NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Q1. Form the pair of linear equations in the following problems and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Answer:

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

Now, according to the question,

Total number of students in the class = 10, i.e.

⇒x+y=10…..(1)

And the number of girls is 4 more than the number of boys,i.e.

x=y+4⇒x−y=4.

Different points (x, y) for equation (1)

Different points (x,y) satisfying (2)

Graph,

As we can see from the graph, both lines intersect at the point (7,3). that is x= 7 and y = 3, which means the number of boys in the class is 7 and the number of girls in the class is 3.

Q1 Form the pair of linear equations in the following problems and find their solutions graphically.

(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Answer:

Let x be the price of 1 pencil and y be the price of 1 pen,

Now, According to the question

5x+7y=50……

And

7x+5y=46……

Now, the points (x,y), that satisfies the equation (1) are

And, the points(x,y) that satisfies the equation (2) are

The Graph,

As we can see from the Graph, both line intersects at point (3,5) that is, x = 3 and y = 5 which means cost of 1 pencil is 3 and the cost of 1 pen is 5.

Q2 On comparing the ratios a1a2,b1b2andc1c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i) 5x−4y+8=07x+6y−9=0

Answer:

Given Equations,

5x−4y+8=07x+6y−9=0

Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=57,b1b2=−46 and c1c2=8−9

As we can see

a1a2≠b1b2

It means that both lines intersect at exactly one point.

Q2 On comparing the ratios a1a2,b1b2andc1c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (ii) 9x+3y+12=018x+6y+24=0

Answer:

Given Equations,

9x+3y+12=018x+6y+24=0

Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=918=12b1b2=36=12 and c1c2=1224=12

As we can see

a1a2=b1b2=c1c2

It means that both lines are coincident.

Q2 On comparing the ratios a1a2,b1b2andc1c2 , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (iii) 6x−3y+10=02x−y+9=0

Answer:

Given Equations,

6x−3y+10=02x−y+9=0

Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=62=3,b1b2=−3−1=3 and c1c2=109

As we can see

a1a2=b1b2≠c1c2

It means that both lines are parallel to each other.

Q3 On comparing the ratios a1a2,b1b2andc1c2 find out whether the lines representing the following pairs of linear equations are consistent, or inconsistent: (i) 3x+2y=5;2x−3y=7

Answer:

Given Equations,

3x+2y=52x−3y=7

Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=32,b1b2=2−3 and c1c2=57

As we can see

a1a2≠b1b2

It means the given equations have exactly one solution and thus pair of linear equations is consistent.

Q3 On comparing the ratios a1a2,b1b2andc1c2, find out whether the lines representing the following pairs of linear equations are consistent, or inconsistent: (ii) 2x−3y=8;4x−6y=9

Answer:

Given Equations,

2x−3y=84x−6y=9

Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=24=12b1b2=−3−6=12 and c1c2=89

As we can see

a1a2=b1b2≠c1c2

It means the given equations have no solution and thus pair of linear equations is inconsistent.

Q3 On comparing the ratios a1a2,b1b2andc1c2, find out whether the lines representing the following pairs of linear equations are consistent, or inconsistent: (iii) 32x+53y=7;9x−10y=14

Answer:

Given Equations,

32x+53y=79x−10y=14

Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=3/29=318=16b1b2=5/3−10=5−30=−16 and c1c2=714=12

As we can see

a1a2≠b1b2

It means the given equations have exactly one solution and thus pair of linear equations is consistent.

Q3 On comparing the ratios a1a2,b1b2andc1c2,find out whether the lines representing the following pairs of linear equations are consistent, or inconsistent: (iv) 5x−3y=11;−10x+6y=−22

Answer:

Given Equations,

5x−3y=11−10x+6y=−22

Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=5−10=−12b1b2=−36=−12 and c1c2=11−22=−12

As we can see

a1a2=b1b2=c1c2

It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.

Q3 On comparing the ratios a1a2,b1b2andc1c2 find out whether the following pair of linear equations are consistent, or inconsistent (v) 43x+2y=8;2x+3y=12

Answer:

Given Equations,

43x+2y=82x+3y=12

Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=4/32=46b1b2=23 and c1c2=812=23

As we can see

a1a2=b1b2=c1c2

It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.

Q4 Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) x+y=52x+2y=10

Answer:

Given Equations,

x+y=52x+2y=10

Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=12b1b2=12 and c1c2=510=12

As we can see

a1a2=b1b2=c1c2

It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.

