Download Quadratic Equation Questions for CAT<\/a><\/p>\n Enroll for CAT 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>A shop owner bought a total of 64 shirts from a wholesale market that came in two sizes, small and large. The price of a small shirt was INR 50 less than that of a large shirt. She paid a total of INR 5000 for the large shirts, and a total of INR 1800 for the small shirts. Then, the price of a large shirt and a small shirt together, in INR, is<\/p>\n a)\u00a0175<\/p>\n b)\u00a0150<\/p>\n c)\u00a0200<\/p>\n d)\u00a0225<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the number of large shirts be l and the number of small shirts be s.<\/p>\n Let the price of a small shirt be x and that of a large shirt be x + 50.<\/p>\n Now, s + l = 64<\/p>\n l (x+50) = 5000<\/p>\n sx = 1800<\/p>\n Adding them, we get,<\/p>\n lx + sx + 50l = 6800<\/p>\n 64x +50l = 6800<\/p>\n Substituting l = (6800 – 64x) \/ 50, in the original equation, we get<\/p>\n $\\frac{\\left(6800-64x\\right)}{50}\\left(x+50\\right)=5000$<\/p>\n (6800 – 64x)(x +\u00a050) = 250000<\/p>\n $6800x+340000-64x^2-3200x = 250000$<\/p>\n $64x^2-3600x-90000=0$<\/p>\n Solving, we get, x=\u00a0$\\frac{225\\pm\\ 375}{8}=\\frac{600}{8}or-\\frac{150}{8}$<\/p>\n SO, x = 75<\/p>\n x +\u00a050 = 125<\/p>\n Answer = 75 + 125 = 200.<\/p>\n Alternate approach: By options.<\/strong><\/p>\n Hint<\/strong>: Each option gives the sum of the costs of one large and one small shirt. We know that large = small + 50<\/p>\n Hence, small + small + 50 = option.<\/p>\n SMALL = (Option – 50)\/2<\/strong><\/p>\n LARGE = Small +\u00a050 = (Option + 50)\/2<\/strong><\/p>\n Option A and Option D gives us decimal values for SMALL and LARGE, hence we will consider them later. Lets start with Option B<\/strong>.<\/p>\n Large = 150 + 50 \/ 2 = 100<\/p>\n Small = 150 – 50 \/ 2 = 50<\/p>\n Now, total shirts = 5000\/100 + 1800\/50 = 50 + 36 = 86 (X – This is wrong)<\/strong><\/p>\n Option C – Large = 200 + 50 \/ 2 = 125<\/p>\n Small = 200 – 50 \/ 2 = 75<\/p>\n Total shirts = 5000\/125 + 1800\/75 = 40 + 24 = 64 ( This is the right answer) Question 2:\u00a0<\/b>Consider the pair of equations: $x^{2}-xy-x=22$ and $y^{2}-xy+y=34$. If $x>y$, then $x-y$ equals<\/p>\n a)\u00a06<\/p>\n b)\u00a04<\/p>\n c)\u00a07<\/p>\n d)\u00a08<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have : Question 3:\u00a0<\/b>For all real values of x, the range of the function $f(x)=\\frac{x^{2}+2x+4}{2x^{2}+4x+9}$ is:<\/p>\n a)\u00a0$[\\frac{4}{9},\\frac{8}{9}]$<\/p>\n b)\u00a0$[\\frac{3}{7},\\frac{8}{9})$<\/p>\n c)\u00a0$(\\frac{3}{7},\\frac{1}{2})$<\/p>\n d)\u00a0$[\\frac{3}{7},\\frac{1}{2})$<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $f(x)=\\frac{x^{2}+2x+4}{2x^{2}+4x+9}$<\/p>\n If we closely observe the coefficients of the terms in the numerator and denominator, we see that the coefficients of the $x^2$ and x in the numerators are in ratios 1:2. This gives us a hint that we might need to adjust the numerator to decrease the number of variables.<\/p>\n $f(x)=\\frac{x^2+2x+4}{2x^2+4x+9}=\\frac{x^2+2x+4.5-0.5}{2x^2+4x+9}$<\/p>\n =\u00a0$\\frac{x^2+2x+4.5}{2x^2+4x+9}-\\frac{0.5}{2x^2+4x+9}$<\/p>\n =\u00a0$\\frac{1}{2}-\\frac{0.5}{2x^2+4x+9}$<\/p>\n Now, we only have terms of x in the denominator.<\/p>\n The maximum value of the expression is achieved when the quadratic expression $2x^2+4x+9$ achieves its highest value, that is infinity.