Class 8: Simultaneous Linear Equations -Exercise 23B

Question 1: The sum of two numbers is $\displaystyle 60$ and their difference is $\displaystyle 14$ . Find the numbers.

Answer:

Let the two numbers be $\displaystyle x \text{ and } y$

$\displaystyle x+y = 60 ... ... ... ... ... ... (i)$

$\displaystyle x-y = 14 ... ... ... ... ... ... (ii)$

Add (i) and (ii)

$\displaystyle 2x = 74 \text{ or }x = 37$

Therefore $\displaystyle y = 37-14 = 23$

Hence the two numbers are $\displaystyle 23 \text{ and } 37$

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Question 2: Twice a number is equal to thrice the other number. If the sum of the numbers is $\displaystyle 85$ , find the numbers.

Answer:

Let the two numbers are $\displaystyle x \text{ and } y$

$\displaystyle 2x = 3y ... ... ... ... ... ... (i)$

$\displaystyle x+y = 85 ... ... ... ... ... ... (ii)$

Solving for $\displaystyle x \text{ and } y$

$\displaystyle 2x-3y = 0$

$\displaystyle \underline {(-)2x+2y = 170 }$

$\displaystyle -5y = -170$

or $\displaystyle y = 34$

Therefore $\displaystyle x = \frac{3}{2} \times 34 = 51$

Hence the two numbers are $\displaystyle 51 \text{ and } 34$ .

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Question 3: Find two numbers such that twice the first added to thrice the second gives $\displaystyle 70$ and twice the second added to thrice the first gives $\displaystyle 75$ .

Answer:

Let the numbers be $\displaystyle x \text{ and } y$

$\displaystyle 2x+3y = 70 ... ... ... ... ... ... (i)$

$\displaystyle 3x+2y = 75 ... ... ... ... ... ... (ii)$

Solving for $\displaystyle x \text{ and } y$

Multiply i) by 2 and ii) by 3 and subtract ii) from i)

$\displaystyle 6x+9y = 210$

$\displaystyle \underline{(-) 6x+4y = 150}$

$\displaystyle 5y = 60$

$\displaystyle \Rightarrow y = 12$

Substituting in i)

$\displaystyle \Rightarrow x = \frac{70-3y}{2} = \frac{70-36}{2} = 17$

Hence the two numbers are $\displaystyle 17 \text{ and } 12$

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Question 4: Find two numbers which differ by $\displaystyle 9$ and are such that four times the larger added to three times the smaller gives $\displaystyle 92$ .

Answer:

Let the numbers be $\displaystyle x \text{ and } y$

$\displaystyle x-y = 9 ... ... ... ... ... ... (i)$

$\displaystyle 4x+3y = 92 ... ... ... ... ... ... (ii)$

Solving by $\displaystyle x \text{ and } y$

Multiply i) by 4 and subtract ii) from i)

$\displaystyle 4x-4y = 36$

$\displaystyle \underline{(-)4x+3y = 92}$

$\displaystyle -7y = -56$

$\displaystyle \Rightarrow y = 8$

Calculating for $\displaystyle x$

$\displaystyle \Rightarrow x = y+9 = 8+9 = 17$

Hence the two numbers are $\displaystyle 8 \text{ and } 17$

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Question 5: The sum of two numbers is $\displaystyle 30$ and the difference of their squares is $\displaystyle 180$ . Find the numbers.

Answer:

Let the two numbers be $\displaystyle x \text{ and } y$

$\displaystyle x+y = 30 ... ... ... ... ... ... (i)$

$\displaystyle x^2-y^2 = 180 ... ... ... ... ... ... (ii)$

Simplifying ii)

$\displaystyle (x-y)(x+y) = 180$

$\displaystyle \Rightarrow x-y = 180/30 = 6$

$\displaystyle x-y = 6 ... ... ... ... ... ... (iii)$

Solving for $\displaystyle x \text{ and } y$

Add i) and iii)

$\displaystyle x+y = 30$

$\displaystyle \underline{(+) x-y = 6}$

$\displaystyle 2x = 36$

$\displaystyle \Rightarrow x = 18$

Substituting in i) $\displaystyle y = 30-18 = 12$

Hence the numbers are $\displaystyle 18 \text{ and } 12$

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Question 6: The sum of the digits of a two-digit number is $\displaystyle 8$ . On adding $\displaystyle 18$ to the number, its digits are reversed. Find the number.

