CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables VBQs Set 06

Solve for 'x' and 'y' by using method of substitution.

Question. \( \frac{1}{2x} - \frac{1}{y} = -1 \), \( \frac{1}{x} + \frac{1}{2y} = 8, x \neq 0, y \neq 0 \)
Answer: \( x = \frac{1}{6}, y = \frac{1}{4} \)

Question. \( \frac{m}{x} - \frac{n}{y} = a, px - qy = 0, x \neq 0, y \neq 0 \)
Answer: \( x = \frac{mp - nq}{ap}, y = \frac{mp - nq}{aq} \)

Question. \( \frac{2}{x} + \frac{3}{y} = \frac{9}{xy}, \frac{4}{x} + \frac{9}{y} = \frac{21}{xy}, x \neq 0, y \neq 0 \)
Answer: \( x = 1, y = 3 \)

Question. \( \frac{2}{y} + \frac{3}{x} = \frac{7}{xy}, \frac{1}{y} + \frac{9}{x} = \frac{11}{xy}, x \neq 0, y \neq 0 \)
Answer: \( x = 2, y = 1 \)

Question. \( \frac{xy}{x + y} = \frac{6}{5}, \frac{xy}{y - x} = 6, xy \neq 0, y - x \neq 0 \)
Answer: \( x = 2, y = 3 \)

Question. \( \frac{a}{x} - \frac{b}{y} = 0, \frac{ab^2}{x} + \frac{a^2b}{y} = a^2 + b^2, x \neq 0, y \neq 0 \)
Answer: \( x = a, y = b \)

Question. \( x + y = 5xy, 3x + 2y = 13xy \)
Answer: \( x = \frac{1}{2}, y = \frac{1}{3} \)

Question. \( 5x + \frac{4}{y} = 9, 7x - \frac{2}{y} = 5, y \neq 0 \)
Answer: \( x = 1, y = 1 \)

Question. \( \frac{5}{x + y} - \frac{2}{x - y} = -1, \frac{15}{x + y} + \frac{7}{x - y} = 10, x + y \neq 0, x - y \neq 0 \)
Answer: \( x = 3, y = 2 \)

Question. \( \frac{2}{3x + 2y} + \frac{3}{3x - 2y} = \frac{17}{5}, \frac{5}{3x + 2y} + \frac{1}{3x - 2y} = 2, 3x + 2y \neq 0, 3x - 2y \neq 0 \)
Answer: \( x = 1, y = 1 \)

Question. \( \frac{5}{x + 1} - \frac{2}{y - 1} = \frac{1}{2}, \frac{10}{x + 1} + \frac{2}{y - 1} = \frac{5}{2}, x \neq -1, y \neq 1 \)
Answer: \( x = 4, y = 5 \)

Question. \( \frac{6}{x + y} = \frac{7}{x - y} + 3, \frac{1}{2(x + y)} = \frac{1}{3(x - y)}, x + y \neq 0, x - y \neq 0 \)
Answer: \( x = -\frac{5}{4}, y = -\frac{1}{4} \)

Question. \( 27x + 31y = 85, 31x + 27y = 89 \)
Answer: \( x = 2, y = 1 \)

Question. \( 37x - 39y = 150, 39x - 37y = 154 \)
Answer: \( x = 3, y = -1 \)

Question. \( \frac{148}{x} + \frac{231}{y} = \frac{527}{xy}, \frac{231}{x} + \frac{148}{y} = \frac{610}{xy} \)
Answer: \( x = 1, y = 2 \)

APPLICATION OF PAIR OF LINEAR EQUATIONS IN TWO VARIABLES (WORD PROBLEMS)

Question. 7 audio cassettes and 3 video cassettes cost Rs. 1110, while 5 audio cassettes and 4 video cassettes cost Rs. 1350. Find the cost of an audio cassette and a video cassette.
Answer: Let the cost of an audio cassette be Rs. \( x \) and that of a video cassette be Rs. \( y \). Then, \( 7x + 3y = 1110 \) ... (1) and \( 5x + 4y = 1350 \) ... (2) Multiplying equation (1) by 4 and equation (2) by 3, we get \( 28x + 12y = 4440 \) ... (3) \( 15x + 12y = 4050 \) ... (4) Subtracting (4) from (3), we have \( 13x = 390 \)
\( \implies x = 30 \) On putting \( x = 30 \) in equation (1), we get \( 7 \times 30 + 3y = 1110 \)
\( \implies 3y = 1110 - 210 = 900 \)
\( \implies y = 300 \) Hence, the cost of an audio cassette \( = x = \) Rs. 30 and the cost of a video cassette \( = y = \) Rs. 300

