Q76. A candidate passes his exam with 515 marks, having obtained $3\%$ above the minimum. If he had obtained 710 marks what percentage would have been above the minimum?
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$\Rightarrow$ Let minimum marks be = $x$
$\therefore 3\%$ above minimum marks = $x + \dfrac{3}{100} \times x$
$\Rightarrow x + \dfrac{3x}{100} = \dfrac{100x + 3x}{100} = \dfrac{103x}{100}$
$\Rightarrow$ Given minimum marks = 515
So we have:
$\Rightarrow \dfrac{103x}{100} = 515$
$\Rightarrow x = 515 \times \dfrac{100}{103} = 5 \times 100 = 500$
When marks obtained is 710, then:
$\Rightarrow$ Mraks above minimum = $710 - 500 = 210$
$\therefore$ Required percentage = $\dfrac{210}{500} \times 100$
$\Rightarrow \dfrac{210}{500} \times 100 = \dfrac{210}{5} = 42\%$ Ans
$\therefore 3\%$ above minimum marks = $x + \dfrac{3}{100} \times x$
$\Rightarrow x + \dfrac{3x}{100} = \dfrac{100x + 3x}{100} = \dfrac{103x}{100}$
$\Rightarrow$ Given minimum marks = 515
So we have:
$\Rightarrow \dfrac{103x}{100} = 515$
$\Rightarrow x = 515 \times \dfrac{100}{103} = 5 \times 100 = 500$
When marks obtained is 710, then:
$\Rightarrow$ Mraks above minimum = $710 - 500 = 210$
$\therefore$ Required percentage = $\dfrac{210}{500} \times 100$
$\Rightarrow \dfrac{210}{500} \times 100 = \dfrac{210}{5} = 42\%$ Ans
Q77. A candidate who gets $35\%$ marks in an examination fails by 40 marks but another candidate who gets $43\%$ marks gets 24 marks more than minimum pass marks. Find pass percentage of marks.
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$\Rightarrow$ Let maximum marks be = $x$
For $1^{st}$ student:
$\Rightarrow$ Marks obtained = $35\%$
$\Rightarrow$ Marks obtained below pass marks = 40 marks
$\therefore$ Pass marks = $x \text{ of } \dfrac{35}{100} + 40 = \dfrac{35x}{100} + 40 = \dfrac{7x}{20} + 40$
For $2^{nd}$ student:
$\Rightarrow$ Marks obtained = $43\%$
$\Rightarrow$ Marks obtained above pass marks = 24 marks
$\therefore$ Pass marks = $x \text{ of } \dfrac{43}{100} - 24 = \dfrac{43x}{100} - 24$
So we have the equation:
$\Rightarrow \dfrac{7x}{20} + 40 = \dfrac{43x}{100} - 24$
$\Rightarrow 40 + 24 = \dfrac{43x}{100} - \dfrac{7x}{20}$
$\Rightarrow 64 = \dfrac{43x - 35x}{100}$
$\Rightarrow \dfrac{8x}{100} = 64$
$\Rightarrow \dfrac{2x}{25} = 64$
$\Rightarrow x = 64 \times \dfrac{25}{2} = 32 \times 25 = 800$
$\therefore$ Maximum marks = 800
$\Rightarrow$ And minimum marks = $(\dfrac{35}{100} \times 800) + 40$
$\Rightarrow (35 \times 8) + 40 = 280 + 40 = 320$
$\therefore$ Required pass percentage = $\dfrac{320}{800} \times 100 = \dfrac{320}{8} = 40\%$ Ans
For $1^{st}$ student:
$\Rightarrow$ Marks obtained = $35\%$
$\Rightarrow$ Marks obtained below pass marks = 40 marks
$\therefore$ Pass marks = $x \text{ of } \dfrac{35}{100} + 40 = \dfrac{35x}{100} + 40 = \dfrac{7x}{20} + 40$
For $2^{nd}$ student:
$\Rightarrow$ Marks obtained = $43\%$
$\Rightarrow$ Marks obtained above pass marks = 24 marks
$\therefore$ Pass marks = $x \text{ of } \dfrac{43}{100} - 24 = \dfrac{43x}{100} - 24$
So we have the equation:
$\Rightarrow \dfrac{7x}{20} + 40 = \dfrac{43x}{100} - 24$
$\Rightarrow 40 + 24 = \dfrac{43x}{100} - \dfrac{7x}{20}$
$\Rightarrow 64 = \dfrac{43x - 35x}{100}$
$\Rightarrow \dfrac{8x}{100} = 64$
$\Rightarrow \dfrac{2x}{25} = 64$
$\Rightarrow x = 64 \times \dfrac{25}{2} = 32 \times 25 = 800$
$\therefore$ Maximum marks = 800
$\Rightarrow$ And minimum marks = $(\dfrac{35}{100} \times 800) + 40$
$\Rightarrow (35 \times 8) + 40 = 280 + 40 = 320$
$\therefore$ Required pass percentage = $\dfrac{320}{800} \times 100 = \dfrac{320}{8} = 40\%$ Ans
Q78. After 30 kgs of water was evaporated from a solution of salt and water which had $15\%$ of salt, the reamining solution had $20\%$ of salt. Find the weight of original solution.
