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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. |
A person on tour has Rs. 4200 for his expenses. If he extends his tour for 3 days, he has to cut down his daily expenses by Rs. 70. Find the original duration of the tour. |
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Answer» Let the original duration of the tour be x days. Then, `(4200)/(x)-(4200)/((x+3))=70` `implies" "4200xx[(1)/(x)-(1)/((x+3))]=70implies((x+3)-x)/(x(x+3))=(70)/(4200)` `implies" "x(x+3)=180impliesx^(2)+3x-180=0` `implies" "x^(2)+15x-12x-180=0impliesx(x+15)-12(x+15)=0` `implies" "(x+15)(x-12)=0impliesx+15=0" or "x-12=0` `implies" "x=-15" or "x=12` `implies" "x=12" "[because" number of days cannot be negative"].` `:." "` original duration of the tour is 12 days. |
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| 102. |
`(x^(2)+1)^(2)-x^(2)=0` has (i) four real roots (ii) two real roots (iii) no real roots (iv) one real root |
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Answer» Given equation is `(x^(2)+1)^(2)-x^(2)=0` `implies(x^(2)+1-x)(x^(2)+1+x)=0` `impliesx^(2)+1-x=0orx^(2)+1+x=0` For first equation `x^(2)+1-x=0` i.e., `x^(2)+1+x=0` `D=(-1)^(2)-4(1)(1)=1-4=-3lt0` `implies` Equation has no real root. For second equation `x^(2)+1+x=0` i.e., `x^(2)+x+1=0` `D=(1)^(2)-4(1)(1)=1-4=-3lt0` `implies` Equation has no real root. Hence, given equation has no real root. |
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| 103. |
The quadratic equation `2x^(2)-sqrt5x+1=0` has (a) two distinct real roots (b) two equal real roots (c) no real roots (d) more than 2 real roots |
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Answer» Given equation is `2x^(2)-sqrt5x+1=0` Here `a=2,b=sqrt5,c=1` `because"Discriminant D"=b^(2)-4ac` `=(-sqrt5)^(2)-4(2)(1)=5-8=-3lt0` `implies` Given equation has no real root. |
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| 104. |
Solve: `x^2-(7-i)x+(18-i)=0ov e rCdot` |
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Answer» `x^2 - (7-i)x - (18-i) = 0` This is a quadratic equation and its roots can be given as, `x = (-b+-sqrt(b^2-4ac))/(2a)` Here, `a = 1, b = -(7-i), c= (18-i)` Now, `b^2-4ac = (-(7-i))^2 - 4(18-i)` `= 49+i^2 - 14i -72+4i`...[As `i^2 = -1`] `= 49-1 - 14i -72+4i` `=-24-10i` `=1 - 25 - 2(1)(5i)` `=1 + 25i^2 - 2(1)(5i)` `= 1^2 + (5i)^2 - 2(1)(5i)` `=(1-5i)^2` `:. sqrt(b^2-4ac) = 1-5i` `:. x = (7-i+-(1-5i))/2` `=>x = (8-6i)/2 and x = (6-4i)/2` `=> x = 4-3i and x = 3-2i`, which are the required solutions for the given equation. |
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| 105. |
The number of real solutions of `|2x-x^2-3|=1`is(a) `0`(b) `2`(c) `3`(d) `4` |
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Answer» `|2x-x^2-3| = 1` `=>2x-x^2-3 = +-1` When `2x-x^2-3 = 1` `=>-x^2+2x-4 = 0` `=>D = b^2-4ac = 2^2-4(-1)(-4) = 4-16 = -12` As, `D lt 0`, no real roots are present for this equation. When `2x-x^2-3 = -1` `=>-x^2+2x-2 = 0` `=>D = b^2-4ac = 2^2-4(-1)(-2) = 4-8 = -4` As, `D lt 0`, no real roots are present for this equation. So, in both cases, no real solution exists. So, the correct option is option - `(a)` that is `0`. |
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| 106. |
Show that `(x^3+x^2+x+1)/(x^3-x^2+x-1)=(x^2+x+1)/(x^2-x+1)`,is not possible for any `x epsilon R`. |
