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# Complete Algebra Previous Questions (CAT PYQs 2017-22) [Download PDF]

Algebra is one of the important topic in Quant Section of the CAT Exam. You can check out these AlgebraÂ Practice a good amount of questions in the CAT Algebra section. This article will look into some important Algebra Questions for CAT. These are good sources for practice; If you want to practice these questions, you can download this CAT Algebra Questions PDF below, which is completely Free.

Algebra, with its intricate equations and formulas, is a cornerstone of the CAT (Common Admission Test) examination. For CAT aspirants, mastering algebra is not just a choice; it’s a necessity. If you’re embarking on your CAT preparation journey and looking to conquer the realm of algebra, you’ve landed in the right place. In this blog post, we’re excited to present a comprehensive collection of Algebra questions from the CAT papers of 2017 to 2022. These questions aren’t mere exercises; they are actual CAT questions, carefully selected to give you a taste of what to expect on the big day.

The world of CAT Algebra can seem daunting, but it’s a challenge that you can overcome with the right guidance and practice. Our blog will unravel the complexity of CAT Algebra by delving into the Previous Year Questions (PYQs) from 2017 to 2022. We’ll provide step-by-step solutions and expert insights, ensuring that you not only understand the problems but also gain the confidence to tackle any Algebra question that comes your way on the CAT exam.

Get ready to dive deep into the world of CAT Algebra, equip yourself with the tools and strategies, and embark on a journey toward CAT success. Let’s get started!

Check out here for Detailed video solutions for Complete Algebra CAT PYQs (2017-22)

Question 1:Â If a and b are integers of opposite signs such that $(a + 3)^{2} : b^{2} = 9 : 1$ and $(a -1)^{2}:(b – 1)^{2} = 4:1$, then the ratio $a^{2} : b^{2}$ is

a)Â 9:4

b)Â 81:4

c)Â 1:4

d)Â 25:4

1)Â AnswerÂ (D)

Solution:

Since the square root can be positive or negative we will get two cases for each of the equation.

For the first one,

a + 3 = 3b .. i

a + 3 = -3b … ii

For the second one,

a – 1 = 2(b -1) … iii

a – 1 = 2 (1 – b) … iv

we have to solve i and iii, i and iv, ii and iii, ii and iv.

Solving i and iii,

a + 3 = 3b and a = 2bÂ  – 1, solving, we get a = 3 and b = 2, which is not what we want.

Solving i and iv

a + 3 = 3b and a = 3 – 2b, solving, we get b = 1.2, which is not possible.

Solving ii and iii

a + 3 = -3b and a = 2b – 1, solving, we get b = 0.4, which is not possible.

Solving ii and iv,

a + 3 = -3b and a = 3 – 2b, solving, we get a = 15 and b = -6 which is what we want.

Thus, $\frac{a^2}{b^2} = \frac{25}{4}$

Question 2: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 $7+3\sqrt{5}$

2)Â AnswerÂ (D)

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 3:Â The number of solutions $(x, y, z)$ to the equation $x – y – z = 25$, where x, y, and z are positive integers such that $x\leq40,y\leq12$, and $z\leq12$ is

a)Â 101

b)Â 99

c)Â 87

d)Â 105

3)Â AnswerÂ (B)

Solution:

x – y – z = 25 andÂ $x\leq40,y\leq12$,Â $z\leq12$
If x = 40 then y + z = 15. Now since both y and z are natural numbers less than 12, so y can range from 3 to 12 giving us a total of 10 solutions.Similarly, if x = 39, then y +Â z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.
If x = 38, then y +Â z = 13. Now y can range from 1 to 12 giving us a total of 12 solutions.
If x = 37 then y + z = 12 which will give 11 solutions.
Similarly on proceeding in the same manner the number of solutions will be 10, 9, 8, 7 and so on till 1.
Hence, required number of solutions will be (1 + 2Â + 3 + 4 . . . . + 12) + 10 + 11
= 12*13/2 + 21
78 + 21 = 99

Question 4:Â For how many integers n, will the inequality $(n – 5) (n – 10) – 3(n – 2)\leq0$ be satisfied?

4)Â Answer:Â 11

Solution:

$(n – 5) (n – 10) – 3(n – 2)\leq0$
=> $n^2 – 15n + 50 – 3n + 6 \leq 0$
=> $n^2 – 18n + 56 \leq 0$
=> $(n – 4)(n – 14) \leq 0$
=> Thus, n can take values from 4 to 14. Hence, the required number of values are 14 – 4 +Â 1 = 11.

Question 5:Â If $f_{1}(x)=x^{2}+11x+n$ and $f_{2}(x)=x$, then the largest positive integer n for which the equation $f_{1}(x)=f_{2}(x)$ has two distinct real roots is

5)Â Answer:Â 24

Solution:

$f_{1}(x)=x^{2}+11x+n$ andÂ $f_{2}(x) = x$
$f_{1}(x)=f_{2}(x)$
=> $x^{2}+11x+n = x$
=> $x^2 + 10x + n = 0$
=> For this equation to have distinct real roots, b$^2$-4ac>0
$10^2 > 4n$
=> n < 100/4
=> n < 25
Thus, largest integral value that n can take is 24.

Question 6:Â $f(x) = \frac{5x+2}{3x-5}$ and $g(x) = x^2 – 2x – 1$, then the value of $g(f(f(3)))$ is

a)Â 2

b)Â $\frac{1}{3}$

c)Â 6

d)Â $\frac{2}{3}$

6)Â AnswerÂ (A)

Solution:

$f(3) = \frac{15 + 2}{9 – 5} = \frac{17}{4}$
$f(f(3)) = \frac{5*17/4 + 2}{3*17/4 – 5} = \frac{93/4}{31/4} = \frac{93}{31} = 3$
$g(f(f(3))) = 3^2 – 3*2 – 1 = 2$

Question 7:Â Let $f(x) = x^{2}$ and $g(x) = 2^{x}$, for all real x. Then the value of f[f(g(x)) + g(f(x))] at x = 1 is

a)Â 16

b)Â 18

c)Â 36

d)Â 40

7)Â AnswerÂ (C)

Solution:

$f[f(g(1)) + g(f(1))]$
= $f[f(2^1) + g(1^2)]$
= $f[f(2) + g(1)]$
= $f[2^2 + 2^1]$
= $f(6)$
= $6^2 = 36$

Question 8:Â The minimum possible value of the sum of the squares of the roots of the equation $x^2+(a+3)x-(a+5)=0$ is

a)Â 1

b)Â 2

c)Â 3

d)Â 4

8)Â AnswerÂ (C)

Solution:

Let the roots of the equationÂ $x^2+(a+3)x-(a+5)=0$ be equal to $p,q$

Hence, $p+q = -(a+3)$ and $p \times q = -(a+5)$

Therefore, $p^2+q^2 = a^2+6a+9+2a+10 = a^2+8a+19 = (a+4)^2+3$

As $(a+4)^2$ is always non negative, the least value of the sum of squares is 3

Question 9:Â How many different pairs(a,b) of positive integers are there such that $a\geq b$ and $\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$?

