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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.
| 51. |
A lady has 25 paise and 50 paise coins in her purse. If in all she has 40 coins and value of her money is Rs. 12.75, how many coins of each type does she have? |
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Answer» Let number of 25 paise coins be x and number of 50 paise coins be y. According to given conditions, we have x + y = 40 …(1) and 25x + 50y = 1275 …(2) Multiplying equation (1) by 50, we get 50x + 50y = 2000 …(3) Subtracting equation (3) from (2), we get -25x = - 725 implies x = 29 Substituting x = 29 in equation (1), we get 29 + y = 40 implies y = 11 Hence, number of 25 paise coins = 29 and number of 50 paise coins = 11. |
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| 52. |
A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days and an additional charge for each day thereafter. Latika paid Rs. 22 for a book kept for six days, while Anand paid Rs. 16 for the book kept for four days. Find the fixed charges and the charge for each extra day. |
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Answer» Let the fixed charged for first two days = Rs x Let the additional charges per day after 2 days = Rs y Latika paid Rs. 22 for six days. (Given) 2 days fixed charges + (6 - 2) days charges = 22 implies x + 4y = 22 …(1) Anand paid Rs. 16 for books kept for four days. 2 days fixed charges + (4 - 2) days additional charges = 16 implies x + 2y = 16 ...(2) x + 2y = 16 [from (2)] `{:(" x + 4y = 22 [from (1)]"),( " "ul "- - -"),(" -2y = - 6 [subtracting equation (1) from (2)]"):}` implies y = Rs. 3 per day Now, x + 2y = 16 [from (2)] implies x + 2(3) = 16 implies x = 16 - 6 implies x = 10 `{:("So, the fixed charges for first 2 days = Rs. 10"),("The additional charges per day after 2 days = Rs. 3 per day "):}}` |
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| 53. |
` ( 1 ) /(x) + (1 ) /( y) = 7`, ` ( 2)/(x) + ( 3) /( y) = 17 (x ne 0, y ne 0 )` . |
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Answer» Correct Answer - ` x = (1)/(4) , y = (1)/(3)` Putting ` (1)/(x)= u and (1)/( y) = v `, we get ` u + v - 7 = 0 and 2u + 3v - 17 = 0 ` . Now, solve for u and v ? |
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| 54. |
Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5, then find the numbers. |
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Answer» Let the numbers be 5x and 6x respectively. So, new numbers after subtracting 8 from each will be (5x - 8) and (6x - 8) respectively. Now, the ratio of new numbers is `4 : 5`. `therefore" " (5x - 8)/(6x - 8) = (4)/(5)` implies 25x - 40 = 24x - 32 implies 25x - 24x = 40 - 32 implies x = 8 `therefore` Required numbers = 5x and 6x become `5xx8, 6xx8` i.e., Required numbers = 40 and 48. |
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| 55. |
` ( 5)/((x + y)) - ( 2 )/(( x - y )) + 1 = 0 `, ` ( 15)/((x + y ) + ( 7) /(( x - y )) - 10 = 0 ( x ne y , x ne - y )` |
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Answer» Correct Answer - ` x = 3, y = 2 ` Putting `(1)/((x + y )) = u and (1)/((x - y)) = v`, we get ` 5u- 2 v + 1 = 0 and 15 u + 7v - 10 = 0 `. Now, solve for u and v. |
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| 56. |
If ` 2^(x + y ) = 2^(x - y ) = sqrt8` then the value of y isA. ` (1)/(2)`B. `(3)/(2) `C. `0 `D. none of these |
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Answer» Correct Answer - C `2^(x+y) = 2^(x-y)=2^(3//2)rArr x +y =(3)/(2)` and `x-y=(3)/(2)`. So, `y =0`. |
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| 57. |
If ` 29 x + 37y = 103 and 37x + 29y = 95 ` thenA. ` x = 1, y = 2`B. ` x = 2, y = 1 `C. ` x = 3, y = 2 `D. ` x = 2, y = 3 ` |
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Answer» Correct Answer - A Adding (i) and (ii), we get `66(x+y)=198 rArr x +y =3`. Subtracting (ii) from (i), we get `8(y-x)=8rArr y -x =1`. |
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| 58. |