Question. Two numbers differ by 3 and their product is 54. Find the numbers.
Answer: Let the two numbers be \( x \) and \( y \). According to the 1st condition, Difference of the two numbers = 3
\( \implies x - y = 3 \)
\( \implies y = x - 3 \) ... (1) According to second condition, Product of two numbers = 54
\( \implies xy = 54 \) ... (2) Substituting \( y = x - 3 \) from (1) in (2), we get \( x(x - 3) = 54 \)
\( \implies x^2 - 3x = 54 \)
\( \implies x^2 - 3x - 54 = 0 \)
\( \implies x^2 - 9x + 6x - 54 = 0 \)
\( \implies x(x - 9) + 6(x - 9) = 0 \)
\( \implies (x - 9)(x + 6) = 0 \)
\( \implies x = 9 \) and \( x = -6 \) On putting \( x = 9 \) in equation (1), we have \( y = 9 - 3 = 6 \) On putting \( x = -6 \) in (1), we get \( y = -6 - 3 = -9 \) Hence, the numbers are 9 and 6 or -9 and -6.

Question. The average score of boys in an examination of a school is 71 and that of girls is 73. The average score of the school in the examination is 71.8. Find the ratio of the number of boys to the number of girls that appeared in the examination.
Answer: Let the number of boys = \( x \) Average score of boys = 71 Total score of boys = \( 71x \) Let the number of girls = \( y \) Average score of girls = 73 Total score of girls = \( 73y \) According to the question, Average score = \( \frac{\text{Total score}}{\text{Total number of students}} \)
\( \implies 71.8 = \frac{71x + 73y}{x + y} \)
\( \implies 71.8x + 71.8y = 71x + 73y \)
\( \implies 0.8x = 1.2y \)
\( \implies \frac{x}{y} = \frac{1.2}{0.8} = \frac{3}{2} \) Hence, the ratio of the number of boys to the number of girls = 3 : 2

Question. The sum of the two numbers is 18. The sum of their reciprocals is \( \frac{1}{4} \). Find the numbers.
Answer: Let the required numbers be \( a \) and \( b \). According to the question; Sum of the two numbers = 18 Sum of their reciprocals = \( \frac{1}{4} \)
\( \implies a + b = 18 \) ... (1)
\( \implies \frac{1}{a} + \frac{1}{b} = \frac{1}{4} \)
\( \implies \frac{a + b}{ab} = \frac{1}{4} \)
\( \implies \frac{18}{ab} = \frac{1}{4} \) [\( \because \) From (1), \( a + b = 18 \)]
\( \implies ab = 72 \) Now, \( (a - b)^2 = (a + b)^2 - 4ab = (18)^2 - (4 \times 72) = 36 \)
\( \implies a - b = 6 \). On solving \( a + b = 18 \) and \( a - b = 6 \), we get \( a = 12 \) and \( b = 6 \). Hence, the numbers are 12 and 6.

Question. Father's age is three times the sum of ages of his two children. After 5 years his age will be twice the sum of ages of two children. Find the age of father.
Answer: Let the age of father = \( x \) years. And the sum of the ages of his two children = \( y \) years According to the question Father's age = 3 \( \times \) (sum of the ages of his two children)
\( \implies x = 3y \) ... (1) After 5 years Father's age = \( (x + 5) \) years Sum of the ages of his two children = \( y + 5 + 5 = y + 10 \) [Age of his each children increases by 5 years] According to the question, After 5 years Father's age = 2 \( \times \) (sum of ages of his two children)
\( \implies x + 5 = 2 \times (y + 10) \)
\( \implies x + 5 = 2y + 20 \)
\( \implies x - 2y = 15 \) ... (2) Putting \( x = 3y \) from (1) in (2), we get \( 3y - 2y = 15 \)
\( \implies y = 15 \) years And \( x = 3y \)
\( \implies x = 3 \times 15 = 45 \)
\( \implies x = 45 \) years. Hence, father's age = 45 years

