BANK & INSURANCE (QUANTITY COMPARISION)

Total Questions: 100

21. In the question below, two statements are given solving which we get two quantities (I) and (II). You have to determine the correct relation between the two quantities and answer accordingly.

Quantity I: Sum of five even numbers is 166. The difference between largest and smallest number is 66. Second smallest and second largest numbers are in the ratio 1:3 respectively. Middle number is 24 which is 33(1/3)% of highest number. Find the second largest number.

Quantity II: A boy covers total 960 km distance by using three vehicles i.e., car, bus and train in 21.2 hours. Boy covers 360 km by train in 7.2 hours and speed of car is 80% of the speed of train. Time taken by boy while travelling with car is 9 hours. What is the speed of bus in km/hr?

Correct Answer: (c) Quantity I = Quantity II or relation can't be established
Solution:

From quantity I:
Let the five numbers are a (smallest), b (second smallest), c (middle), d (second largest) and e (largest). Sum of five even numbers is 166. Then,
a + b + c + d + e = 166

The difference between largest and smallest number is 66, i.e.,
e – a = 66

Middle number is 24 which is 33(1/3)% of highest number, i.e.,
c = 24 = 33(1/3)% of e
e = 72

a = 72 – 66 = 6

Second smallest and second largest numbers are in the ratio 1:3 respectively, i.e.,
b(d) = 1 : 3
d = 3b

Then, 6 + b + 24 + 3b + 72 = 166
b = 16

Then, d = 3 × 16 = 48

Therefore, quantity I = 48

From quantity II:
Total time taken = 21.2 hours
Total distance covered = 960 km
Boy covers 360 km by train in 7.2 hours and speed of car is 80% of the speed of train.

Speed of train = 360/7.2 = 50 km/hr
Speed of car = 80% of 50 = 40 km/hr

Time taken by boy while travelling with car is 9 hours. Then,
Distance covered by car = 40 × 9 = 360 km

Then, distance covered by bus = 960 – 360 – 360
= 240 km

Time taken on travelling with bus = 21.2 – 7.2 – 9
= 5 hours

Then, speed of bus = 240/5 = 48 km/hr

Therefore, quantity II = 48

Hence, Quantity I = Quantity II

22. In the question below, two statements are given solving which we get two quantities (I) and (II). You have to determine the correct relation between the two quantities and answer accordingly.

Quantity I: A person invested Rs.1500 in scheme A at 11.25% per annum rate of simple interest for 4 years and he invested Rs.1200 in scheme B at 25% per annum rate of interest for ‘T’ years. If the interests received from both the schemes are equal, then what is the value of ‘10T’?

Quantity II: Average of ages of Rahul and Soni after 5 years will be 24 years and difference between their ages is 2 years. What is the present age of Rahul?

Correct Answer: (c) Quantity I ≥ Quantity II
Solution:

Quantity I:
Interest received from scheme A = (1500 × 4 × 11.25)/100 = Rs.675

Since the nature of interest in scheme B is not given so, there are two possible cases.

Case 1: When rate of interest is simple.
(1200 × 25 × T)/100 = 675
T = (675)/(12 × 25)
T = 2.25 years

Case 2: When rate of interest is compound.
1200 × [(1.25)ᵀ – 1] = 675
[(1.25)ᵀ – 1] = 675/1200 = 0.5625
(1.25)ᵀ = 0.5625 + 1 = 1.5625
(1.25)ᵀ = (1.25)²
T = 2 years

Value of 10T = 22.5 or 20

Quantity II:
Let present ages of Rahul and Soni be ‘R’ and ‘S’ respectively.

According to question-
[(R + 5) + (S + 5)]/2 = 24
(R + S) + 10 = 48
(R + S) = 38 … (1)

Difference between their ages = R ~ S = 2

Since it is not given that who is elder between Rahul and Soni,

Case 1: When Rahul is elder to Soni:
R – S = 2 … (2)

From (1) and (2)
R = (38 + 2)/2 = 20 years

Case 2: When Soni is elder to Rahul:
S – R = 2 … (2)

From (1) and (2)
R = (38 – 2)/2 = 18 years

Present age of Rahul is either 20 years or 18 years.

Quantity I ≥ Quantity II

23. In the question below, two statements are given solving which we get two quantities (I) and (II). You have to determine the correct relation between the two quantities and answer accordingly.

Quantity I: A shopkeeper sold an article at a discount of 10% on marked price, yet earned a profit of 20% on cost price. Marked price of the article is what percent more than its cost price?

Quantity II: Respective ratio of the incomes of Sujit and Manoj is 5:4 respectively. Ratio of their expenditures is 9:8 respectively. Sujit saves Rs.20000 and Manoj saves Rs.12000. Income of Ragini is Rs.13000 more than the income of Sujit. Income of Ragini is what percent more than the income of Manoj?

