Vaidehi got 92% marks in an examination and Kavyanjali got 78% marks in the same examination. If the difference between the marks obtained by Vaidehi and Kavyanjali is 105, find the marks obtained by Vaidehi in the examination.
690
This question asks us to find the marks obtained by Vaidehi in an examination, given the percentages obtained by two students, Vaidehi and Kavyanjali, and the difference in their marks. To solve this, we first need to determine the total marks of the examination.
We are given the following information:
Let the total marks of the examination be denoted by $T$.
Vaidehi's marks can be expressed as 92% of the total marks, which is $0.92 \times T$.
Kavyanjali's marks can be expressed as 78% of the total marks, which is $0.78 \times T$.
The difference between their marks is the difference between their percentage marks multiplied by the total marks:
Difference in marks = Vaidehi's marks - Kavyanjali's marks
Difference in marks = $(92\% \text{ of } T) - (78\% \text{ of } T)$
Difference in marks = $(0.92T) - (0.78T)$
We are given that this difference is 105 marks. So, we can set up the equation:
$(0.92 - 0.78)T = 105$
Let's calculate the difference in percentages:
$92\% - 78\% = (92 - 78)\% = 14\%$
So, the difference in marks (105) corresponds to 14% of the total marks.
We can write this as:
$14\% \text{ of } T = 105$
In decimal form, this is:
$0.14 \times T = 105$
Now, we can solve for $T$ by dividing 105 by 0.14:
$T = \frac{105}{0.14}$
To simplify the division, we can multiply both the numerator and the denominator by 100 to remove the decimal:
$T = \frac{105 \times 100}{0.14 \times 100} = \frac{10500}{14}$
Now, perform the division:
$T = \frac{10500}{14} = \frac{5250}{7}$
Dividing 5250 by 7:
$5250 \div 7 = 750$
So, the total marks for the examination is 750.
Now that we know the total marks ($T = 750$), we can find Vaidehi's marks. Vaidehi got 92% of the total marks.
Vaidehi's marks = 92% of 750
Vaidehi's marks = $0.92 \times 750$
To calculate this, we can multiply 92 by 75 and then adjust the decimal, or calculate 92% of 750 directly:
$0.92 \times 750 = \frac{92}{100} \times 750 = \frac{92 \times 750}{100}$
We can simplify by dividing 750 and 100 by 50:
$\frac{92 \times (15 \times 50)}{2 \times 50} = \frac{92 \times 15}{2}$
Now, divide 92 by 2:
$\frac{46 \times 2 \times 15}{2} = 46 \times 15$
Now, multiply 46 by 15:
$46 \times 15 = 46 \times (10 + 5) = 46 \times 10 + 46 \times 5 = 460 + 230 = 690$
Alternatively, using the decimal form:
$0.92 \times 750$
We can write 750 as $75 \times 10$:
$0.92 \times 75 \times 10 = 9.2 \times 75$
$9.2 \times 75 = 9.2 \times (70 + 5) = 9.2 \times 70 + 9.2 \times 5$
$9.2 \times 70 = 92 \times 7 = 644$
$9.2 \times 5 = (10 - 0.8) \times 5 = 50 - 4 = 46$
$644 + 46 = 690$
So, Vaidehi obtained 690 marks in the examination.
| Detail | Value |
|---|---|
| Vaidehi's Percentage | 92% |
| Kavyanjali's Percentage | 78% |
| Difference in Percentages | $92\% - 78\% = 14\%$ |
| Difference in Marks | 105 |
| Let Total Marks be $T$ | $14\% \text{ of } T = 105$ |
| Equation | $0.14T = 105$ |
| Total Marks ($T$) | $\frac{105}{0.14} = 750$ |
| Vaidehi's Marks | $92\% \text{ of } 750 = 0.92 \times 750 = 690$ |
The marks obtained by Vaidehi in the examination are 690.
| Concept | Explanation | Formula/Example |
|---|---|---|
| Percentage | A rate, number, or amount in each hundred. | $p\% = \frac{p}{100}$ |
| Finding Percentage of a Quantity | To find $p\%$ of a quantity $Q$, multiply $Q$ by $\frac{p}{100}$. | $p\% \text{ of } Q = \frac{p}{100} \times Q$ |
| Difference between Percentages | Subtracting one percentage from another to find the net percentage. | $p_1\% - p_2\% = (p_1 - p_2)\%$ |
| Using Percentage Difference to Find Total | If $p\%$ of Total = Value, then Total = $\frac{\text{Value}}{p\%} = \frac{\text{Value}}{p/100} = \frac{\text{Value} \times 100}{p}$. | If $14\%$ of $T = 105$, then $T = \frac{105 \times 100}{14} = 750$. |
Percentage calculations are fundamental in many areas, including examination results, finance, and statistics. Understanding how to work with percentages and their differences is crucial for solving various quantitative problems.
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