GRE
Practice Test 24
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The original price of an article was reduced by 20%. During a special sale, the new price was decreased by 25%. The price of the article now has to brought back to the original price by 2 successive mark ups. If the first markup is 25%, what is the second markup approximately?

  • Solution

    Let the original price be $100. After first discount, price is 80% of 100 = $80. After second discount, price is 75% of 80 = $60. To bring this amount back to $100, the first increase is 25%.

    ∴ Price after first increase will be 375

    ∴ The second increase is 100 - 75 = $25.

    ∴ % increase = 25 x 100 = 33.33%

p* \(\frac{p(p + 1)(p- 1)}{3}\) for all integers p and q = 3* then q* =

  • Solution

    \(3* =\frac{3(3 + 1)(3-1)}{3}=8\)

    \(q* =8*=\frac{8\times 9\times 7}{3}=168\)

By weight liquid L makes up 15% of solution I. 18% of solution II and 9% of solution III. If 3kg of solution I is mixed with 2kg of solution II and 1kg of solution III, then liquid L accounts for what % of the weight of resulting solution?

  • Solution

    Kg. Liquid. L

    Solution I 3.45 (15%)

    Solution II 20.36 (18%)

    Solution III \(\underline{10.09}\) (9%)

    60.9G (15%)

In a standard IX class, there are 24 boys having an average height of 165cm which is 3⁄4 of the total number of boys. Total number of boys is equal to 3. of the total number of students. If the average height 2⁄3 of the entire class is 168cm, find the average height of remaining class (i.e. excluding these 24 boys).

  • Solution

    Total number of boys = 24 × 4⁄3 = 32

    Total number of students = 32 × 3⁄2 = 48

    ∴ The average height of 48 students is 168 em. and that of 24 boys is 165 cm.

    ∴ The average height of remaining 24 students is given by \(\frac{(168 \times 48) - (165 \times 24)}{24}\)

    =\(\frac{24(168 \times 2)- (24(165 \times 1)}{24}\)

    = (168 × 2)-(165 × 1) = 171

A rectangular property that is 1600m. by 2000m. has a road frontage along all 4 sides. It is to be sub divided into rectangular plots, so that each plot will have a total of 400m. of road frontage. What is the greatest number of such plots?

  • Solution

    Total road frontage will be equal to the perimeter of the property, which will be 7200 meters. Each plot will have wood frontage of 400 meters.

    ∴ Total plots -\(\frac{7200}{400}\)= 18 plots.

In each production lot for certain shirts, 35% of the shirts are white and 65% are blue. 45% of the shirts are of size Land 55% are of size XL. If 17 out of a lot of 100 shirts are white and size XL, how many of them are blue and size L?

  • Solution

    Such problems must be attempted by tabulating the data as shown below

4 stacks containing equal number of chips are to be made from 11 red chips, 9 blue chips, 13 green chips and 7 white chips. If all of these chips are used and each stack contains at least 1 chip of each color, what is the maximum number of blue chips in anyone stack?

  • Solution

    Each stack will have 10 chips. Also each stack must have at least 1 chip of each colour.

    ∴ Each of the four stacks will have one blue chip. Now the remaining 5 blue chips can be assignmed to one particular stack so that it would have maximum number of blue chips.

    ∴ Maximum blue cries = 5 + 1 = 5

Water is running into a pool from 2 pipes at rate of 1 kiloliter every 6 seconds and every 9 seconds respectively and simultaneously running out from a leak at the rate of 1 kiloliter every 15 seconds. The pool is filling at the rate of 1 kiloliter every

  • Solution

    Method 1:

    Water coming into the pool by two pipes in 1 second is 1⁄6 + 1⁄9 = \(\frac{5}{18}\) kilolitres.

    Water going out per second = \(\frac{1}{15}\)kiloliters.

    ∴ Quantity of water actually filling the pool per second = \(\frac{5}{18}-\frac{1}{15}=\frac{25 - 6}{90}=\frac{19}{90}\)

    \(\frac{19}{90}\) kiloliters is net quantity of water in the pool per second.

    ∴ Water is filled in tank at the rate 1 kiloliter every \(\frac{90}{19}\) seconds = \(4\frac{14}{19}\) seconds

    Method 2:

    LCM of 6, 9 and 15 is 90.

    If all the pipes are working for 90 seconds, 1st pipe will give \(\frac{90}{6}\) = 15 kiloliters, 2nd pipe will give \(\frac{90}{9}\) = 10

    kilolitres and 3rdpipe will pull out \(\frac{90}{15}\) = 6 kilolitres.

    Therefore the net addition is 90 seconds will be 15 + 10 - 6 = 19 kiloliters.

    ∴ Paul will fill 1 kilolitres in \(\frac{90}{19}\) = 4\(\frac{14}{19}\) seconds.

Cost of cleaning a tall chimney of height 100 ft. is $6 for first foot and cost of each foot thereafter is $x more than that of the preceding foot. The total cost of cleaning is $20400. Find x.

  • Solution

    It is an Arithmetic Progression with first term=$6.

    The common difference, d = x.

    Number of terms n = 100

    Sum of n terms Sn = $20400

    ∴ Sn = n⁄2[2a+(n-1)d]

    ∴ 20400 =\(\frac{100}{2}\)[12 +99x]

    ∴ 408 = 12 + 99x ∴ X = \(\frac{396}{99}\) ∴ x = 4

On a team of ‘p’ players, 5 played 12 minutes apiece in each of the 3 games. The others played m minutes in the first game and S minutes each in the second and third game. The average playing time in minutes, per player per game over the 3 games was

  • Solution

    5 players played 12 minutes apiece in each of the three games.

    ∴ 5 × 12 × 3 = 180 minutes was the total playing time for 5 players for 3 games.
    For the remaining p - 5 players, the total playing time was (p - 5)m + (p - 5)2s = (p - 5) (m + 2s)

    ∴ Total playing time for all players for all 3 games = 180 + (p - 5) (m + 2s)

    ∴ Required average = \(\frac{180 + (p - 5)(m + 2s)}{3p}\)

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