Hello Aspirants

Welcome to Online **Quants Test** in AffairsCloud.com. We are starting **Cumulus Course for SBI PO 2018 Main Exam** and we are creating sample questions in Quants section, this type of Question will be asked in SBI PO 2018!!!

**Click Here to View Cumulus Course : SBI PO 2018 Main Exam**

**SBI PO 2018 Main Exam: Reasoning Test**– 7.30 PM Every Day

**SBI PO 2018 Main Exam: Quants Test**– 8.30 PM Every Day

**SBI PO 2018 Main Exam: English Test**– 9.30 PM Every Day

**Help:- **Share Our Cumulus Course for SBI PO 2018 Exam Course page to your Friends & FB Groups

________________________

#### Quiz-summary

0 of 10 questions completed

Questions:

- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10

#### Information

All the Best

You have already completed the quiz before. Hence you can not start it again.

Quiz is loading...

You must sign in or sign up to start the quiz.

You have to finish following quiz, to start this quiz:

#### Results

0 of 10 questions answered correctly

Your time:

Time has elapsed

You have reached 0 of 0 points, (0)

#### Categories

- Quantitative Aptitude 0%

- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10

- Answered
- Review

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

1 pointsCategory: Quantitative Aptitude**The total surface area of a cube, sphere and cylinder is the same. The height of the cylinder is twice its radius.**

Quantity I: Volume of the cube.

Quantity II: Volume of the sphere.

Quantity III: Volume of the cylinder.CorrectAnswer :

**2. Quantity II > Quantity III > Quantity I**

Explanation-

Let the cube’s edge length is a, the surface area of cube = 6a^2 and its volume = a^3.

Let the radius of the sphere is r, then its volume = (4/3) πr^3 and its surface area will be 4πr^2.

Let the height of the cylinder be 2h therefore, its radius will be h.

Surface area of the cylinder = 2πh (2h) +2πh^2 = 6πh^2.

Now since the surface area are same, 6a^2 = 4πr^2 = 6πh^2, therefore, r= (√3/√ (2π)) a.

Volume of sphere = (√6/√π)a^3, this will be always greater than a^3.

Hence Quantity II > Quantity I, Also 6a^2 = 4πr^2 = 6πh^2, a = (√π)/h.

Volume of the cylinder = πh^2(2h) = 2πh^3 = 2π(a/√π)^3 = 2a^3/(√π) > a^3.

Hence Quantity III > Quantity I. Volume of sphere = (√6/√π) a^3 is > volume of the cylinder = 2a^3/ (√π).

Hence, Quantity II > Quantity III > Quantity IIncorrectAnswer :

**2. Quantity II > Quantity III > Quantity I**

Explanation-

Let the cube’s edge length is a, the surface area of cube = 6a^2 and its volume = a^3.

Let the radius of the sphere is r, then its volume = (4/3) πr^3 and its surface area will be 4πr^2.

Let the height of the cylinder be 2h therefore, its radius will be h.

Surface area of the cylinder = 2πh (2h) +2πh^2 = 6πh^2.

Now since the surface area are same, 6a^2 = 4πr^2 = 6πh^2, therefore, r= (√3/√ (2π)) a.

Volume of sphere = (√6/√π)a^3, this will be always greater than a^3.

Hence Quantity II > Quantity I, Also 6a^2 = 4πr^2 = 6πh^2, a = (√π)/h.

Volume of the cylinder = πh^2(2h) = 2πh^3 = 2π(a/√π)^3 = 2a^3/(√π) > a^3.

Hence Quantity III > Quantity I. Volume of sphere = (√6/√π) a^3 is > volume of the cylinder = 2a^3/ (√π).

Hence, Quantity II > Quantity III > Quantity I - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**There are 5 red balls, 6 green balls, 9 blue balls and remaining yellow balls in a lot of 50 balls.**

Quantity I: Probability of picking 2 balls such that one is green and the other is blue.

Quantity II: Probability of picking up three balls such that at least one of them is red.

