# CAT Quadratic Equation Questions (with Notes) PDF

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Quadratic Equations is one of the key topics in the CAT Quants Section (Algebra). CAT Quadratic Equation questions appear in the CAT and other MBA entrance exams every year. You can check out these CAT Quadratic Equation Questions from Previous years. In this article, we will look into some important CAT Quadratic Equation Questions (with Notes) PDF. These are a good source for practice; If you want to practice these questions, you can download the CAT Quadratic Equation Questions PDF below, which is completely Free.

• CAT Quadratic Equation Questions – Tip 1: Be thorough with all the basics of this topic. Understand the important formulas well. If you’re starting the prep, firstly understand the CAT Algebra Syllabus; CAT quadratic equation questions are important and should not be avoided.
• CAT Quadratic Equation Questions – Tip 2: Based on our analysis of the previous year’s CAT questions,Â  this was the CAT Quadratic Equations weightage: 1-2 questions were asked on this topic (in CAT 2021).
• Practice these CAT quadratic equation questions PDF. Learn all the major formulae from these concepts. You can check out the Important CAT quadratic equation questions & Formulas PDF here.

Question 1:Â A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f (x) at x = 10?

a)Â -119

b)Â -159

c)Â -110

d)Â -180

e)Â -105

Solution:

Let the function be $ax^2 + bx + c$.

We know that x=0 value is 1 so c=1.

So equation is $ax^2 + bx + 1$.

Now max value is 3 at x = 1.

So after substituting we get a + b = 2.

If f(x) attains a maximum at ‘a’ then the differential of f(x) at x=a, that is, f'(a)=0.

So in this question f'(1)=0

=> 2*(1)*a+b = 0

=> 2a+b = 0.

Solving the equations we get a=-2 and b=4.

$-2x^2 + 4x + 1$ is the equation and on substituting x=10, we get -159.

Question 2:Â If the roots of the equation $x^3 – ax^2 + bx – c = 0$ are three consecutive integers, then what is the smallest possible value of b?
[CAT 2008]

a)Â $\frac{-1}{\sqrt 3}$

b)Â $-1$

c)Â $0$

d)Â $1$

e)Â $\frac{1}{\sqrt 3}$

Solution:

b = sum of the roots taken 2 at a time.
Let the roots be n-1, n and n+1.
Therefore, $b = (n-1)n + n(n+1) + (n+1)(n-1) = n^2 – n + n^2 + n + n^2 – 1$
$b = 3n^2 – 1$. The smallest value is -1.

Question 3:Â Let p and q be the roots of the quadratic equation $x^2 – (\alpha – 2) x – \alpha -1= 0$ . What is the minimum possible value of $p^2 + q^2$?

a)Â 0

b)Â 3

c)Â 4

d)Â 5

Solution:

Let $\alpha$ be equal to k.

=> f(x) = $x^2-(k-2)x-(k+1) = 0$

p and q are the roots

=> p+q = k-2 and pq = -1-k

We know that $(p+q)^2 = p^2 + q^2 + 2pq$

=> $(k-2)^2 = p^2 + q^2 + 2(-1-k)$

=> $p^2 + q^2 = k^2 + 4 – 4k + 2 + 2k$

=> $p^2 + q^2 = k^2 – 2k + 6$

This is in the form of a quadratic equation.

The coefficient of $k^2$ is positive. Therefore this equation has a minimum value.

We know that the minimum value occurs at x = $\frac{-b}{2a}$

Here a = 1, b = -2 and c = 6

=> Minimum value occurs at k = $\frac{2}{2}$ = 1

If we substitute k = 1 in $k^2-2k+6$, we get 1 -2 + 6 = 5.

Hence 5 is the minimum value that $p^2+q^2$ can attain.

Question 4:Â Ujakar and Keshab attempted to solve a quadratic equation. Ujakar made a mistake in writing down the constant term. He ended up with the roots (4, 3). Keshab made a mistake in writing down the coefficient of x. He got the roots as (3, 2). What will be the exact roots of the original quadratic equation?

a)Â (6, 1)

b)Â (-3, -4)

c)Â (4, 3)

d)Â (-4, -3)

Solution:

We know that quadratic equation can be written as $x^2$-(sum of roots)*x+(product of the roots)=0.
Ujakar ended up with the roots (4, 3) so the equation is $x^2$-(7)*x+(12)=0 where the constant term is wrong.