<\/p>\n In that case, the second term becomes zero and the expression becomes 1\/2. However, at infinity, there is always an open bracket ‘)’.<\/p>\n To obtain the minimum value, we need to find the minimum possible value of the quadratic expression.<\/p>\n The minimum value is obtained when 4x + 4 = 0 [d\/dx = 0]<\/p>\n x=-1.<\/p>\n The expression comes as 7.<\/p>\n The entire expression becomes 3\/7.<\/p>\n Hence,\u00a0$[\\frac{3}{7},\\frac{1}{2})$<\/p>\n Question 4:\u00a0<\/b>Suppose one of the roots of the equation $ax^{2}-bx+c=0$ is $2+\\sqrt{3}$, Where a,b and c are rational numbers and $a\\neq0$. If $b=c^{3}$ then $\\mid a\\mid$ equals.<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Given a, b, c are rational numbers.<\/p>\n Hence a, b, c are three numbers that can be written in the form of p\/q.<\/p>\n Hence if one both the root is 2+$\\sqrt{\\ 3}$ and considering the other root to be x.<\/p>\n The sum of the roots and the product of the two roots must be rational numbers.<\/p>\n For this to happen the other root must be the conjugate of\u00a0$2+\\sqrt{\\ 3}$ so the product and the sum of the roots are rational numbers which are represented by:\u00a0$\\frac{b}{a},\\ \\frac{c}{a}$<\/p>\n Hence the sum of the roots is 2+$\\sqrt{\\ 3}+2-\\sqrt{\\ 3}$ = 4.<\/p>\n The product of the roots is\u00a0$\\left(2+\\sqrt{\\ 3}\\right)\\cdot\\left(2-\\sqrt{\\ 3}\\right)\\ =\\ 1$<\/p>\n b\/a = 4, c\/a = 1.<\/p>\n b = 4*a, c= a.<\/p>\n Since b =\u00a0$c^3$<\/p>\n 4*a =\u00a0$a^3$<\/p>\n $a^2=\\ 4.$<\/p>\n a = 2 or -2.<\/p>\n |a| = 2<\/p>\n Question 5:\u00a0<\/b>If r is a constant such that $\\mid x^2 – 4x – 13 \\mid = r$ has exactly three distinct real roots, then the value of r is<\/p>\n a)\u00a017<\/p>\n b)\u00a021<\/p>\n c)\u00a015<\/p>\n d)\u00a018<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> The quadratic equation of the form\u00a0$\\mid x^2 – 4x – 13 \\mid = r$ has its minimum value at x = -b\/2a, and hence does not vary irrespective of the value of x.<\/p>\n Hence at x = 2 the quadratic equation has its minimum.<\/p>\n Considering the quadratic part :\u00a0$\\left|x^2-4\\cdot x-13\\right|$. as per the given condition, this must-have 3 real roots.<\/p>\n The curve ABCDE represents the function\u00a0$\\left|x^2-4\\cdot x-13\\right|$. Because of the modulus function, the representation of the quadratic equation becomes :<\/p>\n ABC’DE.<\/p>\n There must exist a value, r such that there must exactly be 3 roots for the function. If r = 0 there will only be 2 roots, similarly for other values there will either be 2 or 4 roots unless at the point C’.<\/p>\n The point C’ is a reflection of C about the x-axis. r is the y coordinate of the point C’ :<\/p>\n The point C which is the value of the function at x = 2, =\u00a0$2^2-8-13$<\/p>\n = -17, the reflection about the x-axis is 17.<\/p>\n Alternatively,<\/p>\n $\\mid x^2 – 4x – 13 \\mid = r$ .<\/p>\n This can represented in two parts :<\/p>\n $x^2-4x-13\\ =\\ r\\ if\\ r\\ is\\ positive.$<\/p>\n $x^2-4x-13\\ =\\ -r\\ if\\ r\\ is\\ negative.$<\/p>\n Considering the first case :\u00a0$x^2-4x-13\\ =r$<\/p>\n The quadraticequation becomes :\u00a0$x^2-4x-13-r\\ =\\ 0$<\/p>\n The discriminant for this function is :\u00a0$b^2-4ac\\ =\\ 16-\\ \\left(4\\cdot\\left(-13-r\\right)\\right)=68+4r$<\/p>\n SInce r is positive the discriminant is always greater than 0 this must have two distinct roots.