# Quantitative Aptitude: Data Interpretation Questions Set 59

Data Interpretation on Boats and Streams, Partnership. Chapter based DIs based on SBI PO 2017 Mains Pattern – for IBPS PO/RRB Mains 2017 Exams

Direction (1-5): Study the following bar graph and answer the questions that follow: 1. On a particular day, the time taken by boat A to cover 255 km downstream is 2 hours more than the time taken by boat D to cover 195 km upstream. Find the ratio of upstream speed to downstream speed in case of boat A on that particular day.
A) 4 : 11
B) 3 : 5
C) 15 : 22
D) 8 : 19
E) 7 : 17
Option E
Solution
:
Let x km/hr is the speed of stream on that particular day.
So 255/(12+x) = 195/(20-x) + 2
Solve, x = 5 km/hr
So in case of boat A, ratio of speed upstream : downstream = (12-5) : (12+5) = 7 : 17
2. Boat B covered 756 km downstream on Monday for which it took 6 hours less than that in which it covered half distance upstream on Tuesday. On Tuesday, speed of stream was 2 km/hr more than that on Monday. Find the downstream speed of boat B on Tuesday.
A) 21 km/hr
B) 19 km/hr
C) 23 km/hr
D) 11 km/hr
E) 15 km/hr
Option C
Solution
:
Let on Monday, speed of stream is x km/hr, then on Tuesday it is (x+2) km/hr
On Monday it covered 756 km, so on Tuesday it covered 756/2 =  378 km
So 756/(16+x) = 378/(16-(x+2)) – 6 ………………(1)
126/(16+x) = 63/(14-x) – 1
OR SOlve with options in
378/[16 -(y+2 )] – 756/(16+y )=6
63 [1/(14-y) – 2/(16+y) ] = 1
Solve, x = 5 km/hr
So speed of stream on Tuesday = (5+2) = 7 km/hr
So downstream speed of boat B = (16+7) = 23 km/hr
3. For boat C, its upstream speed is 6 km/hr on a particular day. Find the difference in time in covering 360 km by boats A and C on that particular day.
A) 8 hours
B) 6 hours
C) 4 hours
D) 7 hours
E) 5 hours
Option B
Solution
:
Upstream speed of C = 6 km/hr, so speed of stream: 9 – x = 6, x = 3 km/hr
Downstream speed of boat C = 9+3 = 12 km/hr
Downstream speed of boat A = 12+3 = 15 km/hr
So difference in timings = 360/12 – 360/15 = 30 – 24 = 6 hours
4. On a particular day, ratio of upstream speed to downstream speed of boat D is 3 : 7. It took 20 hours more to cover a distance upstream than same distance downstream by boat D. On that particular day, boat B covered same distance in how much time?
A) 17.5 hours
B) 15 hours
C) 19 hours
D) 19.5 hours
E) 18.5 hours
Option A
Solution
:
Let speed of stream on that day = x km/hr
So (20-x)/(20+x) = 3/7
Solve, x = 8 km/hr
So y/(20-8) – y/(20+8) = 20
Solve, y = 420 km
So required time = 420/(16+8) = 420/24 = 17.5 hours
5. Upstream sped of boat E is 9 km/hr. How many more hours will it take to cover a distance of 315 km upstream than same distance downstream?
A) 18 hours
B) 20 hours
C) 22 hours
D) 25 hours
E) 15 hours
Option B
Solution
:
Upstream speed = 9, speed of boat = 15 km/hr, so speed of stream = 15-9 = 6 km/hr
Downstream speed = 15+6 = 21 km/hr
So required time = 315/9 – 315/21 = 20 hours

