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By how much does $\frac{6}{7 / 8} exceed \frac{6 / 7}{8} = ?$ $\frac{6}{{7/8}}{\text{exceed }}\frac{{6/7}}{8} = ?$

1 $6 \frac{1}{8}$ $6\frac{1}{8}$
2 $6 \frac{3}{4}$ $6\frac{3}{4}$
3 $7 \frac{3}{4}$ $7\frac{3}{4}$
4 $7 \frac{5}{6}$ $7\frac{5}{6}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{6}{7 / 8} - \frac{6 / 7}{8} \\ = 6 \times \frac{8}{7} - \frac{6}{7} \times \frac{1}{8} \\ = \frac{48}{7} - \frac{6}{56} \\ = \frac{384 - 6}{56} \\ = \frac{378}{56} \\ = \frac{27}{4} \\ = 6 \frac{3}{4} \end{aligned}

$\eqalign{ & \frac{6}{{7/8}} - \frac{{6/7}}{8} \cr & = 6 \times \frac{8}{7} - \frac{6}{7} \times \frac{1}{8} \cr & = \frac{{48}}{7} - \frac{6}{{56}} \cr & = \frac{{384 - 6}}{{56}} \cr & = \frac{{378}}{{56}} \cr & = \frac{{27}}{4} \cr & = 6\frac{3}{4} \cr}$

02. If x is a positive number, then which of the following fractions has the greatest value ?

1 $\frac{x}{x}$ $\frac{x}{x}$
2 $\frac{x}{x + 1}$ $\frac{x}{{x + 1}}$
3 $\frac{x + 1}{x}$ $\frac{{x + 1}}{x}$
4 $\frac{x + 2}{x + 3}$ $\frac{{x + 2}}{{x + 3}}$

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Answer:- 1

Explanation:-

Solution:

Clearly, ^(x+1)/_x is the only fraction in which the numerator is greater then the denominator. So, it is greatest fraction.

03. By what least number should 4320 be multiplied so as to obtain a number which a perfect cube ?

140
250
360
490

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} According to question \\ \frac{3 | 4320}{\frac{3 | 1440}{\frac{3 | 480}{\frac{2 | 160}{\frac{2 | 80}{\frac{2 | 40}{\frac{2 | 20}{\frac{2 | 10}{\frac{5 | 5}{1}}}}}}}}} \\ Factors are : \\ 3 \times 3 \times 3 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \\ ∴ It must be multiplied by, \\ 2 \times 5 \times 5  =  50 \end{aligned}

$\eqalign{ & {\text{According to question}} \cr & \,\frac{{3\,\,\,|\,\,\,4320}}{{\frac{{3\,\,\,\,|\,\,\,\,1440}}{{\frac{{3\,\,\,\,|\,\,\,\,\,\,480}}{{\frac{{2\,\,\,\,|\,\,\,\,\,\,\,160}}{{\frac{{2\,\,\,\,|\,\,\,\,\,\,\,\,\,80}}{{\frac{{2\,\,\,\,|\,\,\,\,\,\,\,\,\,40}}{{\frac{{2\,\,\,\,|\,\,\,\,\,\,\,\,\,20}}{{\frac{{2\,\,\,\,|\,\,\,\,\,\,\,\,\,10}}{{\frac{{5\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,5}}{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1}}}}}}}}}}}}}}}}}} \cr & {\text{Factors are}}\,\,:\, \cr & 3 \times 3 \times 3 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \cr & \therefore {\text{It must be multiplied by,}} \cr & {\text{2}} \times {\text{5}} \times {\text{5 = 50}} \cr}$

04. By what least number should 675 be multiplied so as to obtain a perfect cube number ?

13
25
324
440

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} According to question \\ 5 | 675 \\ - - - - - - \\ 5 | 135 \\ - - - - - - \\ 5 | 27 \\ - - - - - - \\ 3 | 9 \\ - - - - - - \\ 3 | 3 \\ - - - - - - \\ | 1 \\ Factors are \\ = 5 \times 5 \times 3 \times 3 \times 3 \end{aligned}

