CAT 2021 Question Paper Slot 3

Given $$x_{n+1}\ =\ x_n\ +\ n\ -1$$ and x1 = -1.

Considering 

x1   = -1.      (1)

x2   = x1+1-1 = x1 + 0    (2)

x3   = x2 + 2 - 1  =x2 + 1     (3)

x4   = x3 + 3 - 1 = x3 + 2        (4)

x100 = x99 + 98      (100)

Adding the LHS and RHS for the hundred equations we have:

(x1+x2+......................x100) = (-1+0+.........98) + (x1+x2+...............x99)

Subtracting this we have :

(x1+...........x100) - (x1+............. x 99) = $$\frac{\left(98\cdot99\right)}{2}$$ - 1.

x100 = 4851 - 1 = 4850

Alternatively

$$x_1=-1$$

$$x_2=x_1+1-1=x_1=-1$$

$$x_3=x_2+2-1=x_2+1=-1+1=0$$

$$x_4=x_3+3-1=x_3+2=0+2=2$$

$$x_5=x_4+4-1=x_4+3=2+3=5$$

......

If we observe the series, it is a series that has a difference between the consecutive terms in an AP.

Such series are represented as $$t\left(n\right)=a+bn+cn^2$$

We need to find t(100).

t(1) = -1

a + b + c = -1

t(2) = -1

a + 2b + 4c = -1

t(3) = 0

a + 3b + 9c = 0

Solving we get,

b + 3c = 0

b + 5c = 1

c = 0.5

b = -1.5

a = 0

Now, 

$$t\left(100\right)=\left(-1.5\right)100+\left(0.5\right)100^2=-150+5000=4850$$

Let sum of marks of students be x 
Now therefore x = 25*50 =1250 
Now to maximize the marks of the toppers 
We will minimize the marks of 20 students 
so their scores will be (30,31,32.....49 ) 
let score of toppers be y
so we get 5y +$$\frac{20}{2}\left(79\right)$$=1250
we get 5y +790=1250
5y=460
y=92
So scores of toppers = 92

Profit per boarder = Total profit / Number of boarders.

Let the number of boarders be n.

Profit/boarder = 1600 - (Total cost/n)

Let the total cost be a + bn, where a = fixed, and b is the variable additional cost per boarder.

Profit/boarder = 1600 - (a + bn)/n

Profit/boarder = 1600 - a/n - b 

1600 - a/50 - b = 200

1600 - a/75 - b = 250

Solving, we get a = 7500, and b = 1250

Hence, total profit with 80 people = 80 ( 1600 - 7500/80 - 1250) = 80 (350 - 7500/80) = 28000 - 7500 = Rs. 20500

Let us draw the rectangle.

Now, definitely, three sides should be fenced at Rs 100/ft, and one side should be fenced at Rs 200/ft.

In this question, we are going to assume that the L is greater than B.

Hence, the one side painted at Rs 200/ft should be B to minimise costs.

Hence, the total cost = 200B + 100B + 100L + 100L = 300B + 200L

Now, L x B = 60000

B = 60000/L

Hence, total cost = 300B + 200L = 18000000/L + 200L

To minimise this cost, we can use AM>=GM,

$$\frac{\frac{18000000}{L}+200L}{2}\ge\sqrt{\ \frac{18000000}{L}\times\ 200L}$$

$$\frac{18000000}{L}+200L\ge2\sqrt{\ 18000000\times\ 200}$$

$$\frac{18000000}{L}+200L\ge2\times\ 60000$$

Hence, minimum cost = Rs 120000.

We can say 40m is the perimeter of the park 
so side of rhombus = 10 
Now $$\frac{1}{2}\times\ d_1\times\ d_2\ =\ 96$$
so we get $$\ d_1\times\ d_2\ =\ 192$$   (1)
And as we know diagonals of a rhombus are perpendicular bisectors of each other :
so $$\ \frac{d_1^2}{4}+\ \frac{d_2^2}{4}=\ 100$$
so we get $$\ d_1^2+\ d_2^2=\ 400$$    (2)
Solving (1) and (2)
We get $$d_1=12$$ and$$d_2=16$$
Now the cost, in INR, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is= (12+16)(125) =3500

