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Class 12 Applied Maths Chapter 1 Solutions
Number, Quantification and Numerical Applications
EXERCISE- 1.4
Q.1 Two pipes A and B can fill the tank in 24 minutes and 36 minutes respectively. If both the pipes are opened together, then find the time at which the tank will be filled.
Ans.
Tank filled by pipe A in 1 min = 1/24
Tank filled by pipe B in min = 1/36 Tank filled by both the pipes = (1/24) + (1/36)
= (3 + 2)/72
= 5/72
Therefore,
Time Taken to fill the tank = 72/5
Q.2 A pipe can fill a tank in 20 minutes and another pipe can empty the same tank in 60 minutes. If both the pipes are opened together, then find out how much time will they take to fill the tank.
Ans. Tank filled by the first pipe in min = 1/20
Tank emptied by the second pipe in 1 min = 1/60
Part of the tank filled by both the pipes = (1/20) – (1/60)
= (3 – 1)/60
= 2/60
Therefore, Time Taken to fill the tank = 60/2
= 30 mins
Q.3 A pipe can fill a tank in 40 minutes. Due to a leakage in the bottom, it took 60 minutes to fill the tank. How much time will it take for the leakage to empty the full tank?
Ans. Let,
Time taken to drain the tank = x
Tank emptied in 1 min = 1/x
Tank filled in 1 min = 1/40
Time is taken by both pump and leakage = 1/60 min
Therefore,
(1/40) – (1/x) = 1/60
1/40 – 1/60 = 1/x
(3 – 2)/120 = 1/x
1/120 = 1/x
SO, x= 120 min (2hours) to drain the full tank.
Q.4 A pump can fill a tank with water in 2 hours. Because of leakage, it took 7/3 hrs to fill the tank. How much time will it take for the leakage to drain all the water in the full tank?
Ans. Let,
Time taken to drain full tank = x
Time taken to fill the tank in 1 hour = 1/2
Time taken to empty the tank in 1 hour = 1/x
Tank filled by both = 7/3 = 3/7 hours
Therefore,
1/2 – 1/x = 3/7
1/2 – 3/7 = 1/x
(7 – 6)/14 = 1/x
x = 14 hours to drain the full tank
Q.5 Two pipes A and B running together ca n fill a tank in 6 minutes. If pipe A takes 5 minutes less than B to fill the tank, find the time taken by pipe B to fill the tank alone.
Ans. Let,
Time taken by B to fill the tank = x
Tank filled by pipe A in 1 min = 1/x-5
Tank filled by pipe B in 1 min = 1/x
Tank filled by both the pipe in 1 min = 1/6
Therefore,
1/x + 1/x-5 = 1/6
x-5+x/x^2 – 5x = 1/6
6(2x – 5) = x^2 – 5x
12x – 30 = x^2 – 5x
X^2 – 5x -12x + 30 = 0
X^2 -17x + 30 = 0
X^2 – (15 + 2)x + 30 = 0
X^2 – 15x – 2x + 30 = 0
X(x – 15) – 2(x – 15) = 0
(X – 2) (x – 15) =0
X = 2 , x = 15
Therefore,
Neglecting x = 2 we get,
X = 15 min
Q.6 Pipe A can fill the tank 3 times faster than pipe B. If both pipes A and B can fill the tank in 36 minutes running together, find how much time will pipe A take to fill the tank alone.
Ans. Let,
Time taken by B be x
So, in 1 hour B = 1/x
A is 3 times faster than B = 3 x 1/x
= 3/x
= 1/x + 3/x = 1/36
= (1 + 3)/x = 1/36
= x = 144
B = 1/144
A = 3 X 1/144
= 1/48 mins
But in 1 hour = 48 mins
Q.7 Two pipes A and B can fill a cistern in 32 minutes and 48 minutes respectively. Both the pipes are opened together for some time and then pipe B is turned off. Find when pipe B must be turned off so that the cistern gets filled in 24 minutes.
Ans.
