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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.
| 1. |
The diagonal of a rectangular field 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides the field. |
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Answer» `sqrt(l^2 + b^2) = b+ 60` `l =b + 30` `l^2 + b^2 = b^2 + 3600 + 120b` `l^2 - 120b - 3600 = 0` `l^2 = b^2 + 900 + 60b` `b^2 + 900 + 60b - 120b - 3600 = 0` `b^2 - 60b - 2700 = 0` `b^2- 90b + 30b - 2700 = 0` `b(b-90) + 30(b-90) = 0` `b= 90, -30` b cannot be negative, so`b=90` `l=b+30= 90+30 = 120 m` Answer |
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| 2. |
The roots of `x^(2)+kx+k=0` are real and equal, find k. |
| Answer» Correct Answer - C | |
| 3. |
A peacock is sitting on the top of a pillar, which is 9m high. From a point 27m away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing the snake the peacock pounces on it. If their speeds are equal, at what distance from the whole is the snake caught? |
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Answer» `s = v xx t` if `SM = x then, MP= x` `v_s = v_p` so, `t_s = t_p` `S_s = S_p` in `/_ PMH` `PM^2 = MH^2 = PH^2` `x^2 = (27-x)^2 + 9^2` `x^2 = 729 - 54x + x^2 + 81` `54x = 810` `x= 15 m` Distance from hole `= 27-15 = 12m` answer |
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| 4. |
Find the value of `k`, so that the equation `2x^2+kx-5=0` and `x^2-3x-4=0` may have one root in common. |
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Answer» `x^2-3x-4=0` `=>x^2-4x+x-4=0` `=>x(x-4)+1(x-4) = 0` `=>(x-4)(x+1) = 0` `=> x=4 and x=-1` Now, as `2x^2+kx-5=0`, have one common root with the previous equation.So, values of `x` will satisfy this equation. When `x = 4` `2(4)^2+4k -5 = 0` `32+4k = 5=>k =-27/4` When `x = -1` `2(-1)^2+(-1)k -5 = 0` `2-k=5=> k = -3` So, for `k=-27/4 and k = -3`, both equations have one root in common. |
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| 5. |
Solve each of the following equations by using the method of completing the square: `x^(2)-4x+1=0` |
| Answer» Correct Answer - `x=(2+sqrt(3))" or "x=(2-sqrt(3))` | |
| 6. |
What is the value of k for which the quadratic equation `3x^(2) - kx+ k = 0 ` has equal roots ?A. 3B. 6C. 9D. 12 |
| Answer» Correct Answer - D | |
| 7. |
If one of the roots of the quadratic equation `kx^(2) + 2x-8=0` is -2, then what is the value of k ?A. 2B. 3C. 1D. 4 |
| Answer» Correct Answer - B | |
| 8. |
Solve each of the following equations by using the method of completing the square: `x^(2)+8x-2=0` |
| Answer» Correct Answer - `x=(-4+3sqrt(2))" or "x=(-4-3sqrt(2))` | |
| 9. |
Solve each of the following quadratic equations: `x^(2)=18x-77` |
| Answer» Correct Answer - `x=11" or "x=7` | |
| 10. |
`3((7x+1)/(5x-3))-4((5x-3)/(7x+1))=11 ; x!=3/5,-1/7` |
| Answer» Correct Answer - `x=1" or "x=0` | |
| 11. |
For solving quadratic equation `x^(2) + 8x = - 15` by completing square method, find the third term. |
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Answer» Third terms `(("middle term")^(2))/(4 xx "first terms")= ((8x)^(2))/( 4 xx x^(2))= ( 64x^(2))/(4x^(2))= 16` The required third term is 16. |
