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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.
| 201. |
23 spoons and 17 forks together cost ₹ 1770, while 17 spoons and 23 forks together cost ₹ 1830. Find the cost of a spoon and that of a fork. |
| Answer» Correct Answer - ₹ 40 , ₹ 50 | |
| 202. |
Solve graphicallyfollowing system of linear equations. Also find the coordinates of the pointswhere the lines meet axis of `y`.`2x-5y+4=0, 2x+y-8=0` |
| Answer» Correct Answer - `(x = 3, y = 2) ; A (3, 2 ), B(0, 0.8), C(0, 8) ; ar ( Delta ABC) = 10.8` sq units | |
| 203. |
For what value(s) of k, the pair of linear equations -2x + 5y = 0 and kx + 3y = 0 has a non-zero solutions? |
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Answer» The system of equations has a non-zero solution, if `(a_(1))/(a_(2)) = (b_(1))/(b_(2))` implies `(-2)/(k) = (5)/(3) implies k = - (6)/(5)` |
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| 204. |
If ` 12 x + 17 y = 53 and 17 x + 12 y = 63 ` then find the values of ` (x + y )` |
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Answer» Adding the given equations, we get ` 29(x + y ) = 116 rArr x + y = (116)/(29) = 4`. |
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| 205. |
Write the value of k for which the system of equations ` 3 x + ky = 0 , 2x - y = 0 ` has a unique solution. |
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Answer» For unique solution, we have ` (a_1)/(a_2) ne (b_1)/(b_2)`. ` therefore (3)/(2) ne ( k )/(-1) rArr k ne (-3)/(2)`. |
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| 206. |
For what value of k does the system of equations `x+2y=3` and `5x + ky + 7 = 0` have no solutionA. ` k = 10 `B. ` k ne 10 `C. ` k = (-7)/(3)`D. ` k = - 21 ` |
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Answer» Correct Answer - A For no solution, we have ` (a_1)/(a_2) = (b_1)/(b_2) ne (c_1)/(c_2)`. ` therefore (1)/(5) = (2)/(k) ne (-3)/(7) rArr k = 10 and k ne (-14)/(3)`. Hence, ` k = 10 `. |
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| 207. |
The value of `k`for which the system ofequations `2x+3y=5, 4x+k y=10`has infinite number ofsolutions, is(a) 1 (b) 3 (c) 6 (d) 0 |
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Answer» For an infinite number of solutions, we have ` (2)/(4) = ( 3)/(k) = (-5)/(10) rArr k = 6`. |
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| 208. |
Find the value of k for which the system of equations ` 3x + 5y = 0 , k x + 10 y = 0 ` has a nonzero solution. |
| Answer» The given system has a nonzero solution when ` ( 3)/(k ) = ( 5)/(10) rArr k = 6`. | |
| 209. |
Show that the system of equations ` 2x + 5y = 17, 5x + 3y = 14` has a unique solution. Find the solution. |
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Answer» The given system of equation is ` 2x + 5y - 17 = 0 , 5x + 3y - 14 = 0 ` These equations are of the form `a_1 x + b_1 y + c _1= 0, a _2 x + b_2 y + c_2 =0 `, where ` a_ 1 = 2, b_ 1 = 5, c_ 1 = - 17 and a_ 2 = 5, b_2 = 3, c_ 2 = -14` ` therefore (a _1 )/(a_ 2) = ( 2)/(5), (b_ 1 )/(b_ 2 ) = ( 5) /(3), (c _ 1 )/(c _ 2 ) = (-17)/(-14) = ( 17)/(14)` Thus, ` (a_ 1 ) /(a _ 2 ) ne ( b_ 1 )/( b_ 2 ) ` Hence, the given system of equations has a unique solution. By cross multiplication, we have ` (##RSA_MATH_X_C03_SLV_042_S01.png" width="80%"> ` therefore (x) /(( - 70 + 51)) = ( y ) /(( - 85 + 28)) = (1)/(( 6- 25 ))` ` rArr (x)/(-19) = ( y ) /( -57) = (1 ) /(-19) ` ` rArr x = (-19)/(-19) = 1 and y = (-57)/(-19) = 3` . Hence, x= 1 and y = 3 is the required solution. |
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| 210. |
(i) Solve 8x - 3y = 5xy and 6x - 5y = - 2xy. How many solutions, this system of equations has? (ii) Solve for x and y, by reducing the following equations in a pair of linear equations : 2x + 3y = 5xy and 3x - 2y = xy. |