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Case 1:
$\Rightarrow$ Let the weight of original solution be = $x$ kg
$\therefore$ Weight of salt in it = $\dfrac{15}{100} \text{ of } x = \dfrac{3x}{20}$
Case 2:
After 30 kg of water evaporated:
$\Rightarrow$ Weight of remaining solution = $(x - 30)$
$\Rightarrow$ Weight of salt in remaining solution = $20\% \text{ of } (x - 30)$
In both cases, the weight of salt will be the same:
$\Rightarrow \dfrac{3x}{20} = \dfrac{20}{100} \times (x - 30)$
$\Rightarrow \dfrac{3x}{20} = \dfrac{1}{5} \times (x - 30)$
$\Rightarrow \dfrac{3x}{20} = \dfrac{x}{5} - \dfrac{30}{5}$
$\Rightarrow \dfrac{3x}{20} = \dfrac{x}{5} - 6$
$\Rightarrow \dfrac{3x}{20} - \dfrac{x}{5} = - 6$
$\Rightarrow \dfrac{3x - 4x}{20} = - 6$
$\Rightarrow \dfrac{x}{20} = 6$
$\Rightarrow x = 6 \times 20 = 120$ kg Ans
$\Rightarrow$ Let the weight of original solution be = $x$ kg
$\therefore$ Weight of salt in it = $\dfrac{15}{100} \text{ of } x = \dfrac{3x}{20}$
Case 2:
After 30 kg of water evaporated:
$\Rightarrow$ Weight of remaining solution = $(x - 30)$
$\Rightarrow$ Weight of salt in remaining solution = $20\% \text{ of } (x - 30)$
In both cases, the weight of salt will be the same:
$\Rightarrow \dfrac{3x}{20} = \dfrac{20}{100} \times (x - 30)$
$\Rightarrow \dfrac{3x}{20} = \dfrac{1}{5} \times (x - 30)$
$\Rightarrow \dfrac{3x}{20} = \dfrac{x}{5} - \dfrac{30}{5}$
$\Rightarrow \dfrac{3x}{20} = \dfrac{x}{5} - 6$
$\Rightarrow \dfrac{3x}{20} - \dfrac{x}{5} = - 6$
$\Rightarrow \dfrac{3x - 4x}{20} = - 6$
$\Rightarrow \dfrac{x}{20} = 6$
$\Rightarrow x = 6 \times 20 = 120$ kg Ans
Q79. A sample of 50 litres of glycerin is adultrated to the extent of $20\%$. Find how much pure glycerin should be added to bring down the impurity level to $5\%$?
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$\Rightarrow$ Amount of glycerin adultrated = $\dfrac{20}{100} \times 50 = 2 \times 5 = 10$ litres
$\Rightarrow$ Let final amount of glycerin be = $x$ litres
$\Rightarrow$ Impurity level is to be brought down by = $5\%$
$\therefore$ Final amount of glycerin = $\dfrac{5}{100} \times x = 10$
$\Rightarrow \dfrac{1}{20} \times x = 10$
$\Rightarrow x = 10 \times 20 = 200$ litres
$\therefore$ Required amount of glycerin = $200 - 50 = 150$ litres Ans
$\Rightarrow$ Let final amount of glycerin be = $x$ litres
$\Rightarrow$ Impurity level is to be brought down by = $5\%$
$\therefore$ Final amount of glycerin = $\dfrac{5}{100} \times x = 10$
$\Rightarrow \dfrac{1}{20} \times x = 10$
$\Rightarrow x = 10 \times 20 = 200$ litres
$\therefore$ Required amount of glycerin = $200 - 50 = 150$ litres Ans
Q80. 5 litres of water is evaporated from 30 litres of salt solution which contains $16\%$ of salt. Find salt percentage in the remaining solution.
