Quantitative Aptitude Practice Question and Answer
8Q: A container contains 50 litres of milk. From that 8 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container ? 2135 05b5cc77fe4d2b419777501ce
5b5cc77fe4d2b419777501ce- 124.52 litresfalse
- 229.63 litrestrue
- 328.21 litresfalse
- 425.14 litresfalse
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Answer : 2. "29.63 litres"
Explanation :
Answer: B) 29.63 litres Explanation: Given that container has 50 litres of milk. After replacing 8 litres of milk with water for three times, milk contained in the container is: ⇒501-8503 ⇒50×4250×4250×4250 = 29.63 litres.
Q: A retailer buys product from a shopkeeper at discount of 40% on the list price (marked price) and sells them to the customer at a discount of 25% on the list price.What is his profit percentage ? 2126 05b5cc7a5e4d2b4197775066f
5b5cc7a5e4d2b4197775066f- 110%false
- 215%false
- 320%false
- 425%true
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Answer : 4. "25%"
Explanation :
Answer: D) 25% Explanation: Let the list price be Rs.100. Therefore, the retailer is buying the products at Rs.60 and selling it to the customer at Rs.75, earning a profit of Rs.15. Therefore, his percentage is = 1560x100 = 25%
Q: What is the ratio of 18 minutes to one hour ? 2120 05b5cc6bfe4d2b4197774da24
5b5cc6bfe4d2b4197774da24- 11/5false
- 23/4false
- 31/7false
- 43/10true
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Answer : 4. "3/10"
Explanation :
Answer: D) 3/10 Explanation: One hour = 60 min 18/60 = 3/10
Q: A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook ? 2118 05b5cc6dde4d2b4197774ea28
5b5cc6dde4d2b4197774ea28- 115false
- 220false
- 310true
- 48false
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Answer : 3. "10"
Explanation :
Answer: C) 10 Explanation: Let the price of each note book be Rs.x. Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y. Hence xy = 300 => y = 300/x (x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy =>5(300/x) - 10x - 50 = 0 => -150+x2+5x multiplying both sides by -1/10x => x2+15x-10x-150=0 => x(x + 15) - 10(x + 15) = 0 => x = 10 or -15 As x>0, x = 10.
Q: Srinivas bought 10 cycles for Rs. 1,750 each. He sold 4 cycles for Rs 8,400, 3 for Rs 1900 each . At what price should he sell the remaining cycles so as to earn an average profit of Rs 320 per cycle ? 2108 05b5cc718e4d2b4197774f4c7
5b5cc718e4d2b4197774f4c7- 1Rs. 1580false
- 2Rs. 2200true
- 3Rs. 1785false
- 4Rs. 1950false
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Answer : 2. "Rs. 2200"
Explanation :
Answer: B) Rs. 2200 Explanation: He sold 4 cycles for Rs. 8400[It means each cycle price is Rs. 2100]And again he sold 3 cycles for Rs. 1900 eachIf he want to gain avg profit on each cycle is Rs. 320 per cycle.So profit for 10 cycles is @ Rs.320 is,10 x 320 = Rs. 3200He already got Rs. 1850 profit by selling 7 cycles.Now, remaining amount is 3200 - 1850 = 1350So from remaining 3 cycles we want to get Rs. 1350 profit ,1350/3 = 450so he must sell remaining 3 cycles at Rs. 2200 each one.
Q: Rs. 69 were divided among 115 family members so that each lady gets 50 paise less than each gentleman, and each gentleman received twice the paise as each lady received. Find the number of gentlemen in the family? 2103 05b5cc67be4d2b4197774c1f0
5b5cc67be4d2b4197774c1f0- 123true
- 292false
- 348false
- 452false
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Answer : 1. "23"
Explanation :
Answer: A) 23 Explanation: Total number of female members = 115 Let the number of males = x Number of females = y => x + y = 115 ......(1) According to the given data, x + y(0.50) = 69 ......(2) From (1) & (2), we get 0.5y = 46 => y = 46 x 2 = 92 => x = 115 - 92 = 23 Hence, required number of gentlemen in the family = x = 23.
Q: Find the missing number in the given number series? 514, 470, ?, 415, 404 2100 05b5cc6a9e4d2b4197774cf19
5b5cc6a9e4d2b4197774cf19- 1441false
- 2437true
- 3426false
- 4420false
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Answer : 2. "437"
Explanation :
Answer: B) 437 Explanation: Given number series is 514 470 ? 415 404 Here the pattern it follows is 514 514 - 44 = 470 470 - 33 = 437 437 - 22 = 415 415 - 11 = 404 Hence, the missing number in the given number series is 437.
Q: A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained ? 2097 05b5cc6dce4d2b4197774e960
5b5cc6dce4d2b4197774e960- 15:3false
- 21:4false
- 34:1false
- 49:1true
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Answer : 4. "9:1"
Explanation :
Answer: D) 9:1 Explanation: Milk = 3/5 x 20 = 12 liters, water = 8 liters If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters. Remaining milk = 12 - 6 = 6 liters Remaining water = 8 - 4 = 4 liters 10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters. The ratio of milk and water in the new mixture = 16:4 = 4:1 If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 x 10 = 8 liters. Amount of water removed = 2 liters. Remaining milk = (16 - 8) = 8 liters. Remaining water = (4 -2) = 2 liters. Now 10 lts milk is added => total milk = 18 lts The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.