CAT 2022 Question Paper Quant Slot 3

Let $$\frac{x}{y}\ be\ t$$

Therefore, $$c=16t\ +\ \frac{49}{t}$$

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.

Therefore $$c\ \in\ \left(-\infty,\ -56\right]\ ∪\ \left[56,\infty\ \right]$$

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

$$2x^{2}+kx+5=0$$ 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

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

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

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

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

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.

$$(\sqrt{\frac{7}{5}})^{3x-y}=\frac{875}{2401}$$

$$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\frac{125}{343}$$

$$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\left(\frac{7}{5}\right)^{-3}$$

3x-y = -6

$$(\frac{4a}{b})^{6x-y}=(\frac{2a}{b})^{y-6x}$$

Therefor, y=6x as the bases are different so the power should be zero for the results to be equal.

3x-y=-6

or, 3x - 6x = -6

or x= 2

y= 6x = 12

x+y = 14

Let the six numbers be a, b, c, d, e, f in ascending order

a+b = 28

e+f =  56

If we want to maximise the average then we have to maximise the value of c and d and maximise e and minimise f

e+f = 56

As e and f are distinct natural numbers so possible values are 27 and 29

Therefore c and d will be 25 and 26 respecitively

So average = $$\frac{\left(a+b+c+d+e+f\right)}{6}=\frac{\left(28+25+26+56\right)}{6}=\frac{135}{6}=22.5$$

Area of ABD : Area of BDC = 1:1

Therefore, area of ABD = 54

Area of ADE : Area of EDB = 1:1

Therefore, area of ADE = 27

O is the centroid and it divides the medians in the ratio of 2:1

Area of BEO : Area of EOD = 2:1

Area of EOD = 9

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

Sum of roots = 6 = -b/a  or b= -6a

Product of roots = 1 = c/a  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

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

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

Let the unit of work done by 1 man in 1 hour and 1 day be 1 MDH unit (Man Day Hour).

Thus, in 7 hours per day for 10 days, the work done by N people =$$N\times\ 7\times\ 10$$ MDH units.

Since this is equal to 35% of the total work,

35% of the total work = $$N\times\ 7\times\ 10$$ MDH units.

Total work = $$\frac{\left(N\times\ 100\times\ 7\times\ 10\right)}{35}=200\times N$$ MDH units.

The work left = $$200N-70N=130N$$ MDH units.

Now, 10 people left the job. So, the number of people left = (N-10)

Since (N-10) people completed the rest of work in 14 days by working 10 hours a day,

$$(N-10)\times\ 14\times\ 10=130N$$

$$10N=1400$$

N = 140

Thus, the correct option is D.

Initially: a glass 500cc milk and a cup 500cc water

Step 1: 150 cc of milk is transferred to the cup from glass

After step 1: Glass - 350 cc milk, Cup - 150 cc milk and 500 cc water

Step 2: 150 cc of this mixture is transferred from the cup to the glass

After step 2: 

Glass - 350 cc milk + 150 cc mixture with milk:water ratio 3:10

Cup - 500 cc mixture with milk:water ratio 3:10

water in glass : milk in cup = $$\frac{10}{13}\times150\ :\ \frac{3}{13}\times500=1:1$$

The answer is option A.

It is given,

$$\ \frac{\ 5.5\times1\times8000}{100}+\ \frac{\ 5.6\times1\times5000}{100}+\ \frac{\ x\times1\times7000}{100}=\frac{5}{100}\times20000$$

$$440+280+70x=1000$$

x = 4%

Interest = $$\ \frac{\ 20000\times4\times1}{100}$$ = Rs 800

The answer is option B.

Both the cars take 1.5 hrs to meet when they travel towards each other.

It is given, speed of slower car is 60 km/hr

Therefore, distance covered by slower car before they meet = 60*1.5 = 90 km

The answer is option B.

The number of 4-digit numbers possible using 1,1,2, and 4 is $$\frac{4!}{2!}=12$$

Number of 1's, 2's and 4's in units digits will be in the ratio 2:1:1, i.e. 6 1's, 3 2's and 3 4's.

Sum = 6(1) + 3(2) + 3(4) = 24

Similarly, in tens digit, hundreds digit and thousands digit as well.

Therefore, sum = 24 + 24(10) + 24(100) + 24(1000) = 24(1111)

Mean = $$\frac{24\left(1111\right)}{12}=2222$$

The answer is option A.

Sum of the three sides of a quadrilateral is greater than the fourth side.

Therefore, let the fourth side be

1+2+4>d or d<7

1+2+d>4 or d>1

Possible values of d are 2, 3, 4, 5 and 6.