$\eqalign{ & {\text{According to question}} \cr & {\text{5}}\,\,{\text{| 675}} \cr & - - - - - - \cr & 5\,\,|\,\,135 \cr & - - - - - - \cr & 5\,\,|\,\,27 \cr & - - - - - - \cr & 3\,\,|\,\,9 \cr & - - - - - - \cr & 3\,\,|\,\,3\quad \cr & - - - - - - \cr & \,\,\,\,\,|\,\,1 \cr & {\text{Factors are}} \cr & = \,5 \times 5 \times 3 \times 3 \times 3 \cr}$ ∴ It must be multiplied by 5 to make a perfect cube.

The smallest fraction which should be subtracted from the sum of $1 \frac{3}{4}$ ${\text{1}}\frac{3}{4}$ , $2 \frac{1}{2}$ ${\text{2}}\frac{1}{2}$ , $5 \frac{7}{12}$ $5\frac{7}{{12}}$ , $3 \frac{1}{3}$ ${\text{3}}\frac{1}{3}$ and $2 \frac{1}{4}$ ${\text{2}}\frac{1}{4}$ to make the result a whole number is = ?

1 $\frac{5}{12}$ $\frac{5}{{12}}$
2 $\frac{7}{12}$ $\frac{7}{{12}}$
3 $\frac{1}{2}$ $\frac{1}{2}$
47

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Sum of given fractions \\ = \frac{7}{4} + \frac{5}{2} + \frac{67}{12} + \frac{10}{3} + \frac{9}{4} \\ = (\frac{21 + 30 + 67 + 40 + 24}{12}) \\ = \frac{85}{12} \\ The whole number just \\ less than \frac{185}{12} is 15 \\ let \frac{185}{12} - x = 15 \\ Then, \\ x = (\frac{185}{12} - 15) \\ = \frac{5}{12} \end{aligned}

$\eqalign{ & {\text{Sum of given fractions}} \cr & {\text{ = }}\frac{7}{4} + \frac{5}{2} + \frac{{67}}{{12}} + \frac{{10}}{3} + \frac{9}{4} \cr & = \left( {\frac{{21 + 30 + 67 + 40 + 24}}{{12}}} \right) \cr & = \frac{{85}}{{12}} \cr & {\text{The whole number just }} \cr & {\text{less than }}\frac{{185}}{{12}}{\text{ is 15}} \cr & {\text{let }}\frac{{185}}{{12}} - x = 15 \cr & {\text{Then,}} \cr & x = \left( {\frac{{185}}{{12}} - 15} \right) \cr & \,\,\,\,\,\, = \frac{5}{{12}} \cr}$

The sum of the first 99 terms of the series $\frac{3}{4} + \frac{5}{16} + \frac{7}{144} + \frac{9}{400} + . . . . .$ $\frac{3}{4} + \frac{5}{{16}} + \frac{7}{{144}} + \frac{9}{{400}} + .....$

1 $\frac{99}{100}$ $\frac{{99}}{{100}}$
2 $\frac{999}{1000}$ $\frac{{999}}{{1000}}$
3 $\frac{9999}{10000}$ $\frac{{9999}}{{10000}}$
41

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given expression, \\ = \frac{4 - 1}{4 \times 1} + \frac{9 - 4}{9 \times 4} + \frac{16 - 9}{16 \times 9} + . . . . . . . . . \\ = (1 - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{9}) + (\frac{1}{9} - \frac{1}{16}) + . . . . . . . . \\ = (\frac{1}{1^{2}} - \frac{1}{2^{2}}) + (\frac{1}{2^{2}} - \frac{1}{3^{2}}) + (\frac{1}{3^{2}} - \frac{1}{4^{2}}) + . . . . \\ ∴ 99 th term of the series \\ = (\frac{1}{99^{2}} - \frac{1}{100^{2}}) \\ ∴ Given expression, \\ (\frac{1}{1^{2}} - \frac{1}{2^{2}}) + (\frac{1}{2^{2}} - \frac{1}{3^{2}}) + (\frac{1}{3^{2}} - \frac{1}{4^{2}}) + . . . . + (\frac{1}{98^{2}} - \frac{1}{99^{2}}) + (\frac{1}{99^{2}} - \frac{1}{100^{2}}) \\ = (1 - \frac{1}{100^{2}}) \\ = (1 - \frac{1}{10000}) \\ = \frac{9999}{10000} \end{aligned}

