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Answer :
Ans.(i)Let numerator =x and let denominator =y
According to given condition,
=>\({{x + 1}\over{y - 1}} = 1\) and \({{x}\over{y + 1}} = {{1}\over{2}}\)
=>\( x + 1 = y – 1\) and \(2x = y + 1\)
=>\( x – y = -2 \)….......(1) and,
\( 2x – y = 1\)…...... (2)
So, we have equations (1) and (2), multiplying equation (1) by 2 we get (3)
\(2x - 2y = -4\)… (3)
\(2x – y = 1\)… (2)
Subtracting equation (2) from (3), we get
\(-y = -5 => y = 5\)
Putting value of y in (1), we get
\(x – 5 = -2 => x = -2 + 5 = 3\)
Therefore, fraction = \({{x}\over{y}} = {{3}\over{5}}\)
(ii) Let present age of Nuri = x years and let present age of Sonu = y years
5 years ago, age of Nuri = \((x – 5)\) years
5 years ago, age of Sonu = \((y – 5)\) years
According to given condition,
\((x - 5) = 3 (y - 5)\)
=> \(x – 5 = 3y – 15\)
=> \(x - 3y = -10\)… (1)
10 years later from present, age of Nuri = \((x + 10)\) years
10 years later from present, age of Sonu = \((y + 10)\) years
According to given condition,
\((x + 10) = 2 (y + 10)\)
=> \(x + 10 = 2y + 20\)
=> \(x - 2y = 10 … (2)\)
Subtracting equation (1) from (2),
\(y = 10 - (-10) = 20 \)years
Putting value of y in (1),
\(x – 3 (20) = -10\)
=> \(x – 60 = -10\)
=> \(x = 50 \)years
Therefore, present age of Nuri = 50 years and present age of Sonu = 20 years
(iii) Let digit at ten’s place = x and Let digit at one’s place = y
According to given condition,
\(x + y = 9 \)… (1)
And 9 (10x + y) = 2 (10y + x)
=>\( 90x + 9y = 20y + 2x\)
=>\( 88x = 11y\)
=>\( 8x = y\)
=>\( 8x – y = 0 \)… (2)
Adding (1) and (2),
\(9x = 9 => x = 1\)
Putting value of x in (1),
\(1 + y = 9\)
=>\( y = 9 – 1 = 8\)
Therefore, number = \(10x + y = 10 (1) + 8 = 10 + 8 = 18\)
(iv)Let number of Rs 100 notes = x and let number of Rs 50 notes = y
According to given conditions,
\(x + y = 25\) … (1)
and \(100x + 50y = 2000\)
=>\( 2x + y = 40\) … (2)
Subtracting (2) from (1),
\(-x = -15 => x = 15\)
Putting value of x in (1),
\(15 + y = 25\)
=>\( y = 25 – 15 = 10\)
Therefore, number of Rs 100 notes = 15 and number of Rs 50 notes = 10
(v) Let fixed charge for 3 days = Rs x
Let additional charge for each day thereafter = Rs y
According to given condition,
\(x + 4y = 27 \)… (1)
\(x + 2y = 21 \)… (2)
Subtracting (2) from (1),
\(2y = 6 => y = 3\)
Putting value of y in (1),
\(x + 4 (3) = 27\)
=>\( x = 27 – 12 = 15\)
Therefore, fixed charge for 3 days = Rs 15 and additional charge for each day after 3 days = Rs 3