Question 3: Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes
, , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 
Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Answer:
(i) Let the numbers be x and y, such that x > y
Therefore, according to question
x - y = 26 …………….(1)
x = 3y…………….(2)
Putting the value of x from equation (2) to equation (1), we get
3y – y = 26
2y = 26
y = 13
Putting the value in equation (2), we get
x = 3 x 13
x = 39
Hence, the numbers are 39 and 13.
(ii) Let one be xâ—¦ and other be yâ—¦ such that (xâ—¦ > yâ—¦)
Therefore, according to question
xâ—¦ + yâ—¦ = 180â—¦…………….(1) (Supplementary angles)
xâ—¦ = 18 + yâ—¦…………….(2)
Putting the value of x from equation (2) to equation (1), we get
18 + yâ—¦ + yâ—¦ = 180â—¦
2yâ—¦ = 162â—¦
Putting the value of y in equation (2), we get
xâ—¦ = 18 + 81
x = 99
(iv) Let the fixed charge be = ₨ x
Let the charge for 1 km distance be = ₨ y
According to first condition,
x + 10y = ₨ 105
x = 105 – 10y …………….(1)
According to second condition,
x + 15y = 155 ………………(2)
Putting the value of x in equation (2), we get
105 – 10y + 15y = 155
5y = 50
y = 10
Putting the value of y in equation (2), we get
x + 15 x 10 = 155
x = 5
Hence, the fixed charge for taxi is ₨ 5 and, the charge for one km distance is ₨ 50.
Charge for 25 km distance
= 25 x 10 + 5
= ₨ 255
(vi) Let the age of Jacob be = x years
Let the age of Jacob’s father be = y years
After 5 years,
Jacob’s age x + 5 years
Son’s age y + 5 years
According to question,
x + 5 = 3 (y + 5)
x + 5 = 3y + 15
x = 3y + 10………………(1)
Five years ago,
(x- 5) = 7 (y - 5)
x – 5 = 7y – 35
x – 7y = -30……………….(2)
Putting the value of x in equation (2), we get
3y + 10 - 7y = -30
-4y = -40
y = 10
Putting the value of y in equation (1), we get
x = 3 (10) + 10
x = 40
Hence, the present age of Jacob is 40 years and the age of his son is 10 years.
(iii) 2x2– 6x + 3 = 0
Add Comment