$\eqalign{ & {\text{Given expression,}} \cr & {\text{ = }}\frac{{4 - 1}}{{4 \times 1}} + \frac{{9 - 4}}{{9 \times 4}} + \frac{{16 - 9}}{{16 \times 9}} + ......... \cr & = \left( {1 - \frac{1}{4}} \right) + \left( {\frac{1}{4} - \frac{1}{9}} \right) + \left( {\frac{1}{9} - \frac{1}{{16}}} \right) + ........ \cr & = \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + .... \cr & \therefore \,99{\text{th term of the series}} \cr & {\text{ = }}\left( {\frac{1}{{{{99}^2}}} - \frac{1}{{{{100}^2}}}} \right) \cr & \therefore \,\,\,{\text{Given expression,}} \cr & \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + .... + \left( {\frac{1}{{{{98}^2}}} - \frac{1}{{{{99}^2}}}} \right) + \left( {\frac{1}{{{{99}^2}}} - \frac{1}{{{{100}^2}}}} \right) \cr & = \left( {1 - \frac{1}{{{{100}^2}}}} \right) \cr & = \left( {1 - \frac{1}{{10000}}} \right) \cr & = \frac{{9999}}{{10000}} \cr}$

The value of $\frac{3}{1^{2} \times 2^{2}} + \frac{5}{2^{2} \times 3^{2}} +$ $\frac{3}{{{1^2} \times {2^2}}} + \frac{5}{{{2^2} \times {3^2}}} +$ $\frac{7}{3^{2} \times 4^{2}} + \frac{9}{4^{2} \times 5^{2}} +$ $\frac{7}{{{3^2} \times {4^2}}} + \frac{9}{{{4^2} \times {5^2}}} +$ $\frac{11}{5^{2} \times 6^{2}} + \frac{13}{6^{2} \times 7^{2}} +$ $\frac{{11}}{{{5^2} \times {6^2}}} + \frac{{13}}{{{6^2} \times {7^2}}} +$ $\frac{15}{7^{2} \times 8^{2}} + \frac{17}{8^{2} \times 9^{2}} +$ $\frac{{15}}{{{7^2} \times {8^2}}} + \frac{{17}}{{{8^2} \times {9^2}}} +$ $\frac{19}{9^{2} \times 10^{2}} is = ?$ $\frac{{19}}{{{9^2} \times {{10}^2}}}\,\,\,\,{\text{is = ?}}$

1 $\frac{1}{100}$ $\frac{1}{{100}}$
2 $\frac{99}{100}$ $\frac{{99}}{{100}}$
31
4 $\frac{101}{100}$ $\frac{{101}}{{100}}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given expression, \\ (\frac{1}{1^{2}} - \frac{1}{2^{2}}) + (\frac{1}{2^{2}} - \frac{1}{3^{2}}) + (\frac{1}{3^{2}} - \frac{1}{4^{2}}) + (\frac{1}{4^{2}} - \frac{1}{5^{2}}) . . . . (\frac{1}{9^{2}} - \frac{1}{10^{2}}) \\ = (\frac{1}{1^{2}} - \frac{1}{2^{2}}) \\ = (1 - \frac{1}{100}) \\ = \frac{99}{100} \end{aligned}

$\eqalign{ & {\text{Given expression,}} \cr & \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + \left( {\frac{1}{{{4^2}}} - \frac{1}{{{5^2}}}} \right)....\left( {\frac{1}{{{9^2}}} - \frac{1}{{{{10}^2}}}} \right) \cr & = \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) \cr & = \left( {1 - \frac{1}{{100}}} \right) \cr & = \frac{{99}}{{100}} \cr}$

$[2 \sqrt{54} - 6 \sqrt{\frac{2}{3}} - \sqrt{96}]$ $\left[ {2\sqrt {54} - 6\sqrt {\frac{2}{3}} - \sqrt {96} } \right]$ is equal to = ?

