Problem 1 :
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 + 20y = 1000
x + 26y = 1180
x + 20y – 1000 = 0 ------(1)
x + 26y – 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
Problem 2 :
A fraction becomes 1/3 when 1 is subtracted from the numerator it becomes 1/4 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
3x – 3 = y
3x – y – 3 = 0 ----------(1)
x/(y + 8) = 1/4
4x = 1(y + 8)
4x = y + 8
4x – 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
So, the required fraction is 5/12
Problem 3 :
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
3x – 1y = 40
4x – 2y = 50
3x – 1y – 40 = 0 ---------(1)
4x – 2y – 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
So, the number of questions answered correctly = 15
Number of questions answered incorrectly = 5
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