Given S = \frac{k}T
So T = \frac{k}S
Now, S is increased by 50%
So S = \frac{1.5k}T
T = \frac{1.5k}S
Applying percentage decrease formula = \frac{Difference}{LargerValue}\times 100
=> \frac{\Large \frac{1.5k}S - \frac{k}S}{\Large \frac{1.5k}S}\times 100 = \frac{1}3 \times 100
= \large33\frac{1}3%
Ans: B
Given: 1, \sqrt2, x and {x}^2
We can write this as 1, 1.4, x and {x}^2
Range = largest - smallest
Consider the possibility of {x}^2 being the largest and 1 is the smallest
4 = {x}^2- 1
{x}^2 = 5 and x = \sqrt5 = 2.23
A is greater
Consider the other possibility, where x to is greater than {x}^2. Here, x should be a decimal value, greater than 0 and less than 1
4 = 1.4 - x
x = -2.6
This is not possible, since it’s given x > 0
Hence, Ans: Quantity A is greater.
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For the first one,
S = k/T
So, ST=k
Now, if S increases by 50%, so value of S(new) = S(1/+50/100) => 1.5S
Since k is constant, so S(old)T(old) = k = S(new)T(new)
T(new) = ST/1.5S => 10T/15 => 2T/3
Decrease in T is => (T-2T/3 )/ T * 100 => 100T/3
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