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Answer» `(x^3+x^2+x+1)/(x^3-x^2+x-1) = (x^2+x+1)/(x^2-x+1)` =>`((x^3+x)+(x^2+1))/((x^3+x)+(x^2+1)) = ((x^2+1)+x)/((x^2+1)-x)` We know, `(a+b)/(a-b) = a/b`.So, `=>(x^3+x)/(x^2+1) = (x^2+1)/x` `=>x(x^3+x) = (x^2+1)^2` `=>x^4+x^2 = x^4+1+2x^2` `=>x^2 = -1` For any real number `x`, square of `x` can not be negative. So, both sides can not be equal. |
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| 107. |
Solve each of the following quadratic equations: `(2)/(x^(2))-(5)/(x)+2=0` |
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Answer» Correct Answer - `x=2" or "x=(1)/(2)` `2x^(2)-5x+2=0implies2x^(2)-4x-x+2=0implies2x(x-2)-(x-2)=0.` |
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| 108. |
Solve each of the following quadratic equations: `2x^(2)+ax-a^(2)=0` |
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Answer» Correct Answer - `x=-a" or "x=(a)/(2)` `2x^(2)+ax-a^(2)=0implies2x^(2)+2ax-ax-a^(2)=0implies2x(x+a)-a(x+a)=0.` |
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| 109. |
Solve the following quadratic equation for `x:4x^2 + 4bx – (a^2-b^2) = 0` |
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Answer» Correct Answer - `x=(-(a+b))/(2)" or "x=(a-b)/(2)` `4x^(2)+4bx+(b^(2)-a^(2))=0implies4x^(2)+2(b+a)x+2(b-a)x+(b^(2)-a^(2))=0.` |
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| 110. |
Solve each of the following quadratic equations: `4x^(2)-4a^(2)x+(a^(4)-b^(4))=0` |
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Answer» Correct Answer - `x=(a^(2)+b^(2))/(2)" or "x=(a^(2)-b^(2))/(2)` `4x^(2)-4a^(2)x+(a^(4)-b^(4))=0implies4x^(2)-2(a^(2)+b^(2))x-2(a^(2)-b^(2))x+(a^(4)-b^(4))=0` `implies" "2x[2x-(a^(2)-b^(2))]-(a^(2)-b^(2))[2x-(a^(2)+b^(2)]=0.` |
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| 111. |
Solve each of the following quadratic equations: `x^(2)+5x-(a^(2)+a-6)=0` |
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Answer» Correct Answer - `x=-(a+3)" or "x=(a-2)` `x^(2)+5x-(a+3)(a-2)=0impliesx^(2)+(a+3)x-(a-2)x-(a+3)x-(a-2)x-(a+3)(a-2)=0.` |
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| 112. |
Solve the following quatratic equation : `4x^(2)-2(a^(2)+b^(2))x+a^(2)b^(2)=0` |
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Answer» Given equation is `4x^(2)-2(a^(2)+b^(2))x+a^(2)b^(2)=0` `implies4x^(2)-2a^(2)x-2b^(2)x+a^(2)b^(2)=0` `implies2x(2x-a^(2))-b^(2)(2x-a^(2))=0` `implies(2x-a^(2))(2x-b^(2))=0` `implies2x-a^(2)=0or2x-b^(2)=0` when `2x-a^(2)=0impliesx=(a^(2))/(2)` and `2x-b^(2)=0impliesx=(b^(2))/(2)` Hence, `a^(2)/(2)and(b^(2))/(2)` are roots of the equation. |
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| 113. |
Solve byfactorization: `2x^2+a x-a^2=0` |
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Answer» Correct Answer - `x=-a" or "x=(a)/(2)` `2x^(2)+ax-a^(2)=0implies2x^(2)+2ax-ax-a^(2)=0` `implies2x(x+a)-a(x+a)=0` `implies(x+a)(2x-a)=0impliesx=-a" or "x=(a)/(2).` |
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| 114. |
Solve byfactorization: `1/(x+4)-1/(x-7)=(11)/(30), x!=4, 7` |
