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Class 12 Applied Maths Chapter 1 Solutions
Number, Quantification and Numerical Applications
EXERCISE- 1.3
Q.1 A woman can swim 8 km/h in still water. If the speed of the stream is 4 km/h, then find the time taken by the woman to cover the distance of 16 km upstream.
Ans.
Speed of still water (x) = 8 km/h
Speed of the stream (y) = 4 km/h
Distance Covered (d) = 16 km
Therefore,
Speed of upstream (v) = x – y
= 8 – 4 = 4 km/h
Time Taken in upstream = d/s = 16/4 = 4 hours
Q.2 A man can swim upstream at 6 km/h and downstream at 10 km/h. Find the speed of stream and speed of man in still water.
Ans. Speed of boat in downstream (v) = 6 km/hr
Speed of boat in downstream (u) = 10 km/hr
Therefore,
Speed of boat in still water (x) = (u+v)/2
= (10 + 6)/2
= 16/2 = 8 km/h
Speed of stream (y) = (u – v)/2
= (10 – 6)/2
= 4/2 = 2 km/h
Q.3 In 2 hours a boat covers a certain distance in a river downstream at 17 km/h and returns at 9 km/h. Find the speed of the stream.
Ans. Speed of boat in downstream (u) = 17 km/h
Speed of boat in upstream (v) = 9 km/h
Time = 2 hrs
Therefore,
Speed of stream (y) = (u – v)/2
= (17 – 9)/2
= 8/2 = 4 km/h
Q.4 A boat goes 20 km upstream and 30 km downstream in 5 hours each time. Find the speed of the boat in still water and the speed of the current.
Ans. Distance of upstream = 20 km
Distance downstream = 30 km
Time = 5 hrs
By using S = D/T
Speed of boat in downstream (u) = 30/5 = 6 km/h
Speed of boat in upstream (v) = 20/5 = 4 km/h
Now,
Speed of stream = (u-v)/2 = (6-4)/2
= 2/2 = 1 km/h
Speed of boat in still water = (6+4)/2
= 10/2 = 5 km/h
Q.5 The ratio of the speed of a motor boat in still water and that of the current of water is 25:4. The motor boat goes a certain distance downstream in 5 hours and 15 minutes. How much time will take to come back?
Ans. Speed of boat in still water: Speed of current
SO,
x/y = 25/4
Time of downstream (t1) = 5 h 15 min , (t2) = ?
= 300 + 15
= 315 min
= t2/t1 = (x + y)/(x – y)
= t2/315 = (25k + 4k)/(25k – 4k)
= t2 = 315 x (29k/21k)
= 435 min
= 435/60
= 17 ¼ hours
Q.6 A boat goes 20 km upstream and 22 km downstream in 6 hours. Also, it goes 25 km upstream and 33 km downstream in 8 hours. Find the speed of the boat in still water and that of the stream.
Ans. Let,
Speed of boat in still water = x km/hr
Speed of stream = y km/hr
Upstream speed = (x – y) km/hr
Downstream speed = (x + y) km/hr
Case1
[22/(x + y)] + [20/(x – y)] = 6
Let,
1/(x + y) = u , 1/(x – y) = v
= 22u + 20v = 6
= 11u + 10v = 3…………….(eq. i)
Case 2
33/(x+y) + 25/x-y = 8
33u + 25v = 8………………..(eq.ii)
By eliminating (i) and (ii)
= 11u+10v = 3 x 3
33u +25v = 8
= 33u + 30v = 9
33u + 25v= 8
(-) (-) (-)
5v = 1
V =1/5
So, x – y = 5
Now,
From eq (i)
=11u + 10 x 1-5 = 3
= 11u = 3 – 2
= u = 1/11
So, x + y = 11
Therefore,
Speed of boat in still water = 11 + 5/2
= 16/2 = 8 km/hr
Distance Covered = 10 km
Time = Distance/Speed = 10/5
= 2 km/hr
Q.7 A boat covers 4 km against the stream in 1 hour and covers the same distance in the direction of the stream in 40 minutes. How long will it take to go 10 km in still water?
Ans. Distance Covered = 4 km
Speed of downstream = 4/1 = 4 km/hr
Speed of upstream = 4/2 x3 = 6 km/hr
Therefore,
Speed of boat in still water = 4 + 6/2 = 5 km/hr
Distance Covered = 10 km
Time = Distance/ Speed = 10/5 = 2 km/hr
Q.8 A man takes twice as long to row a distance against the stream as to row the same distance in the direction of the stream. Find the ratio of the speed of man in still water to the speed of the stream.
Ans. Let,
Time Taken by downstream (t1) = k
Time Taken by upstream (t2) = 2k
Therefore,
x/y = t2 + t1/t2 – t1
= 2k + k/2k – k
= 3k/k
x:y = 3:1
Q.9 A boat takes thrice as long to go upstream to a point as to return downstream to the starting point. If the speed of the stream is 5 km/h, find the speed of the boat in still water.
Ans. Let,
Time Taken by downstream (t1) = k
Time Taken by upstream (t2) = 3k
Speed of stream = 5 km/hr
Therefore,
x/y = t2 + t1/t2 – t1
x/5 = 3k + k/3k – k
x/5 = 4k/2k
x = 10 km/hr
Q.10 A boat moving upstream takes 6 hours 36 minutes to cover a certain distance while it takes 3 hours to cover the same distance downstream. Find the ratio of the speed of the boat in still water to the speed of the stream.