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

Quantity I:
We know that
Marked Price (MP) × (100 - %D) = Cost Price (CP) × (100 + %P)

=> MP × (100 - 10) = CP × (100 + 20)
=> MP × 90 = CP × 120
=> MP/CP = 120/90
=> MP/CP = 4/3

Required percentage = (4 – 3)/3 × 100
= (1/3) × 100
= 33.33%

Quantity II:
Let, incomes of Sujit and Manoj be Rs.5k and Rs.4k respectively

(5k - 20000)/(4k - 12000) = 9/8

=> 40k – 160000 = 36k – 108000
=> 40k – 36k = 160000 – 108000
=> 4k = 52000
=> k = 52000/4
=> k = 13000

Income of Sujit = 5k = 5 × 13000 = Rs.65000
Income of Manoj = 4k = 4 × 13000 = Rs.52000
Income of Ragini = 65000 + 13000 = Rs.78000

Required percentage = [(78000 - 52000)/52000] × 100
= (26000/52000) × 100 = 50%

Hence, Quantity I < Quantity II

24. In the question below, two statements are given solving which we get two quantities (I) and (II). You have to determine the correct relation between the two quantities and answer accordingly.

Jaya and Viren together can complete the whole work in 30 days. Rina is 150% more efficient than Viren, and Viren’s efficiency is 60% of the efficiency of Jaya.

Quantity I: If Viren is 40% more efficient than Puja and efficiency of Viren is 70% of the efficiency of Aradhya, then Puja and Aradhya together can complete the work in how many days?

Quantity II: In how many days Rina alone can complete whole work?

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

The ratio of efficiency of Rina, Viren and Jaya
= 250/100 × 60: 60: 100 = 15: 6: 10

Hence, ratio of time taken by Rina, Viren and Jaya to complete the work = 1/15: 1/6: 1/10 = 2: 5: 3

Therefore, Rina, Viren and Jaya alone can complete the work in 2k days, 5k days, and 3k days, respectively.

According to the question,
1/(5k) + 1/(3k) = 1/30
=> k = 16

Quantity I:
Ratio of efficiency of Viren and Puja = 140:100
= 7:5

Thus, Puja alone can complete the work in = 7/5 × 80 = 112 days

Ratio of efficiency of Viren and Aradhya = 70:100
= 7:10

Thus, Aradhya alone can complete the work in
= 7/10 × 5 × 16 = 56 days

Hence, Aradhya and Puja together can complete the work in = 1/(1/112 + 1/56) = 112/3 = 37.33 days

Quantity II:
Hence, Rina alone can complete the work in = 2 × 16 = 32 days

Hence, Quantity I > Quantity II

25. In the following question, two statements numbered I and II are given. On solving them, we get quantities I and II, respectively. Compare the numeric value of both the quantities and choose the correct option.

Quantity I: An item was marked up by Rs 45 and sold at a discount of 15%. If the profit earned in selling the item was Rs 18, what is the selling price of the item?

Quantity II: The percentage profit earned in selling an item is 10%. It was marked up by 20% and a discount of Rs 15 was provided while selling the item. What is the cost price of the item?

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

Quantity I:
Let the C.P of the item be Rs ‘x’
M.P = Rs (x + 45)
S.P = (x + 45)(85/100) = Rs 17(x + 45)/20

So, 17(x + 45)/20 – x = 18
=> 17x + 765 – 20x = 360
=> 3x = 405
=> x = 135

S.P = (17/20)(135 + 45) = Rs 153

Quantity II:
Let the cost price of the item be Rs ‘y’
S.P = (110/100)y = Rs 11y/10
M.P = (120/100)y = 6y/5
S.P = 6y/5 – 15

So, 6y/5 – 15 = 11y/10
=> 12y – 150 = 11y
=> y = 150

C.P = Rs 150

Hence, Quantity I > Quantity II

26. In the following question, two statements numbered I and II are given. On solving them, we get quantities I and II, respectively. Compare the numeric value of both the quantities and choose the correct option.

Quantity I: The length of a rectangle is 5 cm greater than its breadth. If its area is 300 cm², what is its perimeter?

Quantity II: The area of a semicircle is 616 cm². What is its perimeter? [Use π = 22/7]

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

Quantity I:
Let the breadth of the rectangle = ‘x’ cm

Length = (x + 5) cm

So, x(x + 5) = 300
=> x² + 5x – 300 = 0
=> (x + 20)(x – 15) = 0
=> x = 15

So, perimeter = 2(x + x + 5) = 70 cm

Quantity II:
Let the radius of the semicircle = ‘y’ cm

So, (22/7 × y²)/2 = 616
=> y² = 196 × 2
=> y = 19.796

So, perimeter = 2y + (22/7)xy
= 2×19.796 + (22/7)×19.796 = 102.11 cm

Hence, Quantity I < Quantity II

27. In the following question, two statements numbered I and II are given. On solving them, we get quantities I and II, respectively. Compare the numeric value of both the quantities and choose the correct option.

