Practice Number System

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Which of the following is the smallest fraction?

\[\frac{7}{6},\frac{7}{9},\frac{4}{5},\frac{5}{7}\]

\[\frac{7}{6}\]

\[\frac{7}{9}\]

\[\frac{4}{5}\]

\[\frac{5}{7}\]

\[\frac{7}{6}=1.166\]

\[\frac{7}{9}=0.777\]

\[\frac{4}{5}=0.8\]

\[\frac{5}{7}=0.714\]

Therefore, the smallest number is \[\frac{5}{7}\]

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Which of the following is the largest fraction?

\[\frac{6}{7},\frac{5}{6},\frac{7}{8},\frac{4}{5}\]

\[\frac{6}{7}\]

\[\frac{4}{5}\]

\[\frac{5}{6}\]

\[\frac{7}{8}\]

The decimal equivalents of :

\[\frac{6}{7}=0.857\]

\[\frac{5}{6}=0.833\]

\[\frac{7}{8}=0.875\]

\[\frac{4}{4}=0.8\]

Obviously, \[0.875\] is the greatest.

\[\frac{7}{8}\] is the largest fraction.

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Which of the following is the least number?

\[\frac{4}{9},\ \sqrt{\frac{9}{49}},\ 0.45,\ (0.8)^2\]

\[\frac{4}{9}\]

\[\sqrt{\frac{9}{49}}\]

\[0.45\]

\[(0.8)^2\]

Decimal equivalents :

\[\frac{4}{9}=0.4\]

\[\sqrt{\frac{9}{49}}=\frac{3}{7}=0.43\]

\[0.45\]

\[(0.8)^2=0.64\]

\[\therefore\] Least nmber \[= 0.43\]

\[=\sqrt{\frac{9}{49}}\]

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Which of the following number is the greatest of all ?

\[0.9,\ 0.\overline{9},\ 0.0\overline{9},\ 0.\overline{09}\]

\[0.9\]

\[0.\overline{9}\]

\[0.0\overline{9}\]

\[0.\overline{09}\]

\[0.9=\frac{9}{10}\]

\[0.\overline{9}=\frac{9}{9}=1\]

\[0.0\overline{9}=\frac{9}{10}=\frac{1}{10}\]

\[0.\overline{09}=\frac{9}{99}=\frac{1}{11}\]

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If a number is as much greater than \[31\] as it is less than \[75\], then the number is

106

If the number be \[x\] , then

\[x - 31 = 75 - x\]

\[\Rightarrow 2 x = 75 + 31 = 106\]

\[x = 53\]

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\[\frac{1}{0.04}\] is equal to :

\[\frac{1}{40}\]

\[\frac{2}{5}\]

\[\frac{5}{2}\]

\[25\]

\[\frac{1}{0.04}=\frac{100}{4}=25\]

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When a number is divided by \[56\], the remainder obtained is \[29\]. What will be the remainder when the number is divided by \[8\] ?

Rule : When the second divisor is factor of first divisor, the second remainder is obtained by dividing the first remainder by the second divisor. Hence, on dividing \[29\] by \[8\], the remainder is \[5\].

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If \[5432\ast7\] is divisible by \[9\], then the digit in place of \[\ast\] is :

A number is divisible by \[9\], if sum of its digits is divisible by \[9\].

Let the number be \[x\].

\[\implies 5 + 4 + 3 + 2 + x + 7 = 21 + x\]

\[\therefore x = 6\]

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What is two-third of half of \[369\] ?

\[123\]

\[246\]

\[246\frac{3}{8}\]

\[?=́́369\times\frac{1}{2}\times\frac{2}{3}\]

\[=123\]

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\[\frac{1}{5}\] of a number exceeds \[\frac{1}{7}\] of the same number by \[10\]. The number is :

125

150

175

200

Let the number be \[x\].

\[\therefore\] According to question,

\[\frac{x}{5}-\frac{x}{7}=10\]

\[\Rightarrow\frac{\left\{7x-5x\right\}}{35}=10\]

\[\Rightarrow\frac{2x}{35}=10\]

\[\Rightarrow x=\frac{\left\{10\times35\right\}}{2}=175\]

