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SBI PO Previous Year Question Paper
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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeQ (15) Each of the following question below consists of a question and two statements numbered 1 and 2 are given below it. You have to decide if the data provided in the statements are sufficient to answer the question. Read both statements and mark answer.
In how many days 6 boys can finish the work?
Statement 1. 6 girls can finish the work in 4 days.
Statement 2. If 2 boys work for 4 days and after that the remaining work is finished by 4 girls in 2 days.Correct
Answer – 5) The data in both the statements 1 and 2 together are necessary to answer the question.
Explanation
Statement 1
6 girls can finish the work in 4 days.
12 girls can finish the work in 2 days
24 girls can finish the work in 1 day
1 girl can finish the work in 24 days
Per day work by 1 girl = 1/ 24
Statement 2
Work done by 4 girls in 2 days = 4× 2 × 1/24 = 1/3
Work left = 11/3 = 2/3
Work done by 2 boys in 4 days = 2/3
Work done by 1 boy in 4 days = 1/3
Work done by 1 boy in 1 day = 1/12
Work done by 6 boys in 1 day = ½
Therefore, 6 boys can finish the work in 2 daysIncorrect
Answer – 5) The data in both the statements 1 and 2 together are necessary to answer the question.
Explanation
Statement 1
6 girls can finish the work in 4 days.
12 girls can finish the work in 2 days
24 girls can finish the work in 1 day
1 girl can finish the work in 24 days
Per day work by 1 girl = 1/ 24
Statement 2
Work done by 4 girls in 2 days = 4× 2 × 1/24 = 1/3
Work left = 11/3 = 2/3
Work done by 2 boys in 4 days = 2/3
Work done by 1 boy in 4 days = 1/3
Work done by 1 boy in 1 day = 1/12
Work done by 6 boys in 1 day = ½
Therefore, 6 boys can finish the work in 2 days 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeQ (15) Each of the following question below consists of a question and two statements numbered 1 and 2 are given below it. You have to decide if the data provided in the statements are sufficient to answer the question. Read both statements and mark answer.
Find the height of the cone?
1. The curved surface area of the cone is 440 cm2 and the total surface area of the cone is 594cm2
2. Slant height of the cone is 20cm.Correct
Answer 1. The data in statement 1 alone is sufficient to answer the question while the data in statement 2 alone is not sufficient to answer the question.
Explanation –
Statement 1
πrl = 440
πrl + πr^{2} = 594
πr^{2} = 154
r = 7
l = 20
l^{2 }= h^{2}+ r^{2}
h^{2} = 20^{2} – 7^{2}Incorrect
Answer 1. The data in statement 1 alone is sufficient to answer the question while the data in statement 2 alone is not sufficient to answer the question.
Explanation –
Statement 1
πrl = 440
πrl + πr^{2} = 594
πr^{2} = 154
r = 7
l = 20
l^{2 }= h^{2}+ r^{2}
h^{2} = 20^{2} – 7^{2} 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeQ (15) Each of the following question below consists of a question and two statements numbered 1 and 2 are given below it. You have to decide if the data provided in the statements are sufficient to answer the question. Read both statements and mark answer.
A father distributes a sum of Rs. 333 among three sons Prem, Utkal, and Vinod then who gets the maximum?
Statement 1: Prem gets 4/5th of what Utkal and Vinod together gets
Statement 2: Utkal gets 1/5th of what Prem and Vinod together getsCorrect
Answer 5) The data in both the statements 1 and 2 together are necessary to answer the question.
Explanation
Statement 1
P+U+V = 333
4P + 4U + 4V = 333× 4 ………(1)
P = 4/5 (U+V)
5P = 4U + 4V
Putting , 5P = 4U + 4V in equantion 1
9P = 333 × 4
P = 148
Statement 2
P+U+V = 333
U = 1/5 (P+V)
5U = P+V
6U = 333
U = 55.5Incorrect
Answer 5) The data in both the statements 1 and 2 together are necessary to answer the question.
Explanation
Statement 1
P+U+V = 333
4P + 4U + 4V = 333× 4 ………(1)
P = 4/5 (U+V)
5P = 4U + 4V
Putting , 5P = 4U + 4V in equantion 1
9P = 333 × 4
P = 148
Statement 2
P+U+V = 333
U = 1/5 (P+V)
5U = P+V
6U = 333
U = 55.5 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeQ (15) Each of the following question below consists of a question and two statements numbered 1 and 2 are given below it. You have to decide if the data provided in the statements are sufficient to answer the question. Read both statements and mark answer.
In a school, 55% of boys and 45% of girls participated in a sports. How many girls are there in the school?
