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  • Verbal - 166
  • Quans - 170
  • Total - 336
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GRE Practice Test

GRE Practice: Numeric Entry GRE Practice Test 1

Good you are taking our practice tests. But please make sure you complete all the practice problems from Official GRE Guide. Official GRE Guide is from the test makers(Yes You read it right they conduct the GRE exam worldwide.) ETS.:- The Official Guide to the GRE General Test, Third Edition
Powered by GreGuide Version 2.2.1 Copyrights 2007,

Question: 1

The diagonal of a square is 15*sqrt(6) cm. How many isosceles right triangles of side 1 cm can be cut out from the square?

[sqrt(6) = square root of 6]

Correct Answer: 1350


When 'a' is the side of a square then its diagonal is a*sqrt(2).
Hence, if the diagonal of the square is 15*sqrt(6), then its side will be given by
Side = 15*sqrt(6)/sqrt(2) = 15*sqrt(2)*sqrt(3)/sqrt(2)
= 15*sqrt(3)
Area of square = side*side
= 15*sqrt(3)*15*sqrt(3)
= 675
675 squares of side 1 cm can be cut from the bigger square.
Each square of side 1 cm can be cut into two isosceles right triangles of side 1 cm.
Hence, the square can be divided into 675*2 = 1350 isosceles right triangles of side 1 cm

Question: 2

A semi-circular park has a flowerbed of a certain width running along its arc on the inner side. The area of the park, including the flowerbed is 98*pi sq.m and the area of the park minus the flowerbed is 72*pi sq.m. Find the width of the flowerbed in meters.


Correct Answer: 2m


Area of semi-circle = pi*r*r/2, where r is radius of circle.

Area of the complete park = 98*pi sq.m
Outer radius of the park = sqrt(2*area/pi)
= sqrt(2*98*pi/pi)
= sqrt(196) = 14 m

Area of the park excluding the flowerbed = 72*pi sq.m
Inner radius of the park = sqrt(2*area/pi)
= sqrt(2*72*pi/pi)
= sqrt(144) = 12 m

Width of the flowerbed = Outer radius - Inner radius
= 14-12 = 2 m

[pi = 22/7]
[sqrt(2) = square root of 2]

Question: 3

Find 360% of 360.

Correct Answer: 1296


360% of 360 = 360/100*360
= 1296

Question: 4

Ann and Bill are 120cm and 165cm tall respectively. Their weights are 50kg and 64kg respectively. How much weight in kgs should Ann lose and Bill gain so that the height to weight ratio for both of them is the same?

Correct Answer: 2


Let the weight to be lost by Ann and that to be gained by Bill be x kgs.
Their height to weight ratio will be
120/(50-x) = 165/(64+x)
120*(64+x) = 165*(50-x)
7680+120x = 8250 - 165x
570 = 285x
x = 570/285 = 2

Ann should lose 2 kg and Bill should gain 2 kg

Question: 5

If f(x) = (x-3)/(x-1), then find the value of [x - f{f(x)}]

Correct Answer: 0


[x - f{f(x)}] = x - f{(x-3)/(x-1)}
= x - [{(x-3)/(x-1)-3}/{(x-3)/(x-1)-1}]
= x - {(x-3-3x+3)/(x-1)}/{(x-3-x+1)/(x-1)}
= x - (-2x)/(x-1)*(x-1)/(-2)
= x - x
= 0

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