GRE
Practice Test 35
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A metal container weighs 100 kgs. when filled with chemical. If the container is half filled, it weighs 60 kgs.
Weight of the empty container 20 kgs.
  • Solution

 

The number of time digit 9 appears from 1 to 100 19
  • Solution

    9 appears 10 times as the first digit from 90 to 99

    9 appears 10 times as the second digit from 09, 19,29 up to 99 ∴9 appears 20 times.

At 5 o’clock, the clock ticks 5 times. The time elapsed between 1st and last tick was 16 seconds.
Time in seconds elapsed between the first and last ticks at 12 o’clock 48
  • Solution

    At 5 o'clock, clock ticks 5 times and between first and last tick, there are 4 equal time interval (between 2 successive ticks).

    ∴ One time interval has a duration of \(\frac{16}{4}\)= 4 seconds.

    At 12 o'clock, there are 12 ticks and 11 equal time intervals having a total duration of 11 × 4 = 44 seconds.

Five friends divided some candies equally. After each of them had 3 candies, the total number of candies remaining was equal to (4⁄7)th of the total number of candies before division.
Total number of candies before division 30
  • Solution

    The year lies between 1900 to 1999 (i.e. 20th century).

    The number 1936 is the only perfect square of an integer (i.e. 44)

    ∴ x = 44 years ∴ x2 = 1936

    ∴ Col. A = 1936 - 44 = 1892

A man rides a bicycle for 2rd⁄2 of the distance to be covered. He walks the rest of the distance as the cycle got punctured. He realises that he took twice as much time to walk the remaining distance as was required to cover the first 2⁄3 distance on bicycle.
\(\frac{Cycling\, speed}{Walking\, Speed}\) 2
  • Solution

    Let total distance to be travelled be 3 km.

    ∴ The man cycles down 2 km in two hours and walks the remaining 1 km in four hours.

    ∴ His cycling speed is 1 kmph. and walking speed = 4 kmph.

    ∴ Col.A = 4⁄1 = 4

Danny asks his father, “What is your father’s age?” The father replies, “His age was a square root of the year in the 20th century.” In that year.
birth year of danny’s grand father 1936
  • Solution

    The year lies between 1900 to 1999 (i.e. 20th century).

    The number 1936 is the only perfect square of an integer (i.e. 44)

    ∴ x = 44 years ∴ x2 = 1936

    ∴ Col. A = 1936 - 44 = 1892

ABCD is a square with side 4, P and Q are midpoints of sides. AQ = QR

Length of the semicircular arc ARD π
  • Solution

    Q is midpoint of AD. ∴ AQ = 2

    ∴ radius of semicircle is also 2.

    ∴ Length of semicircle arch = \(\frac{2\pi r}{2}=\frac{2\pi 2}{2}\)= 2π

    Col. A > Col. B

The figure shows a piece of paper in the form of a parallelogram ABCD, having area 80 sq. ems. The paper is cut along QR and QP such that the cutting edges are parallel to the sides of the parallelogram.

\(\frac{QR}{AD}=\frac{QP}{CD}=\frac{3}{4}\)
Area of shaded region cm2 45
  • Solution

    Since the new sides of parallelogram are 3⁄4 of the original sides, are of unshaded region will be \(\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}\) the original area.

    ∴ \(\frac{9}{16}\) × 80 =45 cm2

    ∴ Area of shaded region = 80 - 45 = 35 cm2

    ∴ Col. A = 35, Col. B = 45

K : {4, 7, 11, 18, 23}

L: {x + 4, x + 7, x + 11, x + 18, x + 23}

Positive difference between the mean and median values of elements in set K Positive difference between the mean and median values of elements in set L
  • Solution

    Col. A

    Let the mean be A, median = 11

    ∴ Col. A = |1 - A|

    Col. B:

    The mean is A + x, median = 11 + x

    ∴ Col. B = |11 + x - (A + x)|

    |11 + x - A - x| = |11 - A|

In a class, there are 79 students who like ice skating, roller skating or both. 61 students like ice skating and 35 like roller skating.
The number of students who like one game but not both 58
  • Solution

    n(A ∪ B) = n(A)+ n(B)- n(A ∩ B)

    ∴ 79 = 61+ 35 -n(A ∩ B)

    ∴ n(A ∩ B) = 96 - 79 = 17

    ∴ 61 - 17 = 44 like ice skating only.

    Also 35 - 17 = 18 like roller skating only.

    ∴ Col. A = 44 + 18 = 62, Col. B = 58

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