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Answer» Given equation is `(1)/(x+4)-(1)/(x-7)=(11)/(30)` `implies((x-7)-(x+4))/((x+4)(x-7))=(11)/(30)` `implies(x-7-x-4)/(x^(2)-7x+4x-28)=(11)/(30)` `implies(-11)/(x^(2)-3x-28)=(11)/(30)` `implies11(x^(2)-3x-28)=-11xx30` `impliesx^(2)-3x-28=-30` `impliesx^(2)-3x+2=0` `impliesx^(2)-(2+1)x+2=0` `impliesx^(2)-2x-1x+2=0` `impliesx(x-2)-1(x-2)=0` `implies(x-2)(x-1)=0` `impliesx-2=0orx-1=0` when `x-2=0impliesx=2` and `x-1=0impliesx=1` Hence, 2 and 1 are roots of the equation. |
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| 115. |
Solve each of the following quadratic equations: `x^(2)-2ax-(4b^(2)-a^(2))=0` |
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Answer» Correct Answer - `x=(a-2b)" or "x=(a+2b)` `x^(2)-2ax+(a^(2)-4b^(2))=0impliesx^(2)-2ax+(a+2b)(a-2b)=0` `implies" "x^(2)-(a+2b)x-(a-2b)x+(a+2b)(a-2b)=0.` |
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| 116. |
Solve byfactorization: `4x^2+4b x-(a^2-b^2)=0` |
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Answer» Correct Answer - `x=(-(b+a))/(2)" or "x=((a+b))/(2)` `4x^(2)+4bx+(b^(2)-a^(2))=0` `implies4x^(2)+2(b+a)x+2(b-a)x+(b^(2)-a^(2))=0` `implies2x[2x+(b+a)]+(b-a)[2x+(b+a)]=0` `implies[2x+(b+a)][2x+(b-a)]=0` `impliesx=(-(b+a))/(2)" or "x=(a-b)/(0).` |
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| 117. |
Solve `x^(2)+5x-(a^(2)+a-6)=0`. |
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Answer» Correct Answer - `x=-(a+3)" or "x=(a-2)` `x^(2)+5x-(a+3)(a-2)=0` `impliesx^(2)+(a+3)x-(a-2)x-(a+3)(a-2)=0` `impliesx{x+(a+3)}-(a-2){x+(a+3)}=0` `implies{x+(a+3)}{x-(a-2)}=0` `impliesx=-(a+3)" or "x=(a-2).` |
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| 118. |
Solve for x : `1/(x+1)+3/(5x+1)=5/(x+4), x != -1,-1/5,-4 |
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Answer» We have, `(1)/(x+1)+(3)/(5x+1)=(5)/(x+4)` `(5x+1+3(x+1))/((x+1)(5x+1))=(5)/(x+4)` `implies(5x+1+3x+3)(x+4)=5(x+1)(5x+1)` `implies(8x+4)(x+4)=5(x+1)(5x+1)` `implies8x^(2)+32x+4x+16=5(5x^(2)+x+5x+1)` `implies8x^(2)+36x+16=25x^(2)+30x+5` `implies17x^(2)-17x+11x-11=0` `implies17x(x-1)+11(x-1)=0` `implies(x-1)(17+11)=0` :.Either `x-1=0or17x+11=0` `:.x=1orx=(-11)/(17)` |
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| 119. |
Solve the equation : `(5)/(2-x)+(x-5)/(x+2)+(3x+8)/(x^(2)-4)=0` |
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Answer» We have, `(5)/(2-x)+(x-5)/(x+2)+(3x+8)/(x^(2)-4)=0` `implies(5(x+2)+(2-x)(x-5))/((2-x)(x+2))-(3x+8)/((2+x)(2-x))=0` `implies(5x+10+2x-10-x^(2)+5x-3x-8)/((2-x)(2+x))=0` `implies-x^(2)+9x-8=0impliesx^(2)-9x+8=0` `implies(x-1)(x-8)=0` `:.x=1orx=8` |
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| 120. |
Solve the following equation : `2((2x-1)/(x+3))-3((x+3)/(2x-1))=5,(xne-3,1)` |
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Answer» Given equation is `2((2x-1)/(x+3))-3((x+3)/(2x-1))=5" ".....(i)` Let `(2x-1)/(x+3)=y.....(ii)` Hence, `(x+3)/(2x-1)=(1)/(y)` Now from equation (1) `2y-(3)/(y)=5` `implies2y^(2)-3=5y` `implies2y^(2)-5y-3=0` `implies2y^(2)-(6-1)y-3=0` `implies2y^(2)-6y+1y-3=0` `implies2y(y-3)+1(y-3)=0` `implies(2y+1)(y-3)=0` `implies2y+1=0ory-3=0` when `implies2y+1=0impliesy=-(1)/(2)` and `y-3=0impliesy=3` Substituting values of y in equation (2) when `y=-(1)/(2)` `(2x-1)/(x+3)=-(1)/(2)` `implies2(2x-1)=-1(x+3)implies4x-2=-x-3` `implies5x=-1impliesx=-(1)/(5)` when y=3 `(2x-1)/(x+3)=3` `implies2x-1=3(x+3)implies2x-1=3x+9` `implies-x=10impliesx=-10` Hence, `x=-10orx=-(1)/(5)` are roots of the equation. |
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| 121. |
