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Pair Of Linear Equations In Two Variables Class 10th Mathematics Gujarat Board Solution

Class 10th Mathematics Gujarat Board Solution
Exercise 3.1
  1. Father tells his son, "Five years ago, I was seven times as old as you were.…
  2. The sum of the cost of 1kg apple and 1kg pine-apple is Rs. 150 and the cost of 1…
  3. Nilesh got twice the marks as obtained by Ilesh, in the annual examination of…
  4. Length of a rectangle is four less than the thrice of its breadth. The perimeter…
  5. The sum of the weights of a father and a son is 85 kg. The weight of the son is…
  6. In a cricket match, Sachin Tendulkar makes his score thrice the Sehwag's score.…
  7. In tossing a balanced coin, the probability of getting head on its face is twice…
Exercise 3.2
  1. 2x + y = 8, x + 6y = 15 Solve the following pair of linear equation in two…
  2. x + y = 1, 3x + 3y = 2 Solve the following pair of linear equation in two…
  3. 2x + 3y = 5, x + y = 2 Solve the following pair of linear equation in two…
  4. x-y = 6, 3x - 3y = 18 Solve the following pair of linear equation in two…
  5. (x + 2) (y — 1) = xy, (x — 1) (y + 1) = xy Solve the following pair of linear…
  6. Draw the graphs of lie pair of linear equations 3x + 2y = 5 and 2x-3y = -1.…
  7. 15 students of class X took part in the examination of Indian mathematics…
  8. Examine graphically whether the pair of equations 2x + 3y = 5 and x + 9/6 y =…
Exercise 3.3
  1. x + y = 7, 3x — y = 1 Solve the following pairs of linear equations by the…
  2. 3x — y = 0, x — y + 6 = 0 Solve the following pairs of linear equations by the…
  3. 2x + 3y = 5, 2x + 3y = 7 Solve the following pairs of linear equations by the…
  4. x — y = 3, 3x — 3y = 9 Solve the following pairs of linear equations by the…
  5. 3x/2 - 5y/3 = - 2 , x/3 + y/2 = 13/6 Solve the following pairs of linear…
  6. Solve the pair of linear equations x — y = 28 and x — 3y = 0 and if the solution…
  7. A fraction becomes 4/5 if 3 is added to both the numerator and the denominator.…
  8. The sum of present ages of a father and his son is 50 years. After 5 years, the…
  9. A bus traveller travelling with some of his relatives buys 5 tickets from…
Exercise 3.4
  1. x/5 - y/3 = 4/15 , x/2 - y/9 = 7/18 Solve the following pair of linear…
  2. 4x — 19y + 13 = 0, 13x — 23y = —19 Solve the following pair of linear equations…
  3. x + y = a + b, ax — by = a^2 — b^2 Solve the following pair of linear equations…
  4. 5ax + 6by = 28; 3ax + 4by = 18 Solve the following pair of linear equations by…
  5. The sum of two numbers is 35. Four times the larger number is 5 more than 5…
  6. There are some 25 paise coins and some 50 paise coins in a bag. The total number…
  7. The sum of the digits of two-digit number is 3. The number obtained by…
  8. The length of a rectangle is twice its breadth. The perimeter of the rectangle…
  9. An employee deposits certain amount at the rate of 8% per annum and a certain…
Exercise 3.5
  1. 0.3x + 0.4y = 2.5 and 0.5x — 0.3y = 0.3 Solve the following pairs of equations…
  2. 5x + 8y = 18, 2x — 3y = 1 Solve the following pairs of equations by cross…
  3. x/3 + y/5 = 1 , 7x— 15y = 21 Solve the following pairs of equations by cross…
  4. 3x + y = 5, 5x + 3y = 3 Solve the following pairs of equations by cross…
  5. By cross multiplication method, find such a two digit number such that, the…
  6. The sum of two numbers is 70 and their difference is 6. Find these numbers by…
  7. While arranging certain students of a school in rows containing equal number of…
  8. In AABC, the measure of ∠B is thrice to the measure of ∠C and the measure of ∠A…
Exercise 3.6
  1. 5/2x + 2/3y = 7 , 3/x + 2/y = 12 x ≠ 0, y ≠ 0 Solve the following pairs of…
  2. 2x + 3y = 2xy, 6x + 12y = 7xy Solve the following pairs of linear equations:…
  3. 4/x-1 + 5/y-1 = 2 , 8/x-1 + 15/y-1 = 3 x ≠ 1, y ≠ 1 Solve the following pairs…
  4. 1/3x+y + 1/3x-y = 3/4 , 1/2 (3x+y) - 1/2 (3x-y) = -1/8 3x + y ≠ 0, 3x - y ≠ 0…
  5. 3/root x + 4/root y = 2 , 5/root x + 7/root y = 41/42 x 0, y 0 Solve the…
  6. 5 women and 2 men together can finish an embroidery work in 4 days, while 6…
  7. A boat goes 21 km upstream and 18 km downstream in 9 hours. In 13 hours, it can…
  8. Solve the following pair of equations by cross multiplication method: 4x+7y/xy =…
  9. Mahesh travels 250 km to his home partly by train and partly by bus. He takes 6…
Exercise 3
  1. Obtain a pair of linear equations from the following information: "The rate of…
  2. Draw the graphs of the pair of linear equations in two variables. x + 3y = 6, 2x…
  3. Solve the following pair of equations by the method of elimination: 4/x + 5/y =…
  4. Solve the following pair of linear equations by the method of cross -…
  5. Solve the following pair of equations: 4/x+1 + 7/y+2 = 2 and 10/x+1 + 14/y+2 =…
  6. The difference between two natural numbers is 6. Adding 10 to the twice of the…
  7. The area of a rectangle gets increased by 30 square units, if its length is…
  8. A part of monthly hostel charges is fixed and the remaining depends on the…
  9. A fraction becomes 2/5 when 2 is subtracted from the numerator and denominator…
  10. The solution set of x — 3y = 1 and 3x + y = 3 isA. {(0, 1)) B. {(1, 1)} C.…
  11. The solution set of 2x + y = 6 and 4x + 2y = 5 isA. {(x, Y)| 2x + y = 6; x, y…
  12. To eliminate x, from 3x + y = 7 and —x + 2y = 2 second equation is multiplied…
  13. If 2x + 3y = 7 and 3x + 2y = 3, then x — y =A. 4 B. —4 C. 2 D. —2…
  14. If the pair of linear equations ax + 2y = 7 and 2x + 3y = 8 has a unique…
  15. The pair of linear equations 2x + y — 3 = 0 and 6x + 3y = 9 hasA. a unique…
  16. If in a two digit number, the digit at unit place is x and the digit at tens…
  17. In a two digit number, the digit at tens place is 7 and the sum of the digits…
  18. The sum of two numbers is 10 and the difference of them is 2. Then the greater…
  19. 3 years ago, the sum of ages of a father and his son was 40 years. After 2…
  20. The solution set of 2x + 4y = 8 and x + 2y = 4 isA. {(2, 1)} B. empty set C.…
  21. Equation x/2 - y/3 = 1 can be expressed in the standard form asA. 2x — 3y — 6…

Exercise 3.1
Question 1.