Answer:

Let the two digit numbers be $\displaystyle xy$

$\displaystyle x+y = 8 ... ... ... ... ... ... (i)$

$\displaystyle 10x+y+18 = 10y+x ... ... ... ... ... ... (ii)$

Simplifying ii)

$\displaystyle 9x-9y = -18$

$\displaystyle \Rightarrow x-y = -2 ... ... ... ... ... ... (iii)$

Solving for $\displaystyle x \text{ and } y$

Add i) and iii)

$\displaystyle x+y = 8$

$\displaystyle \underline{(+) x-y = -2}$

$\displaystyle 2x = 6$

$\displaystyle \Rightarrow x = 3$

Substituting in i) $\displaystyle \Rightarrow y = 8-3 = 5$

Hence the numbers is $\displaystyle 35$

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Question 7: Two digit number is three times the sum of its digits. lf $\displaystyle 45$ is added to the number, its digits are reversed. Find the original number.

Answer:

Let the two digit numbers be $\displaystyle xy$

$\displaystyle 3(x+y) = 10x+y ... ... ... ... ... ... (i)$

$\displaystyle 10x+y+45 = 10y+x ... ... ... ... ... ... (ii)$

Simplifying i) and ii)

$\displaystyle 7x-2y = 0 ... ... ... ... ... ... (iii)$

$\displaystyle 9x-9y = -45$

$\displaystyle \Rightarrow x-y = -5 ... ... ... ... ... ... (iv)$

Solving for $\displaystyle x \text{ and } y$

Multiplying iv) by 7 and Subtract iv) from ii)

$\displaystyle 7x-2y = 0$

$\displaystyle \underline{(-)7x-7y = -35}$

$\displaystyle 5y = 35$

$\displaystyle \Rightarrow y = 7$

Hence $\displaystyle x = y-5 = 7-5 = 2$

Therefore the numbers is $\displaystyle 27$

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Question 8: A two-digit number is seven times the sum of its digits. If $\displaystyle 27$ is subtracted from the number, its digits get interchanged. Find the number.

Answer:

Let the two digit number be $\displaystyle xy$

$\displaystyle 10x+y = 7(x+y)$

$\displaystyle \Rightarrow 3x-6y = 0 ... ... ... ... ... ... (i)$

$\displaystyle 10x+y-27 = 10y+x$

$\displaystyle \Rightarrow 9x-9y = 27$

$\displaystyle \Rightarrow x-y = 3 ... ... ... ... ... ... (ii)$

Solving for $\displaystyle x \text{ and } y$

Multiplying ii) by 3 and Subtract ii) from i)

$\displaystyle 3x-6y = 0$

$\displaystyle (-) 3x-3y = 9$

$\displaystyle -3y = -9$

$\displaystyle \Rightarrow y = 3$

Hence $\displaystyle x = y+3 = 6$

Therefore the numbers is $\displaystyle 63$

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Question 9: Find a fraction which reduces to $\displaystyle 2/3$ when $\displaystyle 3$ is added to both its numerator and denominator; and reduces to $\displaystyle 3/5$ when $\displaystyle 1$ is added to both its numerator and denominator.

Answer:

Let the fraction be $\displaystyle \frac{x}{y}$

$\displaystyle \frac{x+3}{y+3} = \frac{2}{3}$

$\displaystyle \Rightarrow 3x+9 = 2y+6$

$\displaystyle \Rightarrow 3x-2y = -3 ... ... ... ... ... ... (i)$

$\displaystyle (x+1)/(y+1) = 3/5$

$\displaystyle \Rightarrow 5x+5 = 3y+3$

$\displaystyle \Rightarrow 5x-3y = 2 ... ... ... ... ... ... (ii)$

Solving for $\displaystyle x \text{ and } y$

Multiplying i) by 5 and ii) by 3 and subtract ii) from i)

$\displaystyle 15x-10y = -15$

$\displaystyle \underline {(-)15x-9y = -6}$

$\displaystyle -y = -9$

$\displaystyle \Rightarrow y = 9$

Hence $\displaystyle x = \frac{2y-3}{3} = \frac{18-3}{3} = 5$

Therefore the fraction is $\displaystyle \frac{5}{9}$

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Question 10: On adding $\displaystyle 1$ to the numerator of a fraction, it becomes $\displaystyle 1/2$ . Also, on adding $\displaystyle 1$ to the denominator of the original fraction, it becomes $\displaystyle 1/3$ . Find the original fraction.

