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02 Oct 2020, 00:00
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Difficulty:
15%
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Question Stats:
89% (01:28) correct
11%
(01:58)
wrong
based on 53
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Because of economies of scale, the cost of production of the second gadget is half the first one which costs c dollars to produce. The production of the third gadget costs half of the second, etc. What is the value of c?
(1) The sum of the costs of the third and fourth gadgets is $9.
(2) The sum of the costs of the second and the ninth gadget is \(12(2^7 + 1)\) dollars.
(1) The sum of the costs of the third and fourth gadgets is $9.
(2) The sum of the costs of the second and the ninth gadget is \(12(2^7 + 1)\) dollars.
globaldesi
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03 Oct 2020, 01:09
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Because of economies of scale, the cost of production of the second gadget is half the first one which costs c dollars to produce. The production of the third gadget costs half of the second, etc. What is the value of c?
(1) The sum of the costs of the third and fourth gadgets is $9.
(2) The sum of the costs of the second and the ninth gadget is \(12(2^7+1) \)dollars.
Definitely the given sequence of cost of production is G.P (geometric progression)
a being the first term.
where nth term =\( a * (r)^{n-1}\)
r here is 1/2 and a = c
(1) The sum of the costs of the third and fourth gadgets is $9.
cost of third =\(\frac{ 1}{2^2}c\)
cost of fourth=\(\frac{ 1}{2^3}c\)
\(\frac{ 1}{2^2}c\)+\(\frac{ 1}{2^3}c\) = 9
solve for c = 24 Sufficient (you actually do not need to calculate value of C equation tells C could easily be found.)
(2) The sum of the costs of the second and the ninth gadget is \(12(2^7+1) \)dollars.
cost of second =\(\frac{ 1}{2}c\)
cost of ninth=\(\frac{ 1}{2^8}c\)
so
\(\frac{ 1}{2}c\) + \(\frac{ 1}{2^8}c\) =\(12(2^7+1) \)
again value of C could easily be found
Thus Sufficient. Hence D
Key to note
(1) The sum of the costs of the third and fourth gadgets is $9.
(2) The sum of the costs of the second and the ninth gadget is \(12(2^7+1) \)dollars.
Definitely the given sequence of cost of production is G.P (geometric progression)
a being the first term.
where nth term =\( a * (r)^{n-1}\)
r here is 1/2 and a = c
(1) The sum of the costs of the third and fourth gadgets is $9.
cost of third =\(\frac{ 1}{2^2}c\)
cost of fourth=\(\frac{ 1}{2^3}c\)
\(\frac{ 1}{2^2}c\)+\(\frac{ 1}{2^3}c\) = 9
solve for c = 24 Sufficient (you actually do not need to calculate value of C equation tells C could easily be found.)
(2) The sum of the costs of the second and the ninth gadget is \(12(2^7+1) \)dollars.
cost of second =\(\frac{ 1}{2}c\)
cost of ninth=\(\frac{ 1}{2^8}c\)
so
\(\frac{ 1}{2}c\) + \(\frac{ 1}{2^8}c\) =\(12(2^7+1) \)
again value of C could easily be found
Thus Sufficient. Hence D
Key to note
04 Oct 2020, 13:47
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Bunuel
The gadgets have a cost of \(c\), \(\frac{c}{2}\), \(\frac{c}{4}\), \(\frac{c}{8}\) etc. We can see they are all linearly scaled with c.
Statement 1:
Using the pattern above we have \(\frac{c}{4} + \frac{c}{8} = 9\), we can factor our c and find the value of c. Sufficient.
Statement 2:
Similar process in statement 1 so this is sufficient. However note the values of c will disagree for the two statements, the value here should be \(12(2^{-7} + 1)\) for a coherent question.
Ans: D