Keshab got the roots as (3, 2) so the equation is $x^2$-(5)*x+(6)=0 where the coefficient of x is wrong .

So the correct equation is $x^2$-(7)*x+(6)=0. The roots of above equations are (6,1).

Question 5:Â If the roots $x_1$ and $x_2$ are the roots of the quadratic equation $x^2 -2x+c=0$ also satisfy the equation $7x_2 – 4x_1 = 47$, then which of the following is true?

a)Â c = -15

b)Â $x_1 = -5$ and $x_2=3$

c)Â $x_1 = 4.5$ and $x_2 = -2.5$

d)Â None of these

Solution:

$x_1 + x_2 = 2$
and $7x_2 – 4x_1 = 47$
So $x_1 = -3$ and $x_2 = 5$
And $c = x_1 \times x_2 = -15$

Question 6:Â If $x+1=x^{2}$ and $x>0$, then $2x^{4}$Â  is

a)Â $6+4\sqrt{5}$

b)Â $3+3\sqrt{5}$

c)Â $5+3\sqrt{5}$

d)Â $7+3\sqrt{5}$

Solution:

We know that $x^2 – x – 1=0$
Therefore $x^4 = (x+1)^2 = x^2+2x+1 = x+1 + 2x+1 = 3x+2$
Therefore, $2x^4 = 6x+4$

We know that $x>0$ therefore, we can calculate the value of $x$ to be $\frac{1+\sqrt{5}}{2}$
Hence, $2x^4 = 6x+4 = 3+3\sqrt{5}+4 = 3\sqrt{5}+7$

Question 7:Â If $U^{2}+(U-2V-1)^{2}$= âˆ’$4V(U+V)$ , then what is the value of $U+3V$ ?

a)Â $0$

b)Â $\dfrac{1}{2}$

c)Â $\dfrac{-1}{4}$

d)Â $\dfrac{1}{4}$

Solution:

Given that $U^{2}+(U-2V-1)^{2}$= âˆ’$4V(U+V)$

$\Rightarrow$ $U^{2}+(U-2V-1)(U-2V-1)$= âˆ’$4V(U+V)$

$\Rightarrow$ $U^{2}+(U^2-2UV-U-2UV+4V^2+2V-U+2V+1)$ =Â âˆ’$4V(U+V)$

$\Rightarrow$ $U^{2}+(U^2-4UV-2U+4V^2+4V+1)$ =Â âˆ’$4V(U+V)$

$\Rightarrow$ $2U^2-4UV-2U+4V^2+4V+1=âˆ’4UV-4V^2$

$\Rightarrow$ $2U^2-2U+8V^2+4V+1=0$

$\Rightarrow$ $2[U^2-U+\dfrac{1}{4}]+8[V^2+\dfrac{V}{2}+\dfrac{1}{16}]=0$

$\Rightarrow$ $2(U-\dfrac{1}{2})^2+8(V+\dfrac{1}{4})^2=0$

Sum of two square terms is zero i.e. individual square term is equal to zero.

$U-\dfrac{1}{2}$ = 0 and $V+\dfrac{1}{4}$ = 0

U = $\dfrac{1}{2}$ andÂ V = $-\dfrac{1}{4}$

Therefore,Â $U+3V$ =Â $\dfrac{1}{2}$+$\dfrac{-1*3}{4}$ =Â $\dfrac{-1}{4}$. Hence, option C is the correct answer.

Question 8:Â If a and b are integers such that $2x^2âˆ’ax+2>0$ and $x^2âˆ’bx+8â‰¥0$ for all real numbers $x$, then the largest possible value of $2aâˆ’6b$ is

Solution:

Let f(x) = $2x^2âˆ’ax+2$. We can see that f(x) is a quadratic function.

For, f(x) > 0, Discriminant (D) < 0

$\Rightarrow$ $(-a)^2-4*2*2<0$

$\Rightarrow$ (a-4)(a+4)<0

$\Rightarrow$ a $\epsilon$ (-4, 4)

Therefore, integer values that ‘a’ can take = {-3, -2, -1, 0, 1, 2, 3}

Let g(x) = $x^2âˆ’bx+8$. We can see that g(x) is also a quadratic function.