Tank filled by pipe A in 1 min = 1/32
Tank filled by pipe B in 1 min = 1/48
After ‘t’ min pipe be turned off then cistern fill in = 24 min
Therefore,
= t(1/32 + 1/48) + (24 – t) 1/32 = 1
= t(3 + 2/96) + (24 – t/32) =1
= 5t/96 + 24 – t/32 = 1
= 5t + (72 – 3t)/96 = 1
= 5t + 72 – 3t = 96
= 2t = 96 – 72
t = 24/2
t = 12 mins
Q.8 Two pipes A and B can fill a tank in 20 hours and 16 hours respectively. Pipe B alone is kept open for ¼ th of the time and both pipes are kept open for the remaining time. Find the time required to fill the tank.
Ans. Let,
The time required to fill the tank = x hours
Pipe A can fill a tank in 1 hour = 1/20
Pipe B can fill a tank in 1 hour = 1/16
Tank filled by A and B = 1/20 + 1/16
= (4 + 5)/80
= 9/80
According to the question,
= 1/16 x 1/4 + 9/80 x 3/4
= 1/4 [1/16 + 27/80]
= 1/4 [5 + 27/80]
= 1/4 x 32/80 = 1/10
Therefore,
Time taken = 10 hours
Q.9 Two pipes A and B can fill a tank in ½ hour and 1 hour respectively. A pipe C can empty the tank in 2 hours. If all three pipes are opened simultaneously, how long does it take to fill the empty tank?
Ans. Tank filled by A in 1 hour = 2
Tank filled by B in 1 hour = 1
Tank emptied by C in 1 hour = 1/2
Tank filled by A, B, and C = 2 + 1 – 1/2 = 3 – 1/2
= (6 – 1)/2 = 5/2
Therefore,
Time taken to fill= 2/5 x 60 = 24 mins
Q.10 Two pipes A and B can fill a tank in 20 minutes and 60 minutes respectively. There is an outlet pipe C at the bottom of the tank. If all three pipes are opened together it took 40 minutes to fill the tank. Find the time taken by outlet C to empty the full tank working alone.
Ans. Let, the time taken by C to empty the tank = t
Tank filled by pipe A in 1 min = 1/20
Tank filled by pipe B in 1 min = 1/60
= (1/20 + 1/60) -1/t = 1/40
= 1/20 + 1/60 – 1/40 = 1/t
= (6 + 2 – 3)/120 = 1/t
= 5/120 = 1/t
T = 24 mins
Q.11 Three pipes A, B, and C can fill a tank in 72 minutes. If all three pipes remain open for 36 minutes and then pipe C is closed it took 1 hour more to fill the tank by pipes A and B. Find the time required to fill the tank by pipe C alone.
Ans. (A+B+C)’s fill the tank in one min = 1/72
(A+B+C)’s fill the tank in 36 min = 1/72 x 36 = 1/2 tank
Remaining tank = 1 – 1/2 = 1/2 tank
Now,
(A+B) fill 1/2 tank in 1 hour
(A+B) fill 1 tank in 2 hour = 120 mins
(A+B) fill in 1 min = 1/120 tank
C’s work alone = 1/72 – 1/20
= (5 – 3)/360
= 2/360
Therefore,
Time taken by C alone is 1/180 = 3 hours
Q.12 Two pipes P and Q can fill a tank in 60 minutes and 30 minutes respectively. Third pipe R drains 5 liters of water per minute. If all three pipes are opened it took 2 hours to fill the tank. Find the capacity of the tank.
Ans.
Pipe A can fill the tank in 60 min
Pipe B can fill the tank in 30 min
Time taken to fill the tank by (A+B+C) = 2 hours = 120 mins
Let, Time taken by pipe R to drain the tank = x
Therefore,
1/60 + 1/30 – 1/x = 1/120
1/60 + 1/30 – 1/120 = 1/x
(2 + 4 – 2)/120 = 1/x
5/120 = 1/x
X = 24
According to the question,
In 1 min pipe R drain 5 litres
In 24 milk pipe R drain 5 x 24 = 120 min
So, the capacity of the tank = 120 L
Q.13 A tank is filled with three pipes with uniform flow. The first two pipes operating simultaneously fill the tank at the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by the first pipe to fill the tank alone.
