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GMAT Pre Questiones

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GMAT Pre Questiones

by maypoon » Thu Jan 29, 2009 7:20 am
Dear all,

Would you please teach me how to do the following questiones?Thank you so much!

Q1.

A certain right triangle has sides of length x, y and z, where x<y<z, if the area of this triangulr region is 1, which of the following indicates all the possible values of y?

A. y> square root 2
B. square root 3/ 2 <y <square root 2

Why the answer is "A"

Q2.

The function f is defined for each positive three-digit n by f (n) = 2 (square x ) 3 (square y) 5 (square z), where x,y and z are hundreds, tens, and units digits of n, respectively. If m & v are three-digit positive integers such that f (m) =9 f(v), then m- v=

A. 8
B. 9
C. 18
D. 20
E. 80.

Why the answer is "D"

Q3.

A sum of money was divided between John & Ann such that the ratio of John's share to Anne's shre was 5:3 . If John's share exceeded 5/9 of the sum of money by $50, what was Anne 's share?

A. $180
B. $270
C. $340
D $450
E. $720 Ans: B

P.s. what is the meaning of "" If John's share exceeded 5/9 of the sum of money by $50"
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by DeepakR » Thu Jan 29, 2009 7:48 am
1. Since its a right triangle 1/2 * b*h= Area of triangle. Now we have x<y<z which implies z greatest i.e. the hypotenuse. We also have A=1=1/2*x*y
Therefore xy=2. Since y>x y=2 and x=1. So answer A.) y> square root 2

3. From the question,
J/A=5/3 --->1
and J=5/9(T) + 50 ---> 2

Equation 1 implies that J=5/8(T) where T=Total ---> 3 Solving 2 and 3 we get T=720 and hence A=3/8(T)= 270


- Deepak
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by DanaJ » Thu Jan 29, 2009 8:05 am
Q1.

Since our triangle is right and x<y<z, this means that z, which is the longest of all the sides, must be the hypotenuse. This in turn means that x and y are both catheti, so the area of our triangle is x*y/2 = 1. So this leads to x*y = 2. Now suppose that x = y. This would mean that x = y = sqrt(2), but since y > x, then y has to be greater than sqrt(2).

Q2. Say x = hundreds digit for m
y = tens digit for m
z = units digit for m.

Then say a = hundreds digit for v
b = tens digit for v
c = units digit for v.

So you have that:
2x^2*3y^2*5z^2 = 9*2a^2*3b^2*5c^2. You can notice that you can immediately eliminate 2, 3 and 5, so you get that:
x^2*y^2*z^2 = 9*a^2*b^2*c^2. This is equivalent to
(x^2*y^2*z^2 - 9*a^2*b^2*c^2) = 0 or (xyz - 3abc)(xyz + 3abc) = 0. But since xyz + 3abc cannot be 0, then xyz = 3abc. I don't know how to continue from here on, unfortunately, because any of x, y, z could equal 3a, 3b or 3c if a, b, care either 1, 2 or 3.
I might have misunderstood the original formula or maybe you wrote it down incorrectly....

Q3. Say j = amount of money john got
a = amount of money Ann got.

We have that: a + j = x
j/a = 5/3, so a = 3/5j
5/9*x = j - 50.

Substitute a in a + j = x with a = 3/5 j and you get that 8/5j = x. This means that j = 5/8x.
Now substitute that in 5/9x = j - 50 and you get that 5/9x = 5/8x - 50 or that 5/72x = 50 or that x/72 = 10. This means that x = 720. Since a = 3/5j = 3/5*5/8x, a = 270.
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by hardik.jadeja » Thu Jan 29, 2009 11:17 am
For Q2 refer this link

https://www.beatthegmat.com/what-s-the-f ... 19361.html
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by DanaJ » Thu Jan 29, 2009 11:29 am
Oh ok then.... I thought it was weird when I first saw it...
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