InterviewSolution
Saved Bookmarks
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.
| 151. |
In a cyclic quadrilateral ABCD, it is being given that ` angle A = (x + y + 10)^(@), angle B = ( y + 20 ) ^(@), angle C = (x + y - 30 ) ^(@) and angle D = (x + y ) ^(@)`. Then, ` angle B ` ?A. `70^(@)`B. `80^(@)`C. `100^(@)`D. `110^(@)` |
|
Answer» Correct Answer - B ` angle A + angle C = 180^(@) and angle B + angle D = 180^(@)` The given ` x + y = 100 and x + 2y = 160`. |
|
| 152. |
In a ` Delta ABC, angle C = 3, angle B = 2 (angle A + angle B)`, then ` angle B ` = ? |
|
Answer» Let ` angle A= x ^(@) and angle B = y^(@)`. Then, ` angle C = 3 angle B = ( 3y) ^(@)` Now, ` angle C = 2 (angle A + angle B )` ` rArr 3y = 2 (x + y ) rArr 2x - y = 0 " " `… (i) We know that the sum of the angles of a triangle is ` 180^(@)` ` therefore angle A + angle B + angle C = 180 ^(@) rArr x + y + 3y = 180` ` " " rArr x + 4y = 180" " `... (ii) On multiplying (i) by 4 and adding the result with (ii), we get ` 8x + x = 180 rArr x = 20`. Putting ` x = 20 ` in ( i ), we get `y = ( 2 xx 20 ) = 40 ` Thus, ` x = 20 and y = 40 ` ` therefore " " angle A = 20 ^(@), angle B = 40 ^(@) and angle C = ( 3xx 40 ^(@)) = 120 ^(@)` |
|
| 153. |
For what value of k will the following pair of linear equations have no solution ? `" " 2x + 3y - 4 = 0 and 2x + 6y - 7 = 0` |
|
Answer» For no solution, we have ` (a_1)/(a_2) = (b_1)/(b_2) ne (c_1)/(c_2)`. ` therefore (2)/(6) = (3)/(k - 2 ) and (2)/(6) ne = (9)/(3k - 2) rArr ( 3)/(k - 2 ) = (1)/(3) and (9)/(3k - 2) ne (1)/(3)` ` rArr k - 2 = 9 and 3k - 2 ne 27 rArr k = 11 and k ne (29)/(3)`. Hence, ` k = 11 ` |
|
| 154. |
If a pair of linear equations is inconsistent then their graph lines will beA. parallelB. always coincidentC. always intersectingD. intersecting or coincident |
| Answer» Correct Answer - A | |
| 155. |
The pair of equations `x + 2y + 5 = 0` and ` -3x - 6y + 1 = 0` hasA. a unique solutionB. exactly two solutionsC. infinitely many solutionsD. no solution |
|
Answer» Correct Answer - D Here, ` (a_1)/(a_2) = (2)/(4)= (1)/(2), (b_1)/(b_2) = (3)/(6) = (1)/(2) and (c_1)/(c_2) = (-5)/(-15) = (1)/(3)`. ` therefore (a_1)/(a_2) = (b_1)/(b_2) ne (c_1)/(c_2)`. So, the given system has no solution. |
|
| 156. |
In a ` Delta ABC, angle C = 3 angle B = 2 (angle A + angle B)`, then ` angle B ` = ?A. `20^(@)`B. `40 ^(@)`C. ` 60 ^(@)`D. `80^(@)` |
|
Answer» Correct Answer - B Let `C = 3B = 2(A+B) = x ^(@)` Then, `C = x ^(@), B = ((x)/(3))^(@) and (A + B) = ((x)/(2)) ^(@)`. `(A + B) +C = 180^(@) rArr (x)/(2) + x = 180 rArr 3x = 360 rArr x = 120 `. ` therefore angle B = ((120)/(3))^(@) = 40 ^(@)` |
|
| 157. |