Solve the following equation : `3^(x+2)+3^(-x)=10` |
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Answer» Given equation is `3^(x+2)+3^(-x)=10` `implies3^(x)xx3^(2)+(1)/(3^(x))=10implies9xx3^(x)+(1)/(3^(x))=10" ".....(2)` Let `3^(x)=a``3^(x)=a" "......(2)` Then from (1) `9a+(1)/(a)=10` `implies9a^(2)+1=10a " "implies9a^(2)-10a+1=0` `implies9a^(2)-(9+1)a+1=0" "implies9a^(2)-9a-a+1=0` `implies9a(a-1)-1(a-1)=0" "implies(9a-1)(a-1)=0` `implies9a-1=0" "or" "a-1=0` when `9a-1=0impliesa=(1)/(9)` and when `a-1=0impliesa=1` Substituting values of a in equation (2) when `a=(1)/(9)` `3^(x)=(1)/(9)" "implies" "3^(x)=(1)/(3^(2)` `implies3^(x)=3^(-2)" "or" "x=-2` when a=1 `3^(x)=1` `implies3^(x)=3^(0)" "implies" "x=0` Hence, 0 and -2 are roots of the equation. |
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| 122. |
Solve the equation : `(a)/(x-b)+(b)/(x-a)=2" "(xneb,a)` |
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Answer» Given equation is `(a)/(x-b)+(b)/(x-a)=2` `implies(a)/(x-b)+(b)/(x-a)=1+1` `implies(a)/(x-b)-1+(b)/(x-a)=1+1` `implies(a)/(x-b)-1+(b)/(x-a)-1=0` `implies(a-x+b)/(x-b)+(b-x+a)/(x-a)=0` `implies(a+b-x)((1)/(x-b)+(1)/(x-a))=0` `impliesa+b-x=0or(1)/(x-b)+(1)/(x-a)=0` when `a+b-x=0impliesx=a+b` and when `(1)/(x-b)+(1)/(x-a)=0implies(x-a+a-b)/((x-b)(x-a))=0` `implies2x-a-b=0` `implies2x=a+b` `impliesx=(a+b)/(2)` Hence, `x=a+bandx=(a+b)/(2)` are roots of the equaton. Alternatively, `implies(1)/(x-b)=-(1)/(a-x)` `impliesx-b=a-x` `2x=a+b` `x=(a+b)/(2)` |
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| 123. |
Solve each of the following quadratic equations: `(a)/((x-b))+(b)/((x-a))=2,xneb,a` |
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Answer» Correct Answer - `x=(a+b)" or "x=((a+b))/(2)` The given equation is `((a)/(x-b)-1)+((b)/(x-a)-1)=0` `implies" "((a-x+b))/((x-b))+((a-x+b))/((x-a))=0` `implies" "(a-x+b).[(1)/((x-b))+(1)/((x-a))]=0` `implies" "(a-x+b)[(2x-(a+b))/((x-a)(x-b))}=0` `implies" "(a-x+b)[2x-(a+b)]=0` `implies" "x=(a+b)" or "x=((a+b))/(2)` |
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| 124. |
Solve each of the following quadratic equations: `(a)/((ax-1))+(b)/((bx-1))=(a+b),xne(1)/(a),(1)/(b)` |
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Answer» Correct Answer - `x=((a+b))/(ab)" or "x=(2)/((a+b))` The given equation is `{(a)/((ax-1))-b}+{(b)/((bx-1))-a}=0` `implies" "((a-abx+b))/((ax-1))+((a-abx+b))/((bx-1))=0` `implies" "(a-abx+b).((1)/(ax-1)+(1)/(bx-1))=0.` |
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| 125. |
Solve each of the following quadratic equations: `3^((x+2))+3^(-x)=10` |
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Answer» Correct Answer - `x=-2" or "x=0` `3^(x)xx3^(2)+(1)/(3^(x))=10implies9y+(1)/(y)=10," where "y=3^(x).` |
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| 126. |
Solve each of the following quadratic equations: `4^((x+1))+4^((1-x))=10` |
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Answer» Correct Answer - `x=(1)/(2)" or "x=(-1)/(2)` `4^(x)xx4+(4)/(4^(x))=10implies2y+(2)/(y)=5," where "y=4^(x).` |
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| 127. |
Solve each of the following quadratic equations: `2^(2x)-3.2^((x+2))+32=0` |
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Answer» Correct Answer - `x=2" or "x=3` `2^(2x)-3xx2^(2)xx2^(x)+32=0impliesy^(2)-12y+32," where "2^(x)=y` `=y^(2)-8y-4y+32=0.` |
