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If y3=0.008, then the value of y5 is:
0.0032
0.000032
0.00032
0.032
0.00032
Recognize 0.008 as (0.2)3. Thus y=0.2. Then y5=(0.2)5=25×10−5=32×10−5=0.00032.
We are given the equation y3=0.008 and need to find the value of y5.
y3=a3⟹y=a, (y3)5/3=y5
Recognize 0.008 as (0.2)3. Thus y=0.2. Then y5=(0.2)5=25×10−5=32×10−5=0.00032.
A common mistake is calculating y2 instead of y5 or miscounting the number of decimal places when raising 0.2 to the fifth power.
Solve for y
Since y3=0.008, we can write 0.008 as (0.2)3. Taking the cube root of both sides, we get y=0.2.
y=30.008=0.2
Calculate y⁵
Substitute y=0.2 into y5: y5=(0.2)5=(2×10−1)5.
y5=25×10−5=32×0.00001
Final Value
Performing the multiplication, 32×0.00001=0.00032.
y5=0.00032
C is correct because calculating y5 when y=0.2 results in 0.00032.
This type of problem frequently appears in RRB exams; always convert decimals to fractions (e.g., 8/1000) to make exponentiation simpler.
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