Solve for x and y : ` 152 x - 378 y = - 74, - 378 x + 152 y = - 604 ` |
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Answer» The given equations are ` 152 x - 378 y = - 74 " "`… (i) ` - 378 x + 152 y = - 604" " `…(ii) Clearly, the coefficients of x and y in one equation are intercharged in the other. Adding (i) and (ii), we get ` { 152 + (- 378)}x + {(-378) + 152 } y = - ( 74 + 604) ` ` rArr ( - 226)x + ( - 226) y = - 678 ` `rArr ( - 226) (x + y ) = - 678 ` ` rArr (x + y ) = (-678)/(- 226) rArr x + y = 3" " `... (iii) Subtracting (ii) from (i), we get `( 152 + 378) x + ( - 378 - 152 ) y = (-74 + 604) ` ` rArr 530 x - 530 y = 530 ` `rArr 530 ( x- y ) = 530 rArr x - y = 1 " " `... (iv) Adding (iii) and (iv), we get ` 2x = 4 rArr x = 2`. Substracting (iv) from (iii), we get `2y = 2 rArr y = 1 ` Hence, ` x = 2 and y = 1`. |
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| 59. |
Express `y`in terms of `x`in the equation `2x-3y=12.`Find the points whether the point `(3,3)`is on the line represented by the equation `3x+y-12=0` |
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Answer» 1)`2x-3y=12` `3y=2x-12` `y=(2x-12)/3=(2/3)x-(12/3)` `y=2/3x-4` 2)`3x+y-12=0` `3*3+3-12=0` (3,3). |
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| 60. |
Draw a graph of the equation: `3x-2y=4` and `x+y-3=0` |
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Answer» `3x-2y=4` When y=0,4/3`3x=4` `0-2y=4` `y=-2` `x+y=3` x=0 `3x-2y=4` `x+y=3` `2x+2y=6` `5x=10` `x=2` `y=1` |
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| 61. |
`(ax)/b-(by)/a=a+b`, `ax-by=2ab` |
| Answer» Correct Answer - ` x = b , y = -a ` | |
| 62. |
Any point on the y- axis is of the formA. (x,y)B. (0,y)C. (x,0)D. (y,y) |
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Answer» Correct Answer - B Any point on the y-axis is of the form (0,y). |
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| 63. |
Any point on the `X-`axis is of the formA. (x,y)B. (0,y)C. (x,0)D. (x,x) |
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Answer» Correct Answer - C Any point on the x-axis is of the form (x,0). |
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| 64. |
If `(2,0)` is a solution of the linear equation `2x+3y=k`, then the value of `k` isA. 6B. 5C. 2D. 4 |
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Answer» Correct Answer - D Putting x=2 and y=0 in `2x+3y=k, " we get " k=(2xx 2 +3 xx0)=(4+0)=4.` |
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| 65. |
A linear equation in two variables x and y is of the form `ax+ by + c = 0,` whereA. `a!=0, b!=0`B. `a!= 0, b=0`C. `a=0,b!=0`D. a=0,c=0 |
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Answer» Correct Answer - A A linear equation in two variables is of the form `ax+by+c = 0, " where " a!= 0, b !=0.` |
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| 66. |
If a linear equation has solutions `(-2,2)`,`(0,0)` and `(2,-2)`, then it is of the formA. `x-y=0`B. `x+y=0`C. `-x+ 2y = 0`D. `x-2y=0` |
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Answer» Correct Answer - B Each of the points `(-2,2),(0,0) and (2,-2)` satisfies the linear equation`x+y=0.` |
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| 67. |
How many linear equations can be satisfied by x=2 and y= 3 ?A. only oneB. only twoC. only threeD. infinitely many |
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Answer» Correct Answer - D Infinitely many equations can be satisfied by `x=2 and y=3`. |
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| 68. |
If the temperature of a liquid can be measured in kelvin units as `x^(@)K` or in fahrenheit units as `y^(@)F`, the relation between the two systems of measurement of temperature is given by the linear equation. `y=(9)/(5)(x- 273)+32` (i) find the temperature of the liquid in fahrenheit, if the temperature of the liquid is 313 K. (ii) If the temperature in kelvin. |
