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16 Apr 2018, 14:15
Kudos
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D
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:
70% (01:42) correct
30%
(02:05)
wrong
based on 282
sessions
History
Date
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The area of field A is 30% greater than that of field B, and 35% less than that of field C. The area of C is what percent greater than the area of field B?
A. 25%
B. 50%
C. 53,8%
D. 100%
E. 200%
Source: ExpertsGlobal
A. 25%
B. 50%
C. 53,8%
D. 100%
E. 200%
Source: ExpertsGlobal
16 Apr 2018, 20:07
Kudos
Bookmarks
Let B =100
Then A = 30% greater than B = 130
A is 35% less than C
That means A is 65 % of C
Hence C =A*100/65 =130*100/65=200
C -B =200-100=100
(C-B)/B *100 = (200-100)/100 *100 =100%
Hence C is greater than B by 100%
Posted from my mobile device
Posted from my mobile device
Then A = 30% greater than B = 130
A is 35% less than C
That means A is 65 % of C
Hence C =A*100/65 =130*100/65=200
C -B =200-100=100
(C-B)/B *100 = (200-100)/100 *100 =100%
Hence C is greater than B by 100%
Posted from my mobile device
Posted from my mobile device
16 Apr 2018, 22:44
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benejo
Find C in terms of B; use A=A
Area of B = B
Area of A = 1.3B
Area of A = .65 C
\(A = A\)
\(1.3 B = .65C\)
\(\frac{C}{B}=\frac{1.3}{.65}\)
\(\frac{C}{B}=\frac{2}{1}\)
C = 2B
C's area is twice the area of B, and "Twice" = 100% greater than.
That is, C is what percent greater than B?
\(C = 2B\)
Percent greater than: \(\frac{New-Old}{Old}*100\)
\((\frac{C-B}{B}*100)=(\frac{2B-B}{B}*100)=\)
\((\frac{B}{B}*100)=(1*100)=100\) % greater than
Answer D
Choose values
Let area of B = 100
Area of A = 30% greater than B's = 130
Area of C? Work from the other direction
A, 130, is 35 percent less than C
130 = .65C
C = \(\frac{130}{.65} = 200\)
200 is what percent greater than 100?
Percent greater than: \(\frac{New-Old}{Old}*100\)
\((\frac{200-100}{100}*100)=(\frac{100}{100}*100)\)
\(=(1*100)=100\) % greater than
Answer D
