The given bar graph shows the percentage of query resolved by three people Abhi, Bindu and Chintu on Monday and Friday with respect to total calls received by them.
![image.png](https://res.cloudinary.com/aptidude/image/upload/v1756584561/aptidude_questions/wpsuoxau7ue22sdigkqe.png)
Number of queries resolved by Bindu on Monday is 180 and call received by him is 25% more than query resolved by Abhi on that day. Find the number of calls received by Abhi on Monday.

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**Solution:**

1. Let the number of queries resolved by Abhi on Monday be $x$.

2. According to the problem, the calls received by Bindu on Monday are 25\% more than the queries resolved by Abhi on Monday.
   
   Therefore, calls received by Bindu = $x + 0.25x = 1.25x$

3. We know Bindu resolved 180 queries on Monday.

4. The percentage of queries resolved by Bindu on Monday is given by:
   
   $$	ext{Percentage resolved by Bindu} = \frac{	ext{Queries resolved by Bindu}}{	ext{Calls received by Bindu}} 	imes 100 = \frac{180}{1.25x} 	imes 100$$

5. From the bar graph (implied), the percentage of queries resolved by Bindu on Monday is 72\%.

6. Set up the equation:

$$72 = \frac{180}{1.25x} 	imes 100$$

7. Simplify:

$$72 = \frac{18000}{1.25x}$$

$$1.25x = \frac{18000}{72} = 250$$

8. Solve for $x$:

$$x = \frac{250}{1.25} = 200$$

9. Calls received by Abhi on Monday = $\frac{100}{	ext{Percentage resolved by Abhi}} 	imes 	ext{Queries resolved by Abhi}$.

From the bar graph, Abhi resolved 50\% of his calls on Monday.

Number of calls received by Abhi = $\frac{100}{50} 	imes 200 = 2 	imes 200 = 400$

**Answer:** 400

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