PE 5 - Mathematical Reasoning | LR - Mathematical Reasoning

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Answer the next 3 questions based on the information given below:

Given below are the marks of 10 students in a test consisting of 10 questions. It is known that marks of one of the students is entered wrong.

For each correct answer a student is awarded 5 marks whereas a penalty of 3 marks is there for each incorrect answer. No marks are awarded or deducted for an un-attempted question.

1. PE 5 - Mathematical Reasoning | LR - Mathematical Reasoning

Identify which students marks were entered wrong?

(a)
F
(b)
G
(c)
E
(d)
I

Answer: Option C

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Explanation :

For a student let the number of correct questions be ‘x’ and incorrect questions be ‘y’.

∴ 5x – 3y = marks obtained.
Also, x + y ≤ 10

For A: 5x – 3y = 15
⇒ x = $\frac{15 + 3 y}{5}$ = 3 + $\frac{3 y}{5}$
∴ Possible values of x and y are (3, 0)

For B: 5x – 3y = 0
⇒ x = $\frac{3 y}{5}$
∴ Possible values of x and y are (0, 0) or (3, 5)

For C: 5x – 3y = 18
⇒ x = $\frac{18 + 3 y}{5}$ = 3 + $\frac{3 + 3 y}{5}$
∴ Possible values of x and y are (6, 4)

For D: 5x – 3y = 34
⇒ x = $\frac{34 + 3 y}{5}$ = 6 + $\frac{4 + 3 y}{5}$
∴ Possible values of x and y are (8, 2)

For E: 5x – 3y = 38
⇒ x = $\frac{38 + 3 y}{5}$ = 7 + $\frac{3 + 3 y}{5}$
∴ There are no possible values of x and y such that x + y ≤ 10.
⇒ Marks of E were entered wrong.

For F: 5x – 3y = 19
⇒ x = $\frac{19 + 3 y}{5}$ =3 + $\frac{4 + 3 y}{5}$
∴ Possible values of x and y are (5, 2)

For G: 5x – 3y = 5
⇒ x = $\frac{5 + 3 y}{5}$ = 1 + $\frac{3 y}{5}$
∴ Possible values of x and y are (1, 0) or (4, 5)

For H: 5x – 3y = 24
⇒ x = $\frac{24 + 3 y}{5}$ = 4 + $\frac{4 + 3 y}{5}$
∴ Possible values of x and y are (6, 2)

For I: 5x – 3y = 29
⇒ x = $\frac{29 + 3 y}{5}$ = 5 + $\frac{4 + 3 y}{5}$
∴ Possible values of x and y are (7, 2)

For J: 5x – 3y = 2
⇒ x = $\frac{2 + 3 y}{5}$
∴ Possible values of x and y are (1, 1) or (4, 6)

Hence, option (c).

2. PE 5 - Mathematical Reasoning | LR - Mathematical Reasoning

Except the student whose marks were entered wrong, what is the maximum number of questions that could have been attempted by other 9 students together.

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3. PE 5 - Mathematical Reasoning | LR - Mathematical Reasoning

Except the student whose marks were entered wrong, if each student attempted maximum number of questions, what is the highest number of questions attempted wrong by any of the remaining 9 students?

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Answer the next 5 questions based on the information given below:

For a given input, a machine performs 6 different mathematical operations and gives a final output. Output of a particular operation becomes the input for next operation until all 6 operations are performed. The machine always performs the same 6 operations but the order of operations may change for different inputs.

Table below gives the initial input and outputs for each intermediate step.

4. PE 5 - Mathematical Reasoning | LR - Mathematical Reasoning

Which of the following cannot be a function performed by any operation?

(a)
x + 2
(b)
x - 1
(c)
√x
(d)
2x

5. PE 5 - Mathematical Reasoning | LR - Mathematical Reasoning

When the input is 6, what is the function performed by 3rd operation?

(a)
x + 2
(b)
x/4
(c)
x - 1
(d)
Cannot be determined

6. PE 5 - Mathematical Reasoning | LR - Mathematical Reasoning

When the input is 9 and the order of operations is same as when the input is 4, what is the final output? [Round off your answer to the nearest integer.]

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7. PE 5 - Mathematical Reasoning | LR - Mathematical Reasoning

If the sequence of operations is same as when the input is 6, for which of these inputs is one of the outputs definitely not an integer?

(a)
18
(b)
12
(c)
38
(d)
None of these

8. PE 5 - Mathematical Reasoning | LR - Mathematical Reasoning

If the initial input is 15, and every intermediate output (including the final output) is an integer, what is the maximum possible value of intermediate output #2?

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PE 5 - Mathematical Reasoning | LR - Mathematical Reasoning

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