BANK & INSURANCE (QUANTITY COMPARISION)

Total Questions: 100

51. In the following questions, two statements numbered I and II are given. On solving them, we get quantities I and II respectively. Solve for both the quantities and choose the correct option.

Quantity I: The speed of a boat in upstream is 75% of the speed of the boat in downstream. What is the speed of the boat in still water if it covers 140 km downstream in 7 hours?

Quantity II: Radha and Mitu are traveling through their scooties from college to hostel which is 60 km apart. Traveling at a certain speed Radha takes one hour more than Mitu to reach hostel. If Radha doubles her speed she will take 30 minutes less than Mitu to reach hostel. what speed was Radha driving the scooty from college to hostel?

Correct Answer: (b) Quantity I < Quantity II
Solution:Quantity I:
Let a km/h be the speed of the boat in still water and b km/h be speed of the current.

Let 3x km/h be the speed of the boat upstream and 4x km/h be the speed of the boat downstream.

So, from the question,
140/4x = 7
=> x = 5

So, speed of the boat upstream = (a + b) = 3 × 5 = 15 ...(1)
And speed of the boat downstream = (a - b) = 4 × 5 = 20 ...(2)

So, adding eq 1 and 2 we get
a + b + a - b = 15 + 20
=> 2a = 35
=> a = 17.5 km/h

Quantity II:
Distance = 60 km

Let x kmph be the Radha’s speed and y kmph be the Mitu’s speed.

So, from the question,
(60/x) - (60/y) = 1 ...(1)
And (60/y) - (60/2x) = 30/60 = 1/2 ...(2)

Adding eq 1 and 2 we get
(60/x) - (60/2x) = 1 + 1/2
=> (60/x) - (30/x) = 3/2
=> 30/x = 3/2
=> x = 20 kmph

Hence, Quantity I < Quantity II

52. In the following question, two statements numbered I and II are given. On solving them, we get two quantities quantity I and quantity II respectively. Compare both the quantities and choose the correct answer:

Quantity I: What is the speed of train A (in m/s) which crosses a pole in 8 sec and crosses another train B, which is having length of 280 metres and is moving at a speed of 20 m/s in direction opposite to that of train A, in 10.4 sec?

Quantity II: A bike crosses a stationary train in 5 minutes, while the same train, when moving at a certain speed, crosses a platform of length 500 m in 60 seconds. A car crosses the same stationary train in 7.5 minutes. The difference between speed of bike and car is 2 km/hr. What is the speed of the train (in m/s)?

Correct Answer: (a) Quantity I > Quantity II
Solution:

Quantity I:
Let length of train A be Y metre
Speed of train A = (Y/8) m/sec

Relative speed of train A w.r.t train B = (Y/8 + 20) m/sec

(Y/8 + 20) = (Y + 280)/10.4
(Y + 160) × 10.4 = 8 × (Y + 280)
2.4Y = 2240 - 1664
Y = 576/2.4

Length of train A = 240 metre
So, speed of the train A = 240/8 = 30 m/s

Quantity II:
Let the length of train = a km

Speed of bike = (a/5) × 60 = 12a km/hr
Speed of car = (a/7.5) × 60 = 8a km/hr
Difference between speed of car and bike = 2 km/hr
12a - 8a = 2
a = 0.5

Length of platform = 500 m = 0.5 km
Time taken by train to cross the 500 m long platform = 60 seconds = 1/60 hour

Speed of train = (a + 0.5)/(1/60) = 60a + 30 = 60 × 0.5 + 30 = 60 km/hr = 60 × 5/18 = 16.66 m/s

Hence, Quantity I > Quantity II

53. In the following question, two statements numbered I and II are given. On solving them, we get two quantities quantity I and quantity II respectively. Compare both the quantities and choose the correct answer:

(18/(X+Y)) + (3/(X-Y)) = 5 and (5/(X-Y)) - (9/(X+Y)) = 4

Quantity I: Value of X
Quantity II: Value of Y

Correct Answer: (a) Quantity I > Quantity II
Solution:

Put (1/(X + Y)) = M and (1/(X - Y)) = N

18M + 3N = 5 ...(1)
5N - 9M = 4 ...(2)

By multiplying (2) by 2 and by adding (1) and (2) we get
13N = 13
N = 1 and M = (1/9)

So, X + Y = 9 ...(3)
X - Y = 1 ...(4)

X = 5 and Y = 4

Quantity I > Quantity II

54. In the following question, two statements numbered I and II are given. On solving them, we get two quantities quantity I and quantity II respectively. Compare both the quantities and choose the correct answer:

A bag contains 19 red balls, 13 blue balls and some green balls. If one ball is drawn at random from the bag, then there are 20% chances that the ball is green.