Question. The age of the father is 3 years more than 3 times the son's age. 3 years hence, the age of the father will be 10 years more than twice the age of the son. Find their present ages.
Answer: Let father's present age = \( x \) years Son's present age = \( y \) years According to the question Father's age = Three times son's age + 3 years
\( \implies x = 3 (y) + 3 \)
\( \implies x = 3y + 3 \)
\( \implies x - 3y = 3 \) ... (1) 3 years hence, father's age will be \( (x + 3) \) years And son's age will be \( (y + 3) \) years According to the question, After 3 years father's age = Twice the age of son then + 10 years \( x + 3 = 2 (y + 3) + 10 \)
\( \implies x + 3 = 2y + 6 + 10 \)
\( \implies x - 2y = 6 + 10 - 3 \)
\( \implies x - 2y = 13 \) ... (2) Subtracting equation (2) from equation (1), we get \( -y = -10 \)
\( \implies y = 10 \) years. Putting the value of \( y \) in equation (1), we get \( x - 3 (10) = 3 \)
\( \implies x - 30 = 3 \)
\( \implies x = 3 + 30 = 33 \) years Hence, Father's age = 33 years. Son's age = 10 years.

Question. The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.
Answer: Let the unit's place digit be \( x \) and the ten's place digit be \( y \). Original number = \( x + 10y \) The number obtained by reversing the digits = \( 10x + y \) According to the question, Original number + Reversed number = 99
\( \implies (x + 10y) + (10x + y) = 99 \)
\( \implies 11x + 11y = 99 \)
\( \implies x + y = 9 \) ... (1)
\( \implies x = 9 - y \) Given the difference of the digits = 3
\( \implies x - y = 3 \) ... (2) On putting the value of \( x = 9 - y \) from equation (1) in equation (2), we get \( (9 - y) - y = 3 \)
\( \implies 9 - 2y = 3 \)
\( \implies 2y = 6 \)
\( \implies y = 3 \) Substituting the value of \( y = 3 \) in equation (1), we get \( x = 9 - y = 9 - 3 = 6 \) Hence, the number is \( x + 10y = 6 + 10 \times 3 = 36 \).

Question. The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Determine the fraction.
Answer: Let Numerator = \( x \) and Denominator = \( y \). Then, fraction = \( \frac{x}{y} \) According to the first condition, Numerator + denominator = twice of the denominator - 3
\( \implies x + y = 2y - 3 \)
\( \implies 2y - y = 3 + x \)
\( \implies y = 3 + x \) ... (1) According to the second condition, Decreased numerator by 1 = \( \frac{1}{2} \) (decreased denominator)
\( \implies (x - 1) = \frac{1}{2} (y - 1) \)
\( \implies 2(x - 1) = y - 1 \)
\( \implies 2x - y = -1 + 2 \)
\( \implies 2x - y = 1 \) ... (2) Substituting \( y = 3 + x \) in equation (2), we have \( 2x - (3 + x) = 1 \)
\( \implies 2x - x = 1 + 3 \)
\( \implies x = 4 \) On putting \( x = 4 \) in equation (1), we get \( y = 3 + 4 \)
\( \implies y = 7 \) Hence, the fraction = \( \frac{x}{y} = \frac{4}{7} \)

Question. A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it has to increase the speed by 250 km/hr from the usual speed. Find its usual speed.
Answer: Let the usual speed of plane = \( x \) km/hr. The increased speed of the plane = \( y \) km/hr.
\( \implies y = (x + 250) \) km/hour. ... (1) Distance = 1500 km. According to the question, (Scheduled time) - (time in increasing the speed) = 30 minutes
\( \implies \frac{1500}{x} - \frac{1500}{y} = \frac{1}{2} \) ... (2) [Time = \( \frac{\text{Distance}}{\text{Speed}} \)]
\( \implies \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2} \) [\( \because \) From (1)]
\( \implies \frac{1500x + 375000 - 1500x}{x(x + 250)} = \frac{1}{2} \)
\( \implies x(x + 250) = 750000 \)
\( \implies x^2 + 250x - 750000 = 0 \)
\( \implies x^2 + 1000x - 750x - 750000 = 0 \)
\( \implies x(x + 1000) - 750 (x + 1000) = 0 \)
\( \implies (x - 750) (x + 1000) = 0 \)
\( \implies x = 750 \) or \( x = -1000 \) But speed can never be negative Hence, the usual speed = 750 km/hr.

