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## Breaking

QUANT PRACTICE SET 37 FOR SBI PO PRELIM 2020

SBI Clerk and SBI PO Prelims exam are going to be held in the upcoming months. We have already provided you with the PDFs of many topics of Quantitative Aptitude like Simplification/Approximation, Number Series,datainterpretation.important Arithmetic questions to prepare for SBI Clerk and PO Pre exam. Practicing these questions will help you to know about the level of the questions. To increase your speed and accuracy, enhance your calculations.

START QUIZ

a.       13 2/3 hours

b.      11 7/8 hours
c.       16 ¾ hours
d.      15 5/9 hours
e.       None of these

a.       14 2/5 hours
b.      12 ¾ hours
c.       11 5/6 hours
d.      13 2/3 hours
e.       None of these

3.      Three pipes A, B and C can fill a tank in 10 hours. After working at it together for 3 hours, C is closed and A and B can fill the remaining part in 14 hours. How much time taken by C to fill the tank alone?
a.       18 hours
b.      20 hours
c.       24 hours
d.      22 hours
e.       None of these

4.      Three pipes A, B, C can fill a cistern in 10, 15, 18 hours respectively. If pipe A is kept open all the time and pipe B and C is opened alternatively for an hour then find in how many hours the tank will be filled?
a.       7 hours 25 mins
b.      5 hours 40 mins
c.       6 hours 12 mins
d.      7 hours 48 mins
e.       None of these

5.      Pipe A and Pipe B can fill a tank in 10 hours and 15 hours respectively while Pipe C can empty the full tank in 18 hours. If pipe A is opened at 7 am, Pipe B opened at 8:30 am and Pipe C at 10 am, then after 10 am, how much more time will be taken by all three pipes together to fill the tank?
a.       7 hour 18 min
b.      8 hour 36 min
c.       6 hour 42 min
d.      5 hour 24 min
e.       None of these

6.      P, Q and R can fill the tank in 12, 16 and 24 min respectively. All three began to fill the tank together but P and Q left 3 and 4 min respectively before filling the tank. Find the total time taken by all of them to fill the tank?
a.       12 min
b.      16 min
c.       8 min
d.      10 min
e.       None of these

7.      Two pipes A and B can fill a tank in 15 minutes and 25 minutes respectively. Both pipes are opened together andpipeBisclosed,5 minutes before the tank is filled completely. Calculate the total time required to fill the tank?
a.       11 ¼ min
b.      13 3/5 min
c.       12 7/8 min
d.      14 5/6 min
e.       None of these

8.      B is 5/4 times as efficient as A. If A can fill the 3/5 of the tank in 15 min, what fraction of the capacity of the tank would remain incomplete if B can fill the tank independently for 10 min only?
a.       2/3
b.      1/3
c.       ¼
d.      ½
e.       None of these

9.      P, Q and R can fill the tank in 15, 18 and 20 min respectively. All three began to fill the tank together but P and Q left 4 and 5 min respectively before filling the tank. In how much time will take to fill the tank?
a.       7 22/25 min
b.      8 30/31 min
c.       7 11/15 min
d.      8 4/7 min
e.       None of these

10.  Two inlet pipes can fill an empty tank in 15 and 18 hours and one outlet pipe can empty the tank in 20 hours. If all the pipes opened simultaneously, then how many hours required fill the full tank?
a.       11 8/11 hours
b.      12 5/7 hours
c.       13 11/13 hours
d.      9 5/9 hours
e.       None of these

a.       5 min
b.      6 min
c.       4 min
d.      4.5 min
e.       None of these

a.       6
b.      14
c.       10
d.      12
e.       8

a.       17 1/2
b.      19 1/3
c.       18 1/3
d.      18 3/5
e.       None of these

a.       33 1/11
b.      32 1/11
c.       33 1/3
d.      32 1/3
e.       11 1/11

15) Three pipes can fill an empty tank in 4 hours. If there is leakage in the tank it takes 1 hour more, then find the no of hours required to empty the full tank by leakage pipe alone
a.       12
b.      15
c.       24
d.      18
e.       20
SOLUTIONS:

