Quantitative Aptitude
Mixtures and Alligation
1,816 Questions
Mixtures and Alligation Questions
-
$50\%$
-
$40\%$
-
$60\%$
-
$70\%$
C
Correct answer
Explanation
Ratio of milk to water is 3:2. Total parts = 5. Percentage of milk = (3/5) * 100 = 60%.
-
$1$:$10$
-
$1:12$
-
$1$:$15$
-
$1:20$
A
Correct answer
Explanation
At a 37.5% profit, the cost of one litre of mixture is 8/1.375 = 64/11 rupees. This requires 10/11 litre of milk costing 6.40 per litre, leaving 1/11 litre of water, so the water-to-milk ratio is 1:10.
-
$Rs\;125$
-
$Rs\;150$
-
$Rs\;140$
-
$Rs\;145$
A
Correct answer
Explanation
Total cost = (25 * 29) + (25 * 20) = 725 + 500 = 1225. Total weight = 50 kg. Selling price = 50 * 27 = 1350. Profit = 1350 - 1225 = 125.
-
$9.50$
-
$10.80$
-
$11.75$
-
$11$
B
Correct answer
Explanation
Selling price = 11, profit 10%, so Cost Price = 11 / 1.1 = 10. Ratio 3:2. Let X = y+2. 3(y+2) + 2y = 5 * 10. 3y + 6 + 2y = 50 => 5y = 44 => y = 8.8. X = 8.8 + 2 = 10.8.
-
$22\dfrac{1}{8}$%
-
$25.75$ %
-
$25.25$ %
-
$25\dfrac{1}{8}$%
D
Correct answer
Explanation
The mixture consists of 5 parts of 22.5% and 7 parts of 27%. The weighted average is ((5 * 22.5) + (7 * 27)) / 12 = (112.5 + 189) / 12 = 301.5 / 12 = 25.125%. 25.125% is 25 1/8%.
-
Rs. $16.00$ gain
-
Rs. $16.00$ Loss
-
Rs. $20.00$ gain
-
Rs. $10$ gain
D
Correct answer
Explanation
Total cost = (25 * 6) + (35 * 7) = 150 + 245 = 395. Total weight = 25 + 35 = 60 kg. Selling price = 60 * 6.75 = 405. Gain = 405 - 395 = 10.
-
$14 : 9$
-
$9 : 14$
-
$16 : 9$
-
$3 : 4$
C
Correct answer
Explanation
Let the ratio of chemical to water be x:y. The cost price of the mixture is 25x / (x+y). Selling at 20 with 25% profit means the cost price is 20 / 1.25 = 16. Setting 25x / (x+y) = 16 gives 25x = 16x + 16y, so 9x = 16y, or x/y = 16/9.
-
Rs.360
-
Rs.270
-
Rs.180
-
None of these
B
Correct answer
Explanation
Total cost = (60 * 5) + (150 * 6) = 300 + 900 = 1200. Total weight = 60 + 150 = 210 kg. Selling price = 210 * 7 = 1470. Gain = 1470 - 1200 = 270.
-
6.0 %
-
6.5 %
-
6.9 %
-
7.2 %
-
7.5 %
D
Correct answer
Explanation
Initial salt = 0.12 * 300 = 36 ml. Total volume = 300 + 200 = 500 ml. New percentage = (36 / 500) * 100 = 7.2%.
A
Correct answer
Explanation
Let x be the amount of 20% solution. (0.20 * x + 0.50 * 40) / (x + 40) = 0.30. 0.2x + 20 = 0.3x + 12. 8 = 0.1x. x = 80.
-
$60$ ml
-
$80$ ml
-
$40$ ml
-
$50$ ml
-
$18.75$%
-
$28.75$%
-
$38.75$%
-
$48.75$%
A
Correct answer
Explanation
Initial alcohol = 30% of 50ml = 15ml. Total volume = 50ml + 30ml = 80ml. New percentage = (15 / 80) * 100 = 18.75%.
-
$20$ mLof solution $A$ and $80$ mL of solution $B$ were used.
-
$80$ mLof solution $A$ and $70$ mL of solution $B$ were used.
-
$30$ mL of solution $A$ and $60$ mL of solution $B$ were used.
-
$40$ mLof solution $A$ and $70$ mL of solution $B$ were used.
C
Correct answer
Explanation
Let the amounts of 20% and 50% acid solutions be a and b millilitres. Since a + b = 90 and 0.20a + 0.50b = 0.40 x 90, solving gives a = 30 and b = 60.
-
$20$ kg
-
$18$ kg
-
$33$ kg
-
$45$ kg
A
Correct answer
Explanation
Let x be the quantity of the second rice. Total cost = 10*10 + 15*x = 100 + 15x. Total weight = 10 + x. Selling price = 14(10 + x). Gain is 5%, so 1.05 * (100 + 15x) = 14(10 + x). Solving for x gives 105 + 15.75x = 140 + 14x, so 1.75x = 35, x = 20.
-
$\displaystyle \frac{ne}{100+e}$
-
$\displaystyle \frac{100-e}{ne}$
-
$\displaystyle \frac{ne}{100-e}$
-
$\displaystyle \frac{100+e}{ne}$
C
Correct answer
Explanation
If the tank is e% empty, it is (100-e)% full. The current volume n represents (100-e)/100 of the total capacity T, so T = 100n / (100-e). The amount needed to fill the remaining e% is (e/100) * T, which simplifies to (e/100) * (100n / (100-e)) = ne / (100-e).