Question Details

Total males in B and C together is what percentage more or less than the total females in D?
Read the following table carefully and answer the questions given below. The table shows total population and percentage of female in four different cities.
Note: Males in A is 700 and females in B & D together is 945.

Options

225.25%

233.33%

246.67%

218.5%

Show Answer

Correct Answer :

Option C

246.67%

Solution :

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246.67%
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`The correct option is 246.67%.`
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Let's do a quick final review of all rules before outputting. Everything matches perfectly. 246.67% The correct option is$ 246.67%.
Step-by-Step Explanation:
Step 1: Understand the Required Percentage Formula
We are required to find by what percentage the total males in cities B and C together is more or less than the total females in city D.
The formula for percentage difference relative to the total females in D is:
$Percentage = \frac{(Males in B + Males in C) - Females in D}{Females in D} \times 100 %$
Step 2: Calculate the Values from the Given Data

From the population and gender distribution parameters:

• Total females in city D = 300

• Total males in city B = 560

• Total males in city C = 480
Sum of males in cities B and C together:
$Total Males (B + C) = 560 + 480 = 1040$
Step 3: Compute the Difference

The difference between the total males in B and C together and total females in D is:
$Difference = 1040 - 300 = 740$
Step 4: Calculate the Required Percentage

Now, express this difference as a percentage of females in D:
$Percentage More = \frac{740}{300}$
Simplifying the fraction:
$Percentage More = \frac{740}{3} % = 246.666 ... % \approx 246.67 %$
Thus, the total males in B and C together is 246.67% more than the total females in D.}}