The article of** Number system Practice Quiz** consists of different models of Number system questions with solutions. The Number System Practice Quiz is extremely important for aspirants of different competitive exams across the globe. In India, the Quantitative Aptitude quiz helps the candidates preparing different competitive exams like **RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.**

ascending order?

**Answer –** Option C

**Explanation –**

\(\frac{5}{16}\) = 0.312

\(\frac{6}{17}\) = 0.352

\(\frac{7}{18}\) = 0.388

**2. If fractions\(\frac{9}{13}\), \(\frac{2}{3}\), \(\frac{8}{11}\), \(\frac{5}{7}\), are arranged in ascending order, then the correct sequenceis**

**Answer –** Option B

**Explanation –**

\(\frac{9}{13}\) = 0.692, \(\frac{2}{3}\) = 0.666,

\(\frac{8}{11}\) = 0.727, \(\frac{5}{7}\) = 0.714

Hence ascending order is \(\frac{2}{3}\) = \(\frac{9}{13}\), \(\frac{5}{7}\), \(\frac{8}{11}\)

**3. Which one of the following is the largest?
2\(\sqrt{5}\), 6\(\sqrt{3}\), 3\(\sqrt{7}\) and 8\(\sqrt{2}\)**

**Answer –** Option A

**Explanation –**

2\(\sqrt{5}\) = 2 * 2.236 = 4.472

6\(\sqrt{3}\) = 6 * 1.732 = 10.392

3\(\sqrt{7}\) = 3 * 2.646 = 7.938

8\(\sqrt{2}\) = 8 * 1.414 = 11.312

Clearly8\(\sqrt{2}\) is largest

**4. I f numer at or and denomi nat or of a pr oper fractions are increased by the same quantity, then the resulting fraction is**

**Answer –** Option A

**Explanation –**

Let proper fraction = \(\frac{2}{3}\) =

therefore Resulting fraction = \(\frac{2 + 1}{3 + 1}\) = \(\frac{3}{4}\)

Hence \(\frac{2}{3}\) < \(\frac{3}{4}\)

\(\frac{1}{2}\) < \(\frac{2}{3}\)

\(\frac{3}{5}\) < \(\frac{3 + 1}{5 + 1}\)

\(\frac{3}{5}\) < \(\frac{4}{6}\) etc.

**5. If x + y > 5 and x â€“ y > 3, then which of the following gives all possible values of x ?**

**Answer –** Option B

**Explanation –**

Solving x + y > 5 and x â€“ y > 3 we get,

x > 4.

**6. I f x and y ar e negat ive, then which of the following statements is/are always true ?**

II. xy is positive

III. x â€“ y is positive

**Answer –** Option B

**Explanation –**

Product of â€“ ve numbers is also +ve.

**7. The value of \(\sqrt{ \sqrt{0.000064}}^{3}\)**

**Answer –** Option B

**Explanation –**

Given expression =\(\sqrt{0.008}^{3}\) = 0.2

**8. If 11, 109, 999 is divided by 1111, then what is the remainder ?**

**Answer –** Option D

**Explanation –**

**9. The value of \(\frac {\frac{1}{2}\frac{1}{2}of \frac{1}{2}}{\frac{1}{2}\frac{1}{2}of \frac{1}{2}}\)**

**Answer –** Option A

**Explanation –**

\(\frac {\frac{1}{2}\frac{1}{2} * \frac{1}{2}}{\frac{1}{2}\frac{1}{2} * \frac{1}{2}}\)

= \(\frac {\frac {1}{2} \frac{4}{1}}{\frac{3}{4}}\) = 2 * \(\frac{4}{3}\)

= \(\frac{8}{3}\)

= 2\(\frac{2}{3}\)

**10. Taking \(\sqrt{2}\) = 1.414, \(\sqrt{3}\) = 1.732, \(\sqrt{5}\) = 2.236 and \(\sqrt{6} \) = 2.449,then the value of \(\frac{9 +\sqrt{2}}{\sqrt{5} +\sqrt{3}}\) + \(\frac{9 -\sqrt{2}}{\sqrt{5} -\sqrt{3}}\) to three places of decimals is**

**Answer –** Option C

**Explanation –**

\(\frac{9 +\sqrt{2}}{\sqrt{5} +\sqrt{3}}\) + \(\frac{9 -\sqrt{2}}{\sqrt{5} -\sqrt{3}}\)

= \(\frac{1}{2}\)[15\(\sqrt{5}\)– 3\(\sqrt{5}\) + 2\(\sqrt{6}\)]

= \(\frac{1}{2}\) [33.540 â€“ 5.196 â€“ 7.898]

= 10.732

**11. The cube root to 1.061208**

**Answer –** Option D

**Explanation –**

Here 1.061208 = \({1.02}^{3}\)

Required cube root = 1.02

**12. The least number having four digits which is a perfect square is**

**Answer –** Option D

**Explanation –**

Required number =1024 = \({32}^{2}\)

**13. The missing number in the series 8, 24, 12, 36, 18, 54, ______ is**

**Answer –** Option A

**Explanation –**

Second term is 3 times of the first term and third term is half of the second term, repeat this process the missing term is half of 54, i.e. 27