Obtain a pair of linear equations in two variables from the following information:

Father tells his son, "Five years ago, I was seven times as old as you were. After five years, I will be three times as old as you will be".


Answer:

Let the present age of father = x


Let the present age of son = y


Now,


Given, Five years ago, father was seven times as old as son.


⇒ (x–5) = 7(y–5)


x – 5 = 7y – 35


= x – 7y + 30 = 0 …[1]


Also, after five years, father will be three times as old as son


⇒ (x + 5) = 3(y + 5)


x + 5 = 3y + 15


= x – 3y – 10 = 0 …[2]


eq[1] and eq[2] are required equations



Question 2.

Obtain a pair of linear equations in two variables from the following information:

The sum of the cost of 1kg apple and 1kg pine-apple is Rs. 150 and the cost of 1 kg apple is twice the cost of 1kg pine-apple.


Answer:

Let the cost of 1 kg apple = x


Let the cost of 1 kg pine-apple = y


Given that, sum of the cost of 1 kg apple and 1 kg pine-apple is Rs. 150


⇒ x + y = 150 …[1]


Also, cost of 1 kg apple is twice the cost of 1kg pine-apple


⇒ x = 2y …[2]


eq[1] and eq[2] are the required equations.



Question 3.

Obtain a pair of linear equations in two variables from the following information:

Nilesh got twice the marks as obtained by Ilesh, in the annual examination of mathematics of standard 10. The sum of the marks as obtained by them is 135.


Answer:

let marks obtained by Ilesh be x and marks obtained by Nilesh by y.

Now,


given, Nilesh got twice the marks as obtained by Ilesh.


⇒ y = 2x …[1]


Also, the sum of the marks as obtained by them is 135.


x + y = 135 …[2]


These are the two required linear equations.



Question 4.

Length of a rectangle is four less than the thrice of its breadth. The perimeter of the rectangle is 110.


Answer:

Let the length of the rectangle be x and breadth be y.

Now,


given, Length of a rectangle is four less than the thrice of its breadth


x = 3y – 4 …[1]


and


also, the perimeter is 110


2(x + y) = 110


⇒ x + y = 55 …[2]


eq[1] and eq[2] are the required equations.



Question 5.

Obtain a pair of linear equations in two variables from the following information:

The sum of the weights of a father and a son is 85 kg. The weight of the son is 1/4 of the weight of his father.


Answer:

let x and y be the weight of father and son respectively.

Now,


Given, The sum of the weights of a father and a son is 85 kg


x + y = 85 …[1]


and,


also, The weight of the son is  of the weight of his father


 …[2]


These are the required linear equations.



Question 6.

Obtain a pair of linear equations in two variables from the following information:

In a cricket match, Sachin Tendulkar makes his score thrice the Sehwag's score. Both of them together make a total score of 200 runs.


Answer:

Let the score of Sachin and Sehwag be x and y respectively

given, Both of them together make a total score of 200 runs.


⇒ x + y = 200 …[1]


also, Sachin Tendulkar makes his score thrice the Sehwag's score


⇒ x = 3y …[2]


These are the two required equations.



Question 7.

Obtain a pair of linear equations in two variables from the following information:

In tossing a balanced coin, the probability of getting head on its face is twice to the probability of getting tail on its face. The sum of both probabilities (head and tail) is 1.


Answer:

Let the probabilities of getting heads and tails are x and y respectively.

Given, the probability of getting head on its face is twice to the probability of getting tail on its face


⇒ x = 2y …[1]


also, the sum of probabilities is 1


⇒ x + y = 1 …[2]


∴ these two equations are the required equations.




Exercise 3.2
Question 1.

Solve the following pair of linear equation in two variables (by graph)

2x + y = 8, x + 6y = 15


Answer:

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on the line, take different values of x related to it, we get different values for y from the equation.


We get the following tables for the given linear equations.


For 2x + y = 8


y = 8 – 2x



For x + 6y = 15


x = 15 – 6y




Plotting these points on a graph and joining them, we get two straight lines.


From the graph, we can see that both lines intersect at (3, 2), hence the solution to this pair is (3, 2).



Question 2.

Solve the following pair of linear equation in two variables (by graph)

x + y = 1, 3x + 3y = 2


Answer:

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on the line, take different values of x related to it, we get different values for y from the equation.

for x + y = 1


⇒ y = 1 – x



for, 3x + 3y = 2



Plotting these points on a graph and joining them, we get two straight lines.



From the graph, we can see that both lines are parallel and have no intersection points.



Question 3.

Solve the following pair of linear equation in two variables (by graph)

2x + 3y = 5, x + y = 2


Answer:

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.


for


2x + 3y = 5



for


x + y = 2



Plot it



Plotting these points on a graph and joining them, we get two straight lines.


From the graph, we can see that both lines intersect at (1, 1), hence solution to this pair is (1, 1).



Question 4.

Solve the following pair of linear equation in two variables (by graph)

x–y = 6, 3x – 3y = 18


Answer:

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.


for


x—y = 6



for, 3x—3y = 18



Plotting these points on a graph and joining them, we get two straight lines.



From the graph, we can see that both lines coincide and hence there are infinite number of solutions.



Question 5.

Solve the following pair of linear equation in two variables (by graph)

(x + 2) (y — 1) = xy, (x — 1) (y + 1) = xy


Answer:

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on the line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.


for


(x + 2) (y — 1) = xy


It is a straight line



for,


(x — 1) (y + 1) = xy




Plotting these points on a graph and joining them, we get two straight lines.


From the graph, we can see that both lines intersect at (4, 3), hence the solution to this pair is (4, 3).



Question 6.

Draw the graphs of lie pair of linear equations 3x + 2y = 5 and 2x–3y = –1. Determine the coordinates of the vertices of the triangle formed by these linear equations and the X-axis.


Answer:

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So, we get two points on the straight line. To find more points on line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.


for


3x + 2y = 5



for, 2x–3y = –1




when y = 0, lines would intersect the x-axis


so the coordinates of triangle at x-axis would be 


So the coordinates of the triangle will be




Question 7.

15 students of class X took part in the examination of Indian mathematics Olympiad. The number of boys participants is 5 less than the number of girls participants. Find the number of boys and girls (using a graph) who took part in the examination of Indian mathematics Olympiad.