The points (x,y) which satisfies in both equations are

Q4 Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (ii) x−y=8,3x−3y=16

Answer:

Given Equations,

x−y=83x−3y=16

Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=13b1b2=−1−3=13 and c1c2=816=12

As we can see

a1a2=b1b2≠c1c2

It means the given equations have no solution and thus pair of linear equations is inconsistent.

Q4 Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (iii) 2x+y−6=0,4x−2y−4=0

Answer:

Given Equations,

2x+y−6=04x−2y−4=0

Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=24=12b1b2=1−2=−12 and c1c2=−6−4=32

As we can see

a1a2≠b1b2

It means the given equations have exactly one solution and thus pair of linear equations is consistent.

Now The points(x, y) satisfying the equation are,

And The points(x,y) satisfying the equation 4x−2y−4=0 are,

GRAPH:

As we can see both lines intersects at point (2,2) and hence the solution of both equations is x = 2 and y = 2.

Q4 Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (iv) 2x−2y−2=0,4x−4y−5=0

Answer:

Given Equations,

2x−2y−2=04x−4y−5=0

Comparing these equations with a1x+b1y+c1=0 and a2x+b2y+c2=0, we get

a1a2=24=12

b1b2=−2−4=12 and

c1c2=−2−5=25

As we can see

a1a2=b1b2≠c1c2

It means the given equations have no solution and thus pair of linear equations is inconsistent.

Q5 Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Answer:

Let l be the length of the rectangular garden and b be the width.
Now, According to the question, the length is 4 m more than its width.i.e.

l=b+4l−b=4

Also given Half Parameter of the rectangle =36 i.e.

l+b=36….(2)

Now, as we have two equations, on adding both equations, we get,

l+b+l−b=4+36⇒2l=40⇒l=20

Putting this in equation (1),

⇒20−b=4⇒b=20−4⇒b=16

Hence Length and width of the rectangle are 20m and 16 respectively.

Q6 Given the linear equation 2x+3y−8=0 , write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) intersecting lines

Answer:

Given the equation,

2x+3y−8=0

As we know that the condition for the intersection of lines a1x+b1y+c1=0 and a2x+b2y+c2=0, is,

a1a2≠b1b2

So Any line with this condition can be 4x+3y−16=0
Here,

a1a2=24=12b1b2=33=1

As

12≠1

the line satisfies the given condition.

Q6 Given the linear equation 2x+3y−8=0 , write another linear equation in two variables such that the geometrical representation of the pair so formed is (ii) parallel lines

Answer:

Given the equation,

2x+3y−8=0

As we know that the condition for the lines a1x+b1y+c1=0 and a2x+b2y+c2=0, for being parallel is,

a1a2=b1b2≠c1c2

So Any line with this condition can be 4x+6y−8=0
Here,

a1a2=24=12b1b2=36=12c1c2=−8−8=1

As
12=12≠1 the line satisfies the given condition.

Q6 Given the linear equation 2x+3y−8=0 , write another linear equation in two variables such that the geometrical representation of the pair so formed is: (iii) coincident lines

Answer:

Given the equation,

2x+3y−8=0

As we know that the condition for the coincidence of the lines a1x+b1y+c1=0 and a2x+b2y+c2=0, is,

a1a2=b1b2=c1c2

So any line with this condition can be 4x+6y−16=0
Here,

a1a2=24=12b1b2=36=12c1c2=−8−16=12

As
12=12=12 the line satisfies the given condition.

Q7 Draw the graphs of the equations x−y+1=0and 3x+2y−12=0 . Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Answer:

Given two equations,

x−y+1=0

And

3x+2y−12=0

The points (x,y) satisfying (1) are

And The points(x,y) satisfying (2) are,

GRAPH:

As we can see from the graph that both lines intersect at the point (2,3), And the vertices of the Triangle are ( -1,0), (2,3) and (4,0). The area of the triangle is shaded with a green color.

Pair of Linear Equations in Two Variables Class 10 Solutions Exercise: 3.2

Q1 Solve the following pair of linear equations by the substitution method. (i) x+y=14x−y=4

Answer:

Given two equations,

x+y=14…….(1)x−y=4…….(2)

Now, from (1), we have

y=14−x

Substituting this in (2), we get

x−(14−x)=4⇒x−14+x=4⇒2x=4+14=18⇒x=9

Substituting this value of x in (3)

⇒y=14−x=14−9=5

Hence, Solution of the given equations is x=9 and y=5.

