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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

Directions.Q(1-5) The line graph below shows the distance(km) travelled by different boats upstream and downstream in same time.

The radar graph shows the speed of stream for boats.

Speed of boat D in still water is how much percentage of speed of the boat A in downstream direction?

Correct

Answer – 3) 80%
Explanation:
As it is given upstream time=downstream time
To find speed of D in still water=110/(x-8) = 330/(x+8);X=16
To find speed of the boat A in still water=120/(Y-5)= 240/(Y+5);Y=15
Speed of A in downstream direction = Speed of stream + speed of boat in still water = 5+15=20
Required answer
16=?*20/100=1600/20=80

Incorrect

Answer – 3) 80%
Explanation:
As it is given upstream time=downstream time
To find speed of D in still water=110/(x-8) = 330/(x+8);X=16
To find speed of the boat A in still water=120/(Y-5)= 240/(Y+5);Y=15
Speed of A in downstream direction = Speed of stream + speed of boat in still water = 5+15=20
Required answer
16=?*20/100=1600/20=80

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

Directions.Q(1-5) The line graph below shows the distance(km) travelled by different boats upstream and downstream in same time.

The radar graph shows the speed of stream for boats.

Distance between the points R and S is 360km then what is the difference between time taken by boat B for downstream and upstream journey to the time taken by D for downstream and upstream journey?

Correct

Answer – 3) 9 hours
Explanation:
Time is equal in downstream and upstream
Speed of the boat D is calculated in question number 6
To find speed of the boat B in still water (X)
100/(X-7)=240/(X+7)
X=17
Time taken by B to travel 360 km both ways = (360/(17+7))+(360/(17-7))=51
Time taken to travel by D in both ways= (360/ (16+8))+(360/(16-8))=60
Difference between points= 60-51=9

Incorrect

Answer – 3) 9 hours
Explanation:
Time is equal in downstream and upstream
Speed of the boat D is calculated in question number 6
To find speed of the boat B in still water (X)
100/(X-7)=240/(X+7)
X=17
Time taken by B to travel 360 km both ways = (360/(17+7))+(360/(17-7))=51
Time taken to travel by D in both ways= (360/ (16+8))+(360/(16-8))=60
Difference between points= 60-51=9

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

Directions.Q(1-5) The line graph below shows the distance(km) travelled by different boats upstream and downstream in same time.

The radar graph shows the speed of stream for boats.

The speed of boat F in still water is 50% more than speed of boat A IN downstream and F travels 62.5km upstream and 210 km downstream in 8 hours and 30 min .what is the speed of stream of F?

Correct

Answer – 2) 5
Explanation:
To find speed of boat A in still water=X
120/(X-5) =240/(X+5)
X=15
Speed of A in downstream=15+5=20
Speed of F in still water=150*speed of A in downstream /100 = 150*20/100=30
Speed of stream of F=Y
(210/(30+Y)) +(62.5/(30-Y))=8.5
Y=5

Incorrect

Answer – 2) 5
Explanation:
To find speed of boat A in still water=X
120/(X-5) =240/(X+5)
X=15
Speed of A in downstream=15+5=20
Speed of F in still water=150*speed of A in downstream /100 = 150*20/100=30
Speed of stream of F=Y
(210/(30+Y)) +(62.5/(30-Y))=8.5
Y=5

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

Directions.Q(1-5) The line graph below shows the distance(km) travelled by different boats upstream and downstream in same time.

The radar graph shows the speed of stream for boats.

What is the ratio of sum of speed of boats C and E in still water to ratio of sum of speed of the boat A and B in still water?

Correct

Answer – 3) 11/8
Explanation:
Speed of boat A in still water =120/x-5=240/x=5
X=15
By the same method speed of the boats are calculated as A=15 ,B=17,C=30,E=14
Ratio=(30+14)/(17+15)=11/8

Incorrect

Answer – 3) 11/8
Explanation:
Speed of boat A in still water =120/x-5=240/x=5
X=15
By the same method speed of the boats are calculated as A=15 ,B=17,C=30,E=14
Ratio=(30+14)/(17+15)=11/8

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

Directions.Q(1-5) The line graph below shows the distance(km) travelled by different boats upstream and downstream in same time.

The radar graph shows the speed of stream for boats.

If the speed of the boat E in still water is increased by 14.28% then find the ratio of time taken by E to travel 414 km downstream to upstream

Correct

Answer – 1) 9/23
Explanation:
Speed of E in still water=140/(x-7)= 220/(x+7); X=14
X is increased by 14.28%
14.28% = 1/7
Hence X will become 16
Time taken by E downstream/time taken by E in upstream = {414/(16+7)}/{414/(16-7)} =9/23

Incorrect

Answer – 1) 9/23
Explanation:
Speed of E in still water=140/(x-7)= 220/(x+7); X=14
X is increased by 14.28%
14.28% = 1/7
Hence X will become 16
Time taken by E downstream/time taken by E in upstream = {414/(16+7)}/{414/(16-7)} =9/23

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

The metro station has a train from Ghatkopar to Varsova and from Varsova to Ghatkopar for every 10 min .The first one is at 5 am. The trip takes 30 min from both sides and all the trains travels at same speed. How many train you will pass while going from Ghatkopar to Varsova if the train start at 12pm?