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Answer» (i) Given equations are 8x - 3y = 5xy and 6x - 5y = -2xy The given equations are not linear in the variable x and y. These can be reduced into linear equations. If we put x = 0 in either of the equations, we get y = 0. Hence, x = 0 and y = 0 is a solution of these equations. To find other solution, dividing each of th given equations by xy. So, `(8)/(y) - (3)/(x) = 5` .... (1) and `(6)/(y) - (5)/(x) = -2` ....(2) Let `(1)/(y)` = a and `(1)/(x)` = b then from equations (1) and (2), we get `{:(8a - 3b = 5),(6a - 5b =-2):}}"linear equations "{:(....(3)),( ....(4)):}` Multiplying equation (3) by 5 and (4) by 3, we get 40a - 15b = 25 ....(5) 18a - 15b = - 6 ....(6) Subtracting equation (6) from (5), we get 22a = 31 implies `a = (31)/(22)` Putting `a = (31)/(22)` in equation (3), we get `8 xx (31)/(22) - 3b = 5` `(124)/(11) - 3b = 5 implies -3b = 5 - (124)/(11)` implies `-3b = - (69)/(11) implies b = (23)/(11)` Now, `(1)/(y) = a implies (1)/(y) = (31)/(22) implies y = (22)/(31)` and `(1)/(x) = b implies (1)/(x) = (23)/(11) implies x = (11)/(23)` Hence, the solution of the given two equations are `{:(x = 0),(y = 0):}}` and `{:(x = (11)/(23)),(y = (22)/(31)):}}`. So, the given system of equations has two solutions. (ii) We have, 2x + 3y = 5xy .... (1) 3x - 2y = xy .... (2) Dividing equations (1) and (2) by xy, we get `(2x)/(xy) + (3y)/(xy) = (5xy)/(xy) implies (2)/(y) + (3)/(x) = 5 " "...(3)` and `(3x)/(xy) - (2y)/(xy) = (xy)/(xy) implies (3)/(y) - (2)/(x) = 1 " "....(4)` Let `(1)/(x) = u` and `(1)/(y) = v`. `therefore` Equations (3) and (4) become, 2v + 3u = 5 ....(5) 3v - 2u = 1 ....(6) Now this is a pair of linear equations,multiplying equation (5) by 2 and equation (6) by 3, we get 4v + 6u = 10 ....(7) 9v - 6u = 3 ....(8) On adding equations (7) and (8), we get 13v = 13 implies v = 1 Putting v = 1 in equation (5), we get 2(1) + 3u = 5 implies 3u = 3 implies u = 1 Now, u = 1 implies `(1)/(x) = 1` implies x = 1 and v = 1 implies `(1)/(y) = 1` implies y = 1 Hence, required solution is `{:(x = 1),(y = 1):}}`. |
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| 211. |
For what value of k does the system of equations ` " " x + 2y = 3, 5x + ky + 7 = 0` have (i) a unique solution, (ii) no solution ? Also, show that there is no value of k for which the given system equations has infinitely many solutions. |
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Answer» Correct Answer - (i) `k ne 10 ` (ii) `k = 10` (i) ` (1)/(5) ne (2)/(k) rArr k ne 10 ` (ii)` (1)/(5) = (2)/(k) ne (-3)/(7) rArr k = 10 ` Clearly, ` (1)/(5) = (2)/(k) = (-3)/(7) ` is never true, as ` (1)/(5) = ( -3)/(7) ` is false. |
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| 212. |
Solve : ` 6x + 3y = 7xy and 3x + 9 y = 11 xy `. |
| Answer» Correct Answer - ` x = 1, y = (3)/(2)` | |
| 213. |
Find the values(s) of `k`for which the system of equations`k x-y=2``6x-2y=3`has (i) a unique solution (ii) no solution.Is there a value of `k`for which the system has infinitely many solutions? |
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Answer» The given system of equations is ` k x - y - 2 = 0, 6x - 2y - 3 = 0 `. This is of the form `a_ 1 x + b_ 1 y + c_1 = 0 and a_ 2 x + b_ 2 y + c_2 = 0 ` , where ` a_ 1 = k, b_ 1 = - 1, c _ 1 = - 2 and a_ 2 = 6, b_ 2 = - 2, c_ 2 = - 3.` (i) For a unique solution, we must have ` (a_ 1 ) /(a_ 2) ne(b_ 1 ) /( b_ 2 )` ` therefore (k ) /( 6) ne (-1) /( - 2) rArr ( k ) /( 6) ne (1)/(2) rArr k ne 3 `. Hence, the given system of equations will have a unique solution when ` k ne 3 `. (ii) For no solution, we must have ` (a_ 1 ) /(a_ 2 ) = ( b_ 1 ) /(b _ 2 ) ne (c_ 1 )/(c _ 2 )` ` therefore ( k ) /( 6) = (-1)/(-2) ne (-2)/(-3) ` ` rArr (k)/( 6) = (1)/(2) ne ( 2 ) /(3 )` ` rArr ( k ) /( 6) = (1)/(2) and (k ) /( 6)ne ( 2) /(3) rArr k = 3 and k =ne 4 `. Clearly, k = 3 also satsfies the condition `k ne 4` Hence, the given system of equations will have no solution when k = 3. (iii) For infinitely many solutions, we must have `(a_ 1)/(a_ 2 ) = (b_ 1 ) /(b_ 2 ) = (c_ 1 ) /(c_ 2 ) , ` i.e., ` ( k ) /( 6) = (1)/(2) = (1)/(3), ` which is never possible, as `(1)/(2) ne (1)/(3)` . Hence, there is no real value of k for which the given system of equations has infinitely many solutions. |