Q1 Solve the following pair of linear equations by the substitution method (ii) s−t=3s3+t2=6

Answer:

Given two equations,

s−t=3…(1)s3+t2=6…(2)

Now, from (1), we have

s=t+3.

Substituting this in (2), we get

t+33+t2=6⇒2t+6+3t6=6⇒5t+6=36⇒5t=30⇒t=6

Substituting this value of t in (3)

⇒s=t+3=6+3=9

Hence, Solution of the given equations is s = 9 and t = 6.

Q1 Solve the following pair of linear equations by the substitution method. (iii) 3x−y=39x−3y=9

Answer:

Given two equations,

3x−y=3…(1)9x−3y=9…(2)

Now, from (1), we have

y=3x−3..

Substituting this in (2), we get

9x−3(3x−3)=9⇒9x−9x+9=9⇒9=9

This is always true, and hence this pair of the equation has infinite solutions.
As we have

y=3x−3

One of many possible solutions is x=1, and y=0.

Q1 Solve the following pair of linear equations by the substitution method. (iv) 0.2x+0.3y=1.30.4x+0.5y=2.3

Answer:

Given two equations,

0.2x+0.3y=1.3…(1)0.4x+0.5y=2.3…(2)

Now, from (1), we have

y=1.3−0.2x0.3

Substituting this in (2), we get

0.4x+0.5(1.3−0.2x0.3)=2.3⇒0.12x+0.65−0.1x=0.69⇒0.02x=0.69−0.65=0.04⇒x=2

Substituting this value of x in (3)

⇒y=(1.3−0.2x0.3)=(1.3−0.40.3)=0.90.3=3

Hence, Solution of the given equations is,

x=2 and y=3

Q1 Solve the following pair of linear equations by the substitution method. (v) 2x+3y=03x−8y=0

Answer:

Given two equations,

2x+3y=0…(1)3x−8y=0…(2)

Now, from (1), we have

y=−2x3……

Substituting this in (2), we get

3x−8(−2x3)=0⇒3x=8(−2x3)⇒3x=−4x⇒7x=0⇒x=0

Substituting this value of x in (3)

⇒y=−2x3=0

Hence, Solution of the given equations is,

x = 0 , y = 0 .

Q1 Solve the following pair of linear equations by the substitution method. (vi) 3x2−5y3=−2x3+y2=136

Answer:

Given

3x2−5y3=−2…(1)x3+y2=136…(2)

From (1) we have,

x=23(5y3−2)

Putting this in (2) we get,

13×23(5y3−2)+y2=13610y27−49+y2=13620y54−49+27y54=13647y54=136+4947y54=11754+245447y=117+2447y=141

y=14147y=3

putting this value in (3) we get,

x=23(5y3−2)x=23(5×33−2)x=23(5−2)x=23×3x=2

Hence x=2 and y=3.

Q2 Solve 2x+3y=11and 2x−4y=−24 and hence find the value of ‘ m’ for which y=mx+3 .

Answer:

Given two equations,

2x+3y=11…(1)2x−4y=−24…(2)

Now, from (1), we have

y=11−2x3

Substituting this in (2), we get

2x−4(11−2x3)=−24⇒6x−44+8x=−72⇒14x=44−72⇒14x=−28⇒x=−2

Substituting this value of x in (3)

⇒y=(11−2x3)=11−2×(−2)3=153=5

Hence, Solution of the given equations is,

x=−2, and y=5

Now,

As it satisfies y=mx+3,

⇒5=m(−2)+3⇒2m=3−5⇒2m=−2⇒m=−1

Hence Value of m is -1.

Q3 Form the pair of linear equations for the following problem and find their solution by substitution method.

(i) The difference between the two numbers is 26 and one number is three times the other. Find them.

Answer:

Let two numbers be x and y and let the bigger number is y.

Now, According to the question,

y−x=26…(1)

And

y=3x…(2)

Now, the substituting value of y from (2) in (1) we get,

3x−x=26⇒2x=26⇒x=13

Substituting this in (2)

⇒y=3x=3(13)=39

Hence the two numbers are 13 and 39.

Q3 Form the pair of linear equations for the following problem and find their solution by substitution method (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Answer:

Let the larger angle be x and smaller angle be y

Now, As we know the sum of supplementary angles is 180. so,

x+y=180…(1)

Also given in the question,

x−y=18…(2)

Now, From (2) we have,

y=x−18

Substituting this value in (1)

x+x−18=180⇒2x=180+18⇒2x=198⇒x=99

Now, Substituting this value of x in (3), we get

⇒y=x−18=99−18=81

Hence the two supplementary angles are

990 and 810

Q3 Form the pair of linear equations for the following problems and find their solution by substitution method. (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Answer:

Let the cost of 1 bat is x and the cost of 1 ball is y.