Question. A boat goes 16 km upstream and 24 km downstream in 6 hours. It can go 12 km upstream and 36 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.
Answer: Let the speed of stream = \( y \) km/hr ; speed of boat in still water = \( x \) km/hr. And the speed of boat in upstream = \( (x - y) \) km/hr The speed of boat in downstream = \( (x + y) \) km/hr According to the question, Time taken in going 16 km upstream + time taken in going 24 km downstream = 6 hours.
\( \implies \frac{16}{x - y} + \frac{24}{x + y} = 6 \) ... (1) [\( \because \) Time = \( \frac{\text{Distance}}{\text{Speed}} \)] Again, according to the question, Time taken in going 12 km upstream + time taken in going 36 km downstream = 6 hours.
\( \implies \frac{12}{x - y} + \frac{36}{x + y} = 6 \) ... (2) Let \( \frac{1}{x - y} = p, \frac{1}{x + y} = q \) Equation (1) becomes \( 16p + 24q = 6 \) ... (3) Equation (2) becomes \( 12p + 36q = 6 \) ... (4) Multiplying equation (3) by 3 and equation (4) by 4, we get \( 48p + 72q = 18 \) ... (5) \( 48p + 144q = 24 \) ... (6) Subtracting equation (5) from equation (6), we get \( 72q = 6 \)
\( \implies q = \frac{6}{72} = \frac{1}{12} \) Putting the value of \( q \) in equation (3), we get \( 16p + 24\left(\frac{1}{12}\right) = 6 \)
\( \implies 16p + 2 = 6 \)
\( \implies 16p = 6 - 2 = 4 \)
\( \implies p = 1/4 \) \( \therefore \frac{1}{x - y} = \frac{1}{4} \) and \( \frac{1}{x + y} = \frac{1}{12} \)
\( \implies x - y = 4 \) ... (7) And, \( x + y = 12 \) ... (8) By adding \( 2x = 16 \implies x = 8 \) Putting \( x = 8 \) in equation (7), we get \( 8 - y = 4 \implies y = 8 - 4 = 4 \) Hence, speed of boat in still water = 8 km/hr. and speed of stream = 4 km/hr.

Question. A passenger train takes 2 hours less for a journey of 300 km, if its speed is increased by 5 km/hour from its usual speed. Find its usual speed.
Answer: Let the usual speed of the train = \( x \) km/hr. Distance travelled = 300 km Time taken = \( \frac{\text{Distance}}{\text{speed}} = \frac{300}{x} \) ... (1) Increased speed of the train = \( y \) km/hr.
\( \implies y = x + 5 \) ... (2) Time taken = \( \frac{300}{y} \) ... (3) According to the question, (Scheduled time) - (time by increasing the speed) = 2 hr.
\( \frac{300}{x} - \frac{300}{y} = 2 \) [From (1) and (3)] Putting the value of \( y \) from (2), we get
\( \implies \frac{300}{x} - \frac{300}{x + 5} = 2 \)
\( \implies \frac{300x + 1500 - 300x}{x(x + 5)} = 2 \implies \frac{1500}{x(x + 5)} = 2 \)
\( \implies 2x(x + 5) = 1500 \implies x^2 + 5x = 750 \)
\( \implies x^2 + 5x - 750 = 0 \implies x^2 + 30x - 25x - 750 = 0 \)
\( \implies x(x + 30) - 25(x + 30) = 0 \implies (x + 30) (x - 25) = 0 \)
\( \implies x = -30 \) or \( x = 25 \) But speed cannot be negative. Hence, the usual speed = 25 km/hr.

Question. A person rowing a boat at the rate of 5 km/hour in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.
Answer: Let the speed of the stream be \( x \) km/h and the speed of the boat in still water be 5 km/h. \( \therefore \) Relative speed of the boat downstream = \( (5 + x) \) km/h and Relative speed of the boat upstream = \( (5 - x) \) km/h \( \therefore \) Time taken by the boat downstream = \( \frac{40}{5 + x} \) h and Time taken by the boat upstream = \( \frac{40}{5 - x} \) h According to the given condition, a boat takes thrice as much time in going 40 km upstream as in going 40 km downstream, we have \( \frac{40}{5 - x} = 3 \left(\frac{40}{5 + x}\right) \)
\( \implies \frac{1}{5 - x} = \frac{3}{5 + x} \)
\( \implies 5 + x = 15 - 3x \)
\( \implies 3x + x = 15 - 5 \)
\( \implies 4x = 10 \)
\( \implies x = 2.5 \) Hence the speed of the stream is 2.5 km/h.