Answer:

Let the number of girls be x and the number of boys be y.

Given, total no. of participants is 15.


⇒ x + y = 15


Also, boys participants is 5 less than the number of girls participants


⇒ x = y – 5


We have two equations we will plot them and their intersection will give the required result.


x + y = 15



x = y – 5




Plotting these points on a graph and joining them, we get two straight lines.


From the graph, we can see that both lines intersect at (5, 10), hence solution to this pair is (5, 10).


∴ no. of boys is 5 and no. of girls is 10



Question 8.

Examine graphically whether the pair of equations 2x + 3y = 5 and  is consistent


Answer:

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.


for


2x + 3y = 5



for 



Plotting these points on a graph and joining them, we get two straight lines.



From the graph, we can see that both lines coincide.


⇒ They don’t have consistent solutions.




Exercise 3.3
Question 1.

Solve the following pairs of linear equations by the method of substitution:

x + y = 7, 3x — y = 1


Answer:

x + y = 7

⇒ x = 7 – y …[1]


By substituting eq[1] in 3x – y = 1


⇒ 3(7 – y) – y = 1


⇒ 21 – 3y – y = 1


⇒ 4y = 20


⇒ y = 5


If we put the above result in eq[1], we get


⇒ x = 7 – 5


⇒ x = 2



Question 2.

Solve the following pairs of linear equations by the method of substitution:

3x — y = 0, x — y + 6 = 0


Answer:

3x – y = 0

⇒ y = 3x …[1]


By substituting eq[1] in x – y + 6 = 0


x – (3x) + 6 = 0


⇒ x = 3


Now, by eq[1]


3x – y = 0


∴ 3(3) – y = 0


⇒ y = 9


⇒ x = 3, y = 9



Question 3.

Solve the following pairs of linear equations by the method of substitution:

2x + 3y = 5, 2x + 3y = 7


Answer:

2x + 3y = 5

⇒ 2x = 5–3y


⇒  …[1]


By substituting eq[1] in 2x + 3y = 7



⇒ 5–3y + 3y = 7


⇒ 5 = 7


it is not true


we got this independent of any variable and it is a false statement, no real solution exists to these linear pair of equations.



Question 4.

Solve the following pairs of linear equations by the method of substitution:

x — y = 3, 3x — 3y = 9


Answer:

x – y = 3

⇒ x = y + 3 …[1]


By substituting eq[1] in 3x – 3y = 9


3(y + 3) – 3y = 9


9 = 9


The equation is independent of any variable and is true and therefore


the pair of linear equations are concurrent and satisfies all real values of x and y.



Question 5.

Solve the following pairs of linear equations by the method of substitution:



Answer:

multiplying 6 to L.H.S and R.H.S to the first equation to remove fraction

we get,


9x – 10y = –12 …[1]


multiplying 6 to L.H.S and R.H.S to the second equation to remove fraction


we get, 2x + 3y = 13


⇒  …[2]


Substituting eq[2] in eq[1]


⇒ 


⇒ 27x – 130 + 20x = –36


⇒ x = 2


Using this result in eq[2]



⇒ y = 3



Question 6.

Solve the pair of linear equations x — y = 28 and x — 3y = 0 and if the solution satisfies, y = mx + 5, then find m.


Answer:

x – y = 28 eq[1] eq[2]

and x – 3y = 0


⇒ x = 3y


Substituting eq[2] in eq[1]


3y – y = 28


⇒ 2y = 28


⇒ y = 14


Now by, x = 3y


⇒ x = 3(14)


⇒ x = 42


∴ in y = mx + 5


putting values of x and y.


14 = 42m + 5


⇒ 14–5 = 42m


⇒ 



Question 7.

A fraction becomes  if 3 is added to both the numerator and the denominator. If 5 is added to the numerator and the denominator, it becomes, Find the fraction.


Answer:

Let the numerator and denominator of the actual fraction be x and y.

Given, fraction becomes  if 3 is added to both the numerator and the denominator



⇒ 5x = 4y – 3


⇒  …[1]


also, If 5 is added to the numerator and the denominator, it becomes .


⇒ 


⇒ 6x – 5y = –5 …[2]


substituting eq[1] in eq[2]



⇒ y = 7


by eq[1] and y = 7



⇒ x = 5


∴ original fraction = 



Question 8.

The sum of present ages of a father and his son is 50 years. After 5 years, the age of the father becomes thrice the age of his son. Find their present ages.


Answer:

Let the present ages of the father be x and the son be y.

given, The sum of present ages of a father and his son is 50 years.


⇒ x + y = 50


⇒ x = 50 – y …[1]


also, After 5 years, the age of the father becomes thrice the age of his son


⇒ x + 5 = 3(y + 5) …[2]


By substituting eq[1] in eq[2]


we get,


50–y + 5 = 3(y + 5)


⇒ y = 10


and by eq[1] and y = 10


x = 40


so the present age of father is 40 years and present ages of son is 10 years.



Question 9.

A bus traveller travelling with some of his relatives buys 5 tickets from Ahmedabad to Anand and 10 tickets from Ahmedabad to Vadodara for Rs. 1100. The total cost of one ticket from Ahmedabad to Anand and one ticket from Ahmedabad to Vadodara is Rs. 140. Find the cost of a ticket from Ahmedabad to Anand as well as the cost of a ticket from Ahmedabad to Vadodara.


Answer:

Let the cost of one ticket from Ahmedabad to Anand be x and the cost of one ticket form Ahmedabad to Vadodara be y.

given, 5 tickets from Ahmedabad to Anand and 10 tickets from Ahmedabad to Vadodara for Rs. 1100.


⇒ 5x + 10y = 1100 …[1]


Also, cost of one ticket from Ahmedabad to Anand and one ticket from Ahmedabad to Vadodara is Rs. 140


⇒ x + y = 140


⇒ y = 140 – x …[2]


By substituting eq[2] in eq[1] we get


5x + 10(140 – x) = 1100


⇒ 5x + 1400 – 10x = 1100


⇒ 5x = 300


x = 60


and by eq[2] and x = 60


⇒ y = 140 – 60


⇒ y = 80




Exercise 3.4
Question 1.

Solve the following pair of linear equations by elimination method:



Answer:

multiply 15 to the first equation to remove fraction

we get, 3x – 5y = 4 …[1]


Now let’s, multiply 18 to the second equation to remove fraction


we get, 9x – 2y = 7 …[2]


Now, let’s find eq[2] – eq[1] × 3


⇒ 9x – 2y – 3(3x – 5y) = 7 – 4×3


⇒ 


Now if we put this result in eq[1]



⇒ 39x = 27


⇒ 



Question 2.

