CRE 4 - Mixtures | Average, Mixture & Alligation

Answer: 60

Explanation :

Total quantity bought by Akanksha = 5 + 10 = 15 kgs

Total cost = 5 × 70 + 10 × 55 = 900

Average Price = $\frac{5\times 70+10\times 55}{5+10}$ = 60. 

Therefore, average price = Rs. 60/-

Hence, 60.

Answer: Option A

Explanation :

Water in 130 lit. mix = $130\times \frac{3}{13}$ =30 lit

∴ milk = 130 - 30 = 100 lit

Let x liters of water is added now such that milk water ratio becomes 8 : 5.

Now, $\frac{100}{30+x}=\frac{8}{5}$

⇒ x = 32.5 lit.

Hence, option (a).

Answer: 36

Explanation :

Solution 1:
64 litres of a mixture, which has milk and water in the ratio 5 : 3
Quantity of milk = $\frac{5}{8}$× 64 = 40 litres
Quantity of water = $\frac{3}{8}$× 64 = 24 litres

Solution 2:
36 litres of a mixture, which has milk and water in the ratio 2 : 1
Quantity of milk = $\frac{2}{3}$× 36 = 24 litres
Quantity of water = $\frac{1}{3}$× 36 = 12 litres

Total quantity of milk in the final mixture = 40 + 24 = 64 liters
Total quantity of water in the final mixture = 24 + 12 = 36 liters

∴ Percentage of water = $\frac{36}{36+64}$ × 100% = 36%

Hence, 36.

Answer: Option C

Explanation :

Let the quantity sold at 6% is x kgs and at 10% profit is (16 - x) kgs.

⇒ Overall profit = 7% = $\frac{6x+10(16-x)}{16}$

⇒ 112 = 6x + 160 – 10x

⇒ x = 12 kgs.

Alternately,
​​​​​​​

$\frac{\mathrm{Quantity} \mathrm{sold} \mathrm{at} 6\%}{\mathrm{Quantity} \mathrm{sold} \mathrm{at} 10\%}$ = $\frac{3}{1}$

Quantity sold at 6% = $\frac{3}{4}$ × 16 = 12 kgs

Hence, option (c).

Answer: Option B

Explanation :

The overall rate of interest = $\frac{130}{2000}\times 100$ = 6.5%

Let the amount lent at 7% be x and at 5% be (2000 - x).

⇒ 6.5 = $\frac{(2000-x)\times 5+x\times 7}{2000}$

⇒ 13000 = (2000 - x) × 5 + x × 7

⇒ 3000 = 2x

⇒ x = 1500.

Alternately,
​​​​​​​

$\frac{\mathrm{Amount} \mathrm{lent} \mathrm{at} 5\%}{\mathrm{Amount} \mathrm{lent} \mathrm{at} 7\%}$ = $\frac{0.5}{1.5}$ = $\frac{1}{3}$

⇒ Amount lent at 7% = $\frac{3}{4}$ × 2000 = Rs. 1500.

Hence, option (b).

Answer: Option D

Explanation :

Proportion of wine in 3 samples is $\frac{1}{3}, \frac{2}{5}, \frac{3}{8}$

Proportion of water in 3 samples is $\frac{2}{3}, \frac{3}{5}, \frac{5}{8}$

Since equal quantities are taken, let 1 liter of each is mixed.

Total quantity of wine is $\frac{1}{3}+\frac{2}{5}+\frac{3}{8}$ = $\frac{133}{120}$

Total quantity of water is $\frac{2}{3}+\frac{3}{5}+\frac{5}{8}$ = $\frac{227}{120}$

∴ Proportion of wine and water in the solution is = 133 : 227.

Alternately,

Proportion of wine in 3 samples is $\frac{1}{3}, \frac{2}{5}, \frac{3}{8}$

Proportion of water in 3 samples is $\frac{2}{3}, \frac{3}{5}, \frac{5}{8}$

Since equal quantities are taken, let 120 liter of each is mixed.  (We’ve chosen 120 liters since it is divisible by the denominators of all the fractions i.e. LCM(3, 5 and 8) = 120)

Total quantity of wine is $\frac{1}{3}\times 120+\frac{2}{5}\times 120+\frac{3}{8}\times 120$ = 40 + 48 + 45 = 133.

Total quantity of water is $\frac{2}{3}\times 120+\frac{3}{5}\times 120+\frac{5}{8}\times 120$ = 80 + 72 + 75 = 227.

∴ Proportion of wine and water in the solution is = 133 : 227.

Hence, option (d).