Quantity I: x² - 28x - 60 = 0

Quantity II: A boat covers 70 km downstream in 2 hours and 100 km upstream in 4 hours. Find the speed of boat in still water.

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

Quantity I:
x² – 28x – 60 = 0
x² – 30x + 2x – 60 = 0
x(x – 30) + 2(x – 30) = 0
(x – 30)(x + 2) = 0
x = 30, –2

Quantity II:
The downstream distance covered by boat = 70 km
Time taken by boat in downstream direction = 2 hours

The downstream speed of boat = 70/2 = 35 km/h

The upstream distance covered by boat = 100 km
Time taken by boat in upstream direction = 4 hours

The upstream speed of boat = 100/4 = 25 km/h

The speed of boat in still water = (35 + 25)/2
= 30 km/h

Hence, Quantity I ≤ Quantity II

28. In the following question, read the given statement and compare the Quantity I and Quantity II on its basis. (Only quantity is to be considered)

A can complete a piece of work in 48 days. B and C together can complete the same work in 30 days. If A and C work on alternate days, 60% of the work gets completed in 36 days.

Quantity I: Number of days required for C alone to complete 40% of the work.

Quantity II: Number of days required for B alone to complete two-third of the work.

Correct Answer: (a) Quantity I = Quantity II or relation can't be determined
Solution:

Let the total work be 240 units (LCM of 48 and 30)
Efficiency of A = 240/48 = 5 units/day

Combined efficiency of B and C = 240/30
= 8 units/day

Let the efficiency of C be ‘t’ units/day and B be (8 – t) units/day

Work done by both A and C in 2 days = (5 + t) units/day

Hence, work done by both A and C in 36 days = 18 × (5 + t) units/day

ATQ, 18 × (5 + t) = 24 × 6
=> 90 + 18t = 144
=> 18t = 54
=> t = 3

Hence, efficiency of C = 3 units/day and efficiency of B = (8 – 3) = 5 units/day

Quantity I:
40% of the work = (2/5) × 240 = 96 units
Time required by C alone to complete 96 units of work = 96/3 = 32 days

Quantity II:
Two-third of the total work = (2/3) × 240 = 160 units
Time required by B alone to complete two-third of the work = 160/5 = 32 days

Hence, Quantity I = Quantity II

29. In the following question, read the given statement and compare the Quantity I and Quantity II on its basis. (Only quantity is to be considered)

Quantity I: Capacity of a tank is 36 L and pipe A can fill the tank in 12 minutes. If flowing rate of water of pipe B is 1 L per minute less than that of pipe A, then in what time (in min) both the pipes together can fill the tank?

Quantity II: A car is running 200 m ahead of a bike and ratio of their speeds is 3:5 respectively. If the bike crosses the car in 40 seconds, then find the speed (in m/s) of the car.

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

Quantity I:
Flowing rate of water of pipe A = 36/12 = 3 L per minute
So, flowing rate of water of pipe B = 3 - 1 = 2 L per minute
Quantity of water filled by pipes A and B together in 1 minute = 3 + 2 = 5 L
So, required time = 36/5 = 7.2 minutes

Quantity II:
Let the speed of car and bike are ‘3x’ m/s and ‘5x’ m/s respectively.
So,
5x - 3x = 200/40
x = 2.5
So, speed of the car = 3 × 2.5 = 7.5 m/s

Hence, Quantity I < Quantity II

30. In the following question, find the value of two quantities given in the question and find the relation between them.

Quantity I: Bakul and Rahul can complete a piece of work in 40 days and k days respectively. With the help of Manoj, Rahul can complete the work in 20 days. Manoj and Bakul together can complete the work in 25 days. Find the value of (70% of 2k + 10).

Quantity II: n% of 600 = 4800 ÷ 40 + 30. Find the value of 2n.

Correct Answer: (e) Quantity I = Quantity II or no relation can be established
Solution:

Quantity I:
The work done by Bakul in 40 days
Let the work done by Rahul in ‘k’ days
The work done by Manoj and Rahul in 20 days
The work done by Manoj and Bakul in 25 days

(1/Manoj) + (1/Rahul) = 1/20 …… (i)
(1/Manoj) + (1/Bakul) = 1/25

=> 1/Manoj + 1/40 = 1/25
=> 1/Manoj = (1/25) - (1/40)
=> 1/Manoj = (8 - 5)/200
=> 1/Manoj = 3/200

Work done by Manoj in (200/3) days

From (i)
(3/200) + 1/k = 1/20
=> 1/k = (1/20) - (3/200)
=> 1/k = (10 - 3)/200
=> 1/k = 7/200
=> k = (200/7) days

Now,
70% of 2k + 10 = (70/100) × 2 × 200/7 + 10
= 40 + 10 = 50

Quantity II:
n% of 600 = 4800 ÷ 40 + 30
=> n/100 × 600 = 120 + 30
=> 6n = 150
=> n = 150/6
=> n = 25

Now, 2n = 2 × 25 = 50

Hence, Quantity I = Quantity II