Quant Quiz on Boats and Streams with concepts

Boats and Streams  

Let us know what is Downstream and what is Upstream. Because these concepts are the heart of Boats and Streams.

Downstream : In water,the direction along the stream is called downstream. I mean, If a boat or a swimmer swims in the same direction as the stream, then it is called downstream. Obviously the boat or swimmer require less efforts to travel using downstream. Because the stream itself helps the objects to move. 

Remember, as the object moves along with the water, the stream helps the object. 

So, the down stream speed (DS) is 

DS =  U+V

where U is the speed of the object in the still (calm) water

V is the speed of the water. 

Upstream : If the boat or the swimmer is swimming in the opposite direction in which the stream is passing is called upstream. In simple words, the direction against the stream is called upstream.

Remember, as the object moves against the water pushes the object in opposite direction. 

So, the upstream speed (US) is

US = U-V

where U is the speed of the object in the still (calm) water

V is the speed of the water. 

Important Formulas on Boats and Streams :

Let us assume that the speed of the boat in still water is U km/hr and the speed of stream is V km/hr then, as mentioned above :

Speed downstream = (u+v)km/hr

Speed upstream = (u-v)km/hr

èIf the speed downstream is a km/hr and the speed upstream is b km/hr then :

Speed in still water = 1/2(a+b) km/hr

Rate of stream = 1/2(a-b)km/hr 

 Questions : - 

1.   A boat can run 13 km/hr in still water and the speed of current 4 km/hr. in how many hours the boat will covered 68 km down streams?

(1) 2 hours                               (2) 3 hours

(3)  4 hours                              (4) 5 hours

2.   A man rows downstream 32 km and 14 km upstream. If he takes 6 hours to cover each distance, then the Velocity (in km/h) of current is:

(1) 1/2                                      (2) 3/2

(3) 2                                         (4) 5/2

3.   A man can row at 5 km/h in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place?

(1) 1.5 km                               (2) 2.4 km

(3) 3.5   km                             (4) 1 km

4.   A man can row 40 km upstream and 55 km downstream in 13 hours. Also, he can row 30 km upstream and 44 km downstream in 10 hours. Find the speed of the man in still water and the speed of the current.

(1) 5                                          (2) 3

(3) 6                                          (4) 4

5.   In a stream running at 2 km/h, a motorboat goes 6 km upstream and back again to the starting point in 33 minutes. Find the speed of the motorboat in still water.

(1) 22 km/hr                            (2) 20 km/hr

(3) 25 km/hr                            (4) 18 km/hr

6.   A man takes 3 hours 45 minutes to row a boat I5 km downstream of a river and 2 hours 30 minutes to cover a distance of 5 km upstream. Find the speed of the river current in km/hr.

(1) 1 km/hr                              (2) 2 km/hr

(3) 3 km/hr                              (4) 4 km/hr

7.   A person can row a boat d km upstream and the same distance downstream in 5 hours 15 minutes. Also, he can row the boat 2d km upstream in 7 hours. How long will it take to row the same distance 2d km downstream?

(1) 3/2 hours                            (2) 7 hours

(3) 7 (¼)                                  (4) 7/2 hours

8.   A man swimming in a stream which flows 1.5 km/h, finds that in a given time he can swim twice as fast with the stream as he can against it. At what rate does he swim?

(1) 4 km/hr                              (2) 4.5 km/hr

(3) 6 km/hr                              (4) 7 km/hr

9.   A motorboat in still water travels at a sped of 36 km/h. It goes 56 km upstream in 1 hour 45 minutes. The time taken by it to cover the same distance down the stream will be:

(1) 2 hours 25 minutes            (2) 3 hours

(3) 1 hour 24 minutes              (4) 2 hours 21 minutes

10.  A man can row 6 km/h in still water. If the speed of the current is 2 km/h, it takes 3 hours more in upstream  than in the downstream for the same distance.

(1) 30 km                                (2) 24 km

(3) 20 km                                (4) 32 km

ANSWER WITH EXPLANATION: 

1.   (3)

The speed downstream = 13+4 = 17 km/hr

∴Time taken to covered 68 km = 68/17 = 4 hours

2.   (2)

Rate downstream = 32/6 km/h; Rate upstream = 14/6 km/h

 Velocity of current = 1/2 (32/6 - 14/6) km/h

= 3/2 km/h

3.   (2)

-Speed downstream = (5 + 1) km/h = 6 km/h.

Speed upstream = (5 – 1) km/h = 4 km/h.

Let the required distance be x km.

Then, x/6 + x/4 = 1

2x + 3x = 12

5x = 12

x = 2.4 km.

4.    (2)

Let rate upstream = x km / hr and rate downstream = y km / hr.

Then,  40/x+ 55/y   =13   ...(i)

and   30/x+44/y=10       ...(ii)

Multiplying (ii) by 4 and (i) by 3 and subtracting, we get : 

11/y=1                   or                     y=11.

Substituting y = 11 in (j), we get : x = 5.
\threrefore Rate in still water =1/2(11+5) km/h=8km/h.

Rateofcurrent = 1/2(11−5) km/h=3km/h.

5. (1)

 Let the speed of the motorboat in still water be x km/h. Then,
Speed downstream = (x + 2) km/h; Speed upstream = (x - 2) km/h.
 6/x+2+6/x−2    =    33/60

<=>  11x2−242x+2x−44=0

«=>   (x - 22) (11x + 2) = 0  <=>   x = 22. Hence, speed of motorboat in still water = 22 km/h.

