Q: On comparing the ratios a2a1,b2b1, and c2c1, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:(i) 5x−4y+8=0 ; 7x+6y−9=0 (ii) 9x+3y+12=0; 18x+6y+24=0(iii) 6x−3y+10=0 ; 2x−y+9=0

(i) a2a1 =75, b2b1 =6−4, c2c1 =−98 ∵a2a1=b2b1, the lines intersect at a point.(ii) a2a1 =189=21,b2b1 =63=21,c2c1 =2412=21∵a2a1=b2b1=c2c1, the lines are coincident. (iii) a2a1 =26=13,b2b1 =−1−3=13,c2c1 =910∵a2a1=b2b1=c2c1, the lines are parallel.

Q: On comparing the ratios a2a1,b2b1 and c2c1, find out whether the following pair of linear equations are consistent, or inconsistent.(i) 3x+2y=5;2x−3y=7(ii) 2x−3y=8;4x−6y=9(iii) 23x+35y=7; 9x−10y=14(iv) 5x−3y=11;−10x+6y=−22(v) 34x+2y=8;2x+3y=12

(i) a2a1 =23,b2b1 =−32,c2c1 =75∵a2a1=b2b1=c2c1, the lines intersect and have an unique consistent solution.(ii) a2a1 =42 =21,b2b1 =−6−3=21,c2c1 =98∵a2a1=b2b1=c2c1, the lines are parallel and have no solutions, i.e. the equations are an inconsistent pair(iii) a2a1 =923=61,b2b1 =−1035=−61,c2c1 =147=21∵a2a1=b2b1=c2c1, the lines intersect and have an unique consistent solution.(iv) a2a1 =−105=−21,b2b1 =6−3=−21,c2c1 =−2211= −21∵a2a1=b2b1=c2c1, the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.(v) a2a1 =234 =32,b2b1 =32,c2c1 =128=32∵a2a1=b2b1=c2c1, the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.

Question 1

Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." (Isnt this interesting?) Represent this situation algebraically and graphically.

Accepted Answer

Consider Aftab's age as x and his daughter's age as y.Then , seven years ago,Aftab's age =x−7His daughter's age =y−7According to the question,x −7=7(y−7) x−7=7y−49 x−7y=−49+7  x−7y=−42                    …(i)After three years,Aftab's age =x+3His daughter's age =y+3According to the question,x+3   =3(y+3)x+3   =3y+9x−3y= 9−3x−3y =6                       …(ii)Representing equation (i) and (ii) geometrically, we plot these equations by finding points on the lines representing these two equationsx−7y=−42⟹x=7y−42 x=7y−42423549 y 121113x−3y=6⟹x=3y+6 x=3y+6 42 3648 y 121014From the graph we can see that two lines will intersect at a point.  x−7y=−42   x−3y =6  On subtracting the two equations, we get 4y=48or y=12Substituting value of y in (2), x−36=6∴x=42

Question 2

The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and solve it geometrically.

Accepted Answer

Let the cost of a bat be Rs. x and cost of a ball be Rs. y.Then algebraically representation of the question is -3x+6y=3900⇒x+2y=1300......(1)x+3y=1300........(2)To represent it geometrically, we plot the two equationsx+3y=1300⇒y=31300−x​ x100 0 1300  y=31300−x​ 40031300​ 0 x+2y=1300⇒y=21300−x​ x100 0 1300  y=21300−x​600650 0 As we can see, the lines intersect at (1300,0) which is the solution for this question.

Question 3

The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Accepted Answer

x 65 55 45 35y 20 40 60 80Let the cost of 1 kg of apples be x and that of 1 kg be ySo the algebraic representation can be as follows:2x+y=1604x+2y=300⇒2x+y=150The situation can be represented graphically by plotting these two equations.2x+y=160 ⇒y=160−2x x 70 60 50 40 y=160−2x 20 40 60 802x+y=150 ⇒y=150−2x x 65 55 45 35 y=150−2x 20 40 60 80We can see that the lines do not intersect anywhere, i.e. they are parallel. Hence we can not arrive at a solution.

Question 4

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.(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.

Accepted Answer

i) Consider number of boys =x and number of girls =y. According to the question,x+y=10or, y=10−x ………(i)y=x+4 ………..(ii) x 1 23  4y 9 8 7 6 x1  2 3 4 y 5 6 7 8Graphical solution : Girls =7, boys =3(ii) Consider price of a pencil is x and price of a pen is y. According to the question,5x+7y=50.....(i)7x+5y=46 .….(ii) x 1 2 3 4 y 7.8 6.4 5 4.2 x 1 2 3 4y 6.4 5.7 54.2Graphical solution : Price of a pencil is Rs 3Price of a pen is Rs 5.