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| 12. |
Write the equation `(x-1)^(2) = 2 x + 3 ` in standard form and write the values of a,b and c. |
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Answer» Correct Answer - `x^(2) - 4x-2=0, a=1, a= 1 , b = - 4, c = - 2 ` `( x-1)^(2) = 2x + 3 ` `:. X^(2) - 2x +1 = 2x + 3 ` `:. x^(2) - 4 x - 2 = 0 ` …...(Standard form ) Comparing with `ax^(2) + bx + c = 0 ` `a= 1, b = - 4, c = - 2 ` |
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| 13. |
Find the values of k for which the given quadratic equation has real and distinct roots: (i)` kx^(2)+6x+1=0" "(ii)" "x^(2)-kx+9=0` (iii)` 9x^(2)+3kx+4=0" "(iv)" "5x^(2)-kx+1=0` |
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Answer» Correct Answer - (i)` klt9" "(ii)" "kgt6" or "klt-6" "(iii)" "kgt4" or "klt-4" "(iv)" "kgt2sqrt(5)" or "klt-2sqrt(5)` (i) `36-4kgt0implies36gt4kimplies4klt36impliesklt9.` (ii) `k^(2)-36gt0impliesk^(2)gt36implieskgt6" or "klt-6.` (iii) `9k^(2)-144gt0impliesk^(2)gt16implieskgt4" or "klt-4.` (iv)` k^(2)-20gt0impliesk^(2)gt20implieskgt2sqrt(5)" or "klt-2sqrt(5).` |
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| 14. |
If a=2,b= -11 , c = 15 , find the value of `b^(2)- 4ac`. |
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Answer» `b^(2) - 4ac = ( - 11)^(2) - 4(2) (15) = 121 - 120 = 1 ` The value of `b^(2) - 4ac ` is 1. |
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| 15. |
The roots of the equation `2x^(2)-6x+7=0` areA. real, unequal and rationalB. real, unequal and irrationalC. real and equalD. imaginary |
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Answer» Correct Answer - D `D=(-6)^(2)-4xx2xx7=(36-56)=-20lt0.` So, the roots of the given equation are imaginary. |
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| 16. |
Solved the equation by using quadratic formula `a(x^(2)+1)=(a^(2)+1)x,ane0`. |
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Answer» Given equation is `a(x^(2)+1)=(a^(2)+1)x` `impliesax^(2)+a=a^(2)x+x` `impliesax^(2)-x(a^(2)+1)+a=0` Comparing with `Ax^(2)+Bx+C=0`, we get `A=a,B=-(a^(2)+1)andC=a` `:.x=(-B+-sqrt(B^(2)-4AC))/(2A)` `impliesx=(-[-(a^(2)+1)]+-sqrt({-(a^(2)+1)}^(2)-4xxaxxa))/(2a)` `impliesx=((a^(2)+1)+-sqrt(a^(4)+1+2a^(2)-4a^(2)))/(2a)` `impliesx=((a^(2)+1)+-sqrt((a^(2)-1))^(2))/(2a)` `impliesx=((a^(2)+1)+-(a^(2)-1))/(2a)` `impliesx=(a^(2)+1+a^(2)-1)/(2a)and(a^(2)+1-a^(2)+1)/(2a)` `impliesx=(2a^(2))/(2a)and(2)/(2a)` `impliesx=aand(1)/(a)` are roots of the equation. |
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| 17. |
If the equation `4x^(2)-3kx+1=0` has equal roots then k=?A. `+-(2)/(3)`B. `+-(1)/(3)`C. `+-(3)/(4)`D. `+-(4)/(3)` |
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Answer» Correct Answer - D Since the roots are equal, we hae D=0. `:." "9k^(2)-16=0impliesk^(2)=(16)/(9)impliesk=(4)/(3)" or "k=(-4)/(3).` |
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| 18. |
In the equation `ax^(2)+bx+c=0`, it is given that `D=(b^(2)-4ac)gt0.` Then, the roots of the equation areA. real and equalB. real and unequalC. imaginaryD. None of these |
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Answer» Correct Answer - B When `Dgt0,`the roots of the given quadratic equation are real and unequal. |