Quantity I: If three balls are drawn at random from the bag one after other without replacement, then find the probability of all three being red balls.

Quantity II: If two balls are drawn at random from the bag one after other with replacement, then find the probability of both being green balls.

Correct Answer: (a) Quantity I > Quantity II
Solution:

Let the number of green balls in the bag = x

Probability of selecting a green ball from the bag = x/(19 + 13 + x) = 20/100 = 1/5

5x = 32 + x
4x = 32
x = 8

Total balls in the bag = 19 + 13 + 8 = 40

Quantity I:
Probability of selecting three red balls from the bag one after other without replacement
= (19/40) × (18/39) × (17/38) = 51/520

Quantity II:
Probability of selecting two green balls from the bag one after other with replacement
= (8/40) × (8/40) = 1/25

Hence, Quantity I > Quantity II

55. In following question, two quantities numbered I and II are given. Calculate values of both quantities and choose the correct option.

Quantity I: P, Q and R started a business in which P and Q initially invested Rs.3000 and Rs.4000 respectively whereas R invested some amount. After 6 months P and Q invested additional Rs.2000 and Rs.1000 respectively. If at the end of 1-year share of P in the profit is Rs.54000 is Rs.16000, then how much amount R invested initially?

Quantity II: If compound interest on certain principal at the rate of 10% per annum compounded annually at the end of 2 years is Rs.1050, then what is the principal?

Correct Answer: (d) Quantity I = Quantity II
Solution:

For Quantity I:
Let R invested Rs.1 (thousand) initially.

Ratio of investments of P, Q and R at the end of 1 year respectively
= ((3 × 2) + 2) : ((4 × 2) + 1) : (2 × 1)
= 8 : 9 : 2

Share of P in total profit = (8/(17 + 2))
= (16000/54000) = (8/27)

17 + 2 = 27
I = 5

Amount invested by R = Rs.5000
Quantity I = Rs.5000

For Quantity II:
Let Principal be P

Amount after 2 years at 10% per annum
= P × (1 + (10/100))²
= P × (1.21)

Interest = 1.21P - P = 0.21P = 1050
P = Rs.5000

So, Quantity I = Quantity II

56. In following question, two quantities numbered I and II are given. Calculate values of both quantities and choose the correct option.

Quantity I: A boy can paint the ceiling of 12 meters long and 15 meters wide room in 1.5 hours. If the height of the room is 10 meters, then how many hours does he take to paint only four walls of the room?

Quantity II: A boy can paint the four walls of 6 meters long, 8 meters wide, and 7.5 meters high room in 23 hours 20 minutes. How many hours does he takes to paint only ceiling of the room?

Correct Answer: (e) Quantity I < Quantity II
Solution:

Quantity I:
The area of the ceiling = 12 × 15 = 180 sq. metres

The efficiency of the boy = 180/1.5 = 120 sq. metres per hour

The area of the four walls of the room = 2(l + b) × h
= 2(12 + 15) × 10 = 540 sq. metres

The time taken by him to paint four walls @ 120 sq. metres per hour
= 540/120 = 4.5 hours = 4 hours 30 minutes

Quantity II:
The area of the four walls of the room = 2(l + b) × h
= 2(6 + 8) × 7.5 = 15 × 14 sq. metres

Time = 23 hours 20 minutes = 70/3 hours

The efficiency of the boy = (15 × 14)/(70/3) = 9 sq. metres per hour

The area of only ceiling = l × b = 6 × 8 = 48 sq. metres

The time taken by him to paint only ceiling @ 9 sq. metres per hour
= 48/9 = 16/3 = 5 hours 20 minutes

Therefore, QI < QII

57. In the following question, two statements numbered I and II are given. On solving them, we get quantities I and II, respectively. Solve for both the quantities and choose the correct option.