If a pair of linear equations is consistent then their graph lines will beA. parallelB. always coincidentC. always intersectingD. intersecting or coincident |
|
Answer» Correct Answer - D If a pair of linear equations is consistent then graph lines will be intersecting or coincident. |
|
| 158. |
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 25 days he has to pay 4500 as hostel charges whereas a student who takes B food for 30 days, pays 5200 as hostel charges. Find the fixed charges per month and the cost of food per day. |
|
Answer» Correct Answer - fixed charges = ₹ 1000, cos of food per day = ₹ 140 Let the fixed charges be ₹ x and the charges for food be ₹ y per day. Then, ` x + 25y = 4500 and x + 30 y = 5200`. |
|
| 159. |
On selling a tea-set 5% loss and a lemon-set at 15% gain, a crockeryseller gains Rs. 7. If he sells the tea-set at 5% gain and the lemon-set at10% gain, he gains Rs. 13. Find the actual price of the tea-set and thelemon-set. |
|
Answer» Correct Answer - tea set = ₹ 100, lemon set = ₹ 80 Let the cost price of the tea set be ₹ x and that of the lemon set by ₹ y. Then, ` ((95x)/(100) + (115y )/(100)) - ( x + y ) = 7 rArr ((19x + 23 y ))/(20) - (x + y ) = 7` ` " " rArr 3y - x = 140 " " `... (i) And , ` ((105x )/(100) + (110y)/(100)) - ( x + y ) = 13 rArr ((21 x + 22y ))/(20 ) - (x + y ) = 13` ` rArr x + 2y = 260 " " `... (ii) Now, solve (i) and (ii). |
|
| 160. |
If a linear equation has solutions (-2,2)(0,0) and (2,-2), then it is of the formA. y-x=0B. x+y=0C. `-2x+y=0`D. `-x+2y=0` |
|
Answer» Let us consider a linear equation ax+by+c=0 Since, `(-2),2)`, (0,0) and (2,-2) are the solutions of linear equation therefore it satisfies the Eq. (i), we get At point `(-2),2), " " -2a+2b+c=0` At point `(0,0), " " 0+0+c=0 implies c=0` and at point `(2,-2), " " 2a-2b+c=0` From Eqs. (ii) and (iii), c=0 and `-2a+2b+0=0, -2a=-2b,a=(2b)/(2) implies a=b` On putting a=b and c=0 in Eq. (i), `bx+by+0=0 implies bx+by=0` `implies " " b(x+y) =0 implies x+y=0, b ne 0` Hence, x+y=0 is the required form of the linear equation. |
|
| 161. |
The equation of X-axis is of the formA. x=0B. y=0C. x+y=0D. x=y |
| Answer» The equation of X-axis is of the form y=0. | |
| 162. |
Any point on the line y=x is of the formA. (a,a)B. (0,a)C. (a,0)D. (a,-a) |
|
Answer» Every point on the line y=x has same value of x and y-coordinates i.e., x=a and y-a. Hence, (a,a) is the required form of the solution of given linear equation. |
|
| 163. |
Any point on the X-axis is of the formA. (x,y)B. (0,y)C. (x,0)D. (x,x) |
|
Answer» Every point on the X-axis has its y-coordinate equal to zero. i.e., ` " " ` y=0 Hence, the general form of every point on X-axis is (x,0). |
|
| 164. |
