GRE
Practice Test 32
Previous Solution Next
In the aptitude test conducted by a company, 35% failed in Maths and 42% in Verbal. 15% failed in both subjects. 5000 candidates appeared for the exam.
Number of candidates who passed in either section but not in both 2450
  • Solution

    \(\frac{35}{100}\) × 5000 = 1750 failed in Maths

    \(\frac{42}{100}\) × 5000 = 2100 failed in verbal

    \(\frac{15}{100}\) × 5000 = 750 failed in both

    ∴ 1750 -750 = 1000 failed only in Maths

    Similarly, 2100 - 750 = 1350 failed only in Verbal

    Number of candidates who failed only in Maths = Number of candidates who passed only in Verbal.

    ∴ Col. A = 1000 + 1350 = 2350

Speed of boat in still water = 10 kmph. The boat can travel 26 km downstream and 14 km upstream in the same time
speed of the stream ill kmph 3.5
  • Solution

    Let speed of stream be y kmph.

    ∴ Upstream speed = 10 - Y

    Downstream speed = 10 + Y

    Time = \(\frac{Distance}{Speed}\)

    t upstream = t downstream

    ∴ \(\frac{14}{10-y}=\frac{26}{10+y}\)

    ∴ 7(10 + y) = 13(10 - y)

    ∴ 70 + 7y = 130 - 13y

    ∴ 20y = 60 ∴ y = 3 km/hr.

a, b, c are positive integers a +b + c = 24, a = c
c 12
  • Solution

    a + b + C = 24 ⇒ 2c + b = 24

    ∴ b = 24 - 2c

    If c = 12, then b = 0, which is not true as b is positive

    ∴ 'c' can take a maximum value which is 11.

Joe can reach a certain place in 30 hours. If she reduces her speed by \(\left (\frac{1}{15} \right )^{th}\), she goes 10 km. less in that time.
speed of joe in kmph 4.5
  • Solution

    Time × Speed = Distance

    Speed = x km per hour.

    ∴ 30x - 30\(\left (\frac{14x}{15} \right )\) = 10

    ∴ 2x = 10,x = 5 = Col.A

Mark lent out a sum of money at a certain rate of simple interest. He found that the sum amounts to $6200 after 3 years and to $7800 after a further period of 7 years.
Sum lent by Mark in $ $4000
  • Solution

    Amount after 3 years and 7 years is $6200 and $7800 respectively

    Amount = P + S.I.

    Let the sum be P

    ∴ P + 8.I3 years = $6200

    ∴ S.I.2 years=$7800

    ∴ S.I·4 years = $7800- $6200 = $1600

    ∴ S·I.3 years = $1200

    ∴ P + $1200 = $6200 ∴ P = $5000

ABC Developers employed 30 men to do a piece of work in 38 days. After 25 days, the company employed 5 more men and the work was finished one day earlier.
The number of days, by which the work would have delayed if these additional number of men had not been employed 2
  • Solution

    In the first 25 days, 30 men do 30 × 25 = 750 units of work. 5 more men were employed and work was completed 1 day earlier.

    ∴ After 25 days, (30 + 5) men worked for (13 - 1) days.

    ∴ They produced 35 × 12 = 420 units of work.

    ∴ Total work content = 750 + 420 = 1170 units of work.

    In the absence of additional number of men, work would be completed in \(\frac{1170}{30}\) = 39 days.

    ∴ The work would take 1 day more

    Col. A = 1, Col. B = 2

m > o
Value of n if m2 – 2mn + n2 = 0 Value of p if m2 + 2mp + p2 = 0
  • Solution

    Col. A:

    m2 - 2m + n2 =0

    ∴ m - n = O ∴ m = n

    ∴ (m-n)2 = 0

    ∴ n > 0

    Col. B:

    m2 + 2mp + p2 = 0

    ∴ (m + p)2 =O ∴ m + p = O

    ∴ m = -p

    But m > 0 ∴ - p > 0 ∴ p < 0 ∴ Col. A is positive, Col. B is negative

a and b are distinct primes. X and y are positive integers.
LCM of ax & by ax × by
  • Solution

    a and b are prime numbers (i.e. they have only number 1 in common)

    ∴ Higher powers of a and b i.e. ax and by will be co-prime numbers (only factor 1 in common).

    E.g. if 3 and 5 are distinct primes, then 33, 52 are co-prime numbers.

    LCM of co-prime numbers = product of numbers.

    ∴ LCM of ax, by = ax x by

In a mathematical problem, the divisor is ~ of the dividend and 3⁄4 times the remainder which is 7.
Value of the quotient if dividend is divided b remainder 21
  • Solution

    The remainder = 7

    ∴ Divisor = 7 × 3 = 21

    ∴ Dividend = 21 × 4⁄3 = 28

    ∴ Col.A =\(\frac{28}{7}\)= 4, Col.B = 21

A square PQRS is inscribed in a circle with centre M. Area of ΔRMS = 18, ∠RMS = 90°

Area of shaded region 48
  • Solution

    RMS is an isosceles right angled triangle.

    ∴ 18 = \(\frac{(RM)^{2}}{2}\) ∴(RM) = 6

    Radius of circle = 6 ∴ Area = 36π

    ∴ Area of square = 18 × 4 = 72

    ∴ Col. A = 361π -72 = 36(π - 2} = 36 × 1.142

    Col. B = 45 = 36 × 1.25

Attempted

0

Correct

0

Wrong

0

Complete
Quantitative Comparison
Attempted
Wrong
Correct