For, g(x)â‰¥0, Discriminant (D) $\leq$ 0

$\Rightarrow$ $(-b)^2-4*8*1<0$

$\Rightarrow$ $(b-\sqrt{32})(b+\sqrt{32})<0$

$\Rightarrow$ bÂ $\epsilon$ (-$\sqrt{32}$, $\sqrt{32}$)

Therefore, integer values that ‘b’ can take = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

We have to find out theÂ largest possible value of $2aâˆ’6b$. The largest possible value will occur when ‘a’ is maximum and ‘b’ is minimum.

a$_{max}$ = 3,Â b$_{min}$ = -5

Therefore, theÂ largest possible value of $2aâˆ’6b$ = 2*3 – 6*(-5) = 36.

Question 9:Â 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

a)Â $A > \frac{1}{16}$

b)Â $A < \frac{1}{16}$

c)Â $A < \frac{1}{8}$

d)Â $A > \frac{1}{8}$

Solution:

The roots ofÂ $x^2 – 4x – log_{2}{A} = 0$ will be real and distinct if and only if the discriminant is greater than zero

16+4*$log_{2}{A}$ > 0

$log_{2}{A}$ > -4

A> 1/16

Question 10:Â 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 $b^2 + c$?

a)Â 3721

b)Â 361

c)Â 427

d)Â 549

Solution:

Given,

The quadratic equation $x^2 + bx + c = 0$ has two roots 4a and 3a

7a=-b

12$a^2$ = c

We have to find the value ofÂ  $b^2 + c$ = 49$a^2$+ 12$a^2$=61$a^2$

Now lets verify the options

61$a^2$ = 3721 ==> a= 7.8 which is not an integer

61$a^2$ = 361 ==> a= 2.42 which is not an integer

61$a^2$ = 427 ==> a= 2.64 which is not an integer

61$a^2$ = 3721 ==> a= 3 which is an integer

Question 11:Â The product of the distinct roots of $\mid x^2 – x – 6 \mid = x + 2$ is

a)Â âˆ’16

b)Â -4

c)Â -24

d)Â -8

Solution:

We have,Â $\mid x^2 – x – 6 \mid = x + 2$

=> |(x-3)(x+2)|=x+2

For x<-2, (3-x)(-x-2)=x+2

=> x-3=1Â  Â =>x=4 (Rejected as x<-2)

For -2$\le\$x<3,Â (3-x)(x+2)=x+2Â  Â  =>x=2,-2

For x$\ge\$3,Â (x-3)(x+2)=x+2Â  Â =>x=4

Hence the product =4*-2*2=-16

Question 12:Â The number of solutions to the equation $\mid x \mid (6x^2 + 1) = 5x^2$ is

Solution:

For x <0, -x($6x^2+1$) =Â $5x^2$

=>Â ($6x^2+1$) = -5x

=>Â ($6x^2 + 5x+ 1$) = 0

=>($6x^2 + 3x+2x+ 1$) = 0

=> (3x+1)(2x+1)=0Â  Â  =>x=$\ -\frac{\ 1}{3}$Â  or x=$\ -\frac{\ 1}{2}$

For x=0, LHS=RHS=0Â  Â  (Hence, 1 solution)

For x >0, x($6x^2+1$) = $5x^2$

=> ($6x^2 – 5x+ 1$) = 0

=>(3x-1)(2x-1)=0Â  Â  =>x=$\ \frac{\ 1}{3}$Â  Â orÂ  Â x=$\ \frac{\ 1}{2}$

Hence, the total number of solutions = 5

Question 13:Â How many disticnt positive integer-valued solutions exist to the equation $(x^{2}-7x+11)^{(x^{2}-13x+42)}=1$ ?

a)Â 8

b)Â 4

c)Â 2

d)Â 6

Solution:

$(x^{2}-7x+11)^{(x^{2}-13x+42)}=1$

ifÂ $(x^{2}-13x+42)$=0 orÂ $(x^{2}-7x+11)$=1 orÂ $(x^{2}-7x+11)$=-1 andÂ $(x^{2}-13x+42)$ is even number

For x=6,7 the value $(x^{2}-13x+42)$=0

$(x^{2}-7x+11)$=1 for x=5,2.

$(x^{2}-7x+11)$=-1 for x=3,4 and for X=3 or 4,Â $(x^{2}-13x+42)$ is even number.

.’. {2,3,4,5,6,7} is the solution set of x.

.’. x can take six values.

Question 14:Â The number of distinct real roots of the equation $(x+\frac{1}{x})^{2}-3(x+\frac{1}{x})+2=0$ equals

Solution:

Let $a=x+\frac{1}{x}$
So, the given equation is $a^2-3a+2=0$
So, $a$ can be either 2 or 1.