Any solution of the linear equation 2x+0y+9=0 in the two variables is of the formA. `(-(9)/(2),m)`B. `(n,-(9)/(2))`C. `(0,-(9)/(2))`D. `(-9,0)` |
|
Answer» The given linear equation is 2x+0y+9=0 `implies " " 2x+9=0` `implies " " 2x=-9` `therefore " " x=-(9)/(2)` and y can be any real number. Hence, `(-(9)/(2),m)` is the required form of the solution of given linear equation. |
|
| 165. |
In a competitive examination, 1 mark is awarded for each correct answer while 1/2 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly? |
|
Answer» Let the number of questions attempted correctly = x Number of questions answered = 120 So, wrong answer attempted = 120 - x Marks awarded for right answer = 1 `xx` x = x marks Marks deducted for (120 - x) wrong answer = `(1)/(2)` (120 - x) To marks awarded = 90 `therefore " "x- (1)/(2) (120 - x) = 90 " implies "x - 60 + (x)/(2) = 90` `implies " "x + (x)/(2) = 90 + 60 " "implies " "(3x)/(2) = 150` `implies " "x = (150 xx 2)/(3) " "implies" "x = 100` Hence, Jayanti answered 100 questions correctly. |
|
| 166. |
`x/a-y/b=0 `, `a x+b y=a^2+b^2` |
|
Answer» Correct Answer - ` x = a , y = b ` ` bx - ay = 0 " " ` … (i) and `a x + by = (a ^(2) + b ^(2))" " `… (ii) Multiply (i) by b and (ii) by a and add. |
|
| 167. |
`{:(ax - by = a^(2) + b^(2)),(x + y = 2a):}` |
|
Answer» Correct Answer - ` x = a + b, y = a - b ` Multiply (ii) by b and add the result with (i). |
|
| 168. |
`6 (ax + by) = 3a + 2b` & `6(bx-ay) = 3b-2a` |
|
Answer» Correct Answer - ` x = (1)/(2), y = (1)/(3)` ` 6ax + 6 by = 3a + 2 b " " `… (i) and ` 6 bx - 6 ay = 3 b - 2a " " `… (ii) Multiply (i) by a and (ii) by b and add. |
|
| 169. |
If (x+1) is a factor of `2x^(3)+ax^(2)+2bx+1`, then find the value of a and b given that 2a-3b=4. |
|
Answer» Let f(x) = `2x^(3) + ax^(2) + 2bx + 1` `because (x + 1)` is a factor of f(x). `therefore f(-1) = 0` implies `2(-1)^(3) + a(-1)^(2) + 2b(-1) + 1 = 0` implies -2 + a - 2b + 1 = 0 implies a - 2b = 1 implies a = 1 + 2b …(1) Given that 2a - 3b = 4 implies 2(1 + 2b) - 3b = 4 [from (1)] implies 2 + 4b - 3b = 4 implies b = 2 From equation (1) a = 2b + 1 implies a = 2 `xx` 2 + 1 = 5 `therefore` a = 5, b = 2 |
|
| 170. |
Solve for x and y : ` ( 2)/(sqrt x ) + (3)/(sqrty ) = 2, ( 4)/(sqrtx ) - ( 9)/(sqrty) = - 1 ( x ne 1, y ne 0 )` |
|
Answer» Putting ` (1)/(sqrtx ) = u and (1)/(sqrty) = v `, the given equations become ` 2u + 3v = 2 " "`… (i) ` 4u - 9v = - 1 " " ` … (ii) Multiplying (i) by 3 and adding the result with (ii), we get ` 6u + 4u = 6- 1 ` ` rArr 10 u = 5` ` rArr u = (5)/(10) rArr u = (1)/(2)` Putting `u = (1)/(2)` in (i), we get ` ( 2 xx (1)/(2)) + 3v = 2 ` ` rArr 1 + 3v = 2 rArr 3 v = 1 rArr v = (1)/(3)` Now, `u = (1)/(2) rArr (1)/(sqrtx) = (1)/(2) rArr sqrtx = 2 rArr = 2 rArr x = 4`. And, `v = (1)/(3) rArr (1)/( sqrty ) = (1)/(3) rArr sqrty = 3 rArr y = 9` Hence, ` x = 4 and x = 9`. |