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| 128. |
The roots of the quadratic equation `2x^2 - x - 6 = 0` are |
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Answer» Correct Answer - `x=2" or "x=(-3)/(2)` `2x^(2)-x-6=0implies2x^(2)-4x+3x-6=0implies2x(x-2)+3(x-2)=0` `implies(x-2)(2x+3)=0impliesx=2" or "x=(-3)/(2).` |
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| 129. |
The sum of two natural numbers is 8 and their product is 15. Find the numbers. |
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Answer» Correct Answer - 3 and 5 `alpha+beta=8" and "alphabeta=15.` The quadratic equation whose roots are `alpha` and `beta` is given by `x^(2)-8x+15=0` `impliesx^(2)-5x-3x+15=0` `impliesx(x-5)-3(x-5)=0` `implies(x-3)(x-5)=0` `:." "x=3" or "x=5.` |
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| 130. |
If `x=-(1)/(2)` is a solution of the quadratic equation `3x^(2)+2kx-3=0`, find the velue of k. |
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Answer» Correct Answer - `k=(-9)/(4)` Since `x=(-1)/(2)` is a solution of `3x^(2)+2kx-3=0,` we have `3xx((-1)/(2))^(2)+2kxx((-1)/(2))-3=0` `implies" "(3)/(4)-k-3=0impliesk=((3)/(4)-3)=(-9)/(4).` |
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| 131. |
For what values of k are the roots of the quadratic equation `3x^(2)+2kx+27=0` real and equal? |
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Answer» Correct Answer - `k=9" or "k=-9` `D=4k^(2)-324`. For real and equal roots, we must have, D=0. |
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| 132. |
The length of a rectangle is thrice as long as the side of a square. The side of the square is 4 cm more than the width of the rectangle. Their areas being equal, find their dimensions. |
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Answer» Correct Answer - `(l=18" cm", b=2" cm")," side"=6" cm "` Let each side of the square be x cm. Then, length of the rectangle `=3x cm.` Width of the rectangle `=(x-4)cm.` `:." "3x(x-4)=x^(2)implies2x^(2)-12x=0implies2x (x-6)=0impliesx=6.` |
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| 133. |
Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals? Why? |
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Answer» Yes. Consider the quadratic equation `sqrt2x^(2)-9sqrt2x+20sqrt2=0` with distinct irrational coefficients. The roots of this equation are 4 and 5 which are rationals. |
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| 134. |
Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number. |
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Answer» Let, the natural number be x. According to given condition `x^(2)-84=3(x+8)` `impliesx^(2)-84=3x+24impliesx^(2)=3x-108=0` `impliesx^(2)-(12-9)x-108=0impliesx^(2)-12x+9x-108=0` `impliesx(x-12)+9(x-12)=0implies(x-12)(x+9)=0` `impliesx-12=0orx+9=0` `impliesx=12orx=-9` But-9 is not a natural number. `because` Requird natural number=12. |
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| 135. |
Find the roots of each of the following equations, if they exist, by applying the quadratic formula: |
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Answer» Correct Answer - `x=(2a+b)" or "x=(2a-b)` `x^(2)-4ax+(2a+b)(2a-b)=0` `impliesx^(2)-(2a+b)x-(2a-b)x+(2a+b)(2a-b)=0` `impliesx{x-(2a+b)}-(2a-b){x-(2a+b)}=0` `implies{x-(2a+b)}{x-(2a-b)}=0` `impliesx=(2a+b)" or "x=(2a-b).` |
