A:
(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 x◦ and other be y◦ such that (x◦ > y◦)
Therefore, according to question
x◦ + y◦ = 180◦…………….(1) (Supplementary angles)
x◦ = 18 + y◦…………….(2)
Putting the value of x from equation (2) to equation (1), we get
18 + y◦ + y◦ = 180◦
2y◦ = 162◦

Putting the value of y in equation (2), we get
x◦ = 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.