General term = 38 + (n-1)17 = 17n + 21 = 17(n+1) + 4 = 17k + 4

Each term is in the form of 17k + 4

Least 3-digit number in the form of 17k + 4 is at k = 6, i.e. 106

Highest 3-digit number in the form of 17k + 4 is at k = 58, i.e. 990

106, 123, 140,..........., 990

990 = 106 + 17(n-1)

n = 53

Sum = $$\frac{53}{2}\left(106+990\right)=53\times548$$

Average = $$53\times\frac{548}{53}=548$$

Let the number of students in section A and B be a and b, respectively.

It is given, a = b - 10

$$\ \ \frac{\ 32a+60b}{a+b}$$ is an integer

$$\ \ \frac{\ 32a+60\left(a+10\right)}{a+a+10}=k$$

$$\ \ \frac{\ 46a+300}{a+5}=k$$

$$k=\ \frac{\ 46\left(a+5\right)}{a+5}+\frac{70}{a+5}$$

$$k=\ \ 46+\frac{70}{a+5}$$

a can take values 2, 5, 9, 30, 65

Difference = 65 - 2 = 63

When x< r

f(x) = r

f(x) = 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

BC is the diameter of circle C2 so we can say that $$\angle BAC=90^{\circ\ }$$ as angle in the semi circle is $$90^{\circ\ }$$

Therefore overlapping area = $$\frac{1}{2}$$(Area of circle C2) + Area of the minor sector made be BC in C1

AB= AC = 8 cm and as $$\angle BAC=90^{\circ\ }$$, so we can conclude that BC= $$8\sqrt{2}$$ cm

Radius of C2 = Half of length of BC = $$4\sqrt{2}$$ cm

Area of C2 = $$\pi\left(4\sqrt{2}\right)^2=32\pi$$ $$cm^2$$

A is the centre of C1 and C1 passes through B, so AB is the radius of C1 and is equal to 8 cm

Area of the minor sector made be BC in C1 = $$\frac{1}{4}$$(Area of circle C1) - Area of triangle ABC = $$\frac{1}{4}\pi\left(8\right)^2-\left(\frac{1}{2}\times8\times8\right)=16\pi-32$$ $$cm^2$$

Therefore,

Overlapping area between the two circles= $$\frac{1}{2}$$(Area of circle C2) + Area of the minor sector made be BC in C1

= $$\frac{1}{2}\left(32\pi\right)\ +\left(16\pi-32\right)=32\left(\pi-1\right)$$ $$cm^2$$

Since the total number of students, when divided by either 9 or 10 or 12 or 25 each, gives a remainder of 4, the number will be in the form of LCM(9,10,12,25)k + 4 = 900k + 4.

It is given that the value of 900k + 4 is less than 5000.

Also, it is given that 900k + 4 is divided by 11.

It is only possible when k = 2 and total students = 1804.

So, the number of 12 students group = 1800/12 = 150.

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

Now, $$\frac{1}{2}$$ and $$-\frac{1}{2}$$ 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 $$\frac{y^2+1}{y}$$ or $$y+\frac{1}{y}$$

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.

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.

Let the speed of Moody be 'x' steps/sec and that of the escalator be 'y' steps/sec.

In 30 seconds, Moody will finish riding the escalator when going in the same direction.

Thus, total steps = 30(x+y)

If Moody's speed becomes twice, the time becomes 20 seconds.

Thus, total steps = 20(2x+y)

Or    30x + 30y = 40x + 20y

Or    x = y

So, total steps = 60y.

Time taken by only escalator= 60y/y = 60s.

Let S be the slower ship and F be the faster ship.

It is given that when S travelled 8 km, the positions of ships with the port is forming a right triangle.

Since one of the angles is 60(since one vertex is still part of the equilateral triangle),

the other two vertexes will have angles of 30 and 90.

The distance between O and S = 24 - 8 = 16

In triangle OFS, $$\cos60^0\ =\ \frac{OF}{OS}$$

Thus, OF = 8.

Thus in the time, S covered 8 km, F will cover 24 - 8 = 16 km.

Thus, the ratio of their speeds is 2:1,

Thus, when F covers 24 km, S will cover 12 km.

The correct option is B.

Let the efficiency of Bob be 3 units/day. So, Alex's efficiency will be 6 units/day, and Cole's will be 2 units/day.

Since Bob can finish the job in 40 days, the total work will be 40*3 = 120 units.

Since Alex and Bob work on the first day, the total work done = 3 + 6 = 9 units.

Similarly, for days 2 and 3, it will be 5 and 8 units, respectively.

Thus, in the first 3 days, the total work done = 9 + 5 + 8 = 22 units.

The work done in the first 15 days = 22*5 = 110 units.

Thus, the work will be finished on the 17th day(since 9 + 5 = 14 units are greater than the remaining work).

Since Alex works on two days of every 3 days, he will work for 10 days out of the first 15 days.

Then he will also work on the 16th day.

The total number of days = 11.