Question. Jamila sold a table and a chair for Rs. 1050, thereby making a profit of 10% on the table and 25% on the chair. If she had taken a profit of 25% on the table and 10% on the chair she would have got Rs. 1065. Find the cost price of each.
Answer: Let the cost price of the table be Rs. \( x \) and the cost price of the chair be Rs. \( y \).
The selling price of the table, when it is sold at a profit of 10% = \( Rs. \left( x + \frac{10}{100}x \right) = Rs. \frac{110}{100}x \)
The selling price of the chair when it is sold at a profit of 25% = \( Rs. \left( y + \frac{25}{100}y \right) = Rs. \frac{125}{100}y \)
So, \( \frac{110}{100}x + \frac{125}{100}y = 1050 \text{ ...(1)} \)
When the table is sold at a profit of 25%, its selling price = \( Rs. \left( x + \frac{25}{100}x \right) = Rs. \frac{125}{100}x \)
When the chair is sold at a profit of 10%, its selling price = \( Rs. \left( y + \frac{10}{100}y \right) = Rs. \frac{110}{100}y \)
So, \( \frac{125}{100}x + \frac{110}{100}y = 1065 \text{ ...(2)} \)
From Equations (1) and (2), we get
\( 110x + 125y = 105000 \)
and \( 125x + 110y = 106500 \)
On adding and subtracting these equations, we get
\( 235x + 235y = 211500 \)
and \( 15x - 15y = 1500 \)
i.e., \( x + y = 900 \text{ ...(3)} \)
and \( x - y = 100 \text{ ...(4)} \)
Solving Equations (3) and (4), we get
\( x = 500, y = 400 \)
So, the cost price of the table is Rs. 500 and the cost price of the chair is Rs. 400.

Question. After covering a distance of 30 km with a uniform speed there is some defect in a train engine and therefore, its speed is reduced to \( \frac{4}{5} \)th of its original speed. Consequently the train reaches its destination late by 45 minutes. Had it happened after covering 18 km further, find the speed of the train and the distance of Journey?
Answer: Let the original speed of train = \( x \) km/hr. and the distance of journey = \( y \) km/hr.
time taken = \( \frac{y}{x} \) hrs.
Case I. When defect in the engine occurs after covering 30 km.
Speed of first 30 km = \( x \) km/hr.
Speed of remaining \( (y - 30) \) km = \( \frac{4x}{5} \) km/hr.
Time taken to cover first 30 km = \( \frac{30}{x} \) hr.
and time taken to cover remaining \( (y - 30) \) km = \( \frac{y - 30}{4x/5} \text{ hrs} = \frac{5(y - 30)}{4x} \text{ hrs} \)
According to given condition,
\( \frac{30}{x} + \frac{5(y - 30)}{4x} = \frac{y}{x} + \frac{45}{60} \)
\( \implies \frac{120 + 5y - 150}{4x} = \frac{y}{x} + \frac{3}{4} \)
\( \implies \frac{5y - 30}{4x} = \frac{4y + 3x}{4x} \)
\( \implies 120 + 5y - 150 = 4y + 3x \)
\( \implies 3x - y = -30 \text{ ...(i)} \)
Case II. When the defect in the engine occurs after covering 48 km then in the same way. (If defect occurs after 48 km, then train reaches 9 minutes earlier, i.e., it is 36 minutes late)
\( \frac{48}{x} + \frac{5(y - 48)}{4x} = \frac{y}{x} + \frac{36}{60} \)
\( \implies \frac{192 + 5y - 240}{4x} = \frac{5y + 3x}{5x} \)
\( \implies 25y - 240 = 20y + 12x \)
\( \implies 12x - 5y = -240 \text{ ...(ii)} \)
Now solving (i) and (ii), we get \( x = 30 \) and \( y = 120 \) km respectively.
Original speed = 30 km/hr. and distance of journey = 120 km.

Question. 90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solutions. Find the amount of each type of acid to be mixed to form the mixture.
Answer: Let 90% of pure acid solution be \( x \) litres and 97% of pure acid solution be \( y \) litres.
Total volume of mixtures = \( x + y \) litres.
According to given data
\( x + y = 21 \text{ ...(i)} \)
Again 90% of \( x \) + 97% of \( y \) = 95% of 21
\( \implies 90x + 97y = 1995 \text{ ...(ii)} \)
Solving (i) and (ii), we get
\( x = 6 \) litres and \( y = 15 \) litres.

