Algebra - Simple Equations - Previous Year CAT/MBA Questions

Answer: Option D

Text Explanation :

Here, AC – Air Conditioning, R – Radio and PW – Power Windows

From the given conditions, we have the above Venn diagram.

When we add up all the values in the Venn diagram, we get 23 (15 + 5 + 1 + 2) cars.

∴ 2 (i.e. 25 − 23) cars don’t have any of the three options.

Hence, option (d).

Answer: Option A

Text Explanation :

Case1: To pay 78 paise using minimum number of coins:

78 = 1(50) + 2(10) + 4(2), i.e. a total of 7 coins were used

Case 2: To pay 69 paise using minimum number of coins:

69 = 1(50) + 1(10) + 1(5) + 2(2), i.e. a total of 5 coins were used

Case 3: To pay 101 paise using minimum number of coins:

101 = 1(50)  + 1(25) + 2(10) + 3(2), i.e. a total of 7 coins were used

∴ Minimum coins used were 7 + 5 + 7 = 19 coins

Hence, option (a).

Answer: Option C

Text Explanation :

5x + 19y = 64 where x, y ∈ I

This means that the values of x have an interval of 19 between each other and the values of y will have an interval of 5 between each other.

Now, there are 2 possible cases; y could either be positive or negative:

Case 1:
When y = 1, then x = 9
When y = 6, then x = −10
When y = 11, then x = −29 and so on
You will notice that the values of y are in intervals of 5 and that of x are in intervals of 19.
Generally speaking, when y is positive, we will get integral values of x when y’s unit’s digit is either 1 or 6.

Case 2:
When y = −4, then x = 28
When y = −9, then x = 47
Again, the values of y are in intervals of 5 and that of x are in intervals of 19.
That is, when y is negative, we will get integral values of x when y’s unit’s digit is either 4 or 9.

Now, let’s evaluate the options:

Option 1: “no solution for x < 300 and y < 0” is False.

∵ According to Case 2, we should get integral values of x when y is −4, −9 or −14 and so on.

Option 2: “no solution for x > 250 and y > –100” is False.

According to Case 2, we should get integral values of x when y is −99, −94, −74 or −69 etc.
Now, when y = −74, x = 294
∴ A solution exists.

Option 3: “a solution for 250 < x < 300” is True.

∵ y = −74, x = 294 is a possible solution

Option 4: “a solution for –59 < y < –56” is False.

∵ From Case 2, when y is negative, we will get integral values of x only when y’s unit’s digit is either 4 or 9.

Hence, option (c).

Answer: Option D

Text Explanation :

Case 1: When the 2^nd letter is m:

The 1^st letter can be any of the 5 vowels.

The 3^rd letter will be any of the 4 remaining vowels (i.e. different from the 1^st one).

Number of possible 3 letter combinations = 5 × 4 = 20

Case 2: When the 2^nd letter is n:

The 1^st letter can be any of the 5 vowels.

The 3^rd letter will be either e or u.

Number of possible 3 letter combinations = 5 × 2 = 10

Case 3: When the 2^nd letter is p:

The 1^st letter can be any of the 5 vowels.

The 3^rd letter will be the same as the 1^st letter.

Number of possible 3 letter combinations = 5 × 1 = 5

∴ Total number of possible 3 letter combinations = 20 + 10 + 5 = 35

Hence, option (d).

Answer: Option C

Text Explanation :

Case 1: The 2^nd letter is m and the 3^rd letter is e:

The 1^st letter may be any of the 4 remaining vowels (i.e. different from e)

Number of possible 3 letter combinations = 4

Case 2: The 2^nd letter is n and the 3^rd letter is e:

The 1^st letter may be any of the 5 vowels.

Number of possible 3 letter combinations = 5

Case 3: The 2^nd letter is p and the 3^rd letter is e:

The 1^st letter will be the same as the 3^rd letter.

Number of possible 3 letter combinations = 1 (i.e. ‘epe’)

∴ Total number of possible 3 letter combinations = 4 + 5 + 1 = 10

Hence, option (c).

