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- Question 1 of 10
##### 1. Question

1 pointsCategory: Quantitative Aptitude**Q (1-5) 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 = 1-1/3 = 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 = 1-1/3 = 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 Aptitude**Q (1-5) 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 Aptitude**Q (1-5) 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 Aptitude**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 Aptitude**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 Aptitude**Directions- Q (6-10): 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 Aptitude**Directions- Q (6-10): 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/h**Quantity 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/h**Quantity 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 Aptitude**Directions- Q (6-10): 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 × (12-k)

4 × 12 : 6 × 12 : 5 × (12-k)

48 : 72 : 5 × (12-k)

48 / (48+72+60-5k) = 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 × (12-k)

4 × 12 : 6 × 12 : 5 × (12-k)

48 : 72 : 5 × (12-k)

48 / (48+72+60-5k) = 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 AptitudeQuantity 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 = (MP-CP) = 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 = (MP-CP) = Rs. 5000

Market Up %age = 5000/30000 = 1/ 6 = 16.66% - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative AptitudeQuantity 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 = 15000-700 = 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 = 15000-700 = 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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