# PE 6 - Mathematical Reasoning | LR - Mathematical Reasoning

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**Answer the next 4 questions based on the information given below:**

Exactly six students, A to F, wrote a test in which, for each question, the number of marks awarded for every correct answer was three times the number of marks deducted for every wrong answer. For any question left un-attempted, no marks were either awarded or deducted. It is also known that the number of marks awarded for a correct answer was an integer. Further, for each of the six students, the number of wrong answers that they marked in the test was three-fifths the number of correct answers that they marked.

The following table provides the total marks scored by each of the six students in the test:

**PE 6 - Mathematical Reasoning | LR - Mathematical Reasoning**

What is the minimum possible number of questions in the test?

- (a)
56

- (b)
64

- (c)
40

- (d)
Cannot be determined

Answer: Option B

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**Explanation** :

Let the marks deducted for correct answer be x, hence the marks awarded for wrong answer = x/3.

Let A answered 5a questions correctly. Hence, he answered 3a questions in correctly.

Total questions answered by A = 8a.

Marks of A: 5a × x – 3a × x/3 = 288

⇒ 4ax = 288

⇒ ax = 72

Similarly, if total questions answered by B, C, D, E and F are 8b, 8c, 8d, 8e and 8f respectively, we will get

bx = 63, cx = 36, dx = 45, ex = 54 and fx = 27

Since a, b, c, d, e and f should be integers ‘x’ should be a factor of 72, 63, 36, 45, 54 and 27. Also ‘x’ itself is an integer.

∴ Possible values of x = Factors of HCF (72, 63, 36, 45, 54 and 27) = Factors of 9 = 1, 3 and 9.

Minimum total questions in the test = 64 [when x = 9 and A attempts all the questions.]

Hence, option (b).

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**PE 6 - Mathematical Reasoning | LR - Mathematical Reasoning**

If at least one student attempted all the questions in the test, what is the maximum possible number of questions left un-attempted by any student?

Answer: 360

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**Explanation** :

Consider the solution to the first question of this set.

The maximum questions un-attempted by any student will be for F when x = 1. Since at least one student attempts all the questions, we can assume that A attempts all the questions in the exam and hence total questions in the exam = 576 while F attempted only 216 questions. Hence the number of un-attempted questions = 576 – 216 = 360.

Hence, 360.

Workspace:

**PE 6 - Mathematical Reasoning | LR - Mathematical Reasoning**

If, for any student, the difference between the number of questions attempted correctly and those attempted incorrectly was not greater than 40, how many marks were awarded for each question answered correctly?

- (a)
9

- (b)
3

- (c)
1

- (d)
Cannot be determined

Answer: Option A

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**Explanation** :

Consider the solution to the first question of this set.

The difference between correct and incorrect attempts of any student is not greater than 40, this is possible when x = 9.

∴ Marks awarded for correct attempt = 9.

Hence, option (a).

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**PE 6 - Mathematical Reasoning | LR - Mathematical Reasoning**

If, for one of the six students, the total number of questions in the test and the number of questions left un-attempted were in the ratio 15 : 7, what is the minimum possible number of questions in the test?

Answer: 75

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**Explanation** :

To find the minimum number of total questions we will again consider the case when x = 9.

A has attempted 64 questions, hence total questions should be greater than or equal to 64.

Given for one of the student, Total questions : Un-attempted questions = 15 : 7

⇒ Attempted questions : Total questions = 8 : 15

⇒ Attempted questions = 8/15 × Total questions

⇒ Total questions should be a multiple of 15 and greater than 64.

∴ Least such possible number is 75.

Checking if total questions = 75 ⇒ Attempted questions for one of the students = 8/15 × 75 = 40.

D attempts 40 questions, hence, a total of 75 questions is possible.

∴ Least number of total questions = 75.

Hence, 75.

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**Answer the next 6 questions based on the information given below:**

Five girls and four boys of a class were to be sent for a sports camp. Their teacher was asked to measure the weight of these students. But the weighing machine available in the school was defective and only measured weights greater than 100 kg. Hence, the teacher decided to measure the weight of three students at a time, but all three students selected would be of the same gender. He took all the possible weight combinations of girls as well as of boys. For girls, the measurements were 146 kg, 151 kg, 154 kg, 154 kg, 157 kg, 158 kg, 161 kg, 162 kg, 166 kg and 169 kg. For boys, the measurements were 198 kg, 194 kg, 189 kg and 187 kg. No two boys or no two girls had the same weight.

**PE 6 - Mathematical Reasoning | LR - Mathematical Reasoning**

What is the average weight of the nine students, rounded off to nearest integer?

- (a)
58 kg

- (b)
56 kg

- (c)
59 kg

- (d)
57 kg

Answer: Option A

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**Explanation** :

Let the weight of the girls, in ascending order (in kgs), be p, q, r, s, t and the weight of the boys, also in ascending order (in kgs), be a, b, c, d.

Thus, p < q < r < s < t and a < b < c < d.

