XAT 2022 Question Paper

The two types of apples sold A and B are bought in the ratio of 5: 8.

Considering the weights to be 5x and 8x.

The cost price of A is 20 percent higher than that of B.

Considering the cost price of B = y, A = 6y/5.

The total cost price of A = $$\left(5x\right)\cdot\left(\frac{6y}{5}\right)$$

The total cost price of B = $$\left(8x\right)\cdot\left(y\right)$$

THe total cost price = 8xy + 6xy = 14xy

14xy = 2800.

xy = 200.

THe cost price of A = 1200.

THe cost price of B = 1600.

A is sold a profit of 15 percent. 15 percent of 1200 = 180.

B is sold at a profit of 10 percent. 10 percent of 1600 = 160.

The total profit is 180 + 160

= 340

Given the ball falls from a height of 3 meters.

The ball reaches a height which 0.8 times the original height every time.

Hence this is in the form of a geometric progression. We need to count distance when the ball flies upward and downward.

Hence considering every time the ball flies upward to a series with terms :

h1, h2,..........................

Every time the ball falls down to be

d1, d2 ,...............

h1 = (0.8)*3, h2 = (0.8)*(0.8)*3 ,.........................

d1 = 3, d2 =3*(0.8), d3 = 3*(0.8)*(0.8)................

h1+h2 ........ = Sum of an infinite geometric progression. = 3*0.8(1+0.8+$$0.64+...............$$)

THe sum of an infinite GP with r less than 1 is : $$\frac{a}{1-r}$$

= $$2.4\cdot\left(\frac{1}{1-0.8}\right)\ =\ 12\ meters$$

The sum of d1+d2++........................

= 3 + (h1+h2+..................) = 15.

The total distance = 15 + 12 = 27 meters.

Considering the two numbers to a, b :

We were given that :

$$a^3+b^3\ =\ 128$$

$$\frac{1}{a^3}+\ \frac{1}{b^3}=\ 2$$

$$\frac{\left(a^3+b^3\right)}{a^3\cdot b^3}=\ 2\ =\ \frac{128}{k}$$

k = 64.

Hence $$a^3\cdot b^3\ =\ 64$$

a*b = 4 and $$a^2\cdot b^2\ =\ 16$$

The age of Nadeem is a two-digit number. When squared yields a three-digit number whose last digit is Y. Y is a prime number.

Using statement 1:

When a number is squared :

The last digit of the number can be :

1, 2, 3, 4, 5, 6, 7, 8, 9. When squared the last digit can be :

For a number ending with 1: 1

For a number ending with 2: 4

For a number ending with 3: 9

For a number ending with 4: 6

For a number ending with 5: 5

For a number ending with 6: 6

For a number ending with 7: 9

For a number ending with 8: 4

For a number ending with 9: 1

The only possible prime number is 5.

Hence the last digit of X is 5 and Y is 5.

Using statement 2 : Y = X/3.

This alone cannot be sufficient to determine the possibilities for Y and X.

Combining both the statements :

Since Y = 5, then the value of X = 15.

The age is equal to 15.

Given the distances are :

AE = 4 meters , EB = 8 meters and EC = 16 meters.

Considering the length of ED = K.

Given the angles DAE and angle DCE  are complementary.

Hence the angles are A and 90 - A.

Tan(90-A) = Cot A

$$\ \tan\ DAE\ =\frac{k}{4}$$ and $$\ \tan\ DCE\ =\frac{1}{\tan\ DAE}=\ \frac{k}{16}$$

Hence $$\frac{k}{16}=\ \frac{4}{k}$$

k = 8 meters.

The angle DBE is given by 

$$Tan\ DBE\ =\ \frac{k}{8}=\ 1$$

Hence the angle is equal to 45 degrees.

Using condition W :

The magnitudes of the net change in enrollment between 2014 and 2021 is closest among any two universities for TRU and MPU.

Going by the color of the lines the net change for different universities is:

Univ 1: 0.7

Univ 2: 4.4

Univ 3: 0.1

Univ 4: 0.2

Univ 5: 3.3

The closest among these are: Univ 3 and Univ 4. They can possibly be : (TRU/MPU)

Using condition X :

The university LTU must have the same enrollment in consecutive years at least twice :

LTU can either be Univ 3 or Univ 1 but since Univ 3 must be among TRU and MPU. LTU is university 1.

Using condition Y :

The increase in the enrollment of JSU between the years 2015 and 2019 is 50 percent of TRU's total enrollment in 2020.

Considering :

TRU = Univ 4 

The enrollment is 5.

TRU = Univ 3

The enrollment is 0.7

For TRU as Univ 3, there is no university whose increase in enrollment between 2015 and 2019 is 50 percent of TRU.

Hence TRU =  Univ 4.

Since the increase in enrollment for JSU is half of TRU. The increase must be half of 5 = 2.5

The only possible case is JSU = Univ 2.

MPU = Univ 3.

LTU = Univ 1

PKU = Univ 5.

Using condition W :

The magnitudes of the net change in enrollment between 2014 and 2021 is closest among any two universities for TRU and MPU.

Going by the color of the lines the net change for different universities is:

Univ 1: 0.7

Univ 2: 4.4

Univ 3: 0.1

Univ 4: 0.2

Univ 5: 3.3

The closest among these are: Univ 3 and Univ 4. They can possibly be : (TRU/MPU)

Using condition X :

The university LTU must have the same enrollment in consecutive years at least twice :

LTU can either be Univ 3 or Univ 1 but since Univ 3 must be among TRU and MPU. LTU is university 1.