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| 136. |
Which of the following is a quadratic eqaution ?A. `6x^(2) = 2 -x^(3)`B. `x^(2) ((1)/(x) -2) = (7)/(2)`C. `(3)/(x) - 3= 4x^(2)`D. `5x+7= 3 x ` |
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Answer» Correct Answer - B (D) is a linear equations. (A) and (C ) are cubic equation. |
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| 137. |
A farmer prepares a rectangular vegatable garden of area 180 sq metres. With 39 netres of barbed wire, he can fence the three sides of the garden, leaving one of the longer sides unfenced. Find the dimensions of the garden. |
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Answer» Correct Answer - `"Either "(l=24 m, b=7.5m)" or "(l=15m,b=12m)` Let the length be x metres. Then, breadth `=(180)/(x)m.` `:." "x+(180)/(x)+(180)/(x)=39impliesx+((360)/(x)=39` `implies" "x^(2)-39x+360=0impliesx^(2)-24x-15x+360=0` `implies" "(x-24)(x-15)=0impliesx=24" or "x=15.` Either `(l=24m,b=7.5m)" or "(l=15m,,b=12m).` |
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| 138. |
A rectangular field is 16 m long and 10 m wide. There is a path of uniform width all around it, having an area of `120 m^(2)`. Find the width of the path. |
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Answer» Correct Answer - 2 m Let the width of the path be x m. So, the length of the field with path `=(16+2x)m` and breadth of the field with path `=(10+2x)m.` `:." "(16+2x)(10+2x)-16xx10=120` `implies" "160+52x+4x^(2)-160=120` `implies" "4x^(2)+52x-120=0impliesx^(2)+13x-30=0` `implies" "x^(2)+15x-2x-30=0impliesx(x+15)-2(x+15)=0` `implies" "(x+15)(x-2)=0impliesx=-15" or "2.` So, width of the path `2m.` |
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| 139. |
A motorboat whose speed is 9 km/hr in still water, goes 15 km downstream and comes back in a total time of 3 hours 45 minutes. Find the speed of the stream. |
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Answer» Correct Answer - 3 km/hr Let the speed of the stream be x km/hr. Then, speed downstream `=(9+x)km//hr` and speed upstream `=(9-x)km//hr.` `:." "(15)/(9+x)+(15)/(9-x)=(1)/(4)implies4[(9-x)+(9+x)]=(9+x)(9-x)` `implies81-x^(2)=72impliesx^(2)=9impliesx=3.` |
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| 140. |
A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work. |
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Answer» Correct Answer - 30 days Suppose B takes x days to finish the work. Then, A takes `(x-10)` days. `:." "(1)/(x)+(1)/((x-10))=(1)/(12)impliesx^(2)-34x+120=0impliesx^(2)-30x-4x+120=0.` |
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| 141. |
Solve the following quadratic equations : (i) `x^(2)-45x+324=0` (ii) `x^(2)-55x+750=0` |
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Answer» (i) Given equation is `x^(2)-45x+324=0` `impliesx^(2)-(36+6)x-324=0` `impliesx^(2)-36x-9x+324=0` `impliesx(x-36)-9(x-36)=0` `implies(x-36)(x-9)=0` `{:("Now"" ",x=36=0impliesx=36),(and,x-9=0impliesx=9):}}` (ii) Given equation is `x^(2)-55x+750=0` `impliesx^(2)-(30+25)x+750=0` `impliesx^(2)-30x-25x+750=0` `impliesx(x-30)-25(x-30)=0` `implies(x-30)(x-25)=0` `{:("Now" ",x-30=0impliesx=30),(and,x-25=0impliesx=25):}}` |