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| 19. |
The roots of `ax^(2)+bx+c=0, ane0` are real and unequal, if `(b^(2)-4ac)`A. `gt0`B. `=0`C. `lt0`D. None of these |
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Answer» Correct Answer - A The roots of `ax^(2)+bx+c=0,ane0` are real and unequal only when `(b^(2)-4ac)gt0.` |
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| 20. |
Solve each of the following equations by using the method of completing the square: `3x^(2)-x-2=0` |
| Answer» Correct Answer - `x=1" or "x=(-2)/(3)` | |
| 21. |
Find the value(s) of `k` so that, the quadratic equation `x^2 - 4kx + k = 0` has equal roots. |
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Answer» Correct Answer - `k=0" or "k=(1)/(4)` `D=(-4k)^(2)-4l=16k^(2)-4k=4k(4k-1).` For equal roots, we must have D = 0. Now, `D=0implies4k(4k-1)=0impliesk=0" or "4k-1=0impliesk=0" or "k=(1)/(4).` |
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| 22. |
Find the roots of each of the following equations, if they exist, by applying the quadratic formula: `3x^(2)-2sqrt(6)x+2=0` |
| Answer» Correct Answer - `x=sqrt((2)/(3)),sqrt((2)/(3))` | |
| 23. |
Solve each of the following equations by using the method of completing the square: `2x^(2)+5x-3=0` |
| Answer» Correct Answer - `x=(1)/(2)" or "x=-3` | |
| 24. |
Find the value of p for which the roots of the equation px (x — 2) + 6 = 0, are equal. |
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Answer» Correct Answer - p=6 Given equation is `px^(2)-2px+6=0.` This is of the form `ax^(2)+bx+c=0,` where `a=p,b=-2p` and `c=6.` `:." "D=(b^(2)-4ac)=4p^(2)-24p.` For equal roots, we have `D=0.` `:." "4p^(2)-24p=0implies4p(p-6)=0impliesp=6." "[becausepne0]` |
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| 25. |
Solve each of the following equations by using the method of completing the square: `x^(2)-(sqrt(2)+1)x+sqrt(2)=0` |
| Answer» Correct Answer - `x=sqrt(2)" or "x=1` | |
| 26. |
Solve each of the following equations by using the method of completing the square: `5x^(2)-6x-2=0` |
| Answer» Correct Answer - `x=(3+sqrt(19))/(5)" or "x=(3-sqrt(19))/(5)` | |
| 27. |
Solve each of the following equations by using the method of completing the square: `(2)/(x^(2))-(5)/(x)+2=0` |
| Answer» Correct Answer - `x=2" or "x=(1)/(2)` | |
| 28. |
If one root of the quadratic equation `3x^(2)-10x+k=0` is reciprocal of the other, find the value of k. |
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Answer» Correct Answer - k=3 Let the roots of `3x^(2)-10x+k=0` be `alpha` and `(1)/(alpha).` Then, product of roots `=(k)/(3).` `:." "(alphaxx(1)/(alpha))=(k)/(3)implies(k)/(3)=1impliesk=3.` |
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| 29. |
Solve each of the following equations by using the method of completing the square: `sqrt(2)x^(2)-3x-2sqrt(2)=0` |
| Answer» Correct Answer - `x=(-1)/(sqrt(2))" or "x=2sqrt(2)` | |
| 30. |
By using the method of completing the square, show that the equation `4x^(2)+3x+5=0` has no real roots. |
