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01 Nov 2014, 07:58
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1. What is the range of Set S (f1, f2, f3, .... f100) where fn = Nn/Dn (f1=N1/D1 , f2=N2/D2 etc) and f(n+1) = Nn+10 and D(n+1)= Dn+10 (0 \(N(100) = N(99+1) = (N1)+99(100)= 1+990 = 991\)
Similary; \(D(n+1)=D1+n(10)\)==>\(D(100) = D(99+1) = D1+99(100)= 10+990 = 1000\)
Below is based on https://gmatclub.com/forum/fractions-faster-calculation-187855.html
We can observe that both numerator and denominator shows constant increase. (N by 10 and D also by 10)
So, increase in Numerator a= 10 and increase in denominator b=10
\(\frac{a}{b}\) = \(\frac{10}{10}\)= 1
Clearly \(1>\frac{1}{10} (\frac{1}{10}\)is the first fraction in the set)
From the theory that when fractions are aranged in such a way that both numerator and denominator increase by constant value and ratio of increase in numerator to increase in denominator is greater than the first fraction then the last fraction is the greatest among all given fractions.
Hence The last fraction is the greatest of all fractions in the set.
last fraction = \(f(100) = \frac{(N99)+10}{(D99)+10}= \frac{991}{1000}\) is greatest of all fractions in the set.
Now range = highest value - least value
= \(\frac{991}{1000} - \frac{1}{10}\)
= \(\frac{891}{1000}\)
=0.891[/spoiler]
2. An equilateral triangle with side x is circumscribed by a circle with radius R and another circle with radius r is inscribed in the same triangle. Find the ratio of the area of the outer circle to that of inner circle
\(A) 1:4\\
B) 4:1\\
C) 1:2\\
D) 2:1\\
E) \sqrt{2}:4\)
Similary; \(D(n+1)=D1+n(10)\)==>\(D(100) = D(99+1) = D1+99(100)= 10+990 = 1000\)
Below is based on https://gmatclub.com/forum/fractions-faster-calculation-187855.html
We can observe that both numerator and denominator shows constant increase. (N by 10 and D also by 10)
So, increase in Numerator a= 10 and increase in denominator b=10
\(\frac{a}{b}\) = \(\frac{10}{10}\)= 1
Clearly \(1>\frac{1}{10} (\frac{1}{10}\)is the first fraction in the set)
From the theory that when fractions are aranged in such a way that both numerator and denominator increase by constant value and ratio of increase in numerator to increase in denominator is greater than the first fraction then the last fraction is the greatest among all given fractions.
Hence The last fraction is the greatest of all fractions in the set.
last fraction = \(f(100) = \frac{(N99)+10}{(D99)+10}= \frac{991}{1000}\) is greatest of all fractions in the set.
Now range = highest value - least value
= \(\frac{991}{1000} - \frac{1}{10}\)
= \(\frac{891}{1000}\)
=0.891[/spoiler]
2. An equilateral triangle with side x is circumscribed by a circle with radius R and another circle with radius r is inscribed in the same triangle. Find the ratio of the area of the outer circle to that of inner circle
\(A) 1:4\\
B) 4:1\\
C) 1:2\\
D) 2:1\\
E) \sqrt{2}:4\)
Show SpoilerAnswer
\(B\)
Show SpoilerSolution
Area of a circle circumscribing an equilateral triangle = \(\frac{pi}{3}x^2\)
Area of a circle that is inscribed in an equilateral triangle = \(\frac{pi}{12}x^2\)
Ratio = \(\frac{pi}{3}x^2 : \frac{pi}{12}x^2 = 4: 1\)
Area of a circle that is inscribed in an equilateral triangle = \(\frac{pi}{12}x^2\)
Ratio = \(\frac{pi}{3}x^2 : \frac{pi}{12}x^2 = 4: 1\)
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