# Algebra - Simple Equations - Previous Year CAT/MBA Questions

The best way to prepare for Algebra - Simple Equations is by going through the previous year **Algebra - Simple Equations OMET questions**.
Here we bring you all previous year Algebra - Simple Equations OMET questions along with detailed solutions.

It would be best if you clear your concepts before you practice previous year Algebra - Simple Equations OMET questions.

**XAT 2023 QADI | Algebra - Simple Equations OMET Question**

Raju and Sarita play a number game. First, each one of them chooses a positive integer independently. Separately, they both multiply their chosen integers by 2, and then subtract 20 from their resultant numbers. Now, each of them has a new number. Then, they divide their respective new numbers by 5. Finally, they added their results and found that the sum is 16. What can be the maximum possible difference between the positive integers chosen by Raju and Sarita?

- (a)
67

- (b)
58

- (c)
49

- (d)
40

- (e)
None of the above

Answer: Option B

**Explanation** :

Let the numbers choosen by Raju and Sarita be r and s respectively.

Multiplying both the numbers with 2, we get the numbers as 2r and 2s respectively.

Subtracting 20 from both the numbers, we get the numbers as 2r - 20 and 2s - 20 respectively.

Dividing both the numbers by 5, we get the numbers as (2r - 20)/5 and (2s - 20)/5 respectively.

∴ (2r - 20)/5 + (2s - 20)/5 = 16

⇒ (2r - 20) + (2s - 20) = 80

⇒ 2r + 2s = 120

⇒ r + s = 60

For maximum difference one number should be least possible and the other maximum possible.

Since r and s both are positive integers, the least value one of them can take is 1 hence the maximum value of other will be (60 - 1 = 59)

⇒ Maximum difference between them = 59 - 1 = 58.

Hence, option (b).

Workspace:

**XAT 2023 QADI | Algebra - Simple Equations OMET Question**

There are three sections in a question paper and each section has 10 questions. First section only has multiple-choice questions, and 2 marks will be awarded for each correct answer. For each wrong answer, 0.5 marks will be deducted. Any unattempted question in this section will be treated as a wrong answer. Each question in the second section carries 3 marks, whereas each question in the third section carries 5 marks. For any wrong answer or un-attempted question in the second and third sections, no marks will be deducted. A student’s score is the addition of marks obtained in all the three sections. What is the sixth highest possible score?

- (a)
92.5

- (b)
94

- (c)
95.5

- (d)
95

- (e)
None of the above

Answer: Option B

**Explanation** :

Marking Scheme:

Section A: Correct Answer = +2; Wrong/Unattempted Question = -0.5

Section B: Correct Answer = +3; Wrong/Question = -0.5

Section C: Correct Answer = +5; Wrong/Question = -0.5

Each section has 10 questions.

∴ Maximum marks = 10 × 2 + 10 × 3 + 10 × 5 = 100.

For each unattempted/wrong question in section A, marks obtained will go down by 2.5.

For each unattempted/wrong question in section B, marks obtained will go down by 3.

For each unattempted/wrong question in section C, marks obtained will go down by 5.

⇒ 2^{nd} highest marks (when 1 Q is wrong in section A) = 100 - 2.5 = 97.5

⇒ 3^{rd} highest marks (when 1 Q is wrong in section B) = 100 - 3 = 97

⇒ 4^{th} highest marks (when 1 Q is wrong in section C) = 100 - 5 = 95

⇒ 5^{th} highest marks (when 1 Q is wrong in section A and B) = 100 - 5.5 = 94.5

⇒ 6^{th} highest marks (when 2 Qs are wrong in section B) = 100 - 6 = 94

Hence, option (b).

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**XAT 2023 VALR | Algebra - Simple Equations OMET Question**

**Read the following statements.**

The leave policy is bound to be unpopular either with the management or among workers. If the leave policy is unpopular with the management, it should be modified. We should adopt a new policy if it is unpopular with workers.

If the above statements are true, which one of the following MUST also be true?