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Answer» We have `4x^(2)+3x+5=0` `implies" "4x^(2)+3x=-5` `implies" "(2x)^(2)+2xx2x xx(3)/(4)+((3)/(4))^(2)=-5+((3)/(4))^(2)` `" "["adding "((3)/(4))^(2)" on both sides"]` `implies" "(2x+(3)/(4))^(2)=(-5+(9)/(16))=((-80+9))/(16)=(-71)/(16)lt0.` But, `(2x+(3)/(4))^(2)` conot be negative for any real value of x. So, there is no real value of x satisfying the given equation. Hence, the given equation has no real roots. |
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| 31. |
For the quadratic equation `x^(2) + 10x - 7 = 0 `, the values of a,b,c areA. `a= -1, b= 10 , c= 7`B. `a= 1 , b = - 10 , c= 7`C. `a= 1 , b = 10 , c = - 7`D. `a = 1 , b = 10 , c = 7` |
| Answer» Correct Answer - C | |
| 32. |
Find the roots of each of the following equations, if they exist, by applying the quadratic formula: `2sqrt(3)x^(2)-5x+sqrt(3)=0` |
| Answer» Correct Answer - `x=((a+b))/(6)" or "x=((a-b))/(6)` | |
| 33. |
Solve the equation `10x-1/x=3` by the method of completing the square. |
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Answer» We have `10x-(1)/(x)=3` `implies" "10x^(2)-1=3x` `implies" "10x^(2)-3x=1` `implies" "100x^(2)-30x=10" "["multiplying each side by 10"]` `implies" "(10x)^(2)-2xx10x xx(3)/(2)+((3)/(2))^(2)=10+((3)/(2))^(2)` `" "["adding "((3)/(2))^(2)" on both sides"]` `implies" "(10x-(3)/(2))^(2)=(10+(9)/(4))=(49)/(4)=((7)/(2))^(2)` `implies" "10x-(3)/(2)=+-(7)/(2)" "["taking square root on both sides"]` `implies" "10x-(3)/(2)=(7)/(2)" or "10x-(3)/(2)=(-7)/(2)` `implies" "10x=((7)/(2)+(3)/(2))=(10)/(2)=5" or "10x=((-7)/(2)+(3)/(2))=(-4)/(2)=-2` `implies" "10x=5" or "10x=-2impliesx(5)/(10)=(1)/(2)" or "x=(-2)/(10)=(-1)/(5).` Hence, `(1)/(2)` and `(-1)/(5)` are the roots of the given equation. |
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| 34. |
Solve the equation `x^(2)-10x-2=0` by the method of completing the square. |
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Answer» We have `x^(2)-10x-2=0` `implies" "x^(2)-10x=0` `implies" "x^(2)-2xx x xx5+5^(2)=2+5^(2)" "["adding "5^(2)" on both sides"]` `implies" "(x-5)^(2)=(2+25)=27` `implies" "x-5=+-sqrt(27)=+-3sqrt(3)" "["taking square root on both sides"]` `implies" "x-5=3sqrt(3)" or "x-5=-3sqrt(3)` `implies" "x=(5+3sqrt(3))" or "x=(5-3sqrt(3)).` Hence, `(5+3sqrt(3))` and `(5-3sqrt(3))` are the roots of the given equation. |
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| 35. |
Solve the following quadratic equation : `10x-(1)/(x)=3` |
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Answer» Given equation is `10x-(1)/(x)=3` `implies10x^(2)-1=3x` `implies10x^(2)-3x-1=0` `implies10x^(2)-(5-2)x-1=0` `implies10x^(2)-5x+2x-1=0` `implies5x(2x-1)+1(2x-1)=0` `implies(5x+1)(2x-1)=0` `implies5x+1=0or2x-1=0` when `5x+1=0impliesx=-(1)/(5)` and `2x-1=0impliesx=(1)/(2)` Hence, `-(1)/(5)and(1)/(2)` are roots of the equation. |
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| 36. |
Salve the following quadratic equations by factorization method.`x^2+2sqrt(2)x-6=0``sqrt(3)x^2+10 x=7sqrt(3)=0` |
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Answer» Correct Answer - `x=sqrt(2)" or "x=-3sqrt(2)` `x^(2)+2sqrt(2)x-6=0impliesx^(2)+3sqrt(2)x-sqrt(2)x-6=0impliesx(x+3sqrt(2))-sqrt(2)(x+3sqrt(2))=0.` |