C can empty the tank in = (21/70)*100= 30 hours
Total units of work= 300
A’s one hour work = 300/20 = 15 units
B’s one hour work = 300/25 = 12 units
C’s capacity to empty in one hour = 300/30 = 10 units
When three pipes are opened simultaneously then in one hour they will fill,
(15+12-10) = 17 units
In 12 hours they will fill = 17*12= 204 units
After 12hrs, C will be closed
Time taken by A& B to fill the remaining tank = 96/27 = 3 15/27 = 3 5/9 hours
Total time will be = 12 + 3 5/9 = 15 5/9 hours

C can empty the tank in =18/60*100= 30 hours
Total units of work= 360
A’s one hour work= 360/18= 20units
B’s one hour work = 360/24= 15 units
C’s capacity to empty in one hour= 360/30= 12 units
When three pipes are opened simultaneously then in one hour they will
(20+15-12)=23 units
In 12 hours they will fill= 23*12= 276 units
After 12 hr, C will be closed
Time taken by A & B to fill the remaining tank = 84/35= 2 2/5 hours
Total time = 12 + 2 2/5 = 14 2/5 hours

Three pipes A, B and C can fill a tank in 8 hours.
A, B and C’s 1 hour work=1/10
A, B and C’s 3 hour work= 3/10
Remaining work= 1 – (3/10) = 7/10
The remaining part will be filled by A and B in 14 hours. Then,
= > (7/10) *(A+B) = 14
= > (A+B)’s whole work= 14*(10/7) = 20 hr
(A+B)’s 1 hour work= 1/20
A, B and C’s 1 hour work = 1/10
C’s 1 hour work= (A+B+C) – (A+B)
= > (1/10) – (1/20)
= > 1/20
C can fill the tank in 20 hours.

Total units of work= 90
A’s one hour work = 9
B’s one hour work = 6
C’s one hour work = 5
Work done by three pipes in 2 hours = 15 + 14 = 29 units
Work done in 6 hours= 29*3 = 87 units
Remaining will be done by pipe A and B = 3/15 = 1/5 hours
Total time required= 6 1/5 hours = 6 hours 12 mins

Let capacity of tank = 90 lit (LCM of 10, 15, 18)
A fills 90/10 = 9 liter/hour
B fills 90/15 = 6 liter/hour
C empties 90/18 = 5 liter/hour
Till 10 am 9*3 + 6*1.5 = 27 + 9 = 36 liters
Remaining volume to be filled =90 – 36 = 54 liters
After 10 am, when all pipes are opened. 9 + 6 – 5= 10 liters/hr is filled.
Required time = 54/10 = 5 4/10 = 5 2/5 hour = 5 hour 24 min

(x-3)/12 + (x-4)/16 + x/24 =1
(4x – 12 + 3x – 12 + 2x)/48 = 1
9x – 24 = 48
9x = 72
X = 8 min

Let total capacity = 75 litres (LCM of 15 and 25)
A = 5 litres/min
B = 3 litres/min
A+B fill 75 litres
A can fill 5*5 = 25 litres in 5 min
But B was closed 5 min before the tank is filled. So A+B together filled (75-25) = 50 litres
50 litres can be filled by A + B in,
= > (50/8) = 6 1/4
Total time = 6 1/4 + 5 = 11 ¼ min

B is 5/4 times as efficient as A. So,
A and B’s time =>5 : 4
A can fill the 3/5 of the tank in 15 min
=> (3/5) * work = 15 min
=> Whole work = 25 min
A takes 25 min to fill the tank, So, B takes,
=> 5’s =25
=>1’s = 5
B can fill the tank in, 20 min
B can fill the tank independently for 10 min only.
B’s 10 min work = 10/20 = 1/2
Remaining = 1- 1/2 = ½
Half of the tank will remain incomplete.

(x – 4)/15 + (x – 5)/18 + x/20 =1
(12x – 48 +10x – 50 + 9x)/180 =1
31x – 98 = 180
31x = 278
X = 278/31 = 8 30/31 min

If all the pipes are opened simultaneously, then in,
= > (1/15) + (1/18) – (1/20)
= > (12 + 10 – 9)/180
= > 13/180
Required hours = 180/13 = 13 11/13 hours

15/20+x/24=1
18+x=24=>x=6 minutes
LCM of 12, 15 and 20 =60 units
Required no of hours = 60/(5+4-3)=60/6=10 hours

LCM of 24 and 16 =48 units
1 cycle (2 hours) = 5 units
9th cycle (18 hours) = 45 units
19th hour =2 units
Required no of hours = 18+1+1/3=19  1/3 hours