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381
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Mixture (Alloy)
medium
Mathematics
In a 50 gm alloy of gold and silver, the gold is 80% by weight. How much pure gold should be mixed with this alloy so that the weight percentage of gold becomes 95%?
A
200 gm
B
150 gm
C
50 gm
D
10 gm
Explanation and memory cue
The original alloy contains 50 gm with 80% gold, so gold weight is 40 gm. Let x be the gold added. The new gold percentage is (40 + x) / (50 + x) = 95%. Solving gives x = 200 gm. Therefore, 200 gm of gold should be added to reach 95% gold content.
382
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Percentage (Expenditure Constant)
medium
Mathematics
The price of a certain item is increased by 15%. If a consumer wants to keep his expenditure on the item the same as before, by what percent must he reduce his consumption of that item?
A
10 20/23 %
B
13 1/23 %
C
16 2/3 %
D
15 %
Explanation and memory cue
If the price increases by 15%, the new price is 115% of the original. To keep expenditure the same, consumption must be reduced so that (new consumption) × (new price) = (original consumption) × (original price). Thus, new consumption = original consumption × (100/115) = original consumption × (20/23). The reduction in consumption is (1 - 20/23) × 100% = (3/23) × 100% ≈ 13.04%. However, the options are given as mixed fractions, and option A is '10 20/23 %' which equals approximately 10.87%, option B is '13 1/23 %' which equals approximately 13.04%, option C is '16 2/3 %' (16.67%), and option D is 15%. The correct reduction is approximately 13.04%, matching option B. Therefore, the original correct_answer 'B' is correct. The explanation was missing and has been added. The difficulty is medium due to the calculation complexity. Tags added for clarity.
A typist uses a paper 30 cm by 15 cm. He leaves a margin of 2.5 cm at the top as well as at the bottom and 1.25 cm on either side. What percentage of the paper area is approximately available for typing?
A
65 %
B
70 %
C
80 %
D
60%
Explanation and memory cue
The total area of the paper is 30 cm × 15 cm = 450 cm². The margins reduce the usable area: top and bottom margins total 2 × 2.5 cm = 5 cm, so the height available for typing is 15 - 5 = 10 cm. The side margins total 2 × 1.25 cm = 2.5 cm, so the width available for typing is 30 - 2.5 = 27.5 cm. Therefore, the typing area is 27.5 cm × 10 cm = 275 cm². The percentage of the paper area available for typing is (275 / 450) × 100 ≈ 61.11%, which is closest to 60%. Hence, option D (60%) is correct.
385
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Ratio & Percentage
easy
Mathematics
A number is first increased by 10% and then reduced by 10%. What is the ratio of the final number to the original number?
A
2:5
B
99:100
C
6:7
D
3:5
Explanation and memory cue
When a number is increased by 10%, it becomes 110% (or 1.10 times) of the original number. Then, reducing this new number by 10% means multiplying by 90% (or 0.90). The final number is therefore 1.10 × 0.90 = 0.99 times the original number, which is 99% of the original. This ratio is 99:100. Among the given options, 4:5 (which is 0.8) is not a correct approximation. However, since none of the options exactly match 99:100, the closest correct ratio is 99:100, and the answer should be corrected to reflect this ratio. The original answer (B: 4:5) is incorrect. The correct ratio is 99:100, which is not listed in the options. Therefore, the question options should be updated to include 99:100 as the correct answer.
386
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Ratio And Proportion (False Weight)
medium
Mathematics
A dishonest dealer professes to sell goods at cost price but uses a false weight and gains 25%. Find his false weight in grams.
A
700 gms
B
750 gms
C
800 gms
D
850 gms
Explanation and memory cue
If a dishonest dealer claims to sell goods at cost price but uses a false weight to gain 25%, it means he gives less than 1 kg while charging for 1 kg. Let the false weight be x grams. The gain percentage is given by ((1000 - x) / x) × 100 = 25%. Solving for x: (1000 - x) / x = 0.25 → 1000 - x = 0.25x → 1000 = 1.25x → x = 800 grams. Therefore, the false weight used is 800 grams to gain 25%. This matches option C.