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| 37. |
Find the roots of each of the following equations, if they exist, by applying the quadratic formula: `2x^(2)+6sqrt(3)x-60=0` |
| Answer» Correct Answer - `x=2sqrt(3)" or "x =-5sqrt(3)` | |
| 38. |
Solve the following quadratic equations by factorisation : (i) `4-11x=3x^(2)` (ii) `x^(2)-(11)/(4)x+(15)/(8)=0` |
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Answer» (i) Given equation is `4-11x=3x^(2)` `implies3x^(2)+11x-4=0` `implies3x^(2)+(12-1)x-4=0` `implies3x^(2)+12x-x-4=0` `implies3x(x+4)-1(x+4)=0` `implies(3x-1)(x+4)=0` `implies3x-1=0orx+4=0` when `3x-1=0impliesx=(1)/(3)` and `x+4=0impliesx=-4` Hence, `(1)/(3)and-4` are roots of equation. (ii) Given equation is `x^(2)-(11)/(4)x+(15)/(8)=0` Multiplying both sides by 8, we get `8x^(2)-22x+15=0` `implies8x^(2)-(12+10)x+15=0` `implies8x^(2)-12-10x+15=0` `implies4x(2x-3)-5(2x-3)=0` `implies(2x-3)(4x-5)=0` :. either `2x-3=0or4x-5=0` `implies2x=3or4x=5` `impliesx=(3)/(2)orx=(5)/(4)` Hence, `(3)/(2)and(5)/(4)` are the roots of given equation. |
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| 39. |
Solve : `sqrt(3)x^(2)+10x-8sqrt(3)=0.` |
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Answer» Correct Answer - `x=-4sqrt(3)" or "x=(2)/(sqrt(3))` `sqrt(3)x^(2)+10x-8sqrt(3)=0impliessqrt(3)x^(2)+12x-2x-8sqrt(3)=0impliessqrt(3)x(x+4sqrt(3))-2(x+4sqrt(3))=0.` |
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| 40. |
Solve each of the following quadratic equations: `4-11x=3x^(2)` |
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Answer» Correct Answer - `x=-4" or "x=(1)/(3)` `3x^(2)+11x-4=0implies3x^(2)+12x-x-4=0.` |
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| 41. |
Solve each of the following quadratic equations: `sqrt(3)x^(2)+11x+6sqrt(3)=0` |
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Answer» Correct Answer - `x=-3sqrt(3)" or "x=(-2)/(sqrt(3))` `sqrt(3)x^(2)+11x+6sqrt(3)=0impliessqrt(3)x^(2)+9x+2x+6sqrt(3)=0impliessqrt(3)x(x+3sqrt(3))+2(x+3sqrt(3))=0.` |
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| 42. |
Find the roots of each of the following equations, if they exist, by applying the quadratic formula: `25x^(2)+30x+7=0` |
| Answer» Correct Answer - `x=((-3+sqrt(2)))/(5)" or "x=((-3-sqrt(2)))/(5)` | |
| 43. |
A dealer sells an article for Rs. 75 and gains as much per cent as the cost price of the article. Find the cost price of the article. |
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Answer» Correct Answer - Rs. 50 Let the CP of the article be Rs. x. Then, gain `=x%`. `:." "SP=Rs. {((100+x)x)/(100)}=Rs. {(x^(2)+100x)/(100)}` `implies" "(x^(2)+100x)/(100)=75impliesx^(2)+100x-7500=0impliesx^(2)+150x-50x-7500=0.` |
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| 44. |
A dealer sells a toyfor Rs 24 and gains as much per cent as the cost price of the toy. Find thecost price of the toy. |
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Answer» Let cost price be=>`Rs.x` Gain=x% of C.P=`x xx x/100=x^2/100` S.P=C.P+Gain `24=x+x^2/100` `x^2+100x-2400=0` `(x+120)(x-20)=0` x cannot be negative so `x`=cost price =`Rs.20` |
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| 45. |