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$\Rightarrow$ Total amount of solution = 30 kgs
$\therefore \dfrac{16}{100} \times 30 = \dfrac{4}{25} \times 30 = \dfrac{4}{5} \times 6 = \dfrac{24}{5}$
$\Rightarrow$ Amount of solution evaporated = 5 litres
$\Rightarrow$ Remaining solution = $30 - 5 = 25$
$\therefore$ Remaining percentage of salt in solution = $\dfrac{\dfrac{24}{5}}{25}$
$\Rightarrow \dfrac{24}{25 \times 5} \times 100 = \dfrac{24}{125} \times 100 = \dfrac{24}{5} \times 4 = \dfrac{96}{5} = 19.2\%$ Ans
$\therefore \dfrac{16}{100} \times 30 = \dfrac{4}{25} \times 30 = \dfrac{4}{5} \times 6 = \dfrac{24}{5}$
$\Rightarrow$ Amount of solution evaporated = 5 litres
$\Rightarrow$ Remaining solution = $30 - 5 = 25$
$\therefore$ Remaining percentage of salt in solution = $\dfrac{\dfrac{24}{5}}{25}$
$\Rightarrow \dfrac{24}{25 \times 5} \times 100 = \dfrac{24}{125} \times 100 = \dfrac{24}{5} \times 4 = \dfrac{96}{5} = 19.2\%$ Ans
Q81. How many litres of $20\%$ alchohol solution should be added to 40 litres of $50\%$ alcohol solution to make a $30\%$ solution?
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$\Rightarrow$ Let $x$ be the quantity of $20\%$ alcohol solution to be added to 40 litres of a $50\%$ alcohol solution.
$\Rightarrow$ And, let $y$ be the quantity of the final $30\%$ solution
So, we have = $x + 40 = y$
Similarly, we have = $y = x + 40$
$\therefore 20\% \text{ of } x + 50\% \text{ of } 40 = 30\% \text{ of } y$
Now, on substituting $y$ by its value of $x + 40$, we have:
$\Rightarrow 20\% \text{ of } x + 50\% \text{ of } 40 = 30\% \text{ of } (x + 40)$
$\Rightarrow \dfrac{20x}{100} + \dfrac{50 \times 40}{100} = \dfrac{30}{100} \times (x + 40)$
$\Rightarrow \dfrac{20x}{100} + \dfrac{50 \times 40}{100} = \dfrac{30x}{100} + \dfrac{30 \times 40}{100}$
$\Rightarrow \dfrac{20x}{100} + \dfrac{2000}{100} = \dfrac{30x}{100} + \dfrac{1200}{100}$
Now, simplify by multiplying all terms by 100, we have:
$\Rightarrow 20x + 2000 = 30x + 1200$
$\Rightarrow 20x - 30x = 1200 - 2000 $
$\Rightarrow -10x = -800 $
$\Rightarrow 10x = 800 $
$\Rightarrow x = \dfrac{800}{10} = 80$
$\therefore$ Required litres of $20\%$ alcohol solution = 80 litres Ans
$\Rightarrow$ And, let $y$ be the quantity of the final $30\%$ solution
So, we have = $x + 40 = y$
Similarly, we have = $y = x + 40$
$\therefore 20\% \text{ of } x + 50\% \text{ of } 40 = 30\% \text{ of } y$
Now, on substituting $y$ by its value of $x + 40$, we have:
$\Rightarrow 20\% \text{ of } x + 50\% \text{ of } 40 = 30\% \text{ of } (x + 40)$
$\Rightarrow \dfrac{20x}{100} + \dfrac{50 \times 40}{100} = \dfrac{30}{100} \times (x + 40)$
$\Rightarrow \dfrac{20x}{100} + \dfrac{50 \times 40}{100} = \dfrac{30x}{100} + \dfrac{30 \times 40}{100}$
$\Rightarrow \dfrac{20x}{100} + \dfrac{2000}{100} = \dfrac{30x}{100} + \dfrac{1200}{100}$
Now, simplify by multiplying all terms by 100, we have:
$\Rightarrow 20x + 2000 = 30x + 1200$
$\Rightarrow 20x - 30x = 1200 - 2000 $
$\Rightarrow -10x = -800 $
$\Rightarrow 10x = 800 $
$\Rightarrow x = \dfrac{800}{10} = 80$
$\therefore$ Required litres of $20\%$ alcohol solution = 80 litres Ans
Q82. John wants to make a 100 ml of $5\%$ alcohol solution mixing a quantity of a $2\%$ alcohol solution with a $7\%$ alcohol solution. What are the quantities of each solution he has to use?