Using condition Y :

The increase in the enrollment of JSU between the years 2015 and 2019 is 50 percent of TRU's total enrollment in 2020.

Considering :

TRU = Univ 4

The enrollment is 5.

TRU = Univ 3

The enrollment is 0.7

For TRU as Univ 3, there is no university whose increase in enrollment between 2015 and 2019 is 50 percent of TRU.

Hence TRU = Univ 4.

Since the increase in enrollment for JSU is half of TRU. The increase must be half of 5 = 2.5

The only possible case is JSU = Univ 2.

MPU = Univ 3.

LTU = Univ 1

PKU = Univ 5.

SInce LTU is univ 1 the university with an enrollment twice that of LTU = 2*(3.2)

= 6.4. 

Hence the only possible case close by is Univ 5 (PKU)

Using condition W :

The magnitudes of the net change in enrollment between 2014 and 2021 is closest among any two universities for TRU and MPU.

Going by the color of the lines the net change for different universities is:

Univ 1: 0.7

Univ 2: 4.4

Univ 3: 0.1

Univ 4: 0.2

Univ 5: 3.3

The closest among these are: Univ 3 and Univ 4. They can possibly be : (TRU/MPU)

Using condition X :

The university LTU must have the same enrollment in consecutive years at least twice :

LTU can either be Univ 3 or Univ 1 but since Univ 3 must be among TRU and MPU. LTU is university 1.

Using condition Y :

The increase in the enrollment of JSU between the years 2015 and 2019 is 50 percent of TRU's total enrollment in 2020.

Considering :

TRU = Univ 4

The enrollment is 5.

TRU = Univ 3

The enrollment is 0.7

For TRU as Univ 3, there is no university whose increase in enrollment between 2015 and 2019 is 50 percent of TRU.

Hence TRU = Univ 4.

Since the increase in enrollment for JSU is half of TRU. The increase must be half of 5 = 2.5

The only possible case is JSU = Univ 2.

MPU = Univ 3.

LTU = Univ 1

PKU = Univ 5.

All the universities can be uniquely determined without using the condition Z.

Boxes 7 to 16 contain a total of 100 pencils each. The minimum number of broken pencils the box can hold is 5 percent of the total pencils and a maximum of 20 percent of the total pencils.

5 percent of 100 = 5 and 20 percent of 100 = 20 pencils.

Hence the number of broken pencils must be in the range of 5 to 20.

The frequency of the different number of broken pencils is :

5 - 1

6 - 2

7 - 4

9 - 3

11 - 1

15 - 2

19 - 1

20 - 3

29 - 1

31 - 1

33 - 1

The boxes cannot contain 29, 31, 33 to be the number of broken pencils because they are beyond 20 percent.

Since boxes 1- 6 can contain only between 2.5 to 10 pencils. The remaining boxes which include broken pencils of numbers less than 10 must be a part of 7 - 16. Because boxes 17 - 20 cannot contain broken pencils of numbers less than 10.

Hence 7 - 16 must have 4 boxes that contain less than 10 broken pencils.

Going through the options :

Option A fails because this includes only 3 boxes with less than 10 pencils.

Option B fails because we only have 2 boxes with 6 broken pencils but this includes 3.

Option D fails because it does not include a box of 15 and a  box of 20 pencils which can only be a part of boxes with 100 or boxes with 200 pencils. Since boxes 17 - 20 can include only one among 15 or 20 because 29, 31, 33 are a part of this group. Hence this case fails.

Option E fails because this includes 5 boxes with broken pencils less than 10 but this is not possible because this must exactly contain 4 boxes with less than 10 pencils.

Option C is a feasible case containing :

1-6 ( 5, 6, 6, 9, 9, 9)  

7 - 16 ( 7, 7, 7, 7, 11, 15, 15, 19, 20, 20 ) 

17 - 20 (20, 29, 31, 33)

Going by the given options :

Option A: The boxes numbered 1- 6 have a capacity of 50 pencils. The maximum number of broken pencils they can contain is 20 percent of 20 pencils = 10 pencils. For boxes numbered 17 to 20 they must contain a minimum of 5 percent which is equivalent to 5 percent of 200. Hence this is true.

Option B: Boxes with broken pencils of 29, 31, 33 must be a part of 17 - 20. There must be one box containing broken pencils in the range of 10 to 20. There are three boxes in total containing exactly 20 pencils. A maximum of only one of the three can be a part of 17 - 20. The remaining must be a part of boxes 7- 16 because they cannot be a part of 1-6. Hence at least one box among 7 - 16 contains 20 percent of broken pencils which is the highest.

Option C: There are a total of ten boxes with less than 10 broken pencils. They can either be a part of boxes 1-6 or 7 - 16. Since boxes 1-6 can only take broken pencils with less than 10 in number. Hence of the 10 six must be a part of 1-6 and the remaining 4 must be a part of 7 - 16.

Option D: The only possibility for containing 20 percent of the broken pencils is only possible for 20 broken pencils which is 20 percent of 100. There must be at least 2 boxes in the range of 7 - 16 which contain 20 broken pencils which is equal to 20 percent. The third box can either be a part of (7-16) to (17-20). If this belongs to 17 - 20 then the case is not possible and hence cannot be concluded.

Option E: Boxes containing 29, 31, and 33 broken pencils must be a part of boxes 17 -20 because they are higher than the 20 percent range of boxes (1-6) and (7-16). Hence this can be concluded.