1. Ratio of total number of boys to no. of girls in school is 3:2.
2. Total 510 students participated in sports.Correct
Answer 5) The data in both the statements 1 and 2 together are necessary to answer the question.
Explanation –
Let Ratio of total number of boys to no. of girls in school is 300x : 200x
Total boys participated = 55% of 300x = 165x
Total girls participated = 45% of 200x = 90x
Total students participated = 255x
255x = 510
X = 2
200x = 400Incorrect
Answer 5) The data in both the statements 1 and 2 together are necessary to answer the question.
Explanation –
Let Ratio of total number of boys to no. of girls in school is 300x : 200x
Total boys participated = 55% of 300x = 165x
Total girls participated = 45% of 200x = 90x
Total students participated = 255x
255x = 510
X = 2
200x = 400 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeQ (15) Each of the following question below consists of a question and two statements numbered 1 and 2 are given below it. You have to decide if the data provided in the statements are sufficient to answer the question. Read both statements and mark answer.
Find the monthly salary of Atul?
Statement 1: Chirag’s monthly salary is Rs. 28000 which is 50% of the monthly salary of Hemant.
Statement 2: Monthly salary of Atul is 16% of the monthly salary received by Atul, Chirag and Hemant together.Correct
Answer 5) The data in both the statements 1 and 2 together are necessary to answer the question.
Explanation
Statement 1
Salary of Chirag(C) = 28000
Salary of Hemant(H) = 56000
Statement 2
A = 16/100 (A+C+H)
100/16 A = (A+C+H)
84A /16 = (28000+56000)
A = 16000Incorrect
Answer 5) The data in both the statements 1 and 2 together are necessary to answer the question.
Explanation
Statement 1
Salary of Chirag(C) = 28000
Salary of Hemant(H) = 56000
Statement 2
A = 16/100 (A+C+H)
100/16 A = (A+C+H)
84A /16 = (28000+56000)
A = 16000 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirections Q (610): Each question contains two sub questions i.e., Quantity 1 and Quantity 2. Quantity 1 will be the answer of the first question and Quantity 2 will be the answer of the second question. Compare the two quantities –
A bag contains 4 blue, 3 green and 6 red pens.
Quantity 1: If two pens are drawn at random, the probability that both the pens are either blue or red.
Quantity 2: If three pens are drawn at random, the probability that 1 is green and 2 are blue.Correct
Answer 1. Quantity1 >Quantity 2
Explanation –
Quantity 1 –
(4C_{2}+6C_{2})/13C_{2 }= 7/ 26
Quantity 2 –
(3C_{1}×4C_{2})/ 13C_{3 }=9/143Incorrect
Answer 1. Quantity1 >Quantity 2
Explanation –
Quantity 1 –
(4C_{2}+6C_{2})/13C_{2 }= 7/ 26
Quantity 2 –
(3C_{1}×4C_{2})/ 13C_{3 }=9/143 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirections Q (610): Each question contains two sub questions i.e., Quantity 1 and Quantity 2. Quantity 1 will be the answer of the first question and Quantity 2 will be the answer of the second question. Compare the two quantities –
Quantity 1: Speed of a boat in still water, if it can travel 45km downstream in 9 hours and the speed of stream is 2 km/h.
Quantity 2: Speed of a boat in still water, if it can travel 45km downstream in 9 hours and 20 km upstream in 5 hours.Correct
Answer 3) Quantity 1 < Quantity 2
Explanation –
Let the speed of boat = x km/h
Speed of stream = y km/h
Upstream speed = x –y
Downstream speed = x + y
Quantity 1
x +y = 45/9
x+2 = 5
X = 3 km/hQuantity 2
x + y = 45/9
x + y = 5km/h
x – y = 20/5
x – y = 4 km/h
x = 4.5 km/hIncorrect
Answer 3) Quantity 1 < Quantity 2
Explanation –
Let the speed of boat = x km/h
Speed of stream = y km/h
Upstream speed = x –y
Downstream speed = x + y
Quantity 1
x +y = 45/9
x+2 = 5
X = 3 km/hQuantity 2
x + y = 45/9
x + y = 5km/h
x – y = 20/5
x – y = 4 km/h
x = 4.5 km/h 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirections Q (610): Each question contains two sub questions i.e., Quantity 1 and Quantity 2. Quantity 1 will be the answer of the first question and Quantity 2 will be the answer of the second question. Compare the two quantities –
Quantity 1: A and B together started a business with an investment of Rs. 3600 and Rs. 5400 respectively. After “k” months, C joined them with an investment of Rs. 4500.If in a year, A received Rs. 2700 out of the total profit of Rs. 9000, find k.
