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Gmat prep test #1 questions

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Gmat prep test #1 questions

by akane » Sun Oct 09, 2011 8:18 am
Hello everyone!

I am about 3weeks away and took the Gmat prep test 1 and got a 590 36V 35 Q, my goal is to get at least 700. I struggled with some of the questions on the prep test and need help with Algebra, as that is my biggest problem, simplifying algebraic expressions. here are some of the questions I could not figure out. thanks for your inputs.

#1: (1/5)^m (1/4)^18=(1/((2(10)^35)) what is the value of m?

#2: The perimeter of a certain isosceles right triangle is 16+16V2. What is the length of the hypothenuse of the triangle?
here I am pretty comfortable with the pythagorean theorem and the special right triangles but this question was a trip. I couldn't figure out why the answer is 16 and not 8V2.

#3:For a finite sequence of non zero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the 2 consecutive term is negative. What is the number of variation in sign for the sequence 1, -3, 2, 5, -4, -6 ?
a) One b) two c) three d) four e)five

#3: A basket contains 5 apples, of which 1 is spoiled and the rest are good. If henry is to select 2 apples simultaneously and at random, what is the probability that the 2 apples selected will include the spoiled apple? The solution says is 2/5 but there is no explanation how we get that answer.

Thanks!
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by MBA.Aspirant » Sun Oct 09, 2011 8:54 am
(1/5)^m (1/4)^18=(1/((2(10)^35)) what is the value of m?


1/5^m * 1/2^36 = 1/(2 * 10^35)

1/5^m * 1/2^35 * 1/2 = 1/10^35 * 1/2

1/5^m * 1/2^35 = 1/10^35


the two bases (5 and 2) will be grouped together under the same exponent giving 10^35, so m should = 35
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by svd.kumar » Sun Oct 09, 2011 9:00 am
First 3 problems I can help you out, but probability I am also working on that.

#1: (1/5)^m (1/4)^18=(1/((2(10)^35)) what is the value of m?

For this problem, we need to make equation in the right hand side similar to left hand side (Make both bases same,then only we can cancel)

right hand side equation can be written as ---

1/ (2^1 * (2^35 * 5 ^35))

1/(2^(1+35) * (5 ^35))

(1/2^36) * (1/5^35)

(1/4^18) * (1/5^35) (2^2 = 4)

i.e (1/5)^m (1/4)^18 = (1/4^18) * (1/5^35)

so m = 35

#2: The perimeter of a certain isosceles right triangle is 16+16V2. What is the length of the hypothenuse of the triangle?

let height and base be x for isosceles triangle, as it is right angled also hypotenuse becomes V(x^2+x^2) = xV2

we have perimeter so, height+base+hypotenuse = perimeter
x + x + xV2 = 16+16V2
2x+v2x = 16+16V2
v2x(v2+1) = 16(1+v2) (Taking v2x common in left and 16 common in right)
x = 16/v2 = 8v2.



#3:For a finite sequence of non zero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the 2 consecutive term is negative. What is the number of variation in sign for the sequence 1, -3, 2, 5, -4, -6 ?
a) One b) two c) three d) four e)five

These kind of questions just read carefully, it is very simple. It is asking to calculate number of times sign is changed

between 1 and -3 once
between -3 and 2 once
between 5 and -4 once

so total 3 times

#4 Problem I am also not good at probability working on that. Please feel free to ask if you need more clarification.


Cheers,
SVD
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by cmal » Mon Oct 17, 2011 1:00 am
For Question on Triangles

Perimeter = 16 +16sqrt2

Let one side of triangle be =x
For Isoless Triangle hypothesis is always equal to y=xsqrt2
hence perimeter
= x+x+y
=2x+xsqrt2=16+16sqrt2
solving equation gives x=11.32 and y=16
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