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$\Rightarrow$ Let $x$ be the quantity of $2\%$ alcohol solution
$\Rightarrow$ And, $y$ be the quantity of $7\%$ alcohol solution
$\therefore$ We have = $x + y = 100$
$\Rightarrow$ Similarly $y = 100 - x$
So, we have the following equation:
$\Rightarrow (2\% \text{ of } x) + (7\% \text{ of } y) = 5\% \text{ of } 100$
Now, on substituting $y$ by its value of $100 - x$, we have:
$\Rightarrow \dfrac{2}{100} \text{ of } x + \dfrac{7}{100} \text{ of } (100 - x) = \dfrac{5}{100} \text{ of } 100$
$\Rightarrow \dfrac{2x}{100} + \dfrac{700}{100} - \dfrac{7x}{100} = \dfrac{500}{100}$
Now, simplify by multiplying all terms by 100, we have:
$\Rightarrow 2x + 700 - 7x = 500$
$\Rightarrow 2x - 7x = 500 - 700$
$\Rightarrow -5x = -200$
$\Rightarrow 5x = 200$
$\Rightarrow x = \dfrac{200}{5} = 40$
$\therefore y = 100 - 40 = 60$ ml each of both solution Ans
$\Rightarrow$ And, $y$ be the quantity of $7\%$ alcohol solution
$\therefore$ We have = $x + y = 100$
$\Rightarrow$ Similarly $y = 100 - x$
So, we have the following equation:
$\Rightarrow (2\% \text{ of } x) + (7\% \text{ of } y) = 5\% \text{ of } 100$
Now, on substituting $y$ by its value of $100 - x$, we have:
$\Rightarrow \dfrac{2}{100} \text{ of } x + \dfrac{7}{100} \text{ of } (100 - x) = \dfrac{5}{100} \text{ of } 100$
$\Rightarrow \dfrac{2x}{100} + \dfrac{700}{100} - \dfrac{7x}{100} = \dfrac{500}{100}$
Now, simplify by multiplying all terms by 100, we have:
$\Rightarrow 2x + 700 - 7x = 500$
$\Rightarrow 2x - 7x = 500 - 700$
$\Rightarrow -5x = -200$
$\Rightarrow 5x = 200$
$\Rightarrow x = \dfrac{200}{5} = 40$
$\therefore y = 100 - 40 = 60$ ml each of both solution Ans
Q83. Sterling silver is $92.5\%$ of pure silver. How many grams of sterling silver must be mixed to a $90\%$ silver alloy to obtain a 500 gram of $91\%$ silver alloy?
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$\Rightarrow$ Let quantity of sterling silver be = $x$
$\Rightarrow$ And, let quantity of silver alloy to be obtained be = $y$
$\therefore$ We have = $x + y = 500$
$\Rightarrow$ Similarly, $y = 500 - x$
So, we have the following equation:
$\Rightarrow (92.5\% \text{ of } x) + (90\% \text{ of } y) = 91\% \text{ of } 500$
Now, on substituting $y$ by its value of $500 - x$, we have:
$\Rightarrow (\dfrac{92.5}{100} \times x) + (\dfrac{90}{100} \times (500 - x)) = \dfrac{91}{100} \times 500$
$\Rightarrow \dfrac{92.5x}{100} + \dfrac{90 \times 500}{100} - \dfrac{ 90x}{100} = \dfrac{91 \times 500}{100}$
Now, simplify by multiplying all terms by 100, we have:
$\Rightarrow 92.5x + 45,000 - 90x = 45,500$
$\Rightarrow 92.5x - 90x = 45,500 - 45,000$
$\Rightarrow 2.5x = 500$
$\Rightarrow \dfrac{25x}{10} = 500$ (Converting into fraction)
$\Rightarrow x = 500 \times \dfrac{10}{25} = 500 \times \dfrac{2}{5} = 100 \times 2 = 200$ grams Ans
$\Rightarrow$ And, let quantity of silver alloy to be obtained be = $y$
$\therefore$ We have = $x + y = 500$
$\Rightarrow$ Similarly, $y = 500 - x$
So, we have the following equation:
$\Rightarrow (92.5\% \text{ of } x) + (90\% \text{ of } y) = 91\% \text{ of } 500$
Now, on substituting $y$ by its value of $500 - x$, we have:
$\Rightarrow (\dfrac{92.5}{100} \times x) + (\dfrac{90}{100} \times (500 - x)) = \dfrac{91}{100} \times 500$
$\Rightarrow \dfrac{92.5x}{100} + \dfrac{90 \times 500}{100} - \dfrac{ 90x}{100} = \dfrac{91 \times 500}{100}$
Now, simplify by multiplying all terms by 100, we have:
$\Rightarrow 92.5x + 45,000 - 90x = 45,500$
$\Rightarrow 92.5x - 90x = 45,500 - 45,000$
$\Rightarrow 2.5x = 500$
$\Rightarrow \dfrac{25x}{10} = 500$ (Converting into fraction)
$\Rightarrow x = 500 \times \dfrac{10}{25} = 500 \times \dfrac{2}{5} = 100 \times 2 = 200$ grams Ans
Q84. How manny kilograms of pure water is to be added to 100 kilograms of a $30\%$ saline solution to make it a $10\%$ saline solution?
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$\Rightarrow$ Let weight of water be = $x$
$\Rightarrow$ And, let weight of salt water be = $y$
$\therefore$ We have = $y = x + 100$
So, we have the following equation:
$\Rightarrow x + 30\% \text{ of } 100 = 10\% \text{ of } y$
Now, on substituting $y$ by its value of $x + 100$, we have:
$\Rightarrow x + 30\% \text{ of } 100 = 10\% \text{ of } (x + 100)$
$\Rightarrow x + \dfrac{30}{100} \times 100 = \dfrac{10}{100} \times (x + 100)$
$\Rightarrow x + \dfrac{30 \times 100}{100} = \dfrac{10x}{100} + \dfrac{10 \times 100}{100}$
Assuming that $x$ contains no salt, we eliminateits value from he equation
$\Rightarrow \dfrac{3000}{100} = \dfrac{10x}{100} + \dfrac{1000}{100}$
Now, simplify by multiplying all terms by 100, we have:
$\Rightarrow \dfrac{3000}{100} - \dfrac{1000}{100} = \dfrac{10x}{100}$
$\Rightarrow 10x = 3000 - 1000 = 2000$
$\Rightarrow x = \dfrac{2000}{10} = 200$ kg Ans
$\Rightarrow$ And, let weight of salt water be = $y$
$\therefore$ We have = $y = x + 100$
So, we have the following equation:
$\Rightarrow x + 30\% \text{ of } 100 = 10\% \text{ of } y$
Now, on substituting $y$ by its value of $x + 100$, we have:
$\Rightarrow x + 30\% \text{ of } 100 = 10\% \text{ of } (x + 100)$
$\Rightarrow x + \dfrac{30}{100} \times 100 = \dfrac{10}{100} \times (x + 100)$
$\Rightarrow x + \dfrac{30 \times 100}{100} = \dfrac{10x}{100} + \dfrac{10 \times 100}{100}$
Assuming that $x$ contains no salt, we eliminateits value from he equation
$\Rightarrow \dfrac{3000}{100} = \dfrac{10x}{100} + \dfrac{1000}{100}$
Now, simplify by multiplying all terms by 100, we have:
$\Rightarrow \dfrac{3000}{100} - \dfrac{1000}{100} = \dfrac{10x}{100}$
$\Rightarrow 10x = 3000 - 1000 = 2000$
$\Rightarrow x = \dfrac{2000}{10} = 200$ kg Ans
Q85. A 50 ml after shave lotion at $30\%$ alcohol is mixed with 30 ml of pure water. Find the percentage of alcohol in the new solution?
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$\Rightarrow$ Quantity of final mixture is given by = 50 ml $+$ 30 ml = 80 ml
Now, the amount of alcohol in the final solution is equal to the amouont of alcohol in pure water (which is zero) plus the amount of alcohol in the final solution.
