Find the value of -n- such that 2-3-4n-1- - 2n-1-n-3-1-3n-4-3-5-4-5-25-45-2

The correct option is D 5/2Given expression: 2/3(4n 1) ( 2n 1+n/3 )=1/3n+4/3 By taking the LCM, nbsp; 2/3(4n 1) 1/3(6n 1 n) nbsp;=1/3n+4/3 The denomina ...

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Standard VIII

Mathematics

Solving Equations with Variables on Both Sides

Find the valu...

Question

$Find the value of 'n' such that$
$\frac{2}{3} (4 n - 1) - (2 n - \frac{1 + n}{3}) = \frac{1}{3} n + \frac{4}{3}$ .

\frac{5}{4}

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\frac{5}{4}

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\frac{5}{2}

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\frac{5}{2}

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Solution

The correct option is D $\frac{5}{2}$
$Given expression: \frac{2}{3} (4 n - 1) - (2 n - \frac{1 + n}{3}) = \frac{1}{3} n + \frac{4}{3}$

By taking the LCM,
$\frac{2}{3} (4 n - 1) - \frac{1}{3} (6 n - 1 - n) = \frac{1}{3} n + \frac{4}{3}$
The denominator of all the terms are same.
So, by removing the denominator,
$2 (4 n - 1) - (5 n - 1) = n + 4$
$8 n - 2 - 5 n + 1 = n + 4$
$(8 n - 5 n) + (- 2 + 1) = n + 4$
$3 n - 1 = n + 4$
$3 n - n = 4 + 1$
$2 n = 5$
$n = \frac{5}{2}$

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Prove that :

$(i) (\sqrt{3 \times 5^{- 3}} \div \sqrt[3]{3^{- 1}} \sqrt{5}) \times \sqrt[6]{3 \times 5^{6}} = \frac{3}{5} (i i) 9^{\frac{3}{2}} - 3 \times 5^{0} - {(\frac{1}{81})}^{\frac{- 1}{2}} = 15 (i i i) {(\frac{1}{4})}^{- 2} - 3 \times 8^{\frac{2}{3}} \times 4^{0} + {(\frac{9}{16})}^{\frac{- 1}{2}} = \frac{16}{3} (i v) \frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 4^{\frac{1}{4}}}{10^{\frac{- 1}{5}} \times 5^{\frac{3}{5}}} \div \frac{3^{\frac{4}{3}} \times 5^{\frac{- 7}{5}}}{4^{\frac{- 3}{5}} \times 6} = 10 (v) \sqrt{\frac{1}{4}} + (0.01)^{\frac{- 1}{2}} - (27)^{\frac{2}{3}} = \frac{3}{2} (v i) \frac{2^{n} + 2^{n - 1}}{2^{n + 1} - 2^{n}} = \frac{3}{2} (v i i) {(\frac{64}{125})}^{\frac{- 2}{3}} + \frac{1}{{(\frac{256}{625})}^{\frac{1}{4}}} + {(\frac{\sqrt{25}}{\sqrt[3]{64}})}^{0} = \frac{61}{16} (v i i i) \frac{3^{- 3} \times 6^{2} \times \sqrt{98}}{5^{2} \times \sqrt[3]{\frac{1}{25}} \times (15)^{\frac{- 4}{3}} \times 3^{\frac{1}{3}}} = 28 \sqrt{2} (i x) \frac{(0.6)^{0} - (0.1)^{- 1}}{{(\frac{3}{8})}^{- 1} {(\frac{3}{2})}^{3} + {(\frac{1}{3})}^{- 1}} = - \frac{3}{2}$

Find the value of 'n' such that

\frac{2}{3} (4 n - 1) - (2 n - \frac{1 + n}{3}) = \frac{1}{3} n + \frac{4}{3}

Find the value of n such that $\frac{2}{3} (4 n - 1) - (2 n - \frac{1 + n}{3}) = \frac{1}{3} n + \frac{4}{3}$ .

Q. Find the value of

n

in the following equation:

\frac{2}{3} (4 n - 1) - (2 n - \frac{1 + n}{3}) = \frac{1}{3} n + \frac{4}{3}

Q. If n is a multiple of

6

, show that each of the series

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot 3 + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot 3^{2} - . . . . .,

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot \frac{1}{3} + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot \frac{1}{3^{2}} - . . . . .,

is equal to zero.

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Find the value of 'n' such that 23(4n−1)−(2n−1+n3)=13n+43.

$Find the value of 'n' such that$
$\frac{2}{3} (4 n - 1) - (2 n - \frac{1 + n}{3}) = \frac{1}{3} n + \frac{4}{3}$ .