SBI PO Quant Mains Questions Solved Problems with Detailed Solutions [Free PDF]
Last updated on Jun 13, 2025
Important SBI PO Quant Mains Questions
SBI PO Quant Mains Questions Question 1:
Comprehension:
Quantity I. 15, 12, X, 9, 21, 6
Quantity II. 24, 12, 12, Y, 36, 90
Quantity III. 3, 2, 10, Z, 17, 16Answer (Detailed Solution Below)
SBI PO Quant Mains Questions Question 1 Detailed Solution
Quantity I:
15 – 3 = 12
12 + 6 = 18
18 – 9 = 9
9 + 12 = 21
21 – 15 = 6
Quantity II:
24 × 0.5 = 12
12 × 1 = 12
12 × 1.5 = 18
18 × 2 = 36
36 × 2.5 = 90
Quantity III:
3 - 13 = 2
2 + 23 = 10
10 - 13 = 9
9 + 23 = 17
17 - 13 = 16
∴ 18 = 18 > 9
Hence, X = Y > Z
SBI PO Quant Mains Questions Question 2:
If one-fifth of a tank is filled by a pipe at the rate of 24 lit/min, one half of a tank is filled by another pipe at the rate of 30 lit/min and rest at the rate of 24 lit/min, the total time taken to fill the tank is 30 minutes. Time taken by pipe A to fill the tank is also 30 minutes. The total capacity of the tank is ________ liters and Time taken by pipe B to fill the tank is 15 minutes. The average rate at which the tank is filled if pipe B and pipe A together opened is ________.
Which of the following options satisfies the two blanks in the question?
A. 800 litre, 80 litre/min
B. 480 litre, 95 litre/min
C. 760 litre, 72 litre/min
D. 900 litre, 60 litre/min
Answer (Detailed Solution Below)
SBI PO Quant Mains Questions Question 2 Detailed Solution
GIVEN :
If one-fifth of a tank is filled by a pipe at the rate of 24 lit/min.
One half of a tank is filled by another pipe at the rate of 30 lit/min.
Rest at the rate of 24 lit/min, the total time taken to fill the tank is 30 minutes.
Time taken by pipe A to fill the tank is also 30 minutes.
CALCULATION :
Let the total capacity of a tank be ‘n’ lit.
Average rate = total distance / time taken
⇒ Time taken by a pipe to fill 1/5th of a cistern = (n/5)/24 = n/120 min
⇒ Time taken by a pipe to fill 1/2nd of a cistern = (n/2)/30 = n/60 min
⇒ Rest part of a cistern remain to fill = n{1 – (1/5 + 1/2)} = n(1 – 7/10)
⇒ Rest part of a cistern remain to fill = 3n/10
⇒ Time taken by a pipe to fill rest of cistern = (3n/10)/24 = n/80 min
Total time taken by a pipe to fill cistern completely = (n/120 + n/60 + n/80) hr = 3n/80min
Given,
⇒ 3n/80 = 30
⇒ n = 800 litres
Pipe A’s 1 minute’s work = 1/30
Pipe B’s 1 Minute’s work = 1/15
Total time taken by pipe A and pipe B = 1/30 + 1/15 = 1/10
∴ Average rate = 800/10 = 80 litre/min.
SBI PO Quant Mains Questions Question 3:
Directions: Information about three values have been given, you have to calculate the value of all three quantities and answer the question accordingly.
X: 60 Men could do a job in 40 days. Due to inclusion of some more Men, the work gets completed in 4/5 time. Fine the number of men added.
Y: A can finish the work in 5 days, while B takes 12 days to complete the same work. If A starts and they work on alternative days, Find the number of days taken to complete the work.
Z: A pipe can fill 40% of the tank in 6 hours. B can empty 30% of the same tank in 12 hours. If the tank is already half filled, find the time taken to fill the tank when both the pipes are opened together.Answer (Detailed Solution Below)
SBI PO Quant Mains Questions Question 3 Detailed Solution
X: Number of men = 60
Time taken = 40
New number of men = a
New time = (4/5) × 40 = 32 days.
Now,
60 × 40 = 32 × a
a = 75
Number of new men = 75 - 60 = 15
Y: Let total work = LCM of 5,12 = 60 units
Work done by A in 1 day = 60/5 = 12 units
Work done by B in 1 day = 60/12 = 5 units
They will complete (5 + 12) units in 2 days.
17 × 3 = 51 units will be completed in 6 days.
