we can write:
h(100) = 2^50 *(1x2x3x4...50)

now, h(100) will be divisible by each nos. from 1 to 50 as it is in the numerator.

but, h(100) + 1 = (2 x 4 x ...100 + 1) = 2^50 (1 x 2 x...50) +1

with 1 added to it these prime nos. cannot divide the nos. in simpler terms if A can divide B, A cannot divide B+1. say 2 can divide 10, but cannot divide 11.

Coming back to the problem, since h(100) is divisible by all prime nos. from 1-50, the no. which can divide h(100)+1 has to be greater than 50.

Hence, the answer is E

[raghuveer_v]

Originally Posted by dissa

hmm.. b/n between? I didn't get it.

My answer is here,
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.

answer (E)

Great ... this question stumped me and your's is a beautiful and simple approach that we can reuse! Thanks a lot, dissa!

The other two questions are relatively much simpler ...

2. (E) ... just a matter of recognizing that sqrt() has two possible solutions
3. (E) ... we can also plug example values for a and b if the answer is not evident

[A.A.A]

Here is a different approach ..

To find the Product of all even integers from 2 to n inclusive means finding the product of all integers from 2 to n inclusive that are multiple of 2.

= { ( Last term - First term ) / 2} +1 , so

h(n) = { (n-2) / 2 } + 1 , now we should find p which = h(100)+1

Using the above formula and plugin 100 for n

h(100) = { ( 100-2 ) / 2 } +1

h(100) = (98/2) +1

h(100) = 49 + 1 = 50

Since P= h(100) + 1 then,

p = 50 + 1 = 51 which is greater than 40 .. E is the answer.