<\/p>\n For the second case :<\/p>\n $x^2-4x-13+r\\ =\\ 0$ the function inside the modulus is negaitve<\/p>\n The discriminant is\u00a0$16\\ -\\ \\left(4\\cdot\\left(r-13\\right)\\right)\\ =\\ 68-4r$<\/p>\n In order to have a total of 3 roots, the discriminant must be equal to zero for this quadratic equation to have a total of 3 roots.<\/p>\n Hence\u00a0$\\ 68-4r\\ =\\ 0$<\/p>\n r = 17, for r = 17 we can have exactly 3 roots.<\/p>\n Question 6:\u00a0<\/b>Suppose hospital A admitted 21 less Covid infected patients than hospital B, and all eventually recovered. The sum of recovery days for patients in hospitals A and B were 200 and 152, respectively. If the average recovery days for patients admitted in hospital A was 3 more than the average in hospital B then the number admitted in hospital A was<\/p>\n 6)\u00a0Answer:\u00a035<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the number of Covid patients in Hospitals A and B be x and x+21 respectively. Then, it has been given that:<\/p>\n $\\frac{200}{x}-\\frac{152}{x+21}=3$<\/p>\n $\\frac{\\left(200x+4200-152x\\right)}{x\\left(x+21\\right)}=3$<\/p>\n $\\frac{\\left(48x+4200\\right)}{x\\left(x+21\\right)}=3$<\/p>\n $16x+1400=x\\left(x+21\\right)$<\/p>\n $x^2+5x-1400=0$<\/p>\n (x+40)(x-35)=0<\/p>\n Hence, x=35.<\/p>\n Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n Question 7:\u00a0<\/b>The number of integers that satisfy the equality $(x^{2}-5x+7)^{x+1}=1$ is<\/p>\n a)\u00a03<\/p>\n b)\u00a02<\/p>\n c)\u00a04<\/p>\n d)\u00a05<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $\\left(x^2-5x+7\\right)^{x+1}=1$<\/p>\n There can be a solution when\u00a0$\\left(x^2-5x+7\\right)=1$ or $x^2-5x\\ +6=0$<\/p>\n or x=3 and x=2<\/p>\n There can also be a solution when x+1 = 0 or x=-1<\/p>\n Hence three possible solutions exist.<\/p>\n Question 8:\u00a0<\/b>Let m and n be positive integers, If $x^{2}+mx+2n=0$ and $x^{2}+2nx+m=0$ have real\u00a0roots, then the smallest possible value of $m+n$ is<\/p>\n a)\u00a07<\/p>\n b)\u00a06<\/p>\n c)\u00a08<\/p>\n d)\u00a05<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n To have real roots the discriminant should be greater than or equal to 0.<\/p>\n So, $m^2-8n\\ge0\\ \\&\\ 4n^2-4m\\ge0$<\/p>\n => $m^2\\ge8n\\ \\&\\ n^2\\ge m$<\/p>\n Since m,n are positive integers the value of m+n will be minimum when m=4 and n=2.<\/p>\n .’. m+n=6.<\/p>\n Question 9:\u00a0<\/b>The number of distinct real roots of the equation $(x+\\frac{1}{x})^{2}-3(x+\\frac{1}{x})+2=0$ equals<\/p>\n 9)\u00a0Answer:\u00a01<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let $a=x+\\frac{1}{x}$ If $a=1$, $x+\\frac{1}{x}=1$ and it has no real roots. So, the total number of distinct real roots of the given equation is 1<\/p>\n Question 10:\u00a0<\/b>How many disticnt positive integer-valued solutions exist to the equation $(x^{2}-7x+11)^{(x^{2}-13x+42)}=1$ ?<\/p>\n a)\u00a08<\/p>\n b)\u00a04<\/p>\n c)\u00a02<\/p>\n d)\u00a06<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $(x^{2}-7x+11)^{(x^{2}-13x+42)}=1$<\/p>\n if\u00a0$(x^{2}-13x+42)$=0 or\u00a0$(x^{2}-7x+11)$=1 or\u00a0$(x^{2}-7x+11)$=-1 and\u00a0$(x^{2}-13x+42)$ is even number<\/p>\n For x=6,7 the value $(x^{2}-13x+42)$=0<\/p>\n $(x^{2}-7x+11)$=1 for x=5,2.<\/p>\n $(x^{2}-7x+11)$=-1 for x=3,4 and for X=3 or 4,\u00a0$(x^{2}-13x+42)$ is even number.<\/p>\n .’. {2,3,4,5,6,7} is the solution set of x.<\/p>\n .’. x can take six values.