Work remaining = 60 - 51 = 9 units
Time taken for 9 units by A = 9/12 = 0.75
Total time = 6.75 days.
Z: Time taken by A to fill 40% tank = 6 hours
Time taken to fill full tank = (6/40) × 100 = 15 hours.
Time taken by B to empty 30% tank = 12 hours.
Time taken to empty 100% tank = (12/30) × 100 = 40 hours
Let the capacity of tank = LCM of 15,40 = 120 units.
A will fill 8 units in an hour while B will empty 3 units in an hour.
Time taken to empty 60 units = 60/(8 - 3) = 12 hours.
X > Z > YSBI PO Quant Mains Questions Question 4:
The question below is followed by three statements I, II and III. You have to determine whether the data given is sufficient for answering the question. You should use the data and your knowledge of mathematics to choose the best possible answer.
Find the total amount paid by the person to the bank.
Statement I∶ A person took the loan of Rs. 7000 from the bank at rate of 5% per annum simple interest.
Statement II∶ He wants to repay the loan in 5 years in some installments.
Statement III∶ At the end of 3 years, he paid Rs. 3000 and cleared the dues after 2 more years.Answer (Detailed Solution Below)
SBI PO Quant Mains Questions Question 4 Detailed Solution
Statement I and III∶
Principal = Rs. 7000 and r = 5%;
∵ At the end of 3 years, he paid Rs. 3000 and cleared the dues after 2 more years;
Calculating the total principal for 5 years∶
P = 7000 + 7000 + 7000 + 4000 + 4000 = Rs. 29000
∴ Interest = [29000 × 5 × 1]/100 = Rs. 1450
∴ To clear the dues after 2 more years, the person must pay Rs. 5450 (4000 + 1450) to the bank.
∴ Total amount paid by the person to the bank = 3000 + 5450 = Rs. 8450
∴ Statement I and III together are sufficient to answer the question.SBI PO Quant Mains Questions Question 5:
The question below is followed by three statements I, II and III. You have to determine whether the data given is sufficient for answering the question. You should use the data and your knowledge of mathematics to choose the best possible answer.
In a live concert, the amount collected by selling the tickets was Rs. 75000. Find the cost of a ticket, if the cost of every ticket is the same.
Statement I: If the price of each ticket is increased by Rs. 50, the amount collected will be Rs. 6250 more.
Statement II: If the number of tickets is decreased by 25 (keeping the cost per ticket constant), the amount collected will be Rs. 15000 more.
Statement III: Ratio of number of tickets and the cost of a ticket is 5 : 24Answer (Detailed Solution Below)
SBI PO Quant Mains Questions Question 5 Detailed Solution
Suppose the price of a ticket is Rs. x;
∴ Number of tickets = 75000/x
Statement I:
(x + 50) × 75000/x = 75000 + 6250
⇒ 3750000/x = 6250
⇒ x = 600
∴ Price of a ticket = Rs. 600
Statement II:
x × [(75000/x) - 25] = 75000 + 15000
⇒ x = 600
∴ Price of a ticket = Rs. 600
Statement III:
Ratio of number of tickets and the cost of a ticket is 5 : 24;
Suppose the number of tickets and the cost of a ticket is 5a & 24a respectively;
⇒ 120a2 = 75000
⇒ a2 = 625
⇒ a = 25
∴ Price of a ticket = Rs. 600
∴ Any of the statements alone is sufficient to answer the question.SBI PO Quant Mains Questions Question 6:
The following question have three statements. Study the question and the statements and decide which of the statement(s) is necessary to answer the question.
In the figure, find the value of EF.
Statement I: DB = 125 cm and ΔDOE is an equilateral triangle.
Statement II: CD = 35 cm and AB = 75
Statement III: AB II EF II DC and DE : EA = CF : FB = 2 : 3.
Answer (Detailed Solution Below)
SBI PO Quant Mains Questions Question 6 Detailed Solution
Statement II and III:
AB II EF II DC and DE : EA = CF : FB = 2 : 3;
∴ It signifies that ΔDEO and ΔDAB are similar and ΔBFO and ΔBCD are similar.
We know CD = 35 cm and AB = 75;
∴ Values of EO and FO can be determined by the property of similar triangles (Ratio of respective sides is equal).
Statement I:
DB = 125 cm and ΔDOE is an equilateral triangle
Combining this with statement III, we can get the value of DO (by the property of similar triangles);
And the value of EO = DO (Given that ΔDOE is an equilateral triangle);
But the value of FO can’t be determined.
∴ Statement II and Statement III together are sufficient to answer the question