Quantitative Aptitude Practice Question and Answer
8Q: Find the odd man out in the following number series? 5, 4, 9, 32, 273 3538 05b5cc695e4d2b4197774c500
5b5cc695e4d2b4197774c500- 14false
- 29false
- 332true
- 4273false
- Show AnswerHide Answer
- Workspace
- SingleChoice
Answer : 3. "32"
Explanation :
Answer: C) 32 Explanation: The given number series follows 5, 4, 9, 32, 308 pattern that 5 5 x 1 - 1 = 4 4 x 3 - 3 = 9 9 x 5 - 5 = 40 (not equals to 32) 40 x 7 - 7 = 273 Hence, the odd man in the given number series is 32.
Q: A garrison of 2000 men has provisions for 54 days. At the end of 15 days, a reinforcement arrives, and it is now found that the provisions will last only for 20 days more. What is the reinforcement ? 3535 05b5cc6e7e4d2b4197774edb0
5b5cc6e7e4d2b4197774edb0- 11900true
- 22100false
- 31700false
- 42000false
- Show AnswerHide Answer
- Workspace
- SingleChoice
Answer : 1. "1900"
Explanation :
Answer: A) 1900 Explanation: Given 2000 ---- 54 days The provisions for 2000 men for 39 days can be completed by 'm' men for only 20 days. i.e, 2000 ----- 39 days == m ---- 20 days => m x 20 = 2000 x 39 m = 3900 So total men for 20 days is 3900 => 2000 old and 1900 new reinforcement. Hence, reinforcement = 1900.
Q: 26 men can complete a piece of work in 17 days. How many more men must be hired to complete the work in 13 days? 3531 15d14995a489f2f135e08df26
5d14995a489f2f135e08df26- 19false
- 28true
- 36false
- 418false
- Show AnswerHide Answer
- Workspace
- SingleChoice
Answer : 2. "8"
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 ? 3510 05b5cc6dce4d2b4197774e960
5b5cc6dce4d2b4197774e960- 15:3false
- 21:4false
- 34:1false
- 49:1true
- Show AnswerHide Answer
- Workspace
- SingleChoice
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.
Q: 40% of the number is equal to three-fourth of another number. What is the ratio between first number and second number? 3501 05d1b13158a754124194a755f
5d1b13158a754124194a755f- 115:16false
- 215:8true
- 39:15false
- 48:17false
- Show AnswerHide Answer
- Workspace
- SingleChoice
Answer : 2. "15:8"
Q: The average age of a couple and their son was 40 years, the son got married and a child was born just two years after their marriage. When child turned to 10 years, then the average age of the family becomes 38 years. What was the age of the daughter in law at the time of marriage ? 3499 05b5cc6c6e4d2b4197774de59
5b5cc6c6e4d2b4197774de59- 112 yearstrue
- 210 yearsfalse
- 314 yearsfalse
- 413 yearsfalse
- Show AnswerHide Answer
- Workspace
- SingleChoice
Answer : 1. "12 years"
Explanation :
Answer: A) 12 years Explanation: Given the average age of couple and their son = 40 => Sum of age (H +W +S) = 40×3 Let the age of daughter in law at the time of marriage = D years Now after 10 years, (H + S +W) + 3×12 + D +12 + 10 = 38×5 178 + D = 190 D = 190 -178 = 12 years
Q: A merchant buys two items for Rs. 7500. One item he sells at a profit of 16% and the other item at 14% loss. In the deal, the merchant makes neither any profit nor any loss. What is the difference between the selling price of both the items? 3485 05b5cc69de4d2b4197774c88d
5b5cc69de4d2b4197774c88d- 1Rs. 620true
- 2Rs. 654false
- 3Rs. 725false
- 4Rs. 747false
- Show AnswerHide Answer
- Workspace
- SingleChoice
Answer : 1. "Rs. 620"
Explanation :
Answer: A) Rs. 620 Explanation: Let the C.P of one item is Rs. P and that of other is Rs. (7500 - P) According to the data given C.P = S.P => Px(116/100) + (7500-P)x(86/100) = 7500 => 30P = 105000 => P = 3500 Required difference between selling prices = Rs. [(3500/100) x 116] - [(4000/100) x 86] = 4060-3440 = Rs. 620
Q: Compound interest earned on a sum for the second and the third years are Rs.1200 and Rs.1440 respectively. Find the rate of interest ? 3484 05b5cc6ebe4d2b4197774eeb9
5b5cc6ebe4d2b4197774eeb9- 120% p.atrue
- 215% p.afalse
- 318% p.afalse
- 424% p.afalse
- Show AnswerHide Answer
- Workspace
- SingleChoice
Answer : 1. "20% p.a"
Explanation :
Answer: A) 20% p.a Explanation: Rs.1440 - 1200 = Rs.240 is the interest on Rs.1200 for one year. Rate of interest = (100 x 240) / (1200) = 20% p.a