GRE
Practice Test 36
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5 artists make 5 paintings in 5 days. Therefore 25 artists can make X paintings in 25 days.
x 25
  • Solution

    Work content = 5 artists × 5 days = 25 man days

    Actual work done = 5 paintings.

    ∴ 25 artists × 25 days = 625 man days.

    ∴ We can produce 125 paintings.

    (5 artists can make 5 paintings in 5 days.)

    (∴ 5 artists can make 25 paintings in 25 days.)

    (∴ 25 artists can make 125 paintings in 25 days.)

A dice has to be manufactured in such a way that 1 & 6,2 & 5,3 & 4 are on the opposite faces.
Number of was in which dice can be manufactured 6
  • Solution

    Consider an unbiased dice with 6 faces as shown.

    Let 1 and 6 occupy the right and left faces respectively. Therefore number of ways in which 1 & 6 can be arranged is 2! ways.

    Then the number of ways in which (1 & 6), (2 & 5), (3 & 4) can be arranged = 2! × 2! × 2! = 8 ways.

    ∴ Col. A = 8, Col. B.= 6

100 applicants applied for a job. 10 of them had never taken a course in Sociology or in Politics. 83 of them had taken at least one course in Sociology, 65 of them had taken at least one. course in Politics.
The number of applicants who had taken at least one course in Sociology or Politics but not in both 36
  • Solution

    n(A ∪ B) = n(A)+ n(B)- n(A ∩ B)

    ∴ n(A ∪ B) = 100 -10 = 90

    90 students have taken Socialogy or Politics or both.

    ∴ n(A) = 83, n(B)= 65

    ∴ 90 = 83 + 65 - n(A ∩ B)

    ∴ n(A ∩ B) = 58 students have taken both subjects.

    ∴ Students who have taken at least one course but not both.

    = 90 - 58 = 32

    ∴ Col. A = 32, Col. B = 36

A rich merchant had collected some gold coins. If he divides the coin into 2 unequal numbers, then 59 times the difference between the two numbers equals the difference between the squares of the two numbers.
The total number of gold coins with the merchant 47
  • Solution

    Let the merchant divide the number of coins into 2 unequal numbers x and y respectively.

    ∴ 59(x -y) = x2 -v2

    ∴ 59 = \(\frac{x^{2}+y^{2}}{x-y}\) = x + y

    ∴ Total number of gold coins = x + Y= 59

A car has a leaking oil tank. At the start of the journey, there are 12 gallons of fuel in the tank. Speed of car is 50 miles per hour.· It travels for 4 hours and tank gets empty. The car travel 25 miles by consuming 1 gallon of fuel.
Quantity of oil in gallons that got leaked 4
  • Solution

    The car travelled for 4 hours at 50 miles per hour.

    ∴ Total distance travelled by car = 50 × 4 = 200 miles.

    It travels 25 miles by consuming 1 gallon of fuel.

    ∴ It consumes 8 gallons travelling 200 miles.

    ∴ 12 - 8 = 4 gallons leaked through the tank.

John is a dealer of consumables. On the Christmas Eve, he decides to give 1 article free on buying 9 articles. He still gains 35% on the total cost.
% b which Selling Price is above the Cost Price 55
  • Solution

    Let Cost Price of each article be $10.

    The Cost Price of 10 articles = 10 × 10 = $100.

    ∴ Selling Price at a gain of 35% = $135.

    But 1 article is free on every 10 articles.

    $135 is the Selling Price of 10 articles.

    ∴ \(\frac{135}{9}\) = $15

    ∴ Col. A = $15 = 50% above 10.

    ∴ Col. A = 50, Col. B = 55

The product of 5 different temperatures is 12. All of them are integers.
Difference between the highest and lowest temperature 4
  • Solution

    The only possible values are -2, -1, 1,2,3.

    Col. A = 3 - (-2) = 5

A ship went on a voyage. After the ship had travelled 180 miles in 9 hours, a plane started with a speed 10 times that of the ship from the same point.
Distance in miles from the starting point when they meet 240
  • Solution

    Ship travelled 180 miles in 9 hours, hence the speed of ship is 20 miles per hour. ∴ Speed of plane will be 200 miles. Plane will travel 200 miles in 1 hour. At the same time ship will travel additional 20 miles, and complete 200 miles and meet the plane.

    Col. A = 180 + 20 = 200 miles.

The diagram represents a sundial, formed by attaching the shortest side of a right triangle to a radius of the circular base. Length of 2 longest sides of triangle are 13 cm and 12 cm.

Area of the sundial minus area of the triangle in cm2 50
  • Solution

    The Pythagoras triplet is 5, 12, 13.

    ∴ Shortest side of the triangle = Radius of sundial = 5 cm.

    ∴ Col.A = \((5)^{2}\pi -\frac{12\times 5}{2}\)

    = 25π - 30 ≈ 8.5 - 30 = 48.5

k < 0
\(\frac{7k-5}{7}\) \(\frac{4k-3}{4}\)
  • Solution

    Col.A= \(\frac{7k}{7}-\frac{5}{7}\) = k - 5⁄7 = k- 0.71

    Col.B = k - 3⁄4 = k - 0.75

    Since k is a negative number k - 0.71 is less negative as compared to k - 0.75.

    ∴ k - 0.71 > k - 0.75

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