(i) The difference between two numbers is 26 and one number is three times the other. Find them.
Solution:
Let the two numbers be x and y respectively, such that y > x.
According to the question,
y = 3x ….(1)
y – x = 26 ….(2)
Substituting the value of (1) into (2), we get
3x – x = 26
x = 13 ….(3)
Substituting (3) in (1), we get; y = 39.
Hence, the numbers are 13 and 39.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution:
Let the larger angle be xo and the smaller angle be yo.
We know that the sum of two supplementary pair of angles is always 180o.
According to the question,
x + y = 180o……………. (1)
x – y = 18o ……………..(2)
From (1), we get x = 180o – y …………. (3)
Substituting (3) in (2), we get
180o – y – y =18o
162o = 2y
y = 81o ………….. (4)
Using the value of y in (3), we get
x = 180o – 81o
= 99o
Hence, the angles are 99o and 81o.
(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.
Solution:
Let the cost of a bat be x and the cost of a ball be y.
According to the question,
7x + 6y = 3800 ………………. (1)
3x + 5y = 1750 ………………. (2)
From (I), we get
y = (3800-7x)/6………………..(3)
Substituting (3) in (2), we get;
3x+5(3800-7x)/6 =1750
⇒ 3x+ 9500/3 – 35x/6 = 1750
⇒ 3x- 35x/6 = 1750 – 9500/3
⇒ (18x-35x)/6 = (5250 – 9500)/3
⇒ -17x/6 = -4250/3
⇒ -17x = -8500
x = 500 ……………………….. (4)
Substituting the value of x in (3), we get
y = (3800-7 ×500)/6 = 300/6 = 50
Hence, the cost of a bat is Rs 500 and the cost of a ball is Rs 50.
(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?
Solution:
Let the fixed charge be Rs x and the per km charge be Rs y.
According to the question,
x + 10y = 105 …………….. (1)
x + 15y = 155 …………….. (2)
From (1), we get x = 105 – 10y ………………. (3)
Substituting the value of x in (2), we get;
105 – 10y + 15y = 155
5y = 50
y = 10 …………….. (4)
Putting the value of y in (3), we get;
x = 105 – 10 × 10 = 5
Hence, the fixed charge is Rs 5, and per km charge = Rs 10
Charge for 25 km = x + 25y = 5 + 250 = Rs 255
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
Solution:
Let the fraction be x/y.
According to the question,
(x+2) /(y+2) = 9/11
11x + 22 = 9y + 18
11x – 9y = -4 …………….. (1)
(x+3) /(y+3) = 5/6
6x + 18 = 5y +15
6x – 5y = -3 ………………. (2)
From (1), we get x = (-4+9y)/11 …………….. (3)
Substituting the value of x in (2), we get;
6(-4+9y)/11 -5y = -3
-24 + 54y – 55y = -33
-y = -9
y = 9 ………………… (4)
Substituting the value of y in (3), we get;
x = (-4+9×9 )/11 = 7
Hence the fraction is 7/9.
(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?
Solution:
Let the age of Jacob and his son be x and y respectively.
According to the question,
(x + 5) = 3(y + 5)
x – 3y = 10 …………………………………….. (1)
(x – 5) = 7(y – 5)
x – 7y = -30 ………………………………………. (2)
From (1), we get x = 3y + 10 ……………………. (3)
Substituting the value of x in (2), we get;
3y + 10 – 7y = -30
-4y = -40
y = 10 ………………… (4)
Substituting the value of y in (3), we get;
x = 3 x 10 + 10 = 40
Hence, the present age of Jacob and his son is 40 years and 10 years respectively.