Solve the following pair of linear equations by elimination method:

4x — 19y + 13 = 0, 13x — 23y = —19


Answer:

multiplying first equation by 13 and second equation by 4 to get same coefficient of x.

we get, 52x – 247 y = –169 …[1]


52x – 92y = –76 …[2]


eq[2] – eq[1]


155y = 93


⇒ 


Put this result in 4x — 19y + 13 = 0 to find the value of x


4x — 19() + 13 = 0


⇒ x = 



Question 3.

Solve the following pair of linear equations by elimination method:

x + y = a + b, ax — by = a2 — b2


Answer:

Multiply the first equation by b


⇒ bx + by = ab + b2 …[1]


⇒ ax — by = a2 — b2 …[2]


Now, eq[2] + eq[1]


x(a + b) = a2 + ab


⇒ x(a + b) = a(a + b)


⇒ x = a


by x + y = a + b and x = a


x + y = a + b (x = a)


⇒ y = b



Question 4.

Solve the following pair of linear equations by elimination method:

5ax + 6by = 28; 3ax + 4by = 18


Answer:

Multiply the first equation by 2 and the second equation by 3.


⇒ 10ax + 12by = 56 …[1]


⇒ 9ax + 12by = 54 …[2]


on subtracting eq[2] from eq [1]


ax = 2


⇒ x = 


by x =  and 3ax + 4by = 18


3a() + 4by = 18


we get 



Question 5.

The sum of two numbers is 35. Four times the larger number is 5 more than 5 times the smaller number. Find these numbers.


Answer:

Let the larger number be x and smaller number y.


given, sum of number is 35


⇒ x + y = 35 …[1]


Also, Four times the larger number is 5 more than 5 times the smaller number.


4x = 5y + 5 …[2]


Multiplying eq[1] by 5 and adding eq[2] to result, we get


5(x + y) + 4x = 175 + 5y + 5


⇒ 9x = 180


⇒ x = 20


From eq[1]


⇒ y = 15


∴ The larger number is 20 and the smaller number is 15.



Question 6.

There are some 25 paise coins and some 50 paise coins in a bag. The total number of coins is 140 and the amount in the bag is Rs. 50. Find the number of coins of each value in the bag.


Answer:

Let the number of 50 paise coins be x, and the number of 25 paise coins be y.

Given, the total no. of coins is 140


⇒ x + y = 140 …[1]


also, the amount in the bag is Rs. 50




⇒ 2x + y = 200 …[2]


eq[2] – eq[1]


x = 60


and by eq[1]


⇒ y = 80


amount of 50 paise cons = 


amount of 25 paise cons = 



Question 7.

The sum of the digits of two–digit number is 3. The number obtained by interchanging the digits is 9 less than the original number. Find the original number.


Answer:

Let the digit at 10s place y and digit at one’s place be x.

∴ 10y + x = The number


Now if we interchange the digits we get a number that is 9 less than the original number.


interchanged number = 10x + y


⇒ 10x + y = 10y + x – 9


⇒ 9x – 9y + 9 = 0


⇒ x – y + 1 = 0 …[1]


and the sum of digits is 3


⇒ x + y = 3 eq[2]


adding eq[1] and …[2]


2x = 2


⇒ x = 1


From eq[1]


x – y + 1 = 0


⇒ y = 2


∴ the original number is 21



Question 8.

The length of a rectangle is twice its breadth. The perimeter of the rectangle is 120 cm. Find the length and breadth of this rectangle. Also find its area.


Answer:

Let the length be

x and breadth by y.


Given, the length of a rectangle is twice its breadth


⇒ x = 2y


⇒ x – 2y = 0 …[1]


and also, perimeter of the rectangle is 120 cm


⇒ 2(x + y) = 120


⇒ 2x + 2y = 120 …[2]


eq[1] + eq[2]


3x = 120


⇒ x = 40


and x = 2y


⇒ y = 20


The area is product of length and breadth


⇒ area = 20×40


⇒ area = 800 cm2



Question 9.

An employee deposits certain amount at the rate of 8% per annum and a certain amount at the rate of 6% per annum at simple interest. He earns Rs. 500 as annual interest. If he interchanges the amount at the same rates, he earns Rs. 50 more. Find the amounts deposited by him at different rates.


Answer:

Let the amount at 8% be x and the amount at 6% be y.



Where I is interest, P is the principal amount N is the time in years and R is the rate per annum.


Now,


I in case 1 of 8%


⇒ I =  …[1]


in the second case of 6%


I =  …[2]


The total interest is 500.


eq[1] + eq[2]


⇒ 


⇒ 4x + 3y = 25000 …[3]


Now if he interchanges the amount he gets 50 more


⇒ 


⇒ 4y + 3x = 27500 …[4]


Now, adding and subtracting equation eq[1] and eq[2].


7x + 7y = 52500


⇒ x + y = 7500 …[5]


x – y = –2500 …[6]


adding [5] and [6]


2x = 5000


⇒ x = 2500


and by [5]


x + y = 7500


⇒ 2500 + y = 7500


⇒ y = 5000




Exercise 3.5
Question 1.

Solve the following pairs of equations by cross multiplication method:

0.3x + 0.4y = 2.5 and 0.5x — 0.3y = 0.3


Answer:

Given equations are


0.3x + 0.4y = 2.5 and 0.5x — 0.3y = 0.3


Change the given equations to the form,


a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 …(i)


⇒ 0.3x + 0.4y – 2.5 = 0 and


0.5x — 0.3y – 0.3 = 0


On comparing with (i) we get,


a1 = 0.3, b1 = 0.4, c1 = – 2.5; a2 = 0.5, b2 = – 0.3, c2 = – 0.3


Applying cross multiplication method which says,



Putting the given values in the above equation we get,







Similarly, 



∴The solution of the pair of equations is (3, 4).



Question 2.

Solve the following pairs of equations by cross multiplication method:

5x + 8y = 18, 2x — 3y = 1


Answer:

Given equations are


5x + 8y = 18, 2x — 3y = 1


Change the given equations to the form,


a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 …(i)


⇒ 5x + 8y – 18 = 0 and


2x — 3y – 1 = 0


On comparing with (i) we get,


a1 = 5, b1 = 8, c1 = – 18; a2 = 2, b2 = – 3, c2 = – 1


Applying cross multiplication method which says,



Putting the given values in the above equation we get,







Similarly, 



∴The solution of the pair of equations is (2, 1).



Question 3.