|
| 171. |
Solve the followingsystem of equations:`2/x+3/y=9/(x y), 4/x+9/y=(21)/(x y), `where `x!=0, y!=0` |
|
Answer» Multiplying each equation throughout by xy, we get ` 2y + 3x = 9 " " `… (i) `4y + 9x = 21 " " `… ( ii) Multiplying (i) by 3 and subtracting (ii) from the result, we get ` ( 6- 4 ) y = ( 27 - 21 ) rArr 2y = 6 rArr y = 3 `. Putting ` y = 3 ` in (i), we get ` (2 xx 3) + 3x = 9 rArr 6 + 3x = 9 rArr 3x = 3 rArr x= 1 ` Hence, ` x = 1 and y = 3`. |
|
| 172. |
Solve for ` x and y ` : ` (2)/(x) + (3)/(y ) = 13, (5)/(x) - ( 4)/( y) = - 2 ( x ne 0 and y ne 0 )` |
|
Answer» Putting ` (1)/(x) = u and (1)/(y) = v `, the given equations become ` 2u + 3 v = 13" " `… (i) ` 5u - 4 v = - 2 " " `… (ii) Multiplying (i) by 4 and (ii) by 3 and adding the results , we get ` 8 u + 15u= 52 - 6` ` rArr 23 u = 46 ` ` rArr u = ( 46)/( 23) = 2`. Putting `u = 2 ` in (i), we get ` ( 2 xx 2 ) + 3v = 13 rArr 3 v = 13 - 4 = 9 rArr v = 3`. Now, ` u = 2 rArr (1)/(x) 2x = 1 rArr x = (1)/(2)` And, ` v = 3 rArr (1)/(y) = 3 rArr 3y = 1 rArr y = (1)/(3)`. Hence, `x = (1)/(2) and y = (1)/(3)`. |
|
| 173. |
`ax+by=c` and `mx+ny=d`. If `an ne bm`, then these simultaneous equations haveA. (0, 2)B. (2, 0)C. `(-2, 0)`D. `(0,-2)` |
| Answer» Correct Answer - B | |
| 174. |
Solve for ` x and y` : ` (1)/(2x) - (1)/(y) = - 1, (1)/(x) + (1)/(2y)= 8 ( x ne 0 , y ne 0 )` |
|
Answer» Putting ` (1)/(x) = u nd (1)/(y) = v `, the given equations become. ` (u)/(2) - v = - 1 rArr u - 2v = - 2 " " `… (i) ` u + (u)/(2) = 8 rArr 2u + v = 16 " " `… (ii) Multiplying (ii) by 2 and adding the result with (i), we get ` u + 4u = - 2 + 32 ` ` rArr 5u = 30 ` ` rArr u = ( 30 )/( 5) = 6 ` Putting `u= 6 ` in (i), we get ` 6- 2 = - 2 rArr 2v = 8 rArr v = 4`. Now, `u = 6 rArr (1)/(x) = 6 rArr 6x = 1 rArr x = (1)/(6)` And, ` v = 4 rArr (1)/(y) =4 rArr 4y = 1 rArr y = (1)/(4)` Hence, ` x = (1)/(6) and y = (1)/(4)`. |
|
| 175. |
Solve for x and y : ` (xy )/(x + y ) = ( 6)/(5), (xy)/( y - x) = 6 (x ne 0, y ne 0 and x ne y )`. |
|
Answer» The given equations may be written as ` (x + y )/( xy) = ( 5)/(6) rArr (1)/(y) + ( 1)/(x) = ( 5)/(6)" " `… (i) ` ( y - x )/( xy ) = (1)/(6) rArr (1)/(x) - (1)/(y) = (1)/(6) " " `… (ii) Adding (i) and (ii), we get ` (2)/(x) = (( 5)/(6) + (1)/(6)) = (6)/(6) = 1 rArr x = 2 ` Subtracting (ii) from (i), we get `(2)/(y)= ((5)/(6) - (1)/(6)) = ( 4)/(6) = (2)/(3) rArr y = 3 ` Hence, ` x = 2 and y = 3 `. |
|
| 176. |
Solve for ` x and y ` : ` ( 3a)/(x) - ( 2b )/(y ) + 5 = 0, (a)/(x) + ( 3b)/( y) - 2 = 0 ( x ne 0 , y ne 0 )` |
|