If $a=1$, $x+\frac{1}{x}=1$ and it has no real roots.
If $a=2$, $x+\frac{1}{x}=2$ and it has exactly one real root which is $x=1$

So, the total number of distinct real roots of the given equation is 1

Question 15:Â Let m and n be positive integers, If $x^{2}+mx+2n=0$ and $x^{2}+2nx+m=0$ have realÂ roots, then the smallest possible value of $m+n$ is

a)Â 7

b)Â 6

c)Â 8

d)Â 5

Solution:

To have real roots the discriminant should be greater than or equal to 0.

So, $m^2-8n\ge0\ \&\ 4n^2-4m\ge0$

=> $m^2\ge8n\ \&\ n^2\ge m$

Since m,n are positive integers the value of m+n will be minimum when m=4 and n=2.

.’. m+n=6.

Question 16:Â The number of integers that satisfy the equality $(x^{2}-5x+7)^{x+1}=1$ is

a)Â 3

b)Â 2

c)Â 4

d)Â 5

Solution:

$\left(x^2-5x+7\right)^{x+1}=1$

There can be a solution whenÂ $\left(x^2-5x+7\right)=1$ or $x^2-5x\ +6=0$

or x=3 and x=2

There can also be a solution when x+1 = 0 or x=-1

Hence three possible solutions exist.

Question 17:Â 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

Solution:

Let the number of Covid patients in Hospitals A and B be x and x+21 respectively. Then, it has been given that:

$\frac{200}{x}-\frac{152}{x+21}=3$

$\frac{\left(200x+4200-152x\right)}{x\left(x+21\right)}=3$

$\frac{\left(48x+4200\right)}{x\left(x+21\right)}=3$

$16x+1400=x\left(x+21\right)$

$x^2+5x-1400=0$

(x+40)(x-35)=0

Hence, x=35.

Question 18:Â 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

a)Â 17

b)Â 21

c)Â 15

d)Â 18

Solution:

The quadratic equation of the formÂ $\mid x^2 – 4x – 13 \mid = r$ has its minimum value at x = -b/2a, and hence does not vary irrespective of the value of x.

Hence at x = 2 the quadratic equation has its minimum.

Considering the quadratic part :Â $\left|x^2-4\cdot x-13\right|$. as per the given condition, this must-have 3 real roots.

The curve ABCDE represents the functionÂ $\left|x^2-4\cdot x-13\right|$. Because of the modulus function, the representation of the quadratic equation becomes :

ABC’DE.

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’.

The point C’ is a reflection of C about the x-axis. r is the y coordinate of the point C’ :

The point C which is the value of the function at x = 2, =Â $2^2-8-13$

= -17, the reflection about the x-axis is 17.

Alternatively,

$\mid x^2 – 4x – 13 \mid = r$ .

This can represented in two parts :

$x^2-4x-13\ =\ r\ if\ r\ is\ positive.$

$x^2-4x-13\ =\ -r\ if\ r\ is\ negative.$

Considering the first case :Â $x^2-4x-13\ =r$

The quadraticequation becomes :Â $x^2-4x-13-r\ =\ 0$

The discriminant for this function is :Â $b^2-4ac\ =\ 16-\ \left(4\cdot\left(-13-r\right)\right)=68+4r$

SInce r is positive the discriminant is always greater than 0 this must have two distinct roots.

For the second case :

$x^2-4x-13+r\ =\ 0$ the function inside the modulus is negaitve

The discriminant isÂ $16\ -\ \left(4\cdot\left(r-13\right)\right)\ =\ 68-4r$

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.

HenceÂ $\ 68-4r\ =\ 0$

r = 17, for r = 17 we can have exactly 3 roots.

Question 19:Â 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.

a)Â 1

b)Â 2

c)Â 3

d)Â 4

Solution:

Given a, b, c are rational numbers.

Hence a, b, c are three numbers that can be written in the form of p/q.

Hence if one both the root is 2+$\sqrt{\ 3}$ and considering the other root to be x.

The sum of the roots and the product of the two roots must be rational numbers.

For this to happen the other root must be the conjugate ofÂ $2+\sqrt{\ 3}$ so the product and the sum of the roots are rational numbers which are represented by:Â $\frac{b}{a},\ \frac{c}{a}$

Hence the sum of the roots is 2+$\sqrt{\ 3}+2-\sqrt{\ 3}$ = 4.

