InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
Solve for ` x and y` : ` 0.4 x - 1.5 y = 6.5 `, ` 0.3 x + 0.2 y = 0.9`. |
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Answer» Multiplying each of the equations by 10 , we get ` 4x - 15y = 65" "`… (i) ` 3x + 2 y = 9" "`… (ii) Multiplying (i) by 2 and (ii) 15 and adding, we get `8x + 45x = 130 + 135 ` ` rArr 53 x = 265 ` ` rArr x = ( 265 )/( 53 ) = 5 ` Putting ` x = 5 `in (ii), we get ` 15 + 2y = 9 rArr 2y = 9 - 15 rArr 2y = - 6 rArr y = - 3 ` Hence, ` x = 5 and y = - 3` |
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| 102. |
`1/(2x)-1/y=-1 , 1/x+1/(2y)=8` |
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Answer» The given equations are `(1)/(2x) - (1)/(y) = - 1` ….(1) and `(1)/(x) + (1)/(2y) = 8` …. (2) Let `(1)/(x) = u` and `(1)/(y) = v` then equation (1) and (2) can be written as `(u)/(2) - v = - 1` implies u - 2v = - 2 ….(3) and `u + (v)/(2) = 8` implies 2u + v = 16 .... (4) Multiplying equation (3) by 1 and (4) by 2, we get u - 2v = - 2. ....(5) 4u + 2v = 32 .... (6) Adding equations (5) and (6), we get 5u = 30 implies u = 6 Substituting u = 6 in equation (3), we get 6 - 2v = - 2 implies -2v = - 8 implies v = 4 Now, `(1)/(x) = u` implies `(1)/(x) = 6` implies `x = (1)/(6)` and `(1)/(y) = v` implies `(1)/(y) = 4` implies `y = (1)/(4)` Hence, the solution is `{:(x = (1)/(6)),(y = (1)/(6)):}}` |
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| 103. |
Solve the followingsystem of equations:`11 x+15 y+23=0, 7x-2y-20=0` |
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Answer» The given equations are ` 11 x + 15y = - 23 " " `… (i) ` 7 x - 2y = 20 " " `… (ii) Multiplying (i) by 2 and (ii) by 15 and adding the results , we get ` 22 x + 105 x = - 46 + 300` ` rArr 127 x = 254 ` ` rArr x = ( 254 )/( 127) = 2 ` Putting ` x = 2 ` in (i), we get ` 22 + 15 y = - 23 ` `rArr 15y = - 23 - 22` `rArr 15y = - 45 rArr y = (-45)/( 15) = - 3 ` Hence, ` x = 2 and y = - 3 ` |
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| 104. |
Solve 3x - y = 23 and `(x)/(3) + (y)/(4)` = 4. |
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Answer» The given equations are 3x - y = 23 ….(1) and `(x)/(3) + (y)/(4) = 4` implies 4x + 3y = 48 ….(2) Multiplying equation (1) by 3 and (2) by 1, we get 9x - 3y = 69 ….(3) 4x + 3y = 48 ….(4) Adding equations (3) and (4), we get 13x = 117 implies x = 9 Substituting x = 9 in equation (1), we get `3 xx 9 - y = 23` implies 27 - y = 23 implies -y -4 implies y = 4 Hence, the solution is `{:(x = 9),(y = 4):}}` |
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| 105. |
Solve `(7)/(x) + (8)/(y) = 2` and `(2)/(x) + (12)/(y) = 20`. |
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Answer» Given equations are `(7)/(x) + (8)/(y) = 2 ….(1)` and `(2)/(x) + (12)/(y) = 20 ….(2)` Let `(1)/(x)` = u and `(1)/(y)` = v then equation (1) and (2) can be written as 7u + 8v = 2 ….(3) and 2u + 12v = 20 ….(4) Multiplying equation (3) by 2 and (4) by 7, we get 14u + 16v = 4 ....(5) and 14u = 84v = 140 ....(6) On subtracting equation (6) from (5), we get -68v =- 136 implies v = 2 Substituting v = 2 in equation (3), we get 7u + 8 `xx` 2 = 2 implies 7u + 16 = 2 implies 7u = - 14 implies u = - 2 When `u = - (1)/(2)` implies `(1)/(x) = - 2` implies `x = - (1)/(2)` and when v = 2 implies `(1)/(y) = 2` implies `y = (1)/(2)` Hence, the solution is `{:(x = -(1)/(2)),(y = (1)/(2)):}}`. |
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| 106. |
Solve:`4x+6/y=15``6x-8/y=14`and hence find p if `y=p x-2` |
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Answer» The given equation are `4x + (6)/(y) = 15` ….(1) and `6x - (8)/(y) = 14 ….(2)` Multiplying equation (1) by 4 and (2) by 3, we get `16x + (24)/(y) = 60` ….(3) `18x - (24)/(y) = 42 ….(4)` Adding equations (3) and (4), we get 34x = 102 implies x = 3 Substituting x = 3 in equation (1), we get `4 xx 3 + (6)/(y) = 15` implies `(6)/(y) = 15 - 12` implies `(6)/(y) = 3` implies 3y = 6 or y = 2 Hence, the solution is `{:(x = 3),(y = 2):}}` |
