Question 1

Question 1Suppose a company needs a computer for some period of time. The company can either hire a computer for ₹ 2,000 per month or buy one for ₹ 25,000. If the company has to use the computer for a long period, the company will pay such a high rent, that buying a computer will be cheaper. On the other hand, if the company has to use the computer for say, just one month, then hiring a computer will be cheaper. Find the number of months beyond which it will be cheaper to buy a computer.

Accepted Answer

Step 1 : Formulation :The relevant factors are the time period for hiring a computer and the two costs given to us. We assume that there is no significant change in the cost of purchasing or hiring the computer. So, we treat any such change as irrelevant. We also treat all brands and generations of computers as the same, i.e., these differences are also irrelevant. The expense of hiring the computer for x months is ₹ 2000x. If this becomes more than the cost of purchasing a computer, we will be better off buying a computer. So, the equation is:⇒ 2000x = 25000 .......(1)Step 2 : Finding the solution :By solving equation (1), we get :⇒ x = 250002000=252\dfrac{25000}{2000} = \dfrac{25}{2}200025000​=225​ = 12.5Step 3 : Interpretation :Since x = 12.5Hence, after 12.5 months it will be cheaper to buy a computer.

Question 2

Question 2Suppose a car starts from a place A and travels at a speed of 40 km/h towards another place B. At the same instance, another car starts from B and travels towards A at a speed of 30 km/h. If the distance between A and B is 100 km, after how much time will the cars meet?

Accepted Answer

Step 1 : Formulation :We will suppose that cars travel at a constant speed. So, any change of speed will be treated as irrelevant. If the cars meet after x hours, the first car would have traveled a distance of 40x km from A and second car would have traveled 30x km from B.So the second car will be at a distance of (100 - 30x) km from A. So, the equation will be:⇒ 40x = 100 - 30x⇒ 40x + 30x = 100⇒ 70x = 100 ....(1)Step 2 : Finding the solution :By solving equation (1), we get :⇒ x = 10070=107\dfrac{100}{70} = \dfrac{10}{7}70100​=710​ = 1.4 hoursStep 3 : Interpretation :Since, x = 1.4 hoursHence, the cars will meet after 1.4 hours.

Question 3

Question 3The moon is about 3,84,000 km from the earth, and its path around the earth is nearly circular. Find the speed at which it orbits the earth, assuming that it orbits the earth in 24 hours. (Use π = 3.14)

Accepted Answer

Step 1 : Formulation :The speed at which the moon orbits the earth is = Length of the orbitTime taken\dfrac{	ext{Length of the orbit}}{	ext{Time taken}}Time takenLength of the orbit​Step 2: Finding the solution :Since the orbit is nearly circular.∴ Its length is its circumference⇒ 2πr = 2 × π × 384000= 2 × 3.14 × 384000 = 2,411,520 kmWe know that,The moon takes 24 hours to complete an orbit.So, Speed = DistanceTime=241152024\dfrac{	ext{Distance}}{	ext{Time}} = \dfrac{2411520}{24}TimeDistance​=242411520​ = 100480 km/hourStep 3 :Interpretation :The speed is 100480 km/h.

Question 4

Question 4A family pays ₹ 1000 for electricity on an average in those months in which it does not use a water heater. In the months in which it uses a water heater, the average electricity bill is ₹ 1240. The cost of using the water heater is ₹ 8.00 per hour. Find the average number of hours the water heater is used in a day.

Accepted Answer

Step 1 : Formulation :An assumption is that the difference in the bill is only because of using the water heater.Let the average number of hours for which the water heater is used in a day be x.Difference in bill per month due to using water heater = ₹ 1240 - ₹ 1000 = ₹ 240.Cost of using water heater for 1 hour = ₹ 8.So, the cost of using the water heater for 30 days = 8 × 30 × x.Also, the cost of using the water heater for 30 days = Difference in bill due to using water heater.So, 240 x = 240 .....(1)Step 2 : Finding the solution :By solving equation (1), we get :⇒ x = 240240\dfrac{240}{240}240240​ = 1.Step 3 : Interpretation :Since, x = 1The water heater is used for an average of 1 hour in a day.

Question 5

Question 1We have given the timings of the gold medalists in the 400-metre race from the time the event was included in the Olympics, in the table below. Construct a mathematical model relating the years and timings. Use it to estimate the timing in the next Olympics.YearTiming (in second)196452.01196852.03197251.08197649.28198048.88198448.83198848.65199248.83199648.25200049.11200449.41

