Case Study Problems - Class 10 CBSE - Old age homes mean for senior citizens who are unable to stay with their families or destitute. These old age homes have special medical facilities for senior citizens such as mobile health care systems, ambulances, nurses and provision of well balanced meals. Himanshu, Gaurav and Gagan start preparing greeting cards for each person of an old age home on new year

 

Old age homes mean for senior citizens who are unable to stay with their families or destitute. These old age homes have special medical facilities for senior citizens such as mobile health care systems, ambulances, nurses and provision of well balanced meals. Himanshu, Gaurav and Gagan start preparing greeting cards for each person of an old age home on new year. In order to complete one card, they take 10, 16 and 20 min respectively.

Based on the above information, solve
the following questions:

Q1.   Co-prime numbers are those numbers which do not have any common factor other than 1. Is this statement true?

Q2.   Find the sum of the powers of all different prime factors of the numbers 10, 16 and 20.

Q3.   If all of them started together, then what time will they start preparing a new card together?

Q4.   What is the common time to make one card?

Answers:

1.      True

2.      By prime factorisation,

10     =2¹× 5¹

16     =2x2x2x2

=2⁴

20     = 2x2x5

=2²x5¹

Required sum = sum of the power of 2 + sum of the power of 5

                   = (1 + 2) + (1+1) + 4

= 7 + 2 = 9

Q3.    The required number of minutes after which they start preparing a new card together is the LCM of 10, 16 and 20 min.

Now,

10     =2 x 5

16     =2 x 2 x 2 x 2

20     =2 x 2 x 5

LCM (10, 16, 20) = 2⁴ x 5¹

= 16 x 5 = 80 min

So, they will start preparing a new card together after 80 min i.e., 1 h 20 min.

Q4.    The common time to make one card =

HCF of (10, 16, 20) = 2 min

Form a quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively. (2020)

 

1.    Form a quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively. (2020)

Soln:

Given that:

Sum of zeros a + b       =  -3

Product of zeros ab     =  2

We know that,

          The quadratic polynomial will be,

          x² – (a + b) x + ab

          i.e., x² – (-3)x + 2

          Þ x² + 3x + 2

Hence the quadratic polynomial is x² + 3x + 2

Find the values of frequencies 𝑥 and 𝑦 in the following frequency distribution table, if 𝑁 = 100 and median is 32. (2019)

 

Find the values of frequencies 𝑥 and 𝑦 in the following frequency distribution table, if 𝑁 = 100 and median is 32. (2019)

Marks	No. of students
0 – 10	10
10 – 20	𝑥
20 – 30	25
30 – 40	30
40 – 50	𝑦
50 – 60	10
Total	100

 

Soln:

First find out the cumulative frequency (cf)

Marks	No. of students (f)	Cumulative frequency (f)
0 – 10	10	10
10 – 20	𝑥	10+x
20 – 30	25	35+x
30 – 40	30	65+x
40 – 50	𝑦	65+x+y
50 – 60	10	75+x+y
Total	100

Now, Given that there are 100 students

          f = 100

          Þ 75+x+y = 100

          Þ x + y = 100 – 75 = 25            …. (1)

Also given that

          Median = 32

          That is, median lies in the range 30-40.

Therefore, 30-40 is the median class.

        

          Here,  L = Lower class containing the median

                   N = Total number of students

                   f  = Frequency of the class containing median

                    cf  = Cumulative frequency before the median class

                   h = class interval = upper limit – lower limit

In our problem

          L = 30

          N = 100

          f = 30

          cf = 35 + x

          h = 10 – 0 = 10

Substitute these values, we get

         

          2 x 3 = 15 – x

          x = 15 - 6

          x = 9

Substituting the value of x in eqn (1) we get

          y = 25 – x

          y = 25 – 9

        y = 16

Hence, the value of x = 9 and value of y = 16.