9)Â Answer:Â 3

Solution:

$\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$
=> $ab = 9(a + b)$
=> $ab – 9(a+b) = 0$
=> $ab – 9(a+b) + 81 = 81$
=> $(a – 9)(b – 9) = 81, a > b$
Hence we have the following cases,
$a – 9 = 81, b – 9 = 1$ => $(a,b) = (90,10)$
$a – 9 = 27, b – 9 = 3$ => $(a,b) = (36,12)$
$a – 9 = 9, b – 9 = 9$ => $(a,b) = (18,18)$
Hence there are three possible positive integral values of (a,b)

Question 10:Â If f(ab) = f(a)f(b) for all positive integers a and b, then the largest possible value of f(1) is

10)Â Answer:Â 1

Solution:

f(1 * 1) = f(1)f(1)
=> f(1) = f(1)f(1)
=> f(1) = 0 or f(1) = 1
Hence maximum value of f(1) is 1

Question 11:Â Let $f(x) =2x-5$ and $g(x) =7-2x$. Then |f(x)+ g(x)| = |f(x)|+ |g(x)| if and only if

a)Â $\frac{5}{2}<x<\frac{7}{2}$

b)Â $x\leq\frac{5}{2}$ or $x\geq\frac{7}{2}$

c)Â $x<\frac{5}{2}$ or $x\geq\frac{7}{2}$

d)Â $\frac{5}{2}\leq x\leq\frac{7}{2}$

11)Â AnswerÂ (D)

Solution:

$|f(x)+ g(x)| = |f(x)| + |g(x)|$ if and only if

case 1: $f(x) \geq 0$ and $g(x) \geq 0$
<=> $2x-5 \geq 0$ and $7-2x \geq 0$
<=> $x \geq \frac{5}{2}$ and $\frac{7}{2} \geq x$

<=> $\frac{5}{2}\leq x\leq\frac{7}{2}$

case 2: $f(x) \leq 0$ and $g(x) \leq 0$

<=> $2x-5 \leq 0$ and $7-2x \leq 0$
<=> $x \leq \frac{5}{2}$ and $\frac{7}{2} \leq x$
So x<=5/2 and x>=7/2 which is not possible.

Hence, answer is

<=> $\frac{5}{2}\leq x\leq\frac{7}{2}$

Question 12:Â Let f(x) = min (${2x^{2},52-5x}$) where x is any positive real number. Then the maximum possible value of f(x) is

12)Â Answer:Â 32

Solution:

f(x) = min (${2x^{2},52-5x}$)
The maximum possible value of this function will be attained at the point in which $2x^2$ is equal to $52-5x$.

$2x^2 = 52-5x$
$2x^2+5x-52=0$
$(2x+13)(x-4)=0$
=> $x=\frac{-13}{2}$ or $x = 4$
It has been given that $x$ is a positive real number. Therefore, we can eliminate the case $x=\frac{-13}{2}$.
$x=4$ is the point at which the function attains the maximum value. $4$ is not the maximum value of the function.
Substituting $x=4$ in the original function, we get, $2x^2 = 2*4^2= 32$.
f(x) = $32$.
Therefore, 32 is the right answer.

Question 13:Â If $f(x + 2) = f(x) + f(x + 1)$ for all positive integers x, and $f(11) = 91, f(15) = 617$, then $f(10)$ equals

13)Â Answer:Â 54

Solution:

$f(x + 2) = f(x) + f(x + 1)$
As we can see, the value of a term is the sum of the 2 terms preceding it.

It has been given thatÂ $f(11) = 91$ and $f(15) = 617$.
We have to find the value ofÂ $f(10)$.

LetÂ $f(10)$ = b
$f(12)$ = b + 91
$f(13)$ = 91 + b +Â 91 = 182 + b
$f(14)$ = 182+b+91+b = 273+2b
$f(15)$ = 273+2b+182+b = 455+3b
It has been given that 455+3b = 617
3b = 162
=> b = 54

Therefore, 54 is the correct answer.

Question 14: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 $\dfrac{1}{4}$

14)Â AnswerÂ (C)

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 15:Â Let f(x)= $\max(5x, 52-2x^2)$, where x is any positive real number. Then the minimum possible value of f(x)

15)Â Answer:Â 20

Solution:

The minimum value of the function will occur when the expressions inside the function are equal.
So, 5$x$ = $52 – 2x^2$
or, $2x^2 + 5x – 52$ = 0
On solving, we get $x$ = 4 or $-\dfrac{13}{2}$
But, it is given that $x$ is a positive number.
So, $x$ = 4
And the minimum value = 5*4 = 20
Hence, 20 is the correct answer.

Question 16:Â The smallest integer $n$ such that $n^3-11n^2+32n-28>0$ is

16)Â Answer:Â 8

Solution:

We can see that at n = 2, $n^3-11n^2+32n-28=0$ i.e. (n-2) is a factor of $n^3-11n^2+32n-28$

$\dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$

We can further factorizeÂ n^2-9n+14 as (n-2)(n-7).

$n^3-11n^2+32n-28=(n-2)^2(n-7)$

$\Rightarrow$Â $n^3-11n^2+32n-28>0$

$\Rightarrow$Â $(n-2)^2(n-7)>0$

Therefore, we can say that n-7>0

Hence, n$_{min}$ = 8

Question 17: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

17)Â Answer:Â 36

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 18: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 $A > \frac{1}{8}$

18)Â AnswerÂ (A)

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 19: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

19)Â AnswerÂ (D)

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$a^2$ = 549 ==> a= 3 which is an integer

Question 20:Â If x is a real number, then $\sqrt{\log_{e}{\frac{4x – x^2}{3}}}$ is a real number if and only if

a)Â $1 \leq x \leq 3$

b)Â $1 \leq x \leq 2$

c)Â $-1 \leq x \leq 3$

d)Â $-3 \leq x \leq 3$

20)Â AnswerÂ (A)

Solution:

$\sqrt{\log_{e}{\frac{4x – x^2}{3}}}$ will be real ifÂ $\log_e\ \frac{\ 4x-x^2}{3}\ \ge\ 0$

$\frac{\ 4x-x^2}{3}\ >=\ 1$

$\ 4x-x^2-3\ >=\ 0$

$\ x^2-4x+3\ =<\ 0$

1=< x=< 3

Question 21:Â If $5^x – 3^y = 13438$ and $5^{x – 1} + 3^{y + 1} = 9686$, then x + y equals

21)Â Answer:Â 13

Solution:

$5^x – 3^y = 13438$ and $5^{x – 1} + 3^{y + 1} = 9686$

$5^{x} + 3^{y}*15 = 9686*5$

$5^{x} + 3^{y}*15 = 48430$

16*$3^y$=34992

$3^y$ = 2187

y = 7

$5^x$=13438+2187=15625

x=6

x+y = 13

Question 22:Â In 2010, a library contained a total of 11500 books in two categories – fiction and nonfiction. In 2015, the library contained a total of 12760 books in these two categories. During this period, there was 10% increase in the fiction category while there was 12% increase in the non-fiction category. How many fiction books were in the library in 2015?

a)Â 6160

b)Â 6600

c)Â 6000

d)Â 5500

22)Â AnswerÂ (B)

Solution:

Let the number of fiction and non-fiction books in 2010 = 100a, 100b respectively

It is given that the total number of books in 2010 = 11500

100a+100b = 11500Â  Â  Â  Â  Â  Â  ——-Eq 1

The number of fiction and non-fiction books in 2015 = 110a, 112b respectively

110a+112b = 12760Â  Â  Â  Â  Â  Â ——-Eq 2

On solving both the equations we get, b=55, a= 60

The number of fiction books in 2015 = 110*60=6600

Question 23:Â Let a, b, x, y be real numbers such that $a^2 + b^2 = 25, x^2 + y^2 = 169$, and $ax + by = 65$. If $k = ay – bx$, then

a)Â $0 < k \leq \frac{5}{13}$

b)Â $k > \frac{5}{13}$

c)Â $k = \frac{5}{13}$

d)Â k = 0

23)Â AnswerÂ (D)

Solution:

$\left(ax+by\right)^2=65^2$

$a^2x^2\ +\ b^2y^2+\ 2abxy\ =\ 65^2$
$k = ay – bx$

$k^2\ =\ a^2y^2+b^2x^2-2abxy$

$(a^2 + b^2)(x^2 + y^2 )= 25* 169$

$a^2x^2+a^2y^2+b^2x^2+b^2y^2=\ 25\times\ 169$

$k^2=\ 65^2\ -\ \left(25\times\ 169\right)$

k = 0

D is the correct answer.