Question. It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled. How long would it take for each pipe to fill the pool separately?
Answer: Let the time taken by the pipe of larger diameter to fill the pool be \( x \) hours and that taken by the pipe of smaller diameter pipe alone be \( y \) hours.
In \( x \) hours, the pipe of larger diameter fills the pool.
So, in 1 hour the pipe of larger diameter fills \( \frac{1}{x} \) part of the pool, and so, in 4 hours, the pipe of larger diameter fills \( \frac{4}{x} \) parts of the pool.
Similarly, in 9 hours, the pipe of smaller diameter fills \( \frac{9}{y} \) parts of the pool.
According to the question,
\( \frac{4}{x} + \frac{9}{y} = \frac{1}{2} \text{ ...(1)} \)
Also, using both the pipes, the pool is filled in 12 hours.
So, \( \frac{12}{x} + \frac{12}{y} = 1 \text{ ...(2)} \)
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then Equations (1) and (2) become
\( 4u + 9v = \frac{1}{2} \text{ ...(3)} \)
\( 12u + 12v = 1 \text{ ...(4)} \)
Multiplying Equation (3) by 3 and subtracting Equation (4) from it, we get
\( 15v = \frac{1}{2} \text{ or } v = \frac{1}{30} \)
Substituting the value of \( v \) in Equation (4), we get \( u = \frac{1}{20} \)
So, \( u = \frac{1}{20}, v = \frac{1}{30} \)
So, \( \frac{1}{x} = \frac{1}{20}, \frac{1}{y} = \frac{1}{30} \)
or \( x = 20, y = 30 \).
So, the pipe of larger diameter alone can fill the pool in 20 hours and the pipe of smaller diameter alone can fill the pool in 30 hours.

Question. Two candles of equal heights but different thickness are lighted. The first burns off in 6 hours and the second in 8 hours. How long after lighting the both, will the first candle be half the height of the second?
Answer: Let after \( t \) hours first candle be the half of the second candle.
Suppose the height of 1st candle which burns in 6 hours = \( H \) cm.
Therefore the height of 1st candle which burns in 1 hour = \( \frac{H}{6} \)
Therefore the height of 1st candle which burns in \( t \) hours = \( \frac{tH}{6} \)
Height of remaining candle after \( t \) hours = \( H - \frac{tH}{6} \)
As the height of second candle which burns in 8 hrs is also \( H \) cm.
The height of second candle in 1 hr = \( \frac{H}{8} \).
The height of second candle after \( t \) hours = \( \frac{tH}{8} \).
Height of remaining candle after \( t \) hrs = \( H - \frac{tH}{8} \)
According to given condition, we have
\( H - \frac{tH}{6} = \frac{1}{2} \left( H - \frac{tH}{8} \right) \)
\( \implies 1 - \frac{t}{6} = \frac{1}{2} \left( 1 - \frac{t}{8} \right) \)
\( \implies \frac{6-t}{3} = \frac{8-t}{8} \)
\( \implies 48 - 8t = 24 - 3t \)
\( \implies 5t = 24 \)
\( \implies t = \frac{24}{5} \)
\( \implies t = 4.8 \text{ hrs} = 4 \text{ hrs and } 0.8 \times 60 \text{ minutes} = 4 \text{ hrs and 48 minutes.} \)

Question. On selling a tea-set at 5% loss and a lemon-set at 15% gain, a crockery seller gains Rs. 7. If he sells the tea-set at 5% gain and the lemon-set at 10% gain, he gains Rs. 13. Find the actual price of the tea-set and the lemon-set.
Answer: Let the cost price of the tea-set be Rs. \( x \) and that of the lemon-set be Rs. \( y \) respectively. We have two cases,
Case I. When tea-set is sold at 5% loss and lemon-set at 15% gain.
Loss on tea-set = \( Rs. \frac{5x}{100} = Rs. \frac{x}{20} \)
Gain on lemon-set = \( Rs. \frac{15y}{100} = Rs. \frac{3y}{20} \)
Net gain = \( \frac{3y}{20} - \frac{x}{20} \)
\( \implies \frac{3y}{20} - \frac{x}{20} = 7 \)
\( \implies 3y - x = 140 \implies x - 3y + 140 = 0 \text{ ...(i)} \)
Case II. When tea-set is sold at 5% gain and the lemon-set at 10% gain.
Gain on tea-set = \( Rs. \frac{5x}{100} = Rs. \frac{x}{20} \)
Gain on lemon-set = \( Rs. \frac{10y}{100} = Rs. \frac{y}{10} \)
Total gain = \( Rs. \frac{x}{20} + Rs. \frac{y}{10} \)
\( \implies \frac{x}{20} + \frac{y}{10} = 13 \implies x + 2y = 260 \implies x + 2y - 260 = 0 \text{ ...(ii)} \)
Subtracting equation (ii) from equation (i), we get
\( -5y + 400 = 0 \)
\( \implies y = 80 \)
Put the value of \( y \) in equation (i), we get
\( x - 240 + 140 = 0 \)
\( \implies x = 100 \)
Hence, cost price of the tea-set is Rs. 100 and that of lemon-set is Rs. 80.