Question 5

On comparing the ratios a2​a1​​,b2​b1​​, and c2​c1​​, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:(i) 5x−4y+8=0 ; 7x+6y−9=0                                   (ii) 9x+3y+12=0; 18x+6y+24=0(iii) 6x−3y+10=0 ; 2x−y+9=0

Accepted Answer

(i) a2​a1​​ =75​, b2​b1​​ =6−4​, c2​c1​​ =−98​ ∵a2​a1​​=b2​b1​​, the lines intersect at a point.(ii) a2​a1​​ =189​=21​,b2​b1​​ =63​=21​,c2​c1​​ =2412​=21​∵a2​a1​​=b2​b1​​=c2​c1​​, the lines are coincident. (iii) a2​a1​​ =26​=13​,b2​b1​​ =−1−3​=13​,c2​c1​​ =910​∵a2​a1​​=b2​b1​​=c2​c1​​, the lines are parallel.

Question 6

On comparing the ratios a2​a1​​,b2​b1​​ and c2​c1​​, find out whether the following pair of linear equations are consistent, or inconsistent.(i) 3x+2y=5;2x−3y=7(ii) 2x−3y=8;4x−6y=9(iii) 23​x+35​y=7; 9x−10y=14(iv) 5x−3y=11;−10x+6y=−22(v) 34​x+2y=8;2x+3y=12

Accepted Answer

(i) a2​a1​​ =23​,b2​b1​​ =−32​,c2​c1​​ =75​∵a2​a1​​=b2​b1​​=c2​c1​​, the lines intersect and have an unique consistent solution.(ii) a2​a1​​ =42​ =21​,b2​b1​​ =−6−3​=21​,c2​c1​​ =98​∵a2​a1​​=b2​b1​​=c2​c1​​, the lines are parallel and have no solutions, i.e. the equations are an inconsistent pair(iii) a2​a1​​ =923​​=61​,b2​b1​​ =−1035​​=−61​,c2​c1​​ =147​=21​∵a2​a1​​=b2​b1​​=c2​c1​​, the lines intersect and have an unique consistent solution.(iv) a2​a1​​ =−105​=−21​,b2​b1​​ =6−3​=−21​,c2​c1​​ =−2211​= −21​∵a2​a1​​=b2​b1​​=c2​c1​​, the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.(v) a2​a1​​ =234​​ =32​,b2​b1​​ =32​,c2​c1​​ =128​=32​∵a2​a1​​=b2​b1​​=c2​c1​​, the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.

Question 7

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:(i) x+y=5,2x+2y=10(ii) x− y=8,3x−3y=16(iii) 2x+y−6=0,4x−2y− 4=0(iv) 2x− 2y− 2=0,4x−4y−5=0

Accepted Answer

(i) x+y=5   ...(i)2x+2y=10    ...(ii)⇒x+y=5⇒y=5−x x 0 3 y 5 2Plot (0,5) and (3,5) on graph and join them to get equation x+y=5.2x+2y=10⇒2y=(10−2x)⇒y=210−2x​=5−x   ...(iii) x 5 2 y 0 3So, the equation is consistent and has infinitely many solution (ii) x−y=8   ....(i)3x−3y=16   ....(ii)⇒x−y=8⇒−x+y=−8⇒y=−8+x⇒y=x−8 x 8 0 y 0 -83x−3y=16   ...(ii)⇒3x=16+3y⇒3x−16=3y⇒y=33x−16​⇒y=x−316​ x 316​ 0 y 0 3−16​Plotting both the equation in graph, we see that the lines are parallel , so inconsistent.(iii) 2x+y−6=0    4x−2y−4=02x+y=6 ....(i)4x−2y=4   ...(ii)For equation (i), 2x+y=6 ⇒y=6−2xx03y60Plot point (0,6) and (3,0) on a graph and join then to get equation 3x+y=6For equation (ii), 4x−2y=4⇒24x−4​=yx10y0−2Plot point (1,0) and (0,−2) on a graph and join them to get equation 4x−2y=0x=2,y=2 is the solution of the given pairs of equation . So. solution is consistent.(iv)  2x−2y=2   ...(i)4x−4y=5  ...(ii)2x−2y=2⇒2x−2=2yy=x−1x01y−10Plot point (0,1) and (1,0) and join to get the equation 2x−2y=2 on a graph 4x−4y=5⇒4x−5=4y⇒44x−5​=yx045​y −45​0Plot point (0,−45​) and (45​,0) and join them to get the equation 4x−4y=5 on a graph.The two lines never intersect, so, the solution is inconsistent.