Answer: Option D

Text Explanation :

The rent of the car = Maximum (number of hours × charge per hour, km travelled × charge per km)

Case 1: Anand drove the Car for 5 hours or less
∴ Rent = Maxima(300, 30 × 12) = 360

Case 2: Anand drove the Car for more than 5 hours
∴ Rent = Maxima(300, 7.5 × 30) = 300

∵ Anand paid Rs. 300 as rent.

∴ Anand drove for 6 hours.

Hence, option (d).

Answer: Option A

Text Explanation :

Let l, m and s be the longest, medium and the shortest lengths of the strings.

l = 3m and s = l – 23

l + s + m = 40

$l+\frac{l}{3}+$ l - 23 = 40

$\frac{7l}{3}$ = 63

 l = 27

Hence, option (a).

Answer: Option B

Text Explanation :

Let the money contributed by Mayank, Mirza, Little and Jagbir be a, b, c and d respectively.

a = $\frac{1}{2}$ (b + c + d)

∴ 2a = b + c + d

∴ 2a = 60 - a

∴ a = 20

b = $\frac{1}{3}$ (a + c + d)

∴ 3b = a + c + d

∴ 3b = 60 - b

∴ b = 15

c = $\frac{1}{4}$ (a + b + d)

∴ 4c = a + b + d

∴ 4c = 60 - c

∴ c = 12

Money contributed by Jagbir = 60 − 20 − 15 − 12 = $13

Hence, option (b).

Answer: Option B

Text Explanation :

Let the man has received a cheque of x rupees and y paise.

∴ The amount on cheque = (100x + y)                                             …(i)

The amount actually received by him = 100y + x

After spending Rs. 5 and 42 paise, the remaining amount = (100y + x – 542)   …(ii)

But, (100y + x – 542) = 6 × (100x + y)                                               …(iii)

Substituting the values from the given options, x = 6 and y = 44

Hence, option (b).

Answer: Option B

Text Explanation :

Assume that the thief had stolen x diamonds.

The 1st watchman got = x/2 + 2

Diamonds left = x − (x/2 + 2) = x/2 – 2

The 2nd watchman got = (x/2 − 2)/2 + 2 = x/4 + 1

Diamonds left = (x/2 − 2) − (x/4 + 1) = x/4 − 3

The 3rd watchman got = (x/4 − 3)/2 + 2 = x/8 + 1/2

Diamonds left = (x/4 − 3) − (x/8 + 1/2) = x/8 − 7/2 = 1

∴ x/8 = 9/2

∴ x = 36

Hence, option (b).

Alternatively,

One may proceed by substituting the options that are even.

Answer: Option C

Text Explanation :

Let y be the number of days that they went for Yoga, t be the number of days that they played Tennis and f be the number of days that they were Free.

On 14 mornings they did not do anything, then they either played tennis in the evening or did nothing.

∴ 14 = t + f                           ...(i)

Similarly, 24 = y + f                ...(ii)

Also, 22 = t + y                     …(iii)

Adding (i) and (ii) and substituting in (iii), we get,

2f = 16

∴ f = 8 days

∴ Total number of days Amar stayed = 22 + f  = 22 + 8 = 30 days

Hence, option (c).

Answer: Option D

Text Explanation :

It is given to us that X₀ = x

∴ X₁ = −x

X₂ = −x

X₃ = x

X₄ = x

There is no trend for odd or even X_n.

∴ No concrete statement can be made.

Hence, option (d).
 

Answer: Option C

Text Explanation :

Let a = no. of males in Chota hazri, b = no. of males in Mota hazri, x = no. of females in Chota hazri and y = no. of females in Mota hazri

∴ a + 4522 = b                               ...(i)

∴ y = b + 4020                              ...(ii)

∴ x = 2a                                        ...(iii)

∴ x = y – 2910                               ...(iv)

Using (iv), (ii)  and (iii), we have,

2a = b + 4020 – 2910

Substituting from (i), we have,

2a = a + 4522 + 4020 – 2910

a  = 5632

Hence, option (c).

