GRE
Practice Test 40
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m is a positive integer. P =\(\frac{18m+81}{m^{3}}\)
Number of values of m which would give an integral value for P 3
  • Solution

    p=\(\frac{18m+81}{m^{3}}=\frac{18}{m^{2}}+\frac{81}{m^{3}}\)

    For P to e an integer, both \(\frac{18}{m^{2}}\) and \(\frac{81}{m^{^{3}}}\) should be integer.

    Consider \(\frac{18}{m^{^{2}}}\) and \(\frac{81}{m^{^{2}}}\).

    m2 be should be a factor of 18 and m3 a factor of 81.

    ∴ m = 1 or 3. Col. A = 2, Col. B = 3

Due to an increase of 50% in the price of pencils, 5 pencils less are available for $3.
Number of pencils available for $3 at present 10
  • Solution

    Let x pencils be available for $3 initially

    ∴ Price of 1 pencil initially = \(\frac{300}{x}\)cents

    Because of 50% price increase, now x - 5 pencils are available for $3 instead of x.

    ∴ New price of 1 pencil = \(\frac{300}{x-5}\) cents

    Also \(\frac{\frac{300}{x-5}}{\frac{300}{x}}=\frac{1.5}{1}\) ∴\(\frac{x}{x-5}=\frac{15}{10}\)

    Initially 15 pencils were available for $3. Today 10 pencils are available for $3.

    ∴ Col. A = Col. B

    Such problems can be tried with options.

    Col. B gives quantity of 10 pencils.

    ∴ Price per pencil today =\(\frac{300}{10}\)= 30 cents.

    The price of 30 cents is 50% more than the previous price.

    ∴ Previous price must be 20 cents.

    ∴ He could buy \(\frac{300}{20}\) = 15 pencils. Hence the conditioned is satisfied.

 

(0.00032)0·6 (0.0081)0·5
  • Solution

    Col. B = [(0.0.0083)4]0.5 = (0.3)4×0.5

    (am)n = am

    Col. A = [(2)5]0.6 = 0.23 = 0.008

    Col. B = (0.3)2 = 0.09

The figure shows a circle with diameter PR, ∠TQP= 65° ,∠UQR = ∠USR = 30°

Measure of ∠QTU in degrees 30
  • Solution

    ∠QUP = 120°

    ∴ ∠QUT = 60°

    ∠TQU = 85°

    ∠QTU = 35°

    ∴ Col. A = 35°, Col. B = 30°

m < 1 n > m
m⁄n n⁄m
  • Solution

    Let m=-1,n=2

    Col.A=m⁄n=-1⁄2, Col.B=n⁄m=-2

    ∴ Col.A > Col.B

    Let m = 0, n = 1

    Col. A =m⁄n= 0, Col.B = n⁄m = infinite

On her birthday, Ana distributed 5 chocolates per student on an average. On the arrival of the teacher and the Principal, to whom she gives 10 and 15 chocolates respectively, the average chocolate distributed per head increases to 6.
strength of the class 20
  • Solution

    Let the strength of the class be x.

    ∴ Total number of chocolates given to x students = 5x

    According to the given condition,

    5x +10 + 15 = 6(x + 2)

    ∴ 5x + 25 = 6x + 12 ∴ x = 13

    ∴ x = 13 = Col. A, Col. B = 20

A painter has 3 big jars containing 57 litres. 133 litres and 171 litres of paint respectively this Paint was transferred to smaller jars of equal size from each of the bigger jar.
Minimum number of smaller jars of equal size in which the paint can be transferred 19
  • Solution

    One has to find the HCF of 57, 133 and 171 which is 19.

    Hence the paint from the first jar will be transferred in 3 smaller jars of 19 lit. each similarly paint from other 2 jars will be transferred into 7 and 9 smaller jars respectively. The total number of jars will be 19.

    Col A. = Col. B.

A rectangular piece of cardboard having dimensions of 24 cm X 18 cm. is made into an open box by cutting a square of 5 cm. side from each comer and building up the side.
Area of the bottom of the box Total area that was cut out from the cardboard
  • Solution

    Col. A = 14 × 8 = 112 cm2

    Col. B = (5 × 5) × 4 = 100 cm

    ∴ Col. A > Col. B

To make a marriage tent, poles are planted along the perimeter of a square field at a distance of 6 meter from each other. The total numbers of poles planted were 20.
Area of the square field in m2 625
  • Solution

    The figure shows, the layout of the square field with 20 equispaced poles.

    The distance between 2 consecutive poles = 6 meter.

    ∴ Distance covered by 20 poles = 120 metres .

    ∴ Each side = 30 metres .

    ∴ Area = 900 m? = Col. A

    Col. B = 625

m > 1
n > 0
nm n(m+n)
  • Solution

    Let n = 1, m = 2

    ∴ Col. A = (1)2 = 1

    Col. B = (1)3 = 1

    Col. A = Col. B

    Let m = 2, n = 2

    ∴ Col. A = (2)2= 4

    Col. B = (2)4 = 16

    ∴ Col. A < Col. B

    Let n = 0.5, m = 2

    ∴ Col. A = (0.5)2 = 0.25

    Col. B = (0.5)3= 0.125

    ∴ Col. A > Col. B

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