Quantitative Aptitude Questions (Data Interpretation) for SBI PO Mains 2018 Day-137

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Directions (1-5): Given below is a pie chart which shows the percentage distribution of time out of 50 hours for which a person covers a certain distance in river in different fragments of the day.

1). A motorboat covers a certain distance downstream in Monday and returns covering the same distance upstream on Thursday. If the speed of the stream is 6 km/hr, then what was the speed of motorboat in still water?

1. 46 km/h
2. 162 km/h
3. 132 km/h
4. 140 km/h
5. None of these

2) If A can row a boat in still water at 8 km/hr. If speed of water current is 3/4th of the speed of boat in still water, and he rows downstream in Sunday.

1. 8.6 km
2. 13.4 km
3. 84 km
4. 22 km
5. None of these

3) The speed of current is what percent of the speed of boat in still water? A boat travels 36 km downstream in the Friday and returns against the current at Tuesday.

1. 8.91%
2. 5.15%
3. 3.09%
4. 13.67%
5. 9.28%

4) A man can row at a speed of 5km/h in still water and the speed of water current is 2 km/h. If he rows to a place downstream during Tuesday, and he has to pay Rs. 11 per kilometer to row through the river, then what is the amount that he pays?

1. Rs. 808.5
2. Rs. 35.8
3. Rs. 92.6
4. Rs. 81.6
5. None of these

5) The ratio of the speed of the boat to that of current is 42 : 5. If the boat goes downstream during Thursday, then it will come back in (approximately)?

1. 6.99 h
2. 8.89 h
3. 9.18 h
4. 5.12 h
5. 6.05 h

Directions (6-10): Study the following and answer the following questions.

The following table shows that the data of 6 employees of ABC company.

6). The employee P is twice as good as the employee R and hence he can finish a work in 10 days less than the employee R, then how much a company has to pay to both of them if they work together to do the same piece of work?

1. Rs 13900
2. Rs 43000
3. Rs 23000
4. Rs 21000
5. Rs 13000

7). The employee Q does 75% of the work in 18 days, and then he assisted the employee O to finish the work in 4 days. Then the amount paid to the employee O per annum if he would have completed the work alone is approximately what percentage more or less than the employee Q’s one year salary.

a.150%

b.130%

c.110%

d.180%

e.187%

8). If N and R together can do a piece of work in 15 days, and they worked for 9 days, then N left. After another 8 days, R finished the remaining work. Then find the total wage received by the employee R if he alone has finished the whole work?

1. Rs 34700
2. Rs 25000
3. Rs 33600
4. Rs 22800
5. Rs 13200

9). If the employees M, Q and R can do a piece of work in 240 hours, 420 hours and 240 hours respectively, then in how many days can the employee M do the work if he is assisted by Q and R on every third day?

1. 14 days
2. 15 days
3. 34 days
4. 18 days
5. 17 days

10). The employee P can do a certain work in the same time period in which N and Q together can do the same work. If N and P together could do it in 30 days and Q alone in 60 days, find the ratio of a number of days taken by the employees P to that of the employee N.

a.1:2

b.2:3

c.1:6

d.1:5

e.1:3

Answers:

Direction (1-5)

1). Answer: b

Time taken to cover the distance downstream in Monday,

t₁ = 13× 50/ 100

= 6.5 hours.

Time taken to cover the distance upstream in Thursday,

t2 = 14× 50 /100 = 7 hours

The speed of stream,  = 6 km/h

The speed of motorboat in still water = y ( t2 + t1/ t2 − t1 )

= 6 ( 6.5 + 7 /6.5 − 7 ) = 6 ( 13.5 /0.5 )

= 162 km/h

2). Answer: c

Speed of boat in still water, x = 8 km/h

Speed of water current, y

= 3/ 4 × 8 = 6 km/h

Required distance = 12/100*50*14=84km

3). Answer: d

Downstream speed = 36/ (time taken in Friday)