- (a)
If the leave policy is popular with the management, then we should adopt a new policy.

- (b)
If the leave policy is popular among workers, then we should adopt a new policy.

- (c)
We should attempt to popularize the leave policy among either the management or workers.

- (d)
We should modify the leave policy only if this will not reduce its popularity among workers.

- (e)
We should modify the leave policy if modification will not reduce its popularity with the management.

Answer: Option A

**Explanation** :

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**XAT 2022 QADI | Algebra - Simple Equations OMET Question**

Sheela purchases two varieties of apples - A and B - for a total of Rupees 2800. Theweights in kg of A and B purchased by Sheela are in the ratio 5 : 8 but the cost perkg of A is 20% more than that of B. Sheela sells A and B with profits of 15% and 10%respectively.

What is the overall profit in Rupees?

- (a)
340

- (b)
600

- (c)
240

- (d)
480

- (e)
380

Answer: Option A

**Explanation** :

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 = (5x) ∙ $\left(\frac{6y}{5}\right)$

The total cost price of B = (8x) ∙ (y)

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

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**XAT 2022 QADI | Algebra - Simple Equations OMET Question**

A marble is dropped from a height of 3 metres onto the ground. After the hitting theground, it bounces and reaches 80% of the height from which it was dropped. Thisrepeats multiple times. Each time it bounces, the marble reaches 80% of the heightpreviously reached. Eventually, the marble comes to rest on the ground.

What is the maximum distance that the marble travels from the time it was droppeduntil it comes to rest?

- (a)
15 m

- (b)
27 m

- (c)
24 m

- (d)
12 m

- (e)
30 m

Answer: Option B

**Explanation** :

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

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

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

The total distance = 15 + 12 = 27 meters

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**XAT 2022 QADI | Algebra - Simple Equations OMET Question**

The sum of the cubes of two numbers is 128, while the sum of the reciprocals oftheir cubes is 2. What is the product of the squares of the numbers?

- (a)
64

- (b)
256

- (c)
16

- (d)
48

- (e)
32

Answer: Option C

**Explanation** :

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

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

k = 64.

Hence a^{3} . b^{3} = 64

a*b = 4 and a^{2} .b^{2} = 16

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**XAT 2022 QADI | Algebra - Simple Equations OMET Question**

Nadeem’s age is a two-digit number X, squaring which yields a three-digit number,whose last digit is Y. Consider the statements below:

**Statement I:**

Y is a prime number

**Statement II:**

Y is one-third of X

To determine Nadeem’s age uniquely:

- (a)
either of I and II, by itself, is sufficient.

- (b)
only II is sufficient, but I is not.

- (c)
only I is sufficient, but II is not.

- (d)
it is necessary and sufficient to take I and II together.

- (e)
even taking I and II together is not sufficient.

Answer: Option D

**Explanation** :

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.

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**XAT 2022 QADI | Algebra - Simple Equations OMET Question**

A tall tower has its base at point K. Three points A, B and C are located at distances of 4 metres, 8 metres and 16 metres respectively from K. The angles of elevation of the top of the tower from A and C are complementary.

What is the angle of elevation (in degrees) of the tower’s top from B?

- (a)
60

- (b)
30

- (c)
45

- (d)
We need more information to solve this.

- (e)
15

Answer: Option C

**Explanation** :

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 = tan $\frac{1}{DAE}$ = $\frac{k}{16}$

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.

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**XAT 2022 QADI | Algebra - Simple Equations OMET Question**

Which amongst the pieces of information mentioned below, if removed, will notprevent us from uniquely identifying the five universities?

- (a)
Either X or Y

- (b)
Y

- (c)
Z

- (d)
NONE, since all four pieces of information are necessary to uniquely identify the five universities.

- (e)
X

Answer: Option C

**Explanation** :

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.

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**Read the following scenario and answer the THREE questions that follow.**

A pencil maker ships pencils in boxes of size 50, 100 and 200. Due to packaging issues, some pencils break.

