Arithmetic Ability Practice Question and Answer8
Q: Mr. Shankar spends 25% of his monthly salary on household expenditure, 20% of the remaining on children’s education, and the remaining is equally invested in three different schemes. If the amount invested in each scheme is Rs.5600, what is the monthly salary of Shankar ? 525 05b5cc74fe4d2b4197774fb305b5cc74fe4d2b4197774fb30
- 1Rs. 34000falseCorrectIncorrect
- 2Rs. 31245falseCorrectIncorrect
- 3Rs. 24315falseCorrectIncorrect
- 4Rs. 28000trueCorrectIncorrect
Answer : 4. "Rs. 28000"
Answer: D) Rs. 28000 Explanation: Let the monthly salary of Shankar be = Rs.xAmount invested on expenditure = 25% = x/4;Remaning amount = 3x/4;Amount invested on children education = 20% i.e = 3x/20;Remaining amount = 3x/4 - 3x/20 = 3x/5;Remaining amount invested in three different schemes i.e is 1/3(3x/5) => x/5 = 5600Therefore x = 28000Hence, Monthly salary of Shankar is Rs. 28,000.
Q: What was the day of the week on 17th June, 1998 ? 2167 65b5cc761e4d2b4197774fd165b5cc761e4d2b4197774fd16
Answer : 3. "Wednesday"
Answer: C) Wednesday Explanation: 17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)Odd days in 1600 years = 0Odd days in 300 years = 197 years has 24 leap years + 73 ordinary years.Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.Jan. Feb. March. April. May. June.(31 + 28 + 31 + 30 + 31 + 17) = 168 days168 days = 24 weeks = 0 odd day.Total number of odd days = (0 + 1 + 2 + 0) = 3.Given day is Wednesday.
Q: Simran started a software business by investing Rs. 50,000. After six months, Nanda joined her with a capital of Rs. 80,000. After 3 years, they earned a profit of Rs. 24,500. What was Simran's share in the profit ? 479 05b5cc731e4d2b4197774f7f15b5cc731e4d2b4197774f7f1
Answer : 4. "10,500"
Answer: D) 10,500 Explanation: The ratio of their investments:50000x36 : 80000x30 = 3 : 4Simran's share of profit = (24500x3/7) = Rs.10,500.
Q: If the true discount on sum due 2 years hence at 14% per annum be Rs. 168, the sum due is ? 3044 45b5cc723e4d2b4197774f6a45b5cc723e4d2b4197774f6a4
Answer : 3. "Rs.768"
Answer: C) Rs.768 Explanation: P.W. = 100×T.DR×T = 100×16814×2 = 600. Sum = (P.W. + T.D.) = Rs. (600 + 168) = Rs. 768.
Q: A milkman has 20 liters of milk. If he mixes 5 liters of water, which is freely available, in 20 liters of pure milk. If the cost of pure milk is Rs. 18 per litre, then the profit of the milkman, when he sells all the mixture at cost price, is: 1543 35b5cc7d7e4d2b4197775151f5b5cc7d7e4d2b4197775151f
Answer : 2. "25%"
Answer: B) 25% Explanation: When the water is freely available and all the water is sold at the price of the milk, then the water gives the profit on the cost of 20 litres of milk. Since, the profit % = ProfitCostPricex100 Therefore, profit percentage = 520x100 = 25%
Q: From a tank of petrol, which contains 200 litres of petrol, the seller replaces each time with kerosene when he sells 40 litres of petrol(or its mixture). Everytime he sells out only 40 litres of petrol(pure or impure). After replacing the petrol with kerosen 4th time, the total amount of kerosene in the mixture is 16759 25b5cc7d7e4d2b4197775151a5b5cc7d7e4d2b4197775151a
- 4None of thesefalseCorrectIncorrect
Answer : 3. "118.08L"
Answer: C) 118.08L Explanation: The amount of petrol left after 4 operations = 200 × 1-402004 = 200 × 454 = 200 × 256625 = 81.92 litres Hence the amount of kerosene = 200 - 81.92 = 118. 08 litres
Q: From a container of wine, a thief has stolen 15 litres of wine and replaced it with same quantity of water. He again repeated the same process. Thus, in three attempts the ratio of wine and water became 343 : 169. The initial amount of wine in the container was: 3645 15b5cc7d7e4d2b419777515105b5cc7d7e4d2b41977751510
- 175 litresfalseCorrectIncorrect
- 2100 litresfalseCorrectIncorrect
- 3150 litresfalseCorrectIncorrect
- 4120 litrestrueCorrectIncorrect
Answer : 4. "120 litres"
Answer: D) 120 litres Explanation: wine(left)wine(added) = 343169 It means wine(left)wine(initial amount) = 343512 (since 343 + 169 = 512) Thus, 343x = 512x1 - 15k3 343512 = 783 = 1 - 15k3 1-15k=78=1-18 Thus the initial amount of wine was 120 liters.
Q: 5 men and 4 women are to be seated in a row so that the women occupy the even places . How many such arrangements are possible? 599 15b5cc7d1e4d2b4197775129f5b5cc7d1e4d2b4197775129f
Answer : 1. "2880"
Answer: A) 2880 Explanation: There are total 9 places out of which 4 are even and rest 5 places are odd. 4 women can be arranged at 4 even places in 4! ways. and 5 men can be placed in remaining 5 places in 5! ways. Hence, the required number of permutations = 4! x 5! = 24 x 120 = 2880