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Class 12 Applied Maths Chapter 1 Solutions
Number, Quantification and Numerical Applications
EXERCISE- 1.5
Q.1 A can run 40 meters while B runs 50 meters at the same time. In a 1000 m race, find by how much distance B beats A.
Ans. When B runs 50 m, then A runs 40 m
When B runs 1 m, then A runs 40/50 m
When B runs 1000 m, then A runs 40/50 x 100 = 800 m
Therefore,
B beats A by (1000 – 800) = 200 m.
Q.2 In a 100 m race, A covers a distance in 36 seconds and B in 45 seconds. Find by how much distance A beats B.
Ans. Speed of B = 100/4 m/s
Time taken by B = 45 – 36 = 9
Distance = speed x time
= 100/45 x 9
= 20 m
Hence, A beats B by 20 m.
Q.3 In a 500 m race, A beats B by 90 m or 18 seconds. Find A’s time to cover the race course.
Ans. Speed of B = 90/18 = 5 m/s
Time taken by B = distance/speed
= 500/5
= 100 sec
Time is taken by A = 100 – 18
= 82 sec
Q.4 A runs 4 times as fast as B. If A gives B a start of 60 m, how far must the goal on the race course be so that A and B reach it at the same time?
Ans. Let, the speed of B be x m/s and A be 4x m/s
In t sec distance covered by B = speed x time = xt
In t sec distance covered by A = 4xt
= 4xt – xt =60
= 3xt = 60
= t = 60/3x = 20/x sec
Therefore,
Total distance of race course = 4x x 20/x = 80 m
Q.5 A and B participate in the 200 m race. A runs at the speed of 6 km/h. A gives B a start of 18 m and still beats him by 6 seconds. Find the speed of B in km/h.
Ans.
Speed of A = 6 km/h = 6 x 5/18
= 5/3 m/s
Time is taken by A to cover 200 m = 200/5 x 3 = 120 secs
Distance covered by B = 200 – 18 = 182 m
Time taken by B = 120 + 6 = 126m
SO,
Speed of B = 182/126 x 18/5 = 26/5
= 5.2 km/h
Q.6 In a 500 m race, the ratio of the speeds of two participants A and B is 3: 4 respectively. If A has a start of 140 m, then find the distance by which A wins.
Ans. Distance covered by A = 500 – 140 = 360 m
Ratio = 3:4
When A covers 3 m, B covers 4 m
When A covers 1 m, B covers 4/3 m
When A covers 3 m, B covers 4/3 x 360 = 480 m
Hence, A wins by(500 – 480) = 20 m.
Q.7 In a 100 m race, A beats B by 20 m and C by 28 m. Find by how much distance B will beat C in a race of 500 m.
Ans.
A : B = 100 : 80 , A : c = 100 : 72
B/C = B/A x A/C
= 80/100 x 100/72 = 10/9
B/C = 10 x 50/9 x 50
= 500/450
Therefore,
B beats C by 500 – 450 = 50 m.
Q.8 In a 100 m race, A can beat B by 20 m and B can beat C by 10 m. In the same race, find by how much distance A beat C.
Ans.
A : B = 100 : 80 , B : C = 100 : 90
A/C = A/B x B/C = 100/80 x 100/90
= 100/72
Therefore,
A beats C by (100 – 72) = 28 m.
Q.9 In a 200 m race, A can give a start of 18 m to B and a start of 31 m to C. In a race of 350 m, how much start can C give to B?
Ans. A : B = 200 : (200 – 31) , A : C = 200 : 182
= 200 : 169
B/C = C/A x A/C = 169/200 x 200/182 = 13/14
B/C = 13/14 x 350/350
= 325/350
Therefore,
C beats B by 25 m.
Q.10 . In a flat race, A beats B by 15 m and C by 29 m. When B and C run over the course together B wins by 15 m. Find the length of the race course.
Ans. Let, the total length of the race course be x m
A : B = x : x-15 , A : C = x : x-29 ,B : C = x : x-15
B/C = B/A x A/C
= x/x-15 = x-15/x X */x-29
= x/x-15 = x-15/x-29
= x^2 – 29 = x^2 – 30x + 225
= x = 225 m
Therefore,
The length of the course is 225 m.
Q.11 A can run 1 km in 4 minutes 54 seconds and B in 5 minutes. How many meters of start can A give B in 1 km races so that the race may end in a dead heat?
Ans.
Total distance = 1 km = 1000 m
Time taken by A = 4 min 54 sec = 294 sec
Time taken by B = 5 min = 300 sec
Speed of A/Speed of B = (1000/294)/(1000/300)
= 50/49
Distance covered by A = 1000 m
Let, ‘s start giving to B be x m.
Distance covered by B = (1000 – x)m
50/49 = 1000/1000 – x
1000 – x = 980
x = 20 m.
Q.12 In a race of 600 m, A can beat B by 60 m, and in a race of 50 m. By many meters will A beat C in a race of 400 m?
Ans. A : B = 600 : 540 , B : C = 500 : 450
A/C = A/B x B/C = 600/540 x 540/450
= 100/81 x 4/4
= 400/324
Therefore,
A beats C by 76 m.
Q.13 In a game of 90 points, A can give 15 points to B and 30 points to C. How many points can B give C in a game of 100 points?
Ans.
A : B = 90 : 75 , A : C = 90 : 60
B/C = B/A x A/C = 75/90 x 90/60
= 5/4
B/C = 5/4 X 20/20 = 100/80
Therefore,
B gives (100 – 80) = 20 points to C.
Q.14 In a game, A can give 20 points to B in 60 and 18 points to C in 90. How many points can C give B in a game of 120?
Ans.
A : B = 60 : 40 , A : C = 90 : 72
C/B = C/A x A/B = 72/90 x 60/40
C/B = 60/50 x 2/2 = 120/100
Therefore,
C can give 20 points to B.
Q.15 In a game, A can give 15 points to B, A can give 30 points to C and B can give 20 points to C. How many points make the game?
Ans.
Let, the total point in a game = x
A : B = x : (x – 15), A : C = x : (x – 30) , B : C = x : (x – 20)
A/B = A/C x C/B
x/x-15 = x/x-30 * x-20/x
x(x – 30) = (x – 20) (x -15)
x^2 – 30x = x^2 – 15x -20x + 300
-30x = – 35x + 300
-30x + 35x = 300
5x = 300
X = 60
Therefore,
Total points = 60.
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.