The product of the roots isÂ $\left(2+\sqrt{\ 3}\right)\cdot\left(2-\sqrt{\ 3}\right)\ =\ 1$

b/a = 4, c/a = 1.

b = 4*a, c= a.

Since b =Â $c^3$

4*a =Â $a^3$

$a^2=\ 4.$

a = 2 or -2.

|a| = 2

Question 20:Â For all real values of x, the range of the function $f(x)=\frac{x^{2}+2x+4}{2x^{2}+4x+9}$ is:

a)Â $[\frac{4}{9},\frac{8}{9}]$

b)Â $[\frac{3}{7},\frac{8}{9})$

c)Â $(\frac{3}{7},\frac{1}{2})$

d)Â $[\frac{3}{7},\frac{1}{2})$

Solution:

$f(x)=\frac{x^{2}+2x+4}{2x^{2}+4x+9}$

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.

$f(x)=\frac{x^2+2x+4}{2x^2+4x+9}=\frac{x^2+2x+4.5-0.5}{2x^2+4x+9}$

=Â $\frac{x^2+2x+4.5}{2x^2+4x+9}-\frac{0.5}{2x^2+4x+9}$

=Â $\frac{1}{2}-\frac{0.5}{2x^2+4x+9}$

Now, we only have terms of x in the denominator.

The maximum value of the expression is achieved when the quadratic expression $2x^2+4x+9$ achieves its highest value, that is infinity.

In that case, the second term becomes zero and the expression becomes 1/2. However, at infinity, there is always an open bracket ‘)’.

To obtain the minimum value, we need to find the minimum possible value of the quadratic expression.

The minimum value is obtained when 4x + 4 = 0 [d/dx = 0]

x=-1.

The expression comes as 7.

The entire expression becomes 3/7.

Hence,Â $[\frac{3}{7},\frac{1}{2})$

Question 21:Â Consider the pair of equations: $x^{2}-xy-x=22$ and $y^{2}-xy+y=34$. If $x>y$, then $x-y$ equals

a)Â 6

b)Â 4

c)Â 7

d)Â 8

Solution:

We have :
$x^2-xy-x\ =22\ \ \ \ \ \ \left(1\right)$
AndÂ $y^2-xy+y\ =34\ \ \ \ \ \ \left(1\right)$ Â  Â  Â  Â  Â  Â  Â  (2)
we get $x^2-2xy+y^2-x+y\ =56$
we get $\left(x-y\right)^2-\ \left(x-y\right)\ =56$
Let (x-y) =t
we get $t^2-t=56$
$t^2-t-56=0$
(t-8)(t+7) =0
so t=8
so x-y =8

Question 22:Â 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

a)Â 175

b)Â 150

c)Â 200

d)Â 225

Solution:

Let the number of large shirts be l and the number of small shirts be s.

Let the price of a small shirt be x and that of a large shirt be x + 50.

Now, s + l = 64

l (x+50) = 5000

sx = 1800

lx + sx + 50l = 6800

64x +50l = 6800

Substituting l = (6800 – 64x) / 50, in the original equation, we get

$\frac{\left(6800-64x\right)}{50}\left(x+50\right)=5000$

(6800 – 64x)(x +Â 50) = 250000

$6800x+340000-64x^2-3200x = 250000$

$64x^2-3600x-90000=0$

Solving, we get, x=Â $\frac{225\pm\ 375}{8}=\frac{600}{8}or-\frac{150}{8}$

SO, x = 75

x +Â 50 = 125

Answer = 75 + 125 = 200.

Alternate approach: By options.

Hint: Each option gives the sum of the costs of one large and one small shirt. We know that large = small + 50

Hence, small + small + 50 = option.

SMALL = (Option – 50)/2

LARGE = Small +Â 50 = (Option + 50)/2

Option A and Option D gives us decimal values for SMALL and LARGE, hence we will consider them later.

Large = 150 + 50 / 2 = 100

Small = 150 – 50 / 2 = 50

Now, total shirts = 5000/100 + 1800/50 = 50 + 36 = 86 (X – This is wrong)

Option C –

Large = 200 + 50 / 2 = 125

Small = 200 – 50 / 2 = 75

Total shirts = 5000/125 + 1800/75 = 40 + 24 = 64 ( This is the right answer)

Question 23:Â For which value of k does the following pair of equations yield a unique solution for x such that the solution is positive?
$x^2 – y^2 = 0$
$(x-k)^2 + y^2 = 1$

a)Â 2

b)Â 0

c)Â $\sqrt2$

d)Â $-2\sqrt2$

Solution:

From 1st equation we know that $(x)^2 = y^2$

Substituting this in 2nd equation. we get , $2*x^2 – 2*x*k + k^2-1 =0$ and for unique solution $b^2-4ac=0$ must satisfy.