10
21
32
4 $\sqrt{6}$ $\sqrt 6$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} According to question \\ 2 \sqrt{54} - 6 \sqrt{\frac{2}{3}} - \sqrt{96} \\ = 6 \sqrt{6} - 2 \sqrt{\frac{2}{3} \times 9} - 4 \sqrt{6} \\ = 6 \sqrt{6} - 2 \sqrt{6} - 4 \sqrt{6} \\ = 0 \end{aligned}

$\eqalign{ & {\text{According to question}} \cr & 2\sqrt {54} - 6\sqrt {\frac{2}{3}} - \sqrt {96} \cr & = \,6\sqrt 6 - 2\sqrt {\frac{2}{3} \times 9} - 4\sqrt 6 \cr & = 6\sqrt 6 - 2\sqrt 6 - 4\sqrt 6 \cr & = 0 \cr}$

$\frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} + \frac{1}{4.5.6}$ $\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \frac{1}{{3.4.5}} + \frac{1}{{4.5.6}}$ is equal to = ?

1 $\frac{7}{30}$ $\frac{7}{{30}}$
2 $\frac{11}{30}$ $\frac{{11}}{{30}}$
3 $\frac{13}{30}$ $\frac{{13}}{{30}}$
4 $\frac{17}{30}$ $\frac{{17}}{{30}}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given expression, \\ \frac{4.5.6 + 5.6 + 2.6 + 2.3}{1.2.3.4.5.6} \\ = \frac{120 + 30 + 12 + 6}{720} \\ = \frac{168}{720} \\ = \frac{7}{30} \end{aligned}

$\eqalign{ & {\text{Given expression,}} \cr & \frac{{4.5.6 + 5.6 + 2.6 + 2.3}}{{1.2.3.4.5.6}} \cr & = \frac{{120 + 30 + 12 + 6}}{{720}} \cr & = \frac{{168}}{{720}} \cr & = \frac{7}{{30}} \cr}$

The value of $1 + \frac{1}{4 \times 3} + \frac{1}{4 \times 3^{2}} + \frac{1}{4 \times 3^{2}} = ?$ ${\text{1 + }}\frac{1}{{4 \times 3}} + \frac{1}{{4 \times {3^2}}} + \frac{1}{{4 \times {3^2}}} = ?$

1 $\frac{121}{108}$ $\frac{{121}}{{108}}$
2 $\frac{3}{2}$ $\frac{3}{2}$
3 $\frac{31}{2}$ $\frac{{31}}{2}$
4None of these

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given expression, \\ \frac{4 \times 3^{3} + 3^{2} + 3 + 1}{4 \times 3^{3}} \\ = \frac{108 + 9 + 3 + 1}{108} \\ = \frac{121}{108} \end{aligned}

$\eqalign{ & {\text{Given expression,}} \cr & \frac{{4 \times {3^3} + {3^2} + 3 + 1}}{{4 \times {3^3}}} \cr & = \frac{{108 + 9 + 3 + 1}}{{108}} \cr & = \frac{{121}}{{108}} \cr}$

Hello Guest

Simplification - 04

By how much does $\frac{6}{7 / 8} exceed \frac{6 / 7}{8} = ?$ $\frac{6}{{7/8}}{\text{exceed }}\frac{{6/7}}{8} = ?$

02. If x is a positive number, then which of the following fractions has the greatest value ?

03. By what least number should 4320 be multiplied so as to obtain a number which a perfect cube ?

04. By what least number should 675 be multiplied so as to obtain a perfect cube number ?

The sum of the first 99 terms of the series $\frac{3}{4} + \frac{5}{16} + \frac{7}{144} + \frac{9}{400} + . . . . .$ $\frac{3}{4} + \frac{5}{{16}} + \frac{7}{{144}} + \frac{9}{{400}} + .....$

$[2 \sqrt{54} - 6 \sqrt{\frac{2}{3}} - \sqrt{96}]$ $\left[ {2\sqrt {54} - 6\sqrt {\frac{2}{3}} - \sqrt {96} } \right]$ is equal to = ?