Quantity 2: P and Q together started a business with total investment of Rs. 2250 in the ration of k:5 respectively. If after a year, P received Rs. 9600 as profit out of a total profit of Rs. 21600, find k.Correct
Answer – 5) Quantity 1 = Quantity 2 or no relation can be established.
Explanation –
Quantity 1
A : B : C
Investment × Months : Investment × Months : Investment × Months
3600 × 12 : 5400 × 12 : 4500 × (12k)
4 × 12 : 6 × 12 : 5 × (12k)
48 : 72 : 5 × (12k)
48 / (48+72+605k) = 2700 / 9000
48 / (180 – 5k) = 3/ 10
16 / (36 – k) = 1/2
32 = 36 – k
K = 4
Quantity 2
K / (k+5) = 9600/21600
K /(k+5) = 4/ 9
9k = 4k + 20
5k = 20
K = 4Incorrect
Answer – 5) Quantity 1 = Quantity 2 or no relation can be established.
Explanation –
Quantity 1
A : B : C
Investment × Months : Investment × Months : Investment × Months
3600 × 12 : 5400 × 12 : 4500 × (12k)
4 × 12 : 6 × 12 : 5 × (12k)
48 : 72 : 5 × (12k)
48 / (48+72+605k) = 2700 / 9000
48 / (180 – 5k) = 3/ 10
16 / (36 – k) = 1/2
32 = 36 – k
K = 4
Quantity 2
K / (k+5) = 9600/21600
K /(k+5) = 4/ 9
9k = 4k + 20
5k = 20
K = 4 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirections Q (610): Each question contains two sub questions i.e., Quantity 1 and Quantity 2. Quantity 1 will be the answer of the first question and Quantity 2 will be the answer of the second question. Compare the two quantities –
Quantity 1: Ravi invested Rs. 4500 in a scheme and received Rs. 1350 as simple interest after 3 years. Find the rate of interest.
Quantity 2: Prakash purchased a bike for Rs. 25000 and he spent Rs. 5000 on its maintenance. If he marked its price up by x% and sold it for Rs. 31500 after offering a discount of 10%. Find the value of x.Correct
Answer – 3) Quantity 1 < Quantity 2
Explanation –
Quantity 1
PRT/100 = 1350
(4500 × R × 3) / 100 = 1350
(10 × R × 3) / 100 = 3
(10 × R × 1) / 100 = 1
R = 10%
Quantity 2
CP = 25000 + 5000 = 30000
90/100 × MP = SP
9/10 × MP = SP
9/10 × MP = 31500
MP = 35000
Market Up = (MPCP) = Rs. 5000
Market Up %age = 5000/30000 = 1/ 6 = 16.66%Incorrect
Answer – 3) Quantity 1 < Quantity 2
Explanation –
Quantity 1
PRT/100 = 1350
(4500 × R × 3) / 100 = 1350
(10 × R × 3) / 100 = 3
(10 × R × 1) / 100 = 1
R = 10%
Quantity 2
CP = 25000 + 5000 = 30000
90/100 × MP = SP
9/10 × MP = SP
9/10 × MP = 31500
MP = 35000
Market Up = (MPCP) = Rs. 5000
Market Up %age = 5000/30000 = 1/ 6 = 16.66% 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirections Q (610): Each question contains two sub questions i.e., Quantity 1 and Quantity 2. Quantity 1 will be the answer of the first question and Quantity 2 will be the answer of the second question. Compare the two quantities –
Quantity 1: Maximum Retail Price of a mobile is Rs. 15000. After giving a discount of 14/3% a gain of 10% is achieved. Find the Cost Price of Mobile.
Quantity 2: Find the selling price of a device after two successive discount of 5% and 15%. The marked price of the device is Rs. 16000.Correct
Answer 1) Quantity1 >Quantity 2
Explanation –
Quantity 1
MRP = 15000
Discount = 14/3% of 15000 = 700
SP = 15000700 = 14300
For 10% gain,
110× CP = 100 × SP
110CP = 100 × 14300
CP = 13000
Quantity 2
16000 × 95 / 100 × 85/ 100 = 12920Incorrect
Answer 1) Quantity1 >Quantity 2
Explanation –
Quantity 1
MRP = 15000
Discount = 14/3% of 15000 = 700
SP = 15000700 = 14300
For 10% gain,
110× CP = 100 × SP
110CP = 100 × 14300
CP = 13000
Quantity 2
16000 × 95 / 100 × 85/ 100 = 12920
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