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Solved Examples(Set 4) - Mixture and Alligation

16. In what ratio must a grocer mix two varieties of pulses costing Rs.15 and Rs. 20 per kg respectively to obtain a mixture worth Rs.16.50 per kg?
A. 7 : 3B. 2 : 1
C. 3 : 7D. 1 : 2
Discuss
answer with explanation

Answer: Option A

Explanation:

By rule of alligation,

CP of 1 kg of 1st variety pulseCP of 1 kg of 2nd variety pulse
1520
Mean Price
16.5
20-16.5 = 3.516.5-15=1.5

Required Ratio = 3.5 : 1.5 = 35 : 15 = 7 : 3

17. A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is
A. 400B. 500
C. 300D. 600
Discuss
answer with explanation

Answer: Option D

Explanation:

By rule of alligation,

Profit% by selling 1st partProfit% by selling 2nd part
818
Net % profit
14
18-14=414-8=6

=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3

Total quantity is given as 1000 kg. So quantity of part2 (quantity sold at 18% profit)
$=1000 × \dfrac{3}{5}$ = 600 kg

18. A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture?
A. 24%B. 22%
C. 25%D. 20%
Discuss
answer with explanation

Answer: Option D

Explanation:

Solution 1

If a trader professes to sell his goods at cost price, but uses false weights, then
Gain% $=\left[\dfrac{\text{Error}}{(\text{True Value- Error})} × 100 \right]\%$

Here Gain= 25%
error = quantity of water he mixes in the milk $=x$
true value = true quantity of milk = T

So the formula becomes, $25=\dfrac{x}{(T - x)} × 100$
$\Rightarrow 1= \dfrac{x}{(T - x)}× 4\\\Rightarrow T - x = 4x\\\Rightarrow T = 5x$

Percentage of water in the mixture
$=\dfrac{x}{T}× 100 = \dfrac{x}{5x}× 100 \\= \dfrac{1}{5}× 100 = 20\%$


Solution 2

Let CP of 1 litre milk = Rs.1
SP of 1 litre mixture = CP of 1 litre milk = Rs.1
Gain = 25%

Hence CP of 1 litre mixture
$=\dfrac{100} {\left(100 + \text{Gain} \% \right)}× \text{SP}\\= \dfrac{100}{(100+25)}× 1 = \dfrac{100}{125}= \dfrac{4}{5}$

By rule of alligation,
CP of 1 litre milkCP of 1 litre water
10
CP of 1 litre mixture
$\dfrac{4}{5}$
$\dfrac{4}{5}-0=\dfrac{4}{5}$$1-\dfrac{4}{5}=\dfrac{1}{5}$

=> Quantity of milk : Quantity of water $=\dfrac{4}{5}:\dfrac{1}{5}=4:1$

Hence percentage of water in the mixture $=\dfrac{1}{5}×100=20\%$

19. In what ratio must water be mixed with milk to gain $16\dfrac{2}{3}\%$ on selling the mixture at cost price?
A. 1 : 4B. 6 : 1
C. 1 : 6D. 4 : 1
Discuss
answer with explanation

Answer: Option C

Explanation:

Let CP of 1 litre milk = Rs.1

SP of 1 litre mixture = CP of 1 litre milk = Rs.1

Gain $=16\dfrac{2}{3}\%=\dfrac{50}{3}\%$

CP of 1 litre mixture $=\dfrac{100}{\left(100 + \text{Gain}\% \right)}× \text{SP}$
$=\dfrac{100}{\left(100+\dfrac{50}{3}\right)}× 1=\dfrac{100}{\left(\dfrac{350}{3}\right)}\\=\dfrac{300}{350}=\dfrac{6}{7}$

By rule of alligation,
CP of 1 litre waterCP of 1 litre milk
01
CP of 1 litre mixture
$\dfrac{6}{7}$
$1-\dfrac{6}{7}=\dfrac{1}{7}$$\dfrac{6}{7}-0=\dfrac{6}{7}$

Quantity of water : Quantity of milk $=\dfrac{1}{7}:\dfrac{6}{7}=1:6$

20. In what ratio must rice at Rs.7.10 be mixed with rice at Rs.9.20 so that the mixture may be worth Rs.8 per kg?
A. 5 : 4B. 2 : 1
C. 3 : 2D. 4 : 3
Discuss
answer with explanation

Answer: Option D

Explanation:

By rule of alligation,

CP of 1 kg rice of 1st kindCP of 1 kg rice of 2nd kind
7.19.2
Mean Price
8
9.2 - 8 = 1.28 - 7.1 = .9

Required ratio = 1.2 : .9 = 12 : 9 = 4 : 3

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