Answer:

Let the fraction be $\displaystyle \frac{x}{y}$

$\displaystyle \frac{x+1}{y} = \frac{1}{2}$

$\displaystyle \Rightarrow 2x-y = -2 ... ... ... ... ... ... (i)$

$\displaystyle \frac{x}{y+1} = \frac{1}{3}$

$\displaystyle \Rightarrow 3x-y = 1 ... ... ... ... ... ... (ii)$

Solving for $\displaystyle x \text{ and } y$

Multiplying i) by 3 and ii) by 2 and subtract ii) from i)

$\displaystyle 6x-3y = -6$

$\displaystyle \underline{ (-) 8x-2y = 2}$

$\displaystyle -y = -8$

$\displaystyle \Rightarrow y = 8$

Hence $\displaystyle x = \frac{1+8}{3} = 3$

Therefore the fraction is $\displaystyle \frac{3}{8}$

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Question 11: In a given fraction, if the numerator is multiplied by $\displaystyle 2$ and the denominator is reduced by $\displaystyle 5$ , we get $\displaystyle 6/5$ . But, if the numerator of the given fraction is increased by $\displaystyle 8$ and the denominator is doubled, we get $\displaystyle 2/5$ . Find the fraction.

Answer:

Let the fraction be $\displaystyle \frac{x}{y}$

$\displaystyle \frac{2x}{y-5} = \frac{6}{5}$

$\displaystyle \Rightarrow 10x-6y = -30 ... ... ... ... ... ... (i)$

$\displaystyle \frac{x+8}{2y} = \frac{2}{5}$

$\displaystyle \Rightarrow 5x-4y = -40 ... ... ... ... ... ... (ii)$

Solving for $\displaystyle x \text{ and } y$

Multiplying ii) by 2 and subtract ii) from i

$\displaystyle 10x-6y = -3$

$\displaystyle \underline{(-)10x-8y = -80}$

$\displaystyle 2y = 50$

$\displaystyle \Rightarrow y = 25$

Hence $\displaystyle x = (4y-40)/5 = (100-40)/5 = 12$

Therefore the fraction in $\displaystyle \frac{12}{25}$

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Question 12: $\displaystyle 5$ years ago, a lady was thrice as old as her daughter. $\displaystyle 10$ years hence, the lady would be twice as old as her daughter. What are their present ages?

Answer:

Let the present age of lady $\displaystyle = x$

Let the Present age of Daughter $\displaystyle = y$

$\displaystyle x-5 = 3(y-5)$

$\displaystyle \Rightarrow x-3y = -10 ... ... ... ... ... ... (i)$

$\displaystyle x+10 = 2(y+10)$

$\displaystyle \Rightarrow x-2y = 10 ... ... ... ... ... ... (ii)$

Solving for $\displaystyle x \text{ and } y$

Subtract ii) from i)

$\displaystyle x-3y = -10$

$\displaystyle \underline{(-) x-2y = 10}$

$\displaystyle -y = -20$

$\displaystyle \Rightarrow y = 20$

Hence $\displaystyle x = 2y+10 = 50$

Therefore :

Lady’s Age $\displaystyle = 50 \text{ years }$

Daughter’s Age $\displaystyle = 20 \text{ years }$

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Question 13: The sum of the ages of A and B is $\displaystyle 39$ years. In $\displaystyle 15$ years’ time, the age of A will be twice the age of B. Find their present ages.

Answer:

Let A’s Age $\displaystyle = x$

Let B’s Age $\displaystyle = y$

$\displaystyle x+y = 39 ... ... ... ... ... ... (i)$

$\displaystyle x+15 = 2(y+15)$

$\displaystyle \Rightarrow x-2y = 15 ... ... ... ... ... ... (ii)$

Solving for $\displaystyle x \text{ and } y$

Subtract ii) from i)

$\displaystyle x+y = 39$

$\displaystyle \underline{(-) x-2y = 15}$

$\displaystyle 3y = 24$

or $\displaystyle y = 8$

$\displaystyle \Rightarrow x = 39-y = 39-8 = 31$

Therefore

A’s Age $\displaystyle = 31 \text{ years }$

B’s Age $\displaystyle = 8 \text{ years }$

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Question 14: A is $\displaystyle 15$ years elder than B. $\displaystyle 5$ years ago A was four times as old as B. Find their present ages.