Question 8

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 graphically.Solution:

Accepted Answer

Let the length of the garden be x m and its breadth be y m.Then according to the question, we have:​x=y+4⇒x−y=4…(i)​ and x+y=36For graphical representation:From equation (i), we have: y=x-4When x=10, then y=10-4=6When x=12 then y-12-4=8When x=20, then y=20-4=16Thus, we have the following table of solutions:From equation (ii), we have: y=36When x=20, then y=36-20=16When x-24 then y=36-24=12When x=12, then y=36-12=24

Question 9

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        (ii) parallel lines             (iii) coincident lines

Accepted Answer

a) Intersecting linesSolution: For intersecting line, the linear equations should meet following condition:a2​a1​​=b2​b1​​For getting another equation to meet this criterion, multiply the coefficient of x with any number and multiply the coefficient of y with any other number. A possible equation can be as follows:4x+9y−8=0(b) Parallel linesSolution: For parallel lines, the linear equations should meet following condition:a2​a1​​= b2​b1​​=c2​c1​​For getting another equation to meet this criterion, multiply the coefficients of x and y with the same number and multiply the constant term with any other number. A possible equation can be as follows:4x+6y–24=0(c) Coincident linesSolution: For getting coincident lines, the equations should meet following condition;a2​a1​​= b2​b1​​=c2​c1​​For getting another equation to meet this criterion, multiply the whole equation with any number. A possible equation can be as follows:4x+6y–16=0

Question 10

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

Accepted Answer

x−y+1=0x−y=−1    ...(i)3x+2y−12=03x+2y=12   ...(ii)x−y=−1y=1+xx0−1y10Plot (0,1) and (1,2) on graph and join them to get eqn (i)For 3x+2y=12⇒2y=12−3x⇒r=212−3x​ x04 y60Plot point (0,6) and (4,0) on graph and join them to get eqn (ii)△ABC is the required triangle with coordinates of A as (2,3),B(−1,0) and C(4,0).

Question 11

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

Accepted Answer

2x+3y=112x=11−3ySubstituting this in 2nd equation, we get(11−3y)−4y=−2411−7y=−247y=11+24y=735​=5Substituting value of y 2x=11−3×5x=2−4​=−2Now, to find the value of m, we substitute the values of x and y in the equation5=m×(−2)+3m=−25−3​=−1

Question 12

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

Accepted Answer

firstangle=xosecondangle=xo+15firstangle+secondangle=180 [sum of supplementary angles is 180 degree]x+x+15=1802x+15=180x=82.5firstangle=xo=82.5osecondangle=xo+15=97.5o

Question 13

Solve the following pair of linear equations by the eliation method and the substitution method :(i) x+y=5 and 2x−3y=4 (ii) 3x+4y=10 and 2x−2y=2(iii) 3x−5y−4=0 and 9x=2y+7(iv) 2x​+32y​=−1 and x−3y​=3

