BANK & INSURANCE (DATA SUFFICIENCY) PART 2

Total Questions: 30

21. The question below consists of a question and three statements numbered I, II and III given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read all the three statements and give answers:

In a group, there are 6 managers, some architects, 4 teachers and some doctors. A committee of six members is to be formed such that the committee contains 1 manager, 2 architects, 1 teacher and 2 doctors. Find the total number of ways in which the committee can be formed.

Statement I: A committee of 4 members such that the committee contains 2 managers, 1 architect and 1 teacher can be formed in 120 different numbers of ways.

Statement II: Total number of members in the group is 18.

Statement III: Probability of selecting one doctor from the group is 1/3.

Correct Answer: (c) Any two of the three.
Solution:

Managers = 6
Teachers = 4

Let,
Architects = m
Doctors = n

Total = 6 + 4 + m + n = 10 + m + n

From I and II:
⁶C₂ × ᵐC₁ × ⁴C₁ = 120
⇒ 15 × m × 4 = 120
⇒ m = 120/60
⇒ m = 2

10 + m + n = 18
⇒ 10 + 2 + n = 18
⇒ n = 18 - 12
⇒ n = 6

Required number of ways = ⁶C₁ × ²C₂ × ⁴C₁ × ⁶C₂
= 6 × 1 × 4 × 15 = 360

From I and III:
⁶C₂ × ᵐC₁ × ⁴C₁ = 120
⇒ 15 × m × 4 = 120
⇒ m = 120/60
⇒ m = 2

And
n/(10 + m + n) = 1/3
⇒ n/(10 + 2 + n) = 1/3
⇒ n/(12 + n) = 1/3
⇒ 3n = 12 + n
⇒ 3n - n = 12
⇒ 2n = 12
⇒ n = 12/2
⇒ n = 6

Required number of ways = ⁶C₁ × ²C₂ × ⁴C₁ × ⁶C₂
= 6 × 1 × 4 × 15 = 360

From II and III:
10 + m + n = 18
n/18 = 1/3
⇒ n = 18/3
⇒ n = 6

Now, 10 + m + 6 = 18
⇒ m = 18 - 16
⇒ m = 2

Required number of ways = ⁶C₁ × ²C₂ × ⁴C₁ × ⁶C₂
= 6 × 1 × 4 × 15 = 360

Hence, any two of the three statements are sufficient

22. In the question below, the question is followed by three statements. Read all the statements carefully and find which of the following statement(s) is/are sufficient to answer the question.

Who is older between A and C?

I. Age of A 6 years hence is equal to the sum of 2/3rd of present age of C and 1/5th of present age of B.

II. B is 18 years older than his son and A is 12 years older than B’s son.

III. Double the present age of A is equal to the age of C 12 years hence.

Correct Answer: (d) All three statements together.
Solution:

From statement I:
Age of A 6 years hence is equal to the sum of 2/3rd of present age of C and 1/5th of present age of B.

A + 6 = 2/3 of C + 1/5 of B
15A + 90 = 10C + 3B

From statement II:
Let present age of B’s son = y years
B is 18 years older than his son and A is 12 years older than B’s son.

B = 18 + y
A = 12 + y

So, B - A = (18 + y) - (12 + y)
B = A + 6

From statement III:
Double the present age of A is equal to the age of C 12 years hence.

2A = C + 12
C = 2A - 12

Here, we can see that no statement alone is sufficient to find the answer.
Also, by taking any two statements together we cannot find the answer.

From statements I, II and III together:
15A + 90 = 10C + 3B
15A + 90 = 10(2A - 12) + 3(A + 6)
15A + 90 = 20A - 120 + 3A + 18
8A = 192
A = 24

C = 2 × 24 - 12 = 36
This means C is older than A.

Hence, all three statements together are necessary to find the answer

23. The question below consists of a question and three statements numbered I, II and III given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read all the statements and give answers.

Find the total amount invested by Raghav.

Statement I: He invested half of the amount on simple interest at 6% per annum for five years from which he will get an interest of Rs.10500.

Statement II: He invested 1/4th of the amount on compound interest at 10% per annum for two years from which he will get an interest of Rs.3675.

