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

1 pointsCategory: Quantitative Aptitude**Find the wrong number in the series.**

235, 1235, 2576, 9425, 17425CorrectAnswer –

**2) 2566**

Explanation

235 + (2+3+5)^{3}= 1235

1235 + (1+2+3+5)^{3}= 2566

2566 + (2+5+6+6)^{3}= 9425

9425 + (9+4+2+5)^{3}= 17425IncorrectAnswer –

**2) 2566**

Explanation

235 + (2+3+5)^{3}= 1235

1235 + (1+2+3+5)^{3}= 2566

2566 + (2+5+6+6)^{3}= 9425

9425 + (9+4+2+5)^{3}= 17425 - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**What approximate value should be in question mark ?**

? = 1/9 of 3/7 of 7/10 of 1857 + 4/5 of 35/48 of 9/20 of 242 + 1/7 of 7/1 + 9/2 of 2/9 – (1024)^{1/10 }CorrectAnswer –

**4) 125**

Explanation

= 1/9 x 3/7 x 1857 + 4/5 x 35/48 x 9/20 x 242 +1/7 x 7/1 + 9/2 x 2/9 -2 10 x1/10

= 62 +63 +1 +1 -2

= 125IncorrectAnswer –

**4) 125**

Explanation

= 1/9 x 3/7 x 1857 + 4/5 x 35/48 x 9/20 x 242 +1/7 x 7/1 + 9/2 x 2/9 -2 10 x1/10

= 62 +63 +1 +1 -2

= 125 - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**If A is divisible by 99 and X both and B is divisible by 45 and Y both then what is the average of X and Y where A, B, X and Y are natural numbers?**

1) A lies between 550 and 600 and Q lies between 500 and 550.

2) Difference between Y and X is 6.CorrectAnswer –

**4) If data in both 1 and 2 together are sufficient to answer**

Explanation

From statement 1

A = multiple of X and 99

Value of A between 550 and 600 = 594 which is multiple of 6

X can be 1,2,3 or 6

B = Multiple of Y and 45

Value of Y = 1,2,3,4,5,6 or 12.

So 1 alone is not sufficient to answerFrom statement 2

Data itself is not sufficient because range of A and B is not given.

From statement 1 and 2 together we have Y= 12 and X=6

So average of X and Y = (18/2) = 9

So data in both 1 and 2 together is sufficient to answerIncorrectAnswer –

**4) If data in both 1 and 2 together are sufficient to answer**

Explanation

From statement 1

A = multiple of X and 99

Value of A between 550 and 600 = 594 which is multiple of 6

X can be 1,2,3 or 6

B = Multiple of Y and 45

Value of Y = 1,2,3,4,5,6 or 12.

So 1 alone is not sufficient to answerFrom statement 2

Data itself is not sufficient because range of A and B is not given.

From statement 1 and 2 together we have Y= 12 and X=6

So average of X and Y = (18/2) = 9

So data in both 1 and 2 together is sufficient to answer - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude**The average marks obtained by 45 students in a class is 80. The difference between the marks who got the highest mark and the student got the lowest mark is 99. If both of these students are not considered, the average of the class falls by 1 mark. Find the highest mark**CorrectAnswer –

**2) 151**

Explanation

Total marks 80 x 45 = 3600

Highest(H) – Lowest (L) = 99 (i)

Total marks of 43 students = 43 x 79 = 3397

3397+H+L/45 = 3600

H+ L = 203 (ii)

From (i) and (ii)

H = 151IncorrectAnswer –

**2) 151**

Explanation

Total marks 80 x 45 = 3600

Highest(H) – Lowest (L) = 99 (i)

Total marks of 43 students = 43 x 79 = 3397

3397+H+L/45 = 3600

H+ L = 203 (ii)

From (i) and (ii)

H = 151 - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative Aptitude**The difference between S.I and C.I for two years on a sum of $ 200000 at 12% p.a is is $x. On what sum is the difference between S.I and C.I for two years at 16% p.a. equals S.I and C.I for two years at 16% p.a. equal to $x?**CorrectAnswer –

**1. $112500**

Explanation

As the difference between S.I and C.I is $x we have

Difference = P(r/100)2

200000(12/100)2 = x= P(16/100)2

200000 x 9/16 = $ 112500IncorrectAnswer –

**1. $112500**

Explanation

As the difference between S.I and C.I is $x we have

Difference = P(r/100)2

200000(12/100)2 = x= P(16/100)2

200000 x 9/16 = $ 112500 - Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**The average monthly salary per head of the entire staff of a college is $20,400. The average salary per head of the four HOD is $24000 while those who are not HOD have an average salary of $18000 per head. The entire staff of the college, barring the HOD contribute a monthly salary to orphanage of 10,000 children. Out of this contribution if each boys get $12 and each girl gets $8, then how many boys are there in the orphanage?**CorrectAnswer –

**1) 7000**

Explanation

IncorrectAnswer –

**1) 7000**

Explanation

- Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative AptitudeQuantity I: What is the average score of Virat after 60 matches when he scored 120 runs in his 60th ODI and thus his average increased by 1.2 runs?