Answer: Option A

Text Explanation :

Let the cost of one burger, one shake and one order of fries be b, s and f respectively.

∴ 3b + 7s +1f = 120                       ...(i)

∴ 4b + 10s + 1f = 164.5                 ...(ii)

Subtracting (i) from (ii), we get,

 1b + 3s = 44.5

Multiplying by 2, we get,

 2b + 6s = 89                                    ...(iii)

Subtracting equation (iii) from equation (i), we get,

 b + s + f = 31

Hence, option (a).

Answer: Option B

Text Explanation :

Let the number of Re. 1, Re. 2 and Re. 5 coins be x, y and z respectively.

x + y + z = 300                           ...(i)

x + 2y + 5z = 960                       ...(ii)

2x + y + 5z = 920                       ...(iii)

Adding equation (ii) and equation (iii), we get,

3x + 3y + 10z = 1880                 ...(iv)

Multiply equation (i) by 3, we get,

3x + 3y + 3z = 900                     ...(v)

Subtracting equation (v) from equation (iv), we get,

7z = 980

∴ z = 140

Hence, option (b).

Answer: Option D

Text Explanation :

Statement A simply gives the ratio of apples between Ram and Gopal.

∴ Statement A is not alone sufficient to answer the question.

Statement B doesn’t provide any information related to number of apples bought by Ram and Gopal.

Even after combining both the statements A and B together, the question cannot be answered.

Hence, option (d).

Answer: Option A

Text Explanation :

Sum of A and B = (Average of A and B) × 2

=  /(@(A, B), 2)

∴ Hence, option (a).

Answer: Option D

Text Explanation :

We need to find average of A, B and C i.e. (A + B + C)/3

The only operator that can give 3 in the denominator is “X”

Option 1 cannot be the answer, as there is no “X”.

Option 2 also cannot be the answer as X(@(/(@(/(B, A), 2), C), 3), 2) gives 2 in the denominator and not 3.

Similarly option 3 is not the answer.

Hence, option (d) must be the answer.

Option 4:

/ ( X (@ (/ (@ (B, A), 2), C), 3), 2)

$=/\left(x\left(@\left(/\left(\frac{A+B}{2},2\right),c\right),3\right),2\right)$

=/(X(@((A + B), C), 3), 2)

$=/\left(x\left(\left(\frac{A+B+C}{2}\right),3\right),2\right)=/\left(\frac{A+B+C}{6},2\right)$

$=\frac{2(A+B+C)}{6}=\frac{A+B+C}{3}$

Hence, option (d).

Answer: Option D

Text Explanation :

x, y and z  are distinct real numbers.

∴ Without loss of generality, let x < y < z

Then,

f(x, y, z) = y; g(x, y, z) = y; h(x, y, z) = z

j(x, y, z) = x; m(x, y, z) = z; n(x, y, z) = x

Substituting these values in the given options,

Option 1 = (z – y)/x may or may not be greater than 1.

Option 2 = x/z < 1

Option 3 = y/y = 1

Option 4 = (y + z − y)/x > 1

Hence, option (d).

Answer: Option C

Text Explanation :

There are 8 teams in each group.

Stage 1:

Within a group each team played with every other team.

∴ Total number of matches = 7 + 6 + 5 + 4 + 3 + 2 + 1 = (7 × 8) / 2 = 28 matches in one group

∴ Total number of matches in both the groups = 56 matches

Stage 2:

Number of matches = 4 matches (Round 1) + 2 matches (Round 2) + 1 Match (Round 3) = 7

∴ Total number of matches played in the tournament = 63

Hence, option (c).

Answer: Option B

Text Explanation :

28 matches are played between 8 teams.

Consider option 1:

There is a possibility that 5 teams win exactly 5 matches each (5 × 5 = 25 matches played), one team wins 2 matches, one team wins exactly one match and one team lose all.

