CAT 2021 Question Paper Quant Slot 3

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 + 50) = 250000

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

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

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

SO, x = 75

x + 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 + 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)

Let Rahul work at a units/hr and Gautam at b units/hour
Now as per the condition :
8a+6b =7.5a+7.5b
so we get 0.5a=1.5b
or a=3b
Therefore total work = 8a +6b = 8a +2a =10a
Now Rahul alone takes 10a/10 = 10 hours.

Initially number of matches = 40 
Now matches won = 12 
Now let remaining matches be x 
Now number of matches won = 0.6x
Now as per the condition :
$$\frac{\left(12+0.6x\right)}{40+x}=\frac{1}{2}$$
24 +1.2x=40+x
0.2x=16
x=80
Now when they won 90% of remaining = 80(0.9) =72
So total won = 84

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) < 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.

We have :$$\frac{\log_{15}{a}+\log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})}=4$$
We get $$\frac{\left(\frac{\log a}{\log\ 15}+\frac{\log a}{\log32}\right)}{\frac{\log a}{\log\ 15}\times\ \frac{\log a}{\log32}\ \ }=4$$
we get $$\log a\left(\log32\ +\log\ 15\right)=4\left(\log\ a\right)^2$$
we get $$\left(\log32\ +\log\ 15\right)=4\log a$$
=$$\log480=\log a^4$$
=$$a^4\ =480$$
so we can say a is between 4 and 5 .

Let us solve this question by assuming values(multiples of 100) and not variables(x).

Since we know that the female population was twice the male population in 1990, let us assume their respective values as 200 and 100.

Note that while assuming numbers, some of the population values might come out as a fraction(which is not possible, since the population needs to be a natural number). However, this would not affect our answer, since the calculations are in ratios and percentages and not real values of the population in any given year.

Now, we know that the female population became 1.25 times itself in 1990 from what it was in 1980.

Hence, the female population in 1980 = 200/1.25 = 160

Also, the female population became 1.2 times itself in 1980 from what it was in 1970.

Hence, the female population in 1970 = 160/1.2 = 1600/12 = 400/3

Let the male population in 1970 be x. Hence, the male population in 1980 is 1.4x.

Now, the total population in 1980 = 1.25 times the total population in 1970.

Hence, 1.25 (x + 400/3) = 1.4x + 160

Hence, x = 400/9.

Population change = 300 - 400/9 - 400/3 = 300 - 1600/9 = 1100/9

percentage change = $$\frac{\frac{1100}{9}}{\frac{1600}{9}}\times\ 100\ =\ \frac{1100}{16}\%=68.75\%$$

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

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

Now Anil Paints in 12 Days 
Barun paints in 16 Days 
Now together Arun , Barun and Chandu painted in 6 Days 
Now let total work be W 
Now each worked for 6 days 
So Anil's work = 0.5W 
Barun's work = $$\frac{6W}{16}=\frac{3W}{8}$$
Therefore Charu's work = $$\frac{W}{2}-\frac{3W}{8}=\ \frac{W}{8}$$
Therefore proportion of charu =$$\frac{24000}{8}=\ 3,000$$

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

$$(\sqrt[7]{10})^{1+2+...+n}>999$$

$$10^{\frac{1+2+...+n}{7}}>999$$

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.

The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once.

The different possibilities include :

Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are :

$$\frac{4!}{3!}$$ = 4.

Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are :

$$\frac{4!}{2!\cdot2!}=\ 6$$

Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are :

$$\frac{4!}{3!}$$ = 4

Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are :

$$\frac{4!}{2!}=\ 12$$

Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are :

$$\frac{4!}{2!}=\ 12$$

Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are :

$$\frac{4!}{2!}=\ 12$$

A total of 12 + 12 + 12 + 4 + 6 + 4 = 50 possibilities.

Alternatively


  We have to form 4 digit numbers using 1,2,3 such that 2,3 appears at least once 
So the possible cases :

Now we get $$\frac{4!}{2!}\times\ 3$$= 36 ( When one digit is used twice and the remaining two once )
$$\frac{4!}{3!}\times\ 2$$ = 8 ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time )
$$\frac{4!}{2!\times\ 2!}=\ 6$$( When 2 and 3 is used 2 times each )
So total numbers = 36+8+6 =50

Applying cosine rule in triangle ACD, 

$$100+X^2-2\times\ 10\times\ X\cos30=400$$

$$X^2-10X\sqrt{\ 3}-300=0$$

Solving, we get X = $$\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)$$

Hence, area = 10Xsin 30 = $$\frac{\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)10}{2}$$

= $$25(\sqrt{3}+\sqrt{15})$$

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 with 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 test 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.