6.    (1) 
Rate downstream=15334 km/hr

=15×415km/hr=4km/hr.

Rate upstream=(5212) km/hr

=(5×25)=2km/hr

Speed of current=12(4−2)km/hr=1km/hr.

7.    (4)

 Let the speeds of boat and stream be x and y km/hr respectively.

Then, Rate downstream = (x + y) km/hr and Rate upstream = (x - y) km/hr

Given d / (x+y) + d / ( x-y ) =5 hrs 15 minutes = 21/4 hours

and 2d / (x+y) = 7 => d / (x+y)=7/4 => 2d / (x+y)=7/2

Hence, he takes 7/2 hours to row 2d km distance downstream.

8.     (2)

 Rate of stream =1.5 km/hr

Let speed of man in still water = x km/hr and Distance = d

Upstream speed = (x - 1.5) km/hr Downstream speed = (x + 1.5) km/hr

Given, 2d / (x+1.5) = d/ (x-1.5) => 2x - 3 = x + 1.5 => x = 4.5 km/hr

9.   (3)

 Speed of the motorboat upstream

= 56km/1 ¾ hours

=32 km/h

Let the speed of the current = x km/h

36 – x    = 32

à 36 – 32  = 4 km/h

speed of motorboat downstream = 36 + 4 = 40 km/h

Time taken to cover 56 km at 40 km/h =     56/40 = 7/5 hours

= 1 hour 24 minutes

10.  (2)

Let he required distance be x km.

x/6-2 + x/ 6 + 2 = 3

à x/4 – x/8 = 3

à 2x –x/8 = 3

= x=3 x 8 = 24 km.

Give your answer in the comment section along with your timing and judge where you stand in the competition.

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Formulae and Questions of Percentage

1.    Concept of percentage or Percentage formula

To express x% as a fraction: We have, x% x/100 Thus, 30%=30/100= 3/10

In simple terms we can conclude that percentage symbol "%" means 1/100

In case we are having a fraction and we have to calculate its percentage then we will multiply it      by 100.

that is, 

a/b=(a/b)×100)%

2.    Increase or decrease in percentage with price 
(1) If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure will be = [R/ (100+R)×100]%

( 2) If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure will be = [R/(100−R)×100]

3.    Results on population
Formula's on result of population are very important when we have to calculate the population n years after or n years before,
Let the population of a town be P now and suppose it increases at the rate of R% per annum, then
(1) Population after n years =P(1+R/100)ⁿ

(2)  Population before n years =P/(1+R/100)ⁿ

4.    Results on Depreciation
We know that value of a machine depreciate with time, so it will decrease with the time. To calculate the value of machine after n years or before n years, we use these formula's

Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. (1) Population after n years   = P(1−R/100)ⁿ

(2) Population before n years =P/(1−R/100)ⁿ

5.    Important results for Percentage
The formulas we are going to mention below is same as of increase or decrease in consumption with increase or decrease in the commodity price, just here we are in a bit different context.
(1) If A is R% more than B, then B is less than A by = [R/(100+R)×100]%

(2) If A is R% less than B, then B is more than A by = [R/(100−R)×100]%

Some Questions:- 

1.    Two numbers are less than third number by 30% and 37% respectively. How much percent is the second number less than by the first
(1) 20%
(2) 30 %
(3) 10%
(4) 40%

2.    In an examination, 34% of the students failed in mathematics and 42% failed in English. If 20% of the students failed in both the subjects, then find the percentage of students who passed in both the subjects.

(1) 40%                 (2) 41%

(3) 43%                 (4) 44%

3.    In the new budget , the price of kerosene oil rose by 25%. By how much percent must a person reduce his consumption so that his expenditure on it does not increase ?

(1) 20%                 (2) 30%

(3) 22%                 (4) 24%

4.    The value of a machine depreciates at the rate of 10% per annum. If its present is Rs.1,62,000.? What was the value of the machine 2 years ago ?

(1) 200000             (2) 20000

(3) 2000                 (4) None of these

5.    A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation?

(1) 60%                 (2) 62%

(3) 64%                 (4) 65%

 ANSWERS WITH EXPLANATIONS:

1.    (3) : Let the third number is x.
then first number = (100-30)% of x
= 70% of x = 7x/10

Second number is (63x/100)
Difference = 7x/10 - 63x/100 = 7x/10
So required percentage is, difference is what percent of first number
=> (7x/100 * 10/7x * 100 )% = 10%

2.    (4) : Failed in mathematics, n(A) = 34Failed in English, n(B) = 42

n(AUB) = n(A) + n (B) – n (A∩ B )

= 34 + 42- 20 = 56
Failed in either or both subjects are 56

Percentage passed = (100−56)%=44%

3.    (1) : Reduction in consumption = [((R/(100+R))*100]%? [(25/125)*100]%=20%.

4.    (1): Value of the machine 2 years ago
= Rs.[162000/(1-(10/100)²)]=Rs.[162000*(10/9)*(10/9)]= Rs.200000

5.    (3) : Let the number be x

Then, error =  (5/3)x – (3/5)x =(16/15)x

Error% = [(16x/15) /(5x/3)] * 100% = 64%

Give your answer in the comment section along with your timing and judge where you stand in the competition. 

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