BANK & INSURANCE (QUANTITY COMPARISION)

Total Questions: 100

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

Quantity I: Simple interest on Rs.5800 at r% per annum for (r - 2) years is Rs.1200 more than the simple interest on Rs.256 at 15% per annum for 5 years. What is the value of ‘r’?

Quantity II: If 3p(p - 3) = 48 + p, then find the value of ‘p

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

From quantity I:
Simple interest = sum × rate × time/100

Then, Simple interest on Rs.5800 at r% per annum for (r - 2) years = 1200 + simple interest on Rs.256 at 15% per annum for 5 years

5800 × r × (r - 2)/100 = 1200 + 256 × 5 × 15/100
58 × r × (r - 2) = 1392
r² - 2r - 24 = 0
r² - 6r + 4r - 24 = 0
(r - 6)(r + 4) = 0
r = 6, -4

But rate cannot be negative. So, r = 6

From quantity II:
3p(p - 3) = 48 + p
3p² - 9p = 48 + p
3p² - 10p - 48 = 0
3p² - 18p + 8p - 48 = 0
(3p + 8)(p - 6) = 0
p = -8/3, 6

For r = 6, p = -8/3, quantity I > quantity II
For r = 6, p = 6, quantity I = quantity II

Hence, Quantity I ≥ Quantity II

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

Quantity I: The area of a rectangular park is 80 cm² and perimeter of the rectangular park is 36cm. The difference between length and breadth of the park is (in cm)-

Quantity II: y² - 11y + 18 = 0

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

Quantity I:
Area of a rectangular park = 80 cm²
l × b = 80 ...(1)

Perimeter of the rectangular park = 36 cm
2(l + b) = 36 ...(2)

On solving (1) and (2)
L = 10 cm, b = 8 cm
Required difference = 10 - 8 = 2

Quantity II:
y² - 11y + 18 = 0
y² - 9y - 2y + 18 = 0
(y - 9)(y - 2) = 0
y = 2, 9

Quantity I = 2 and Quantity II = 2, Quantity I = Quantity II
Quantity I = 2 and Quantity II = 9, Quantity I < Quantity II

Hence, Quantity I ≤ Quantity II

43. In the following question two statements (Quantity - I and Quantity - II) are given. You have to solve both the statements and give answer

Quantity I: What is the greatest integral value of m + n, where 2 < m < 7 and 3 < n < 8?

Quantity II: A number of two digits is 4 times the sum of its digits. If 18 is added to the number, the digits are interchanged. What is the number?

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

Quantity I:
Maximum value that m have = 6
Maximum value that n have = 7

So, (m + n) would be 13, but m and n can take values in fractions also e.g. 6.5 and 7.5. So the maximum value of (m + n) is 14.

Quantity II:
Let the number is 10x + y

So,
10x + y = 4(x + y)
6x = 3y
2x = y ...(1)

And,
10x + y + 18 = 10y + x

9x + 18 = 9y
x + 2 = y
x = 2 [from equation (1)]
So, y = 4

Therefore, the number is 24.

Hence, Quantity I < Quantity II

44. In the following questions two statements (Quantity - I and Quantity - II) are given. You have to solve both the statements and give answer. If abc = 42, where a, b and c are prime numbers.

Quantity I: Difference between a and b if c has maximum value.

Quantity II: If ‘a’ is not an even prime number, and b < c, then find the maximum value of c - b - 3.

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

Since a, b and c are prime numbers. Then,
abc = 42 = 2 × 3 × 7
where 2, 3 and 7 are values of a, b and c but not necessarily in the same order.

From quantity I:
c has maximum value. So, c = 7

Then, either (a = 3, b = 2) or (a = 2, b = 3)

So, when a = 3 & b = 2,
The difference between a and b = 3 - 2 = 1

Also, when a = 2 & b = 3
The difference between a and b = 2 - 3 = -1

So, quantity I = 1 & -1

From quantity II:
a is not an even prime number. So, values of a is not 2. Then either b = 2 or c = 2.

But b < c
So, b = 2

And either (a = 3, c = 7) or (a = 7, c = 3)

So, when b = 2, c = 7, then c - b - 3 = 7 - 2 - 3 = 2 (maximum value)

And when b = 2, c = 3, then c - b - 3 = 3 - 2 - 3 = -2

So, quantity II = 2

Hence, quantity I < quantity II

45. In the following questions two statements (Quantity - I and Quantity - II) are given. You have to solve both the statements and give answer.

Quantity I: A and B started a business by investing amount in the ratio 3: 8 respectively. After 9 months, B left. At the end of one-year partnership, total profit earned by them is Rs.840. What is the share (in rupees) of B in profit?

Quantity II: If 3z² - 20z - 7 = 0 and z > 0, then what is the value of 80z?

Correct Answer: (e) Quantity I = Quantity II or relationship cannot be established.
Solution:

From quantity I:
Let the amount invested by A and B are Rs.3a and Rs.8a respectively. Then,

Profit ratio, A : B = (3a × 12) : (8a × 9) = 1 : 2

Total profit earned = Rs.840

Therefore, share of B in profit = 840 × 2/(1 + 2) = Rs.560

So, quantity I = 560

From quantity II:
3z² - 20z - 7 = 0
3z² - 21z + z - 7 = 0
(3z + 1)(z - 7) = 0
z = -1/3, 7

But z > 0, so z = 7

And 80z = 80 × 7 = 560

So, quantity II = 560

Hence, Quantity I = Quantity II

46. In the following question, find the value of two quantities given in the question and find the relation between them. Rakesh, Manoj and Suraj entered into a business with investment in the ratio 8:5:4 respectively. After one year, Shyam joined them with investment equal to average of the initial investment of Rakesh and Suraj. After one more year, Suraj doubled his investment. At the end of three years of business they earned a total profit of Rs.134000.

