GRE
Practice Test 26
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If a person sells an article for $x, he loses 20% but if he sells it for $250 he gains 25%.
The value of x expressed as a fraction of $250 3⁄5
  • Solution

    When Selling Price = $250, gain is 25%.

    ∴ The Cost Price = $200

    At 20% loss, the S. P. = $x

    ∴ S.P. = \(\frac{200\times 80}{100}=\$ 160\)

    ∴ Required fraction = \(\frac{160}{250}\)

    ∴ Col.A = \(\frac{16}{25}\)

    Col. B = 3⁄5 = \(\frac{15}{25}\)

A pen costs $2 and pencil costs $1.
Number of ways in which exactly $9 can be spent to buy atleast one pen and one pencil \(\frac{9c_{3}}{14}\)
  • Solution

    In $9, one can buy

    4 pens and 1 pencil or

    3 pens and 3 pencils or

    2 pens and 5 pencil or

    1 pens and 7 pencils

    ∴ Col. A = 4

    Col. B = \(\frac{9c_{3}}{14}=\frac{9\times 8\times 7}{3\times 2\times 1\times 14}=6\)

    ∴ Col. B > Col.A

m⁄n = o⁄p = 7⁄q

Where m, n, 0, p, q are positive real numbers.

14p + 7n mg + 2og
  • Solution

    o⁄p = 7⁄q ∴ 7p = oq ∴ 14p = 20q

    Also m⁄n = 7⁄q ∴ mq = 7n

    ∴ Col. A = 14p + 7n

    Col. B = mq + 2oq = 7n + 14p

Linda has decided to fence 4 sides of her farmhouse. She has estimated the cost of wire required for fencing to be $1200. Since she has to manage within a budget, she decides to fence 80% of length and 75% of width.
Cost of wire required to fence the farmhouse now $720
  • Solution

    It is possible to calculate the percentage reduction in area. However, to workout the cost of the wire, it is required to workout the percentage change in perimeter and it cannot be calculated from the given information. It is not possible to find out length and width of the farmhouse.

In a campus selection test of 96 students, ratio of number of girls to boys appeared for the test is 7 : 5. After the test (1⁄7)th of girls and (1⁄5)th of boys were declared as failed.

Out of every 4 failure, 3 reappeared for the test and passed.

Ratio of boys and girls who passed the test 2⁄3
  • Solution

    Number of girls and boys appeared for the test is 56 and 40 respectively. 8 girls and 8 boys failed. But out of them, 6 girls and 6 boys reappeared for the test and passed. Now total number of boys who passed the test would be (40 - 8 + 6) = 38. Total number of girls who passed the test would be (56 - 8 + 6) = 54.

    ∴ Now the ratio of boys to girls would be \(\frac{38}{54}\) which is more than 2⁄3 = \(\frac{36}{54}\)

 

\(\sqrt{75}+\sqrt{12}+\sqrt{50}\) \(\sqrt{27}+\sqrt{8}+\sqrt{48}+\sqrt{18}\)
  • Solution

    Col. A:

    5√3 + 2√3 + 5√2 = 7√3 + 5√2

    Col. B:

    3√3 + 2√2 + 4√3 + 3√2 = 7√3 + 5√2

    ∴ Col. A = Col. B

A window cleaner cleans square window of certain size in time ‘t’. Time taken to clean a window varies directly with area of window. Each side of the window is now increased by 50%.
Additional amount of time taken to clean the new window in terms of ‘t’ 1.5t
  • Solution

    Let 'a' be the side of the window.

    ∴ Area = a2 and the time required is 't'

    Now the side of the window is 1.5 a

    ∴ New area will be 2.25 a2

    ∴ Time required will be 2.25t

    ∴ Additional amount of time required will be

    2.25t - t = 1.25t

    ∴ Col. B > Col. A

Tom, Dick and Harry take 12,6 and 44⁄5 days respectively to complete a job.
Efficiency of Tom and Dick together Efficiency of harry
  • Solution

    Col. A:

    Tom and Dick respectively finish \(\left (\frac{1}{12} \right )^{th}\) and (1⁄6)th of the job in 1 day.

    ∴ They together finish \(\frac{1}{12}+\frac{1}{6}+\frac{3}{12}=\left ( \frac{1}{4} \right )^{th}\) of the job in 1 day i.e. they take 4 days to finish the job completely.

    Col. B:

    Harley can finish the same job in 44⁄5 days.

    ∴ Tom and Dick are more efficient then Harley.

    ∴ Col. A> Col. B

The ratio of the speeds of 3 cars A, B, C is 3:5:7. Car A covers 27 kms. in 11⁄2 hours.
Column A Column B
  • Solution

    Speed = \(\frac{Distance}{Time}\)

    ∴ Speed of car A =\(\frac{27}{1.5}\)= 18 kmph.

    ∴ In the same proportion, speed of car C

    \(\frac{18\times 7}{3}\)= 42 kmph.

    ∴ Time required to cover 168 kms. =\(\frac{168}{42}\)= 4 hrs.

A {3, 7, 9} B = {9, 11, 13, 15}
Number of subsets of A and B having only two elements, one from set A and one from Set B 12
  • Solution

    Each element of A can be combined with each element of B e.g. (3,9), (3, 1), (3, 13), (3, 15)

    ∴ We can have 12 subsets.

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