Answer» Putting ` (1)/(x) = u and (1)/(y) = v `, the given equations become ` 3au - 2 bv = - 5" " `… (i) ` au + 3 bv = 2" "`… (ii) Multiplying (ii) by and substracting (i) from the result, we get ` 9bv + 2 bv = 6 + 5` ` rArr 11bv = 11 rArr v = (11)/(11 b ) = (1)/(b)` Putting ` v = (1)/(b) ` in (ii), we get `au + 3= 2 rArr au = - 1 rArr u = (-1)/(a)` Now, `u = (-1)/(a) rArr (1)/(x) = (-1)/(a) rArr - x = a rArr x = - a ` And, ` v = (1)/(b) rArr (1)/(y) = (1)/(b) rArr y = b ` . Hence,` x = - a and y = - b ` |
|
| 177. |
Solve for ` x and y ` : ` (1)/( 7x ) + ( 1) /( 6y) = 3, (1)/( 2x) - (1)/( 3y ) = 5 (x ne 0, y ne 0)` |
|
Answer» Putting `(1)/(x) = u and (1) /( y) = v ` , the given equations become ` (u)/( 7) + (v)/( 6) = 3 rArr 6u + 7v = 126 " " ` … (i) ` (u)/(2 ) - (v)/( 3) = 5 rArr 3u - 2v = 30 " " `… (ii) Multiplying (i) by 2 and (ii) by 7 and adding the results, we get ` 12 u + 21 u = 252 + 210 ` ` rArr 33u = 462 ` ` rArr u = ( 462 )/(33) = 14 ` Putting `u = 14 `in (i), we get ` ( 6xx 14) + 7v = 126` ` rArr 7v = 126 - 84 = 42 rArr v = ( 42 )/(7) = 6` Now, ` u = 14 rArr (1)/(x) = 14 rArr 14 x = 1 rArr x = (1)/( 14)` And ,` v = 6 rArr (1)/(y) = 6 rArr 6y = 1 rArr y = (1)/(6)`. Hence, ` x = (1)/( 14) and y = (1)/( 16 )` |
|
| 178. |
` x + y = 5 xy `, ` 3x + 2y = 13 xy ( x ne 0, y ne 0 )` |
| Answer» Correct Answer - ` x= (1)/(2), y = (1)/(3)` | |
| 179. |
` ( 9)/(x) - ( 4)/( y) = 8`, ` ( 13)/(x) + ( 7)/(y) = 101 ( x ne 0, y ne 0 )` |
| Answer» Correct Answer - ` x = (1)/(4), y = (1)/(7)` | |
| 180. |
Find the value of `k`for which the followingsystem of linear equations has infinite solutions:`x+(k+1)y=5, (k+1)x+9y=8k-1` |
|
Answer» The given system of equations is ` x + ( k + 1 ) y - 5 = 0 " " `… (i) `(k + 1) x + 9y + (1 - 8k ) = 0 " " `… (ii) These equations are 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 = 1 , b_ 1 = ( k + 1 ), c_ 1 = - 5` and ` a_ 2 = ( k + 1 ), b_ 2 = 9, c_ 2 = ( 1 - 8k )`. ` therefore ( a_ 1 ) /( a_ 2) = (1)/(( k +1)) , (b _ 1 ) /( b_ 2 ) = (( k + 1 ) )/(9 ) and (c _ 1 ) /(c _ 2) = (-5)/(( 1- 8k )) = ( 5)/(( 8k - 1 ))` Let the given system of equations have infinitely many solutions. Then, ` (a_ 1 )/(a_ 2 ) = (b_ 1)/(b_ 2 ) = (c_ 1 )/(c_ 2)` ` rArr ( 1)/(( k + 1 )) = ((k + 1 ))/( 9) = ( 5)/(( 8 k - 1 )) ` ` rArr (1)/(( k + 1 )) = ((k + 1 ) )/(9) and (( k +1 ))/( 9 ) = ( 5)/( ( 8 k - 1))` ` rArr ( k + 1 ) ^(2) = 9 and ( k + 1 ) ( 8 k - 1 ) = 45` ` rArr ( k + 1 = 3 or k + 1 = - 3 ) and 8 k ^(2) + 7k - 46 = 0 ` ` rArr ( k = 2 or k = - 4 ) and ( k - 2 ) ( 8k + 23)= 0 ` ` rArr ( k = 2 or k = - 4) and ( k = 2 or k = (-23)/(8)) ` `rArr k = 2`. Hence, the given system of equations will have infinitely many solutions when k = 2. |
|
| 181. |
` ( 5)/(x) + 6y = 13,` ` (3)/(x) + 4y = 7 ( x ne 0 )` |