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Answer» Given relation is `" " y=(9)/(5)(x-273)+32` (i) Given, `x=313 .^(@)K` then from Eq. (i), we get `y=(9)/(5)(313-273)+32` `=(9)/(5)xx40+32=9xx8+32=72+32=104^(@)F` (ii) Given, `y=158^(@)F`, then from Eq. (i), we get `158=(9)/(5)(x - 273) +32` `implies " " 158=(9(x - 273)+32xx5)/(5)` `implies " " 158xx5=9(x-273)+160` `implies " " 790=9(x-273)+160` `implies " " 790-160=9(x-273)` `implies " " 9(x - 273) = 630` `implies " " x- 273 = (630)/(9)= 70 implies x- 273=70` `because " " x = 70+273= 343 ^(@)K` |
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| 69. |
Let y varies directly as x. If y=12 when x=4, then write a linear equation. What is value of y when x=5 ? |
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Answer» Given that, y varies directly as x. i.e., `y prop x implies y =kx` Given, y=12 and x=4 12=4k `implies " " k=(12)/(4)` `therefore " " k=3` On putting the value of k in Eq. (i), we get y=3x When x=5, then from Eq. (ii), we get `y=3xx5 implies y=15` Hence, the value of y is 15. |
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| 70. |
The linear equation that converts Fahrenheit (F) to Celsius (C ) is given by the relation, `C=(5F-160)/(9)`. (i) If the temperature is `86^(@)` F , what is the temperature in Celsius ? (ii) If the temperature is `35^(@)`C, what is the temperature in Fahrenheit ? (iii) If the temperature is `0^(@)`C, what is the temperature in Fahrenheit and if the temperature is `0^(@)`F, what is the temperature in Celsius ? (iv) What is the numerical value of the temperature which is same in both the scales ? |
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Answer» Given relation is , `C=(5F-160)/(9)` `implies " " 9C=5F-160 implies 5F=9C+160` `implies " " F=(9C+160)/(5)` (i) Given, `F=86^(@)F`, then from Eq. (i), we get `C=(5xx86-160)/(9)=(430-160)/(9)=(270)/(9)=30^(@)C` (ii) Given, `C=35^(@)C`, then from Eq. (ii), we get `F=(9xx35+160)/(5)=(315+160)/(5)=(475)/(5)=95^(@)F` (iii) Given, `C=0^(@)C`, then from Eq. (ii), we get `F=(9xx0+160)/(5)=(160)/(5)=32^(@)F` `F=0^(@)F`, then from Eq. (i), we get `C=(5xx0-160)/(9)=(-160)/(9)=(-(160)/(9)).^(@)C` (iv) By given condition, C=F Put this value in Eq. (i), we get `C=(5C-160)/(9) implies 9C=5C- 160` `implies 9C- 5C= - 160` `implies 4C= - 160 implies C= (-160)/(4) implies C = - 40 = F` |
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| 71. |
Draw the graph of the equation y=3x. Form your graph, find the value of y when (i) x=2 (ii) x=-2. |
| Answer» Correct Answer - (i) y=6 (ii)` y=-6` | |
| 72. |
The graph given below represents the linear equation x=3. |
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Answer» True Since, given graph is a line parallel to Y-axis at a distance 3 units to thr right of the orgin. Hence, it represents a linear equation x=3. |
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| 73. |
Draw the graph of the equation `x+2y-3=0`. From your graph, find the value for y when (i) x=5 (ii) x=-5. |
| Answer» Correct Answer - (i) y=1 (ii) y=4 | |
| 74. |
The graph given below represents the linear equation x+y=0. |
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Answer» True If the given points `(-1,1) " and " (-3,3)` lie on the linear equation x+y=0, then both points will satisfy the equation. So, at point `(-1,1)` we put x=-1 and y=1 in LHS of the given equation, we get LHS`=x+y= - 1+1=0`=RHS Again, at point `(-3,3)` put x=-3 and y=3 in LHS of the given equation, we get LHS `=x+y=-3+3=0`= RHS Hence, `(-1,1) " and " (-3,3)` both satisfy the given linear equation x+y=0. |
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| 75. |
Find the solution of the linear equation x+2y =8 which represents a point on (i) X-axis (ii) Y-axis |
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Answer» We have, x+2y=8 (i) When the point is on the X-axis, then put y=0 in Eq. (i) we get `x+2xx0=8` `implies` x=8 Hence, the required point is (8,0). (ii) When the point is on the Y-axis, then put x=0 in Eq. (i), we get 0+2y=8 `implies " " y=(8)/(2)=4` Hence, the required point is (0,4). |
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| 76. |