Quantity I: Find the difference between the shares of Rakesh and Shyam in the profit.

Quantity II: Find the share of Manoj in the profit.

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

Let, initial amounts invested by Rakesh, Manoj and Suraj be Rs.8k, Rs.5k and Rs.4k respectively.

Ratio of shares in the profit:
Rakesh : Manoj : Suraj : Shyam = (8k × 3) : (5k × 3) : (4k × 2 + 8k) : (6k × 2)
= 24k : 15k : 16k : 12k
= 24 : 15 : 16 : 12

Quantity I:
Required difference = ((24 - 12)/(24 + 15 + 16 + 12)) × 134000
= (12/67) × 134000 = Rs.24000

Quantity II:
Share of Manoj in the profit = [15/(24 + 15 + 16 + 12)] × 134000
= (15/67) × 134000 = Rs.30000

Hence, Quantity I < Quantity II

47. The question below contains a statement followed by two quantities Quantity I and Quantity II. Read the following information carefully and answer the question accordingly.

Quantity I: Difference between ages of P and Q is 6 years. Three years later, the sum of their ages will be 40 years. How old was P 4 years ago?

Quantity II: Two years later, Madhu’s age will be two-third of Suman’s age at that time. Sum of their present ages is 26 years. What is the present age of Madhu?

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

From quantity I:
P - Q = ±6
(P + 3) + (Q + 3) = 40
P + Q = 34

Then, P = (34 ± 6)/2 = 20 years or 14 years
Age of P 4 years ago = 16 years or 10 years

From quantity II:
Madhu + Suman = 26 => Suman = 26 - Madhu
(Madhu + 2) = (2/3) × (Suman + 2)
Suman = (3/2) × Madhu + 1

Then, 26 - Madhu = (3/2) × Madhu + 1
Madhu = 10 years

Hence, Quantity I ≥ Quantity II

48. The question below contains a statement followed by two quantities Quantity I and Quantity II. Read the following information carefully and answer the question accordingly. The speed of a boat in still water is 15 km/hr. The boat can cover a distance of 126 km in a river and return to its initial position in altogether 20 hours.

Quantity I: Distance covered by the boat (in km) with the river current in 5 hours when the speed of the boat in still water is increased by 20%.

Quantity II: Distance covered by the boat (in km) against the river current in 18 hours when the speed of the boat in still water is decreased by 20%.

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

Speed of the boat in still water = 15 km/hr
Let the speed of the river current be ‘t’ km/hr

ATQ,
126/(15 + t) + 126/(15 - t) = 20

=> 63 [1/(15 + t) + 1/(15 - t)] = 10
=> 189 = (15 + t)(15 - t)
=> 189 = (15 + t)(15 - t)
=> t² = 36
=> t = +6 and -6

Ignoring negative value of river current, we get ‘t’ = 6 km/hr

Quantity I:
Increased speed of the boat in still water = (6/5) × 15 = 18 km/hr
Downstream speed of the boat = 18 + 6 = 24 km/hr

Hence, distance travelled by the boat with the current in 5 hours = 24 × 5 = 120 km

Quantity II:
Decreased speed of the boat in still water = (4/5) × 15 = 12 km/hr
Upstream speed of the boat = 12 - 6 = 6 km/hr

Hence, distance travelled by the boat against the current in 18 hours = 18 × 6 = 108 km

Therefore, Quantity I > Quantity II

49. 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: 8

Quantity II: In a rack, there are 10 physics books, 6 chemistry books and x maths books. One book is picked up randomly, the probability that is neither physics book nor maths book is 1/5. What is the value of x?

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

Quantity I: 8

Quantity II:
Total no. of books = 10 + 6 + x = 16 + x

Let E = event that the book drawn is neither physics nor maths,
n(E) = 6

Therefore P(E) = n(E)/n(s)
=> 1/5 = 6/(16 + x)
=> 16 + x = 30 => x = 14

Hence, Quantity I < Quantity II

50. 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: Krishna started a catering business. After 4 months from the start of the business, Nitin and Piyush joined. The respective ratio between the investments of Krishna, Nitin and Piyush was 4 : 6 : 5. If Krishna's share in annual profit was Rs. 250 more than Piyush's share, what was the total annual profit earned?

Quantity II: Raman and Sitha started a business in partnership with initial investments in the ratio of 5: 3 and after 6 months, Amit joined them and Sitha's initial investment is 66(2/3)% of the initial investment of Amit. If they got Rs. 12300 as total annual profit, then what was the total sum of Sitha's share and Amit's share in the profit?

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

Quantity I:
Ratio of the equivalent capitals of Krishna, Nitin and Piyush for 1 year = (12 × 4) : (6 × 8) : (5 × 8)
= 6 : 6 : 5

Let Rs. X be the total annual profit.

Then from the question,
Krishna’s share in profit - Piyush’s share in profit = 250

=> (6/17) × X + (5/17) × X = 250
=> X/17 = 250
=> X = 250 × 17 = Rs. 4250

Quantity II:
From the question,
Raman’s share : Sitha’s share = 5 : 3 = 10 : 6
Sitha’s share : Amit’s share = 2 : 3 = 6 : 9

So, Raman’s share : Sitha’s share : Amit’s share = 10 : 6 : 9

Ratio of the capitals of Raman, Sitha and Amit = (10 × 12) : (6 × 12) : (9 × 6) = 20 : 12 : 9

Total sum of Sitha’s share and Amit’s share in profit = [(12 + 9)/41] × 12300 = Rs. 6300

Hence, Quantity I < Quantity II