Class 10 Maths · Appendix 2

Mathematical Modelling

4 Class 10 Maths questions on Mathematical Modelling from the NCERT textbook, each with its answer and a step-by-step solution.

CBSE Class 10 Maths previous year papers, with answers →

Questions

1. Q1 NCERT · Exercise A2.1
   Consider the following situation. A problem dating back to the early ​13th century, posed by Leonardo Fibonacci, asks how many rabbits you would have if you started with one pair ​and let them reproduce. Assume that a pair of rabbits produces a pair of offspring each month, and that each pair of rabbits produces their first offspring at the age of $2$ months. Month by month, the number of pairs of rabbits is given by the sum of the pairs in the two preceding ​months, except for the $0$th and the $1$st months. ​The table alongside lists the number of pairs for months $0$ to $16$; after about $16$ ​months you have nearly $1600$ pairs of rabbits. Clearly state the problem and the different stages of ​mathematical ​modelling in this situation.

   Pairs of rabbits month by month (months 0 to 16)

   Month	Pairs of Rabbits
   0	1
   1	1
   2	2
   3	3
   4	5
   5	8
   6	13
   7	21
   8	34
   9	55
   10	89
   11	144
   12	233
   13	377
   14	610
   15	987
   16	1597

   Show answer

   Problem: starting from one pair of rabbits, find the number of pairs present at the end of each month up to month $16$. Stages of modelling: (1) State the problem precisely; (2) Make assumptions (no deaths, one new pair per mature pair per month, maturity at $2$ months); (3) Formulate the recurrence $a_n=a_{n-1}+a_{n-2}$ with $a_0=a_1=1$; (4) Iterate to get $a_{16}=1597$; (5) Interpret and validate against the stated near-$1600$ pairs.

   Step-by-step solution →
2. Q2 NCERT · Exercise A2.2
   In each of the ​problems below, ​show the different stages of ​mathematical modelling for solving ​the problem. ​An ornithologist ​wants to estimate the number of parrots in a large field. She uses a ​net to catch some, and ​catches $32$ parrots, which she rings and sets free. The following week she manages to net $40$ parrots, of which $8$ are ringed.

   1. (i)What fraction of her second catch is ringed?
   2. (ii)Find an estimate of the total number of parrots in the field.
   Show answer

   (i) The fraction of the second catch that is ringed is $\frac{8}{40}=\frac{1}{5}$. (ii) Using $\frac{8}{40}=\frac{32}{N}$, the estimated total is $N=160$ parrots.

   Step-by-step solution →
3. Q3 NCERT · Exercise A2.2
   Suppose ​the ​adjoining figure represents an aerial photograph of a forest with each dot representing a ​tree. ​Your ​purpose is to find ​the number of trees there are on this tract ​of land as part of ​an environmental census.

   Show answer

   Estimate by sampling: cover the photograph with a grid of equal squares (here $6\times 4=24$), count the trees in a few squares chosen at random ($8$, $8$ and $7$), and multiply the average, $\frac{23}{3}$ trees a square, by the number of squares: $24\times \frac{23}{3}=184$, so about $184$ trees. This is the textbook's method: count the dots in a unit area and multiply by the total area.

   Step-by-step solution →
4. Q4 NCERT · Exercise A2.2
   A T.V. can be purchased for Rs 24000 cash or for Rs 8000 cashdown payment and six monthly instalments of Rs 2800 each. Ali goes to market to buy a T.V., and ​he has Rs 8000 with him. He has now two ​options. One ​is to buy TV under instalment scheme or to make ​cash ​payment by taking loan from some financial society. The ​society charges simple interest at the rate of 18% per ​annum simple interest. Which option is ​better for Ali?
   Show answer

   The instalment scheme is better. Following NCERT Example 2: balance after cash-down = 24000 - 8000 = 16000; instalments total 6 x 2800 = 16800, so interest = 800. Principal outstanding month by month = 16000, 13200, 10400, 7600, 4800, 2000, total 54000 rupee-months, so 54000 x r / (100 x 12) = 800 gives r = 17.78 % per annum (NCERT prints 17.74 %). That is less than the society’s 18 %, so the instalment scheme is cheaper.

   Step-by-step solution →