Statement III: He invested 1/4th of the amount on simple interest at 8% per annum for three years from which he will get a total amount of Rs.21700

Correct Answer: (d) Any one of the three
Solution:

Let, total amount invested by Raghav be Rs.P

From I:
We know that
SI = (P × r × t)/100

⇒ 10500 = (P/2 × 6 × 5)/100
⇒ 10500 = 3P/20
⇒ P = 10500 × 20/3
⇒ P = Rs.70000

From II:
We know that
CI = P × (1 + r/100)ᵗ - P

⇒ 3675 = P/4 × (1 + 10/100)² - P/4
⇒ 3675 = P/4 × (1 + 1/10)² - P/4
⇒ 3675 = P/4 × (11/10)² - P/4
⇒ 3675 = P/4 × 121/100 - P/4
⇒ 3675 = 121P/400 - P/4
⇒ 3675 = (121P - 100P)/400
⇒ 3675 = 21P/400
⇒ P = 3675 × 400/21
⇒ P = Rs.70000

From III:
We know that
Amount on SI = (P × r × t)/100 + P

⇒ 21700 = (P/4 × 8 × 3)/100 + P/4
⇒ 21700 = 3P/50 + P/4
⇒ 21700 = (6P + 25P)/100
⇒ 21700 = 31P/100
⇒ 31P = 2170000
⇒ P = 2170000/31
⇒ P = Rs.70000

Hence, any one of the three statements is sufficient.

24. The question below consists of a question and three statements numbered I, II and III given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read all the statements and give answers.

Four pipes A, B, C and D are used to fill a tank and all the pipes can be used as both inlet and outlet pipes. Time taken by any pipe while working as an inlet and while working as outlet is the same, then what is the time taken to fill the tank when all the four pipes are opened out of which A and B are outlets and C and D are inlets? If the nature of the pipe is not given, then consider them as inlet pipes.

I: Time taken by all the four pipes together to fill the tank is 2.5 hours and time taken by pipes A and B together to fill the tank is 7.5 hours.

II: Time taken by pipes A and C together to fill the tank is 12 hours and time taken by pipes B and D together to fill the tank is 20 hours where pipes A and B are outlets.

III: Pipes C and D together are twice as efficient as pipes A and B together and pipe A alone takes 12 hours to fill the tank while pipe D alone takes 10 hours to fill that tank.

Correct Answer: (d) Either I alone or II alone is sufficient
Solution:

Let, time taken by pipes A, B, C and D to fill/empty the tank alone is ‘A’, ‘B’, ‘C’ and ‘D’ respectively.

From I:
According to question:
(1/A) + (1/B) + (1/C) + (1/D) = (1/2.5) .......... (1)

(1/A) + (1/B) = (1/7.5) .......... (2)

From (1) and (2):
(1/7.5) + (1/C) + (1/D) = (1/2.5)
(1/C) + (1/D) = (1/2.5) - (1/7.5)
(1/C) + (1/D) = (3 - 1)/7.5
(1/C) + (1/D) = (1/3.75) .......... (3)

Time taken to fill the tank when all the four pipes are opened out of which A and B are outlets and C and D are inlets =
1 / [((1/C) + (1/D)) - ((1/A) + (1/B))]
= 1 / [(1/3.75) - (1/7.5)]
= 1 / [(2 - 1)/7.5]
= 7.5 hours

Statement I alone is sufficient.

From II:
According to question:
(1/C) - (1/A) = (1/12) .......... (4) [Since, A is an outlet pipe.]

(1/D) - (1/B) = (1/20) .......... (5) [Since, B is an outlet pipe.]

From (4) and (5):
[(1/C) - (1/A)] + [(1/D) - (1/B)] = (1/12) + (1/20)
(1/C) + (1/D) - (1/A) - (1/B) = (5 + 3)/60
(1/C) + (1/D) - (1/A) - (1/B) = (1/7.5)

Hence, time taken to fill the tank when all the four pipes are opened out of which A and B are outlets and C and D are inlets = 7.5 hours

Statement II alone is sufficient.

From III:
Let time taken by pipes C and D together to fill the tank = ‘N’ hours

(1/C) + (1/D) = (1/N)
(1/C) + (1/10) = (1/N)
(1/C) = (1/N) - (1/10) .......... (6)

Time taken by pipes A and B together to fill the tank = ‘2N’ hours

(1/A) + (1/B) = (1/2N)
(1/12) + (1/B) = (1/2N)
(1/B) = (1/2N) - (1/12) .......... (7)

Since, we can’t calculate the time taken either pipe alone to fill/empty the tank.