Number of weight combinations for three girls at a time = ^{5}C_{3} = 10

Hence, the weight combinations for three girls at a time are:

p + q + r; p + q + s; p + q + t; p + r + s; p + r + t; p + s + t; q + r + s; q + r + t; q + s + t and r + s + t

Addition of all these combinations gives: 6(p + q + r + s + t)

Total of all these combinations = 146 + 151 + 154 + 154 + 157 + 158 + 161 + 162 + 166 + 169 = 1578 kgs

∴ 6(p + q + r + s + t) = 1578

∴ Total weight of the girls = p + q + r + s + t = 1338/6 = 263 kgs

Similarly, for the boys:

3(a + b + c + d) = 198 + 194 + 189 + 187 = 768

∴ Total weight of the boys = a + b + c + d = 768/3 = 256 kgs

∴ Average weight of the nine students = (263 + 256)/(5 + 4) = 519/9 = 57.67 kgs ≈ 58 kgs.

Hence, option (a).

Workspace:

**PE 6 - Mathematical Reasoning | LR - Mathematical Reasoning**

What is the difference in the weights of the second heaviest boy and the second lightest girl?

- (a)
14 kg

- (b)
15 kg

- (c)
18 kg

- (d)
20 kg

Answer: Option C

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**Explanation** :

Consider the solution to the first question.

p < q < r < s < t and a < b < c < d

Consider the boys.

a + b + c < a + b + d < a + c + d < b + c + d

∴ a + b + c = 187; a + b + d = 189; a + c + d = 194 and b + c + d = 198

From the earlier solution, a + b + c + d = 256 ... (i)

Hence, subtracting each of the above equations from (i), a = 58, b = 62, c = 67 and d = 69

Similarly, consider the combinations for the girls.

The three lightest girls will have the least total and the three heaviest girls will have the greatest total.

∴ p + q + r = 146 and r + s + t = 169

Adding the two equations above, (p + q + r + s + t) + r = 146 + 169

Considering the total weight of the girls as obtained in the earlier question,

263 + r = 315

∴ r = 52

∴ p + q = 94 and s + t = 117

Considering the eight remaining combinations, it can be said that the second lowest sum is for p + q + s.

∴ p + q + s = 151

∴ s = 151 − 94 = 57

∴ t = 117 − 57 = 60

Similarly, the second highest sum is from q + s + t.

∴ q + s + t = 166

∴ q = 166 − 117 = 49

∴ p = 94 − 49 = 45

Weight of girls = 60, 57, 52, 49, 45

∴ Second heaviest boy - Second lightest girl = c − q = 67 − 49 = 18 kg

Hence, option (c).

Workspace:

**PE 6 - Mathematical Reasoning | LR - Mathematical Reasoning**

What is the weight of the heaviest boy?

- (a)
62 kg

- (b)
64 kg

- (c)
67 kg

- (d)
69 kg

Answer: Option D

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**Explanation** :

Consider the solution to the earlier questions.

Weight of the heaviest boy = d = 69 kg.

Hence, option (d).

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**PE 6 - Mathematical Reasoning | LR - Mathematical Reasoning**

How many girls weigh less than all the boys?

Answer: 4

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**Explanation** :

Consider the solution to the earlier questions.

Lightest boy = a = 58 kg

Among the girls, (p, q, r, s) < a

Hence, four girls weigh less than all the boys.

Hence, 4.

Workspace:

**PE 6 - Mathematical Reasoning | LR - Mathematical Reasoning**

Independent of gender, the ten students stand one behind the other in a straight line with the heaviest person first and the lightest person last. Which of these is an invalid representation of a group of three consecutively standing students?

- (a)
Two boys followed by a girl

- (b)
One boy between two girls

- (c)
Two girls followed by a boy

- (d)
Three girls one behind the other

Answer: Option C

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**Explanation** :

Consider the solution to the earlier questions.

Arranged from heaviest to lightest, the sequence of boys and girls is:

Boy: 69

Boy: 67

Boy: 62

Girl: 60

Boy: 58

Girl: 57

Girl: 52

Girl: 49

Girl: 45

Hence, there can be:

1) Two boys followed by a girl (59-54-52)

2) One boy between two girls (52-50-49)

4) Three girls one behind the other (49-44-41 or 44-41-37)

The combination in option (c) (two girls followed by a boy) is not possible.

Hence, option (c).

Workspace:

**PE 6 - Mathematical Reasoning | LR - Mathematical Reasoning**

One more boy joins this group. What is his least integral weight (in kg) if he is heavier than at least one person of each gender?

Answer: 58

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**Explanation** :

Consider the solution to the earlier questions.

The weight of the lightest boy is 58 kg. Also, this boy is heavier than four other girls.

Hence, to be heavier than at least one person of each gender, the new boy has to weigh at least 59 kg.

Once he is heavier than the lightest boy, he is automatically heavier than the four girls.

Hence, option (c).

Note: Since the question does not specify whether his weight is unique compared to the rest, he can weigh 58 kg or 60 kg (equal to the heaviest girl) or any other value.

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