Question 24:Â Let f be a function such that f (mn) = f (m) f (n) for every positive integers m and n. If f (1), f (2) and f (3) are positive integers, f (1) < f (2), and f (24) = 54, then f (18) equals

24)Â Answer:Â 12

Solution:

Given, f(mn) = f(m)f(n)
when m= n= 1, f(1) = f(1)*f(1) ==> f(1) = 1

when m=1,Â  n= 2, f(2) = f(1)*f(2) ==> f(1) = 1

when m=n= 2, f(4) = f(2)*f(2) ==> f(4) = $[f(2)]^2$

Similarly f(8) =Â f(4)*f(2) =$[f(2)]^3$

f(24) = 54

$[f(2)]^3$ *Â $[f(3)]$ = $3^3*2$

On comparing LHS and RHS, we get

f(2) = 3 and f(3) = 2

Now we have to find the value of f(18)

f(18) =Â $[f(2)]$ * $[f(3)]^2$

= 3*4=12

Question 25: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

25)Â AnswerÂ (A)

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 26:Â Consider a function f satisfying f (x + y) = f (x) f (y) where x,y are positive integers, and f(1) = 2. If f(a + 1) +f (a + 2) + … + f(a + n) = 16 (2$^n$ – 1) then a is equal to

26)Â Answer:Â 3

Solution:

f (x + y) = f (x) f (y)

Hence, f(2)=f(1+1)=f(1)*f(1)=2*2=4

f(3)=f(2+1)=f(2)*f(1)=4*2=8

f(4)=f(3+1)=f(3)*f(1)=8*2=16

…….=> f(x)=$2^x$

Now,Â f(a + 1) +f (a + 2) + … + f(a + n) = 16 (2$^n$ – 1)

On putting n=1 in the equation we get, f(a+1)=16Â  Â => f(a)*f(1)=16Â  (It is given thatÂ f (x + y) = f (x) f (y))

=>Â $2^a$*2=16

=> a=3

Question 27:Â For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd. If m is a positive integer such that 8f(m + 1) – f(m) = 2, then m equals

27)Â Answer:Â 10

Solution:

Assuming m is even, then 8f(m+1)-f(m)=2

m+1 will be odd

So,Â 8(m+1+3)-m(m+1)=2

=> 8m+32-$m^2-m$=2

=> $m^2-7m-30=0$

=> m=10,-3

Rejecting the negative value, we get m=10

Assuming m is odd, m+1 will be even.

then, 8(m+1)(m+2)-m-3=2

=> 8($m^2+3m+2$)-m-3=2

=>Â $8m^2+23m+11=0$

Solving this, m = -2.26 and -0.60

Hence, the value of m is not integral. Hence this case will be rejected.

Question 28:Â The number of the real roots of the equation $2 \cos (x(x + 1)) = 2^x + 2^{-x}$ is

a)Â 2

b)Â 1

c)Â infinite

d)Â 0

28)Â AnswerÂ (B)

Solution:

$2 \cos (x(x + 1)) = 2^x + 2^{-x}$

The maximum value of LHS is 2 when $\cos (x(x + 1))$ is 1 and the minimum value of RHS is 2 using AM $\geq$ GM

Hence LHS and RHS can only be equal when both sides are 2. For LHS, cosx(x+1)=1Â  Â => x(x+1)=0Â  Â => x=0,-1

For RHS minimum value, x=0

Hence only one solution x=0

Question 29:Â Let S be the set of all points (x, y) in the x-y plane such that $\mid x \mid + \mid y \mid \leq 2$ and $\mid x \mid \geq 1.$ Then, the area, in square units, of the region represented by S equals

29)Â Answer:Â 2

Solution:

Sum of the area of region I and II is the required area.

Now, required area =Â Â $\ 4\times\frac{\ 1}{2}\times\ 1\times\ 1$ = 2

Question 30:Â The number of solutions to the equation $\mid x \mid (6x^2 + 1) = 5x^2$ is

30)Â Answer:Â 5

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 31:Â A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially?

31)Â Answer:Â 62

Solution:

Let the initial number of chocolates be 64x.

First child gets 32x+1 and 32x-1 are left.

2nd child gets 16x+1/2 and 16x-3/2 are left

3rd child gets 8x+1/4 and 8x-7/4 are left

4th child gets 4x+1/8 and 4x-15/8 are left

5th child gets 2x+1/16 and 2x-31/16 are left.

Given, 2x-31/16=0=> 2x=31/16 => x=31/32.

.’. Initially the Gentleman has 64x i.e. 64*31/32 =62 chocolates.

Question 32:Â The number of real-valued solutions of the equation $2^{x}+2^{-x}=2-(x-2)^{2}$ is:

a)Â 1

b)Â 2

c)Â infinite

d)Â 0

32)Â AnswerÂ (D)

Solution:

The graphs ofÂ $2^{x}+2^{-x} and 2-(x-2)^{2}$ never intersect. So, number of solutions=0.

Alternate method:

We notice that the minimum value of the term in the LHS will be greater than or equal toÂ 2 {at x=0;Â LHS = 2}. However, the term in the RHS is less than or equal to 2 {at x=2; RHS = 2}. The values of xÂ at which both the sides become 2 are distinct; hence, there are zero real-valued solutions to the above equation.

Question 33: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

33)Â AnswerÂ (D)

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 34:Â The area of the region satisfying the inequalities $\mid x\mid-y\leq1,y\geq0$ and $y\leq1$ is

34)Â Answer:Â 3

Solution:

The area of the region contained by the lines $\mid x\mid-y\leq1,y\geq0$ and $y\leq1$ is the white region.

Total area = Area of rectangle + 2 * Area of triangle =Â $2+\left(\frac{1}{2}\times\ 2\times\ 1\right)\ =3$

Hence, 3 is the correct answer.

Question 35:Â Among 100 students, $x_1$ have birthdays in January, $X_2$ have birthdays in February, and so on. If $x_0=max(x_1,x_2,….,x_{12})$, then the smallest possible value of $x_0$ is

a)Â 8

b)Â 9

c)Â 10

d)Â 12

35)Â AnswerÂ (B)

Solution:

$x_0=max(x_1,x_2,….,x_{12})$

$x_0$ will be minimum if x1,x2…x12 are close to each other

100/12=8.33

.’.Â max$(x_1,x_2,….,x_{12})$ will be minimum ifÂ $(x_1,x_2,….,x_{12})$=(9,9,9,9,8,8,8,8,8,8,8,8,)

.’. Option B is correct.

