# SBI PO 2018 Main Exam: Quants Test Day 3

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- 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)

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