Question. P takes 3 hours more than Q to walk 30 km. But, if P doubles his pace, he is ahead of Q by \( 1\frac{1}{2} \) hours. Find their speed of walking.
Answer: Let the speed of P and Q be \( x \) km/hr and \( y \) km/hr respectively. Then,
Time taken by P to cover 30 km = \( \frac{30}{x} \) hrs,
and Time taken by Q to cover 30 km = \( \frac{30}{y} \) hrs
By the given conditions, we have
\( \frac{30}{x} - \frac{30}{y} = 3 \implies \frac{10}{x} - \frac{10}{y} = 1 \text{ ...(i)} \)
If P doubles his pace, then speed of P = \( 2x \) km/hr
Times taken by P to cover 30 km = \( \frac{30}{2x} \) hrs.
Times taken by Q to cover 30 km = \( \frac{30}{y} \) hrs
According to the given conditions, we have
\( \frac{30}{y} - \frac{30}{2x} = 1\frac{1}{2} \)
\( \implies \frac{30}{y} - \frac{15}{x} = \frac{3}{2} \)
\( \implies \frac{10}{y} - \frac{5}{x} = \frac{1}{2} \)
\( \implies -\frac{10}{x} + \frac{20}{y} = 1 \text{ ...(ii)} \)
Putting \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \), in equations (i) and (ii) we get
\( 10u - 10v = 1 \implies 10u - 10v - 1 = 0 \text{ ...(iii)} \)
\( -10u + 20v = 1 \implies -10u + 20v - 1 = 0 \text{ ...(iv)} \)
Adding equations (iii) and (iv), we get
\( 10v - 2 = 0 \)
\( \implies v = \frac{1}{5} \)
Putting \( v = \frac{1}{5} \) in equation (iii), we get
\( 10u - 2 = 1 \)
\( \implies 10u = 3 \)
\( \implies u = \frac{3}{10} \)
Now, \( u = \frac{3}{10} \implies \frac{1}{x} = \frac{3}{10} \implies x = \frac{10}{3} \)
and, \( v = \frac{1}{5} \implies \frac{1}{y} = \frac{1}{5} \implies y = 5 \)
Hence, P's speed = \( \frac{10}{3} \) km/hr and Q's speed = 5 km/hr.

Question. 8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14 days. Find the time taken by one man alone and that by one boy alone to finish the work.
Answer: Suppose that one man alone can finish the work in \( x \) days and one boy alone can finish it in \( y \) days. Then,
One man's one day's work = \( \frac{1}{x} \)
One boy's one day's work = \( \frac{1}{y} \)
Eight men's one day's work = \( \frac{8}{x} \)
12 boy's one day's work = \( \frac{12}{y} \)
Since 8 men and 12 boys can finish the work in 10 days
\( 10 \left( \frac{8}{x} + \frac{12}{y} \right) = 1 \implies \frac{80}{x} + \frac{120}{y} = 1 \text{ ...(i)} \)
Again, 6 men and 8 boys can finish the work in 14 days.
\( 14 \left( \frac{6}{x} + \frac{8}{y} \right) = 1 \implies \frac{84}{x} + \frac{112}{y} = 1 \text{ ...(ii)} \)
Putting \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \) in equations (i) and (ii), we get
\( 80u + 120v - 1 = 0 \)
\( 84u + 112v - 1 = 0 \)
By using cross-multiplication, we have
\( \frac{u}{-120 + 112} = \frac{-v}{-80 + 84} = \frac{1}{80 \times 112 - 120 \times 84} \)
\( \implies \frac{u}{-8} = \frac{v}{-4} = \frac{1}{-1120} \)
\( \implies u = \frac{-8}{-1120} = \frac{1}{140} \text{ and } v = \frac{-4}{-1120} = \frac{1}{280} \)
Now, \( u = \frac{1}{140} \implies \frac{1}{x} = \frac{1}{140} \implies x = 140 \)
and \( v = \frac{1}{280} \implies \frac{1}{y} = \frac{1}{280} \implies y = 280 \).
Thus, one man alone can finish the work in 140 days and one boy alone can finish the work in 280 days.

