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Answer :
Ans.(i)Let fixed monthly charge = Rs x and let charge of food for one day = Rs y
According to given conditions,
\(x + 20y = 1000 \)… (1),
\( x + 26y = 1180 \)… (2)
Subtracting equation (1) from equation (2),
\(6y = 180\)
=>\( y = 30\)
Putting value of y in (1),
\(x + 20 (30) = 1000\)
=>\( x = 1000 – 600 = 400\)
Therefore, fixed monthly charges = Rs 400 and, charges of food for one day = Rs 30
(ii) Let numerator = x and let denominator = y
According to given conditions,
\({{x - 1}\over {y}} = {{1}\over {3}}\)… (1) and \({{x}\over {y + 8}} = {{1}\over {4}}\)… (2)
=>\( 3x – 3 = y \)… (1) \(4x = y + 8\) … (2)
=>\( 3x – y = 3 \)… (1) \(4x – y = 8 \)… (2)
Subtracting equation (1) from (2),
\(4x – y - (3x - y) = 8 – 3\)
\( x = 5\)
Putting value of x in (1),
\(3 (5) – y = 3\)
\(=> 15 – y = 3\)
\(=> y = 12\)
Therefore, numerator = 5 and, denominator = 12
It means fraction = \({{5}\over {12}}\)
(iii) Let number of correct answers = x and let number of wrong answers = y
According to given conditions,
\(3x – y = 40\) … (1)
And, \(4x - 2y = 50\) … (2)
From equation (1), \(y = 3x - 40\)
Putting this in (2),
\(4x – 2 (3x - 40) = 50\)
=>\( 4x - 6x + 80 = 50\)
=>\( -2x = -30\)
=>\( x = 15\)
Putting value of x in (1),
\(3 (15) – y = 40\)
=>\( 45 – y = 40\)
=>\( y = 45 – 40 = 5\)
Therefore, number of correct answers = x = 15 and number of wrong answers = y = 5
Total questions = x + y = 15 + 5 = 20
(iv)Let speed of car which starts from part A = x km/hr
Let speed of car which starts from part B = y km/hr
According to given conditions,(Assuming x > y)
=>\({{100} \over {x-y}} = 5\)
=>\( 5x - 5y = 100\)
=>\(x – y = 20 \)… (1)
And, \({{100} \over {x+y}} = 1\)
=>\( x + y = 100\) … (2)
Adding (1) and (2),
\(2x = 120\)
=>\( x = 60\)
Putting value of x in (1),
\(60 – y = 20\)
=>\( y = 60 – 20 = 40\) km/hr
Therefore, speed of car starting from point A = 60 km/hr
And, Speed of car starting from point B = 40 km/hr
(v) Let length of rectangle = x units and Let breadth of rectangle = y units
Area =xy square units. According to given conditions,
\(xy – 9 = (x - 5) (y + 3)\)
=>\( xy – 9 = xy + 3x - 5y – 15\)
=>\( 3x - 5y = 6 \)… (1)
And, \(xy + 67 = (x + 3) (y + 2)\)
=>\( xy + 67 = xy + 2x + 3y + 6\)
=>\( 2x + 3y = 61 \)… (2)
From equation (1),
=>\(3x = 6 + 5y\)
=>\( x = {{6 + 5y} \over {3}}\)
Putting this in (2),
=>\(2({{6 + 5y} \over {3}}) + 3y = 61\)
=>\( 12 + 10y + 9y = 183\)
=>\( 19y = 171\)
=>\( y = 9 \) units
Putting value of y in (2),
\(2x + 3 (9) = 61\)
=>\( 2x = 61 – 27 = 34\)
=>\( x = 17 \) units
Therefore, length = 17 units and, breadth = 9 units