Solve the following pairs of equations by cross multiplication method:

, 7x— 15y = 21


Answer:

Given equations are


, 7x— 15y = 21


Change the given equations to the form,


a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 …(i)


⇒ 


And7x — 15y – 21 = 0


⇒ 5x + 3y – 15 = 0 and 7x — 15y – 21 = 0


On comparing with (i) we get,


a1 = 5, b1 = 3, c1 = – 15; a2 = 7, b2 = – 15, c2 = – 21


Applying cross multiplication method which says,



Putting the given values in the above equation we get,







Similarly, 



∴The solution of the pair of equations is (3, 0).



Question 4.

Solve the following pairs of equations by cross multiplication method:

3x + y = 5, 5x + 3y = 3


Answer:

Given equations are


3x + y = 5, 5x + 3y = 3


Change the given equations to the form, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 …(i)


⇒ 3x + y – 5 = 0 and


5x + 3y – 3 = 0


On comparing with (i) we get,


a1 = 3, b1 = 1, c1 = – 5; a2 = 5, b2 = 3, c2 = – 3


Applying cross multiplication method which says,



Putting the given values in the above equation we get,







Similarly, 



∴The solution of the pair of equations is (3, – 4).



Question 5.

By cross multiplication method, find such a two digit number such that, the digit at unit's place is twice the digit at tens place and the number obtained by interchanging the digits of the number is 36 more than the original number.


Answer:

Let the required two digit number be (10x + y) where x is the digit at tens place and y is the digit at unit place.


On interchanging the digits the two digit number formed will be (10y + x).


According to the question,


y = 2x and (10y + x) = 36 + (10x + y)


⇒ 2x – y = 0 and 10x + y – 10 y – x + 36 = 0


⇒ 2x –y = 0 and 9x – 9y + 36 = 0


⇒ 2x – y = 0 and x – y + 4 = 0


On comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0we get,


a1 = 2, b1 = – 1, c1 = 0; a2 = 1, b2 = – 1, c2 = 4


Applying cross multiplication method which says,



Putting the given values in the above equation we get,






Similarly,



∴The solution of the pair of equations is (4, 8).


So, the required two digit number will be (10x + y) = (10× 4 + 8) = 48



Question 6.

The sum of two numbers is 70 and their difference is 6. Find these numbers by cross – multiplication method.


Answer:

Let the two numbers be x and y.


According to the question,


x + y = 70 and x – y = 6


⇒ x + y – 70 = 0 and x –y – 6 = 0


On comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0we get,


a1 = 1, b1 = 1, c1 = – 70; a2 = 1, b2 = – 1, c2 = – 6


Applying cross multiplication method which says,



Putting the given values in the above equation we get,






Similarly,



∴The solution of the pair of equations is (38, 32).


So, the required numbers will be 38 and 32.



Question 7.

While arranging certain students of a school in rows containing equal number of students; if three rows are reduced, then three more students have to be arranged in each of the remaining rows. If three more rows are formed, then two students have to be taken off from each previously arranged rows. Find the number of students arranged.


Answer:

Let the number of rows be x and number of students in each row = y.


⇒ Number of students arranged = x × y


According to the question,


If three rows are reduced, number of rows = (x – 3)


Then three more students have to be arranged in each row, number of students in each row = (y + 3)


Number of students arranged = (x – 3)(y + 3)


If three more rows are formed, number of rows = (x + 3)


Then two students have to be taken off from each row, number of students in each row = (y – 2)


Number of students arranged = (x + 3)(y – 2)


In above both the cases the number of students arranged will remain same.


⇒ (x – 3)(y + 3) = xy and (x + 3)(y – 2) = xy


⇒ xy + 3x – 3y – 9 = xy


And xy – 2x + 3y – 6 = xy


⇒ 3x – 3y – 9 = 0 and – 2x + 3y – 6 = 0


⇒ x – y – 3 = 0 and – 2x + 3y – 6 = 0


On comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0we get,


a1 = 1, b1 = – 1, c1 = – 3; a2 = – 2, b2 = 3, c2 = – 6


Applying cross multiplication method which says,



Putting the given values in the above equation we get,






Similarly,



∴The solution of the pair of equations is (15, 12).


So, Number of students arranged = x×y = 15× 12 = 180.



Question 8.

In AABC, the measure of ∠B is thrice to the measure of ∠C and the measure of ∠A is the sum of the measures of ∠B and ∠C. Find the measures of all the angles of AABC and also state the type of this triangle.


Answer:

Given ∠ B = 3 ∠ C …(i)


And  …(ii)


Putting (i) in (ii),



⇒ ∠A = 2∠C


We know that the angle sum property of a triangle states that ∠ A + ∠ B + ∠ C = 180°


⇒ ∠ A + 3∠C + ∠ C = 180°


⇒ ∠ A + 4∠ C – 180° = 0


Consider ∠A – 2∠C = 0 and ∠ A + 4∠ C – 180 = 0


On comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0we get,


a1 = 1, b1 = – 2, c1 = 0; a2 = 1, b2 = 4, c2 = – 180


Applying cross multiplication method which says,



Putting the given values in the above equation we get,






Similarly,



∴ ∠ B = 3∠C = 3× 30° = 90°


∵ One of the angles of this triangle is 90° i.e. ∠ B


∴ ∆ABC is a right angled triangle.




Exercise 3.6
Question 1.

Solve the following pairs of linear equations:

 x ≠ 0, y ≠ 0


Answer:



Putting these value in given equations, we get




So, the given equations transforms into linear equation in two variables i.e.


15a + 4b = 42 … (i)


3a + 2b = 12 …..(ii)


Multiply equation (ii) by 5, so


15a + 4b = 42 … (iii)


15a + 10b = 60 …..(ii)


Subtract (iii) from (ii),


15a + 10b – 15a – 4b = 60 – 42


⇒ 6b = 18


⇒ b = 3


Putting above value in (ii),


3a + 2× 3 = 12


⇒ 3a = 12 – 6


⇒ 3a = 6


⇒ a = 2




Question 2.

Solve the following pairs of linear equations:

2x + 3y = 2xy, 6x + 12y = 7xy


Answer:

Given: 2x + 3y = 2xy, 6x + 12y = 7xy


One of the solutions for given pair of equations will be x = 0 and y = 0


But let us find out some other solution as well.


Divide both the equations with xy,





Putting these value in given equations, we get


2b + 3a = 2, 6b + 12a = 7


So, the given equations transforms into linear equation in two variables i.e.


2b + 3a = 2… (i)


6b + 12a = 7….. (ii)


Multiply equation (i) by 3, so


6b + 9a = 6 … (iii)


6b + 12a = 7….. (ii)


Subtract (iii) from (ii),


6b + 12a – 6b – 9a = 7 – 6


⇒ 3a = 1



Putting above value in (ii),



⇒ 2b = 2 – 1





Question 3.

