XAT 2021 - QUANT

$$\log_4mn=\log_2(m+n)$$ 

$$\sqrt{\ mn}=(m+n)$$

Squarring on both sides

$$m^2+n^2+mn\ =\ 0$$

Since m, n are positive real numbers, no value of m and n satisfy the above equations.

Let the number of eggs bought = 9x+2

number of eggs left after throwing away 2 = 9x

number of eggs kept in fridge = 5x

number of eggs brought to his mothers' house = 4x

number of eggs left after cooking 2 which are kept in fridge = 4x-2

Given, 4x-2 <=5

=> x$$\le\frac{7}{4}$$

Hence the max value of x is 1

Max number of eggs bought = 11

Let the total amount be 3x

Case 1: 

Smaller amount = x, rate of interest = 10

Larger amount = 2x, rate of interest = 5

Total amount received at the end of two years( smaller amount) = $$x\left(1+\frac{10}{100}\right)^2\ =\ 1.21x$$. CI = 0.21x

Total amount received at the end of two years( larger amount) = $$2x\left(1+\frac{5}{100}\right)^2\ =\ 2.205x$$   CI = 0.205x

Given, 0.21x + 0.205x = 830

=> x = 2000

2x= 4000

Case 2:

Smaller amount = x, rate of interest = 20

Larger amount = 2x, rate of interest = 10

Total amount received at the end of two years( smaller amount) = $$x\left(1+\frac{20}{100}\right)^2\ =\ 1.44x$$. CI = 0.44x

Total amount received at the end of two years( larger amount) = $$2x\left(1+\frac{10}{100}\right)^2\ =\ 2.42x$$ CI = 0.42x

Given, 0.44x+0.42x = 830

=> x = 965.11 which is not valid since it should be greater than 1000

Number of white phones = 2

Number of black phones = 2

Number of red phones = 1

customer 1 will have 3 choices

customer 2 will have 3 choices

customer 3 will have 3 choices

Hence total choices = 3 x 3 x 3 = 27

The cases not possible = BBB, RRR,WWW, RRB,RBR,BRR, RRW,RWR, WRR

Possible cases = 18

Probability = 18/27 = 2/3

The difference between the hour and minute hand of a clock is given by $$\left|30H-5.5m\right|$$. Here H is the current hour and m represents the number of completed minutes in the current hour.

In the given time frame of 2: 30 to 3: 00 pm.

At 2 : 30 pm the angle = $$\left|30\cdot2-5.5\cdot30\right|\ =\ 105\ \deg rees$$

At 3: 00 pm the angle = $$\left|30\cdot3-5.5\cdot0\right|\ =\ 90\ \deg rees$$

The function of $$\left|30\cdot H-5.5\cdot m\right|\ =\ $$ constantly increases as the value of m increases from 31, 32................ 59.

Because of the modulus function, the net value of the function remains positive

Between 2: 30 to 2: 59 the angle is constantly increasing. The minimum value is 2: 30 which is equal to 105 degrees which is greater than the 90 degrees when the time is 3: 00.

Hence 90 degrees is the minimum angle.

Let us assume there are 9 buses.

3 of them stop at C and 6 go non-stop

Given, One-third of the passengers, in the buses stopping at City C, continue to City B, while the rest alight at City C

=> Since all buses have equal capacity. we can say 2 will elite at C and 1 will proceed to B.

Hence required proportion = 1/7

Given, CE=0.5, BC = 1.3 and ED=2.5

Triangle CEB is a right-angled triangle => EB = 1.2

Triangles ECB is similar to triangle EDA

EB/EC = AE/ED  => AE = 6

Hence total distance travelled = AB + BC + CD = 7.2 + 1.3 + 3.5 = 11.5km

$$\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(x^2+1\right)}{x^2-1}\cdot\frac{\left(x-1\right)}{x+1}=\frac{\left(x^2+1\right)}{\left(x+1\right)^2}$$

This function is definitely greater than 0

$$let\ y=\frac{\left(x^2+1\right)}{\left(x+1\right)^2}$$

=> $$x^2\left(y-1\right)+2yx+\left(y-1\right)=0$$ which is quadratic in x

Disctiminant should be greater than 0

$$4y^2-4\left(y-1\right)^2\ge0$$

=> y>=1/2

When x =1, f(x)/g(x) = 1/3

Hence either the value should be greater than 1/2 or should be equal to 1/3

Let the speed of the stream be x

$$\frac{12}{5-x}=\frac{12}{5+x}+1$$

The value of x satisfying the above equation is 1

Now,

$$\frac{N}{5+1}+\frac{N}{5-1}\ge2$$

$$\frac{2N+3N}{12}\ge2$$

=> $$N\ge4.8$$

Sum of 3rd row = sum of 2nd column

=> 2x+4y = y+2z-1 

=> 2x+3y-2z= -1 ------- (A)

Sum of diagonals are also equal

=> 3x+4y+z-1 = y+z+2x+y+z

=> x+2y-z=1 -----(B)