About the 20 boxes he has supplied to a shop, the following information is available:

* Box no. 1 through 6 have 50 pencils, Box no. 7 through 16 have 100 pencils and Box no. 17 through 20 have 200 pencils.

* No box has less than 5% or more than 20% broken pencils.

Following is the frequency table of the number of broken pencils for the twenty boxes:

**XAT 2022 QADI | Algebra - Simple Equations OMET Question**

Which of the following can possibly be the sequence of the number of brokenpencils in Boxes 7-16?

- (a)
6,7,9, 11,15,19,20,20,20,29

- (b)
5,6,6,6,11,15,15,20,20,20

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

- (d)
7,7,9,9,11,13,15,19,20,20

- (e)
5,7,7,7,9,11,15,20,20,20

Answer: Option C

**Explanation** :

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)

Workspace:

**XAT 2022 QADI | Algebra - Simple Equations OMET Question**

Suppose that additionally it is known that the number of broken pencils in Boxes 17-20 are in increasing order. Which among the following additional information, if true, is not sufficient to uniquely know the number of defective pencils in each of the boxes numbered 17-20?

- (a)
Boxes no. 7-16 contains a total of 124 defective pencils.

- (b)
Boxes no. 17-20 contain a total of 108 defective pencils.

- (c)
Boxes no. 11-16 contain a total of 101 defective pencils.

- (d)
Box no. 17 contains more defective pencils than any box from among boxes no. 1 - 14.

- (e)
Boxes no. 7 - 16 contain a total of 133 defective pencils.

Answer: Option E

**Explanation** :

Going by the options :

Option A: Boxes 7- 16 contain a total of 124 pencils. Boxes (1-6) has 6 boxes with broken pencils which can be included from :

(5, 6, 6, 7, 7, 7, 7, 9, 9, 9). The minimum possible sum of the 6 pencils is : (5+6+6+7) = 24 and the maximum possible sum is (7+9+9+9) = 34.

Boxes 7 - 16 contains all the boxes with broken pencils except one among the boxes with broken pencils among 11 - 20 and hence ;

This can contain : (11+15+15+19+20+20) or (15+15+19+20+20+20) or (11+15+15+20+20+20) or (11+15+19+20+20+20 ) =100/109/101/105.

The only possible case to contain 124 pencils is by considering the case : (24+100 ) = (5, 6,6,7, 11+15+15+19+20+20) = 124 .

Hence box 17 - 20 must contain (20, 29, 31, 33).

Option B: Boxes 17 - 20 contain a total of 108 pencils. Since 29, 31, 33 pencils must be a part of 17 - 20 boxes. The remaining box must contain 108 - (29 +31+33) = 15 pencils. Hence the order is (15, 29, 31, 33).

Option C: Boxes 11- 16 contain a total of 101 defective pencils. This is only possible if the boxes here contain : ( 20, 20, 20, 15, 15, 11) pencils. Hence the box containing 19 pencils must be a part of boxes 17 - 20 and the remaining three contain 29, 31, 33. (19, 29, 31, 33)

Option D : Box number 17 containing more pencils than any box from box number 1- 14. Hence this only possible if Box 15, 16, 17 contains 20 pencils each and 18, 19, 20 contain (29, 31, 33).

Option E : Box 7-16 containing 133 broken pencils :

Boxes (1-6) has 6 boxes with broken pencils which can be included from :

(5, 6, 6, 7, 7, 7, 7, 9, 9, 9). The minimum possible sum of the 6 pencils is : (5+6+6+7) = 24 and the maximum possible sum is (7+9+9+9) = 34.

Boxes 7 - 16 contains all the boxes with broken pencils except one among the boxes with broken pencils among 11 - 20 and hence ;

This can contain : (11+15+15+19+20+20) or (15+15+19+20+20+20) or (11+15+15+20+20+20) or (11+15+19+20+20+20 ) =100/109/101/105.

This has multiple possibilities which include (109 + 24) or (101+32) or (105+28). Hence cannot be uniquely determined.