Ans. Time Taken by downstream (t1) = 3h
Time Taken by upstream (t2) = 6h 36/60min
= 33/5 h
Now,
x/y = t2 + t1/t2 – t1
x/y = (33/5 + 3)/(33/5 – 3)
= (48/5)/(18/5)
= 48/18 = 8/3
x:y = 8:3
Q.11 A motor boat takes 3 hours to cover a certain distance upstream and returns the same distance downstream in 1 hour 15 minutes. If the speed of the stream is 7 km/h, find the speed of the boat in still water.
Ans. Time Taken by downstream (t1) = 1h 15/60min = 5/4 h
Time Taken by upstream (t2) = 3 h
Speed of stream (y) = 7 km/hr
Now,
x/y = t2 +t1/t2 – t1
x/7 = 3 + 5/4/3 – 5/4
x/7 = (17/4)/(7/4)
x/7 = 17/7
x= 17 km/hr
Q.12 A boat rows a distance of 12 km downstream at 13 km/h and covers the same distance upstream at 7 km/h. Find the average speed of the boat.
Ans. Distance covered = 12 km
Speed of boat in downstream (u) = 13 km/hr
Speed of boat in upstream (v) = 7 km/hr
Therefore,
Average speed = x^2 – y^2/x
X = u + v/2 , y = u – v/2
= 13 + 7/2 = 13 – 7/2
= 10 km/hr = 3 km/hr
Average speed = (10)^2 – (3)^2/10
= 100-9/10
= 91/100 = 9.1 km/hr
Q.13 A man can row at 8 km/h in still water. If the speed of the current is 2 km/h and it takes 4 hours to row to a place and return, how far is the place?
Ans. Speed of boat in still water (x) = 8 km/hr
Speed of stream (y) = 2 km/hr
Total Time Taken (t2 + t1) = 4 h
Now,
D = t(x^2 + y^2)/2x
= 4 [(8)^2 + (2)^2]/2×8
= 60/4 = 15 km
Q.14 The speed of a boat in still water is 12 km/h and the speed of the stream is 2 km/h. A person rows to a place at a distance of 70 km and returns to the starting point. Find the total time taken by him.
Ans. Speed of boat in still water = 12 km/hr
Speed of stream = 2 km/hr
Total distance = 70 km
Now,
D = t [x^2 – y^2/2x]
70 = t [144 – 4/2×12]
70 x 24/140 = t
T = 12 hours
Q.15 A boat goes 30 km downstream and comes back to the starting point in 4 hours and 30 minutes. If the speed of the boat in still water is 15 km/h, find the speed of the stream.
Ans. Speed of boat in still water (x) = 15 km/hr
Distance Covered (d) = 30 km
Time = 4 ½ = 9/2 h
Now,
D = t [x^2 – y^2/2x]
30 = 9/2 [(15)^2 – y^2/2×15 ]
2×900/9 = 225 – y^2
y^2 = 225 – 200
y^2 = 25
y = 5 km/hr
Q.16 A boat takes 90 minutes less to travel 36 km downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 km/h, find the speed of the stream.
Ans. Distance Covered (d) = 36 km
Time (t2 – t1) = 90 min
= 90/60 = 3/2 h
Speed of boat in still water (x) = 10 km/hr
Now,
D= t [x^2 – y^2/2y]
36 = 3/2 [100 – y^2/2y]
36 x 4y/3 = 100 – y^2
y^2 + 48y – 100 = 0
y^2 + (50 – 2)y – 100 = 0
y^2 + 50y – 2y – 100 = 0
y(y + 50) – 2(y + 50) = 0
(y – 2) (y + 50) = 0
y = 2 , y = -50
Therefore, neglecting y = -50 we get
y = 2 km/hr
Q.17 A man rows to a place at a distance of 48 km and returns in 14 hours. He finds that he can row 4 km with the stream at the same time as 3 km against the stream. Find the speed of the stream.
Ans. Distance Covered = 48 km
Time (t2 + t1) = 14
Speed of downstream (u)/Speed of upstream (v) = 4/3
Let,
U = 4k , v = 3k
Speed of boat in still water (x) = (u+v)/2
=(4k + 3k)/2
Speed of stream (y) = (u – v)/2
= (4k – 3k)/2
= k/2
Now,
D = t [x^2 – y^2/2x]
48 = 14[49/4k^2 – k^2/4/2 x7k/2]
48 x 7k/14 = 48 k/4
48 x k/2 x 12 k^2
K = 2
Speed of stream (y) = k/2 = 2/2 = 1 km/h
FAQ’s related to Class 12 Applied Maths Chapter 12 on Number, Quantification and Numerical Applications:
Q.1 What is the primary focus of Chapter 12 in Applied Mathematics for Class 12?
Ans. The chapter primarily deals with numerical applications like time and work, pipes and cisterns, clocks, calendars, and quantitative estimation. It also includes topics such as numbers and numerical problems in real-life situations.
Q.2 What are the types of numbers covered in this chapter?
Ans. This chapter covers different types of numbers, including natural numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers.
Q.3 What is Quantification?
Ans. Quantification refers to the process of expressing a numerical value or a quantity. It is often used in mathematical modeling, calculations, and real-life applications like estimating cost, time, or distances.
Q.4 What is the importance of Pipes and Cisterns problems?
Ans. Pipes and cisterns problems are analogous to time and work problems. Here, pipes filling or emptying a cistern at different rates are analyzed. It helps in understanding flow rates and capacities.
These are a few Frequently Asked Questions relating to Class 12 Applied Maths Chapter 13
In Class 12 Applied Maths chapter 1, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 12 Applied Maths Chapter 1 promises an enriching learning experience that will set you on the path to success. Class 12 Applied Maths Chapter 1, we delve deep into advanced mathematical concepts that are crucial for understanding.