Solve the following pairs of linear equations:

 x ≠ 1, y ≠ 1


Answer:



Putting these value in given equations, we get


4a + 5b = 2, 8a + 15b = 3


So, the given equations transforms into linear equation in two variables i.e.


4a + 5b = 2 … (i)


8a + 15b = 3 …..(ii)


Multiply equation (i) by 2, so


8a + 10b = 4 … (iii)


8a + 15b = 3 …..(ii)


Subtract (iii) from (ii),


8a + 15b – 8a – 10b = 3 – 4


⇒ 5b = – 1



Putting above value in (i),



⇒ 4a = 2 + 1









⇒ – y + 1 = 5


⇒ y = – 4



Question 4.

Solve the following pairs of linear equations:

 3x + y ≠ 0, 3x – y ≠ 0


Answer:



Putting these value in given equations, we get




So, the given equations transforms into linear equation in two variables i.e.


4a + 4b = 3 … (i)


4a – 4b = 1 …..(ii)


Add (i) and (ii),


4a + 4b + 4a – 4b = 3 + 1


⇒ 8a = 4



Putting above value in (i),



⇒ 4b = 3 – 2


⇒ 4b = 1





 … (iii)



 …. (iv)


Add (iii) and (iv),


3x + y + 3x – y = 2 + 4


⇒ 6x = 6


⇒ x = 1


Putting the above value in (iii),


3(1) + y = 2


⇒ y = 3 – 2


⇒ y = 1



Question 5.

Solve the following pairs of linear equations:

 x > 0, y > 0


Answer: 



Putting these value in given equations, we get




So, the given equations transforms into linear equation in two variables i.e.


3a + 4b = 2 … (i)


60a + 84b = 41 …..(ii)


Multiply (i) by 20,


60a + 80b = 40 …. (iii)


60a + 84b = 41 …..(ii)


Subtract (iii) from (ii),


60a + 84b – 60a – 80b = 41 – 40


⇒ 4b = 1



Putting above value in (i),



⇒ 3a = 2 – 1


⇒ 3a = 1





Squaring both the sides,


x = 9


Similarly, 


Squaring both the sides,


y = 16



Question 6.

5 women and 2 men together can finish an embroidery work in 4 days, while 6 women and 3 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work. Also, find the time taken by 1 man alone to finish the work.


Answer:

Let W and M represent the work done by a woman and a man respectively.


It is given that 5 women and 2 men can do the work in 4 days.


So, work done by 5 women and 2 men in a day will be 



⇒ 20W + 8M = 1 …. (i)


Also,


It is given that 6 women and 3 men can do the same work in 3 days.


So, work done by 6 women and 3 men in a day will be 



⇒ 18W + 9M = 1 …. (ii)


From (i) and (ii) equation,


20W + 8M = 18W + 9M


⇒ 2W = M


This means that the work done by 2 women in a day is equal to the work done by a man in a day. Substituting the value of M in (i) equation, we get


20W + 8M = 1


⇒ 20W + 8× 2W = 1


⇒ (20 + 16) W = 1



So, a woman do  work in a day.


This means a woman will finish the work in 36 days if working alone.


Now, put the value of W in (ii),


18W + 9M = 1






So, a man do  work in a day.


This means a man will finish the work in 18 days if working alone.



Question 7.

A boat goes 21 km upstream and 18 km downstream in 9 hours. In 13 hours, it can go 30 km upstream and 27 km downstream. Determine the speed of the stream and that of the boat in still water. (Speed of boat in still water is more than the speed of the stream of river.)


Answer:

Let the speed of the boat in still water be x km/hr and the speed of stream be y km/hr. It is necessary that x > y.


The speed of the boat in downstream = (speed of boat in still water + speed of stream) = (x + y) km/hr.


The speed of boat in upstream = (speed of boat in still water – speed of stream) = (x – y) km/hr.


Also, we know that



In the first case, it is given that boat goes 21 km upstream and 18 km downstream in 9 hours



⇒ Time taken by boat in upstream and downstream = 9 hr



In the second case, it is given that boat goes 30 km upstream and 27 km downstream in 13 hours



⇒ Time taken by boat in upstream and downstream = 13 hr




Putting these value in (i) and (ii), we get


21a + 18b = 9


30a + 27b = 13


So, the given equations transforms into linear equation in two variables


7a + 6b = 3 …. (iii)


30a + 27b = 13 … (iv)


Multiply (iii) by 9 and (iv) by 2,


63a + 54b = 27 …. (v)


60a + 54b = 26 … (vi)


Subtract (vi) from (v),


63a + 54b – 60a – 54b = 27 – 26


⇒ 3a = 1



Putting above value in (iii),








 … (vii)



 …. (viii)


Add above two equations,


x – y + x + y = 3 + 9


⇒ 2x = 12


⇒ x = 6


Putting x = 6 in (viii),


y = 9 – 6


⇒ y = 3



Question 8.

Solve the following pair of equations by cross multiplication method:



Answer:





Putting these value in given equations, we get


4b + 7a = 16… (i), 10b + 3a = 11 ….. (ii)


So, the given equations transforms into linear equation in two variables.


Multiply equation (i) by 5 and (ii) by 2,


20b + 35a = 80 … (iii)


20b + 6a = 22….. (iv)


Subtract (iv) from (iii),


20b + 35a – 20b – 6a = 80 – 22


⇒ 29a = 58


⇒ a = 2


Putting above value in (ii),


10b + 3× 2 = 11


⇒ 10b = 11 – 6





Question 9.

Mahesh travels 250 km to his home partly by train and partly by bus. He takes 6 hours if he travels 50 km by train and remaining distance by bus. If he travels 100 km by train and remaining distance by bus, he takes 7 hours. Find the speed of the train and the bus separately.


Answer:

Let the speed of the train and the bus be x km/hr and y km/hr respectively.


We know that



Total distance travelled by Mahesh = 250 km


In the first case it is given that he travels 50 km by train and 200 km by bus.




Given that total time taken by Mahesh to reach in this case = 6 hr



In the second case it is given that he travels 100 km by train and 150 km by bus.




Given that total time taken by Mahesh to reach in this case = 7 hr




Putting these value in above equations, we get


50a + 200b = 6 … (i), 100a + 150b = 7 ….. (ii)


So, the given equations transforms into linear equation in two variables.


Multiply equation (i) by 2,


100a + 400b = 12 … (iii)


100a + 150b = 7 ….. (ii)


Subtract (ii) from (iii),


100a + 400b – 100a – 150b = 12 – 7


⇒ 250b = 5



Putting above value in (ii),








Exercise 3
Question 1.

Obtain a pair of linear equations from the following information:

"The rate of tea per kg is seven times the rate of sugar per kg. The total cost of 2 kg tea and 5 kg sugar is Rs. 570."