Workspace:

**XAT 2022 QADI | Algebra - Simple Equations OMET Question**

If both the sequences x, a1, a2, y and x, b1, b2, z are in A.P. and it is given that y > x and z < x, then which of the following values can $\left\{\frac{(a1-a2)}{(b1-b2)}\right\}$ possibly take?

- (a)
2

- (b)
5

- (c)
-3

- (d)
1

- (e)
0

Answer: Option C

**Explanation** :

The two given sequences in AP are:

x, a1, a2, y and x, b1, b2, z.

Additionally, it is given that : y > x and z < x.

Hence the common difference is not zero for both the series :

Since y > x the common difference is positive for the first series. (Considering the common difference to be d1)

Similarly z < x the common difference is negative for the given series. (Considering the common difference to be d2)

Now for the given value:

$\frac{(a1-a2)}{(b1-b2)}$

The value of a1 - a2 is negative and b1 - b2 is positive.

Hence the value of $\frac{(a1-a2)}{(b1-b2)}$ takes a negative value.

The only possible option is -3.

The answer is option C.

Workspace:

**XAT 2022 QADI | Algebra - Simple Equations OMET Question**

The Madhura Fruits Company is packing four types of fruits into boxes. There are126 oranges, 162 apples, 198 guavas and 306 pears. The fruits must be packed insuch a way that a given box must have only one type of fruit and must contain thesame number of fruit units as any other box.

What is the minimum number of boxes that must be used?

- (a)
21

- (b)
18

- (c)
44

- (d)
42

- (e)
36

Answer: Option C

**Explanation** :

The number of oranges, apples, guavas, and pears = 126, 162, 198, and 306.

Each box must contain an equal number of fruits with only one type of fruit. The additional condition provided is that there should be a minimum number of boxes in total.

The distribution is possible in multiple ways in such a way that distribution in each box is placed in such that each box contains a certain number of fruits n which is a factor for all the four given number of fruits :

Arrangement of 1 fruit of one kind in a basket.

2 is a factor of 126, 162, 198, and 306. So we can place 2 fruits of a particular kind in a basket.

Since we were asked for the minimum number of boxes this is possible when a maximum number of fruits of a kind are placed in a box.

Hence each box must contain the Highest common factor for the four numbers :

The prime factorization for the four numbers:

126 : 2 ∙ 7 ∙ 9, 162 : 2 ∙ 9 ∙ 9, 198 : 2 ∙ 9 ∙ 11, 306 = 2 ∙ 9 ∙ 17

The HCF is 18.

The number of boxes required for each :

$\frac{126}{18}$, $\frac{162}{18}$, $\frac{198}{18}$, $\frac{306}{18}$

7 + 9 + 11 + 17 = 44.

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**XAT 2022 QADI | Algebra - Simple Equations OMET Question**

Consider the real-valued function f(x) = $\frac{\mathrm{log}(3x-7)}{\sqrt{2{x}^{2}-7x+6}}$. Find the domain of f(x).

- (a)
$\left(\frac{7}{3},\infty \right)$

- (b)
R - $\left\{\frac{7}{3}\right\}$

- (c)
R - $\left\{\frac{3}{2},2\right\}$

- (d)
R - $\left\{\frac{3}{2},2,\frac{7}{3}\right\}$

- (e)
$\left(-\infty ,\frac{7}{3}\right)$

Answer: Option A

**Explanation** :

The function f(x) = $\frac{\mathrm{log}(3x-7)}{\sqrt{2{x}^{2}-7x+6}}$ is only defined when both the numerator and the denominator of the function are defined are the denominator is not equal to zero.

The logarithm of the function is only defined for positive values :

Hence 3x - 7 is greater than zero. Hence x > 7/3.

The value inside square root are defined for positive values. The value of the quadratic equation in the square root must be positive.

Hence 2x^{2} - 7x + 6 = 0 has the roots:

$\frac{(7+\sqrt{49-48})}{4}$, $\frac{(7-\sqrt{49-48})}{4}$ : 2, 3/2

The quadratic equation is positive for:

$\left(-\infty ,\frac{3}{2}\right)$ ∪ (2, ∞)

Since in order to be a part of the domain the values of x must be greater than 7/3 and 7/3 is greater than 2 all values of x which are greater than 7/3 must be a part of the domain for x.