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| 107. |
Draw the graph of the equation y=3x From your graph, find the value of y when x=-2. |
| Answer» after drawing the graph it can be seen that it intersect at point (-2,-6). | |
| 108. |
x+y= 14 x-y= 4 |
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Answer» adding both equations `x+y + x -y = 14 + 4 =18` `2x = 18` `x= 9` `y = 14-x` `y = 14-9` `y= 5` Answer |
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| 109. |
Ram Phal sowed wheat and paddy in two fields of total area 5000 square metres. Write a linear equation which satisfies the data. Draw the graph of the same |
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Answer» Ram phul sowed wheat in=`x m^2` Ram phul sowed Paddy in=`y m^2` x+y=5000 Let x=0,y=5000 Let y=0,x=5000 joining these 2 points we get our desired graph. |
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| 110. |
Five years age a manwas seven times as old as his son. Five years hence, the father will be threetimes as old as his son. Find their present ages. |
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Answer» Let present age of son be x years and the present age of father by y years. Five years hence, the age of son = x + 5 and the age of father = y + 5 Using given information 3(x + 5) = y + 5 implies 3x - y = - 10 ….(1) Five years ago, age of son = (x - ) years and age of father = (y - 5) years `therefore 7(x - 5) = y - 5` implies 7x - y = 30 ...(2) Subtracting equation (1) from (2), we get 4x = 40 implies x = 10 On substituting x = 10 in equation (1) `3 xx 10 - y = - 10` implies 30 - y = - 10 implies - y = - 40 implies y = 40 Hence, present age of son is 10 years and present age of father is 40 years. |
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| 111. |
Five years hence,fathers age will be three times the age of his son. Five years ago, fatherwas seven times as old as his son. Find their present ages. |
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Answer» Correct Answer - 40 years, 10 years Let the present ages of the man and his son by x years and y years respectively. Then, ` x + 5 = 3( y + 5) rArr x - 3y = 10" "`… (i) `and x - 5 = 7( y - 5) rArr x - 7y = - 30." " `… (ii) |
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| 112. |
5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 times the age of his son. The present age of the man isA. 45 yearsB. 50 yearsC. 47 yearsD. 40 years |
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Answer» Correct Answer - D Let the present ages of the man and his son be x years and y years repectively. Then, `(x + 5) = 3( y + 5) rArr x - 3y = 10" " `...(i) ` (x - 5) = 7(y - 5) rArr x - 7y = - 30 " " `... (ii) |
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| 113. |
The coach of a cricket team buys 7 bats and 6 balls for ₹ 13200. Later, he buys 3 bats and 5 balls for ₹ 5900. Find the cost of each bat and each ball. |
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Answer» Let the cost each bat be ₹ x and the cost of each ball be ₹ y . Then, ` 7x + 6y = 13200" "` ...(i) And, ` 3x + 5y = 5900" "`...(ii) On multiplying (i) by 5, (ii) by 6 and subtracting the results, we get ` 35 x - 18 x = 66000 - 35400 rArr 17x = 30600 ` ` rArr x = ( 30600) /( 17 ) = 1800` Putting ` x = 1800 ` in (ii), we get ` 5400 + 5y = 5900 rArr 5y = 5900 - 5400 ` `rArr 5y = 500` ` rArr y = 100` ` therefore ` cost of each bat = ₹ 1800 and cost of each ball = ₹ 100. |
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| 114. |
`{:(217x + 131y = 913),(131x + 217y = 827):}` |
| Answer» Correct Answer - ` x = 3, y = 2` | |
| 115. |
A sailor goes 8 kmdownstream in 40 minutes and returns in 1 hours. Determine the speed of thesailor in still water and the speed of the current. |