Answer:

Let the rate of tea per kg be Rs x and the rate of sugar per kg be Rs y.


According to the question,


x = 7y


⇒ x – 7y = 0 … (i)


Also, cost of 2 kg of tea = 2x and cost of 5 kg of sugar = 5y


⇒ 2x + 5y = 570 … (ii)


Hence, equation (i) and (ii) is the required pair of linear equation.



Question 2.

Draw the graphs of the pair of linear equations in two variables. x + 3y = 6, 2x — y = 5. Find its solution set.


Answer:

For drawing the given pair of linear equations we need few coordinates to plot of these equations.



In x + 3y = 6,


If x = 0 then 0 + 3y = 6


⇒ y = 2


B(0, 2)


If y = 0 then x + 0 = 6


⇒ x = 6


D(6, 0)


In 2x — y = 5,


If x = 0 then 0 – y = 5


⇒ y = – 5


C(0, – 5)


If y = 0 then 2x + 0 = 5


⇒ x = 2.5


E(2.5, 0)


Now plot B, C, D and E on graph,



We find that A(3, 1) is the intersecting point of the lines. Hence, {(3, 1)} is the solution set of the pair of linear equations.



Question 3.

Solve the following pair of equations by the method of elimination:



Answer:



Using elimination method:


Multiply (i) by 4 and (ii) by 5,




Subtract (iii) from (iv),





⇒ x = 2


Putting this value of x in (i),







Question 4.

Solve the following pair of linear equations by the method of cross – multiplication:

(a + b)x + (a — b)y = a2 + 2ab — b2, a ≠ b

(a — b) (x + y) = a2 — b2, a ≠ b


Answer:

Given equations are


(a + b)x + (a — b)y = a2 + 2ab — b2


And (a — b) (x + y) = a2 — b2


Change the given equations to the form,


a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 …(i)


⇒ (a + b)x + (a — b)y – a2 – 2ab + b2 = 0


And


(a – b)x + (a – b)y – a2 + b2 = 0


On comparing with (i) we get,


a1 = (a + b),


b1 = (a — b),


c1 = – a2 – 2ab + b2;


a2 = (a – b),


b2 = (a – b),


c2 = – a2 + b2 = b2 – a2 = (b + a)(b – a) {Using a2 – b2 = (a + b)(a – b)}


Applying cross multiplication method which says,



Putting the given values in the above equation we get,







Similarly, 



∴The solution of the pair of equations is (a, b).



Question 5.

Solve the following pair of equations:

 and  x ≠ 1, y ≠ 2


Answer:



Putting these value in given equations, we get


4a + 7b = 2, 2(10a + 14b) = 9


So, the given equations transforms into linear equation in two variables i.e.


4a + 7b = 2… (i)


20a + 28b = 9 …..(ii)


Multiply equation (i) by 4, so


16a + 28b = 8 … (iii)


20a + 28b = 9…..(ii)


Subtract (iii) from (ii),


20a + 28b – 16a – 28b = 9 – 8


⇒ 4a = 1



Putting above value in (i),



⇒ 7b = 2 – 1









⇒ y + 2 = 7


⇒ y = 5



Question 6.

The difference between two natural numbers is 6. Adding 10 to the twice of the larger number, we get 2 less than 3 times of the smaller number. Find these numbers.


Answer:

Let the two natural numbers be x and y such that x is the larger number and y is the smaller number.


According to the question,


x – y = 6


and 2x + 10 = 3y – 2


⇒ x – y – 6 = 0 …(i)


And 2x – 3y + 12 = 0 ..(ii)


Multiply (i) by 2,


2x – 2y – 12 = 0 .. (iii)


2x – 3y + 12 = 0 ..(ii)


Subtract (iii) from (ii),


2x – 3y + 12 – 2x + 2y + 12 = 0


⇒ – y + 24 = 0


⇒ y = 24


Putting this value in (i),


x – 24 – 6 = 0


⇒ x = 30


So the numbers are 30 and 24.



Question 7.

The area of a rectangle gets increased by 30 square units, if its length is reduced by 3 units and breadth is increased by 5 units. If we increase the length by 5 units and reduce the breadth by 3 units then the area of a rectangle reduces by 10 square units. Find the length and breadth of the rectangle.


Answer:

Let the length be x units and breadth be y units of a rectangle.


⇒ Area of rectangle = x × y


According to the question,


If length is reduced by 3, length = (x – 3)


And breadth is increased by 5, breadth = (y + 5)


Area = (xy + 30)


⇒ (x – 3)(y + 5) = (xy + 30)


⇒ xy + 5x – 3y – 15 = xy + 30


⇒ 5x – 3y = 45 …(i)


If length is incresed by 5, length = (x + 5)


And breadth is reduced by 3, breadth = (y – 3)


Area = (xy – 10)


⇒ (x + 5)(y – 3) = (xy – 10)


⇒ xy – 3x + 5y – 15 = xy – 10


⇒ – 3x + 5y = 5 …(ii)


Multiply (i) by 3 and (ii) by 5,


15x – 9y = 135 .. (iii)


– 15x + 25y = 25 …(iv)


Add (iii) and (iv),


15x – 9y – 15x + 25y = 135 + 25


⇒ 16y = 160


⇒ y = 10


Putting this in (i),


5x – 3×10 = 45


⇒ 5x = 45 + 30


⇒ 5x = 75


⇒ x = 15


So, the length is 15 units and breadth is 10 units.



Question 8.

A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. Yash takes food for 25 days. He has to pay Rs. 2200 as hostel charges where as Niyati takes food for 20 days. She has to pay Rs. 1800 as hostel charges. Find the fixed charges and the cost of food per day.


Answer:

Let the fixed charge be Rs x and the cost of food per day be Rs y.


According to the question,


For Yash,


He took food for 25 days


⇒ x + 25 y = 2200..(i)


For Niyati,


She took food for 20 days


⇒ x + 20 y = 1800….(ii)


Subtract (ii) from (i),


x + 25y – x – 20 y = 2200 – 1800


⇒ 5y = 400


⇒ y = 80


Putting this value in (i),


x + 25× 80 = 2200


⇒ x = 2200 – 2000


⇒ x = 200


Hence, the fixed cost = Rs 200 and the cost for food per day = Rs 80



Question 9.

A fraction becomes  when 2 is subtracted from the numerator and denominator it becomes  when 5 is added to its denominator and numerator, find the fraction.