Solving A and B we get y= 3

Putting it in A, we get x-z = -5 ----- (C)

Sum of 1st row = sum of 2nd column

5x+5y+z = 3x+4y+2z

=> 2x +y - z =0

Since y=3, 2x-z = -3 ------ (D)

Solving C and D we get x=2 and z=7

Hence N = 36

Let speed of deer = 5steps/second and speed of tiger = 6 steps/sec

Let deer cover 1 m in a step => tiger covers 2/3 m in a step

Hence speed of deer = 5m/s and spped of tiger = 6 x 2/3 m/s = 4m/s

Hence time taken by a deer to catch tiger = 40 seconds

Distance travelled by deer in 40 seconds = 5 x 40 =200 steps

Total percentage incentive when number of team members = 1 = 100%

Total percentage incentive when the number of team members =2 =160%

Total percentage incentive when the number of team members =3=180%

Total percentage incentive when the number of team members =4= 190%

Total percentage incentive when the number of team members >4 = 200%

From 1, Number of people in 8 different projects = 6, 3, 3,3,3, 2,2,2 respectively

From 2, Given, exactly three projects are rated C and 4.8 lakh is paid in total

A minimum of 3 lakhs has to be paid for rating C => 3 *1.6 = 4.8lakhs => All 2 member teams have been rated C

From 3, one project has been rated A*. Let that project be handled by the team of 3 members => Incentives = 180% of 6 = 10.8 lakh

Now remaining 6,3,3,3 should be either rated A or B and the total incentives should be equal to 45 - 10.8-4.8 = 29.4 lakhs

Let us assume 6 has been rated B => Incentives = 200% of 3 = 6 lakhs

The remaining 23.4 lakhs should come from 180%. $$\frac{23.4}{1.8}=13\ lakhs$$

Hence the remaining 3,3,3 can be rated as A, A, B

Hence final ratings are and total payouts are

6 - B - 6lakhs

3- A - 9 lakhs

3-A - 9 lakhs

3-B - 5.4 lakhs

3-A* - 10.8lakhs

2-C - 1.6 lakhs

2-C - 1.6 lakhs

2-C - 1.6 lakhs

Total percentage incentive when number of team members = 1 = 100%

Total percentage incentive when the number of team members =2 =160%

Total percentage incentive when the number of team members =3=180%

Total percentage incentive when the number of team members =4= 190%

Total percentage incentive when the number of team members >4 = 200%

From 1, Number of people in 8 different projects = 6, 3, 3,3,3, 2,2,2 respectively

From 2, Given, exactly three projects are rated C and 4.8 lakh is paid in total

A minimum of 3 lakhs has to be paid for rating C => 3 *1.6 = 4.8lakhs => All 2 member teams have been rated C

From 3, one project has been rated A*. Let that project be handled by the team of 3 members => Incentives = 180% of 6 = 10.8 lakh

Now remaining 6,3,3,3 should be either rated A or B and the total incentives should be equal to 45 - 10.8-4.8 = 29.4 lakhs

Let us assume 6 has been rated B => Incentives = 200% of 3 = 6 lakhs

The remaining 23.4 lakhs should come from 180%. $$\frac{23.4}{1.8}=13\ lakhs$$

Hence the remaining 3,3,3 can be rated as A, A, B

Hence final ratings are and total payouts are

6 - B - 6lakhs

3- A - 9 lakhs

3-A - 9 lakhs

3-B - 5.4 lakhs

3-A* - 10.8lakhs

2-C - 1.6 lakhs

2-C - 1.6 lakhs

2-C - 1.6 lakhs

Total percentage incentive when number of team members = 1 = 100% 

Total percentage incentive when the number of team members =2 =160%

Total percentage incentive when the number of team members =3=180%

Total percentage incentive when the number of team members =4= 190%

Total percentage incentive when the number of team members >4 = 200%

From 1, Number of people in  8 different projects = 6, 3, 3,3,3, 2,2,2 respectively

From 2, Given, exactly three projects are rated C and 4.8 lakh is paid in total

A minimum of 3 lakhs has to be paid for rating C => 3 *1.6 = 4.8lakhs => All 2 member teams have been rated C

From 3, one project has been rated A*. Let that project be handled by the team of 3 members => Incentives = 180% of 6 = 10.8 lakh

Now remaining 6,3,3,3 should be either rated A or B and the total incentives should be equal to 45 - 10.8-4.8 = 29.4 lakhs

Let us assume 6 has been rated B => Incentives = 200% of 3 = 6 lakhs

The remaining 23.4 lakhs should come from 180%. $$\frac{23.4}{1.8}=13\ lakhs$$

Hence the remaining 3,3,3 can be rated as A, A, B

Hence final ratings are and total payouts are 

6 - B - 6lakhs

3- A - 9 lakhs

3-A - 9 lakhs

3-B - 5.4 lakhs

3-A* - 10.8lakhs

2-C - 1.6 lakhs

2-C - 1.6 lakhs

2-C - 1.6 lakhs

From the given data the following table can be created:

Hence the value of x=10

From the given data the following table can be created:

Hence the value of x=10

From the given options, number of neutral first time female shoppers are the least

From the given data the following table can be created:

Hence the value of x=10

All the values can be uniquely determined

Since A, B and C are adjacent faces. If we remove them, the resultant solid will also be a cube with side 5.