Accepted Answer

(i) x+y=5    ...(i)             2x−3y=4      ...(ii)     (elimination method)Multiply eqn (i) by 2    2x+2y=10    2x−3y=4−+ −____________             5y=6⇒y=56​Substitute y=56​in eqn (i)x+y=5x+56​=5⇒x=55×5​−56​=519​Substitution method x+y=5          ....(i)2x−3y=4     ....(ii)y=5−x        ...(iii)Putting eqn. (3) in (2)2x−3(5−x)=42x−15+3x=4x=519​y=55×5​−519​y=56​(ii)Elimination method 3x+4y=10   ...(i)2x−2y=2     ...(ii)Multiply eqn (2) by 24x−4y=43x+4y=10____________7x=14x=22×2−2y=2⇒4−2=2yy=1Substitution method 3x+4y=10   ...(i)2x−2y=2  ⇒x−y=1 x−1=y    ...(ii)Putting eqn (ii) in (i) 3x+4(x−1)=103x+4x−4=10_________________7x =14x=2y=x−1y=2−1⇒y=1(iii) 3x−5y=4      ...(i)9x−2y=7     ...(ii)  (elimination method)Multiply eqn (i) by 3(ii)-(i)×39x−2y=79x−15y=12_____________     13y=−5⇒y=13−5​3x−5×13−5​=4⇒3x=4−1325​3x=1352−25​x=13×327​⇒x=139​Substitution method 3x−5y=4   ...(i)9x−2y=7   ⇒9x−7=2y29x−7​=y3x−5y=4⇒3x−5(29x−7​)=4⇒3x−245​x+235​=4⇒26x​−245​x=4−235​⇒2−39x​=28−35​x=3927​⇒x=139​Putting x=139​ in 3x−5y=4133×9​−5y=41327​−5y=4⇒1327​−4=5y⇒1327−52​=5y13−25​=5y⇒y=13−5​(iv)Elimination method 2x​+32y​=−163x+4y​=−1⇒3x+4y=−6    ...(i)x−3y​ =3⇒3x−y=9    ...(ii)(i)−(ii)3x+4y=−63x−y=−9−+−______________          5y=−15⇒y=−3Putting y=−3 in eqn (i)3x−12=−6⇒3x=−6+12⇒3x=6⇒x=2Substitution method 3x+4y=−6   ...(i)3x−y=9    ...(ii)3x−9=y   ....(iii)Putting enq (iii) in (i)3x+4(3x−9)=−6⇒3x+12x−36=−6⇒15x=−6+36⇒15x=30⇒x=2Putting x=2 in eq (iii)6−9=y⇒y=−3

Question 14

Form the pair of linear equations in the following problems, and find their solutions(if they exist) by the eliation method :(i) If we add 1 to the numerator and subtract 1 from the denoator, a fraction reduces to 1. It becomes 21​ if we only add 1 to the denoator. What is the fraction?(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?(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.(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.(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.

Accepted Answer

(i) Let the fraction be yx​Given, y−1x+1​=1x+1=y−1x−y=−2     .....(1)Also,y+1x​=21​2x=y+12x−y=1        .....(2)Substracting (1) from (2)x=3Substituting value of x in (2), 6−y=1y=5Hence, the fraction is 53​(ii) Let the age of Nuri be x and that of Sonu be y.Given, (x−5)=3(y−5)3y−x=10    ..... (1)And,(x+10)=2(y+10)2y−x=−10    ....(2)Subtracting (2) from (1), we gety=20Substituting value of y in (1)60−x=10x=50Hence, the ages of Nuri and Sonu are 50 and 20 respectively.(iii) Let the two digit number be  10x+yGiven, x+y=9         .....(1)And, 9(10x+y)=2(10y+x)90x+9y=20y+2x88x−11y=0           .....(2)Multiplying (1) by 11, we get11x+11y=99       ....(3)Adding (1) and (3)99x=99x=1Substituting this value of x in (1),1+y=9y=8Hence, the number is 18.(iv) Let the number of Rs 100 notes be x and the number of Rs 50 notes be y.Given,x+y=25     ...(1)And, 100x+50y=2000   ...(2)Multiplying (1) by 50, we get50x+50y=1250     ....(3)Subtracting (3) from (2), 50x=750x=15Substituting value of x in (1), 15+y=25y=10(v) Let the fixed charge for first 3days be  Rsx and after 3days the charge for each day be  Rsy.Given, x+4y=27       ....(1) [first 3days + 4days]And,x+2y=21       ...(2) [first 3days + 2days]Subtracting (2) from (1), we get2y=6y=3Substituting value of y in (1)x+4×3=27x+12=27x=15Hence, the fixed charge is Rs 15 and the per day charge is Rs 3.

Question 15

Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions.(i) x−3y−3=0; 3x−9y−2=0(ii) 2x+y=5; 3x+2y=8(iii) 3x−5y=20; 6x− 10y=40(iv) x−3y−7=0; 3x− 3y−15=0

Accepted Answer

(i) x–3y–3=0 3x–9y–2=0 Comparing these with pair of equations of the form a1​x+b1​y+c1​=0 and a2​x+b2​y+c2​=0,   a2​a1​​ =31​,b2​b1​​ =−9−3​=31​,c2​c1​​ =23​∵a2​a1​​=b2​b1​​=c2​c1​​, the lines have no solution.(ii) 2x+y=53x+2y=8a2​a1​​ =32​,b2​b1​​ = 21​,c2​c1​​ = 85​∵a2​a1​​=b2​b1​​, the lines have unique solution. (iii) 3x−5y=206x− 10y=40a2​a1​​ =63​=21​,b2​b1​​ =−10−5​=21​,c2​c1​​ =4020​= 21​∵a2​a1​​=b2​b1​​=c2​c1​​, the lines have infinitely many solution.  (iv) x−3y−7=0;3x− 3y−15=0a2​a1​​ =31​,b2​b1​​ = −3−3​= 1,c2​c1​​ = 157​∵a2​a1​​=b2​b1​​, the lines have unique solution.

