10th cbse maths solution for exercise 3.5 part 3

This page 10th cbse maths solution for exercise 3.5 part 3 is going to provide you solution for every problems that you find in the exercise no 3.5

10th CBSE maths solution for Exercise 3.5 part 3

(4) Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charge and the cost of food per day.

Solution:

Let “x” be the fixed charge

Let “y” be the charge of food per day

x + 20 y = 1000

x + 26 y = 1180

x + 20 y – 1000 = 0  ------(1)

x + 26 y – 1180 =0   ------(2)

x/(-23600 + 26000) = y/(-1000 + 1180) = 1/(26 - 20)

x/2400 = y/180 = 1/6

x/2400 = 1/6           y/180 = 1/6

x = 2400/6                y = 180/6

x = 400                      y = 30

Therefore the fixed charge = Rs.400

Charge of food per day = Rs.30


(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator it becomes ¼ when 8 is added to its denominator. Find the fraction.

Solution:

Let “x/y” be the required fraction

(x – 1)/ y = 1/3

3(x – 1) = 1 y

3 x – 3 = y

3 x – y – 3 = 0 ----------(1)

x/(y + 8) = ¼

4 x = 1( y + 8)

4 x = y + 8

4 x – y – 8 = 0  ----------(2)

x/(8 – 3) = y/(-12 + 24) = 1/(-3 + 4)

x/5 = y/12 = 1/1

x/5 = 1         y/12 = 1

x = 5               y = 12

Therefore the required fraction is 5/12


(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

Solution:

Let “x” be the number of questions answered correctly

Let “y” be the number of questions answered incorrectly

3 x – 1 y = 40

4 x – 2 y = 50

3 x – 1 y – 40 = 0  ---------(1)

4 x – 2 y – 50 = 0   ---------(2)

x/(50 – 80) = y/(-160 + 150) = 1/(-6 + 4)

x/(-30) = y/10 = 1/(-2)

x = -30/(-2)         y = 10/(-2)

x = 15                  y = 5

Therefore the number of questions answered correctly = 15

Number of questions answered incorrectly = 5    10th cbse maths solution for exercise 3.5 part 3




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