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| 142. |
In a two-digit natural number, the digit at th etens place is equal to the square of the digit at units place. If 54 is subtracted from the number, the digits get interchanged . Find the number. |
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Answer» Correct Answer - The required natural number is 93 Let the digit at the units place be x. Then the digit at the tens place is `x^(2)`. The number is `10x^(2) + x ` . The number obtained by interchanging the digits is `10x + x^(2)`. From the given condition, `10x^(2) +x- 54 = 10 x + x^(2) ` `: 10x^(2) - x^(2) + x- 10 x - 54 = 0 ` ` :. 9 x^(2) - 9x - 54 =0 ` `:. x^(2) - x -6 = 0 ` .....(Dividing by 9 ) `:. x^(2)-3x + 2 x - 6 = 0` `:. x( x-3) + 2 (x-3) = 0 ` `:. ( x-3) ( x+ 2) = 0 ` `:. x-3 =0` or ` x+ 2 = 0 ` `:. x = 3 ` or `x = - 2 ` But the natural cannot negative `:. x = - 2 ` is unacceptable `:. x = 3 ` and `x^(2) = ( 3)^(2) =9 ` The number is `10x^(2) + x = 10 ( 9) + 3 = 90 + 3 = 93 ` |
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| 143. |
In an orchard , the number of trees in each column is 5 more than that in each row. Find the number of trees in each column, if the total number of trees is 1400. Flow chart. |
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Answer» Correct Answer - The number of trees in a column is 40. Let the number of trees in a row be x. The number of trees in a column `= ( ul ( x+5))` Total number of trees `= ul(x ( x+5))` Total number of trees `= ul ( 1400)` ….(Given ) `:. ul ( x( x+5) = 1400)` `:. x^(2) + 5x - 1400 = 0 ` `:. x^(2) + 40 x - 35x - 1400 = 0 ` `:. x( x+ 40 ) - 35( x+ 40) = 0 ` ` :. ( x+40) ( x-35) = 0 ` `:. x + 40 0` or `x- 35=0` `:. x = - 40 ` or `x = 35 ` But the number of trees cannot be negative. `:. x - 40 ` is unacceptable `:. x = 35 ` `:. x + 5 = 35 + 5 = 40` |
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| 144. |
The length of a hypotenuse of a right triangle exceeds the length of its base by 2cm and exceed twice the length of the altitude by 1cm .Find the length of each side of the triangle (in cm) |
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Answer» Correct Answer - base = 15 cm, altitude = 8 cm , hypotenuse = 17 cm Let the base be x cm. Then, hypotenuse `=(x+2)cm.` `(x+2)-(2xx"altitude")=1implies2xx" altitude"=(x+1)` `implies" altitude "=(1)/(2)(x+1)cm.` `:." "(x+2)^(2)=x^(2)+(1)/(4)(x+1)^(2)implies4(x+2)^(2)=4x^(2)+(x+1)^(2)` `implies" "4(x^(2)+4x+4)=5x^(2)+2x+1impliesx^(2)-14x-15=0` `implies" "(x-15)(x+1)=0impliesx=15.` |
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| 145. |
Write the roots of the following quadratic equations `:` (i) `x^(2) + x-6 0 ` (ii) `( x + 6 ) (x-3) = 0 ` |
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Answer» (i) `x^(2) + x - 6 = 0 ` `:. (x +3) ( x -2) = 0 ` `:. X + 3 = 0 ` or ` x - 2 = 0 ` ` :. X = - 3 ` or ` x = 2 ` (ii) `( x+ 6) ( x - 3) = 0 ` ` :. X + 6 = 0 ` or ` x - 3 = 0 ` ` :. X = - 6 ` or ` x = 3 ` |
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| 146. |
Write the quadratic equations, if (i) `alpha + beta = -6, alpha beta = 4 ` (ii) ` alpha + beta = 8 , alpha beta = -3` |
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Answer» (i) `x^(2) - ( alpha + beta) x + alpha beta = 0 ` `:. X^(2) - ( - 6 ) x + 4 = 0 ` ` :. X^(2) + 6 x + 4 = 0` (ii) ` x^(2) - ( alpha + beta ) x + alpha beta = 0 ` `:. X^(2) - ( 8)x + ( -3) = 0 ` `:. X^(2) - 8x -3 = 0 ` |
