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Tenths and Hundredths

Question

$3. ¯ ¯¯¯¯ ¯ 25$ is equal to
$2 + \frac{25}{99}$
$3 + \frac{24}{99}$
$3 + \frac{25}{99}$
$3 + \frac{25}{98}$

A

3 + \frac{24}{99}

B

3 + \frac{25}{99}

C

3 + \frac{25}{98}

D

2 + \frac{25}{99}

Open in App

Solution

Verified by Toppr

$3. ¯ 25 = 3 + 25 * 0. ¯ 01 = 3 + 25 * \frac{1}{99} = 3 + \frac{25}{99}$

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Similar Questions

Q1

If

[x]

denotes the greates integer

\leq x

, then

[\frac{2}{3}] + [\frac{2}{3} + \frac{1}{99}] + [\frac{2}{3} + \frac{2}{99}] + . . . . + [\frac{2}{3} + \frac{98}{99}] =

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Q2

If [x] denotes the greatest integers

\leq

x, then find the value of

[\frac{2}{3}] + [(\frac{2}{3}) + (\frac{1}{99})] + [(\frac{2}{3}) + (\frac{2}{99})] + \dots \dots + [(\frac{2}{3}) + \frac{98}{99}]

.

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Q3

Factorise:

(1) x² − 2x − 99
(2) 3p² − 10p − 25
(3) 125x³ − 216y³
(4) 128 + 2x³

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Q4

What is the remainder when p - q is divided by 100, if

p = 2^{99} + 4^{99} + 6^{99} + 8^{99} + . . . . . . . . . + 100^{99}

and

q = 1^{99} + 3^{99} + 5^{99} + 7^{99} + . . . . . . . . . . + 99^{99} ?

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Q5

${lim}_{π \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \dots \dots n^{99}}{n^{100}} =$ [EAMCET 1994]

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