GRE
Practice Test 19
Previous Solution Next
By selling a pen for $24 to John, Paul gains as much % as the number of dollars he paid for it.
Gain % of paul 16
  • Solution

    Let C. P. of pen be $x ∴ Profit % = x%

    ∴ Profit = $(24 - x)

    Profit % = \(\frac{Profit}{C.P.}\times 100\) x 100 ∴ \(\frac{24-x}{x}\times 100=x\)

    ∴ 2400 - 100 x = x2

    ∴ x2 + 100x - 2400 = 0 ∴ (x + 120)(x - 20) = 0

    ∴ x = -120 or x = 20

    x cannot be negative. ∴ x = 20

 

Maximum area of a triangle with perimeter 12 cm. Maximum area of a quadrilateral with perimeter 10 cm.
  • Solution

    Col. A-

    A triangle has maximum area when it is an equilateral triangle.∴ Side = 4 cm.

    Area = √3⁄4 × (4)2 = 4√3 ≈ 6.8 sq. cm.

    Col. B-

    A quadrilateral has maximum area when it is a square.

    ∴ Side = 2.5 em. Area = 6.25 cm2

All page numbers of a book starting from page 1 are added. One of the page numbers has been added twice and thus total obtained was 1000.
Pa e number that has been added twice 15
  • Solution

    Addition of page numbers

    = addition of first n natural numbers = \(\frac{n(n + 1)}{2}\)

    Let × be the page number i.e. added twice.

    ∴ \(\frac{n(n + 1)}{2}\) + x = 1000 ∴ \(\frac{n(n + 1)}{2}\)< 1000

    ∴ n = 44 ⇒ \(\frac{n(n + 1)}{2}\) = 990 ∴ x = 10

PT = TU = 1⁄3PQ

\(\frac{Area\, of \Delta TUR}{Area\, of \Delta PRQ}\) \(\frac{Area\, of \Delta TRQ}{Area\, of □ PQRS}\)
  • Solution

    Let us assume that PT = TU = UQ = 1

    Let us assume that SP = 2

    Col. A = Area of ΔTUR = 1⁄2 × 1 × 2 = 1

    Area of ΔPRQ = 1⁄2 × 3 × 2 = 3

    Col. B = Area of ΔTRQ= 1⁄2 × 2 × 2 = 2

    Area of rectangle = 3 × 2 = 6

    ∴ Col. B = 2⁄6 = 1⁄3

P takes 4 days to do a work.
Q takes twice as much time as P.
R takes twice as much time as Q.
S takes twice as much time as R.
Work done by fastest and slowest workers together in one day Work done by 2 other workers together in one day.
  • Solution

    P takes 4 days, Q takes 8 days, R takes 16 days and Stakes 32 days to do a work.

    P does 1⁄4 of work in 1 day.

    ∴ Q, R, S do 1⁄8,\(\frac{1}{16}\),\(\frac{1}{32}\) of work in 1 day respectively.

    ∴ Col. A = \(\frac{1}{4}+\frac{1}{32}=\frac{9}{32}\)

    Col. B =\(\frac{1}{8}+\frac{1}{16}=\frac{6}{32}\)

Six pipes are fitted to a water tank. Some of these are filling pipes and others are emptying pipes. Each filling pipe can fill the tank in 9 hours and each emptying pipe can empty the tank in 6 hours. On opening all pipes, an empty tank is filled in 9 hours.
Number of filling pipes 4
  • Solution

    Each filling pipe can fill (1⁄9)th of the tank in one hour.

    Each emptying pipe can (1⁄6)th of the tank in one hour.

    This means that the emptying pipe does more amount of work than the filling pipe.

    ∴ To fill the tank there must be more number of filling pipes and less number of emptying pipes.

    With 4 filling pipes and 2 emptying pipes, we (4 × 1⁄9) - (2 × 1⁄6) = 4⁄9 - 3⁄9 = 1⁄9

    ∴ The tank can filled with 4 filling and 2 emptying pipes in 9 hours.

In a polygon, two of the external angles are 55° and 35° and others are equal, each equal to 27°.
sum of internal angles of the polygon 1900°
  • Solution

    Sum of the measures of all exterior angles of any polygon = 3600

    Number of sides =\(\left [ \frac{360 - (55 + 35)}{27} \right ]+2\)

    =\(\frac{270}{27}+2\)=12

    ∴ Number of sides = 12

    ∴ Sum of all internal angles = (2n - 4)90°

    = 20 × 90° = 1800°

Circumference of a circle = Area of circle
Side of largest square that can be inscribed in this circle 3√2
  • Solution

    2πr = πr2 ∴ r = 2

    Diagonals of square = diameter of the circle = 4.

    ∴ Side = 2√2

A hostel has provisions for 25 cadets for 35 days. After 5 days, a fresh batch of 5 cadets were admitted to hostel. Again after 10 days, a batch of 5 cadets left the hostel.
Number of days the remaining provisions will last 22
  • Solution

    Initially, the hostel has provisions for 25 × 35 man days.

    After 5 days the available stock was 25 × 30 = 750 man days.

    Number of cadets were 25 + 5 = 30

    In next 10 days, 30 cadets consumed 30 × 10 = 300 man days' worth of stock.

    The balance available will be 750 - 300 = 450 man days.

    Now again 5 cadets left. Hence we have only 25 cadets.

    They will consume the stock in \(\frac{450}{25}\) = 18 days.

ab > 1
4⁄a + 7⁄b \(\frac{7a+4b}{a+b}\)
  • Solution

    Col. A \(\frac{4b+7a}{ab}\)                         

    Col. B \(\frac{4b+7a}{a+b}\)

    Cancelling 4b + 7a from both sides, we have

                                         
    Col. A\(\frac{1}{ab}\)

    Col. B \(\frac{1}{a+b}\)

    Since ab > 1 either a and b both are positive or both are negative.

    Negative values for both gives Col. A > Col. B

    For positive values, put a = b = 2

    ∴ Col. A = Col. B

    Put a = 4, b = 2

    ∴ Col. A < Col. B

Attempted

0

Correct

0

Wrong

0

Complete
Quantitative Comparison
Attempted
Wrong
Correct