Now, According to the question,

7x+6y=3800…(1)3x+5y=1750…(2)

Now, From (1) we have

y=3800−7x6

Substituting this value of y in (2)

3x+5(3800−7x6)=1750⇒18x+19000−35x=1750×6⇒−17x=10500−19000⇒−17x=−8500⇒x=850017⇒x=500

Now, Substituting this value of x in (3)

y=3800−7x6=3800−7×5006=3800−35006=3006=50

Hence, The cost of one bat is 500 Rs and the cost of one ball 50 Rs.

Q3 Form the pair of linear equations for the following problems and find their solution by substitution method. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?

Answer:

Let the fixed charge is x and the per km charge is y.

Now According to the question

y=3800−7x6=3800−7×5006=3800−35006=3006=50

Hence, the fixed charge is 5 Rs and the per km charge is 10 Rs.

Now, Fair For 25 km :

⇒x+25y=5+25(10)=5+250=255

Hence fair for 25km is 255 Rs.

Q3 Form the pair of linear equations for the following problems and find their solution by substitution method. (v) A fraction becomes 911,, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 56, . Find the fraction.

Answer:

Let the numerator of the fraction be x and denominator of the fraction is y

Now According to the question,

x+2y+2=911⇒11(x+2)=9(y+2)⇒11x+22=9y+18⇒11x−9y=−4…(1)

Also,

x+3y+3=56⇒6(x+3)=5(y+3)⇒6x+18=5y+15⇒6x−5y=−3…(2)

Now, From (1) we have

y=11x+49.

Substituting this value of y in (2)

6x−5(11x+49)=−3⇒54x−55x−20=−27⇒−x=20−27⇒x=7

Substituting this value of x in (3)

y=11x+49=11(7)+49=819=9

Hence the required fraction is

xy=79

Q3 Form the pair of linear equations for the following problems and find their solution by substitution method. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer:

Let x be the age of Jacob and y be the age of Jacob's son,

Now, According to the question

x+5=3(y+5)⇒x+5=3y+15⇒x−3y=10…(1)

Also,

x−5=7(y−5)⇒x−5=7y−35⇒x−7y=−30…(2)

Now,
From (1) we have,

x=10+3y

Substituting this value of x in (2)

10+3y−7y=−30⇒−4y=−30−10

⇒4y=40⇒y=10

Substituting this value of y in (3),

x=10+3y=10+3(10)=10+30=40

Hence, Present age of Jacob is 40 years and the present age of Jacob's son is 10 years.

Pair of Linear Equations in Two Variables Class 10 Solutions Exercise: 3.3

Q1 Solve the following pair of linear equations by the elimination method and the substitution method :

(i) x+y=5 and 2x−3y=4

Answer:
Elimination Method:
Given equations

x+y=5………..(1) and 2x−3y=4…….(2)

Now, multiplying (1) by 3, we get

3x+3y=15

Now, Adding (2) and (3), we get

2x−3y+3x+3y=4+15⇒5x=19⇒x=195

Substituting this value in (1), we get

195+y=5⇒y=5−195⇒y=65

Hence,

x=195 and y=65

Substitution method :
Given equations

x+y=5…2x−3y=4.

Now, from (1) we have,

y=5−x……(3)

substituting this value in (2)

2x−3(5−x)=4⇒2x−15+3x=4⇒5x=19⇒x=195

Substituting this value of x in (3)

⇒y=5−x=5−195=65

Hence,

x=195 and y=65

Q1 Solve the following pair of linear equations by the elimination method and the substitution method :

(ii) 3x+4y=10 and 2x−2y=2

Answer:
Elimination Method:
Given equations

3x+4y=10……….(1) and 2x−2y=2……….(2)

Now, multiplying (2) by 2, we get

4x−4y=4

Now, Adding (1) and (3), we get

3x+4y+4x−4y=10+4⇒7x=14⇒x=2

Putting this value in (2), we get

2(2)−2y=2⇒2y=2⇒y=1

Hence,

x=2 and y=1

Substitution method :
Given equations

3x+4y=10………….(1) and 2x−2y=2………….(2)