Statement III alone is not sufficient

25. Each question below is followed by three statements I, II and III. You have to determine whether the data given in the statement is sufficient for answering the question.

A mixture contains y litres mixture of milk and water which contains milk and water in the ratio a:b respectively. The milkman sold z litres of the mixture and added 10 litres of pure milk and 8 litres of water to the remaining mixture such that the ratio of milk and water in the final mixture became 5:4 respectively. Find the respective ratio of amount of milk in the initial mixture and amount of milk in the final mixture.

Statement I: A 54 litres mixture of syrup and water contains syrup and water in the ratio a:b respectively. If 10 litres syrup and 6 litres water is added to the mixture, ratio of syrup and water becomes 4:3 respectively.

Statement II: A 60 litres mixture of milk and water contains milk and water in the ratio 7:5 respectively. If z litres of the mixture is replaced with water, ratio of milk and water in the mixture becomes 7:23 respectively.

Statement III: y = 3z

Correct Answer: (d) All I, II and III together
Solution:

From I:
Syrup = a/(a + b) × 54
Water = b/(a + b) × 54

[a/(a + b) × 54 + 10] / [b/(a + b) × 54 + 6] = 4/3

⇒ 3 × [a/(a + b) × 54 + 10] = 4 × [b/(a + b) × 54 + 6]

⇒ 162a/(a + b) + 30 = 216b/(a + b) + 24
⇒ (216b - 162a)/(a + b) = 30 - 24
⇒ (216b - 162a)/(a + b) = 6
⇒ 216b - 162a = 6a + 6b
⇒ 216b - 6b = 6a + 162a
⇒ 210b = 168a
⇒ a/b = 210/168
⇒ a/b = 5:4

From II:
A 60 litres mixture of milk and water contains milk and water in the ratio 7:5 respectively. If y litres of the mixture is replaced with water, ratio of milk and water in the mixture becomes 7:23 respectively.

Amount of milk = 7/(7 + 5) × 60
= 7/12 × 60 = 35 litres

Amount of water = 5/(7 + 5) × 60
= 5/12 × 60 = 25 litres

(35 - 7z/12)/(25 - 5z/12 + z) = 7/23
⇒ 23 × (35 - 7z/12) = 7 × (25 - 5z/12 + z)
⇒ 23 × (420 - 7z)/12 = 7 × (300 - 5z + 12z)/12
⇒ 23 × (420 - 7z) = 7 × (300 + 7z)
⇒ 9660 - 161z = 2100 + 49z
⇒ 161z + 49z = 9660 - 2100
⇒ 210z = 7560
⇒ z = 7560/210
⇒ z = 36

From III:
y = 3z

From I, II and III:
y = 3 × 36 = 108

Amount of milk in the initial mixture = 5/(5 + 4) × 108 = 5/9 × 108 = 60 litres

Amount of milk in the final mixture =
5/9 × 36 + 10 = 60 - 20 + 10 = 50 litres

Required ratio = 60 : 50 = 6 : 5

Hence, all I, II and III together are sufficient.

26. A bag contains some balls of three different colours. Probability of drawing a blue colour ball is (5/24) and the ratio of red to green colour balls is 9 : 10. What is the total number of balls in the bag?

I. Average of total number of balls of all the three colours is 8.

II. Two balls are drawn randomly from the bag and the probability that both the balls are green is 15/92.

III. One ball is drawn at random from the bag and the probability that the ball is either blue or red is (7/12).

Correct Answer: (b) Either I alone or II alone is sufficient
Solution:

Let total number of balls in the bag = 24x

Probability of drawing a blue colour ball = (5/24)

Total blue colour balls = 5x
Total red and green colour balls = 24x - 5x = 19x

Total red colour balls = 19x × (9/19) = 9x
Total green colour balls = 19x × (10/19) = 10x

From I:
Average of total number of balls of all the three colours =
(5x + 9x + 10x)/3 = 8

24x = 24
x = 1

Total balls in the bag = 24x = 24

Statement I alone is sufficient.

From II:
Probability that two drawn balls are of green colour = 10xC₂ / 24xC₂ = 15/92

5x × (10x - 1) / [12 × (24x - 1)] = 15/92
23 × (10x - 1) = 9 × (24x - 1)
230x - 23 = 216x - 9
14x = 14
x = 1

Total balls in the bag = 24x = 24

Statement II alone is sufficient.

From III:
Probability that one ball is drawn and that ball is either red or blue.