Quantity II: Rohan completed 4/5th of the total work in 16 days when he is working at 1/3rd of his efficiency. In how many days he will complete the entire work when he is working at his normal efficiency ?CorrectAnswer –

**1) Quantity I > Quantity II**

Explanation

Quantity I

Virat average = 59n + 120 = 60 (n + 1.2)

N = 48

Average after 60th ODI = 48+1.2 = 49.2

Quantity II

Total number of days Rohan completes work at 1/3rd of his efficiency = 16 x 5/4 = 20days

Rohan completes at normal efficiency = 20/3 = 6.67 daysIncorrectAnswer –

**1) Quantity I > Quantity II**

Explanation

Quantity I

Virat average = 59n + 120 = 60 (n + 1.2)

N = 48

Average after 60th ODI = 48+1.2 = 49.2

Quantity II

Total number of days Rohan completes work at 1/3rd of his efficiency = 16 x 5/4 = 20days

Rohan completes at normal efficiency = 20/3 = 6.67 days - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**Manish buys a certain number of oranges at 12 per Rupee 9 and the same number at 18 per Rupee 9. If she sells them at 18 per Rupee 15, does she gain or lose and by what percentage ?**CorrectAnswer –

**1) 33.33%**

Explanation

Let the number of oranges that Manish bought be 2X

i.e. X oranges at each of the 2 prices.

=X (9/12) = 3/4X

Similarly C.P of X oranges at 18 per Rupee 9

= X (9/18) = X/2

C.P of the oranges = 3X/4 + X/2 = 5X/4

S.P of the oranges = 2X.15/18 = 5X/3

Profit % = 5X/3-5X/4 / 5X/4 x 100(because 5X/3 > 5X/4)

= 33.33 %IncorrectAnswer –

**1) 33.33%**

Explanation

Let the number of oranges that Manish bought be 2X

i.e. X oranges at each of the 2 prices.

=X (9/12) = 3/4X

Similarly C.P of X oranges at 18 per Rupee 9

= X (9/18) = X/2

C.P of the oranges = 3X/4 + X/2 = 5X/4

S.P of the oranges = 2X.15/18 = 5X/3

Profit % = 5X/3-5X/4 / 5X/4 x 100(because 5X/3 > 5X/4)

= 33.33 % - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude**In a tank there is a leak, which can empty the tank in 20 hours, at 3/4th of the height above the base. There are two pipes filling the tank which can fill the tank in 6 and 12 hours respectively. How much time does it take, if both the filling pipes are opened simultaneously .**CorrectAnswer –

**2) 4 ¼ hours**

Explanation

The two pipes can fill 1/6 + 1/12 = 1/4 th of the tank in one hour 3/4th of the tank in 3 hours.

At a height which is 3/4 th of the total height of the tank there is a leak, which empties 1/20th of the tank in 1 hour.

i.e. 1/4th is filled at (1/6+1/12-1/20)th per hour i.e. (1/5)th per hour

The last (1/4)th is filled in 1 hour 15 minutes = 4 ¼ hoursIncorrectAnswer –

**2) 4 ¼ hours**

Explanation

The two pipes can fill 1/6 + 1/12 = 1/4 th of the tank in one hour 3/4th of the tank in 3 hours.

At a height which is 3/4 th of the total height of the tank there is a leak, which empties 1/20th of the tank in 1 hour.

i.e. 1/4th is filled at (1/6+1/12-1/20)th per hour i.e. (1/5)th per hour

The last (1/4)th is filled in 1 hour 15 minutes = 4 ¼ hours - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude**A person covered the first 60 km of his journey at 80 km/hr, the next 50 km of his journey at 40 km/hr and the next 60 km of his journey at 50 km/hr and the remaining distance in 1 hour. If the average speed for the entire journey is 55 km/hr, find the distance covered in the last stretch of the journey.**CorrectAnswer –

**3) 61 km**

Explanation

The sum of time taken for the first, second and third parts of the journey =

(60/80+50/40+60/50)hrs = 3 1/5 hrs

Total time taken = 3 1/5 + 1 = 4 1/5 hrs

Total distance = 55 x 4 1/5 = 231 km

Distance covered in last stretch

= 231- (60+50+60)

= 61 kmIncorrectAnswer –

**3) 61 km**

Explanation

The sum of time taken for the first, second and third parts of the journey =

(60/80+50/40+60/50)hrs = 3 1/5 hrs

Total time taken = 3 1/5 + 1 = 4 1/5 hrs

Total distance = 55 x 4 1/5 = 231 km

Distance covered in last stretch

= 231- (60+50+60)

= 61 km

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