Directions (67-69):The radar graph given below shows the number of buckets (plastic + wooden) sold by five different shops (P , Q, R, S and T) in two different months (May and June). Read the data carefully and answer the questions.

In May, if total number of plastic buckets sold by T are four times of the difference between total number of buckets sold by Q in both months, then find the minimum possible average of total wooden buckets sold by T in May and total plastic buckets sold by T in June (Given, number of wooden buckets sold by T in June > total wooden buckets sold by T in May and it also a perfect square of even number).

The correct answer is 60.

Let's solve the problem step-by-step by extracting the data from the radar graph and applying the given conditions:

Step 1: Extract the data for Shop Q
From the radar graph, the total number of buckets sold by shop Q in both months are:
- Total buckets sold by Q in May (represented by the blue line with square markers) = 120
- Total buckets sold by Q in June (represented by the orange line with circle markers) = 100 (or approximately 99)

The difference between the total number of buckets sold by Q in both months is:
$\text{Difference}=\; 120\; -\; 100\; =\; 20$ (or 21 if using 99 for June).

Step 2: Find the number of wooden buckets sold by T in May
In May, the total number of plastic buckets sold by T is four times the difference for shop Q:
$T_{\text{plastic, May}}=\; 4\; *\; 20\; =\; 80$ (or 84 if the difference is 21).

From the radar graph, the total buckets sold by T in May is 168.
Therefore, the number of wooden buckets sold by T in May is:
$T_{\text{wooden, May}}=\; 168\; -\; 80\; =\; 88$ (or 84 if the difference is 21).

Step 3: Analyze the conditions for T in June
Let $T_{\text{wooden, June}}$ be the number of wooden buckets sold by T in June. We are given:
1. $T_{\text{wooden, June}}>{\mathrm{T<mtext>wooden,\; May</mtext>}}_{}$ (so $T_{\text{wooden, June}}>\; 88$ or 84).
2. $T_{\text{wooden, June}}$ is a perfect square of an even number.

The perfect squares of even numbers greater than 84 or 88 are:
- ${10}^2=\; 100$
- ${12}^2=\; 144$
- ${14}^2=\; 196$
- ${16}^2=\; 256$, etc.

From the radar graph, the total buckets sold by T in June is 180 (or 176 depending on the exact visual scale). Since the wooden buckets cannot exceed the total buckets sold by T in June, the maximum possible value of $T_{\text{wooden, June}}$ is 144.

Step 4: Find the minimum possible average
We want to find the minimum possible average of the total wooden buckets sold by T in May and total plastic buckets sold by T in June:
$\text{Average}=$
$\frac{T_{\text{wooden, May}}+T_{\text{plastic, June}}}{2}$

To minimize the average, we must minimize the number of plastic buckets sold by T in June ($T_{\text{plastic, June}}$).
Since $T_{\text{plastic, June}}=T_{\text{total, June}}-T_{\text{wooden, June}}$, minimizing $T_{\text{plastic, June}}$ requires maximizing $T_{\text{wooden, June}}$.
By choosing the maximum possible value of $T_{\text{wooden, June}}=\; 144$:

- If $T_{\text{total, June}}=\; 176$ and $T_{\text{wooden, May}}=\; 88$:
$T_{\text{plastic, June}}=\; 176\; -\; 144\; =\; 32$
$\text{Average}=$
$\frac{88+32}{2}=60$

- If $T_{\text{total, June}}=\; 180$ and $T_{\text{wooden, May}}=\; 84$:
$T_{\text{plastic, June}}=\; 180\; -\; 144\; =\; 36$
$\text{Average}=$
$\frac{84+36}{2}=60$

Thus, the minimum possible average is 60.