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| 147. |
If `alpha ` and `beta ` are the roots of the quadratic equation `x^(2) - 4x - 6 = 0 ` , find the values of `alpha^(2) + beta^(2)`. |
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Answer» Correct Answer - `alpha^(2) + beta^(2) = 28` `x^(2) - 4x - 6 = 0 ` Here, `a= 1, b = - 4, c = - 6 ` `alpha + beta = ( - b)/( a) = ( - ( - 4))/( 1)= 4 ` ....(1) `alpha beta = ( c )/( a ) = ( -6)/( 1) = - 6 ` ....(2) `alpha^(2) + beta^(2) = ( alpha + beta)^(2) - 2alpha beta ` .....(identity ) `= ( 4)^(2) - 2( - 6)` ...[From (1) and (2) ] ` = 16 + 12 = 28` |
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| 148. |
Which of the following is a quadratic equation? (i) `x^(2)+2x+1=(4-x)^(2)+3` (ii) `-2x^(2)=(5-x)(2x-(2)/(5))` (iii) `(k+1)x^(2)+(3)/(2)x=7" wherek"=-1` (iv) `x^(3)-x^(2)=(x-1)^(3)` |
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Answer» (i) `x^(2)+2x+1=(4-x)+3` `impliesx^(2)+2x+1=16+x^(2)-8x+3` `implies10x-8=0` which is not in the form of `ax^(2)+bx+c,ane0` Thus, the equation is not quadratic. (ii) `-2x^(2)=(5-x)(2x-(2)/(5))` `implies-2x^(2)=10x-2-2x^(2)+(2x)/(5)` `implies50x+2x-10=0` `implies52x-10=0` which is not a quadratic equation. (iii) `(k+1)x^(2)+(3)/(2)x=7` Given k=-1 `implies(-1+1)x^(2)+(3)/(2)x=7` `implies3x-14=0` which is not a quadratic equation. (iv) `x^(3)-x^(2)=(x-1)^(3)` `impliesx^(3)-x^(2)=x^(3)-1-3x(x-1)" "[because(a-b)^(3)=a^(3)-b^(3)-3ab(a-b)]` `impliesx^(3)-x^(2)=x^(3)-1-3x^(2)+3x` `implies2x^(2)-3x+1=0` which is in the form of `ax^(2)+bx+c=0,ane0` Hence, it is a quadratic equation. |
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| 149. |
Find the values of k for each of the following quadratic equations, so that they have two equal roots.(i) `2x^2+k x+3=0` (ii) `k x(x-2)+6=0` |
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Answer» (i) The given equation is `2x^(2)+kx+3=0` Comparing with `ax^(2)+kx+c=0` a=2, b=k and c=3 `because` `because" ""Discriminant D"=b^(2)-4ac` `impliesD=k^(2)-4xx2xx3` `impliesD=k^(2)-24` Since the equation has two equal roots Therefore, D=0 Hence, `k^(2)-24=0` `impliesk^(2)=24` `impliesk=+-sqrt24=+-2sqrt6` (ii) Given equation is `kx(x-2)+6=0` `implieskx^(2)-2kx+6=0` Comparing with `ax^(2)+bx+c=0` a=k, b=-2k and c=6 `because"Discriminant D"=b^(2)-4ac` `impliesD=(-2k)^(2)-4xxkxx6` `impliesD=4k^(2)-24k` For equal roots `D=0implies4k^(2)-24k=0` |
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| 150. |
The difference of squares of twonumbers is 180. The square of the smaller number is 8 times the largernumber. Find the two numbers. |
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Answer» Let, the required number be x and y, where `xgty` Given `x^(2)-y^(2)=180" ".....(1)` and `y^(2)=8x" "......(2)` From equation (1) and (2) we have `x^(2)-8x=180` `impliesx^(2)-8x-180=0impliesx^(2)-(18-10)x-180=0` `impliesx^(2)-18x+10x-180=0impliesx(s-18)+10(x-18)=0` `implies(x-18)(x+10)=0` `impliesx=18orx=-10` when `x=18impliesy^(2)=8xx18=144ory=+-12` and when `x=-10impliesy^(2)=8xx-10=-80` which is imaginary i.e., not possible. Hence, numbers are (18 and 12) or (18 and -12). |
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