Quantity III: Probability of picking 3 balls such that at least one is blue.CorrectAnswer :

**4. Quantity III > Quantity II > Quantity I**

**Explanation-**

Quantity I: 6*9/(50C2) = 54/1225.

Quantity II = 1- 45C3/(50C3)= 1 – 14190/19600 = 541/1960 > Quantity I .

Quantity III = 1 – 41C3/(50C3)= 1 – 10660/19600 = 894/1960 > Quantity II > Quantity I.

Hence, Quantity III > Quantity II > Quantity IIncorrectAnswer :

**4. Quantity III > Quantity II > Quantity I**

**Explanation-**

Quantity I: 6*9/(50C2) = 54/1225.

Quantity II = 1- 45C3/(50C3)= 1 – 14190/19600 = 541/1960 > Quantity I .

Quantity III = 1 – 41C3/(50C3)= 1 – 10660/19600 = 894/1960 > Quantity II > Quantity I.

Hence, Quantity III > Quantity II > Quantity I - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**what will be the increase percent in volume?**

Statement I: An ice cube is dip into 10 litre water

Statement II: If radius of a cylinder increases by 20% and height increases by 10%

Statement III: If the height of ice cube is 10 cm.CorrectAnswer :

**2. II alone and I and III together**

Explanation-

Volume = π*r*r*h

so by successive formula, the increase in volume can be found

First: r*r

So 20 + 20 + (20) (20)/100 = 44%

Next: 44 + 10 + (44)*(10)/100 – final increase in volume

From I and III:

volume of given shape (cylinder, cube, cone etc.) = 10 l (initial volume). And ice cube of height 10 cm is dip into it.

So after finding new volume or increase in volume after the ice cube is dipped, % increase in volume can be found.IncorrectAnswer :

**2. II alone and I and III together**

Explanation-

Volume = π*r*r*h

so by successive formula, the increase in volume can be found

First: r*r

So 20 + 20 + (20) (20)/100 = 44%

Next: 44 + 10 + (44)*(10)/100 – final increase in volume

From I and III:

volume of given shape (cylinder, cube, cone etc.) = 10 l (initial volume). And ice cube of height 10 cm is dip into it.

So after finding new volume or increase in volume after the ice cube is dipped, % increase in volume can be found. - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude**How many students failed in class 6?**

Statement I: 400 students passed in class.

Statement II: Number of students failed in class 6 is 10% of those of failed in class 7?

Statement III: 75% of the students who appeared in examination have passed either in class 8.Correctanswer :

**4. None of these**

Explanation-

Data is not sufficient to find the number of failed students in class 6.Incorrectanswer :

**4. None of these**

Explanation-

Data is not sufficient to find the number of failed students in class 6. - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative Aptitude**A bag contains 3 red, 4 green and 2 blue balls. Two balls are drawn at random.**

Quantity I: Probability that none of the ball drawn is blue.

Quantity II: Fraction of work completed by A in 7 days if he is 20% more efficient than B who can complete the work in 12 days.CorrectAnswer :

**3. Quantity II > Quantity I**

Explanation-

Solution: I: 7C2/9C2=7/12

II: A-> 10 days => fraction of work in 7 days = 7/10

Hence II > IIncorrectAnswer :

**3. Quantity II > Quantity I**

Explanation-

Solution: I: 7C2/9C2=7/12

II: A-> 10 days => fraction of work in 7 days = 7/10

Hence II > I - Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**Q. (6-10) in college A, B, C, D, and E students watch different-different serials, there are three serials GOT, SUITS and FRIENDS. Number of students who watch particular serial and from particular college are shown above, Answer the following questions-**

**From college D, out of the students watching FRIENDS, 40% got selected for State level’ quiz competition. Out of which 25% further got selected for National level’s quiz competition. From college E, out of the students watching FRIENDS, 33 1/3% got selected for State level’s quiz competition, out of which two-fifth further got selected for National level’s quiz competition. What is the total number of students watching FRIENDS from these two colleges who got selected for National level’s quiz competition?**CorrectAnswer :