Question 36:Â The number of distinct real roots of the equation $(x+\frac{1}{x})^{2}-3(x+\frac{1}{x})+2=0$ equals

36)Â Answer:Â 1

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 37:Â If $f(5+x)=f(5-x)$ for every real x, and $f(x)=0$ has four distinct real roots, then the sum of these roots is

a)Â 0

b)Â 40

c)Â 10

d)Â 20

37)Â AnswerÂ (D)

Solution:

Let ‘r’ be the root of the function. It follows that f(r) = 0. We can represent this as $f\left(r\right)=f\left\{5-\left(5-r\right)\right\}$

Based on the relation: $f\left(5-x\right)=f\left(5+x\right)$; $f\left(r\right)=f\left\{5-\left(5-r\right)\right\}=f\left\{5+\left(5-r\right)\right\}$

$\therefore\ f\left(r\right)=f\left(10-r\right)$

Thus, every root ‘r’ is associated with another root ‘(10-r)’ [these form a pair]. For even distinct roots, in this case four, let us assume the roots to be as follows: $r_1,\ \left(10-r_1\right),\ r_2,\ \left(10-r_2\right)$

The sum of these roots = $r_1\ +\left(10-r_1\right)+\ r_2+\ \left(10-r_2\right)\ =\ 20$

Hence, Option D is the correct answer.

Question 38:Â Let A, B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is

a)Â 5

b)Â 4

c)Â 6

d)Â 7

38)Â AnswerÂ (C)

Solution:

Given

A+(B+C)/2=5 => 2A+B+C=10….(i)

(A+C)/2 +B=7 => A+2B+C=14…..(ii)

(i)-(ii)=> B-A=4 => B=4+A.

Given, A, B, C are positive integers

If A=1=>B=5 => C=3

If A=2=>B=6 => C=0 but this is invalid as C is positive.

Similarly if A>2 C will be negative and cases are not valid.

Hence, A+B=6.

Question 39: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

39)Â AnswerÂ (B)

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 40:Â Dick is thrice as old as Tom and Harry is twice as old as Dick. If Dick’s age is 1 yearÂ less than the average age of all three, then Harry’s age, in years, is

40)Â Answer:Â 18

Solution:

Let tom’s age = x

=>Â Dick=3x

=>harry = 6x

Given,

3x+1 = (x+3x+6x)/3

=> x= 3

Hence, Harry’s age = 18 years

Question 41:Â Let k be a constant. The equations $kx + y = 3$ and $4x + ky = 4$ have a unique solution ifÂ and only if

a)Â $\mid k\mid\neq2$

b)Â $\mid k\mid=2$

c)Â $k\neq2$

d)Â $k=2$

41)Â AnswerÂ (A)

Solution:

Two linear equations ax+by= c and dx+ ey = f have a unique solution ifÂ $\frac{a}{d}\ne\ \frac{b}{e}$

Therefore, $\frac{k}{4}\ne\ \frac{1}{k}$ =>Â $k^2\ne\ 4$

=> kÂ $\ne\$ |2|

Question 42:Â The area, in sq. units, enclosed by the lines $x=2,y=\mid x-2\mid+4$, the X-axis and the Y-axis is equal to

a)Â 10

b)Â 6

c)Â 8

d)Â 12

42)Â AnswerÂ (A)

Solution:

The required figure is a trapezium with vertices A(0,0), B(2,0), C(2,4) and D(0,6)

AB = 2 BC = 4 and AD = 6

Area of trapezium =Â $\frac{1}{2}\left(sum\ of\ the\ opposite\ sides\right)\cdot height$ =Â $\frac{1}{2}\left(4+6\right)\cdot2\ =\ 10$

Question 43:Â If $f(x+y)=f(x)f(y)$ and $f(5)=4$, then $f(10)-f(-10)$ is equal to

a)Â 14.0625

b)Â 0

c)Â 15.9375

d)Â 3

43)Â AnswerÂ (C)

Solution:

The given function is equivalent to f(x) = $a^x$

Given, f(5) = 4

=> $a^5=4=>\ a=4^{\frac{1}{5}}$

=> f(x) = $4^{\frac{x}{5}}$

f(10) – f(-10) = 16 – 1/16 = 15.9375

Question 44:Â Let $f(x)=x^{2}+ax+b$ and $g(x)=f(x+1)-f(x-1)$. If $f(x)\geq0$ for all real x, and $g(20)=72$. then the smallest possible value of b is

a)Â 16

b)Â 4

c)Â 1

d)Â 0

44)Â AnswerÂ (B)

Solution:

$f\left(x\right)=\ x^2+ax+b$

$f\left(x+1\right)=x^2+2x+1+ax+a+b$

$f\left(x-1\right)=x^2-2x+1+ax-a+b$

$g(x)=f(x+1)-f(x-1)= 4x+2a$

Now $g(20) = 72$ from this we get $a = -4$Â ;Â $f\left(x\right)=x^2-4x\ +b$

For this expression to be greater than zero it has to be a perfect squareÂ which is possible for $b\ge\ 4$

Hence the smallest value of ‘b’ is 4.

Question 45:Â In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of theÂ highest 9 scores is 47. For the entire group of 10 students, the maximum possibleÂ mean exceeds the minimum possible mean by

a)Â 5

b)Â 4

c)Â 3

d)Â 6

45)Â AnswerÂ (B)

Solution:

Let x(1) be the least number and x(10) be the largest number. Now from the condition given in the question , we can say that

x(2)+x(3)+x(4)+……..x(10)= 47*9=423……………….(1)

Similarly x(1)+x(2)+x(3)+x(4)…………….+x(9)= 42*9=378……………(2)

Subtracting both the equations we get x(10)-x(1)=45

Now, the sum of the 10 observations from equation (1) is 423+x(1)

Now the minimum value of x(10) will be 47 and the minimum value of x(1) will be 2 . Hence minimum average 425/10=42.5

Maximum value of x(1) is 42. Hence maximum average will be 465/10=46.5

Hence difference in average will be 46.5-42.5=4 which is the correct answer

Question 46: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

46)Â AnswerÂ (A)

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 47:Â The number of pairs of integers $(x,y)$ satisfying $x\geq y\geq-20$ and $2x+5y=99$

47)Â Answer:Â 17

Solution:

We have 2x + 5y = 99 orÂ $x=\frac{\left(99-5y\right)}{2}$

Now $x\ge\ y\ \ge\ -20$ ;Â SoÂ $\frac{\left(99-5y\right)}{2}\ge\ y\ ;\ 99\ge7y\ or\ y\le\ \approx\ 14$

So $-20\le y\le14$. Now for this range of “y”, we have to find all the integral values of “x”. As the coefficient of “x” is 2,

then (99 –Â 5y) must be even, which will happen when “y” is odd. However, there are only 17 odd values of “y” be -20 and 14.

Hence the number of possible values is 17.

Question 48:Â In May, John bought the same amount of rice and the same amount of wheat as he hadÂ bought in April, but spent â‚¹ 150 more due to price increase of rice and wheat by 20%Â and 12%, respectively. If John had spent â‚¹ 450 on rice in April, then how much did heÂ spend on wheat in May?

a)Â Rs.560

b)Â Rs.570

c)Â Rs.590

d)Â Rs.580

48)Â AnswerÂ (A)

Solution:

Let John buy “m” kg of rice and “p” kg of wheat.

