What will be the vulgar fraction of 0.75 ?
Answer: B
\(0.75 = \dfrac {75}{100} =\dfrac { 3}{4}\)
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The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
Answer: B
Suppose commodity X will cost 40 paise more than Y after z years.
Then,
\((4.20 + 0.40z) - (6.30 + 0.15z) = 0.40\\
\Rightarrow 0.25z = 0.40 + 2.10\\
\Rightarrow z= \dfrac {2.50}{0.25} = \dfrac {250}{25} = 10\)
X will cost 40 paise more than Y 10 years after 2001 i.e., 2011
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The expression \((12.86 \times 12.86 + 12.86 \times p + 0.14 \times 0.14)\) will be a perfect square for \(p\) equal to:
Answer: A
\((a + b)^2 = a^2 + 2ab + b^2\)
\(12.86 \times 12.86 + 12.86 \times p + 0.14 \times 0.14\)
\(= (12.86)^2 + 12.86 \times p + (0.14)^2\)
This can be written as,
\((12.86 + 0.14)^2 = 13^2\)
If, \(12.86 \times p = 2 \times 12.86 \times 0.14\)
i.e., if \(p = 2 \times 0.14 = 0.28\)
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Evaluate : \(\dfrac{(2.39)^2 - (1.61)^2}{2.39 - 1.61}\)
Answer: B
\(\dfrac{2.39^2 - 1.61^2}{2.39 - 1.61} \\~\\= \dfrac{(2.39 + 1.61)(2.39 - 1.61)}{(2.39 - 1.61)}\\~\\ = 2.39 + 1.61 \\~\\= 4\)
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Arrange the fractions in ascending order.
Answer: D
\(\dfrac {5}{8} = .625\)
\(\dfrac {7}{12} = .5833\)
\(\dfrac {3}{4} = .75\)
\(\dfrac {13}{16} = .8125\)
So order will be
\(\dfrac {7}{12} < \dfrac {5}{8} < \dfrac {3}{4}< \dfrac {13}{16}\)
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