Answer:

Let the required fraction be


According to the question,



⇒ 5x – 10 = 2y – 4


⇒ 5x – 2y = 6 …(i)


Also, 


⇒ 4x + 20 = 3y + 15


⇒ 4x – 3y = – 5 …(ii)


Multiply (i) by 3 and (ii) by 2,


15x – 6y = 18 ..(iii)


8x – 6y = – 10 ….(iv)


Subtract (iv) from (iii),


15x – 6y – 8x + 6y = 18 + 10


⇒ 7x = 28


⇒ x = 4


Putting this in (i),


5× 4 – 2y = 6


⇒ – 2y = 6 – 20


⇒ – 2y = – 14


⇒ y = 7


Hence the required fraction is 



Question 10.

The solution set of x — 3y = 1 and 3x + y = 3 is
A. {(0, 1))

B. {(1, 1)}

C. {(1, 0)}

D. {, 0)}


Answer:

Given: x — 3y = 1…(i)


3x + y = 3 …(ii)


Multiply (i) by 3,


3x – 9y = 3 ..(iii)


3x + y = 3…(ii)


Subtract (ii) from (iii),


3x – 9y – 3x – y = 3 – 3


⇒ – 10y = 0


⇒ y = 0


Putting this in (i),


x – 0 = 1


⇒ x = 1


So, the solution set will be {1, 0}.


Question 11.

The solution set of 2x + y = 6 and 4x + 2y = 5 is
A. {(x, Y)| 2x + y = 6; x, y ∈ R}

B. {(x, y) | 2x + y = 0; x, y y ∈ R}

C. Φ

D. infinite set


Answer:

Given: 2x + y = 6 and 4x + 2y = 5


Change the given equations to the form,


a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 …(i)


⇒ 2x + y – 6 = 0


And


4x + 2y – 5 = 0


On comparing with (i) we get,


a1 = 2,


b1 = 1,


c1 = – 6;


a2 = 4,


b2 = 2,


c2 = – 5;





This means that the equations are inconsistent and have no solution.


Question 12.

To eliminate x, from 3x + y = 7 and —x + 2y = 2 second equation is multiplied by
A. 1

B. 2

C. 3

D. —1


Answer:

Given: 3x + y = 7 ..(i)and —x + 2y = 2 …(ii)


Using elimination method,


Multiply (ii) by 3,


3x + y = 7 …(i)


– 3x + 6y = 6 …(iii)


Add (i) and (iii),


3x + y – 3x + 6y = 7 + 6


⇒ 7y = 13


Hence, x is eliminated by multiplying second equation by 3.


Question 13.

If 2x + 3y = 7 and 3x + 2y = 3, then x — y =
A. 4

B. —4

C. 2

D. —2


Answer:

Given: 2x + 3y = 7 …(i)


3x + 2y = 3 …(ii)


Subtract (i) from (ii),


3x + 2y – 2x – 3y = 3 – 7


⇒ x – y = – 4


Question 14.

If the pair of linear equations ax + 2y = 7 and 2x + 3y = 8 has a unique solution, then a ≠ ………….
A. 

B. – 

C. 

D. – 


Answer:

Given: ax + 2y = 7 and 2x + 3y = 8


Change the given equations to the form,


a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 …(i)


⇒ ax + 2y – 7 = 0


And


2x + 3y – 8 = 0


On comparing with (i) we get,


a1 = a,


b1 = 2,


c1 = – 7;


a2 = 2,


b2 = 3,


c2 = – 8;



For unique solution,






Question 15.

The pair of linear equations 2x + y — 3 = 0 and 6x + 3y = 9 has
A. a unique solution

B. two solutions

C. no solution

D. infinitely many solutions


Answer:

Given: 2x + y — 3 = 0 and 6x + 3y = 9


Change the given equations to the form,


a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 …(i)


⇒ 2x + y – 3 = 0


And


6x + 3y – 9 = 0


On comparing with (i) we get,


a1 = 2,


b1 = 1,


c1 = – 3;


a2 = 6,


b2 = 3,


c2 = – 9;




This means that the equations are consistent and have infinitely many solutions.


Question 16.

If in a two digit number, the digit at unit place is x and the digit at tens place is 5, then the number is
A. 50x + 5

B. 30x + 5

C. x + 50

D. 5x


Answer:

A two digit number is of the form 10y + x where x is the digit at unit place and y is the digit at tens place.


According to the question,


y = 5


Then number will be 10 × 5 + x = 50 + x


Question 17.

In a two digit number, the digit at tens place is 7 and the sum of the digits is 8 times the digit at unit place. Then the number is
A. 70

B. 71

C. 17

D. 78


Answer:

Let the number be 10x + y where x is the digit at tens place and y is the digit at unit place.


According to the question,


x = 7


Sum of the digits = 7 + y


⇒ 7 + y = 8y


⇒ 7y = 7


⇒ y = 1


So, the number = 71


Question 18.

The sum of two numbers is 10 and the difference of them is 2. Then the greater number of these two is
A. 2

B. 4

C. 6

D. 8


Answer:

Let the two numbers be x and y.


According to the question,


x + y = 10 ..(i)


x – y = 2 ..(ii)


Adding (i) and (ii),


x + y + x –y = 10 + 2


⇒ 2x = 12


⇒ x = 6


Putting x = 6 in (i),


6 + y = 10


⇒ y = 4


Then the greater number is 6.


Question 19.

3 years ago, the sum of ages of a father and his son was 40 years. After 2 years the sum of ages of the father and his son will be
A. 40

B. 46

C. 50

D. 60


Answer:

Let the present ages of father and son be x years and y years respectively.


Age of father three years ago = x – 3


Age of son three years ago = y – 3


According to the question,


x – 3 + y – 3 = 40


⇒ x + y = 40 + 6


⇒ x + y = 46 .. (i)


After 2 years,


Age of father = x + 2


Age of son = y + 2


Sum of their ages after 2 years = x + 2 + y + 2 = (x + y) + 4 = 46 + 4 = 50 {Using (i)}


Question 20.

The solution set of 2x + 4y = 8 and x + 2y = 4 is
A. {(2, 1)}

B. empty set

C. infinite set

D. {(0, 0)}


Answer:

Given: of 2x + 4y = 8 …(i) and


x + 2y = 4 …(ii)


Using elimination method,


Multiply (ii) by 2,


2x + 4y = 8 …(i)


2x + 4y = 8 …(iii)


Since both the equations are same therefore they have infinite solution.


Question 21.

Equation  can be expressed in the standard form as
A. 2x — 3y — 6 = 0

B. 3x — 2y — 6 = 0

C. 3x — 2y = 1

D. 2x — 3y = 3


Answer:


Standard form of linear equation in two variable: ax + by + c = 0 where a, b and c are the constants.


Taking LCM (2, 3) = 6 in the denominator,



⇒ 3x – 2y = 6


⇒ 3x – 2y – 6 = 0


This is the standard form of the given equation.


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