The graph of the linear equation 2x+3y=6 cuts the Y-axis at the point |
| Answer» Correct Answer - (0,3) | |
| 77. |
Draw the graph of the equation, `2x-3y=5`. From your graph , find (i) the value of y when x=4 and (ii) the value of x when y=3. |
| Answer» Correct Answer - (i) y=1) (ii) x=7 | |
| 78. |
The graph of the line x=3 passes through the pointA. (0,3)B. (2,3)C. (3,2)D. none of these |
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Answer» Correct Answer - C The graph of the line x=3 passes through the point (3,2). |
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| 79. |
`x=5` and `y=2` is a solution of the linear equationA. `x+2y=7`B. `5x+2y = 7`C. `x+y=7`D. `5x+y=7` |
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Answer» Correct Answer - C Clearly, `x+y=7` is satisfied by x=5,y=2. |
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| 80. |
The positive solutions of the equation ax+by+c=0 always lies in theA. Ist quadrantB. Iind quadrantC. IIIrd quadrantD. Ivth quadrant |
| Answer» We know that, if a line passes through the Ist quadrant, then all solution lying on the line in first quadrant must be positive because the coordinate of all points in the Ist quadrant are positive. | |
| 81. |
If the point `(3, 4)` lies on the graph of `3y=ax+7`, then find the value of `a`.A. `2/5`B. `5/3`C. `3/5`D. `2/7` |
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Answer» Correct Answer - B Since `3y=ax+7` is satisfied by (3,4) we have `3 xx 4=a xx 3+7 rArr 3a + 7 = 12 rArr 3a = 5 rArr a = 5/3.` |
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| 82. |
How many linear equations in x and y can be satisfied by x=1 and y=2 ?A. Only oneB. TwoC. Infinitely manyD. Three |
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Answer» Let the linear equation be ax+by+c=0. On putting x=1 and y=2, in above equation we get `implies " " a+2b+c=0`, where a, b and c, are real number Here, different values of a,b and c satisfy a +2b+c=0. Hence, infinitely many linear equations in x and y can be satisfied by x=1 and y=2. |
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| 83. |
The graph of `y=5` is a lineA. making an intercept 5 on the x-axisB. making an intercept 5 on the y- axisC. parallel to the x-axis at a distance of 5 units from the originD. parallel to the y-axis at a distance fo 5 units from the origin |
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Answer» Correct Answer - C The graph of y=5 is a line parallel to the x-axis at a distance of 5 units from the origin. |
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| 84. |
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) `3x+5y=7.5` (ii) `2x-y/5+6=0` (iii) `3y - 2x =6` (iv) `4x=5y`(v) `x/5-y/6=1` (vi) `sqrt(2x) + sqrt(3y) = 5` |
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Answer» (i)`6x+10y-15=0,(a=6,b=10,c=-15)` (ii) `10x-y+30=0,(a=10,b=-1,c=30)` (iii) `-2x+ 3y - 6 = 0,(a=-2,b=3,c=-6)` (iv) `4x-5y = 0,(a=4,b=-5,c=0)` (v) `6x-5y-30=0,(a=6,b=-5,c=-30)` (vi) `sqrt(2x)+sqrt(3y)-5=0,(a=sqrt(2),b=sqrt(3),c=-5)` |
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| 85. |
The graph of the linear equation y=x passes through the pointA. `(3//2,-3//2)`B. `(0,3//2)`C. (1,1)D. `(-(1)/(2),(1)/(2))` |
| Answer» The linear equation y=x has same value of x and y-coordinates are same. Therefore, the point (1,1) must lie on the line y=x. | |
| 86. |
The graph of the linear equation y=x passes through the pointA. `((-1)/2,1/2)`B. `(3/2,(-3)/2)`C. `(0,-1)`D. (1,1) |
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Answer» Correct Answer - D The graph of the line `x-y=0`, passes through the point (1,1). |
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| 87. |
The graph of the line `y=-3` does not pass through the pointA. (2,-3)B. (3,-3)C. (0,-3)D. (-3,2) |
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Answer» Correct Answer - D The line `y=-3` does not pass through the point(-3,2). |
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| 88. |
The graph of the line y=3 passes through the pointA. (3,0)B. (3,2)C. (2,3)D. none of these |
