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(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 \({{9}\over{11}}\) , if 2 is added to both the numerator and the 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 his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer :

Ans (i)Let first number and second number be x and y respectively.

According to given question

\(x – y = 26\) (assuming x > y) … (1)

\(x = 3y \)(Since,x > y)… (2)

On putting equation (2) in (1),

\(3y – y = 26\)

=>\( 2y = 26\)

=>\( y = 13\)

On putting value of y in equation (2),

=>\(x = 3y = 3(13) = 39\)

Therefore, two numbers are 13 and 39.

**(ii)** Let smaller angle and larger angle be x and y respectively.

According to given question

\(y = x + 18\) … (1)

\(x + y = 180 \)(Sum of supplementary angle)… (2)

On putting equation (1) in (2),

\(x + x + 18 = 180\)

=>\( 2x = 180 - 18 = 162\)

=>\( x = 81^\circ\)

On putting value of x in equation (1),

=>\(y = x + 18 = 81 + 18 = 99^\circ \)

Therefore, two numbers are \(81^\circ\) and \(99^\circ \)

**(iii)** Let cost of one bat and cost of one ball be x and y respectively.

According to given question

\(7x + 6y = 3800\) .....(1)

\(3x + 5y = 1750 \) .....(2)

Using equation (1):

\(7x = 3800 - 6y \)

=>\( x = {{3800 - 6y} \over{7}}\) .....(3)

On putting equation (3) in (1),

\(3({{3800 - 6y} \over{7}}) + 5y = 1750\)

=>\( {{11400 - 18y} \over{7}} + 5y = 1750\)

=>\( {{5y} \over{1}} - {{18y} \over{7}} = {{1750} \over{1}} -{{11400 } \over{7}}\)

=>\({{35y - 18y} \over{7}} = {{12250 - 11400} \over{7}}\)

=>\(17y = 850\)

=>\(y = 50\)

On putting value of y in equation (2),

=>\(3x + 250 = 1750\)

=>\(3x = 1500\)

=>\(x = 500\)

Therefore, cost of one bat and cost of one ball is Rs.500 and Rs.50 respectively.

**(iv)** Let fixed charge and charge for every km be Rs.x and Rs.y respectively.

According to given question

\(x + 10y = 105\) … (1)

\(x + 15y = 155 \) … (2)

Using equation (1)

\(x = 105 - 10y\) … (3)

On putting equation (3) in (2),

\(105 - 10y + 15y = 155\)

=>\( 5y = 50\)

=>\( y = 10\)

On putting value of y in equation (1),

=>\(x + 10 (10) = 105\)

=>\(x = 105 – 100 = 5\)

Therefore, fixed charge and charge for every km is Rs.5 and Rs.10 respectively.

To travel distance of 25 Km, person will have to pay

=> Rs \(x + 25y\)

=> Rs \(5 + 25 × 10\)

=> Rs \(5 + 250\) = Rs 255

**(v)** Let numerator and denominator be x and y respectively.

According to given question

\({{x + 2} \over{y + 2}} = {{9} \over{11}}\) .........(1)

\({{x + 3} \over{y + 3}} = {{5} \over{6}}\) .........(2)

Using equation (2):

\(6(x + 3) = 5(y + 3)\).......(4)

Using equation (1):

\(11(x + 2) = 9(y + 2)\)

=>\(11x + 22 = 9y + 18\)

=>\(11x = 9y - 4\)

=>\(x = {{9y - 4} \over{11}}\)........(3)

On putting equation (3) in (4),

\(6({{9y - 4} \over{11}} + 3) = 5(y + 3) \)

=>\( {{54y} \over{11}} - {{24} \over{11}} + 18 = 5y + 15\)

=>\(-{{24} \over{11}} + {{33} \over{11}} = {{55y} \over{11}} - {{54y} \over{11}}\)

=>\(-{{24 + 33} \over{11}} = {{55y - 54y} \over{11}}\)

=>\(y = 9\)

On putting value of y in equation (1),

=>\({{x + 2} \over{9 + 2}} = {{9} \over{11}}\)

=>\( x + 2 = 9\)

=>\(x = 7\)

Therefore, numerator and denominator is 7 and 9 respectively.

Fraction = \({{7} \over{9}}\)

**(vi)** Let present age of Jacob and his son be x and y respectively.

According to given question

\(x + 5 = 3(y + 5)\) .....(1)

\(x - 5 = 7(y - 5) \) .....(2)

Using equation (1):

\(x + 5 = 3y + 15\)

=>\( x = 10 + 3y \) .....(3)

On putting equation (3) in (2),

\(10 + 3y – 5 = 7y - 35\)

=>\( -4y = -40\)

=>\(y = 10\)

On putting value of y in equation (3),

=>\(x = 10 + 3(10)\)

=>\(x = 40\)

Therefore, present age of Jacob and his son be 40 years and 10 years respectively.

- Solve the following pair of linear equations by the substitution method. (i) \(x + y = 14,x – y = 4\)(ii) \(s – t = 3,{{s}\over{3}} + {{t}\over{2}} = 6\)(iii) \(3x – y = 3,9x - 3y = 9\)(iv) \(0.2x + 0.3y = 1.3,0.4x + 0.5y = 2.3\)(v) \(\sqrt{2}x + \sqrt{3}y = 0,\sqrt{3}x - \sqrt{8}y = 0\)(vi) \({{3x}\over{2}} - {{5y}\over{3}} = -2,{{x}\over{3}} + {{y}\over{2}} = {{13}\over{6}}\)
- Solve \(2x + 3y = 11\) and \(2x - 4y = -24\) and hence find the value of ‘m’ for which \(y = mx + 3\).

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