Now let the price of rice be “r” in April. Price in May will be “1.2(r)”

Now let the price of wheat be “w” in April . Price in April will be “1.12(w)”.

Now he spent â‚¹150 more in May , so 0.2(rm)+0.12(wp)=150

Its also given that he had spent â‚¹450 on rice in April. So (rm)=450

So 0.2(rm)= (0.2)(450)=90 Substituting we get (wp)=60/0.12 or (wp)=500

Amount spent on wheat in May will be 1.12(500)=â‚¹560

Question 49:Â If x and y are non-negative integers such that $x + 9 = z$, $y + 1 = z$ and $x + y < z + 5$, then the maximum possible value of $2x + y$ equals

49)Â Answer:Â 23

Solution:

We can write x=z-9 and y=z-1 Now we have x+y< z+5

Substituting we get z-9+z-1<z+5 or z<15

Hence the maximum possible value of z is 14

Maximum value of “x” is 5 and maximum value of “y” is 13

Now 2x+y = 10+13=23

Question 50:Â Aron bought some pencils and sharpeners. Spending the same amount of money asÂ Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of oneÂ sharpener is â‚¹ 2 more than the cost of a pencil, then the minimum possible number ofÂ pencils bought by Aron and Aditya together is

a)Â 33

b)Â 27

c)Â 30

d)Â 36

50)Â AnswerÂ (A)

Solution:

Let the number of pencils bought by AronÂ be “p” and the cost of each pencil be “a”.

Let the number of sharpeners bought Aron be “s” and the cost of each sharpener be “b”.

Now amount spent by Aron will be (pa)+(sb)

Aditya bought (2p) pencils and (s-10) sharpeners. Amount spent will be (2pa)+(s-10)b

Amount spent in both the cases is same

pa + sb = 2pa + (s-10)b or pa=10b

Now its given in the question that cost of sharpener is 2 more than pencil i.e. b=a+2

pa= 10a+20 or a=20/(p-10)

Now the number of pencils has to be minimum, for that we have to find smallest “p” such that both “p” and “a” are integers. The smallest such value is p=11 . Total number of pencils bought will be p+2p=11+22=33

Question 51:Â For real x, the maximum possible value of $\frac{x}{\sqrt{1+x^{4}}}$ is

a)Â $\frac{1}{2}$

b)Â $1$

c)Â $\frac{1}{\sqrt{3}}$

d)Â $\frac{1}{\sqrt{2}}$

51)Â AnswerÂ (D)

Solution:

NowÂ $\frac{x}{\sqrt{\ 1+x^4}}=\ \frac{\ 1}{\sqrt{\ \ \frac{\ 1+x^4}{x^2}}}=\frac{1}{\sqrt{\ \frac{1}{x^2}+x^2}}$

Applying A.M>= G.M.

$\frac{\left(\frac{1}{x^2}+x^2\right)}{2}\ge\ 1\ or\ \ \frac{1}{x^2}+x^2\ge\ 2$ Substituting we get the maximum possible value of the equation asÂ $\frac{1}{\sqrt{\ 2}}$

Question 52:Â In how many ways can a pair of integers (x , a) be chosen such that $x^{2}-2\mid x\mid+\mid a-2\mid=0$ ?

a)Â 6

b)Â 5

c)Â 4

d)Â 7

52)Â AnswerÂ (D)

Solution:

$x^{2}-2\mid x\mid+\mid a-2\mid=0$

where x>= 0 and x>=2

$x^2-2x+a-2\ =0$ Using quadratic equation we have $x=\ 1+\sqrt{\ 3-a}\ and\ x=1-\sqrt{\ 3-a}$ Only two integer values are possible

a=2 and a=3. So corresponding “x” values are x=1 and a=3, x=2 and a=2, x=0 and a=2

where x>=0 and x<2

Applying the above process we get x=1 and a=1

where x<0 and x>=2 we get a=3 and x=-1 , a=2 and x=-2

where x<0 and x<2 we get a=1 and x=-1

Hence there are total 7 values possible

Question 53:Â if x and y are positive real numbers satisfying $x+y=102$, then the minimum possible valus of $2601(1+\frac{1}{x})(1+\frac{1}{y})$ is

53)Â Answer:Â 2704

Solution:

Now we haveÂ $2601\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=2601\left(\frac{xy+y+x+1}{xy}\right)$

Now we know that x+y=102. Substituting it in the above equation

$2601\left(\frac{xy+y+x+1}{xy}\right)=2601\left(\frac{103}{xy}+1\right)$

Maximum value of xyÂ  can be found out by AM>= GM relationship

$\ \frac{\ x+y}{2}\ge\ \sqrt{xy}\ or\ \ \sqrt{\ xy}\le\ 51\ or\ xy\le\ 2601$

Hence the maximum value of “xy” is 2601. Substituting in the above equation we get

$2601\left(\ \frac{\ 103+2601}{2601}\right)=2704$

Question 54:Â A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is

a)Â 11

b)Â 13

c)Â 10

d)Â 12

54)Â AnswerÂ (B)

Solution:

Let the cost of an apple, an orange and a mango be a, o, and m respectively.

Then it is given that:

2a+4o+6m = a+4o+8m

or a = 2m.

Also, a+4o+8m = 8o + 7m

10m-7m = 4o

3m = 4o.

We can now express the cost of a basket in terms of mangoes only:

2a+4o+6m = 4m+3m+6m = 13m.

Question 55:Â The number of integers n that satisfy the inequalities $\mid n – 60 \mid < \mid n – 100 \mid < \mid n – 20 \mid$ is

a)Â 21

b)Â 19

c)Â 18

d)Â 20

55)Â AnswerÂ (B)

Solution:

We haveÂ $\mid n – 60 \mid < \mid n – 100 \mid < \mid n – 20 \mid$.

Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line. This means that when the absolute difference from a number is larger, n would be further away from that number.

Example: The absolute difference of n and 60 is less than that of the absolute difference between n and 20. Hence, n cannot be $\le\ 40$, as then it would be closer to 20 than 60, and closer on the number line would indicate lesser value of absolute difference.Â Thus we have the condition that n>40.

The absolute difference of n and 100 is less than that of the absolute difference between n and 20. Hence, n cannot be $\le\ 60$, as then it would be closer to 20 than 100. Thus we have the condition that n>60.

The absolute difference of n and 60 is less than that of the absolute difference between n and 100. Hence, n cannot be $\ge80$, as then it would be closer to 100 than 60. Thus we have the condition that n<80.

The number which satisfies the conditions are 61, 62, 63, 64……79. Thus, a total of 19 numbers.

Alternatively

as per the given condition :Â $\mid n – 60 \mid < \mid n – 100 \mid < \mid n – 20 \mid$.

Dividing the range of n into 4 segments. (n < 20, 20<n<60, 60<n<100, n > 100 )

1) For n < 20.

|n-20| = 20-n,Â |n-60| = 60- n,Â |n-100| = 100-n

considering the inequality part :$\left|n-100\right|<\ \left|n\ -20\right|$

100 -n < 20 -n,

No value of n satisfies this condition.

2)Â For 20 < n < 60.

|n-20| = n-20, |n-60| = 60- n, |n-100| = 100-n.

60- n < 100 – n and 100 – n < n – 20

For 100 -n < n – 20.

120 < 2n and n > 60. But for the considered range n is less than 60.

3)Â For 60 < n < 100

|n-20| = n-20, |n-60| = n-60, |n-100| = 100-n

n-60 < 100-n and 100-n < n-20.

