Arithmatical - Percentage
DIRECTIONS : Problems based on Percentage.
1. |
In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was |
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A. 2500 |
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B. 2700 |
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C. 3000 |
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D. 3100 |
Solution |
Number of valid votes = 80% of 7500 |
= 6000. |
Valid votes polled by other candidates | = 45% of 6000 |
(45/100×6000) |
= 2700. |
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2. |
If a number x is 10% less than another number y and y is 10% more than 125, then x is equal to |
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A. 143 |
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B. 150 |
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C. 123.75 |
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D. 140.55 |
Solution |
y = 125 + 10% of 125 |
= 125 + 12.50 = 137.50. |
x = 137.50 - 10% of 137.50 |
= 137.50 - 13.75 |
= 123.75. |
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3. |
In an examination, 35% of the students passed and 455 failed. How many students appeared for the examiantion? |
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A. 490 |
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B. 620 |
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C. 700 |
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D. 845 |
Solution |
Let the number of students appeared be x. |
Then, 65% of x = 455 |
‹=›65 / 100 x = 455 |
‹=› x= [455 x 100 / 65] |
= 700. |
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4. |
If the price of sugar rises from Rs. 6 per kg to Rs. 7.50 per kg, a person, to have no increase in his expenditure on sugar, will have to reduce his consumption of sugar by |
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A. 10% |
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B. 20% |
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C. 25% |
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D. 30% |
Solution |
Let the original consumption = 100 kg and new consumption = x kg. |
So, 100 x 6 = x × 7.50 |
x = 80 kg. |
Reduction in consumption = 20%. |
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5. |
From a container having pure milk, 20% is replaced by water and the process is repeated thrice. At the end of the third operation, the milk is |
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A. 40% pure |
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B. 50% pure |
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C. 51.2% pure |
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D. 58.8% pure |
Solution |
Let total quantity of original milk = 100 gm. |
Milk after first operation = 80% of 1000 |
‹=› 800gm. |
Milk after second operation = 80% of 800 |
‹=› 640gm. |
Milk after third operation = 80% of 640 |
‹=› 512gm. |
Strength of final mixtures = 51.2%. |
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