**3. 50**

ExplanationTotal student in D who watch FRIENDS are 300, student selected for state level quiz is 40% of 300=120

For national level is 25% of 120=30

Total student who watch FRIENDS are 150, for state level quiz 33(1/3)% of 150= 50, for national level are two fifth of 50 is 20

Total student from both colleges selected for national level quiz are 50IncorrectAnswer :

**3. 50**

ExplanationTotal student in D who watch FRIENDS are 300, student selected for state level quiz is 40% of 300=120

For national level is 25% of 120=30

Total student who watch FRIENDS are 150, for state level quiz 33(1/3)% of 150= 50, for national level are two fifth of 50 is 20

Total student from both colleges selected for national level quiz are 50 - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative Aptitude**Q. (6-10) in college A, B, C, D, and E students watch different-different serials, there are three serials GOT, SUITS and FRIENDS. Number of students who watch particular serial and from particular college are shown above, Answer the following questions-**

**Total number of students watch suits from all colleges together is approximately what percent of the total number of students watch GOT from all colleges together?**CorrectAnswer :

**3. 95.23**

Explanation

Total no of student who watch SUITS= 200+300+250+150+100=1000

Total no of students who watch GOT=250+100+150+200+350=1050

1000/1050=95.23IncorrectAnswer :

**3. 95.23**

Explanation

Total no of student who watch SUITS= 200+300+250+150+100=1000

Total no of students who watch GOT=250+100+150+200+350=1050

1000/1050=95.23 - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**Q. (6-10) in college A, B, C, D, and E students watch different-different serials, there are three serials GOT, SUITS and FRIENDS. Number of students who watch particular serial and from particular college are shown above, Answer the following questions-**

**If out of the students watching SUITS from colleges B, D and E 16.66 %, 6.66%, and 20% respectively got selected for state level’s quiz competition, what was the total number of students watching SUITS got selected for State level’s quiz competition from these colleges together?**CorrectAnswer :

**1. 80**

Explanation

Students selected for state level from college B = 16(2/3)% of 300=50

Students selected for state level from college D = 6(2/3)% of 150=10

Students selected for state level from college E = 20% of 100= 20

Total = 50+10+20= 80IncorrectAnswer :

**1. 80**

Explanation

Students selected for state level from college B = 16(2/3)% of 300=50

Students selected for state level from college D = 6(2/3)% of 150=10

Students selected for state level from college E = 20% of 100= 20

Total = 50+10+20= 80 - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude

**If the number of students watching each serial in college C is increased by 15% and the number of students watching each serial in college E is decreased by 5%, then what will be the difference between number of students in college C and E?**CorrectAnswer :

**2. 120**

Explanation-

Total no of student after increment in C is =600(115/100)=690

Total no of student after decrement in E is=600(95/100)=570

Difference = 690-570= 120IncorrectAnswer :

**2. 120**

Explanation-

Total no of student after increment in C is =600(115/100)=690

Total no of student after decrement in E is=600(95/100)=570

Difference = 690-570= 120 - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude

**Total number of students watching GOT and SUITS together from college A is what percent of the total number of students watching these two serials together from college D?**CorrectAnswer :

**1. 128(4/7)**

Explanation-

Student from A who watch these two serials =450

Student from D who watch these two serials =350

Required answer=(450/350)×100=128(4/7)IncorrectAnswer :

**1. 128(4/7)**

Explanation-

Student from A who watch these two serials =450

Student from D who watch these two serials =350

Required answer=(450/350)×100=128(4/7)

**Click view Questions button to view Explanation****Note:-**We are providing unique questions for you to practice well, have a try !!- Ask your doubt in comment section, AC Mod’s ll clear your doubts in caring way.
- If you find any mistake, please tell us in the comment section.

**Bank Jobs Notification**

**Central Govt Jobs Notification**