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\[\frac{2}{3}\] of three-fourth of a number is:

\[\frac{1}{2}\] of the number

\[\frac{1}{3}\] of the number

\[\frac{8}{9}\] of the number

\[\frac{17}{12}\] of the number

\[\frac{2}{3}\times\frac{3}{4}=\frac{1}{2}\]

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\[0.\overline{001}\] is equal to :

\[\frac{1}{1000}\]

\[\frac{1}{999}\]

\[\frac{1}{99}\]

\[\frac{1}{9}\]

\[0.\overline{001}=\frac{1}{999}\]

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If the difference between the reciprocal of a positive proper fraction and the fraction itself be \[\frac{9}{20}\], then the fraction is

\[\frac{3}{5}\]

\[\frac{3}{10}\]

\[\frac{4}{5}\]

\[\frac{5}{4}\]

The required fraction is \[\frac{4}{5}\],

because \[\frac{5}{4}-\frac{4}{5}=\frac{\left\{25-16\right\}}{20}=\frac{9}{20}\]

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A tree increases annually by \[\frac{1}{8}\] of its height. By how much will it increase after \[2\] years, if it stands today \[64\] cm high?

\[72\] cm

\[74\] cm

\[75\] cm

\[81\] cm

Height of tree after 1 year

\[=64+\Bigg(64\times\frac{1}{8}\Bigg)=72\]

Height of tree after 2 year

\[=72+\Bigg(72\times\frac{1}{8}\Bigg)\]

\[=72+9=81\] cm

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In a class, there are '\[z\]' students. Out of them '\[x\]' are boys. What part of the class is composed of girls ?

\[\frac{x}{z}\]

\[\frac{z}{x}\]

\[1-\frac{x}{z}\]

\[\frac{x}{z}-1\]

Boys \[= x\]

Girls \[= z - x\]

\[\therefore\] Part of girls \[=\frac{\left\{z-x\right\}}{z}=1-\frac{x}{z}\]

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\[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{7}+\frac{1}{14}+\frac{1}{28}\] is equal to :

\[2\]

\[2.5\]

\[3\]

\[3.5\]

\[?=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{7}+\frac{1}{14}+\frac{1}{28}\]

\[=\frac{28+14+7+4+2+1}{28}\]

\[=\frac{28+28}{28}=2\]

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Six numbers are arranged in decreasing order. The average of the first five numbers is \[30\] and the average of the last five numbers is \[25\]. The difference of the first and the last numbers is :

Numbers are :

\[a > b > c > d > e > f\]

According to the question,

\[a + b + c + d + e = 5 × 30 = 150 \rightarrow (i)\]

\[b + c + d + e + f = 5 × 25 = 125 \rightarrow (ii)\]

By equation \[(i) - (ii)\]

\[a – f = 150 – 125 = 25\]

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The digit in unit’s place of the product \[81 × 82 × 83 × ... × 89\] is

The digit in unit’s place \[=\] unit’s digit in the product \[1 × 2 × 3 × .... × 9 = 0\].

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The sum of three consecutive odd natural numbers is \[147\]. Then, the middle number is :

\[x + (x + 2) + (x + 4) = 147\]

\[\implies 3 x + 6 = 147\]

\[\implies 3 x = 147 – 6 = 141\]

\[\implies x=\frac{141}{3}=47\]

\[\therefore\] Middle Number

\[=x + 2 = 47 + 2 = 49\]

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The product of two positive numbers is \[2500\]. If one number is four times the other, then the sum of the two numbers is :

125

225

250

Let one of the positive numbers be \[x\].

The other will be \[4x\]

Now, \[4x \times x = 2500\]

\[\implies x^2 = 2500 \div 4 = 625\]

\[x = \sqrt{625} = 25\]

\[\therefore\] Sum of the two numbers

\[= 5x = 5 \times 25 = 125\]