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Answer» Correct Answer - 10 kmph, 2 kmph Let the speed of the sailor in still water be x kmph and the speed of the current be y kmph. Then, speed downstream = ` (x + y ) ` km/hr. And, speed upstream = ` (x - y ) ` km/hr ` therefore (8)/(x + y ) = ( 40)/(60) = (2)/(3)rArr x + y = 12 " " `… (i) and `(8)/(x - y ) = 1 rArr x - y = 8. " "`… (ii) Solve (i) and (ii). |
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| 116. |
A boat which travels at the rate of 10.5 km/hr downstream takes 3 times as long time to go to a certain distance up a river as to go the same distance down. Find the rate at which stream flows. |
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Answer» Let the speed of stream = x km/hr and the speed of boat in still water = y km/hr `therefore` Speed of boat downstream = (x + y) km/hr and speed of boat upstream = (y - x) km/hr Let the distance be D km. `therefore` According to the problem, x + y = 10.5 ...(1) Also, time taken by boat upstream = 3 `xx` time taken by boat downsteam implies `(D)/(y - x) = 3 xx (D)/(10.5)` `implies y - x = 3.5` Adding equations (1) and (2), we get 2y = 14 implies y = 7 Putting y = 7 in equation (1), we get x + 7 = 10.5 implies x = 3.5 Hence, speed of stream = 3.5 km/hr |
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| 117. |
`{:(2x + (k - 2)y = k),(6x + (2k - 1) y = 2k + 5):}` |
| Answer» Correct Answer - `k = 5` | |
| 118. |
If `x = 2k-1` and `y=k` is a solution the equation `3x-5y-7=0`; find the value of k. |
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Answer» The given equation is ` 3x - 5y - 7 =0." "`…(i) Since `x=2k -1 and y=k` is solution of (i), we have `3 xx (2k - 1) - 5k - 7 =0` `rArr 6k - 3 - 5K - 7 = 0 rArr k- 10 = 0 rArr k = 10.` Hence, k=10. |
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| 119. |
Find the value of k, if `x=2, y=1`is a solution of the equations `2x+3y=k`. |
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Answer» The given equation is `2x+ 3y - k = 0." "`…(i) Since x=2, y=1 is a solution of (i) , we have `2 xx 2 + 3 xx1 - k = 0 rArr 4+3- k=0 rArrk=7.` Hence, k=7. |
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| 120. |
Write each of the following equations in the form `ax+by+c=0` and indicate the values of a,b,c in each case. (i) `3 = 2x + y` (ii) `3x-8 = 5y` (iii) `x=4y` (iv) `x/3- y/2=5` (v) `4y-3 = sqrt(2) x` (vi) `pix+ y = 6` |
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Answer» We have `3=2x + y rArr 2x + y-3 =0.` This is of the form `ax+by+c=0`, where `a=2, b=1 and c=-3.` (ii)` 3x - 8 = 5y rArr 3x = 5y + 8` ` rArr 3x - 5y -8=0.` This is of the form `ax+by+c=0, " where " a =3, b=-5 and c=-8.` (iii) `x=4y rArr x-4y = 0` `rArr x- 4y + 0 = 0.` This is of the form` ax+by + c=0 " where " a=1,b=-4 and c=0.` (iv) `x/3-y/2=5 rArr 2x-3y = 30.` This is of the form `ax+by+c=0," where " a=2, b=-3 and c=-30`. (v) `4y - 3 = sqrt(2x) rArr sqrt (2x) - 4y + 3 = 0.` This is of the form `ax+by+ c =0," where " a = sqrt(2), b=-4 and c=3.` (vi) `pix + y= 6 rArr pi x +y = 6 =0.` This is of the form `ax+by+c=0," where " a=pi, b=1 and c=-6.` |
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| 121. |
Show that x=3 and y=2 is a solution of the linear equation `2x+3y=12`. |
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Answer» The given linear equation is `2x+3y - 12=0." "` …(i) Substituting x=3 and y=2 in (i) , we get `LHS = (2xx 3 + 3 xx2 -12) = ( 6+6-12) = 0 = RHS.` `:. x= 3 and y=2 " is a solution of the equation " 2x+3y=12` NOTE We may say that (3,2) is a solution of the equation `2x+3y = 12.` |
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| 122. |
Write each of the following as an equation of the form `ax+by+c=0` and write the values of a,b,c in each case. (i) `x=-3` (ii) `y=5` (iii) `3x=2` (iv) `5y=4` |