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| 214. |
Find the value of `k`for which the followingsystem of equations has a unique solution:`4x+k y+8=0, 2x+2y+2=0` |
| Answer» Correct Answer - `k ne 4` | |
| 215. |
Find the value of `k`for which the system`k x+2y=5``3x+y=1`has (i) a unique solution, and (ii) no solution. |
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Answer» Correct Answer - (i) ` k ne 6` (ii) ` k = 6` |
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| 216. |
For what value of `k`, will the system ofequations `x+2y=5, 3x+k y-15=0`has (i) a uniquesolution? (ii) no solution |
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Answer» The given system of equations is x + 2y - 5 = 0 ….(1) 3x + ky - 15 = 0 ….(2) Here, `a_(1) = 1, b_(1) = 2, c_(1) = - 5` [from (1)] `a_(2) = 3, b_(2) = k, c_(2) = - 15` [from (2)] If the equations have no solution then `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) ne (c_(1))/(c_(2))` implies `(1)/(3) = (2)/(k) ne (-5)/(-15)` implies `(1)/(3) = (2)/(k)` and `(2)/(k) ne (-5)/(-15)` implies k = 6 and `k ne 6` which is impossible. Hence, there is no such value of k for which the given system has no solution. |
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| 217. |
The system ` kx - y = 2 and 6x - 2y = 3` has a unique solution only whenA. ` k = 0 `B. ` k ne 0 `C. ` k =3 `D. ` k ne 3 ` |
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Answer» Correct Answer - D For a unique solution, we have ` (a_1)/(a_2) ne (b_1)/(b_2 ) ` . ` therefore (k)/(6) ne (-1)/(2) rArr k ne 3`. |
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| 218. |
The graphs of the equations `2x +3y-2=0` and `x-2y-8=0` lines which are (a) coincident (b) parallel (c) intersecting exactly at one point (d) perpendicular to each otherA. coincidentB. parallelC. intersecting exactly at one pointD. perpendicular to each other |
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Answer» Correct Answer - C `(a_1) /(a_2) = (2)/(1), (b_1)/(b_2) = (3)/(-2)`. So, `(a_1)/(a_2) ne (b_1)/(b_2)`. Thus, the system has a unique solution and therefore the lines intersect exactly at one point. |
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| 219. |
For what value of k does the system of equations `" " kx + 2y = 5, 3x - 4y = 10 ` have (i) a unique solution, (ii)no solution ? |
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Answer» Correct Answer - (i)` k ne (-3)/(2)` (ii) ` k = (-3)/(2)` (i) ` (k)/(3) ne ( 2)/(-4) rArr (k ) /(3) ne (-1)/(2) rArr k ne (-3)/(2)` (ii) `(k)/(3) = ( 2)/(-4) ne (5)/(10) rArr (k)/(3) = (-1)/(2) and (k)/(3) ne 1/2 rArr k = (-3)/2` |
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| 220. |
For what value of k do the equations `kx-2y = 3` and `3x + y = 5` represent two lines intersecting at a unique point?A. `k = 3 `B. `k = -3 `C. `k = 6`D. all real values except ` - 6` |
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Answer» Correct Answer - D For a unique intersecting point, we have ` (a_1)/(a_2) ne (b_1)/(b_2)` ` therefore (k)/(3) ne (-2)/(1) rArr k ne - 6 `. Hence, correct answer is (d). |
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| 221. |
Solve the followingsystem of equations by method of cross-multiplication: `2x+y=35 , 3x+4y=65` |
| Answer» Correct Answer - ` x = 15 , y = 5` | |
| 222. |
The cost of 5 pens and 8 pencils is ₹ 120, while the cost of 8 pens and 5 pencils is ₹ 153. Find the cost of 1 pen and that of 1 pencil. |
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Answer» Let the cost of 1 pen be ₹ x and that of 1 pencil be ₹ y. Then, ` 5x + 8y = 120 " " `… (i) and ` 8x + 5y = 153" " `… (ii) Adding (i) and (ii) , we get ` 3(x - y ) = 273 rArr x + y = 21" " `... (iii) Subtracting (i) from (ii), we get ` 3 ( x - y ) = 33 rArr x - y = 11" " `... (iv) From (iii) and (iv), we get ` x = 16 and y = 5`. |
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