Direction (6-10): Study the following table and answer the questions that follow: 1. In 2012, if out of the total profit, R got Rs 9750 as his share, what is the number of months for which R invested his money?
A) 7 months
B) 8 months
C) 6 months
D) 4 months
E) 5 months
Option C
Solution
:
Ratio of share of profit of P : Q : R is
4*4 : 3*6 : 5*x
16 : 18 : 5x
So 5x/(34+5x) * 20800 = 9750
Solve, x = 6 months
2. In 2013, if out of the total profit, P and R got a share of Rs 10,400 and Rs 19,500 respectively, what is the ratio of investment of P, Q and R respectively?
A) 3 : 3 : 5
B) 4 : 3 : 2
C) 2 : 5 : 5
D) 2 : 3 : 5
E) 2 : 5 : 3
Option D
Solution
:
x*8 : 3*5 : y*6
8x : 15 : 6y
So 8x/(8x+15+6y) * 39650 = 10400
Solve, 225x – 60y = 150
15x – 4y = 10……………………………………(1)
Also
6y/(8x+15+6y) * 39650 = 19500
Solve, 61y = 75 + 40x + 30y
40x – 31y = -75………………………………….(2)
Solve equations (1) and (2)
x = 2, y = 5
So ratio of investment is x : 3 : y = 2 : 3 : 5
3. In 2014, out of the total profit, Q got Rs 12,000 as his share. R invested Rs 3600. If ratio of share of R in total profit to total profit is 3 : 10, find the total profit.
A) Rs 40, 000
B) Rs 35, 000
C) Rs 45, 000
D) Rs 50, 000
E) Rs 30, 000
Option E
Solution
:
In 2014, P invested Rs 5x, Q – 4x and R 3600
So ratio of share of profits of P, Q and R:
5x*3: 4x*5 : 3600*5
3x : 4x : 3600
Let total profit = Rs y
So 4x/(7x+3600) * y = 12000………………………(1)
xy/(7x+3600) = 3000
Also,
[3600/(7x+3600) * y]/y = 3/10
y gets cancelled
Solve, x = 1200
Put in (1)
Solve, y = Rs 30, 000 = total profit
4. In 2015, P and Q invested Rs 3900 and Rs 2600 respectively. Also P and Q invested for same number of months. If difference in the shares of P and Q out of total profit is Rs 2750, find the ratio of number of months of investment of P to investment of R.
A) Other than given in options
B) 2 : 265
C) 1 : 260
D) 1 : 1040
E) 2 : 1045
Option D
Solution
:
Let investment of R is 4x, and months of investment of P = Q = y months
So ratio of profit share of P : Q : R is
3900*y : 2600*y = 4x*4
975y : 650y : 4x
So (975y-650y)/(975y+650y+4x) * 22550 = 2750
325y/(1625y+4x) = 2750/22550
1625y + 4x = 2665y
4x = 1040y
So y/4x = 1/1040
5. The total investment of P, Q and R in 2016 is Rs 16,800. If difference in shares of profit of Q and P is Rs 5250, find the investment of Q.
A) Rs 3600
B) Rs 7200
C) Rs 6000
D) Rs 4800
E) Rs 2400
Option B
Solution
:
Ratio of share of profit of P : Q : R is
3*4 : x*7 : 5*6
12 : 7x : 30
So (7x-12)/(7x+42) * 14700 = 5250
(7x-12)*14 = (7x+42)*5
98x – 168 = 35x + 210
Solve, x = 6
So ratio of investments of P : Q : R is 3 : x : 5 = 3 : 6 : 5
So investment of Q = 6/(3+6+5) * 16800 = Rs 7200

## 54 Thoughts to “Quantitative Aptitude: Data Interpretation Questions Set 59”

1. Jellyfish

6/10
Thx mam pls post similar pattern DI

2. chinnu

good level di

3. %%%

Thanku mam …high level di

4. %%%

jo 9th ques acche se nhi padhega ..wo jaroor fasega jase mere sath hua…..dimag ka sara jung hta diya isne…TQ mam :))

5. think tank

Kisi ne first 2 questions Ki equation solve Ki Ho to plz ek bar detail me solve Kr de

1. %%%

378/16 -(y+2 ) – 756/(16+y )=6
63 [1/(14-y) – 2/(16+y) ] = 1
ab optn me jo value di h usse karo
jaise