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Answer» We have `x=-3 rArrx+ 3 = 0` `rArr 1* x+0*y + 3 =0.` This is of the form `ax+ by + c= 0," where " a= 1, b=0 and c=3.` (ii) `y=5 rArr y-5=0` `rArr 0* x+ 1. y- 5 = 0.` This is of the form `ax+by+ c=0, " where " a=0, b=1 and c=-5.` (iii) `3x=2 rArr 3x- 2 = 0.` `rArr 3x+ 0* y-2=0.` This is of the form `ax+by + c=0, " where " a= 3, b=0 and c=-2.` (iv) `5y= 4 rArr 5y-4=0` `rArr0*x+5y-4=0.` This is of the form `ax+by+c=0," where " a=0,b=5 and c=-4.` |
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| 123. |
The graph of the linear equation x+2y=7 passes through the point (0,7). |
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Answer» False If we put x=0 and y=7 in LHS of the given equation, we get LHS`=(0)+2(7)=0+14=14 ne 7`= RHS Hence, (0,7) does not lie the line x+2y=7. |
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| 124. |
Check which of the following are solutions of the equations `x-2y=4`and which are not:(i) `(0,2)` (ii) `(2,0)` (iii) `(4,0)` (iv) `(sqrt(2),4sqrt(2))` (v) `(1, 1)` `` |
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Answer» The given equation may be written as `x- 2y - 4 =0." "`…(A) (i) Putting x=2 and y=0 in (A), we get `LHS=(2-2 xx 0-4) =(2-0-4) = -2 !=0= RHS.` `:. (2,0) " is not a solution of " x- 2y =4.` (ii) Putting `x=0 and y=-2` in (A), we get `LHS = { 0- 2 xx (-2) -4 } = {0+ 4-4} =0= RHS.` `:. (0,2) " is a solution of " x-2y=4.` (iii) Putting `x=4sqrt(2) and y= sqrt(2)` in (A), we get `LHS = ( 4 sqrt(2) - 2 xx sqrt(2) - 4)=(4 sqrt(2) -2 sqrt(2)-4)` `=(2sqrt(2) - 4) != 0 = RHS.` `:. ( 4 sqrt(2), sqrt(2) )" is not a solution of "x- 2y = 4`. (iv) Putting x=4 and y=0 in (A), we get `LHS = (4-2 xx 0-4) = (4-0-4) = 0=RHS.` `:. (4,0)" is a solution of " x-2y=4.` |
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| 125. |
Express each of the following equations in the form `ax+by+ c = 0` and indicate the values of a,b,c in each case. (i) `x=6` (ii) `3x - y=x-1` (iii) `2x+9=0` (iv) `4y=7` (v) `x+y=4` (vi) `x/2-y/3=1/6+y` |
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Answer» (i) `x+0*y-6 = 0, ( a+1, b=0,c=-6)` (ii) `2x- y+1 = 0, ( a = 2, b=-1, c= 1)` (iii) `2x + 0*y+9=0, ( a = 2, b= 0, c=9)` (iv) `0*x+4y-7=0,(a=0,b=4,c=-7)` (v) `x+y-4=0,(a=1,b=1,c=-4)` (vi) `3x-8y-1=0,(a=3,b=-8,c=-1)` |
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| 126. |
Check which of the following are the solutions of the equation `5x-4y = 20`. (i) (4,0) (ii) (0,5) (iii) `(-2,5/2)`(iv) `(0,-5)` (v) `(2,-5/2) ` |
| Answer» On by `(4,0),(0,-5),(2,(-5)/2)` are the solutions. | |
| 127. |
The point (0,3) lies on the graph of the linear equation 3x+4y=12. |
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Answer» True If we put x=0 and y=3 in LHS of the given equation, we find LHS`=3xx0+4xx3=0+12=12`=RHS Hence, (0,3) lies on the linear equation 3x+4y=12. |
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| 128. |
The point of the form `(a,a)` always lies onA. the x-axisB. the y-axisC. the line y=xD. the line `x+y=0` |
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Answer» Correct Answer - C The point` (a,a), a!=0` lies on the line y=x. |
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| 129. |
The point of the form (a,-a) always lies on the lineA. x=aB. y= -aC. y=xD. x+y=0 |
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Answer» Taking option (d), x+y=a(-a)=a-a=0 [since, give point is of the form (a,-a)] Hence, the point (a,-a) always lies on the line x+y=0. |
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| 130. |
The point of the form (a,a) always lies onA. X-axisB. Y-axisC. the line y=xD. the line x+y=0 |
| Answer» Since, the given point (a,a) has same value of x and y-coordinates. Therefore, the point (a,a) must be lie on the line y=x. | |
| 131. |