$\frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} + \frac{1}{4.5.6}$ $\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \frac{1}{{3.4.5}} + \frac{1}{{4.5.6}}$ is equal to = ?

The value of $1 + \frac{1}{4 \times 3} + \frac{1}{4 \times 3^{2}} + \frac{1}{4 \times 3^{2}} = ?$ ${\text{1 + }}\frac{1}{{4 \times 3}} + \frac{1}{{4 \times {3^2}}} + \frac{1}{{4 \times {3^2}}} = ?$

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Hello Guest

Simplification - 04

01. By how much does 67/8exceed 6/78=?67/8exceed 6/78=?\frac{6}{{7/8}}{\text{exceed }}\frac{{6/7}}{8} = ?

02. If x is a positive number, then which of the following fractions has the greatest value ?

03. By what least number should 4320 be multiplied so as to obtain a number which a perfect cube ?

04. By what least number should 675 be multiplied so as to obtain a perfect cube number ?

05. The smallest fraction which should be subtracted from the sum of 134134{\text{1}}\frac{3}{4}, 212212{\text{2}}\frac{1}{2}, 571257125\frac{7}{{12}}, 313313{\text{3}}\frac{1}{3} and 214214{\text{2}}\frac{1}{4} to make the result a whole number is = ?

06. The sum of the first 99 terms of the series 34+516+7144+9400+.....34+516+7144+9400+.....\frac{3}{4} + \frac{5}{{16}} + \frac{7}{{144}} + \frac{9}{{400}} + .....

08. [254−−√−623−−√−96−−√][254−623−96]\left[ {2\sqrt {54} - 6\sqrt {\frac{2}{3}} - \sqrt {96} } \right] is equal to = ?

09. 11.2.3+12.3.4+13.4.5+14.5.611.2.3+12.3.4+13.4.5+14.5.6\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \frac{1}{{3.4.5}} + \frac{1}{{4.5.6}} is equal to = ?

10. The value of 1 + 14×3+14×32+14×32=?1 + 14×3+14×32+14×32=?{\text{1 + }}\frac{1}{{4 \times 3}} + \frac{1}{{4 \times {3^2}}} + \frac{1}{{4 \times {3^2}}} = ?

Subject Wise

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NCERT 6 to 12

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Practice Batch

By how much does $\frac{6}{7 / 8} exceed \frac{6 / 7}{8} = ?$ $\frac{6}{{7/8}}{\text{exceed }}\frac{{6/7}}{8} = ?$

The sum of the first 99 terms of the series $\frac{3}{4} + \frac{5}{16} + \frac{7}{144} + \frac{9}{400} + . . . . .$ $\frac{3}{4} + \frac{5}{{16}} + \frac{7}{{144}} + \frac{9}{{400}} + .....$

$[2 \sqrt{54} - 6 \sqrt{\frac{2}{3}} - \sqrt{96}]$ $\left[ {2\sqrt {54} - 6\sqrt {\frac{2}{3}} - \sqrt {96} } \right]$ is equal to = ?

$\frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} + \frac{1}{4.5.6}$ $\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \frac{1}{{3.4.5}} + \frac{1}{{4.5.6}}$ is equal to = ?

The value of $1 + \frac{1}{4 \times 3} + \frac{1}{4 \times 3^{2}} + \frac{1}{4 \times 3^{2}} = ?$ ${\text{1 + }}\frac{1}{{4 \times 3}} + \frac{1}{{4 \times {3^2}}} + \frac{1}{{4 \times {3^2}}} = ?$

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6 to 12