Question 16

For which values of a and b does the following pair of linear equations have an infinite number of solutions?2x+3y=7 ; (a− b)x+(a+b)y=3a+b− 2(ii) For which value of k will the following pair of linear equations have no solution?3x+y=1 ; (2k−1)x+(k−1)y=2k+1

Accepted Answer

i) For the pair of linear equations to have infinite number of solutions, a2​a1​​=b2​b1​​=c2​c1​​Given equation is 2x+3y=7Thus a1​=2,b1​=3,c1​=−7We know a2​=a−b,b2​=a+b,c2​=−3a−b+2Therefore, a−b2​=a+b3​=(3a+b−2)7​⇒−6a−2b+4=−7a+7b⇒ a−9b=−4         ...(1)and −9a+3b+6=−7a−7b⇒ −2a+4b=−6       ...(2)Solving (1) and (2), we geta=5,b=1ii) For the pair of linear equations to have infinite number of solutions, a2​a1​​=b2​b1​​=c2​c1​​Given equation is 3x+y=1Thus a1​=3,b1​=1,c1​=1We know a2​=2k−1,b2​=k−1,c2​=2k+1Therefore, 2k−13​=(k−1)1​=(2k+1)1​⇒3k−3=2k−1,2k+1=k−1⇒k=2,k=−2∴whenk=2 then equations will have no solution

Question 17

Solve the following pair of linear equations by the substitution and cross-multiplication methods :8x+5y=9; 3x+2y=4

Accepted Answer

Substitution method:8x+5y=9      ∴y=59−8x​       ....(1)3x+2y=4         ...(2)Substitute eq(1) in eq (2), we get3x+2(59−8x​)=4Multiply by 5 on both sides, we get15x+18−16x=20⇒−x=2⇒x=−2Substitute x=−2 in eq (1), we gety=59−8(−2)​⇒y=525​=5So, (x,y)=(−2,5)Cross-multiplication method:8x+5y=9         ...(1)So, a1​=8,b1​=5,c1​=−93x+2y=4         ...(2)So, a2​=3,b2​=2,c2​=−4∴x=a1​b2​−a2​b1​b1​c2​−b2​c1​​⇒x=8(2)−3(5)5(−4)−2(−9)​=16−15−20+18​=−2y=a1​b2​−a2​b1​c1​a2​−c2​a1​​⇒y=8(2)−3(5)−9(3)−(−4)8​ =16−15−27+32​=5So, (x,y)=(−2,5)

Question 18

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the cost of food per day.(ii) A fraction becomes 31​ when 1 is subtracted from the numerator and it becomes 41​ when 8 is added to its denoator. Find the fraction.(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Accepted Answer

