Algebra - Simple Equations - Previous Year CAT/MBA Questions

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).

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).

Answer: Option A

Explanation :

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

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

Answer: Option C

Explanation :

Considering the two numbers to a, b :
We were given that:

a³ + b³ = 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 a³ . b³ = 64

a*b = 4 and a² .b² = 16

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.

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.

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.

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)

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.

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.

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.

Answer: Option A

Explanation :

The function f(x) = $\frac{\log (3x-7)}{\sqrt{2x^2-7x+6}}$  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 2x² - 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.

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

Answer: Option D

Explanation :

From the given data the following table can be created:

​​​​​​​

Hence the value of x=10

​​​​​​​​​​​​​​

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

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

Answer: Option E

Explanation :

Since x² - y² = 20 and x, y, z are positive integers, the only values possible are x = 6 and y = 4.

y³ - 2x² - 4z 12 ≥ -12

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

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

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

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

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

Answer: Option D

Explanation :

αf(x) = (1 - α)x² - 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

36α² - 32α (1 - α) > 0

68α² - 32α > 0

α(68α - 32) > 0

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

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

Answer: Option C

Explanation :

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

x² - 552 = $\sqrt{552+\sqrt{52+\sqrt{552+...}}}$

x² - x - 552 = 0

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

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

Answer is option C.