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UPSC CDS I 2025 Mathematics Paper

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Question : 77 of 100

Marks: +1, -0

If p =

0.0\bar{9}

, then what is the value of

70p^{2} + 43p - 5

:

$-1$
0
1
10

Solution:

Concept:

To find the value of

p = 0.0\bar{9}

, convert the repeating decimal into a fraction using algebraic manipulation.

Explanation:

Let

p = 0.0\bar{9} = 0.09999\ldots

Multiply by 10:

10p = 0.9999\ldots

Multiply by 100:

100p = 9.9999\ldots

Subtract:

100p - 10p = 9.9999\ldots - 0.9999\ldots

So

90p = 9

, hence

p = \frac{9}{90} = \frac{1}{10} = 0.1

.

Now substitute

p = 0.1

into

70p^{2} + 43p - 5

:

70(0.1)^{2} + 43(0.1) - 5 = 70 \times 0.01 + 4.3 - 5 = 0.7 + 4.3 - 5 = 0

.

Answer:

0

(Option B)

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