Naresh purchased a TV set for Rs. 11,250 after getting discount of 10% on the labelled price. He spent Rs. 150 on transport and Rs. 800 on installation. At what price should it be sold so that the profit earned would have been 15% if no discount was offered?

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BANK PO Exam (United Bank of India) 2009 click here to view all questions

Naresh purchased a TV set for Rs. 11,250 after getting discount of 10% on the labelled price. He spent Rs. 150 on transport and Rs. 800 on installation. At what price should it be sold so that the profit earned would have been 15% if no discount was offered?

ARs. 12937.50

BRs. 14,030

CRs. 13,450

DRs.15467.50

Nithin ? undefined

Correct answer is: Rs.15467.50

Explanation:

Let original price was X.

X - 10% of X = 11,250

X - X/10 = 11, 250

9X/10 = 11, 250

9X = 112500

X = 12500.

Total expences,

12500 + 150 + 800 = 13,450.
Selling Price with 15% profit,

= 13450 + 15% of 13450 = Rs. 15467.5

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Deepa bought a calculator at 30% discount on the listed price. Had she not got the discount, she would have paid Rs. 82.50 extra. At what price did she buy the calculator?

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BANK PO Exam (Canara Bank) 2003 click here to view all questions

Deepa bought a calculator at 30% discount on the listed price. Had she not got the discount, she would have paid Rs. 82.50 extra. At what price did she buy the calculator?

A192.50

B275

C117.85

DCant be determined

Nithin ? undefined

Correct answer is: 192.50

Explanation:

Let the original price be x, then
30% of x = 82.50
x = \(\frac{82.50}{30} \times 100\) = Rs. 275
Deepa bought calculator in 275 – 82.50 = Rs. 192.50

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A container has 30 litres of water. If 3 litres of water is replaced by 3 litres of spirit and this operation is repeated twice, what will be the quantity of water in the new mixture?

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INSURANCE EXAMS (LIC) 2013 click here to view all questions

A container has 30 litres of water. If 3 litres of water is replaced by 3 litres of spirit and this operation is repeated twice, what will be the quantity of water in the new mixture?

A24 litres

B23 litres

C24.3 litres

D23.3 litres

Nithin ? undefined

Correct answer is: 24.3 litres

Explanation:

Suppose a container contains x units of liquid from which y Units are taken out and replaced by water.
After n operations. the quantity of pure liquid

\(= x \left(1-\frac{y}{x}\right)^{n} units\)

Therefore, Remaining water
\(=30 \left(1-\frac{3}{30}\right)^{2}\)

\(= \frac{30 \times 9 \times 9}{100}\)

= 24.3 litres

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A and B are two alloys of gold and copper prepared by mixing metals in the proportions 7 : 2 and 7 : 11 respectively. If equal quantities of the alloys are melted to form a third alloy C, proportion of gold and copper in C will be

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INSURANCE EXAMS (OIC) 2012 click here to view all questions

A and B are two alloys of gold and copper prepared by mixing metals in the proportions 7 : 2 and 7 : 11 respectively. If equal quantities of the alloys are melted to form a third alloy C, proportion of gold and copper in C will be

A5 : 7

B5 : 9

C9 : 5

D7 : 5

Nithin ? undefined

Correct answer is: 7 : 5

Explanation:

Gold in C = \(\left( \frac{7}{9} + \frac{7}{18} \right)\)

\(= \frac{21}{18} = \frac{7}{6}\)

Copper in C =  \( \left( \frac{2}{9} + \frac{11}{18} \right)\)

\(= \frac{15}{18} = \frac{5}{6}\)

Gold : Copper = \(\frac{7}{6} : \frac{5}{6}\)

= 7 : 5

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To x litres of an x% solution of acid, y litres of water is added to get (x -10)% solution of acid. If x > 20, then value of y is

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INSURANCE EXAMS (OIC) 2012 click here to view all questions

To x litres of an x% solution of acid, y litres of water is added to get (x -10)% solution of acid. If x > 20, then value of y is

A$\frac{x^2}{100}$

B$\frac{10x}{x-10}$

C$\frac{10x}{x+10}$

D$\frac{10x^2}{x-10}$

Nithin ? undefined

Correct answer: \(\frac{10x}{x-10}\)

Explanation:

In litres of solution,

Acid = \(\frac{x^2}{100}\)

Water = \(x- \frac{x^2}{100}\)

\(\therefore \frac{\frac{x^2}{100}}{x+y} \times 100 = x-10\)

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

\(\implies x^2 =y^2-10x-y(x-10)\)

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

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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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