Number System Practice Quiz 2

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Number System Practice Quiz 2

Introduction

Number Series is one of the important topic in the Quantitative Aptitude section. Number Series is the arrangement of numbers in a certain order where some numbers are wrongly kept or some numbers are missing from that series. So accurate series are to be found. Number Series in competitive exams are divided into two. One is missing series and the other is wrong series. A number series is given in which a number is wrongly placed is the wrong series. Candidates are asked to identify that particular wrong number. A number series in which a specific number is missing is the missing series. Candidates have to identify the missing number. The article Number System Practice Quiz 2 lists important number series practice questions for competitive exams like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.

Quiz

1. What is the place value of 7 in the numeral 2734?

A. 70
B. 7
C. 700
D. 7.00

Explanation: 7 × 100 = 700

2. What is the place value of 3 in the numeral 3259

A. 300
B. 30
C. 3
D. 3000

Explanation: 3 × 1000 = 3000

3. What is the difference between the place value of 2 in the numeral 7229?

A. 20
B. 200
C. 180
D. 18

Explanation: 200 – 20 = 180

4. What is the place value of 0 in the numeral 2074?

A. 100
B. 70
C. 7.0
D. 0

Explanation: The place value of zero (0) is always 0. It may hold any place in a number,
its value is always 0.

5. What is the difference between the place value and face value of 3 in the numeral 1375?

A. 300
B. 3
C. 297
D. 303

Explanation: place value of 3 = 3 × 100 = 300
face value of 3 = 3
300 – 3 = 297

6. A number, when divided by a divisor, leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?

A. 73
B. 37
C. 64
D. 53

Explanation: Let the original number be ‘a’
Let the divisor be ‘d’
Let the quotient of the division of aa by dd be ‘x’
Therefore, we can write the relation as $$\frac{a}{d}$$ = x and the remainder is 24.
i.e., a=dx+24 When twice the original number is divided by d, 2a is divided by d.
We know that a=dx+24. Therefore, 2a = 2dx + 48
The problem states that $$\frac{2dx+48}{d}$$leaves a remainder of 11.
2dx2dx is perfectly divisible by d and will, therefore, not leave a remainder.
The remainder of 11 was obtained by dividing 48 by d.
When 48 is divided by 37, the remainder that one will obtain is 11.
Hence, the divisor is 37.

7. How many numbers from 10 to 50 are exactly divisible by 3.

A. 13
B. 12
C. 14
D. 11

Explanation: 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45,48.
13 Numbers.
$$\frac{10}{3}$$ = 3 and $$\frac{50}{3}$$ = 16 ==> 16 – 3 = 13. Therefore 13 digits.

8. In an election, candidate A got 75% of the total valid votes. If 15% of the total votes were declared invalid and the total numbers of votes is 560000, find the number of valid votes polled in favor of the candidate.

A. 357600
B. 356000
C. 367000
D. 357000

Explanation: Total number of invalid votes = 15 % of 560000
= 15/100 × 560000
= $$\frac{8400000}{100}$$
= 84000
Total number of valid votes 560000 – 84000 = 476000
Percentage of votes polled in favour of candidate A = 75 %
Therefore, the number of valid votes polled in favour of candidate A = 75 % of 476000
= $$\frac{75}{100}$$ × 476000
= $$\frac{35700000}{100}$$
= 357000

9. Aravind had $2100 left after spending 30 % of the money he took for shopping. How much money did he take along with him? A.$ 3600
B. $3300 C.$ 3000
D. $3100 Answer: Option C Explanation: Let the money he took for shopping be m. Money he spent = 30 % of m = 30/100 × m = 3/10 m Money left with him = m – $$\frac{3}{10}$$ m = $$\frac{(10m – 3m)}{10}$$ = $$\frac{7 m}{10}$$ But money left with him =$ 2100
Therefore 7m/10 = $2100 m =$ 2100× $$\frac{10}{7}$$
m = $$$\frac{21000}{7}$$ m =$ 3000
Therefore, the money he took for shopping is \$ 3000.

10. A shopkeeper bought 600 oranges and 400 bananas. He found 15% of oranges and 8% of bananas were rotten. Find the percentage of fruits in good condition.

A. 87.8%
B. 86.8%
C. 85.8%
D. 84.8%

Explanation:

Total number of fruits shopkeeper bought = 600 + 400 = 1000
Number of rotten oranges = 15% of 600
= 15/100 × 600 = $$\frac{9000}{100}$$ = 90
Number of rotten bananas = 8% of 400
= 8/100 × 400 = $$\frac{3200}{100}$$ =32
Therefore, total number of rotten fruits = 90 + 32 = 122
Therefore Number of fruits in good condition = 1000 – 122 = 878
Therefore Percentage of fruits in good condition = ($$\frac{878}{1000}$$ × 100)%
= ($$\frac{87800}{1000}$$)% = 87.8%

11. On dividing 109 by a number, the quotient is 9 and the remainder is 1. Find the divisor.

A. 13
B. 12
C. 14
D. 11

Explanation:

d = $$\frac{(D-R)}{Q}$$
$$\frac{(109 – 1)}{9}$$
= $$\frac{(108)}{9}$$ = 12

12. What is the dividend. divisor 17, the quotient is 9 and the remainder is 5.

A. 153
B. 156
C. 158
D. None of these

Explanation:

D = d × Q + R
D = 17 × 9 + 5
= 153 + 5
D = 158

13. In a division sum, the divisor is ten times the quotient and five times the remainder. If the remainder is 46, the dividend is:

A. 5336
B. 5347
C. 5337
D. None of these

Explanation:
Divisor = (5 × 46) = 230
= 10 × Quotient = Divisor
=> Quotient = $$\frac{230}{10}$$ = 23
Dividend = (Divisor × Quotient) + Remainder
Dividend = (230 × 23) + 46 = 5336

14. In a division sum, the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 2 to the thrice of the remainder. The dividend is:

A. 89
B. 88
C. 86
D. 85

Explanation:

Divisor = (6 × 3) + 2 = 20
5 × Quotient = 20
Quotient = 4.
Dividend = (Divisor × Quotient) + Remainder
Dividend = (20 × 4) + 6 = 86

15. In a question on division with zero remainders, a candidate took 12 as divisor instead of 21. The quotient obtained by him was 35. The correct quotient is:

A. 26
B. 25
C. 0
D. 20

Explanation:

Number = (35 × 12) = 420
Correct quotient = $$\frac{420}{21}$$ = 20

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