|
Answer» Correct Answer - ` x = (1)/(5) , y = - 2 ` Multiply (i) by 3 and (ii) by 5 and subtract. |
|
| 182. |
For the followingsystem of equations determine the value of `k`for which the givensystem has infinitely many solutions:`k x+3y=k-3, 12 x+k y=k` |
|
Answer» The given system of equations is ` k x + 3y + ( 3 - k ) = 0 `, ` 12 x + k y - k = 0 `. These equations are 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 = 3, c_ 1 = ( 3 - k ) and a_ 2 = 12, b_2 = k , c _ 2 = - k ` Let the given system of equations have infinitely many solutions. Then, ` (a_ 1 ) /(a _ 2) = (b _ 1 ) /( b _ 2 ) = (c_1 ) /( c_ 2 ) ` ` rArr (k )/( 12) = ( 3) /(k ) = (( 3- k ))/( - k ) ` `rArr (k)/( 12) = ( 3) /(k ) = ( k - 3 ) /( k )` ` rArr ( k ) /(12) = ( 3) /(k ) and ( 3 ) /( k ) = ( k - 3 ) /( k ) ` `rArr k ^(2) = 36 and k ^(2) - 6k = 0 ` ` rArr ( k = 6 or k =- 6) and k (k - 6) = 0 ` ` rArr ( k = 6 or k = - 6) and ( k = 0 or k = 6 )` `rArr k = 6 ` Hence, the given system of equations has infinitely many solutions when ` k = 6 `. |
|
| 183. |
Find the values of k for which the following pair of linear equations has infinitely many solutions : ` 2x - 3y = 7, ( k + 1 ) x + ( 1 - 2k) y = ( 5k - 4)`. |
|
Answer» The given equations are ` 2x - 3y - 7 =0`, ` ( k + 1 ) x + ( 1 - 2k ) y + ( 4 - 5k ) = 0 ` These equations are 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 = 2, b_ 1 = - 3, c_ 1 = - 7 ` and `a_ 2 = ( k + 1 ), b_ 2 = (1 - 2k ), c_ 2 = ( 4- 5k )` Let the given system of equations have infinitely many solutions. Then, `(a_ 1 ) /(a_ 2 ) = (b_ 1 ) /( b_ 2 ) = (c _ 1 ) /(c_2)` `rArr ( 2)/(( k+ 1 )) = (-3)/((1- 2k )) = ( -7)/(( 4- 5k ))` `rArr (2)/(( k + 1 ) ) = (3)/( ( 2k - 1 )) = ( 7)/(( 5k - 4))` ` rArr ( 2)/(( k + 1 )) = ( 3)/(( 2k - 1 )) = ( 7)/(( 5k - 4))` `rArr ( 2)/((k + 1 )) = (3)/(( 2k - 1 )) and (3)/((2k -1 )) = ( 7)/((5 k - 4))` `rArr 4k - 2 = 3k + 3 and 15k - 12 = 14k - 7` ` rArr k = 5 and k = 5` Hence, ` k = 5` |
|
| 184. |
In a `DeltaABC, angleC = 3 angleB = 2(angleA + angleB)` find the three angles. |
|
Answer» Let `angleA = x^(@)` and `angleB = y^(@)` `therefore angleC = 3 angleB implies angleC = 3y^(@)` Also `angleC = 2 (angleA + angleB)` implies 3y = 2 (x + y) implies 2x + 2y - 3y = 0 implies 2x - y = 0 …(1) Since `angleA + angleB + angleC = 180^(@)` (angle sum property of a triangle) implies x + y + 3y = 180 implies x + 4y = 180 ...(2) Multiplying equation (2) by 2, we get 2x + 8y = 360 ...(3) Subtracting equation (3) from (1), we get -9y = - 360 implies `y = 40^(@)` Substituting `y = 40^(@)` in equation (1), we get 2x - 40 = 0 implies 2x = 40 implies `x = 20^(@)` Hence, `angleA = 20^(@), angleB = 40^(@)` and `angle C = 3 xx 40 = 120^(@)` |