i) Let the fixed monthly hostel charges be Rs. x and the cost of food per day taken from the mess be y.As per the given condition,x+20y=1000        ...(1)x+26y=1180         ...(2)Subtracting eq(1) from eq (2), we get⇒6y=180⇒y=30Substitute y=30 in eq (1), we get⇒x+20(30)=1000⇒x+600=1000⇒x=400So, the fixed monthly hostel charge is Rs. 600 and the cost of food per day is Rs. 30.ii) Let the numerator be x and the denominator be y, so the fraction will be yx​.yx−1​=31​   ⇒3x−y=3⇒3x−3=y       ...(1)    y+8x​=41​       ⇒4x=y+8       ...(2)Substituting for y in (2) we get,4x=3x−3+8x=5putting x=5 in eq(1) we get y=12∴fraction=yx​=125​iii) Let x be the number of correct answers and y be the number of wrong answers.So, 3x−y=40   ...(1)4x−2y=50       ...(2)Multiplying (1) by 2, we get6x−2y=80       ...(3)4x−2y=50      ...(4)Subtracting eq(4) from eq(3), we get⇒2x=30⇒x=15Putting x=15 in eq (1), we get ⇒3(15)−y=40⇒y=5So, Correct answers =15 and wrong answers =5.Total number of questions in the test =x+y=20.iv) Let the speeds of the first and second cars be x km/hr and y km/hr respectively.Distance between places A and B =100km.Moving in the same direction, relative speed will be =x−y5×(x−y)=100⇒x−y=20            ...(1)  Moving in opposite directions, relative speed will be =x+y⇒1×(x+y)=100⇒x+y=100         ...(2)Adding (1) and (2), we get ⇒2x=120⇒x=60Substitute in eq (1), we get⇒y=40The speeds of the first and second cars are 60 km/hr and 40 km/hr respectively.v) Let the length of rectangle be l and the breadth be b. So, the area will be A=l×b.Case 1: l′=l−5b′=b+3A′=A−9=lb−9⇒(l−5)(b+3)=lb−9⇒lb+3l−5b−15=lb−9⇒3l−5b=6    ....(1)    Case 2:l′′=l+3b′′=b+2A′′=A+67=lb+67(l+3)(b+2)=lb+67⇒lb+2l+3b+6=lb+67⇒2l+3b=61  ...(2)Multiplying (1) by 3 and (2) by 5 and adding both equations, we get9l−15b=18        ...(3)10l+15b=305   ...(4)Adding eq (3) and eq (4), we get19l=323⇒l=17Putting l=17 in eq(1), we get ⇒3(17)−5b=6⇒45=5b⇒b=9The length and breadth of the rectangle are 17 and 9 units respectively.

Question 19

Solve the following pairs of equations by reducing them to a pair of linear equations:(i) 2x1​+3y1​=2;3x1​+2y1​=613​(ii) x​2​+y​3​=2;x​4​−y​9​=−1(iii) x4​+3y=14;x3​−4y=23(iv) x−15​+y−21​=2;x−16​−y−23​=1(v) xy7x−2y​=5;xy8x+7y​=15(vi) 6x+3y=6xy;2x+4y=5xy(vii) x+y10​+x−y2​=4;x+y15​−x−y5​=−2(viii) 3x+y1​+3x−y1​=43​;2(3x+y)1​−2(3x−y)1​=8−1​

Accepted Answer

(i)2x1​+3y1​=23x1​+2y1​=613​Lets assume x1​=p and y1​=q, both equations will become2p​+3q​=2⇒3p+2q=12    ... (A)3p​+2q​=613​⇒2p+3q=13   ...(B)Multiply (A) by 3 and (B) by 2, we get9p+6q=364p+6q=26Subtracting these both, we get5p=10⇒p=2⇒x=p1​=21​∴q=626−8​=3⇒y=q1​=31​(ii) x​2​+y​3​=2x​4​−y​9​=−1Lets assume x​1​=p and y​1​=q, both equations will become2p+3q=2   ...(A)4p−9q=−1    ...(B)Multiply (A) by 3, we get6p+9q=64p−9q=−1Adding these both, we get10p=5⇒p=21​⇒  x=p21​=4∴q=31​⇒y=q21​=9(iii)x4​+3y=14  x3​−4y=23Lets assume x1​=p, both equations become4p+3y=14   ...(A)3p−4y=23   ...(B)Multiply  (A) by 4 and (B) by 3, we get16p+12y=569p−12y=69Adding these both, we get25p=125⇒p=5⇒x=p1​=51​⇒y=−2(iv) x−15​+y−21​=2x−16​−y−23​=1Lets assume x−11​=p and y−21​=q, both equations become5p+q=26p−3q=1Multiply (A) by 3, we get15p+3q=66p−3q=1Adding both of these, we get21p=7⇒p=31​⇒   x=4∴q=31​⇒   y=5(v)xy7x−2y​=5⇒y7​−x2​=5xy8x+7y​=15⇒   y8​+x7​=15Let t=x1​ and r=y1​ equation will be7r−2t=5;8r+7t=15On solving we gett=1r=1∴x=1andy=1(vi)Dividing both side by xy then we havey6​+x3​=6;y2​+x4​=5Let t=x1​andr=y1​ equation will be6r+3t=6;2r+4t=5 on solving we gett=1r=21​∴x=1andy=2(vii)Let t=x+y1​andr=x−y1​ then equation will be10t+2r=4;15t−5r=−2 on solving we havet=51​andr=1∴x+y=5andx−y=1Hence, x=4,y=1(viii)Let t=3x+y1​andr=3x−y1​ then equation will bet+r=43​;2t​−2r​=8−1​On solving we havet=41​andr=21​∴3x+y=4and3x−y=2Hence, x=1,y=1

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