Sneha2021 wrote:
I am not able to understand why 5/3 was introduced in the equation?
100(a+1)=5/3(100a+10b)
Is there any alternate approach?
Bunuel MathRevolutionnick1816 wrote:
Assume x='abc'= 100a+b+c
x when rounded to nearest hundred > x when rounded to nearest ten
x when rounded to nearest hundred= 100(a+1)
There are 2 possible cases
1. c<5100(a+1)=5/3(100a+10b)
4a+b=6
when a=0, b=6.......(1)
60,61,62,63,
64when a=1, b=2........(2)
120, 121,122,123,124
2. c≥5100(a+1)=5/3[100a+10(b+1)]
4a+b=5
if a=0, b=5.......(2)
55,56,57,58,59
if a=1, b=1.......(4)
115,116,117,118,119
We can solve it using options too
Bunuel wrote:
If the integer x is rounded to the nearest hundred, the result represents a \(66 \frac{2}{3}\)% increase over the value obtained when x is rounded to the nearest ten. which of the following is a possible value of x?
A. 64
B. 67
C. 99
D. 133
E. 147
Are You Up For the Challenge: 700 Level QuestionsHi, I do not think you should try to do it in the algebraic way. The algebraic way is shown to give the students an idea that it can be done in any way. But during the test, the value testing way should be your approach for this particular problem. Please follow Dillesh's solution as that is the fastest way for this problem.
Since you asked the question, I will try to answer.
The 5/3 is simplification of this -
Let x be the integer.
\(66\frac{2}{3}\)% increase = \(x + 66\frac{2}{3}\)% of \(x\)
\(= x(1+\frac{200}{300})\)
\(= x(1+\frac{2}{3})\)
\(=x(\frac{5}{3})\)
I hope this helps.