|
| 185. |
` 3 x + 2y = 12 , ` ` 5x - 2y = 4`. |
| Answer» Correct Answer - ` x = 2, y = 3 ` | |
| 186. |
If x=3 and y=4 is a solution of the equation `5x-3y=k`, find the value of k. |
| Answer» Correct Answer - `k=3` | |
| 187. |
Find the angles of a cyclic quadrilateral `ABCD` in which `angleA= (4x+20)^@, angleB = (3x - 5)^@, angleC = (4y)^@ and angleD = (7y + 5)^@.` |
| Answer» Correct Answer - `angle A = 120^(@), angle B = 70^(@), angle C = 60^(@), angle D = 110^(@)` | |
| 188. |
Solve the followingsystem of equations graphically:`3x+y+1=0; 2x-3y+8=0` |
| Answer» Correct Answer - ` x = - 1, y = 2 ` | |
| 189. |
Draw the graphs of the equations `x-y=1 a n d 2x+y=8`. Shade the area bounded by these two lines and y-axis. Also, determine this area. |
| Answer» Correct Answer - 13.5 sq units | |
| 190. |
` 2x - ( 3y)/(4) = 3`, ` 5x = 2y + 7`. |
| Answer» Correct Answer - `x = 3, y = 4 ` | |
| 191. |
5 chairs and 4 tables together cost ₹ 5600, while 4 chairs and 3 tables together cost ₹ 4340. Find the cost of a chair and that of a table. |
| Answer» Correct Answer - ₹ 560, ₹ 760 | |
| 192. |
The sum of two numbers is `137` and their difference is `43.` Find the numbers. |
| Answer» Correct Answer - 90, 47 | |
| 193. |
Solve the following system of equations graphically: `" " 2 x - 3y = 1, 4x - 3y + 1 = 0 `. |
| Answer» Correct Answer - ` x = - 1, y = - 1 ` | |
| 194. |
Solve the followingsystem of equations graphically:`2x-3y+6=0, 2x+3y-18=0`Also, find the area ofthe region bounded by these two lines and y-axis. |
| Answer» Correct Answer - `(x= 3, y = 4); A(3, 4 ), B(0, 2 ), C (0, 7) ; ar(Delta ABC) = 7` sq units. | |
| 195. |
Solve the followingsystem of equations graphically:`4x-y=4, 3x+2y=14`. Shade the regionbetween the lines and the y-axis. |
| Answer» Correct Answer - ` (x = 2 , y = 4) ; A(2, 4), B (0, -4 ), C(0, 7) ; ar (Delta ABC) = 6 ` sq units | |
| 196. |
` 5x - y - 7 = 0 , x - y + 1 = 0 ` |
| Answer» Correct Answer - `(x = 2, y = 3) ; A(2, 3), B( 0, - 7), C(0, 1); ar (Delta ABC) = 8 ` sq units | |
| 197. |
` 8x + 5y = 9 , kx + 10 y = 15`. |
| Answer» Correct Answer - ` k = 16` | |
| 198. |
Find two solutions for each of the following equations:(i) `4x+3y=12` (ii) `2x+5y=0` (iii) `3y+4=0` |
|
Answer» 1)4x+3y=12 let x=o then y=4 let y=0 then x=3 2)2x+5y=0 let x=0 then y=0 let x=1 then y=-2/5 3)3y+4=0 let x=0 then y=-4/3 let x=1 then y=-4/3 |
|
| 199. |
A two digit number is such that the product of its digits is 18. When63 is subtracted from the number, the digits interchange their places. Findthe number. |
| Answer» Correct Answer - 92 | |
| 200. |
`{:(2x - 3y = 17),(4x + y = 13):}` |
| Answer» Correct Answer - ` x = 4, y = - 3 ` | |