For the first part 2n < 160 and for the second part 120 < 2n.

n takes values from 61 …………….79.

A total of 19 values

4)Â For n > 100

|n-20| = n-20, |n-60| = n-60, |n-100| = n-100

n-60 < n – 100.

No value of n in the given rangeÂ satisfies the given inequality.

Hence a total of 19 values satisfy the inequality.

Question 56: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

56)Â Answer:Â 35

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 57:Â $f(x) = \frac{x^2 + 2x – 15}{x^2 – 7x – 18}$ is negative if and only if

a)Â -5 < x < -2 or 3 < x < 9

b)Â x < -5 or -2 < x < 3

c)Â -2 < x < 3 or x > 9

d)Â x < -5 or 3 < x < 9

57)Â AnswerÂ (A)

Solution:

$f(x) = \frac{x^2 + 2x – 15}{x^2 – 7x – 18}$<0

$\frac{\left(x+5\right)\left(x-3\right)}{\left(x-9\right)\left(x+2\right)}<0$

We have four inflection points -5, -2, 3, and 9.

For x<-5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative. Hence, the overall expression will be positive. Similarly, when x>9, all four terms will be positive.

When x belongs to (-2,3), two terms are negative and two are positive. Hence, the overall expression is positive again.

We are left with the range (-5,-2) and (3,9) where the expression will be negative.

Question 58: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

58)Â AnswerÂ (A)

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 59: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

59)Â AnswerÂ (B)

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 60: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 $[\frac{3}{7},\frac{1}{2})$

60)Â AnswerÂ (D)

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 61:Â For a real number x the condition $\mid3x-20\mid+\mid3x-40\mid=20$ necessarily holds if

a)Â $10<x<15$

b)Â $9<x<14$

c)Â $7<x<12$

d)Â $6<x<11$

61)Â AnswerÂ (C)

Solution:

Case 1 : $x\ge\frac{40}{3}$
we get 3x-20 +3x-40 =20
6x=80
x=$\frac{80}{6}$=$\frac{40}{3}$=13.33
Case 2 :$\frac{20}{3}\le\ x<\frac{40}{3}\$
we get 3x-20+40-3x =20
we get 20=20
So we get x$\in\ \left[\frac{20}{3},\frac{40}{3}\right]$
Case 3 $x<\frac{20}{3}$
we get 20-3x+40-3x =20
40=6x
x=$\frac{20}{3}$
but this is not possible
so we get from case 1,2 and 3
$\frac{20}{3}\le\ x\le\frac{40}{3}$
Now looking at options
we can say only option C satisfies for all x .
Hence 7<x<12.

Question 62: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

62)Â AnswerÂ (D)

Solution:

We have :
$x^2-xy-x\ =22\ \ \ \ \ \ \left(1\right)$
AndÂ $y^2-xy+y\ =34\ \ \ \ \ \ \left(2\right)$
Adding (1) and (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 63: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

63)Â AnswerÂ (C)

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

Adding them, we get,

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.

Lets start with Option B.

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 64:Â The number of distinct pairs of integers (m,n), satisfying $\mid1+mn\mid<\mid m+n\mid<5$ is:

64)Â Answer:Â 12

Solution:

Let us break this up into 2 inequations [ Let us assume x as m and y as n ]

| 1 + mn | < | m + n |

| m + n | < 5

Looking at these expressions, we can clearly tell that the graphs will be symmetrical about the origin.

Let us try out with the first quadrant and extend the results to the other quadrants.

We will also consider the +X and +Y axes along with the quadrant.

So, the first inequality becomes,

1 + mn < m + n

1 + mn – m – n < 0

1 – m + mn – n < 0

(1-m) + n(m-1) < 0

(1-m)(1-n)Â < 0

(m – 1)(n – 1) < 0

Let us try to plot the graph.

If we consider only mn < 0, then we get

But, we have (m – 1)(n – 1) < 0, so we need to shift the graphs by one unit towards positive x and positive y.

So, we have,

But, we are only considering the first quadrant and the +X and +Y axes. Hence, if we extend, we get the following region.

So, if we look for only integer values, we get

(0,2), (0,3),…….

(0,-2), (0, -3),……

(2,0), (3,0), ……

(-2,0), (-3,0), …….

Now, let us consider the other inequation as well, in which |x + y| < 5

Since one of the values is always zero, the modulus of the other value is less than or equal to 4.

Hence, we get

(0,2), (0,3), (0,4)

(0,-2), (0, -3), (0, -4)

(2,0), (3,0), (4,0)

(-2,0), (-3,0), (-4,0)

Hence, a total of 12 values.

Question 65:Â If $f(x)=x^{2}-7x$ and $g(x)=x+3$, then the minimum value of $f(g(x))-3x$ is:

a)Â -20

b)Â -12

c)Â -15

d)Â -16

65)Â AnswerÂ (D)

Solution:

Now we have :
$f(g(x))-3x$
so we get f(x+3)-3x
=Â $\left(x+3\right)^2-7\left(x+3\right)-3x$
=$x^2-4x-12$
Now minimum value of expression = $-\frac{D}{4a}$Â $\frac{\left(4ac-b^2\right)}{4a}$
We get – (16+48)/4
= -16

Question 66:Â If n is a positive integer such that $(\sqrt[7]{10})(\sqrt[7]{10})^{2}…(\sqrt[7]{10})^{n}>999$, then the smallest value of n is

66)Â Answer:Â 6

Solution:

$(\sqrt[7]{10})(\sqrt[7]{10})^{2}…(\sqrt[7]{10})^{n}>999$

$(\sqrt[7]{10})^{1+2+…+n}>999$

$10^{\frac{1+2+…+n}{7}}>999$

For minimum value of n,

$\frac{1+2+…+n}{7}=3$

1 + 2 + … + n = 21

We can see that if n = 6, 1 + 2 + 3 + … + 6 = 21.

Question 67:Â If $3x+2\mid y\mid+y=7$ and $x+\mid x \mid+3y=1$ then $x+2y$ is:

a)Â $-\frac{4}{3}$

b)Â $\frac{8}{3}$

c)Â $0$

d)Â $1$

67)Â AnswerÂ (C)

Solution:

We need to check for all regions:

x >= 0, y >= 0

x >= 0, y < 0

x < 0, y >= 0

x < 0, y < 0

However, once we find out the answer for any one of the regions, we do not need to calculate for other regions since the options suggest that there will be a single answer.

Let us start withÂ x >= 0, y >= 0,

3x + 3y = 7

2x + 3y = 1

Hence, x = 6 and y = -11/3

Since y > = 0, this is not satisfying the set of rules.

Next, let us testÂ x >= 0, y < 0,

3x – y = 7

2x + 3y = 1

Hence, y = -1

x = 2.

This satisfies both the conditions. Hence, this is the correct point.

WE need the value of x + 2y

x + 2y = 2 + 2(-1) = 2 – 2 = 0.

Question 68:Â Let $0 \leq a \leq x \leq 100$ and $f(x) = \mid x – a \mid + \mid x – 100 \mid + \mid x – a – 50\mid$. Then the maximum value of f(x) becomes 100 when a is equal to

a)Â 25

b)Â 100

c)Â 50

d)Â 0

68)Â AnswerÂ (C)

Solution:

x>=a, so |x-a| = x-a

x<100, so |x-100| = 100-x

f(x) = (x-a) + (100-x) + |x-a-50| =100

or, |x-a-50| = a

From the graph we can can see that when x=a then

|x-a-50|=a

or, a= 50

Similarly when x=a+100

|x-a-50|=a

or, a= 50

So value of a is 50 when f(x) is 100.