Workspace:

**Read the following scenario and answer the three questions that follow.**

A company awards incentives to its employees for successful project performances. It rates successful project performance in categories A*, A, B, and C. Employees, in solo projects rated A*, A, B, and C, are awarded incentives ₹6 lakh, ₹5 lakh, ₹3 lakh, and ₹1 lakh respectively. When a project has multiple team members, the following scheme is used to award the incentives:

For example, for a project rated A, with three members, the team lead gets ₹4 lakh, and the other team members get ₹2.5 lakh each. A project always has a single team lead. Six employees: Altaf, Bose, Chakrabarthi, Dipa, Ernie, and Fatima receive a total of ₹45 lakh in incentives by participating in a total of eight different projects that does not involve any other person. Not all six employees are involved in all eight projects.

The following are additionally known about these eight projects:

1. One project involves all six employees. Four projects involve three each, and the rest, two each.

2. Exactly three projects are rated C, for which a total of ₹4.8 lakh is paid.

3. Only one project is rated A*

**XAT 2021 QADI | Algebra - Simple Equations OMET Question**

Total amount of money paid for projects rated A (in lakhs of Rupees) is:

- (a)
19

- (b)
15

- (c)
16

- (d)
17

- (e)
18

Answer: Option E

**Explanation** :

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

Workspace:

**Read the following scenario and answer the three questions that follow.**

A quick survey at the end of a purchase at buyagain.com asks the following three questions to each shopper:

1. Are you shopping at the website for the first time? (YES or NO)

2. Specify your gender: (MALE or FEMALE)

3. How satisfied are you? (HAPPY, NEUTRAL or UNHAPPY)

240 shoppers answer the survey, among whom 65 are first time shoppers. Furthermore:

i. The ratio of the numbers of male to female shoppers is 1 : 2 while the ratio of the numbers of unhappy, happy and neutral shoppers is 3 : 4 : 5

ii. The ratio of the numbers of happy first-time male shoppers, happy returning male shoppers, unhappy female shoppers, neutral male shoppers, neutral female shoppers and happy female shoppers is 1 : 1 : 4 : 4 : 6 : 6

iii. Among the first-time shoppers, the ratio of the numbers of happy male, neutral male, unhappy female and the remaining female shoppers is 1 : 1 : 1 : 2, while the number of happy first-time female shoppers is equal to the number of unhappy first-time male shoppers

**XAT 2021 QADI | Algebra - Simple Equations OMET Question**

What is the number of happy male shoppers?

- (a)
10

- (b)
15

- (c)
5

- (d)
20

- (e)
40

Answer: Option D

**Explanation** :

From the given data the following table can be created:

Hence the value of x=10

Workspace:

**Read the following scenario and answer the three questions that follow.**

A quick survey at the end of a purchase at buyagain.com asks the following three questions to each shopper:

1. Are you shopping at the website for the first time? (YES or NO)

2. Specify your gender: (MALE or FEMALE)

3. How satisfied are you? (HAPPY, NEUTRAL or UNHAPPY)

240 shoppers answer the survey, among whom 65 are first time shoppers. Furthermore:

i. The ratio of the numbers of male to female shoppers is 1 : 2 while the ratio of the numbers of unhappy, happy and neutral shoppers is 3 : 4 : 5

ii. The ratio of the numbers of happy first-time male shoppers, happy returning male shoppers, unhappy female shoppers, neutral male shoppers, neutral female shoppers and happy female shoppers is 1 : 1 : 4 : 4 : 6 : 6

iii. Among the first-time shoppers, the ratio of the numbers of happy male, neutral male, unhappy female and the remaining female shoppers is 1 : 1 : 1 : 2, while the number of happy first-time female shoppers is equal to the number of unhappy first-time male shoppers

**XAT 2021 QADI | Algebra - Simple Equations OMET Question**

Which among the following is the lowest?