Hence total number of cubes unpainted = $$5^3$$ = 125

We will try to maximize the value of the angle ACD:

For a fixed triangle ABC, the angle ACD can be maximized when we take the median CD to be perpendicular to AB and the value of AC is as small as possible, so that the sine of angle ACD, and hence, the angle ACD itself if maximized, as the value of AD is fixed at half of AB at 0.5.

Now, the least possible value of AC is 2. The triangle will be of sides (1,2,2).

Value of sin(ACD)=$$\frac{0.5}{2}=0.25$$

$$\angle\ ACD=\sin^{-1}\left(0,25\right)=14.78\approx\ 15.$$

Since $$x^2-y^2=20$$ and x,y,z are positive integers, 

(x+y)*(x-y) = 20, Hence x-y, x+y are factors of 20.

Since x, y are positive integers, x+y is always positive, and for the product of (x+y)*(x-y)  to be positive x-y must be positive.

x, y are positive integers and x-y is positive x must be greater than y.

The possible cases are : (x+y = 10, x-y = 2), (x+y = 5, x-y = 4).

The second case fails because we get x =9/2, y = 1/2 but x, y are integral values

For case one x = 6, y = 4.

$$y^3 - 2x^2 - 4z \geq -12$$

Substituting  the values of x and y, we have :

64 - 72 - 4*z $$\ge\ -12$$

 -8 - 4*z $$\ge\ -12$$

z$$\le\ 1$$

Since x, y, z are positive integers, the only possible value for z is 1.

Flight become znmorl

Let's assume $$k>3$$

So flight will become thgilf -> znmrol. Hence the value of k will be 6

There are 5 ladies whose weights are 50 or above

There ages are 50,50, 70,60 and 80 

Average = 310/5 = 62

For the highest BMI, weight should be as high as possible and height as little as possible.

Hence it is possible with the person with a weight of 69 kg and a height of 1.6m

His BMI will be $$\frac{69}{\left(1.6\right)^2}=27$$

The BMI of 1st oldest person= $$\frac{40}{\left(1.5\right)^2}=\ 17.77$$

The BMI of next oldest person = $$\frac{61}{\left(1.75\right)^2}=19.9$$

Let ED = $$\sqrt{\ 3}x$$
In triangle CDE, $$\tan\ 30\ =\ \frac{CD}{\sqrt{\ 3}x}$$ => CD = x

In triangle BDE, $$\tan\ 45\ =\ \frac{BD}{\sqrt{\ 3}x}$$ => BD = $$\sqrt{\ 3}x$$ => BC = $$\sqrt{\ 3}x\ -x$$

In triangle ADE, $$\tan\ 60\ =\ \frac{AD}{\sqrt{\ 3}x}$$ => AD = 3x => AB = $$3x\ -\ \sqrt{\ 3}x$$

AB : BC : CD = $$(3 - \sqrt 3) : (\sqrt 3 - 1) : 1$$

We know that radius of circle P is 2 cm 

Length of MO = 1 cm

radius of circle R = AO = $$\sqrt{\ 2^2-1^2}$$ = $$\sqrt{\ 3}$$

Area of overlap between the circle R and P(shaded region) = semi circle area of R + area of segment ANBOA

Area of segment ANBOA = area of sector ANBM - area of triangle AMB

                                        = $$\frac{120}{360}\pi\ \left(2\right)^2$$ - $$\frac{1}{2}\left(1\right)\left(2\sqrt{\ 3}\right)$$

                                        = $$\frac{4\pi}{3}-\sqrt{\ 3}$$

Area of overlap between circle R and P = $$\frac{\pi\ \left(\sqrt{\ 3}\right)^2}{2}$$+ $$\frac{4\pi}{3}-\sqrt{\ 3}$$

                                                               = $$\ \frac{\ 17\pi\ }{6}-\sqrt{\ 3}$$

Answer is option E.

All the four friends will meet at the starting point after LCM(60,90,40,80) = 720 seconds.

Number of laps by A in 720 seconds = 12

Number of laps by B in 720 seconds = 8

Number of laps by C in 720 seconds = 18

Number of laps by D in 720 seconds = 9

Together they complete = 47 laps

Let the 6 digit number be  _ _ _ 42_

It is divisible by 2,3,5,7,11,13

Since the number is divisible by both 2 and 5 the last digit of the number must be 0.

The number is also divisible by 3, 7, 11, and 13.

Hence the number must also be divisible by 7*11*13.

=7*11*13 = 1001.

A number which is a multiple of 1001 is of the form abcabc.

This is because abc*(1001) = abc*(1000+1) = abc000 + abc = abcabc.

Hence the number is 420420.

The first digit is 4.