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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

NDTV is a very popular TV channel. It telecasts the programmes from 8 : 00 a.m. to 12 : 00 p.m. It telecasts 60 advertisements each of 8 seconds and 16 advertisements each of 30 seconds. What is the percentage of time devoted in a day for the advertisements?

Correct

Explanation:
Percentage of time devoted in a day for advertisement
= (60*8+16*30)/(16*60*60)= 1.66%

Incorrect

Explanation:
Percentage of time devoted in a day for advertisement
= (60*8+16*30)/(16*60*60)= 1.66%

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

The marked price of an article is increased by 25% and the selling price is increased by 16.66%, then the amount of profit doubles. If the original marked price be Rs. 400 which is greater than the corresponding cost price by 33.33%, what is the increased selling price?

Correct

Explanation:
Let C.P=100 and profit= X………..S.P=100+X
Marked price =133.33
After change, C.P=100 and profit=2X
S.P.= (100+X)*1.1666
(100+X)*1.1666—100=2X
After solving X=20%
Increased selling price= 400*100/125+ 2* 20% 0f 300=420

Incorrect

Explanation:
Let C.P=100 and profit= X………..S.P=100+X
Marked price =133.33
After change, C.P=100 and profit=2X
S.P.= (100+X)*1.1666
(100+X)*1.1666—100=2X
After solving X=20%
Increased selling price= 400*100/125+ 2* 20% 0f 300=420

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

A property dealer bought a rectangular plot (of land) in Noida 5 years ago at the rate of Rs. 1000 per m2. The cost of plot is increases by 5% in every 6 years and the worth of a rupee falls down at a rate of 2% in every 5 years. What is the approximate value of the land per sq meter 25 years hence?

Correct

Explanation:
Total time (25+5) years
No. of time period for constant increment =30/6=5
No. of time for rupee depreciation=30/5=6
Value of plot=100*1.05^5*0.98^6=Rs. 1134/-

Incorrect

Explanation:
Total time (25+5) years
No. of time period for constant increment =30/6=5
No. of time for rupee depreciation=30/5=6
Value of plot=100*1.05^5*0.98^6=Rs. 1134/-

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

Kaushalya can do a work in 20 days, while Kaikeyi can do the same work in 25 days. The started the work jointly. Few days later Sumitra also joined them and thus all of them completed the whole work in 10 days. All of them were paid total Rs. 700. What is the share of Sumitra?

Correct

Explanation:
Efficiency of kaushalya=5%
Efficiency of kaikeyi=4%
Thus, in 10 days working together they will complete only 90% of the work
= (5+4)*10=90
Hence, the remaining work will surely done by sumitra i.e. 10% of work
Sumitra will get 10% of Rs. 700 i.e. Rs. 70

Incorrect

Explanation:
Efficiency of kaushalya=5%
Efficiency of kaikeyi=4%
Thus, in 10 days working together they will complete only 90% of the work
= (5+4)*10=90
Hence, the remaining work will surely done by sumitra i.e. 10% of work
Sumitra will get 10% of Rs. 700 i.e. Rs. 70

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

The speeds of Vimal and Kamal are 30 km/h and 40 km/h. Initially Kamal is at a place L and Vimal is at a place M. The distance between L and M is 650 km. Vimal started his journey 3 hours earlier than Kamal to meet each other. If they meet each other at a place P somewhere between L and M, then the distance between P and M is :

Correct

Explanation:
In first 3 hours, vimal covers=30*3=90km
Rest distance=650—90=560
Both travel together, towards each other, time= 560/(40+30)=8 h
Vimal travels total time= 3+8=11h
Distance travelled by vimal=11*30=330km

Incorrect

Explanation:
In first 3 hours, vimal covers=30*3=90km
Rest distance=650—90=560
Both travel together, towards each other, time= 560/(40+30)=8 h
Vimal travels total time= 3+8=11h
Distance travelled by vimal=11*30=330km

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

Antriksha fires two bullets from the same place at an interval of 6 minutes but Bhagwat sitting in a car approaching the place of firing hears the second fire 5 minute 32 seconds after the first firing. What is the speed of car, if the speed of sound is 322 m/s?

Correct

Explanation:
Speed of wind/speed of car= time utilized/time saved
332/X=(5*60+32)/{6*60–(5*60+32)}
X=28m/s

Incorrect

Explanation:
Speed of wind/speed of car= time utilized/time saved
332/X=(5*60+32)/{6*60–(5*60+32)}
X=28m/s

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

A rectangular water reservoir is 15 m by 12 m at the base. Water flows into it through a pipe whose cross section is 5 cm by 3 cm at the rate of 16 m per second. Find the height to which the water will rise in the reservoir in 25 minutes:

Correct

Explanation:
Volume of water which flow in 25 min=25*60*0.05*0.03*16=36
Rise in water level=36/(15*12)=0.2m

Incorrect

Explanation:
Volume of water which flow in 25 min=25*60*0.05*0.03*16=36
Rise in water level=36/(15*12)=0.2m

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

Directions (7-10): Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between ‘x’ and ‘y’ and give answer.
(1) if x > y
(2) if x < y
(3) if x >= y
(4) if x=< y
(5) if x = y or no relation can be established between ‘x’ and ‘y’.
I. 6x2 - 19x + 15 = 0
II. 10y2 - 29y + 21 = 0

Correct

Explanation:
X=5/3, 3/2 and Y=3/2,7/5………….X=

Incorrect

Explanation:
X=5/3, 3/2 and Y=3/2,7/5………….X=

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

Directions (7-10): Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between ‘x’ and ‘y’ and give answer.
(1) if x > y
(2) if x < y
(3) if x >= y
(4) if x=< y
(5) if x = y or no relation can be established between ‘x’ and ‘y’.
I. 12x2 + 11x - 56 = 0
II. 4y2 - 15y + 14 = 0

Correct

Explanation:
X=-8/3, 7/4 and y= 2,7/4……….. X=

Incorrect

Explanation:
X=-8/3, 7/4 and y= 2,7/4……….. X=

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

Directions (7-10): Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between ‘x’ and ‘y’ and give answer.
(1) if x > y
(2) if x < y
(3) if x >= y
(4) if x=< y
(5) if x = y or no relation can be established between ‘x’ and ‘y’.
I. 3x2 + 13x + 12 = 0
II. y2 + 9y + 20 = 0 64