= 36 /(16 × 50/ 100) =36/ 8 = 4.5 km/h

Upstream speed = 36/ (time taken during Tuesday)

= 36 /( 21 /100 × 50) = 36/ 10.5 = 3.42km/h

Speed of boat in still water = ½* (downstream speed + upstream speed)

= 1/2 (4.5 + 3.42)

= 1/2 (7.9)

= 3.95 km/h

Speed of stream = 1/ 2 (down speed – up speed)

= 1/ 2 (4.5 – 3.42)

=0.54

Required percentage = 0.54/3.95*100= 13.67%

4). Answer: a

The speed of the man, x = 5 km/h

The speed of water current, y = 2 km/h

Distance covered = 21/100*50*(5+2)= 73.5 km

Required amount = 73.5*11= Rs.808.5

5). Answer: b

Let the speed of the boat be 42x km/h and speed of current be 5x km/h

The boat goes downstream in (14 × 50 /100) h

= 7 h

Distance = speed × time

Distance = 7 × (42x + 5x)

= 47x × 7

Upstream speed = 42x – 5x

= 37x km/h

Time taken = 47x × 7/ 37x

= 8.89 h

Direction (6-10)

6). Answer: a

Given that, the ratio of time taken by the employees P and R= 1:2

If R takes 20 days to complete the work, then the number of days taken by P to complete the work= 10 days

Now, the part of the work completed by employees P and R in 1 day= (1/10+1/20) = 3/20

(i.e) The employees P and R can complete the work in 20/3 days.

Now, the Total amount received by P=945*20/3=Rs 6300

And one day wage of the employee R= 27360/24=Rs 1140

Then Total amount received by the employee R= 1140*20/3=Rs 7600

Hence total amount paid by the company= 6300+7600= Rs 13900

7). Answer: e

Number of days taken by the employee Q to finish the whole work= 18*100/75 =18* 4/3

= 24 days

Now remaining work= 1- ¾= ¼

Thus, ¼ of the work is done by O and Q in 4 days.

Then Whole work can be completed by O and Q in 16 days

Now, work done by O in one day= 1/16-1/24

= 1/48

Now the wage of the employee O for one day= 31860/27=Rs 1180

Then total annual salary received by the employee O for 48 days = 1180*48*12

= Rs 679680

One year salary of the employee Q = 810*24*12 = 233280

Required percentage= (436680/233280)*100

=187%

8). Answer: d

Since N and R worked together for 9 days,

The part of the work completed by N and R in 9 days= (1/15)*9=3/5

Now remaining work= 1-3/5= 2/5

(i.e) 2/5 part of the work is done by R in 8 days.

Then whole work will be done by R= 8*5/2

= 20 days

From the table, we have, the Monthly salary of R= Rs 27360

Then salary of the employee per day= 27360/24=Rs 1140

Hence total wage received by the employee R for 20 days= 1140*20=Rs 22800

9). Answer: b

Given that the employee M can finish the work in 240 hours. (i.e) 240/12= 20 days

Similarly, the employee Q can finish the work in 420 hours (i.e) 420/7 = 60 days

And the employee R can finish the work in 240 hours. (i.e) 240/8 = 30 days

The work was done by the employee in 2 days= 1/20*2= 1/10

Then the work was done by the employees M, Q and R=1/20+1/60+1/30

= 1/10

Thus work done in 3 days= 1/10+1/10

= 1/5

Hence The whole work can be completed in 3*5 days.(i.e) 15 days.

10). Answer: e                                                 

Given that, Work done by the employees P and N in one day= 1/30 and Q’s one day work= 1/60

Then work done by all the three employees in one day= (1/30+1/60)

= 1/20

Given that, P’s one-day work= work done by N and Q together in one day.

Then we have,

2* (P’s one day work) = 1/20

Then P’s one day work= 1/40

From this we have,

N’s one day work = 1/30-1/40

= 1/120

(i.e)  N alone can complete the work in 120 days.

Then required ratio= 40:120

= 1:3

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