The graph of x=4 is a lineA. making an intercept 4 on the x=axisB. making an intercept 4 on the y- axisC. parallel to the x-axis at a distance of 4 units from the originD. parallel to the y-axis at a distance fo 4 units from the origin |
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Answer» Correct Answer - D The graph of x=4 is a line parallel to the x-axis at a distance of 4 units from the origin. |
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| 132. |
The point of the form (a,-a),`a!=0` lies onA. the x-axisB. the y=axisC. the line y=xD. the line `x+y=0` |
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Answer» Correct Answer - D The point `(a,-a),a!=0` lies on the line `x+y=0` |
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| 133. |
The linear equation `3x-5y=15` hasA. a unique solutionB. two solutionsC. infinitely many solutionsD. no solution |
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Answer» Correct Answer - C The linear equation `3x-5y=15` has infinitely many solutions. |
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| 134. |
The graph of `x+3 =0` is a lineA. making an intercept -3 on the x=axisB. making an intercept -3 on the y- axisC. parallel to the y-axis at a distance of 3 units from the originD. parallel to the x-axis at a distance fo 3 units from the origin |
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Answer» Correct Answer - C The graph of `x+3 = 0` is a line parallel to the y - axis at a distance of 3 units to the left of y-axis. |
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| 135. |
If the point (3, 4) lies on the graph of 3y=ax+7, then find the value of a. |
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Answer» Since, the point (x=3, y=4) lies on the equation 3y=ax+7, then the equation will be satisfied by the point . Now, put x=3 and y=4 in given equation , we get `3(4) =a (3) +7 implies 12 =3a+7` `implies " " 3a=12-7 implies 3a=5` `therefore " " a=(5)/(3)` Hence, the value of a is `5//3`. |
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| 136. |
The graph of the linear equation `2x+ 5y = 10` meets the x-axis at the pointA. (0,2)B. (2,0)C. (5,0)D. (0,5) |
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Answer» Correct Answer - C The graph of `2x+5y = 10` cuts the x-axis at the point where y=0, i.e., at the point (5,0). |
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| 137. |
The graph of `y+2 = 0` is a lineA. making an intercept -2 on the x=axisB. making an intercept -2on the y- axisC. parallel to the x-axis at a distance of 2 units from the originD. parallel to the y-axis at a distance fo 2 units from the origin |
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Answer» Correct Answer - C The graph of `y+2 =0` is a line parallel to the x-axis at a distance of 2 units below the x-axis. |
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| 138. |
The graph of the linear equation `2x+3y=6` cuts the `Y-`axis at the pointA. (2,0)B. (3,0)C. (0,2)D. (0,3) |
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Answer» Correct Answer - C The graph of `2x+ 3y =6` cuts the y-axis at the point where x=0, i.e., at the point (0,2). |
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| 139. |
Draw the graph of the equation, `2x+y=6` find the coordinates of the point where the graph cuts the x-axis. |
| Answer» Correct Answer - (3,0) | |
| 140. |
Draw the graph of line `4x+3y=24`Write the coordinatesof points where this line intersects the x-axis and y-axis.Use this graph to findthe area of the triangle formed by the line and the coordinates axes. |
| Answer» Correct Answer - (i) A(6,0) and B(0,8) (ii) 24 sq units | |
| 141. |
The coordinates of points in the table represent some of the solutions of the equation x-y+2=0. |