This is possible only when k = $\sqrt{2}$

Question 24:Â If the equation $x^3 – ax^2 + bx – a = 0$ has three real roots, then it must be the case that,

a)Â b=1

b)Â b $\neq$ 1

c)Â a=1

d)Â a $\neq$ 1

Solution:

It can be clearly seen that if b=1 then $x^2(x – a) + (x – a) = 0$ an the equation gives only 1 realÂ valueÂ of x

Question 25:Â The number of real roots of the equation $A^2/x + B^2/(x-1) = 1$ , where A and B are real numbers not equal to zero simultaneously, is

a)Â 3

b)Â 1

c)Â 2

d)Â Cannot be determined

Solution:

The given equation can be written as : $A^2 * (x-1) + B^2 * x = x^2 – x$
=> $x^2 + x(-1 – A^2 – B^2) + A^2 = 0$
Discriminant of the equation = $(-1 – A^2 – B^2) ^2 – 4A^2$
=$A^4 + B^4 + 1 – 2A^2 + 2B^2 + 2A^2B^2$
= $A^4 + B^4 + 1 – 2A^2 – 2B^2 + 2A^2B^2 + 4B^2$
= $(A^2 + B^2 – 1)^2 + 4B^2$
>= 0, 0 when B =0 and A =1
Hence, the number of roots can be 1 or 2.
Option d) is the correct answer.

Question 26:Â Davji Shop sells samosas in boxes of different sizes. The samosas are priced at Rs. 2 per samosa up to 200 samosas. For every additional 20 samosas, the price of the whole lot goes down by 10 paise per samosa. What should be the maximum size of the box that would maximise the revenue?

a)Â 240

b)Â 300

c)Â 400

d)Â None of these

Solution:

Let the optimum number of samosas be 200+20n

So, price of each samosa = (2-0.1*n)

Total price of all samosas = (2-0.1*n)*(200+20n) = $400 – 20n + 40n – 2n^2$ = $400 + 20n – 2n^2$

This quadratic equation attains a maximum at n = -20/2*(-2) = 5

So, the number of samosas to get the maximum revenue = 200 + 20*5 = 300

Question 27:Â The value of $\frac{(1-d^3)}{(1-d)}$ is

a)Â > 1 if d > -1

b)Â > 3 if d > 1

c)Â > 2 if 0 < d < 0.5

d)Â < 2 if d < -2

Solution:

$\frac{(1-d^3)}{(1-d)}$ = $1+d^2+d$ (where $d \neq 1$)
Let’s say $f(d)=1+d^2+d$
Now $f(d)$ will always be greater than 0 and have its minimum value at d = -0.5. The value is $\frac{3}{4}$.
For $d<-1$ ; $f(d) >1$
$-1<d<0$ ; $\frac{3}{4} <f(d)<1$
$0<d<1$ ; $1<f(d)<3$
$d>1$ ; $f(d)>3$

So, for d > 1, f(d) > 3. Option b) is the correct answer.

Question 28:Â The roots of the equation $ax^{2} + 3x + 6 = 0$ will be reciprocal to each other if the value of a is

a)Â 3

b)Â 4

c)Â 5

d)Â 6

Solution:

If roots of given equation are reciprocal to each other than product of roots should be equal to 1.
i.e. $\frac{6}{a} = 1$
hence a=6

Question 29:Â If $xy + yz + zx = 0$, then $(x + y + z)^2$ equals

a)Â $(x + y)^2 + xz$

b)Â $(x + z)^2 + xy$

c)Â $x^2 + y^2 + z^2$

d)Â $2(xy + yz + xz)$

Solution:

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$
as $xy+yz+xz = 0$
so equation will be resolved to $x^2 + y^2 + z^2$

Question 30:Â Given the quadratic equation $x^2 – (A – 3)x – (A – 2)$, for what value of $A$ will the sum of the squares of the roots be zero?

a)Â -2

b)Â 3

c)Â 6

d)Â None of these