GRE
Practice Test 27
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Out of 200 students in a class, the number of students appearing for GRE, GMAT, SAT were 69, 55, 53 respectively. 23 appeared for both GRE and GMAT, 19 appeared for GMAT and SAT. 21 appeared for both GRE and SAT. If 11 appeared for all three, find the number of students who didn’t appear for any exam.

  • Solution

    Total number of students appearing for any exam = 36 + 24+ 24 + 12 + 10 + 8 + 11 = 125

    ∴ Number' of students who didn't appear for any exam = 200 - 125 = 75

The lengths of the sides of the equiangular polygon are as shown above. Find total area of the shaded region.

  • Solution

    Area of shaded region = Area of 4 triangles + Area of square. Area of square = 2 × 2 = 4

    All the four triangles are isosceles right angled triangles with arm length = 2

    ∴ Area 4 triangles = 4 × 1⁄2 × (2 × 2) = 8

    ∴ Area of shaded region = 8 + 4 = 12

Mark sells goods at a profit of 20%. If the cost of goods doubles and profit % becomes 1⁄5th its present %, then the actual profit now is what % of the previous profit?

  • Solution

    Let the cost price be $100

    ∴ Selling price is $120 and profit is $20. The new cost = $200. New profit = 4%

    ∴ Profit = $8

    Required % = \(\frac{8}{20}\)= 40%

The speed of a boat when it travels downstream is 5 times the speed when it travels upstream. Speed of current is what % of speed of the boat in still water?

  • Solution

    Let speed of boat in still water be x and that of current be y.

    ∴ Speed of boat upstream = x - y and speed of boat downstream = x + y

    According to the given condition

    x + y = 5(x - y) ∴ x + y = 5x - 5y

    ∴ 6y = 4x ∴ y =\(\frac{4x}{6}=\frac{2x}{3}\)

    ∴ Speed of current = 67% of speed of boat in still water.

A tank has a leak at the bottom that can empty the entire tank in 6 hours. The tank is entirely full and a tap that fills 250 litres per hour is turned on due to which tank gets emptied in 9 hours. Find capacity of tank in litres.

  • Solution

    Tank is emptied in 6 hrs, But since the inlet tab is on, tank is now emptied in 9 hrs. It means that 3 extra hours are required to drain water poured by inlet tab in 9 hours. In other words in 3 hours leakage drains 250 lit × 9 Hrs = 2250 lit

    ∴ In 6 hours leakage will drain 4500 lit. Since tank is emptied in 6 hrs capacity of the tank must be 4500 lit.

George travels from point A to point B, which are 120 miles apart at 22 miles per hour, starting at 10 a.m. in the morning. His friend Jimmy leaves B for A at 12 noon at 16 miles per hour. At what time do they meet?

  • Solution

    George travels alone from 10 a.m. to 12 noon.

    He covers 22 × 2 hrs. = 44 miles

    Distance covered by George and Jimm together after 12 noon would be (120 - 44) 76 miles.

    Since they are moving in opposite direction, their combined speed would be 22 + 16 = 38 miles per hour.

    ∴They will cover 76 miles in 2 hours and meet each other at 2 p.m.

Two taps A and B can individually fill a tank in 4 hours and 5 hours respectively. Both taps are opened and after 1 hour, tap A is closed. How much more time does it take to fill the tank?

  • Solution

    In one hour tap A fills (1⁄4)th and tap B fills (1⁄5)th of the tank.

    ∴ In one hour, they together fill 1⁄4 + 1⁄5 = \(\frac{9}{20}\) of the tank.

    After one hour, 1 - \(\frac{9}{20}\) = \(\frac{11}{20}\) of the tank is to be filled by pipe B alone.

    ∴ Time, required to fill that portion = \(\frac{11}{20}/\frac{1}{5}\)

    = \(\frac{11}{4}\) = 2 hours 45 minutes.

A certain cash register contains only coins having a face value of 10 cents (dimes) and 25 cents (quarters). If the register contains exactly 450 coins and there are 8 times as many dimes as quarters in register, what is the total face value of register?

  • Solution

    Let there be x quarters in the register. :. There are 8x dimes. Total number of coins = 9x = 450

    ∴ x = 50 ∴ The register has 50 quarters and 400 dimes value of 50 quarters = $12.5 and that 400 dimes = $40

    ∴ Total face value of register = 40 + 12.5 = $52.5

q=\(\frac{2m+ 3n}{\frac{3mn}{p^{2}}},m=\frac{3p}{4},n=\frac{4p}{3}\) Find q in terms of p

  • Solution

    q = \(\frac{2\times \left ( \frac{3p}{4} \right )+3\times \left ( \frac{4p}{3} \right )}{\frac{3\times \frac{3p}{4}\times \frac{4p}{3}}{p^{2}}}=\frac{\frac{3p}{2}+4p}{\frac{3p^{2}}{p^{2}}}\)

    =\(\frac{\frac{11p}{2}}{3}=\frac{11p}{6}\)

Rectangle region ABCO shown above is partitioned into 14 identical small rectangles, each of which has width x. What is the perimeter of ABCO in terms of x?

  • Solution

    AD = PQ = 7x

    Also AP = 7x As area of all 14 rectangle is the same i.e. each rectangle has length 7x and width x

    ∴ AB = 14x, AD = 7x

    ∴ Perimeter = 2(14x + 7x) = 42x

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