Correct

Answer – 2) 5
Explanation:
Every 10 min train is going from both stations and it takes 30 minutes for travel
When train starts from Ghatkopar at 12pm it will cross 1st train coming from varsova at time 11.40am and so on it will cross trains by the time of 12pm
So the number of trains from varsova from 11.40am, 11.50am, 12pm, 12.10pm, 12.20pm. So it will cross 5 trains.

Incorrect

Answer – 2) 5
Explanation:
Every 10 min train is going from both stations and it takes 30 minutes for travel
When train starts from Ghatkopar at 12pm it will cross 1st train coming from varsova at time 11.40am and so on it will cross trains by the time of 12pm
So the number of trains from varsova from 11.40am, 11.50am, 12pm, 12.10pm, 12.20pm. So it will cross 5 trains.

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

The ratio of ages of Yamuna and Jamuna is 18/13 and Tanu and Manu is 15/18. Average age of the four is 32 years. And ratio of sum of ages of Yamuna and Jamuna to sum of ages of Tanu and Manu is 31/33. Find out the average age of Yamuna, Tanu and Manu after 10 years from now.

Correct

Answer – 3) 44
Y/J=18x/13x — (1)
T/M=15y/18 y — (2)
Y+J/T+M =31/33 –(3)
By 3rd equation we can conclude x=y
(Y+T+M+J)/4 =32
Y+T+M+J= 128 (4)
By using 1, 2,3 in 4
18x+13x+18y+15y=128
X=2 as (x=y)
18x+15y+18Y=51x
Average of Y ,T ,M after 10 years
51 x+10*3/3
51*2+10*3/3=44

Incorrect

Answer – 3) 44
Y/J=18x/13x — (1)
T/M=15y/18 y — (2)
Y+J/T+M =31/33 –(3)
By 3rd equation we can conclude x=y
(Y+T+M+J)/4 =32
Y+T+M+J= 128 (4)
By using 1, 2,3 in 4
18x+13x+18y+15y=128
X=2 as (x=y)
18x+15y+18Y=51x
Average of Y ,T ,M after 10 years
51 x+10*3/3
51*2+10*3/3=44

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

800 students appeared for IBPS PO exam. 60% of the boys and 80% of the girls cleared the cut off. Total percentage of student qualifying exam is 65%.What is the ratio of girls who cleared the cut off to number of boys who could not clear cut off.

Correct

Answer – 3) 2/3
By using alligation and mixture

So ratio of boys /girls=15/5=3/1
No of boys =800*3/4=600
No of girls=800*1/4=200
No of girls clearing cut off=200*80/100=160
No of boys not clearing cut off=600*40/100 =24
Ratio=160/240

Incorrect

Answer – 3) 2/3
By using alligation and mixture

So ratio of boys /girls=15/5=3/1
No of boys =800*3/4=600
No of girls=800*1/4=200
No of girls clearing cut off=200*80/100=160
No of boys not clearing cut off=600*40/100 =24
Ratio=160/240

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

In an examination there are three subjects hindi, English and maths.8% of the total students passed in all the subjects ,15% passed in only maths and hindi,10% passed in only hindi and English ,5% passed in only maths and English. 43% passed in maths , 40% passed in English and 60% passed in hindi .135 students failed in all the subjects then find the no of students who passed in all the subjects.

Correct

Answer – 2) 360
Explanation:

So the percentage of student failed in all the subjects is 3%
By diagram
3=135
8=?
8*135/3=360

Incorrect

Answer – 2) 360
Explanation:

So the percentage of student failed in all the subjects is 3%
By diagram
3=135
8=?
8*135/3=360

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

Two jars having capacity of 6 and 10 liters respectively are filled with mixture of milk and water.In smaller jar 25% of mixture is milk and in larger 25% of mixture is water. The jars are emptied into a 20L drum and remaining capacity of drum is filled up with water. Find the percentage of milk in the drum.

Correct

Answer – 2) 45%
Explanation:
From smaller jar 25%is milk=6*25/100=6/4
In bigger jar 75% is milk=10*75/100=30/4
Quantity of milk in drum=6/4+30/4=9
Total capacity of both jars is 16L
Water quantity in both jars is 16-9=7
Total water in drum will be 7+4=11
% of milk in drum =9*100/20=45%

Incorrect

Answer – 2) 45%
Explanation:
From smaller jar 25%is milk=6*25/100=6/4
In bigger jar 75% is milk=10*75/100=30/4
Quantity of milk in drum=6/4+30/4=9
Total capacity of both jars is 16L
Water quantity in both jars is 16-9=7
Total water in drum will be 7+4=11
% of milk in drum =9*100/20=45%

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