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Answer» Correct Answer - C The graph of the line y=3 passes through the point (5,3). |
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| 89. |
Solve for x and y : `2^(y-x) (x + y) = 1` and `(x + y)^(x - y) = 2`. |
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Answer» We have, `2^(y-x) (x + y) = 1 implies x + y = (1)/(2^(y-x))` implies `x + y = 2^(x-y) " ...(1) and " (x + y)^(x-y) = 2 " "....(2)` `implies (2^(x-y))^(x-y) = 2` ` implies 2^((x-y)^(2)) = 2^(1)` `implies x - y = pm 1` `{:("If x -y = 1 then x + y = 2"),("Solving these two, we get"),(x = (3)/(2). y = (1)/(2)):}:|{:("If x - y = - 1 then (x + y)"^(-1)=2),("So, x - y = - 1"),("and x + y ="(1)/(2)),("Solving these two, we get "),(x = - (1)/(4). y = (3)/(4)):}` `therefore` `{:(x = (3)/(2)),(y = (1)/(2)):}}` or `{:(x = - (1)/(4)),(y = (3)/(4)):}}` |
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| 90. |
Solve x + y = 7 and 3x - 2y = 11. |
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Answer» Given x + y = 7 …(1) and 3x - 2y = 11 …(2) Form equation (1) we get y = 7 - x …(3) Substituting value of y from equation (3) in (2), we get 3x - 2 (7 - x) = 11 implies 3x - 14 + 2x = 11 implies 5x - 14 = 11 implies 5x = 25 or x = 5 Substituting value of x in equation (3), we get y = 7 - 5 implies y = 2 `therefore` Solution is `{:(x = 5),(y = 2):}}`. |
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| 91. |
Solve 3x - 4y = 20 and x + 2y = 5. |
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Answer» The given equations are 3x - 4y = 20 …(1) and x + 2y = 5 …(2) Multiplying equation (1) by 1 and (2) by 2, we get 3x - 4y = 20 ….(3) 2x + 4y = 10 ….(4) Adding these equations, we get 5x = 30 implies x = 6 Substituting x = 6 in equation (1), we get `3 xx 6 - 4y = 20` implies 18 - 4y = 20 implies - 4y = 2 implies `y = - (1)/(2)` Hence, the solution is `{:(x = 6),(y = -(1)/2):}}`. |
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| 92. |
Solve 15x - 8y = 29 and 17x + 12y = 75. |
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Answer» Given equations are 15x - 8y = 29 …(1) and 17x + 12y = 75 …(2) From equation (1), we get `x = (29 + 8y)/(15) " "...(3)` Substituting value of x from equation (3) in (2), we get `17 ((29 + 8y)/(15)) + 21y = 75` implies 493 + 136y + 180y = 1125 implies 316y = 1125 - 493 implies 316y = 632 implies y = 2 Substituting value of y in equation (3) we get `x = (29 + 8 xx 2)/(15)` `x = (45)/(15)` implies x = 3 Hence, the solution is `{:(x = 3),(y = 2):}}` |
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| 93. |
Solve the following equations for `(x - 3)^(2)` and `(y + 2)^(2)` : `2x^(2) + y^(2) - 12x + 4y + 16 = 0` and `3x^(2) - 2y^(2) - 18x - 8y + 3 = 0`. |
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Answer» When it is said that solve for x and y then we have to find such values of x and y which satisfy both the equations. Similarly if we are asked that solve for `(x - 3)^(2)` and `(y - 2)^(2)`, then we have to find the value of `(x - 3)^(2)` and `(y + 2)^(2)` which satisfy the above two equations. First take the first equation as `2(x^(2) - 6x) + (y^(2) + 4y) = - 16` implies `2(x^(2)-6x+9) + (y^(2) + 4y + 4) = - 16 + 18 + 4` implies `2(x - 3)^(2) + (y + 2)^(2) = 6 " "....(1)` Now, take the second equation as `3(x^(2) - 6x) - 2 (y^(2) + 4y) = - 3` `implies 3(x^(2) - 6x + 9) - 2 (y^(2) + 4y + 4) = - 3 + 27 - 8` `implies 3(x - 3)^(2) - 2 (y-2)^(2) = 16" ....(2)"` Let `(x - 3)^(2)` = u and `(y + 2)^(2) = v`. So, equations (1) and (2) become 2u + v = 6 ....(3) and 3u - 2v = 16 ....(4) Multiplying equation (3) by 2, we get 4u + 2v = 12 ....(5) Adding equations (5) and (4), we get 4u + 2v = 12 `(3u - 2v = 16 )/(7u = 28)` `therefore u = 4 implies (x - 3)^(2) = 4` Putting u = 4 in equation (3), we get 2(4) + v = 6 implies v = - 2 implies `(y + 2)^(2) = - 2` But square of any real number cannot be negative. So, `(x - 3)^(2) = 4` and `(y + 2)^(2)` does not exist. |