Answer:

Let the age of B $\displaystyle = x,$

$\displaystyle \Rightarrow Age of A = x+15$

$\displaystyle +15-5 = 4(x-5)$

$\displaystyle \Rightarrow x+10 = 4x-20$

$\displaystyle 3x = 30 \text{ or }x = 10$

Hence A’s Age $\displaystyle = 10+15 = 25$

Therefore

B’s Age $\displaystyle = 10 \text{ years }$

A’s Age $\displaystyle = 25 \text{ years }$

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Question 15: Six years ago, the ages of Geeta and Seema were in the ratio $\displaystyle 3 : 4$ . Nine years hence, their ages will be in the ratio $\displaystyle 6 : 7$ . Find their present ages.

Answer:

Let the age of Geeta $\displaystyle = x$

Let the age of Seema $\displaystyle = y$

$\displaystyle \frac{x-6}{y-6} = \frac{3}{4}$

$\displaystyle \Rightarrow 4x-24 = 3y-18$

$\displaystyle \Rightarrow 4x-3y = 6 ... ... ... ... ... ... (i)$

$\displaystyle \frac{x+9}{y+9} = \frac{6}{7}$

$\displaystyle \Rightarrow 7x+63 = 6y+54$

$\displaystyle 7x-6y = -9 ... ... ... ... ... ... (ii)$

Solving for $\displaystyle x \text{ and } y$

Multiplying i) by 7 and ii) by 4 and Subtract ii) from i)

$\displaystyle 28x-21y = 42$

$\displaystyle \underline{(-)28x-24y = -36}$

$\displaystyle 3y = 78$

$\displaystyle \Rightarrow y = 26$

Hence $\displaystyle x = \frac{6+3\times 26}{4} = 21$

Therefore

Geeta’s Age $\displaystyle = 21 \text{ years }$

Seema’s Age $\displaystyle = 26 \text{ years }$

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Question 16: $\displaystyle 4$ knives and $\displaystyle 6$ forks cost Rs. $\displaystyle 200$ , while $\displaystyle 6$ knives and, $\displaystyle 7$ forks together cost Rs. $\displaystyle 264$ . Find the cost of a knife and that of a fork.

Answer:

Let the cost of knives $\displaystyle = x$

Let the Cost of fork $\displaystyle = y$

$\displaystyle 4x+6y = 200 ... ... ... ... ... ... (i)$

$\displaystyle 6x+7y = 264 ... ... ... ... ... ... (ii)$

Solving for $\displaystyle x \text{ and } y$

Multiplying i) by 3 and ii) 2

$\displaystyle 12x+18y = 600$

$\displaystyle \underline{(-)12x+14y = 528}$

$\displaystyle 4y = 72$

$\displaystyle \Rightarrow y = 18$

Hence $\displaystyle x = \frac{200-6\times 18}{4} = 23$

Hence Cost of :

Knive $\displaystyle = 23 \text{ Rs. }$

Fork $\displaystyle = 18 \text{ Rs. }$

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Question 17: The cost of $\displaystyle 13$ cups and $\displaystyle 16$ spoons is Rs. $\displaystyle 296$ , while the cost of $\displaystyle 16$ cups and $\displaystyle 13$ spoons is Rs. $\displaystyle 284$ . Find the cost of $\displaystyle 2$ cups and $\displaystyle 5$ spoons.

Answer:

Let the cost of Cup $\displaystyle = x$

Let the Cost of Spoon $\displaystyle = y$

$\displaystyle 13x+16y = 296 ... ... ... ... ... ... (i)$

$\displaystyle 16x+13y = 284 ... ... ... ... ... ... (ii)$

Solve for $\displaystyle x \text{ and } y$

Multiply i) by 16 and ii) by 13 and subtract ii) from i)

$\displaystyle 208x+256y = 4736$

$\displaystyle \underline{(-) 208x+169y = 3692}$

$\displaystyle 87y = 1044$

$\displaystyle \Rightarrow y = 12$

Hence $\displaystyle x = \frac{296-16\times 12}{13} = 8$

Hence Cost of:

Cup $\displaystyle = 8 \text{ Rs. }$

Spoon $\displaystyle = 12 \text{ Rs. }$

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Question 18: Rahul covered a distance of $\displaystyle 128$ km in $\displaystyle 5$ hours, partly on bicycle at $\displaystyle 16$ kmph and partly on moped at $\displaystyle 32$ kmph. How much distance did he cover on moped?