Ans. Let,
Time taken by B be x hours
Time is taken by A be (x + 5) hours
Time is taken by C be (x – 5) hours
(A+B)’s work = C’s work
= 1/x + 1/(x+5) = 1/(x-4)
= (x+5+x)/[x(x+5)] = 1/(x-4)
= 2x + 5/x^2 + 5x = 1/(x–4)
= 2x^2 – 8x + 5x – 20 = x^2 + 5x
= x^2 – 8x – 20 = 0
= (x-10) (x+2) = 0
= x = 10 , x = -2
Therefore,
Time taken by pipe A alone = 10 + 5
= 15 hours
Q.14 Three pipes A, B, and C can fill a tank in 5 hours working simultaneously. Pipe C is twice as faster as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
Ans. Let,
Time taken by A be x hours
Time taken by B be 1/2 hours
Time taken by C be 1/2 x 1/2 = 1/4 hours
According to the question,
= 1/x + 1/ ½x + 1/ ¼x = 1/5
= 1/x + 2/x + 4/x = 1/5
= 7/x = 1/5
X= 35
Therefore,
The time of A is 35 hours.
Q.15 Two taps A and B can fill a tank in 4 and 6 hours respectively. The pipes are opened simultaneously and it takes 6 minutes more to fill the tank due to leakage. If the tank is full, then find the time taken by the leakage to empty the tank.
Ans. Part of a tank filled by A and B in 1 hour = 1/4 + 1/6
= (3 + 2)/12 = 5/12
Time taken by A and B = 12/5 = 2 2/5hrs = 2h 2/5 x 60
= 2 h 24 min
Due to leakage, time taken = 2 h 24 min + 6 min
= 2 ½ h = 5/2 hours.
Let,
Time taken by leakage be x hours
= 5/12 – 1/x = 1/5/2
= 5/12 – 1/x = 2/5
= 5/12 – 2/5 = 1/x
= 25 – 24/60 = 1/x
= 1/60 = 1/x
Therefore, x = 60 hours.
Q.16 A tank can be filled by two inlet pipes P and Q in 15 minutes and 20 minutes respectively. Another outlet pipe R can empty the full tank in 24 minutes. If outlet pipe R is opened 6 minutes after pipes P and Q are opened, find the total time to fill the tank.
Ans. Part of a tank filled by P and Q in 1 min = 1/15 + 1/20
= (4 + 3)/60 = 7/60
Part of tank filled by P and Q in 6 min = 6 x 7/60 = 7/10
Tank remaining = 1 – (7/10) = 3/10
Now,
Part of a tank filled in 1 min
= 1/15 + 1/20 – 1/24
= (8+6–5)/120
= 9/120 = 3/40
So, 3/40 tank filled in 1min
1 tank filled in 40/3 min
3/10 tank filled in 40/3 min
3/10 tank filled in 40/3 x 3/10 = 4 min
Therefore,
Total time = 6 + 4 = 10 mins.
Q.17 Three pipes A, B, and C can fill a tank together in 2 hours. Pipes B and C take 3 hours to fill the tank, while pipes A and C take 4 hours to fill the tank. Find the time taken by each pipe to fill the tank alone.
Ans. Part of a tank filled by A+B+C in 1 hour = 1/2
Part of a tank filled by B+C in 1 hour = 1/3
Part of a tank filled by A+C in 1 hour = 1/4
So,
Part of a tank filled by A alone in 1 hour = 1/2 – 1/3
= (3 – 2)/6 = 1/6
Part of a tank filled by B alone in 1 hour = 1/2 – 1/4 = 1/4
Part of a tank filled by C alone in 1 hour = 1/4 – 1/6 = 1/12
Therefore,
Time is taken by A = 6 hours
Time is taken by B = 4 hours
Time is taken by C = 12 hours.
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.
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