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth. |
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Answer» let length=xm let breath=ym According to Question Perimeter=80m 2(x+y)=80 x+y=40 y=40-x area=400m xy=400 x(40-x)=400 `x^2-40x+400=0` Solving this x=20m putting this value in equation 1 y=20m |
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| 46. |
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km h more than that of the passenger train, find the average speed of the two trains. |
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Answer» speed of passenger train`= x`km/hr speed of express train `=(x+11)`km/hr time taken by passenger train`= 132/x = 132/x` time taken by express train`=132/(x+11) = 132/(x+11)` acc to question `132/(x+11) =132/x - 1 ` `132[1/x - 1/(x+11)] = 1` `132[(x+11-x)/(x(x+11))]=1` `132*11 = x(x+11)` `x^2+11x-1452=0` `x^2 + 44x-3x- 1452=0 ` `x(x+44) - 33(x+44)=0` `(x-33)(x+44)=0` `x=33 , -44` x=-44 cant be possible as x cant be zero so, x=33 & `x+11 = 44` answer |
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| 47. |
Determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. |
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Answer» let the age of 1 be`x` le the age of 2 be `y` acc to question `x+y= 20` & `(x-4)(y-4) = 48` `xy-4x-4y+16=48` `xy-4(x+y) = 32` `xy-4(20) = 32` `xy-80=32` `xy=112` `x+ 112/x = 20` `x^2- 20x +112=0` D`=b^2-4ac` `=(-20)^2 - 4(112)` `=-48` age can not be imaginary and negative so solution is not possible. |
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| 48. |
If the equation `(1+m^2)x^2+2m c x+(c^2-a^2)=0`has equal roots, prove that `c^2=a^2(1+m^2)dot` |
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Answer» We have, `(1+m^(2))x^(2)+2mcx+(c^(2)-a^(2))=0` It has equal roots, if D=0 `impliesB^(2)-4AC=0` `implies(2mc)^(2)-4(1+m^(2))(c^(2)-a^(2))=0` `implies4m^(2)c^(2)-4(c^(2)-a^(2)+m^(2)c^(2)-m^(2)a^(2))=0` `impliesm^(2)c^(2)-c^(2)+a^(2)-m^(2)c^(2)+m^(2)a^(2)=0` `impliesc^(2)=a^(2)+m^(2)a^(2)=a^(2)(1+m^(2))` |
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| 49. |
For what value of n, the quadratic equation `3^(n)x^(2)+54x+81^(n)=0` have coincident roots? |
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Answer» We have, `3^(n)x^(2)+54x+81^(n)=0` For coincident roots, D=0 `implies(54)^(2)-4(3^(n))(81^(n))=0` `implies4xx3^(n)xx3^(4n)=(54)^(2)implies3^(5n)=(54xx54)/(4)=729` `implies3^(5n)=3^(6)implies5n=6impliesn=(6)/(5)`. |
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| 50. |
If -4 is a root of the equation `x^(2)+px-4=0` and the equation `x^(2)+px+q=0` has coincident roots, find the values of p and q. |
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Answer» Since -4 is a root of `x^(2)+px-4=0` Hence, (-4) will satisfy the equation. Therefore, `(-4)^(2)+p(-4)-4=0` `implies16-4p-4=0` `implies-4p+12=0` `implies-4p=-12` `impliesp=3" "....(1)` Given that `x^(2)+px+q=0` has coincident roots. `:.D=b^(2)-4ac=0` `impliesD=p^(2)-4xx1xxq=0` `impliesp^(2)-4q=0` `implies3^(2)-4q=0" "["form"(1)]` `implies9-4q=0` `implies-4q=-9` `impliesq=(9)/(4)` Hence, p=3 and `q=(9)/(4)` |
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