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.**

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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