RRB ALP 2018 Practice Test Papers | Arithmetic Questions (Day-13)

Dear Aspirants, Here we have given the Important RRB ALP & Technicians Exam 2018 Practice Test Papers. Candidates those who are preparing for RRB ALP 2018 can practice these Arithmetic Questions to get more confidence to Crack RRB 2018 Examination.

[WpProQuiz 1610]

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1. The HCF of two numbers “a” and “b” is 18. What can be the LCM of “a” and “b”?

1. 42
2. 9
3. 12
4. None of these

2. How many pairs of numbers are there such that their HCF and LCM are 12 and 924, respectively?

1. 1
2. 2
3. 3
4. 4

3. Ajay cycles at 3 km/h more than his usual speed and reached the destination 60 km away 1 h earlier. What is his usual speed?

1. 15 km/h
2. 12 km/h
3. 25 km/h
4. 10 km/h

4. 12 is divided into two parts such that 3 times the first part added to 4 times the second part makes 43. The difference between both parts is

1. 1
2. 2
3. 3
4. 7

5. What will be the minimum value of (x – 5) (x – 13)?

1. – 8
2. 4
3. 8
4. – 16

6. How many sequences of answers are possible in a test having 12 multiple choice questions if each question has five options?

1. 60
2. 12! × 5!
3. 12! + 5!
4. 5¹²

7. 27, 125, 343, 729, 1331,?

1. 2197
2. 1864
3. 2469
4. 1965

8. If 3888 is divided by the square of a number and the answer so obtained is multiplied by 21, then the final answer is 252. What is the number?

1. 17
2. 19
3. 18
4. 16

9. A pipe can fill a tank in 32 h. After 8 h, 7 more similar pipes are used. What is the additional time required to fill the tank?

1. 3 h
2. 2 h
3. 5 h
4. 4 h

10. Find the number, if 4/5 of that number exceeds its 2/3 by 4.

1. 15
2. 27
3. 30
4. 45

Solution:

1. Answer: (D)

HCF is always a factor of LCM. None of the given options is a multiple of 18. Therefore, the correct option is “None of these”.

2. Answer: (B)

Since HCF is a factor of both numbers, these numbers must be 12a and 12b, where a and b are relatively prime. According to the given condition, 12ab = 924. Therefore, ab = 77 So, the required possible pairs are (12, 924) and (84, 132).

3. Answer: (B)

Let the usual speed be x km/h.

∴ Usual time = 60/x h

New time = 60/(x+3) h

∴60/x -60/(x + 3) = 1

x² + 3x – 180 = 0

x = 12 or –15 Disregarding the negative value, the usual speed of Ajay is 12 km/h.

4. Answer: (B)

Let the first part be x, then the second part is 12 – x. According to the question, 3x + 4(12 – x) = 43

3x + 48 – 4x = 43

– x = 43 – 48

– x = – 5

x = 5

First number = 5

Second number = 12 – 5 = 7 Difference = 7 – 5 = 2

5. Answer: (D)

(x – 5) (x – 13) = x² – 18x + 65

For a quadratic equation, ax² + bx + c = 0, the minimum value of the equation is calculated as             (4ac- b²)/4a . Therefore, for the quadratic equation, x² – 18x +65 = 0, the minimum value will be

4(1)(65) -(-18)²/ 4(1)

(260-324)/ 4 =-64/4=-16

6. Answer: (D)

Since each question can be answered in 5 ways. Therefore total number of the possible sequence = 5¹²

7. Answer: (A)

The terms of the given series are 3³, 5³, 7³, 9³ and 11³. By following the same pattern, the required term is 13³ = 2197

8. Answer: (C)

Let the required number be x. According to the question, 3888 ÷ x² = a and a × 21 = 252 On solving, we get a = 12 and x2 = 324 Hence, the required number is 18.

9. Answer: (A)

The pipe can fill 1/32 of the tank in 1 h. After 8 h, remaining part = 3/4

Remaining time = (3/4)/(8×1/32) = 3 h

10. Answer: (C)

Let the required number be x. So, according to the question,  4/5 x – 2/3x =4

Therefore x = 30

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