Now, from (2) we have,

y=2x−22=x−1

substituting this value in (1)

3x+4(x−1)=10

⇒3x+4x−4=10⇒7x=14⇒x=2

Substituting this value of x in (3)

⇒y=x−1=2−1=1

Hence,

x=2 and y=1

Q1 Solve the following pair of linear equations by the elimination method and the substitution method: (iii) 3x−5y−4=0 and 9x=2y+7

Answer:

Elimination Method:

Given equations

3x−5y−4=0…(1)9x=2y+7⇒9x−2y−7=0…(2)

Now, multiplying (1) by 3, we get

9x−15y−12=0

Now, Subtracting (3) from (2), we get

9x−2y−7−9x+15y+12=0⇒13y+5=0⇒y=−513

Putting this value in (1) we get

3x−5(−513)−4=0

⇒3x=4−2513⇒3x=2713⇒x=913

Hence,

x=913 and y=−513

Substitution method :
Given equations

3x−5y−4=0……….(1) and 9x=2y+7⇒9x−2y−7=0…….(2)

Now, from (2) we have,

y=9x−72……

substituting this value in (1)

3x−5(9x−72)−4=0⇒6x−45x+35−8=0⇒−39x+27=0⇒x=2739=913

Substituting this value of x in (3)

⇒y=9(9/13)−72=81/13−72=−513

Hence,

x=913 and y=−513

Q1 Solve the following pair of linear equations by the elimination method and the substitution method :(iv) x2+2y3=−1 and x−y3=3

Answer:

Elimination Method:

Given equations

x2+2y3=−1…(1)x−y3=3…(2)

Now, multiplying (2) by 2, we get

2x−2y3=6

Now, Adding (1) and (3), we get

x2+2y3+2x−2y3=−1+6⇒5x2=5⇒x=2

Putting this value in (2), we get

2−y3=3

⇒y3=−1⇒y=−3

Hence,

x=2 and y=−3

Substitution method:
Given equations

x2+2y3=−1……(1) and x−y3=3………(2)

Now, from (2) we have,

y=3(x−3)

substituting this value in (1)

x2+2(3(x−3))3=−1⇒x2+2x−6=−1⇒5x2=5⇒x=2

Substituting this value of x in (3)

⇒y=3(x−3)=3(2−1)=−3

Hence,

x=2 and y=−3

Q2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?

Answer:

Let the numerator of the fraction be x and denominator is y,

Now, According to the question,

x+1y−1=1⇒x+1=y−1…(1)⇒x−y=−2…(2)

Also,

xy+1=12⇒2x=y+1⇒2x−y=1

Now, Subtracting (1) from (2) we get

x=3

Putting this value in (1)

3−y=−2

⇒y=5

Hence

x=3 and y=5

And the fraction is

35

Q2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Answer:

Let the age of Nuri be x and age of Sonu be y.

Now, According to the question

x−5=3(y−5)⇒x−5=3y−15…(1)⇒x−3y=−10…(2)

Also,

x+10=2(y+10)⇒x+10=2y+20…(1)⇒x−2y=10…(2)

Now, Subtracting (1) from (2), we get

y=20

putting this value in (2)

x−2(20)=10⇒x=50

Hence the age of Nuri is 50 and the age of Nuri is 20.

Q2 Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Answer:

Let the unit digit of the number be x and 10's digit be y.

Now, According to the question,

x+y=9……(1)

Also

9(10y+x)=2(10x+y)⇒90y+9x=20x+2y⇒88y−11x=0⇒8y−x=0…..(2)

Now adding (1) and (2) we get,

⇒9y=9⇒y=1

now putting this value in (1)

x+1=9⇒x=8

Hence the number is 18.

Q2 Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

Answer:

Let the number of Rs 50 notes be x and the number of Rs 100 notes be y.

Now, According to the question,

x+y=25…(1).

And

50x+100y=2000⇒x+2y=40…(2)

Now, Subtracting(1) from (2), we get

y=15

Putting this value in (1).

x+15=25⇒x=10

Hence Meena received 10, 50 Rs notes and 15, 100 Rs notes.

Q2 Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer:

Let fixed charge be x and per day charge is y.

Now, According to the question,

x+4y=27…(1)

And

x+2y=21…(2)

Now, Subtracting (2) from (1). we get,

4y−2y=27−21⇒2y=6⇒y=3

Putting this in (1)

x+4(3)=27⇒x=27−12=15

Hence the fixed charge is 15 Rs and per day charge is 3 Rs.