(9xC₁ + 5xC₁)/24xC₁ = 7/12
(9 + 5)/24 = 7/12
14/24 = 7/12
1 = 1

Statement III alone is not sufficient.

27. In the following questions, which of the given statement(s) are necessary for determining the answer.

What is the investment made by R if the sum of investments of P, Q and R is Rs.24000?

I. P and Q start a business with Rs.p and Rs.q for 15 months and 18 months respectively. Profit earned by Q and P is in the ratio 18 : 25.

II. R and P are partners in a business. R contributes Rs.r and P contributes Rs.p for 12 months and 15 months respectively. Profit earned by P and R is in the ratio 25 : 16.

III. Q and R enter into a business with Rs.q and Rs.r for 18 months and 12 months respectively. Profit earned by Q and R is in the ratio 9 : 8.

Correct Answer: (e) None of these
Solution:

p + q + r = 24000

From I:
15 × p : 18 × q = 25 : 18
p : q = 5 : 3

From II:
12 × r : 15 × p = 16 : 25
r : p = 4 : 5

From III:
18 × q : 12 × r = 9 : 8
q : r = 3 : 4

From I and II:
q = 3p/5
r = 4p/5

p + (3p/5) + (4p/5) = 24000
p = Rs. 10000

r = 4 × 10000/5 = Rs. 8000

From I and III:
p = 5q/3
r = 4q/3

(5q/3) + q + (4q/3) = 24000
q = Rs. 6000

r = 4 × 6000/3 = Rs. 8000

From II and III:
p = 5r/4
q = 3r/4

(5r/4) + (3r/4) + r = 24000
r = Rs. 8000

Hence, the answer can be determined by using any two of the given three statements.

28. Each question below is followed by three statements I, II and III. You have to determine whether the data given in the statement is sufficient for answering the question.

Find the difference between compound interest and simple interest on Rs.P at 12% per annum after 3 years.

Statement I: Sum of the simple interests on Rs.P/2 at r% per annum after 4 years and on Rs.P/4 at 5% per annum after three years will be Rs.12600.

Statement II: Compound interest on Rs.40000 at r% per annum after two years will be Rs.4944.

Statement III: Difference between compound interest and simple interest on Rs.P/4 at r% per annum after two years will be Rs.72.

Correct Answer: (c) Any two of the three
Solution:

From I:
We know that
SI = (P × r × t)/100

(P/2 × r × 4)/100 + (P/4 × 5 × 3)/100 = 12600

⇒ Pr/50 + 3P/80 = 12600
⇒ (8Pr + 15P)/400 = 12600
⇒ P × (8r + 15) = 12600 × 400
⇒ P × (8r + 15) = 5040000 .......... (i)

From II:
We know that
CI = P × (1 + r/100)ᵗ - P

⇒ 4944 = 40000 × (1 + r/100)² - 40000
⇒ 4944 + 40000 = 40000 × (1 + r/100)²
⇒ 44944/40000 = (1 + r/100)²
⇒ 2809/2500 = (1 + r/100)²
⇒ (53/50)² = (1 + r/100)²
⇒ 53/50 = 1 + r/100
⇒ r/100 = 53/50 - 1

⇒ r/100 = (53 - 50)/50
⇒ r = 100 × 3/50
⇒ r = 6% .......... (ii)

From III:
We know that, for two years
CI - SI = P × (r/100)²

⇒ 72 = P/4 × (r/100)²
⇒ 288 = P × r²/10000
⇒ Pr² = 2880000
⇒ P = 2880000/r² .......... (iii)

From I and II:
P × (8 × 6 + 15) = 5040000
⇒ P × (48 + 15) = 5040000
⇒ P × 63 = 5040000
⇒ P = 5040000/63
⇒ P = Rs. 80000

We know that, for three years
CI - SI = P × (r/100)² × (300 + r)/100

= 80000 × (12/100)² × (300 + 12)/100
= 80000 × (3/25)² × 312/100
= 80000 × 9/625 × 312/100
= Rs. 3594.24

From I and III:
P × (8r + 15) = 5040000
2880000/r² × (8r + 15) = 5040000
⇒ 4/r² × (8r + 15) = 7
⇒ 7r² - 32r - 60 = 0
⇒ 7r² - 42r + 10r - 60 = 0
⇒ 7r(r - 6) + 10(r - 6) = 0
⇒ (7r + 10)(r - 6) = 0
⇒ r = -10/7 (rejected), 6
⇒ r = 6%

