CAT 2021 Question Paper Slot 2

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 of $$2+\sqrt{\ 3}$$ so the product and the sum of the roots are rational numbers which are represented by: $$\frac{b}{a},\ \frac{c}{a}$$

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

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

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

b = 4*a, c= a.

Since b = $$c^3$$

4*a = $$a^3$$

$$a^2=\ 4.$$

a = 2 or -2.

|a| = 2

Let initial volume be V, final be F for milk.

The formula is given by : $$F\ =\ V\cdot\left(1-\frac{K}{V}\right)^n$$ n is the number of times the milk is drawn and replaced.

so we get $$F=\ V\left(1-\frac{K}{V}\right)^{^2}$$
here K =9
we get
$$\frac{16}{25}V\ =\ V\ \left(1-\frac{9}{V}\right)^{^2}$$
we get $$1-\frac{9}{V}=\ \frac{4}{5}or\ -\frac{4}{5}$$

If considering $$1-\frac{9}{V}=-\frac{4}{5}$$

V =5, but this is not possible because 9 liters is drawn every time.

Hence : $$1-\frac{9}{V}=\frac{4}{5},\ V\ =\ 45\ liters$$

All the sides of the rhombus are equal.

The area of a rhombus is $$12\ cm^2$$

Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal.

The area of a rhombus is $$\left(\frac{1}{2}\right)\left(d1\right)\cdot\left(d2\right)\ =\ 12$$

d1*d2 = 24.

The length of the side of a rhombus is given by $$\frac{\sqrt{\ d1^2+d2^2}}{2}$$. This is because the two diagonals and a side from a right-angled triangle with sides d1/2, d2/2 and the side length.

$$\frac{\sqrt{\ d1^2+d2^2}}{2}=\ 5$$

Hence $$\sqrt{\ d1^2+d2^2}\ =\ 10$$

$$d1^2+d2^2\ =\ 100$$

Using d1*d2 = 24, 2*d1*d2 = 48.

$$d1^2+d2^2\ +2\cdot d1\cdot d2=\ 100+48\ =\ 148$$

$$d1^2+d2^2\ -2\cdot d1\cdot d2=\ 100-48\ =\ 52$$

$$d1+d2\ =\ \sqrt{\ 148}$$  (1)

d1-d2 = $$\sqrt{52}$$   (2)

(1) + (2)= 2*(d1) = 2*($$\sqrt{\ 37}+\sqrt{\ 13}$$)

d1 = $$\sqrt{\ 37}+\sqrt{\ 13}$$

or 

In a rhombus the area of a Rhombus is given by :

The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be 2a, 2b.

The area of a Rhombus is : $$\left(\frac{1}{2}\right)\cdot\left(2a\right)\cdot\left(2b\right)\ =\ 12$$

ab =6.

The length of each side is : $$\sqrt{\ a^2+b^2}$$ = 5, $$a^2+b^{2\ }=\ 25,\ $$

$$\left(a+b\right)^2=\ 37,\ \left(a+b\right)\ =\ \sqrt{\ 37}$$

($$\left(a-b\right)^2=\ 13,\ a-b\ =\ \sqrt{\ 13}$$

$$2a\ =\ \left(\sqrt{\ 37}+\sqrt{\ 13}\right)$$,  $$2b\ =\ \left(\sqrt{\ 37}-\sqrt{\ 13}\right)$$.

2a is longer diagonal which is equal to $$\ \left(\sqrt{\ 37}+\sqrt{\ 13}\right)$$

We have :
$$\log_2\left\{3+\log_3\left\{4+\log_4\left(x-1\right)\right\}\right\}=2$$
we get $$3+\log_3\left\{4+\log_4\left(x-1\right)\right\}=4$$
we get $$\log_3\left(4+\log_4\left(x-1\right)\ =\ 1\right)$$
we get $$4+\log_4\left(x-1\right)\ =\ 3$$
$$\log_4\left(x-1\right)\ =\ -1$$
x-1 = 4^-1 
x = $$\frac{1}{4}+1=\frac{5}{4}$$
4x = 5

We are given that :

Let DP =x 
So AB =x 
Now DP=CP 
So CD = 2x 
Now let height of trapezium be h
we can say A(Parallelogram ABPD ) = xh
And A (BPC) =$$\frac{1}{2}xh$$
Now by condition$$xh-\frac{1}{2}xh=10$$
$$\frac{xh}{2}=10$$
so xh =20
Now therefore area of trapezium ABCD = $$\frac{1}{2}\left(x+2x\right)h\ =\ \frac{3}{2}xh\ =\ 30$$

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

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.

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

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

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

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, $$[\frac{3}{7},\frac{1}{2})$$

Now as per the given series :
we get $$x_1=1+2\ =3$$
Now $$x_1-x_2=\ 8$$
so$$x_2=-5$$
Now $$x_1-x_2+x_3\ =\ 15$$
so $$x_3\ =7$$
so we get $$x_n\ =\left(-1\right)^{n+1}\left(2n+1\right)$$
so $$x_{49}\ =\ 99$$ and $$x_{50}\ =-101$$
Therefore $$x_{49\ }+x_{50}\ =-2$$

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.

Let Entire work be W 
Now Anil worked for 24 days 
Bimal worked for 14 days and Charu worked for 14 days .
Now Anil Completes W in 60 days 
so in 24 days he completed 0.4W 
Bimal completes W in 84 Days
So in 14 Days Bimal completes = $$\frac{W}{6}$$
Therefore work done by charu = $$W-\frac{W}{6}-\frac{4W}{10}$$= $$\frac{26W}{10}$$=$$\frac{13W}{30}$$
Therefore proportion of Charu = $$\frac{13}{30}\times\ 21000$$=9100

Let the number of white balls be x and black balls be y 
So we get x+y =450       (1)
Now metallic black balls = 0.5y
Metallic white balls = 0.4x
From condition 0.4x=0.5y
we get 4x-5y=0     (2)
Solving (1) and (2) we get
x=250 and y =200
Now number of Non Metallic balls = 0.6x+0.5y = 150+100 = 250