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Answer» False The coordinates of points are `(0,2),)(1,3),(2,4),(3,-5) " and " (4,6)`. Given equation is x-y+2=0 At point (0,2) `0-2+2=0 implies 0=0`, it satisfies. At point (1,3), `1-3+2=3-3=0 implies 0=0`, it satisfies. At point (2,4), `2-4+2=4-4=0 implies 0=0`, it satisfies. At point `(3,-5), " " 3-(-5)+2=3+5+2=10 ne 0`, it does not satisfy. At point `(4,6), " " 4-6+2=6-6=0 implies 0=0`, it satsisfies. Hence, point (3,-5) does not satisfy the equation. |
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| 142. |
The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph. |
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Answer» total distance covered=x km total fair=yRs y=8+(x-1)5 y=8+5x-5 y=5x+3 y-5x=3 let x=0, y=3 let x=1, y=8 putting this value in graph we get our desired graph |
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| 143. |
If ` x = - y and y gt 0`, which of the following is wrong ?A. ` x ^(2) y gt 0 `B. ` x + y = 0 `C. ` xy lt 0 `D. `(1)/(x) - (1)/(y) = 0 ` |
| Answer» Correct Answer - D | |
| 144. |
`{:(x - ky = 2),(3x + 2y =-5):}` |
| Answer» Correct Answer - ` k ne = (-2)/(3)` | |
| 145. |
Solve the followingsystem of equations:`2/(3x+2y)+3/(3x-2y)=(17)/5, 5/(3x+2y)+1/(3x-2y)=2` |
| Answer» Correct Answer - ` x =1 , y = 1 ` | |
| 146. |
The area of a rectangle gets reduced by 67 square meters, when its length is incrased by 3m and breadth is decred by 4m. If the lenght is reduced by 1 m breadth is incrased by 4m, the area is incrased by 89 square meters. Find the dimesions of the rectangle. |
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Answer» Correct Answer - length = 28 m, breath = 19 m Let length = x metres and breadth = y metres. Then, ` x y - (x + 3) ( y - 4) = 67 rArr 4x - 3y = 55`. `and " " (x -1) ( y + 4) - xy = 89 rArr 4x - y = 93.` |
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| 147. |
Write the number ofsolutions of the following pair of linear equations: `x+2y-8=0, 2x+4y=16` |
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Answer» Given equations are ` x + 2y - 8 = 0 and 2x + 4y - 16= 0 `. ` therefore (a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)` . So, there are infinitely many solutions. |
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| 148. |
The pair of equations ` 2 x + y = 5, 3x + 2y = 8` hasA. a unique solutionB. two solutionsC. no solutionD. infinitely many solutions |
| Answer» Correct Answer - A | |
| 149. |
If the length of a rectangle is reduced by 5 units and its breadth is increased by 2 units then the area of the rectangle is reduced by 80 sq units. However, if we increase its length by 10 units and decrease the breadth by 5 units, its area increased by 50 sq units. Find the length and breadth of the rectangle. |
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Answer» Let the length and breadth of the rectangle be x units and y units respectively. Then, area of the rectangle = xy sq units. Case I When the length is reduced by 5 units and the breadth is increased by 2 units. Then, new length = `( x - 5)` units. and new breadth ` = ( y +2 )` units. ` therefore ` new area = ` (x - 5) ( y + 2 )` sq units. ` therefore xy - (x - 5 ) ( y +2 ) = 80 rArr 5y - 2 x = 70" " `... (i) Case II When the length is increased by 10 units and the breadth is decreased by 5 units. Then, new length = ` ( x + 10 ) ` units. and new breadth `= ( y - 5) ` units. ` therefore ` new area = ` (x + 10) ( y - 5) `sq units. ` therefore (x+ 10 ) ( y - 5) - xy = 50 ` `rArr 10 y - 5 x = 100 rArr 2y - x = 20 " " `... (ii) On multiplying (ii) by 2 and subtracting the result from (i), we get ` y = 30 `. Putting ` y = 30 ` in (ii), we get ` ( 2 xx 30 ) - x = 20 rArr 60 - x = 20 rArr x = ( 60 - 20 ) = 40 ` ` therefore x = 40 and y = 30 ` Hence, length = 40 units and breadth = 30 units. |
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| 150. |
The graphical representation of the equations ` x + 2y = 3 and 2x + 4 y + 7 = 0 ` gives a pair ofA. parallel linesB. intersecting linesC. coincident linesD. none of these |
| Answer» Correct Answer - A | |