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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.