- (a)
Number of neutral first-time female shoppers

- (b)
Number of unhappy first-time female shoppers

- (c)
Number of unhappy first-time male shoppers

- (d)
Number of neutral first-time male shoppers

- (e)
Number of happy returning male shoppers

Answer: Option A

**Explanation** :

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

Workspace:

**Read the following scenario and answer the three questions that follow.**

A quick survey at the end of a purchase at buyagain.com asks the following three questions to each shopper:

1. Are you shopping at the website for the first time? (YES or NO)

2. Specify your gender: (MALE or FEMALE)

3. How satisfied are you? (HAPPY, NEUTRAL or UNHAPPY)

240 shoppers answer the survey, among whom 65 are first time shoppers. Furthermore:

i. The ratio of the numbers of male to female shoppers is 1 : 2 while the ratio of the numbers of unhappy, happy and neutral shoppers is 3 : 4 : 5

ii. The ratio of the numbers of happy first-time male shoppers, happy returning male shoppers, unhappy female shoppers, neutral male shoppers, neutral female shoppers and happy female shoppers is 1 : 1 : 4 : 4 : 6 : 6

iii. Among the first-time shoppers, the ratio of the numbers of happy male, neutral male, unhappy female and the remaining female shoppers is 1 : 1 : 1 : 2, while the number of happy first-time female shoppers is equal to the number of unhappy first-time male shoppers

**XAT 2021 DM | Algebra - Simple Equations OMET Question**

Which among the following cannot be determined uniquely?

- (a)
The number of first-time happy male shoppers

- (b)
The number of returning male shoppers

- (c)
All the numbers can be determined uniquely

- (d)
The number of returning unhappy female shoppers

- (e)
The number of first-time neutral male shoppers

Answer: Option C

**Explanation** :

From the given data the following table can be created:

Hence the value of x=10

All the values can be uniquely determined

Workspace:

**XAT 2021 DM | Algebra - Simple Equations OMET Question**

Find z, if it is known that:

a: -y^{2} + x^{2} = 20

b: y^{3} - 2x^{2} - 4z ≥ -12 and

c: x, y and z are all positive integers

- (a)
Any integer greater than 0 and less than 24

- (b)
24

- (c)
We need one more equation to find z

- (d)
6

- (e)
1

Answer: Option E

**Explanation** :

Since x^{2} - y^{2} = 20 and x, y, z are positive integers, the only values possible are x = 6 and y = 4.

y^{3} - 2x^{2} - 4z 12 ≥ -12

Keeping the values of x and y, we get z ≤ 1

⇒ z = 1

Workspace:

**XAT 2021 DM | Algebra - Simple Equations OMET Question**

An encryption system operates as follows:

Step 1. Fix a number k (k ≤ 26).

Step 2. For each word, swap the first k letters from the front with the last k letters from the end in reverse order. If a word contains less than 2k letters, write the entire word in reverse order.

Step 3. Replace each letter by a letter k spaces ahead in the alphabet. If you cross Z in the process to move k steps ahead, start again from A.

Example: k = 2: zebra → arbez → ctdgb.

If the word “flight” becomes “znmorl” after encryption, then the value of k:

- (a)
5

- (b)
4

- (c)
7

- (d)
Cannot be determined uniquely from the given information

- (e)
6

Answer: Option E

**Explanation** :

Flight become znmorl

Let's assume k > 3

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

Workspace:

**Read the following scenario and answer the three questions that follow.**

The following plot describes the height (in cm), weight (in kg), age (in years) and gender (F for female, M for male) of 20 patients visiting a hospital.

A person’s body mass index (BMI) is calculated as weight (in kg) divided by squared height (measured in square metres). For example, a person weighing 100 kg and of height 100 cm (1m) will have a BMI of 100. A person with BMI less than or equal to 18.5 is considered as underweight, above 18.5 but less than or equal to 25 as normal weight, above 25 but less than or equal to 30 as overweight, and above 30 as obese.

