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27 Dec 2021, 06:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
73% (02:02) correct
27%
(02:28)
wrong
based on 299
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Two varieties of sugar are mixed together in a certain ratio. The cost of the mixture per Kg is 0.50 less than that of the superior and 0.75 more than the inferior variety. The ratio in which the superior and infereior varieties of sugar have been mixed is:
(A) 3:2
(B) 5:2
(C) 5:1
(D) 1:1
(E) None of these
(A) 3:2
(B) 5:2
(C) 5:1
(D) 1:1
(E) None of these
27 Dec 2021, 06:55
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Let the cost of superior= x
then,
the Cost of mixture= x-0.50
and the cost of inferior= x-0.50-0.75
= x-1.25
Formula,\( \frac{Quantity of superior}{Quantity of inferior}\)= \(\frac{cost of mixture - cost of inferior}{ cost of superior - Cost of mixture}\)
= \(\frac{(x-0.50)-(x-1.25)}{x-(x-0.50)}\)
=\( \frac{0.75}{0.50}\)
= \(\frac{3}{2}\) or 3:2
Answer will be A
then,
the Cost of mixture= x-0.50
and the cost of inferior= x-0.50-0.75
= x-1.25
Formula,\( \frac{Quantity of superior}{Quantity of inferior}\)= \(\frac{cost of mixture - cost of inferior}{ cost of superior - Cost of mixture}\)
= \(\frac{(x-0.50)-(x-1.25)}{x-(x-0.50)}\)
=\( \frac{0.75}{0.50}\)
= \(\frac{3}{2}\) or 3:2
Answer will be A
05 Mar 2022, 04:14
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Let,
Amount of Superior Sugar = x
Amount of Inferior Sugar = y
Cost of Superior Sugar = p
Cost of Inferior Sugar = q
Cost of Mixture = p-0.5 = q+0.75
p-0.5 = q+0.75
q = p-1.25
Using Weighted Avg. = \(\frac{px + qy}{(x+y)}\) = \(\frac{xp + y(p-1.25)}{(x+y)}\) = \(p-0.5\)
xp + yp - 1.25y = xp - 0.5x + yp - 0.5y
0.5x = 0.75y
\(\frac{x}{y}\) = \(\frac{0.75}{0.5}\) = \(\frac{3}{2}\) (A)
Amount of Superior Sugar = x
Amount of Inferior Sugar = y
Cost of Superior Sugar = p
Cost of Inferior Sugar = q
Cost of Mixture = p-0.5 = q+0.75
p-0.5 = q+0.75
q = p-1.25
Using Weighted Avg. = \(\frac{px + qy}{(x+y)}\) = \(\frac{xp + y(p-1.25)}{(x+y)}\) = \(p-0.5\)
xp + yp - 1.25y = xp - 0.5x + yp - 0.5y
0.5x = 0.75y
\(\frac{x}{y}\) = \(\frac{0.75}{0.5}\) = \(\frac{3}{2}\) (A)