From (iii):
P = 2880000/36 = Rs.80000

We know that, for three years
CI - SI = P × (r/100)² × (300 + r)/100

= 80000 × (12/100)² × (300 + 12)/100
= 80000 × (3/25)² × 312/100
= 80000 × 9/625 × 312/100
= Rs. 3594.24

From II and III:
P = 2880000/36 = Rs.80000

We know that, for three years
CI - SI = P × (r/100)² × (300 + r)/100

= 80000 × (12/100)² × (300 + 12)/100
= 80000 × (3/25)² × 312/100
= 80000 × 9/625 × 312/100
= Rs. 3594.24

Hence, any two of the three are sufficient

29. In each of the following questions, each question is followed by three statements. Read all the statements carefully and find which of the following statement(s) is/are redundant to answer the question.

Product of three positive numbers A, B and C is 1248 and their LCM is 156. What is the difference between B and C?

Statement I: B² - AC - 40 = 0

Statement II: HCF of B and C is 4 and their sum is 16.

Statement III: Difference between HCF of A and B and HCF of A and C is 0.

Correct Answer: (c) Only statement III is redundant
Solution:

ABC = 1248

LCM (A,B,C) = (ABC × HCF (A,B,C))/(HCF (A,B) × HCF (B,C) × HCF (A,C)) = 156

HCF (A,B) × HCF (B,C) × HCF (A,C)/(HCF (A,B,C)) = 8

From statement I: B² - AC - 40 = 0
B² - AC - 40 = 0
AC = 1248/B
B² - (1248/B) - 40 = 0
B = 12

From statement II: HCF of B and C is 4 and their sum is 16.
B + C = 16
HCF of B and C = 4

HCF (A,B) × 4 × HCF (A,C)/(HCF (A,B,C)) = 8

HCF (A,B) × HCF (A,C)/(HCF (A,B,C)) = 2

From statement III: Difference between HCF of A and B and HCF of A and C is 0.

Then, HCF (A,B) = HCF (A,C)

So, (HCF (A,B))² × HCF (B,C)/(HCF (A,B,C)) = 8

From statements I and II:
B = 12, C = 16 - 12 = 4

Then, difference between B and C = 12 - 4 = 8

From statements II and III:
(HCF (A,B))²/(HCF (A,B,C)) = 2

From statements I and III:
B = 12

(HCF (A,B))² × HCF (B,C)/(HCF (A,B,C)) = 8

Hence, only statement III is redundant.

30. Mahesh sold a cycle to Rahul, then what is the selling price of the cycle for Rahul if he sold it at 12% profit?

I: Mahesh marked up the price of cycle by 15% and offered Rs.260 as discount when sold to Rahul.

II: Total amount of profit gained by Mahesh when he sold the cycle to Rahul is Rs.100.

III: Ratio of amount of discount offered to the marked-up amount by Mahesh when he sold the cycle to Rahul is 13 : 18.

Correct Answer: (d) Either I and II together or I and III together are sufficient.
Solution:

From I and II:
Let CP of cycle for Mahesh = 100x

MP for Mahesh = 115% of 100x = 115x
SP for Mahesh = (115x - 260) = (100x + 100)

(115x - 260) = (100x + 100)
15x = 360
x = 24

Selling price of cycle for Mahesh = (100x + 100)
= Rs.2500

Selling price of cycle for Rahul = 112% of 2500
= Rs.2800

Statements I and II together are sufficient.

From I and III:
Let CP of cycle for Mahesh = 100x

MP for Mahesh = 115% of 100x = 115x
SP for Mahesh = (115x - 260)

Marked-up amount by Mahesh = 260 × (18/13)
= Rs.360

According to the question:
115x - 100x = 360
15x = 360
x = 24

Selling price of cycle for Mahesh = (115x - 260)
= Rs.2500

Selling price of cycle for Rahul = 112% of 2500
= Rs.2800

Statements I and III together are sufficient.

From II and III:
Let CP of cycle for Mahesh = 100x
SP for Mahesh = (100x + 100)

Let amount of discount and marked up amount by Mahesh is 13y and 18y respectively.

MP for Mahesh = (100x + 18y)
SP for Mahesh = (100x + 18y - 13y)
= (100x + 100)

5y = 100
y = 20

Since, we can’t calculate the value of ‘x’.
Statements II and III together are not sufficient.