**XAT 2021 DM | Algebra - Simple Equations OMET Question**

The highest BMI among all patients is approximately

- (a)
20

- (b)
33

- (c)
30

- (d)
27

- (e)
23

Answer: Option D

**Explanation** :

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 (${\left(\frac{69}{1.6}\right)}^{2}$) = 27

Workspace:

**Read the following scenario and answer the three questions that follow.**

The following plot describes the height (in cm), weight (in kg), age (in years) and gender (F for female, M for male) of 20 patients visiting a hospital.

A person’s body mass index (BMI) is calculated as weight (in kg) divided by squared height (measured in square metres). For example, a person weighing 100 kg and of height 100 cm (1m) will have a BMI of 100. A person with BMI less than or equal to 18.5 is considered as underweight, above 18.5 but less than or equal to 25 as normal weight, above 25 but less than or equal to 30 as overweight, and above 30 as obese.

**XAT 2021 DM | Algebra - Simple Equations OMET Question**

The BMI of the oldest person considered as normal weight is approximately

- (a)
20

- (b)
25

- (c)
22

- (d)
24

- (e)
19

Answer: Option A

**Explanation** :

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

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

Workspace:

**IIFT 23 Dec 2021 QA | Algebra - Simple Equations OMET Question**

The last two digits of the expression 1(1!)^{1!} + 2(2!)^{2!} + 3(3!)^{3!} + .... + 121(121!)^{121!}

- (a)
61

- (b)
71

- (c)
81

- (d)
91

Answer: Option C

**Explanation** :

From the 5th term onwards the last two of all the terms will 00.

The last two digits of 24 to the power of an even number will be 76 always.

So, last two digits of the above expression will be the last two digits of 1 + 2(2)^{2 }+ 3(6)^{6} + 4(76)

= 1 + 8 + 139968 + 304 = 140281

So last two digits is 81

Workspace:

**IIFT 23 Dec 2021 QA | Algebra - Simple Equations OMET Question**

Find the set S that denotes the set of all values of 'α' for which the roots of the equation (1 - α)x^{2} - 6αx + 8α = 0 is greater than 2.

- (a)
$\left(\frac{2}{5},\frac{1}{2}\right)$

- (b)
$\left(\frac{2}{5},\frac{32}{68}\right)$

- (c)
$\left(\frac{32}{68},1\right)$

- (d)
$\left(\frac{32}{68},\frac{1}{2}\right)$

Answer: Option D

**Explanation** :

αf(x) = (1 - α)x^{2} - 6αx + 8α = 0

Now roots are greater than 2 therefore,

$-\frac{b}{2a}$ > 2

f(2) > 0

D > 0

$-\frac{b}{2a}$ > 0

2(1 - α) > 0

$\frac{\alpha}{\alpha -1}$ < 0

α ∈ (0, 1)

f(2) > 0

(1 − α ) 4 − 12α + 8α > 0

4 − 8α > 0

a < $\frac{1}{2}$

D > 0

36α^{2} - 32α (1 - α) > 0

68α^{2} - 32α > 0

α(68α - 32) > 0

α ∈ (-∞, 0) ∪ $\left(\frac{32}{68},\infty \right)$

Taking the intersection of all we get α ∈ $\left(\frac{32}{68},\frac{1}{2}\right)$

Workspace:

**IIFT 05 Dec 2021 QA | Algebra - Simple Equations OMET Question**

Find the value of $\sqrt{552+\sqrt{552+\sqrt{552+...}}}$

- (a)
26

- (b)
-24

- (c)
24

- (d)
-26

Answer: Option C

**Explanation** :

x = $\sqrt{552+\sqrt{52+\sqrt{552+...}}}$

x^{2} - 552 = $\sqrt{552+\sqrt{52+\sqrt{552+...}}}$

x^{2} - x - 552 = 0

(x - 24)(x + 23) = 0

x cannot be negative. Therefore, $\sqrt{552+\sqrt{52+\sqrt{552+...}}}$= 24

Answer is option C.

Workspace:

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