<\/p>\n Question 11:\u00a0<\/b>The number of solutions to the equation $\\mid x \\mid (6x^2 + 1) = 5x^2$ is<\/p>\n 11)\u00a0Answer:\u00a05<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n For x <0, -x($6x^2+1$) =\u00a0$5x^2$<\/p>\n =>\u00a0($6x^2+1$) = -5x<\/p>\n =>\u00a0($6x^2 + 5x+ 1$) = 0<\/p>\n =>($6x^2 + 3x+2x+ 1$) = 0<\/p>\n => (3x+1)(2x+1)=0\u00a0 \u00a0 =>x=$\\ -\\frac{\\ 1}{3}$\u00a0 or x=$\\ -\\frac{\\ 1}{2}$<\/p>\n For x=0, LHS=RHS=0\u00a0 \u00a0 (Hence, 1 solution)<\/p>\n For x >0, x($6x^2+1$) = $5x^2$<\/p>\n => ($6x^2 – 5x+ 1$) = 0<\/p>\n =>(3x-1)(2x-1)=0\u00a0 \u00a0 =>x=$\\ \\frac{\\ 1}{3}$\u00a0 \u00a0or\u00a0 \u00a0x=$\\ \\frac{\\ 1}{2}$<\/p>\n Hence, the total number of solutions = 5<\/p>\n Question 12:\u00a0<\/b>The product of the distinct roots of $\\mid x^2 – x – 6 \\mid = x + 2$ is<\/p>\n a)\u00a0\u221216<\/p>\n b)\u00a0-4<\/p>\n c)\u00a0-24<\/p>\n d)\u00a0-8<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have,\u00a0$\\mid x^2 – x – 6 \\mid = x + 2$<\/p>\n => |(x-3)(x+2)|=x+2<\/p>\n For x<-2, (3-x)(-x-2)=x+2<\/p>\n => x-3=1\u00a0 \u00a0=>x=4 (Rejected as x<-2)<\/p>\n For -2$\\le\\ $x<3,\u00a0(3-x)(x+2)=x+2\u00a0 \u00a0 =>x=2,-2<\/p>\n For x$\\ge\\ $3,\u00a0(x-3)(x+2)=x+2\u00a0 \u00a0=>x=4<\/p>\n Hence the product =4*-2*2=-16<\/p>\n Question 13:\u00a0<\/b>The quadratic equation $x^2 + bx + c = 0$ has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of $b^2 + c$?<\/p>\n a)\u00a03721<\/p>\n b)\u00a0361<\/p>\n c)\u00a0427<\/p>\n d)\u00a0549<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n The quadratic equation $x^2 + bx + c = 0$ has two roots 4a and 3a<\/p>\n 7a=-b<\/p>\n 12$a^2$ = c<\/p>\n We have to find the value of\u00a0 $b^2 + c$ = 49$a^2$+ 12$a^2$=61$a^2$<\/p>\n Now lets verify the options<\/p>\n 61$a^2$ = 3721 ==> a= 7.8 which is not an integer<\/p>\n 61$a^2$ = 361 ==> a= 2.42 which is not an integer<\/p>\n 61$a^2$ = 427 ==> a= 2.64 which is not an integer<\/p>\n 61$a^2$ = 3721 ==> a= 3 which is an integer<\/p>\n Question 14:\u00a0<\/b>Let A be a real number. Then the roots of the equation $x^2 – 4x – log_{2}{A} = 0$ are real and distinct if and only if<\/p>\n a)\u00a0$A > \\frac{1}{16}$<\/p>\n b)\u00a0$A < \\frac{1}{16}$<\/p>\n c)\u00a0$A < \\frac{1}{8}$<\/p>\n d)\u00a0$A > \\frac{1}{8}$<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The roots of\u00a0$x^2 – 4x – log_{2}{A} = 0$ will be real and distinct if and only if the discriminant is greater than zero<\/p>\n 16+4*$log_{2}{A}$ > 0<\/p>\n $log_{2}{A}$ > -4<\/p>\n A> 1\/16<\/p>\n Question 15:\u00a0<\/b>If a and b are integers such that $2x^2\u2212ax+2>0$ and $x^2\u2212bx+8\u22650$ for all real numbers $x$, then the largest possible value of $2a\u22126b$ is<\/p>\n 15)\u00a0Answer:\u00a036<\/b><\/p>\n
\n<\/strong><\/p>\n
\n<\/strong><\/p>\n
\n<\/strong><\/p>\n
\n$x^2-xy-x\\ =22\\ \\ \\ \\ \\ \\ \\left(1\\right)$
\nAnd\u00a0$y^2-xy+y\\ =34\\ \\ \\ \\ \\ \\ \\left(1\\right)$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (2)
\nAdding (1) and (2)
\nwe get $x^2-2xy+y^2-x+y\\ =56$
\nwe get $\\left(x-y\\right)^2-\\ \\left(x-y\\right)\\ =56$
\nLet (x-y) =t
\nwe get $t^2-t=56$
\n$t^2-t-56=0$
\n(t-8)(t+7) =0
\nso t=8
\nso x-y =8<\/p>\n
\n![](\"https:\/\/media-cdn.cracku.in\/uploads\/image_7Uarbxv.png\")
<\/p>\n
\nSo, the given equation is $a^2-3a+2=0$
\nSo, $a$ can be either 2 or 1.<\/p>\n
\nIf $a=2$, $x+\\frac{1}{x}=2$ and it has exactly one real root which is $x=1$<\/p>\n