**14. What is the eighth term of the sequence 1, 4, 9, 16, 25,_______ ?**

**Answer –** Option B

**Explanation –**

Given sequence can be written as \({1}^{2}, {2}^{2}, {3}^{2}, {4}^{2}, {5}^{2}\), …

Hence itâ€™s eighth term

=\({8}^{2}\)

= 64

**15. Which of the following is the best approximation for the following expression \(\sqrt{{7.9986 / 0.115} + 19.97}\)?**

**Answer –** Option B

**Explanation –**

Rounding off, we get \(\sqrt{{\frac{8}{0.1}} + 20}\) = \(\sqrt{100}\)

**Answer –** Option A

**Explanation –**

1 + 0.1000 + 0.110 + 0.111 = 1.321

**2. When a number is divided by 5, it gives remainder 3. What is the remainder when square of that number is divided by 5?**

**Answer –** Option C

**Explanation –**

Let the number be 8.

Thus, when \( {8}^{2}\) = 64 will be divided by 5, then remainder will be 4.

**3. \(\sqrt{10}\) = 3.1623(approx). What is the approx, value of \(\frac{1}{\sqrt{10}}?\)**

**Answer –** Option A

**Explanation –**

\(\frac{1}{3.1623}\) will be less than 0.3333

**4. . Find the value of \({(2744)}^{\frac{1}{3}}\)?**

**Answer –** Option B

2744 is a multiple of 7.

Hence, the answer has to be 14.

**Explanation –**

**5. Find the L.C.M. of 148 and 185.**

**Answer –** Option B

**Explanation –**

148 = 37 Ã— 4 and 185 = 37 Ã— 5

LCM = 37 Ã— 4 Ã— 5 = 740

**6. If \({2}^{2n – 1}\) = \(\frac{1}{{8}^{2n – 1}}\) then the value of ‘n’ is:**

- 1.

**Answer –** Option B

**Explanation –**

\({2}^{2n – 1}\) = \(\frac{1}{3n – 9}\)

\({2}^{2n – 1}\) = \({2}^{3n – 9}\) = \({2}^{0}\)

5n â€“ 10 = 0

n = \(\frac{10}{5}\) = 2

**7. What is the largest possible length of a scale that can be used to measure exactly the lengths 3 m, 5 m 10 cm and 12 m 90 cm ?**

**Answer –** Option A

**Explanation –**

Required scale has to be of length 10 cm because 10 cm is the shortest length in in given question.

**8. After measuring 120 metres of a rope, it was discovered that the metre rod was 3 cm longer. The true length of the rope measured is :**

**Answer –** Option D

**Explanation –**

Actual length has to be

120m + (120 Ã— 3) cm = 120 m + 360 cm

= 123 m 60 cm.

**9. Solve \(\sqrt{0.000064}^{3}\)**

**Answer –** Option B

**Explanation –**

\( \sqrt{0.000064}^{3}\) = \( \sqrt{\frac{4}{100} * \frac{4}{100} * \frac{4}{100}}^{3}\) = 0.4

**10. The HCF of two numbers is 6 and their LCM is 72. If one number is 24, the other number is**

- 1.

**Answer –** Option B

**Explanation –**

We know that, x Ã— y = LCM Ã— HCF (x, y are the two distinct numbers)

x Ã— 24 = 72 Ã— 6

x = 18.

**11. The largest number which divides by 72 and 125, leaving remainders 7 and 8 respectively is **

**Answer –** Option A

**Explanation –**

In such a questions it is better to check the options. In this case 13 satisfies the given condition. Rest of the numbers are large enough to be eliminated easily.

**12. The HCF of two numbers is 12 and their LCM is 72. If one number is 36, the other number is**

**Answer –** Option B

**Explanation –**

set the number be x

x Ã— 36 = 12 Ã— 72

x = 24

**13. The largest number which divides 81 and 108, leaving remainders 6 and 3 respectively is **

**Answer –** Option B

**Explanation –**

Checking the options we get the correct answer as 15.

**14. The HCF of two numbers is 15 and their LCM is 270. If one number is 45, the other number is**

**Answer –** Option B

**Explanation –**

HCF Ã— LCM = Product of numbers

Other number =\(\frac{15 * 270}{45}\) = 90

**15. The largest number which divides 247 and 319, leaving remainders 7 and 4 respectively is **

**Answer –** Option A

Numbers = 247, 319

i.e, Remainders = 7 & 4,

i.e, 247 – 7 = 240, 319 – 4 = 315 are divisible

HCF (240, 315) = 15

**Answer –** Option A

**Explanation –**

\(2^{2^{4}}\)divided by \(2^{2^{3}}\)

=\(2^{8}\)divided by \(2^{6}\)

= \(2^{2}\)

**2. Arrange the following fractions in ascending order. \(\frac{7}{10}\), \(\frac{3}{8}\), \(\frac{4}{5}\)**

**Answer –** Option A

**Explanation –**

\(\frac{7}{10}\) = 0.7

\(\frac{3}{8}\) = 0.375

i.e, \(\frac{3}{8}\) < \(\frac{7}{10}\) < \(\frac{4}{5}\)

\(\frac{4}{5}\) = 0.8

**3. By what least number should 192,000 be divided so as to become a perfect cube?**

**Answer –** Option C

**Explanation –**

\(\frac{192000}{2}\) = 96,000 Not a perfect cube.