Question. In a competitive examination, one mark is awarded for each correct answer while \( \frac{1}{2} \) mark is deducted for each wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?
Answer: Let the number of correctly answered questions be \( x \), then the number of wrongly answered questions be \( y \).
It is given that "the total number questions answered is 120".
\( x + y = 120 \text{ ...(1)} \)
Also, it is given that "one mark is awarded for each correct answer and \( \frac{1}{2} \) mark is deducted for each wrong answer and Jayanti has scored 90 marks".
\( 1 \cdot x - \frac{1}{2} \cdot y = 90 \)
\( \implies 2x - y = 180 \text{ ...(2)} \)
Adding (1) and (2), we get
\( (x + y) + (2x - y) = 120 + 180 \)
\( \implies 3x = 300 \)
\( \implies x = 100 \)
Substituting \( x = 100 \) in (1), we get: \( y = 120 - 100 = 20 \).
Hence, the number of correctly answered questions by Jayanti are 100.

Question. A boat covers 32 km upstream and 30 km downstream in 7 hours. Also, it covers 40 km upstream and 48 km downstream in 9 hours. Find the speed of the boat in still water and that of the stream.
Answer: Let the speed of the boat in still water be \( x \) km/h and that of the stream be \( y \) km/h. Then,
Speed upstream = \( (x - y) \) km/h
Speed downstream = \( (x + y) \) km/h
Now, time taken to cover 32 km upstream = \( \frac{32}{x - y} \) h
time taken to cover 30 km downstream = \( \frac{30}{x + y} \) h
the total time of journey is 7 hours
\( \frac{32}{x - y} + \frac{30}{x + y} = 7 \text{ ...(i)} \)
time taken to cover 40 km upstream = \( \frac{40}{x - y} \) h
time taken to cover 48 km downstream = \( \frac{48}{x + y} \) h
In this case, total time of journey is 9 hours.
\( \frac{40}{x - y} + \frac{48}{x + y} = 9 \text{ ...(ii)} \)
Put \( \frac{1}{x - y} = u \) and \( \frac{1}{x + y} = v \) in equations (i) and (ii), we get
\( 32u + 30v = 7 \implies 32u + 30v - 7 = 0 \)
\( 40u + 48v = 9 \implies 40u + 48v - 9 = 0 \)
By cross-multiplication, we have
\( \frac{u}{-270 + 336} = \frac{-v}{-288 + 280} = \frac{1}{1536 - 1200} \)
\( \implies \frac{u}{66} = \frac{v}{8} = \frac{1}{336} \)
(Using correct values from the calculation on the page)
\( \implies \frac{u}{12} = \frac{v}{8} = \frac{1}{96} \)
\( \implies u = \frac{12}{96} = \frac{1}{8} \) and \( v = \frac{8}{96} = \frac{1}{12} \)
We have \( \frac{1}{x - y} = \frac{1}{8} \implies x - y = 8 \text{ ...(iii)} \)
and \( \frac{1}{x + y} = \frac{1}{12} \implies x + y = 12 \text{ ...(iv)} \)
Solving equations (iii) and (iv), we get \( x = 10 \) and \( y = 2 \).
Hence, speed of the boat in still water is 10 km/h and speed of the stream is 2 km/h.

Question. Solve:
\( x + y + 2z = 9 \)
\( 2x - y + 2z = 6 \)
\( 3x + y + 4z = 17 \)
Answer: The given equations are
\( x + y + 2z = 9 \text{ ...(1)} \)
\( 2x - y + 2z = 6 \text{ ...(2)} \)
\( 3x + y + 4z = 17 \text{ ...(3)} \)
From equation (1), we have \( z = \frac{1}{2}(9 - x - y) \text{ ...(4)} \)
Substituting the value of \( z \) in equations (2) and (3), we get
\( 2x - y + 9 - x - y = 6 \implies x - 2y = -3 \text{ ...(5)} \)
\( 3x + y + 2(9 - x - y) = 17 \implies x - y = -1 \text{ ...(6)} \)
Subtracting equation (6) from equation (5), we have
\( -y = -2 \implies y = 2 \)
Putting the value of \( y = 2 \) in equation (6), we get
\( x - 2 = -1 \implies x = 1 \)
Putting \( x = 1 \) and \( y = 2 \) in equation (4), we get
\( z = \frac{1}{2}(9 - 1 - 2) = \frac{6}{2} = 3 \)
Hence, the solution is \( x = 1, y = 2, z = 3 \).