Let price of smallest cup be 2x and medium be 5x and large be y 
Now by condition  1 
we get $$2x\ \times\ \ 5x\ \times\ y\ =800$$
we get $$x^2y\ =80$$    (1)
Now as per second condition ;
$$\left(2x+6\right)\times\ \left(5x+6\right)\ y\ =3200$$        (2)
Now dividing (2) and (1)
we get $$\frac{\left(\left(2x+6\right)\times\ \left(5x+6\right)\right)}{x^2}=40$$
we get $$10x^2+42x+36\ =\ 40x^2$$
we get $$\ 30x^2-42x-36=0$$
$$5x^2-7x-6=0$$
we get x=2
So 2x=4 and 5x=10
Now substituting in (1) we get y =20
Now therefore sum = 4+10+20 =34
 

Considering the distance travelled by Mira in one minute = M,

The distance traveled by Amal in one minute = A.

Given if they walk in the opposite direction it takes 3 minutes for both of them to meet. Hence 3*(A+M) = C. (1)

C is the circumference of the circle.

Similarly, it is mentioned that if both of them walk in the same direction Amal completes 3 more rounds than Mira :

Hence 45*(A-M) = 3C. (2)

Multiplying (1)*15 we have :

45A + 45M = 15C.

45A - 45M = 3C.

Adding the two we have A = $$\frac{18C}{90}$$

Subtracting the two M = $$\frac{12C}{90}$$

Since Mira travels $$\frac{12C}{90}$$ in one minute, in one hour she travels :$$\frac{12C}{90}\cdot60\ =\ 8C$$

Hence a total of 8 rounds.

Alternatively, 

Let the length of track be L 
and velocity of Mira be a and Amal be b 
Now when they meet after 45 minutes Amal completes 3 more rounds than Mira
so we can say they met for the 3rd time moving in the same direction
so we can say they met for the first time after 15 minutes
So we know Time to meet = Relative distance /Relative velocity
so we get $$\frac{15}{60}=\frac{L}{a-b}$$     (1)
Now When they move in opposite direction
They meet after 3 minutes
so we get $$\frac{3}{60}=\frac{L}{a+b}$$    (2)
Dividing (1) and (2)
we get $$\frac{\left(a+b\right)}{\left(a-b\right)}=5$$
or 4a =6b
or a = 3b/2
Now substituting in (1)
we get :
$$\frac{L}{b}\times\ 2=\ \frac{15}{60}$$
so $$\frac{L}{b}\ =\frac{1}{8}$$
So we can say 1 round is covered in $$\frac{1}{8}$$ hours
so in 1-hour total rounds covered = 8.

Let the alloy contain x Kg silver and y kg copper 
Now when mixed with 3Kg Pure silver 
we get $$\frac{\left(x+3\right)}{x+y+3}=\frac{9}{10}$$
we get 10x+30 =9x+9y+27
9y-x=3    (1)
Now as per condition 2
silver in 2nd alloy = 2(0.9) =1.8
so we get$$\frac{\left(x+1.8\right)}{x+y+2}=\frac{21}{25}$$
we get 21y-4x =3   (2)
solving (1) and (2) we get y= 0.6 and x =2.4
so x+y = 3

We need to find out p.

Angle ADE = 180 - 2x

Angle BDF = 180 - 2y

Now, 180 - 2y + p + 180 - 2x = 180 [Straight line = 180 deg]

p = 2x + 2y - 180

Also, x + y + 50 = 180 [Sum of the angles of triangle = 180]

x + y = 130

p = 260 - 180 = 80 degrees.

Bank A: 6% p.a. 1/2 yearly (CI)

Bank B: x% p.a (SI)

Bank C: 2x% p.a (SI)

Let Raju invest Rs P in bank B for t years. Hence, Rupa invests Rs 10,000 in bank C for 2t years.

Now, 

$$P\left(\frac{x}{100}\right)t\ =\ P\left(1+\frac{3}{100}\right)^2-P$$

$$\left(\frac{x}{100}\right)t\ =\ 1.0609-1$$

$$\left(\frac{x}{100}\right)t\ =\ 0.0609$$

We need to calculate

SI = $$10000\times\ 2t\times\ \left(\frac{2x}{100}\right)=40000\left(\frac{x}{100}\right)t=40000\times\ 0.0609=2436$$