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| 94. |
Solve for ` x and y ` using substitution method : ` ( 3x )/( 2) - ( 5y )/( 3) = - 2, ( x)/(2) + ( y)/(2) = (13)/(6)` |
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Answer» The given system of equations may be written as ` 9x - 10 y + 12 = 0 " " ` … (i) ` 2x + 3y - 13 = 0 " " `… (ii) From (ii) we get `y = (13 - 2x )/( 3 )` Substituting `y = ( 13 - 2x )/(3 )` in (i), we get `9x - (10 (13 - 2x ))/(3) + 12 = 0 ` `rArr 27x - 10 ( 13 - 2x ) + 36 = 0 ` ` rArr 27 x - 130 + 20 x + 36 = 0 ` `rArr 47 x - 94 = 0 rArr 47 x = 94 rArr x = (94)/( 47 ) = 2`. Substituting ` x = 2 `in (i), we get ` 9 xx 2 - 10 y + 12 = 0 rArr 10 y = 30 rArr y = (30)/( 10 ) = 3.` Hence, ` x = 2 and y = 3 ` is the required solution. |
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| 95. |
Solve for x and y, using substitution method : ` 2x + y = 7, 4x - 3y + 1 =0 ` |
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Answer» The given system of equation is ` 2x + y = 7" " `… (i) `4x - 3y = - 1 " " `… (ii) From (i), we get `y = ( 7 - 2x )` Substituting `y = ( 7 - 2x ) ` in (ii), we get ` 4x- 3 ( 7 - 2x ) = -1 ` `rArr 4x - 21 + 6x = -1 ` `rArr 10 x = 20` ` rArr x = 2 ` Substituting `x = 2 ` in (i), we get ` 2xx 2 + y = 7 rArr y = 7 - 4 = 3`. Hence, the solution is ` x = 2, y = 3 ` . |
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| 96. |
Show that ` x = 3 , y = 2 ` is not a solution of the system of linear equations ` 3x - 2y = 5, 2 x + y = 7` |
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Answer» The given equations are ` 3x - 2y = 5 " " ` … (i) ` 2x + y = 7 " " `… (ii) Putting ` x = 3 and y = 2 ` in (i), we get LHS = ` ( 3 xx 3 - 2 xx 2 ) = 5 = ` RHS Putting ` x = 3 and y = 2 ` in (ii), we get LHS =` ( 2xx 3 + 2) = 8 ne ` RHS. Thus, the values ` x = 3, y = 2 ` do not saisfy (ii) Hence, ` x = 3, y = 2 ` is not a solution of the given system of equations. |
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| 97. |
Show tha ` x = 5, y = 2 ` is a solution of the system of linear equation ` 2x + 3y = 16, x - 2y = 1 ` |
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Answer» The given equations are ` 2x + 3y = 16" " `… (i) ` x -2y = 1 " "` … (ii) Putting ` x = 5 and y = 2 ` in (i), we get LHS `= ( 2xx 5+ 3 xx 2 ) = 16 ` = RHS. Putting ` x = 5 and y = 2 ` in (ii), we get LHS `= ( 5 - 2 xx 2 ) = 1 = ` RHS Thus, ` x = 5 and y = 2 ` satisfy both (i) and (ii). Hence, ` x= 5, y = 2 `is a solution of the given system equations. |
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| 98. |
Show that ` x = 3 and y = 2 ` is a solution of ` 5x - 3y = 9` |
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Answer» Substituting ` x = 3 and y = 2 ` in the given equation, we get LHS = ` 5xx 3 - 3 xx 2 = (15 - 6) = 9` RHS ` therefore x = 3 and y = 2 ` is a solution of ` 5x - 3y = 9` |
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| 99. |
Two years ago, a father was five times as old as his son. Two yearslater, his age will be 8 more than three times the age of the son. Find thepresent ages of father and son. |
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Answer» Correct Answer - 42 years, 10 years Let the present ages of the man and his son be x years and y years respectively. Then, `(x - 2) = 5 ( y - 2 ) rArr x - 5y = - 8 " " `… (i) ` and (x + 2) = 3 ( y + 2) + 8 rArr x - 3y = 12" "`... (ii) |
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A lending library has afixed charge for the first three days and an additional charge for each daythereafter. Saritha paid Rs 27 for a book kept for seven days, while Susypaid Rs 21 for the book she kept for five days. Find the fixed charge and thecharge for each extra day. |
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Answer» Correct Answer - ₹ 15, ₹ 3 per day Let the fixed charge be ₹ x and charge for each extra day be ₹ y . Then, ` x + ( 7- 3)y = 27 " " `… (i) and, ` x + ( 5 -3 ) y = 21 rArr x + 2y = 21" " `… (ii) |
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