Quantity I: [p(pq - r) + q(qr + p)]
Quantity II: [p² + pq - r² - rp]

Given, -2 ≤ r ≤ -1, 4 ≤ q ≤ 5 and 6 ≤ p ≤ 8

Correct Answer: (d) Quantity I > Quantity II
Solution:

Given, -2 ≤ r ≤ -1, 4 ≤ q ≤ 5 and 6 ≤ p ≤ 8

Quantity I:
For minimum value p = 6, q = 4, and r = -2

[p(pq - r) + q(qr + p)]
= 6 × (6 × 4 - (-2)) + 4 × (4 × (-2) + 6)
= 6 × (24 + 2) + 4 × (-8 + 6)
= 156 + (-8) = 148

And for maximum value p = 8, q = 5, and r = -1

[p(pq - r) + q(qr + p)]
= 8 × (8 × 5 - (-1)) + 5 × (5 × (-1) + 8)
= 8 × (40 + 1) + 5 × (-5 + 8)
= 328 + 15 = 343

Thus, 148 ≤ Quantity I ≤ 343

Quantity II:
For minimum value p = 6, q = 4, and r = -2

[p² + pq - r² - rp]
= 36 + 24 - 4 + 12 = 68

And for maximum value p = 8, q = 5, and r = -1

[p² + pq - r² - rp]
= 64 + 40 - 1 + 8 = 111

Hence, 68 ≤ Quantity II ≤ 111

Therefore, Quantity I > Quantity II

58. In the following question, two statements numbered I and II are given. On solving them, we get quantities I and II, respectively. Solve for both the quantities and choose the correct option.

Quantity I: In an exam paper, Virat solves 25 questions out of 45 questions. In the same exam, Sachin solves 35 questions out of 45. Sachin's probability of solving exam paper is what percent of Virat's probability of solving exam paper?

Quantity II: In a restaurant, there are four types of mouth fresheners (A, B, C, and D). The probability of a customer choosing A, B, C and D are 0.15, 0.25, 0.55 and 0.05 respectively. The probability of choosing either B or C is what percentage of the probability of choosing either A or D?

Correct Answer: (d) Quantity I < Quantity II
Solution:

Quantity I:
Virat’s probability of solving paper = 25/45 = 5/9
Sachin’s probability of solving paper = 35/45 = 7/9

Required percentage = 7 × 100/5 = 140%

Quantity II:
[P(A ∩ B) = P(A ∩ C) = P(A ∩ D) = P(B ∩ C) = P(B ∩ D) = P(C ∩ D) = 0]

Probability of choosing either A or D = 0.20
Probability of choosing either B or C = 0.80

Required percentage = 0.80 × 100/0.20 = 400%

Hence, Quantity I < Quantity II

59. In following question, two quantities numbered I and II are given. Calculate values of both quantities and choose the correct option.

Quantity I: If 48 girls leave a group of students then the percent of girls in the entire group becomes 42 percent. What is the original percent of the girls in the group?

Quantity II: What is the original percent of the boys in the group if the percent of boys in group becomes 64% when 56 girls leave the group?

Correct Answer: (e) Quantity I = Quantity II or the relationship cannot be determined
Solution:

Quantity I
We can’t find the original % of the girls in the group based on the information given in the question.

Quantity II:
We can’t find the original % of the boys in the group based on the information given in the question.

60. In following question, two quantities numbered I and II are given. Calculate values of both quantities and choose the correct option

Quantity I: If A² - 12A + 35 = 0 and breadth of a rectangle is equal to the value of A (in cm), then what is the area of rectangle if length of rectangle is double of its breadth? (in cm²)

Quantity II: If area of a circle is 77 cm², then what is the area of the square which is inscribed in the circle? (in cm²)

Correct Answer: (a) Quantity I > Quantity II
Solution:

For Quantity I:
A² - 12A + 35 = 0
A² - 7A - 5A + 35 = 0
A (A - 7) - 5 (A - 7) = 0
(A - 7)(A - 5) = 0
A = 5 or A = 7

If breadth = 5 cm, then length = 10 cm
Area = 50 cm²

If breadth = 7 cm, then length = 14 cm
Area = 98 cm²

Quantity I = 50 or 98

For Quantity II:
Let radius of the circle be R
πR² = 77
(22/7) × R² = 77
R² = 49/2
R = 7/√2 cm

Diameter of the circle = 2 × (7/√2) = 7√2 cm
Diameter of the circle is equal to the diagonal of the square inscribed in it

Diagonal of the square = 7√2 cm
Side of square = (7√2)/(√2) = 7 cm

Area of square = 49 cm²

Quantity II = 49

Hence, Quantity I > Quantity II