Form the pair of linear equations for the following problems and find their solution by substitution method.

Question :  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 charges per kilometer ? How much does a person have to pay for travelling a distance of 25 km ?

(v)  A fraction becomes \(9\over 11\), if 2 is added to both the numerator and denominator. If 3 is added to both the numerator and denominator it becomes \(5\over 6\). Find the fraction.

(vi)  Five years hence, the age of Jacob will be three times that of son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages.

Solution :

(i)  Let x and y be the numbers.

And the difference of two numbers is 26.

i.e.  x – y = 26          ………(1)

One number is three times the other.

i.e.  x = 3y               ………(2)

Putting x = 3y in equation (1)  we get

3y – y = 26  \(\implies\)  2y = 26  \(\implies\)  y =13

Now, put the value of y = 13 in equation (2), we get  x = 39

Hence, the numbers are x = 39 and y = 13.

(ii)  Let x and y be the angles. Then,

ATQ,

x + y = 180      …….(1)

and  x = y + 18        ……(2)

Putting x = y + 18 in equation (1)  we get

y + 18 + y = 180

2y = 180 – 18   \(\implies\)   2y = 162   \(\implies\)  y = 81

Now, put the value of y = 81 in equation (2), we get

x = 81 + 18 = 90

Hence, the angles are x = 99 degrees and y = 81 degrees.

(iii)  Let x be the cost of one bat and y be the cost of one ball respectively. Then,

7x + 6y = 3800          ………(1)

i.e.  3x + 5y = 1750               ………(2)

From equation (2), y = \(1750 – 3x\over 5\)

Substituting the value of y = \(1750 – 3x\over 5\) in equation (1), we get

35x + 10500 – 18x = 19000

\(\implies\)  17x = 8500  \(\implies\)  x = 500

Now, put the value of x = 500 in equation (2), we get

3(500) + 5y = 1750  \(\implies\)  5y = 250  \(\implies\)  y = 50

Hence, the cost of one bat is 500 Rupees and the cost of one ball is 50 Rupees.

(iv)  Let x Rs be the fixed charges of taxi and

the y Rs be the running charges of taxi per km.

ATQ,

Expenses of travelling 10 km = 105 Rs

i.e.   x + 10y = 105          ……(i)

Again the expenses of travelling 15 km = 155 Rs

i.e.   x + 15y = 155          ……..(ii)

Putting the value of x = 155 – 15y from equation (2) in equation (1), we get

155 – 15y + 10y = 105

155 – 5y = 105

\(\implies\)  5y = 50  \(\implies\)  y = 10

Now, put the value of y = 10 in equation (2), we get

x + 15(10) = 155

\(\implies\) x = 155 – 150 = 5

Hence, the fixed charges of taxi is 5 Rupees and the running charges of taxi per km is 10 Rupees.

A person have to pay for travelling 25 km = 5 + 25(10) = 5 + 250 = 255 Rupees.

(v)  Let x be the numerator and y be the denominator. Then, according to the question,

CASE 1 :  \(x+ 2\over y + 2\) = \(9\over 11\)

\(\implies\)  11(x + 2) = 9(y + 2)   \(\implies\)   11x + 22 = 9y + 18

\(\implies\)  11x – 9y = -4    ……(1)

CASE 2 :  \(x + 3\over y + 3\) = \(5\over 6\)

\(\implies\)  6(x + 3) = 5(y + 3)  \(\implies\)  6x + 18 = 5y + 15

\(\implies\)  6x – 5y = – 3

\(\implies\)  x = \(5y – 3\over 6\)           ……(2)

Put the value of x = \(5y – 3\over 6\)  in equation (1), we get

11(\(5y – 3\over 6\) ) – 9y = -4

\(\implies\)  11(5y – 3) – 54y = -24

\(\implies\)  55y – 33 – 54 = -24

\(\implies\)   y = 33 – 24 = 9

Putting the value of  y = 9 in (1), we get

11x – 9(9)  = – 4

11x = -4 + 81 = 77

x = 7

Hence, \(7\over 8\) is the required fraction.

(vi)  Let Jacob’s present age be x and his son be y.

CASE 1 : After five years age of Jacob = (x + 5),

After five years the age of his son = (y + 5).

According question,

x + 5 = 3(y + 5)

\(\implies\)  x – 3y = 10

CASE 2 : Five years ago Jacob’s age = x – 5, his son’s age = y – 5. Then,

ATQ,

x – 5 = 7(y – 5)   \(\implies\)   x = 7y – 30

Putting x = 7y – 30  from equation (2) in (1), we get

7y – 30 – 3y = 10

\(\implies\)  4y = 40   \(\implies\)  y = 10

Now, put y = 10 in equation (1), we get

x – 3(10) = 10

\(\implies\)  x = 10 + 30 = 40

Hence, the age of Jacob is 40 years, and age of his son is 10 years.

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