Question 69:Â Let a, b, c be non-zero real numbers such that $b^2 < 4ac$, and $f(x) = ax^2 + bx + c$. If the set S consists of all integers m such that f(m) < 0, then the set S must necessarily be

a)Â the set of al!I positive integers

b)Â the set of all integers

c)Â either the empty set or the set of all integers

d)Â the empty set

69)Â AnswerÂ (C)

Solution:

$b^2 < 4ac$ means that the discriminant is less than 0. Therefore, f(x)>0 for all x if the coefficient ofÂ $x^2$ is positive, and f(x)<0 for all x if the coefficient of $x^2$ is negative.

We are given that f(m)<0 and m is an integer.

So the set containing values of m will either be empty ifÂ the coefficient of $x^2$ is positive, orÂ it will be a set of all integers if theÂ coefficient of $x^2$ is negative.

Question 70:Â Let a and b be natural numbers. If $a^2 + ab + a = 14$ and $b^2 + ab + b = 28$, then $(2a + b)$ equals

a)Â 8

b)Â 7

c)Â 10

d)Â 9

70)Â AnswerÂ (A)

Solution:

a(a + b + 1) = 14 …… (1)

b(a + b + 1) = 28 …… (2)

$\frac{a}{b}=\frac{1}{2}$

b = 2a

Substituting in (1), we get

a(3a + 1) = 14

$3a^2+a-14=0$

$3a^2-6a+7a-14=0$

$3a\left(a-2\right)+7\left(a-2\right)=0$

Given, a and b are natural numbers.

Therefore, a = 2 and b = 4

2a + b = 2(2) + 4 = 8

The answer is option A.

Question 71:Â The largest real value of a for which the equation $\mid x + a \mid + \mid x – 1 \mid = 2$ has an infinite number of solutions for x is

a)Â -1

b)Â 0

c)Â 1

d)Â 2

71)Â AnswerÂ (C)

Solution:

In the question, it is given thatÂ the equation $\mid x + a \mid + \mid x – 1 \mid = 2$ has an infinite number of solutions for any value of x. This is possible when x in |x+a| and x in |x-1| cancels out.

Case 1:

x + a < 0, x – 1Â $\ge\$ 0

– a – x + x – 1 = 2

a = -3

Case 2:

x + aÂ $\ge\$ 0 and x – 1 < 0

x + a – x + 1 = 2

a = 1

Largest value of a is 1.

The answer is option C.

Question 72:Â For natural numbers x, y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is

72)Â Answer:Â 34

Solution:

It is given,Â y(x + z) = 19

y cannot be 19.

If y = 19, x + z = 1 which is not possible when both x and z are natural numbers.

Therefore, y = 1 and x + z = 19

It is given, z(x + y) = 51

z can take values 3 and 17

Case 1:

If z = 3, y = 1 and x = 16

xyz = 3*1*16 = 48

Case 2:

If z = 17, y = 1 and x = 2

xyz = 17*1*2 = 34

Minimum value xyz can take is 34.

Question 73:Â For any real number x, let [x] be the largest integer less than or equal to x. If $\sum_{n=1}^N \left[\frac{1}{5} + \frac{n}{25}\right] = 25$ then N is

73)Â Answer:Â 44

Solution:

It is given,

$\Sigma_{n=1}^N\ \left[\frac{1}{5}+\frac{n}{25}\right]=25$

$\Sigma_{n=1}^N\ \left[\frac{5+n}{25}\right]=25$

For n = 1 to n = 19, value of function is zero.

For n = 20 to n = 44, value of function will be 1.

44 = 20 + n – 1

n = 25 which is equal to given value.

This implies N = 44

Question 74:Â The number of integer solutions of the equation $\left(x^{2} – 10\right)^{\left(x^{2}- 3x- 10\right)} = 1$ is

74)Â Answer:Â 4

Solution:

Case 1: WhenÂ $x^2-3x-10=0$ and $x^2-10\ne\ 0$

$x^2-3x-10=0\$or, $(x-5)(x+2) = 0$

or, x= 5 or -2

Case 2:Â $x^2-10=1$

$x^2-11=0$

No integer solutions

Case 3:Â $x^2-10=-1\ and\ x^2-3x-10\ is\ even$

$x^2-9=0$

or, (x+3)(x-3)=0

or, x= -3 and 3

for x= -3 and +3Â $x^2-3x-10$ is even

In total 4 values of x satisfy the equations

Question 75:Â Let $f(x)$ be a quadratic polynomial in $x$ such that $f(x) \geq 0$ for all real numbers $x$. If f(2) = 0 and f( 4) = 6, then f(-2) is equal to

a)Â 12

b)Â 24

c)Â 6

d)Â 36

75)Â AnswerÂ (B)

Solution:

$f(x) \geq 0$for all real numbers $x$, so D<=0

Since f(2)=0 therefore x=2 is a root of f(x)

Since the discriminant of f(x) is less than equal to 0 and 2 is a root so we can conclude that D=0

Therefore f(x) =Â $a\left(x-2\right)^2$

f(4)=6

or, 6 =Â $a\left(x-2\right)^2$

a= 3/2

$f\left(-2\right)=\ -\frac{3}{2}\left(-4\right)^2=24$

Question 76:Â Let r and c be real numbers. If r and -r are roots of $5x^{3} + cx^{2} – 10x + 9 = 0$, then c equals

a)Â $-\frac{9}{2}$

b)Â $\frac{9}{2}$

c)Â -4

d)Â 4

76)Â AnswerÂ (A)

Solution:

Let the roots of the given equationÂ $5x^{3} + cx^{2} – 10x + 9 = 0$ be r, -r and p

r – r + p =Â $-\frac{c}{5}$

p =Â $-\frac{c}{5}$ …… (1)

$-r^2-pr+pr=-2$

$r^2=2$ …… (2)

$-r^2p=-\frac{9}{5}$

$p=\frac{9}{10}$ …… (3)

Substituting p in (1), we get

$\frac{9}{10}=-\frac{c}{5}$

$-\frac{9}{2}=c$

The answer is option A.

Question 77:Â Suppose for all integers x, there are two functions f and g such that $f(x) + f (x – 1) – 1 = 0$ and $g(x ) = x^{2}$. If $f\left(x^{2} – x \right) = 5$, then the value of the sum f(g(5)) + g(f(5)) is

77)Â Answer:Â 12

Solution:

Given,

$f\left(x\right)+f\left(x-1\right)=1$ …… (1)

$f\left(x^2-x\right)=5$ ……Â  (2)

$g\left(x\right)=x^2$

Substituting x = 1 in (1) and (2), we get

f(0) = 5

f(1) + f(0) = 1

f(1) = 1 – 5 = -4

f(2) + f(1) = 1

f(2) = 1 + 4 = 5

f(n) = 5 if n is even and f(n) = -4 if n is odd

f(g(5)) + g(f(5)) = f(25) + g(-4) = -4 + 16 = 12

Question 78:Â In an examination, there were 75 questions. 3 marks were awarded for each correct answer, 1 mark was deducted for each wrong answer and 1 mark was awarded for each unattempted question. Rayan scored a total of 97 marks in the examination. If the number of unattempted questions was higher than the number of attempted questions, then the maximum number of correct answers that Rayan could have given in the examination is

78)Â Answer:Â 24

Solution:

Let the number of questions attempted be x+y out of which x are correct and y are incorrect and the number of questions unattempted be z.

