Question  3. Form the Pair of Linear Equations For the Following Exercise 3.3 Chapter 3 Pair of Linear Equations in

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 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.
(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)The difference between two numbers is 26 and one number is three times the other. Find them.

Let larger number    = x
Smaller number       = y
The difference between two numbers is 26
x – y   = 26
x         = 26 + y
Given that one number is three times the other
So                               x          = 3y
plug the value of x we get

26 + y       = 3y
-2y         = - 26
y         = 13
So value of x  = 3y
Plug value of y we get
X        = 3* 13            = 39
Answer smaller number = 13 bigger number is 39
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Let        first angle     = x
And second number = y

x + y   = 180
x         = 180 - y                     ... (1)

x – y   = 18
Plug the value of x we get

180– y – y    = 18
2 y     = -162
y        = -162/-2
y         = 81
Plug the value back in equation first we get
X         = 180 – 81     =  99
Hence angles are 81 and 99

(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.
Let        cost of each bat      =  Rs x
Cost of each ball      = Rs y
Given that coach of a cricket team buys 7 bats and 6 balls for Rs 3800.
So that                       7x + 6y           = 3800
6y        = 3800 – 7x
Divide by 6 we get
y         = (3800 – 7x) /6         … (1)
Given that she buys 3 bats and 5 balls for Rs 1750.so that
3x + 5y = 1750
Plug the value of y
3x + 5 ((3800 – 7x) /6)       = 1750
Multiply by 6 we get
18 x  + 19000 – 35 x            = 10500
-17x    =10500 - 19000
-17x    = -8500
x        = - 8500 / - 17
x        =  500
Plug this value in equation first we get
Y         = ( 3800 – 7 * 500) / 6
Y         = 300/6
Y         = 50
Hence cost of each bat = Rs 500 and cost of each balls is Rs 50

(iii) 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?

Let        the fixed charge  for taxi    = Rs x
And      variable cost per Km          = Rs y
Total cost      = fixed charge + variable charge
Given that for a distance of 10 km, the charge paid is Rs 105
x + 10 y        = 105                          … (1)
x         = 105 – 10 y
Given that for a journey of 15 km, the charge paid is Rs 155
x + 15 y          = 155
Plug the value of x we get
105 – 10 y + 15 y      = 155
5y        = 155 – 105
5y        = 50
Divide by 5  we get
y         = 50 / 5           = 10
Plug this value in equation (1) we get
x         = 105 – 10 * 10
x         = 5
People have to pay for travelling a distance of 25 km
= X + 25 y
= 5 + 25*10
= 5 + 250
=255
Answer pay for 25 Km is 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.
Let Numerator    = x
Denominator  = y
Fraction will              = x/ y
A fraction becomes 9/11, if 2 is added to both the numerator and the denominator
(x+2)  =          9
y+2              11

Cross multiply we get
11 x + 22        = 9y + 18
Subtract 22 both side we get
11 x     = 9 y – 4
Divide by 11 we get
x       = 9 y – 4
             11                          … (1)
Given that 3 is added to both the numerator and the denominator it becomes 5/ 6.
If, 3 is added to both the numerator and the denominator it becomes 5/6
(x+3)  =          5                      … (2)
y + 3             6

Cross multiply we get
6x + 18 = 5y + 15
Subtract the value of x now we get
6(9 y – 4 )   + 18 = 5y + 15
   11                  
Subtract 18 both side we get
6(9 y – 4 )     = 5y  - 3 
     11                   

54 – 24         =  55 y  -  33

-y        =  -9
y        = 9
Plug this value of y in equation (1)  we get
x              = 9 y – 4
                      11                                … (1)
x         = (81 – 4)/77
x         = 77/11
x         = 7

Hence our 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?
Let        present age of Jacob                     = x year
And      present Age of his son is             = y year
Five years hence,
Age of Jacob will                              = x + 5 year
Age of his son will                                       = y + 5year
Given that the age of Jacob will be three times that of his son
x + 5    =3(y+5)
Add 5 both side we get
x         = 3y + 15 - 5
x          = 3y + 10                    … (1)
Five years ago,
Age of Jacob will                              = x - 5 year
Age of his son will                                       = y - 5year
Jacob’s age was seven times that of his son
x – 5    = 7(y -5)
Plug the value of x from equation (1)  we get
3y + 10 – 5     = 7 y – 35
3y + 5             = 7y – 35
3y – 7 y          = - 35 – 5
-4y     =  - 40
y        = - 40 / - 4
y         = 10 year
Plug the value of y in equation first we get
x         = 3* 10 + 10
x         = 40 years
Answer is Present age of Jacob = 40 years Present age of his son = 10 years