Accepted Answer

Let us first convert the problem into a mathematical problem.Step 1 : Formulation :Let us take 1964 as 0th year, and write 1 for 1968, 2 for 1972 and so on. So table will be :YearTiming (in second)052.01152.03251.08349.28448.88548.83648.65748.83848.25949.111049.41The difference in timings of gold medalist in 400 meters race in Olympics is given in the following table :YearTiming (in second)Difference052.010152.030.02251.08-0.95349.28- 1.8448.88-0.4548.83-0.05648.65-0.18748.830.18848.25-0.58949.110.861049.410.3TotalΣDifference = -2.6At the end of 4 years period from 1964 - 1968 the timing has increased by 0.02 second from 52.01 to 52.03 second.At the end of second Olympic the reduction in timing is 0.95 second from 52.03 to 51.08. from the table above we cannot find a definite relationship between the number of years and change in timing.By formula,Mean difference = Sum of all observationNumber of observation\dfrac{\text{Sum of all observation}}{\text{Number of observation}}Number of observationSum of all observation​Mean of differences = −2.610\dfrac{-2.6}{10}10−2.6​ = -0.26We have assumed that the timing in 400 m race of Olympic reduced at the rate of 0.26 per year.So, timing in the first year = 52.01 - 0.26 = 51.75Timing in the second year = 52.01 - 2 x 0.26 = 52.01 - 5.2 = 51.49Similarly in the nth year⇒ t = 52.01 - 0.26 n for n ≥ 1 ......(1)Now we have to find t for n = 11.Step 2 : Finding a solution :Substituting n = 11 in equation (1) we get :⇒ t = 52.01 - 0.26 x 11⇒ t = 52.01 - 2.86⇒ t = 49.15 seconds.Step 3 : Interpretation :Since, we are dealing with a real life situation, we have to see to what extent this value matches with the real situation.Step 4 : Validation :Let us find the values for the years we already know, using formula/equation (1) and compare it with known values by finding the difference. the values are given in the table :YearTiming (in second)Timing using equation (1)Deviation052.0152.010152.0351.750.28251.0851.49-0.41349.2851.23-1.95448.8850.97-2.09548.8350.71-1.88648.6550.45-1.8748.8350.19-1.36848.2549.93-1.68949.1149.67-0.561049.4149.410.00TotalΣ Deviation = -11.45Suppose we decide that this error is negligible. In this case formula/equation (1) is our mathematical model and if we decide that this error is not acceptable. Then we have to go back to step 1, the formulation and change equation (1).Step 1 : Reformulation :We still assume that the values decrease steadily by 0.26, but we will now introduce a correction factor to reduce the error. For this, we find the mean of all the deviations. This is :⇒Mean deviation=Sum of deviationNo. of observation⇒−11.4510=−1.145\Rightarrow\text{Mean deviation} = \dfrac{\text{Sum of deviation}}{\text{No. of observation}} \[1em] \Rightarrow -\dfrac{11.45}{10} = -1.145⇒Mean deviation=No. of observationSum of deviation​⇒−1011.45​=−1.145We take the mean of the errors and correct our formula by this value.Step 2 : Revised Mathematical Description :Let us now add the mean of the deviations to our formula/equation (1). So, our corrected formula is :⇒ t = 52.01 - 0.26 n - 1.145⇒ t = 50.865 - 0.26 n ......(2) for n ≥ 1Step 3 : Altered solution :By solving equation (2) for t,Putting n = 1 in equation (2) we get :⇒ t = 50.865 - 0.26(1)⇒ t = 50.865 - 0.26 = 50.605Putting n = 2 in equation (2) we get :⇒ t = 50.865 - 0.26(2)⇒ t = 50.865 - 0.52 = 50.345Putting n = 11 in equation (2) we get :⇒ t = 50.865 - 0.26(11)⇒ t = 50.865 - 2.86⇒ t = 48.005 seconds.Step 3 : Interpretation :So in 11th year(in next Olympics) t = 48.005 secondsStep 4 : Validation :Once again, let us compare the values got by using formula/equation (2) with the actual values. Table is given below for comparison.YearTiming (in second)Timing using equation (2)Deviation052.0152.010152.0350.6051.425251.0850.3450.735349.2850.085-0.805448.8849.825-0.945548.8349.565-0.735648.6549.305-0.655748.8349.045-0.215848.2548.785-0.535949.1148.5250.5851049.4148.2651.145TotalΣ Deviation = 0As we can see many of the values that formula/equation (2) gives are closer to the actual value than the values that formula/equation (1) gives the mean of the deviations is 0. So, formula/equation (2) is our mathematical description that gives a mathematical relationship between years and timing for race.Hence, timing in next olympics is 48.005 seconds.

Question 6

Question 1How are the solving of word problems that you come across in textbooks different from the process of mathematical modelling?

Accepted Answer

S.No.Word ProblemsMathematical Modelling1Formulation is easier.Formulation is comparatively difficult.2We validate the problems.No validation is done.3We always get a correct solution.There is no such guarantee of correct solution.

Question 7

Question 2Suppose you want to minimize the waiting time of vehicles at a traffic junction of four roads. Which of these factors are important and which are not?(i) Price of petrol.(ii) The rate at which the vehicles arrive in the four different roads.(iii) The proportion of slow-moving vehicles like cycles and rickshaws and fast moving vehicles like cars and motorcycles.

Accepted Answer

The important factors are :(ii) The rate at which the vehicles arrive in the four different roads, because using this we can alter the amount of waiting time on each of the four different roads.(iii) The proportion of slow-moving vehicles like cycles and rickshaws and fast moving vehicles like cars and motorcycles.Here, price of petrol is not an important factor as it has nothing to do with the waiting time of vehicles at the traffic junction.

Mathematical Modelling

Exercise A21

Exercise A22

Exercise A23