\(\frac{192000}{3}\) = 64,000 A perfect cube

**4. Find the value of: 3 + 0.03 + 0.003 + 0.0003**

**Answer –** Option B

**Explanation –**

3 + 0.03 + 0.003 + 0.0003

= 3.0333

**5. Value of \(\pi\) (approx. value 3.14) is :**

**Answer –** Option B

**6. Which one of the following is not a prime number?**

**Answer –** Option B

**Explanation –**

91 is not a prime number

91 = 7 Ã— 13

**7. Find the value of : \(\frac{{489 + 375}^{2}-{489 – 375}^{2}}{489*375}\)**

**Answer –** Option D

**Explanation –**

\(\frac{{(489 + 375)}^{2}-{(489 – 375)}^{2}}{489*375}\)

\(\frac{{(a + b)}^{2}-{(a – b)}^{2}}{a * b}\)

\(\frac{4ab}{ab}\) = 4

**8. Find the L.C.M. of \(\frac{1}{3}\), \(\frac{5}{6}\), \(\frac{2}{9}\), \(\frac{4}{27}\)**

**Answer –** Option C

**Explanation –**

= \(\frac{(LCM 1, 5, 2, 4)}{(HCF 3, 6, 9, 27)}\)

\(\frac{20}{3}\)

**9. . Arrange the fractions \(\frac{5}{7}\), \(\frac{13}{16}\), \(\frac{8}{12}\), \(\frac{16}{29}\) and \(\frac{3}{4}\) in ascending order of magnitude :**

**Answer –** Option A

**Explanation –**

\(\frac{5}{8}\) = 0.625

\(\frac{7}{12}\) = 0.0.583

\(\frac{13}{16}\) = 0.8125

\(\frac{16}{29}\) = 0.551

**10. If \(\frac{a}{b}^{x-1}\) = \(\frac{b}{a}^{x-3}\), then the value of ‘x’ is :**

**Answer –** Option C

**Explanation –**

\(\frac{a}{b^{x – 1}}\) = \(\frac{b}{a^{x – 3}}\)

\(\frac{a}{b^{x – 1}}\) = \(\frac{b}{a^{3 – x}}\)

x – 1 + 3 – x

2x = 4

x = 2

**11. . Find the value of (0.000216)3 **

**Answer –** Option B

**Explanation –**

\({0.000326}{\frac{1}{3}}\) = \({216}^{\frac{1}{3}}\) * \({216}^{{(-6)}^\frac{1}{3}}\)

= 6 Ã— \({10}^{-2}\)= 0.06

**12. If a =\({2}^{129}\) * \({3}^{81}\) * \({5}^{128}, \), b = \({2}^{125}\) \({3}^{81}\) * \({5}^{128} \), c = \({2}^{126}\) * \({3}^{82}\) * \({5}^{128} \), d = \({2}^{125}\) * \({3}^{82}\) * \({5}^{129}\)then HCF OF a, b, c and d is**

**Answer –** Option B

**Explanation –**

Highest power of 2 common in a1, b1, c1, d1= 125

Highest power of 3 common in a1, b1, c1, d1 = 81

Highest power of 5 common in a1, b1, c1, d1 = 128

i.e, HCF = \({2}^{125}\) * \({3}^{81}\) * \({5}^{128} \),

**13. Let x be a least number which when divided by 21,33,35 and 55 leaves in each case a remainder 3, but is exactly divisible by 67. The sum of digits of x is**

**Answer –** Option D

**Explanation –**

= LCM (211, 331, 351, 55)k + 3

= 1155k + 3

\(\frac{A}{Q}\) x = 67 k5

= 67k = 1155k + 3

at k = 4

Number = 4623 which satisfies all conditions.

i.e, sum of digits = 4 + 6 + 2 + 3 = 15

**14. HCF of two numbers, each consisting of four digits, is 103 and their LCM is 19261. The difference of the numbers is**

**Answer –** Option A

**Explanation –**

Product of two Numbers = LCM Ã— HCF

x Ã— y = 103 Ã— 19261

= 103 Ã— 103 Ã— 11 Ã— 17 = 1133 Ã— 1751

Difference = \(\frac{6}{8}\)

**15. Let a =\({3}^{129}\) * \({5}^{128}\) * \({7}^{22}, \), b = \({3}^{128}\) \({5}^{129}\) * \({7}^{22} \), c = \({3}^{128}\) * \({5}^{128}\) * \({7}^{128} \), d = \({2}^{129}\) * \({5}^{128}\) * \({7}^{33}\)then HCF OF a, b, c and d is**

**Answer –** Option D

**Explanation –**

Highest power of 3 common in a1, b1, c1, d1= 129

Highest power of 5 common in a1, b1, c1, d1 = 129

Highest power of 7 common in a1, b1, c1, d1 = 23