Answer:

Total Distance $\displaystyle = 128 km$

Total Time $\displaystyle = 5 Hrs.$

Let the distance covered by cycle $\displaystyle = x km$

Let the Distance covered by moped $\displaystyle = (120-x) Km.$

$\displaystyle \Rightarrow \frac{x}{16} + \frac{120x}{32} = 5$

Simplify $\displaystyle \Rightarrow 2x+128-x = 160$

$\displaystyle \Rightarrow x = 160-128 = 32$

Hence Distance Covered by:

Cycle $\displaystyle = 32 \text{ Km. }$

Moped $\displaystyle = 96 \text{ Km. }$

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Question 19: A boat can go $\displaystyle 75$ km downstream in $\displaystyle 5$ hours and, $\displaystyle 44$ km upstream in $\displaystyle 4$ hours. Find i) the speed of the boat in still water (ii) the rate of the current.

Answer:

Let the Speed of boat in still water $\displaystyle = x$

Speed of Stream $\displaystyle = y$

$\displaystyle 75/(x+y) = 5$

$\displaystyle \Rightarrow x+y = 17 ... ... ... ... ... ... (i)$

$\displaystyle 4x/(x-y) = 4$

$\displaystyle \Rightarrow x-y = 11 ... ... ... ... ... ... (ii)$

Solve for $\displaystyle x \text{ and } y$ , add i) and ii)

$\displaystyle 2x = 28 \text{ or }x = 14 \text{ Km /hr }$

Speed of Stream $\displaystyle = 3 \text{ Km/hr }$

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Question 20: The monthly incomes of A and B are in the ratio $\displaystyle 4 : 3$ and their monthly savings are in the ratio $\displaystyle 9 : 5$ . If each spends Rs. $\displaystyle 3500$ per month, find the monthly income of each.

Answer:

Let A’s Income $\displaystyle = x Rs.$

Let B’s Income $\displaystyle = y Rs.$

$\displaystyle \frac{x}{y} = \frac{4}{3}$

$\displaystyle \Rightarrow 3x = 4y ... ... ... ... ... ... (i)$

$\displaystyle \frac{x-3500}{y-3500} = \frac{9}{5}$

$\displaystyle \Rightarrow 5x-17500 = 9y-31500$

$\displaystyle \Rightarrow 5x-9y = -14000 ... ... ... ... ... ... (ii)$

Substituting $\displaystyle x = \frac{4}{3} y in ii)$

$\displaystyle \Rightarrow 5( \frac{4}{3} y)-9y = -14000$

$\displaystyle -7y = -14000\times 3 \text{ or }y = 6000 \text{ Rs. }$

Hence $\displaystyle x = \frac{4}{3} \times 6000 = 8000 \text{ Rs. }$

Therefore :

A’s Income $\displaystyle = 8000 \text{ Rs. }$

B’s Income $\displaystyle = 6000 \text{ Rs. }$

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Question 21: $\displaystyle 6$ nuts and $\displaystyle 5$ bolts weigh $\displaystyle 278$ grams, while $\displaystyle 8$ nuts and $\displaystyle 3$ bolts weigh $\displaystyle 268$ grams. Find the weight of each nut and that of each bolt. How much do $\displaystyle 3$ nuts and $\displaystyle 3$ bolts weigh together?

Answer:

Let the Weight of Nut $\displaystyle = x gm.$

Let the Height of Bolt $\displaystyle = y gm.$

$\displaystyle 6x+5y = 278 ... ... ... ... ... ... (i)$

$\displaystyle 8x+3y = 268 ... ... ... ... ... ... (ii)$

Solving for $\displaystyle x \text{ and } y$

Multiply i) by 8 and ii) by 6 and Subtract ii) from i)

$\displaystyle 48x+40y = 2224$

$\displaystyle \underline{(-)48x+18y = 1608}$

$\displaystyle 22y = 616)$

$\displaystyle \Rightarrow y = 28$

Hence $\displaystyle x = \frac{278-5\times 28}{6} = 23$

Weight of:

Nut $\displaystyle = 23 \text{ gm }$

Bolt $\displaystyle = 28 \text{ gm }$

Therefore 3 Nuts and 3 Bolts will weight $\displaystyle = 3\times 23+3\times 28 = 153 \text{ gm }$

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Question 22: There are some girls in two classrooms, A and B. If $\displaystyle 12$ girls are sent from room A to room B, the number of girls in both the rooms will become equal. If $\displaystyle 11$ girls are sent from room B to room A, then the number of girls in room A would be double the number of girls in room B. How many girls are there in each class room?