387
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Fractions And Percent Change
medium
Mathematics
If the numerator of a fraction is increased by 140% and the denominator is increased by 150%, the resultant fraction is 4/15. What is the original fraction?
A
3/5
B
5/16
C
2/9
D
None of these
Explanation and memory cue
Let the original fraction be x/y. Increasing numerator by 140% means numerator becomes x + 1.4x = 2.4x. Increasing denominator by 150% means denominator becomes y + 1.5y = 2.5y. The new fraction is (2.4x)/(2.5y) = 4/15. Simplifying, (2.4x)/(2.5y) = 4/15 => (2.4/2.5)(x/y) = 4/15 => (0.96)(x/y) = 4/15 => x/y = (4/15)/0.96 = (4/15)*(25/24) = 100/360 = 5/18, which is not among options. Rechecking calculation: 2.4x/2.5y = 4/15 => (2.4/2.5)(x/y) = 4/15 => 0.96(x/y) = 4/15 => x/y = (4/15)/0.96 = (4/15)*(25/24) = (100/360) = 5/18. Since 5/18 is not an option, check options for closest fraction: 2/9 = 4/18, which is close to 5/18. Testing 2/9: numerator increased by 140%: 2 + 2*1.4 = 2 + 2.8 = 4.8; denominator increased by 150%: 9 + 9*1.5 = 9 + 13.5 = 22.5; new fraction = 4.8/22.5 = 0.2133, original fraction 2/9 = 0.2222. New fraction should be 4/15 = 0.2667, so 2/9 is not exact. Testing 3/5: numerator increased: 3 + 4.2 = 7.2; denominator increased: 5 + 7.5 = 12.5; new fraction = 7.2/12.5 = 0.576, not 4/15. Testing 5/16: numerator increased: 5 + 7 = 12; denominator increased: 16 + 24 = 40; new fraction = 12/40 = 3/10 = 0.3, not 4/15. Since none matches exactly, the original fraction is 5/18, which is not listed, so correct answer is D (None of these).
388
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Percentage
medium
Mathematics
In two successive years, 100 and 75 students of a school appeared at the final examination. Respectively 75% and 60% of them passed. The average rate of pass is: _______?
A
68 4/7 %
B
78 %
C
80 %
D
80 4/7 %
Explanation and memory cue
The average pass rate is calculated by dividing the total number of students who passed in both years by the total number of students who appeared in both years. (75% of 100) + (60% of 75) = 75 + 45 = 120 passed out of 175 total students, so the average pass rate = (120/175)*100 = 68 4/7%.
389
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Profit and Loss
medium
Mathematics
Oranges are bought at 11 for a rupee and an equal number more at 9 for a rupee. If these are sold at 10 for a rupee, find the loss or gain percent?
A
1% gain
B
2% gain
C
1% loss
D
2% loss
Explanation and memory cue
The cost price for 11 oranges is 1 rupee, and the same number more (11) are bought at 9 for a rupee, so 11 at 1 rupee and 11 at 1.22 rupees (11/9). Total cost price = 1 + 1.22 = 2.22 rupees for 22 oranges. Selling price at 10 for a rupee means 22 oranges sold for 2.2 rupees. Since selling price (2.2) is less than cost price (2.22), there is a loss. Loss percent = (0.02/2.22)*100 ≈ 0.9%, approximately 1% loss.
390
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Time & Percent
easy
Mathematics
A period of 4 hours 30 minutes is what percent of a day?
A
18 3/4 %
B
20 %
C
16 3/4 %
D
19 %
Explanation and memory cue
A day has 24 hours. 4 hours 30 minutes equals 4.5 hours. To find the percentage: (4.5 / 24) × 100 = 18.75%, which is 18 3/4%.