CAT 2022 Question Paper Slot 3

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

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