{For example: Let A, B, C, D, E, F, G, and H are the eight teams. A wins matches against B, C, D, E, F; B wins matches against C, D, E, F, G; C wins matches against D, E, F, G, H; G wins matches against A, D, E, F, H and H wins matches against A, B, D, E, F. Also E wins match against D. F wins matches against D and E. D loses all the matches}

Thus, for a team, there is no guarantee of its advancement to the next stage if it wins 5 matches.

 But if a team wins six matches, there is no possibility that 5 teams will win 6 matches each. (As there are only 28 matches) Hence, the team will definitely advance to the next stage.

 Hence, option (b).

Answer: Option A

Text Explanation :

We need to find, maximum value of x, such that despite winning x matches, a team gets definitely eliminated at the end of the first stage.

If x = 4; the team can reach in second stage.

{For example: Let A, B, C, D, E, F, G, and H are the eight teams. A wins matches against B, C, D, E, F, G(i.e. 6 matches); B wins matches against C, D(i.e. wins 2 matches);; C wins matches against D, E(i.e. wins 2 matches); D wins match against E; E wins match against B; F wins matches against B, C, D, E(i.e. wins 4 matches); G loose only one match against A (i.e. wins 6 matches);  and H loose only one match against G (i.e. wins 6 matches).}

If x = 3; the team can reach in second stage.

{For example: Let A, B, C, D, E, F, G, and H are the eight teams. A wins matches against B, C, D, E, F, G(i.e. 6 matches); B wins matches against C, D; C wins matches against D, E; D wins match against E and F(i.e. wins 2 matches); E wins match against B; F wins matches against B, C, E(i.e. wins 3 matches); G loose only one match against A (i.e. wins 6 matches);  and H loose only one match against G (i.e. wins 6 matches).}

If x = 2; the team can reach in second stage.

{For example: Let A, B, C, D, E, F, G, and H are the eight teams. A wins matches against B, C, D, E, F, G(i.e. 6 matches); B wins matches against C, D; C wins matches against D, E; D wins match against E and F(i.e. wins 2 matches); E wins match against B and F(i.e. wins 2 matches); F wins matches against B, C,(i.e. wins 2 matches); G loose only one match against A (i.e. wins 6 matches);  and H loose only one match against G (i.e. wins 6 matches).}

Thus, for x = 2, 3 and 4; we definitely cannot say that the team will not reach the second stage.

We can say that the value of x must be 1, because only four teams reach the second stage and it is not possible that more than two teams win exactly one match each.

Hence, option (a).

Answer: Option C

Text Explanation :

When the second stage starts, the first round will have 4 matches between 8 teams. From this round, only 4 teams will advance to the second round. So, second round will have 2 matches. In the third round, two teams will play 1 match and after that the winner will be declared.

Thus, the number of rounds in the second stage of the tournament = 3

Hence, option (c).

Answer: Option C

Text Explanation :

We have seen that even if a team wins 2 matches, it can be in the second stage of the tournament. Consider Team A wins 2 matches and enters the second round and wins the tournament. Then its total score will be 5 points. While there can be a team having more than 6 points but not the winner of the tournament. Thus, option 1 may not be true.

We have seen that even if a team wins 5 matches is can be eliminated and a team winning only two matches can enter second round. Thus, option 2 may not be true.

Consider a team in group I eliminated and the score is 5. A team in group II advances to the second stage with score 2. If this team (i.e. the team with score 2) is a winner then the score will be 2 + 3 = 5. Thus, option 3 is true.

Eight teams enter second stage. Of these, 4 teams advances to the second round scoring 1 point each. In the second round, only two teams win and their score in the second round so far will be 2. Thus, only two teams will have score 1. Thus, option 4 is not true.

Hence, option (c).

Answer: Option C

Text Explanation :

​​​​​​​​​​​​​​

Starting from the fourth place of worship and moving backwards, we find that number of flowers before entering the first place of worship is $\frac{15}{8}y.$

Hence, number of flowers before doubling = $\frac{15}{16}y$

(but this is equal to 30)
Hence, y = 32

Hence, option (c).