Two barrels contain a mixture of ethanol and gasoline. The content of the ethanol is 60% in the first barrel and 30% in the second barrel. In what ratio must the mixtures from the first and the second barrels be taken to form a mixture containing 50% ethanol ?

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INSURANCE EXAMS (GIC) 2011 click here to view all questions

Two barrels contain a mixture of ethanol and gasoline. The content of the ethanol is 60% in the first barrel and 30% in the second barrel. In what ratio must the mixtures from the first and the second barrels be taken to form a mixture containing 50% ethanol ?

A1 : 2

B2 : 1

C2 : 3

D3 : 2

Nithin ? undefined

Correct answer is: 2 : 1

Explanation:

By allegation,
Mixture-I Mixture-II
Therefore, Required ratio
\(= \frac{1}{5}:\frac{1}{10}\)

\(= 2:1\)

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When one litre of water is added to a mixture of acid and water, the new mixture contains 20% acid. When one litre of acid is added to the new mixture, then the resulting mixture contains $33\frac{1}{3}$% acid. The percentage of acid in the original mixture was

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INSURANCE EXAMS (New India Assurance) 2011 click here to view all questions

When one litre of water is added to a mixture of acid and water, the new mixture contains 20% acid. When one litre of acid is added to the new mixture, then the resulting mixture contains $33\frac{1}{3}$% acid. The percentage of acid in the original mixture was

A20 %

B22 %

C24 %

D25 %

Nithin ? undefined

Correct answer is: 25 %

Explanation:

If there be 1 litre of acid in 4 litres of mixture, then in case 1,
Percentage of acid

\(= \frac{1}{4+1} \times 100 = 20 \text{%}\)

Case ll.
Percentage of acid

\(= \frac{2}{6} \times 100 = \frac{100}{3}=33\frac{1}{3}\text{%}\)

Therefore, Percentage of acid in original 1 mixture = \(= \frac{1}{4} \times 100 \)

\(= 25 \text{%}\)

 

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To m litres of a m% solution of acid. x litres of water is muted to yield (m - 10)% solution of acid. If m > 25. then x equals

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INSURANCE EXAMS (New India Assurance) 2009 click here to view all questions

To m litres of a m% solution of acid. x litres of water is muted to yield (m - 10)% solution of acid. If m > 25. then x equals

A$\frac{10m}{m-10}$

B$\frac{5m}{m-10}$

C$\frac{2m}{m-10}$

D$\frac{m}{m-10}$

Nithin ? undefined

Correct answer is: \(\frac{10m}{m-10}\)

Explanation:

Tricky approach

\(\text{ Acid = }\frac{m}{100} \times m = \frac{m^{2}}{100} \)

\(\text{New percentage =} \frac{m^{2} \times 100}{ \left(m+x\right) \times 100 }\)

\(\implies \frac{m^{2}}{m+x}=m-10 \)

\(\implies m^{2} - 10m + mx - 10x = m^{2}\)
\(\implies x \left(m-10\right)=10m\)

\(\implies x = \frac{10m}{m-10}\)

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A man purchased 35 kg of rice at the rate of Rs.9.50 per kg and 30 kg at the rate of Rs.10.50 per kg. He mixed the two. Approximately, at what price (in Rupees) per kg. should he sell the mixture to make 35 per cent profit in the transaction?

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INSURANCE EXAMS (LIC) 2009 click here to view all questions

A man purchased 35 kg of rice at the rate of Rs.9.50 per kg and 30 kg at the rate of Rs.10.50 per kg. He mixed the two. Approximately, at what price (in Rupees) per kg. should he sell the mixture to make 35 per cent profit in the transaction?

ARs.12

BRs.12.50

CRs.13

DRs.13.50

Nithin ? undefined

Correct answer is: Rs.13.50

Explanation:

CP of 65 kg of the mixture

= Rs. \(\left(35 \times 9.50 + 30 \times 10.50\right)\)

= Rs. \(\left(332.5 + 315\right)\)

= Rs.647.5

Therefore, Rate per kg of the mixture = Rs.  \(\left(\frac{647.5}{65}\right)\)

Therefore, Required rate = \(\frac{647.5}{65} \times \frac{135}{100}\) =  Rs.133.50/kg.

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A grocer purchased 2 kg. of rice at the rate of Rs.15 per kg. and 3 kg. of rice at the rate of Rs.13 per kg. At what price per kg. should he sell the mixture to earn $33\frac{1}{3}$% profit on the cost price?

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BANK PO Exam (SBI) 2006 click here to view all questions

A grocer purchased 2 kg. of rice at the rate of Rs.15 per kg. and 3 kg. of rice at the rate of Rs.13 per kg. At what price per kg. should he sell the mixture to earn $33\frac{1}{3}$% profit on the cost price?

ARs.28.00

BRs.20.00

CRs.18.40

DRs.17.40

Nithin ? undefined

Correct answer is: Rs.18.40

Explanation:

Mixture = 2 kg of rice at Rs.15/kg + 3 kg. of rice at Rs.13/kg
Total weight = 2 + 3 = 5 kg

Total cost price = \(\left(2 \times 15\right)+ \left(3 \times 13\right)\)

= 30 + 39 = Rs.69
Cost price per kg of the mixture = \(\frac{69}{5}\) = Rs.13.80

Selling price to get \(33\frac{1}{3}\) % profit

\(=\frac{100+33\frac{1}{3}}{100}\) = Rs.13.80

\(= \frac{400}{3 \times 100}\) = Rs.13.80

\(= \frac{4}{3}\) = Rs.13.80

= Rs.18.40

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The wheat sold by a grocer contained 10% low quality wheat. What quantity of good quality wheat should be added to 150 kgs of wheat so that the percentage of low quality wheat becomes 5%?

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The wheat sold by a grocer contained 10% low quality wheat. What quantity of good quality wheat should be added to 150 kgs of wheat so that the percentage of low quality wheat becomes 5%?

A150 kgs

B135 kgs

C50 kgs

D85 kgs

Nithin ? undefined

Correct answer is: 150 kgs

Explanation:

Good quality content in 150 kgs. of wheat = 90% of 150 = 135 kgs.
In new mixture, low quality wheat is 5%, so good quality wheat 95%
and 5% of the new mixture = 15 kgs.

Therefore, New mixture = \(\frac{15 \times 100}{5}\) = 300 kg.

Therefore, Good quality of wheat added =  \(\left(300 - 150\right)\)= 150 kg.

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In 1 kg of a mixture of sand and iron, 20% is iron .How such sand should be added so that the proportion of iron becomes 5%?

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BANK PO Exam (SBI) 1999 click here to view all questions

In 1 kg of a mixture of sand and iron, 20% is iron .How such sand should be added so that the proportion of iron becomes 5%?

A3 kg

B4 gms

C5 gms

D6 kg

Nithin ? undefined

Correct answer is: 3 kg

Explanation:

Total mixture of sand and iron = 1 kg

Quantity of iron = \(\frac{20}{100} \times 1 = 0.2 \text{ kg}\)

Let x kg of sand should be added, thus total iron in the mixture

\(\implies 0.2=\frac{5}{100} \times (x+1)\)

\(\implies 4=x+1\)

\(\implies x=4-1=3 \text{ kg}\)

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