Answer:

Let No. of girls in classroom $\displaystyle A = x \text{ and in } B = y$

$\displaystyle x-12 = y+12$

$\displaystyle \Rightarrow x-y = 24 ... ... ... ... ... ... (i)$

$\displaystyle 2(y-11) = x+11$

$\displaystyle \Rightarrow x-2y = -33 ... ... ... ... ... ... (ii)$

Solving $\displaystyle x \text{ and } y$ ,

Subtract ii) from i)

$\displaystyle x-y = 24$

$\displaystyle \underline{(-) x-2y = -33}$

$\displaystyle y = 57$

Hence $\displaystyle x = 24+57 = 81$

No. of Girls in Class Room:

$\displaystyle A = 81$

$\displaystyle B = 57$

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Question 23: $\displaystyle 4$ men and $\displaystyle 4$ boys can do a piece of work in $\displaystyle 3$ days, while $\displaystyle 2$ men and $\displaystyle 5$ boys can finish it in $\displaystyle 4$ days. How long would it take $\displaystyle 1$ man alone to do it?

Answer:

Suppose 1 man finished the work in $\displaystyle x$ days and 1 Boy finished the work in $\displaystyle y$ days

There 1 man’s 1 day’s $\displaystyle = \frac{1}{x}$

And, 1 Boy’s 1 Day’s work $\displaystyle = \frac{1}{y}$

Now 4 men and 4 Boys finished the work in 3 days

$\displaystyle \Rightarrow \frac{ 4}{x} + \frac{5}{y} = \frac{1}{3} ... ... ... ... ... ... (i)$

Similarly 2 men and 5 boys finished in 4 days

$\displaystyle \Rightarrow \frac{ 2}{x} + \frac{5}{y} = \frac{1}{4} ... ... ... ... ... ... (ii)$

Solving for $\displaystyle x \text{ and } y$

Multiply ii) by 2 and Subtract ii) from 2

$\displaystyle \Rightarrow \frac{ 4}{x} + \frac{4}{y} = \frac{1}{3}$

$\displaystyle \Rightarrow \underline{(-) \frac{ 4}{x}+\frac{10}{y} = \frac{1}{2}}$

$\displaystyle \frac{-6}{y} = \frac{-1}{6}$

or $\displaystyle y = 36$

Similarly $\displaystyle x = \frac{4}{\frac{1}{3}-\frac{4}{36}} = \frac{4 \times 36}{8} = 18$

So one men can finished the work in $\displaystyle 18 days.$

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Question 24: A takes $\displaystyle 3$ hours longer than B to walk $\displaystyle 30$ km. But, if A doubles his pace, he is ahead of B by $\displaystyle 1$ hour $\displaystyle 30$ minutes. Find the speeds of A and B.

Answer:

Let the Speed of A $\displaystyle = x \text{ KM/Hr. }$

Let the Speed of B $\displaystyle = y \text{ KM/Hr. }$

Time taken By A $\displaystyle = \frac{30}{x}$

Time Taken By B $\displaystyle = \frac{30}{y}$

$\displaystyle \Rightarrow \frac{30}{x} = \frac{30}{y} +3 ... ... ... ... ... ... (i)$

If A Double him Race Time taken by A $\displaystyle = \frac{30}{2x}$

Therefore

$\displaystyle \frac{30}{2x} + \frac{3}{2} = \frac{30}{y}$

$\displaystyle \Rightarrow \underline{(-) \frac{30}{x} = \frac{30}{y}+3}$

$\displaystyle \frac{-30}{2x} + \frac{3}{2} = -3$

$\displaystyle \Rightarrow x = \frac{10}{3} = 3 \frac{1}{3} \text{ KM/Hr. }$

Hence $\displaystyle \frac{30}{y} = \frac{30}{10} \times 3-3 = 6$

$\displaystyle \Rightarrow y = \frac{30}{6} = 5 \text{ KM/Hr. }$

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Question 25: If the length of a rectangle is reduced by $\displaystyle 1$ m and breadth increased by $\displaystyle 2$ m, its area increases by $\displaystyle 32$ m2. If however, the length is increased by $\displaystyle 2$ m and breadth reduced by $\displaystyle 3$ m, then the area is reduced by $\displaystyle 49$ m2. Find the length and breadth of the rectangle.