A) 21 km/hr
21 = 16 +y
y =5…..now check satisfy or not

1. think tank

Tq

2. ~CASIMIRO~ NITROUS~ OXIDE~

TQ

1. %%%

🙁 ye tq vq mt bola karo …….aise hi method aate h bs…..isse accha h to btao

1. ~CASIMIRO~ NITROUS~ OXIDE~

sums cahiye ja solution 😛

2. %%%

nhi issi ka pucha h ….koi or way ho to

3. ~CASIMIRO~ NITROUS~ OXIDE~

kk bta duga wese mene yeh sums kiye nhi 😛
ab kruga nikla koi new to bta duga 🙂

4. %%%

ok….

5. ~CASIMIRO~ NITROUS~ OXIDE~

🙂

6. ~CASIMIRO~ NITROUS~ OXIDE~

option se hi easy hoga :p

7. %%%

hw r u:)

8. ~CASIMIRO~ NITROUS~ OXIDE~

fine u say and ur prep kessi gng

9. %%%

f9 thik h

10. ~CASIMIRO~ NITROUS~ OXIDE~

exam date urs 10 and 24 ?

11. %%%

9 , 24

12. ~CASIMIRO~ NITROUS~ OXIDE~

po bhar lena ty pe :p

13. %%%

ji

14. ~CASIMIRO~ NITROUS~ OXIDE~

:))

15. %%%

bye

16. ~CASIMIRO~ NITROUS~ OXIDE~

ta ta

17. %%%

suno issi dp me tum Godzilla lg rhe ho…..

18. ~CASIMIRO~ NITROUS~ OXIDE~

badia hai phir do sabko darauga 😛

19. %%%

hmm vesse bhi punjab wale ajkl sbko shocked kr rhe h :p

20. ~CASIMIRO~ NITROUS~ OXIDE~

o-o essa nhi hum bhut sareef hai wese 😛

21. %%%

hehe tumse compare ni kiya wese :)………going now bye

22. ~CASIMIRO~ NITROUS~ OXIDE~

kk atb ta ta :))

23. %%%

hehe not compare with u

24. ~CASIMIRO~ NITROUS~ OXIDE~

yes yes LP

25. %%%

LP ??

26. ~CASIMIRO~ NITROUS~ OXIDE~

:p KRNA THA LP CHLA GYA

27. %%%

ok ….wese kisi ne mjk kiya hoga apke sath ….ki ap shareef ho :/

28. ~CASIMIRO~ NITROUS~ OXIDE~

nhi ji sab kehte hai yeh :

29. %%%

🙂

30. ~CASIMIRO~ NITROUS~ OXIDE~

:)) /

2. Shubhra

yes solve by options

1. think tank

Ok mam I will try next time through option

6. hi

Tq

7. hi

Mam/sir……
what i do for such type di boat and stream pattern my basic concept is not clear….

1. Shubhra

Solve basic questions
It has only basic concepts::::::::::

time = distance/speed
speed of boat = b, speed of stream = s
upstream speed, u = b – s
downstream speed, u = b + s
just these formulas, nothing else – Questions are based only on these.

1. hi

Thank u mam;)

8. Divaker

Baht calculation Hai yr koi tarika calculation kam karne ke 1 St and 2nd one mein

1. Shubhra

use options

378/[16 -(y+2 )] – 756/(16+y )=6

63 [1/(14-y) – 2/(16+y) ] = 1.

2. Parag Chavda

use the options dude, avoid quadratic method, quadratic is time consuming and lengthy..

9. Ashish Chouhan

Thank you

10. hope@IBPS po..

how to solve question 1,2 equation facing difficulty in solving..@@Shubhra_AspirantsZone:disqus

1. Shubhra

use options

378/[16 -(y+2 )] – 756/(16+y )=6

63 [1/(14-y) – 2/(16+y) ] = 1

1. heena

mam 1st vala kaise kare..bahut calculative h..So 255/(12+x) = 195/(20-x) + 2

11. Adamant

ty ma’am.
Post more questions like this..

12. jaga

thank u mam….in q9 P=Q= y not 5 🙂