It is given,

x + y + z = 75 …… (1)

3x – y + z = 97 …… (2)

(2)-(1) -> x – y = 11

(1)+(2) -> 2x + z = 86

z > x + y

z > 75 – z

z > 37.5

Minimum possible value of z is 38

2x + 38 = 86

2x = 48

x = 24

The maximum number of correct answers is 24.

Question 79:Â The number of distinct integer values of n satisfying $\frac{4-\log_{2}n}{3-\log_{4}n} < 0$, is

79)Â Answer:Â 47

Solution:

LetÂ $\ \log_2n=y$

$\ \ \frac{\ 4-y}{3-\frac{y}{2}}<0$

$\ \ \left(4-y\right)\left(3-\frac{y}{2}\right)<0$

$\ \ \left(4-y\right)\left(6-y\right)<0$

$\ \ \left(y-4\right)\left(y-6\right)<0$

$4 < y < 6$

$4<\log_2n<6$

$2^4<n<2^6$

$16<n<64$

n can take values from 17 to 63(inclusive).

The number of n values possible = 47

Question 80:Â If $c=\frac{16x}{y}+\frac{49y}{x}$ for some non-zero real numbers x and y, then c cannot take the value

a)Â $60$

b)Â $-50$

c)Â $-70$

d)Â $-60$

80)Â AnswerÂ (B)

Solution:

LetÂ $\frac{x}{y}\ be\ t$

Therefore,Â $c=16t\ +\ \frac{49}{t}$

Applying AM>= GM

$\frac{\left(16t\ +\ \frac{49}{t}\right)}{2}\ge\ \left(16t\times\frac{49}{t}\right)^{\frac{1}{2}}$

$16t\ +\ \frac{49}{t}\ge56$

When t is positive then c is greater than equal to 56.

When t is negative then c is less than equal to -56.

ThereforeÂ $c\ \in\ \left(-\infty,\ -56\right]\ âˆª\ \left[56,\infty\ \right]$

As -50 is not in the range of c so it is the answer

Question 81:Â Suppose k is any integer such that the equation $2x^{2}+kx+5=0$ has no real roots and the equation $x^{2}+(k-5)x+1=0$ has two distinct real roots for x. Then, the number of possible values of k is

a)Â 9

b)Â 7

c)Â 8

d)Â 13

81)Â AnswerÂ (A)

Solution:

$2x^{2}+kx+5=0$ has no real roots so D<0

$k^2-40\ <0$

$\left(k-\sqrt{40}\right)\left(k+\sqrt{40}\right)<0$

$k\in\left(-\sqrt{40},\sqrt{40}\right)$

$x^{2}+(k-5)x+1=0$ has two distinct real roots so D>0

$\left(k-5\right)^2-4>0$

$k^2-10k+21>0$

$\left(k-3\right)\left(k-7\right)>0$

$k\in\left(-\infty\ ,3\right)âˆª\left(7,\infty\ \right)$

Therefore possibe value of k are -6, -5, -4, -3, -2, -1, 0, 1, 2

In 9 total 9 integer values of k are possible.

Question 82:Â If $(3+2\sqrt{2})$ is a root of the equation $ax^{2}+bx+c=0$ and $(4+2\sqrt{3})$ is a root of the equation $ay^{2}+my+n=0$ where a, b, c, m and n are integers, then the value of $(\frac{b}{m}+\frac{c-2b}{n})$ is

a)Â 0

b)Â 1

c)Â 3

d)Â 4

82)Â AnswerÂ (D)

Solution:

a, b, c, m and n are integers so if one root isÂ $3+2\sqrt{2}$ then the other root isÂ $3-2\sqrt{2}$

Sum of roots = 6Â = -b/aÂ  or b= -6a

Product of roots = 1 = c/aÂ  or c=a

a, b, c, m and n are integers so if one root is $4+2\sqrt{3}$ then the other root is $4-2\sqrt{3}$

Sum of roots = 8 = -m/a or m = -8a

product of roots = 4 = n/a or n = 4a

$(\frac{b}{m}+\frac{c-2b}{n})$

= $\frac{6a}{8a}+\frac{\left(a+12a\right)}{4a}=\frac{3}{4}+\frac{13}{4}=\frac{16}{4}=4$

Question 83:Â Let r be a real number and $f(x) = \begin{cases}2x -r & ifx \geq r\\ r &ifx < r\end{cases}$. Then, the equation $f(x) = f(f(x))$ holds for all real values of $x$ where

a)Â $x > r$

b)Â $x \leq r$

c)Â $x \neq r$

d)Â $x \geq r$

83)Â AnswerÂ (B)

Solution:

When x< r

f(x) = r

f(x) =Â f(f(x))

r = f(r)

r= 2r-r

r=r

When x>=r

f(x) = 2x-r

f(x) = f(f(x))

2x-r = f(2x-r)

2x-r = 2(2x-r) – r

2x-r = 4x-3r

or, x=r

Therefore x<= r

Question 84:Â The minimum possible value of $\frac{x^{2} – 6x + 10}{3-x}$, for $x < 3$, is

a)Â $-\frac{1}{2}$

b)Â 2

c)Â $\frac{1}{2}$

d)Â -2

84)Â AnswerÂ (B)

Solution:

Let $\frac{x^2-6x+10}{3-x}=p$

$x^2-6x+10=3p-px$

$x^2-\left(6-p\right)x+10-3p=0$

Since the equation will have real roots,

$\left(6-p\right)^2-4\times\left(10-3p\right)\ge0$

$p^2-12p+12p+36-40\ge0$

$p^2\ge4$

$p\ge2\ ,\ p\le-2$

Now, when $p=-2$, $x = 4$.Â  Since it is given that $x<3$, thus this value will be discarded.

Now,Â $\frac{1}{2}$ andÂ $-\frac{1}{2}$ do not come in the mentioned range.

when $p=2$, $x = 2$

Thus, the minimum possible value of p will be 2.

Thus, the correct option is B.

Alternate explanation:

Since $x<3$,

$3-x>0$

Let $3-x=y$. So, $y>0$.

Now, $\frac{x^2-6x+10}{3-x}=\frac{x^2-6x+9+1}{3-x}$

=> $\frac{\left(3-x\right)^2+1}{3-x}$

Since $3-x=y$, the equation will transform to $\frac{y^2+1}{y}$ orÂ $y+\frac{1}{y}$

The minimum value of the expression $y+\frac{1}{y}$ for $y>0$ will at $y=1$

i.e., Minimum value = $1+1=2$

Thus, the correct option is B.

Question 85:Â A donation box can receive only cheques of â‚¹100, â‚¹250, and â‚¹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of â‚¹15250. Then, the maximum possible number of cheques of â‚¹500 that the donation box may have contained, is

85)Â Answer:Â 12

Solution:

Let the number of 100 cheques, 250 cheques and 500 cheques be x, y and z respectively.

We need to find the maximum value of z.

x + y + z = 100 …… (1)

100x + 250y + 500z = 15250

2x + 5y + 10z = 305 …… (2)

2x + 2y + 2z = 200 ……. (1)

(2) – (1), we get

3y + 8z = 105

At z = 12, x = 3

Therefore, maximum value z can take is 12.