$\Rightarrow$ Let percentage of alcohol in final solution be = $x$
$\therefore 0 + 30\% \text{ of } 50 = x\% \text{ of } 80$
$\Rightarrow \dfrac{30}{100} \times 50 = \dfrac{x}{100} \times 80$
$\Rightarrow \dfrac{30 \times 50}{100} = \dfrac{80x}{100}$
Now, simplify by multiplying all terms by 100, we have:
$\Rightarrow 1500 = 80x$
$\Rightarrow x = \dfrac{1500}{80} = \dfrac{150}{8} = 18.75\%$ Ans
Now, the amount of alcohol in the final solution is equal to the amouont of alcohol in pure water (which is zero) plus the amount of alcohol in the final solution.
$\Rightarrow$ Let percentage of alcohol in final solution be = $x$
$\therefore 0 + 30\% \text{ of } 50 = x\% \text{ of } 80$
$\Rightarrow \dfrac{30}{100} \times 50 = \dfrac{x}{100} \times 80$
$\Rightarrow \dfrac{30 \times 50}{100} = \dfrac{80x}{100}$
Now, simplify by multiplying all terms by 100, we have:
$\Rightarrow 1500 = 80x$
$\Rightarrow x = \dfrac{1500}{80} = \dfrac{150}{8} = 18.75\%$ Ans
Q86. You add $x$ ml of a $25\%$ alcohol solution to a 200 ml of a $10\%$ alcohol solution to obtain another solution. Find the amount of alcohol in the final solution in terms of $x$. Find:
(i) Find the ratio, in terms of $x$, of the alcohol in the final solution to the total amount of the solution. (ii) Also, what do you think will happen if the value of $x$ is very large? (iii) Find $x$ so that the final solution has a percentage of $15\%$?
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(i) Amount of alcohol in 200 ml solution = $\dfrac{10}{100} \times 200 = 20 \text{ ml}$
$\Rightarrow$ Amount of alcohol in $x$ ml solution = $\dfrac{25}{100} \times x = \dfrac{25x}{100} = 0.25x \text{ ml}$
$\therefore$ Ratio of alcohol in final solution to the total amount:
$\Rightarrow \dfrac{20 + 0.25x}{x + 200}$ Ans
(ii) Now, if $x$ becomes very large in the above formula, then the ratio becomes close to 0. This means that if you increase the amount of $x$ in $25\%$ solution, it will dominate and the final solution will be very close to $25\%$ solution.
(iii) In order to have a $15\%$ percentage, we need to have:
$\Rightarrow 10\% \text{ of } 200 + 0.25x = 15\%(x + 200)$
$\Rightarrow \dfrac{10 \times 200}{100} + 0.25x = \dfrac{15x}{100} + (\dfrac{15 \times 200}{100})$
$\Rightarrow \dfrac{2000}{100} + \dfrac{25x}{100} = \dfrac{15x}{100} + \dfrac{3000}{100}$
Now, simplify by multiplying all terms by 100, we have:
$\Rightarrow 2000 + 25x = 15x + 3000$
$\Rightarrow 25x - 15x = 3000 - 2000$
$\Rightarrow 10x = 1000$
$\Rightarrow x = \dfrac{1000}{10} = 100$ ml Ans
$\Rightarrow$ Amount of alcohol in $x$ ml solution = $\dfrac{25}{100} \times x = \dfrac{25x}{100} = 0.25x \text{ ml}$
$\therefore$ Ratio of alcohol in final solution to the total amount:
$\Rightarrow \dfrac{20 + 0.25x}{x + 200}$ Ans
(ii) Now, if $x$ becomes very large in the above formula, then the ratio becomes close to 0. This means that if you increase the amount of $x$ in $25\%$ solution, it will dominate and the final solution will be very close to $25\%$ solution.
(iii) In order to have a $15\%$ percentage, we need to have:
$\Rightarrow 10\% \text{ of } 200 + 0.25x = 15\%(x + 200)$
$\Rightarrow \dfrac{10 \times 200}{100} + 0.25x = \dfrac{15x}{100} + (\dfrac{15 \times 200}{100})$
$\Rightarrow \dfrac{2000}{100} + \dfrac{25x}{100} = \dfrac{15x}{100} + \dfrac{3000}{100}$
Now, simplify by multiplying all terms by 100, we have:
$\Rightarrow 2000 + 25x = 15x + 3000$
$\Rightarrow 25x - 15x = 3000 - 2000$
$\Rightarrow 10x = 1000$
$\Rightarrow x = \dfrac{1000}{10} = 100$ ml Ans