Quantitative Aptitude Practice Question and Answer
8Q: If 78K928L is divisible by 6, then which of the following can Q and R take ? 6773 05b5cc741e4d2b4197774f995
5b5cc741e4d2b4197774f995- 1K=2 & L=3false
- 2K=1 & L=4true
- 3K=1 & L=2false
- 4K=3 & L=3false
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Answer : 2. "K=1 & L=4"
Explanation :
Answer: B) K=1 & L=4 Explanation: As per divisibility rule a number is divisible by 6 means it should be divisible by 2 and 3 case 1 : the divisibility rule for 2 is the number should be end with even number in this case R =2 & R=4 case 2 : The divisibility rule for 3 is the sum of numbers should be divisible by 3so if we take option (2) Q=1 & R=4 the sum is 39 if we take option (3) Q=1 & R=2 the sum is 37 So Option (2) is correct 39 is divided by 3
Q: The average temperature of Monday to Wednesday was 37 C and of Tuesday to Thursday was 34 C. If the temperature on Thursday was 4/5 th of that of Monday, the temperature on Thursday was ? 2591 05b5cc741e4d2b4197774f986
5b5cc741e4d2b4197774f986- 135 cfalse
- 236 ctrue
- 334 cfalse
- 432 cfalse
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Answer : 2. "36 c"
Explanation :
Answer: B) 36 c Explanation: Monday + Tuesday + Wednesday = 37 x 3 -----------(1)Tuesday + Wednesday + Thursday = 34 x 3---------(2)Thursday = 4/5 of Monday ------------------(3)subtract eqn (1) from (2) we get,Thursday - Monday = -9 => Monday - Thursday = 9.......(4)From (3) & (4), we get So,Thursday = 36 C
Q: On a ruler's tombstone, it is said that one sixth of his life was spent in childhood and one twelfth as a teenager. One seventh of his life passed between the time he became an adult and the time he married; five years later, his son was born. Alas, the son died four years before he did. He lived to be twice as old as his son did. How old did the ruler live to be ? 3487 05b5cc741e4d2b4197774f981
5b5cc741e4d2b4197774f981- 184 yearstrue
- 272 yearsfalse
- 382 yearsfalse
- 464 yearsfalse
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Answer : 1. "84 years"
Explanation :
Answer: A) 84 years Explanation: Let the age of ruler is x so that of son = x/2 (given)Now according to the given condition (x/6) + (x/12) + (x/7) + 5 + (x/2) + 4 = x=> x = 84
Q: The Manager of a company accepts only one employees leave request for a particular day. If five employees namely Roshan, Mahesh, Sripad, Laxmipriya and Shreyan applied for the leave on the occasion of Diwali. What is the probability that Laxmi priya’s leave request will be approved ? 6547 05b5cc741e4d2b4197774f97c
5b5cc741e4d2b4197774f97c- 11false
- 21/5true
- 35false
- 44/5false
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Answer : 2. "1/5"
Explanation :
Answer: B) 1/5 Explanation: Number of applicants = 5On a day, only 1 leave is approved.Now favourable events = 1 of 5 applicants is approvedProbability that Laxmi priya's leave is granted = 1/5.
Q: A Certain sum of money an amounts to Rs 2500 in a span Of 5 years and further to Rs.3000 in a span of 7 years at simple interest The sum is ? 4149 05b5cc73de4d2b4197774f972
5b5cc73de4d2b4197774f972- 1Rs. 1800false
- 2Rs. 2000false
- 3Rs. 1400false
- 4Rs. 1250true
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Answer : 4. "Rs. 1250"
Explanation :
Answer: D) Rs. 1250 Explanation: 2500 in 5th year and 3000 in 7th year So in between 2 years Rs. 500 is increased => for a year 500/2 = 250So, per year it is increasing Rs.250 then in 5 years => 250 x 5 = 1250Hence, the initial amount must be 2500 - 1250 = Rs. 1250
Q: There are several peacocks and deers in a cage (with no other types of animals). There are 72 heads and 200 feet inside the cage. How many peacocks are there, and how many deers ? 3188 05b5cc73de4d2b4197774f95d
5b5cc73de4d2b4197774f95d- 142 & 24false
- 238 & 26false
- 336 & 24false
- 444 & 28true
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Answer : 4. "44 & 28"
Explanation :
Answer: D) 44 & 28 Explanation: Letx = peacocks y = deers 2x + 4y = 200 x + y = 72 y = 72 - x 2x + 288 - 4x = 200 x = 44 peacocks y = 28 deers
Q: Three taps I, J and K can fill a tank in 20,30and 40 minutes respectively. All the taps are opened simultaneously and after 5 minutes tap A was closed and then after 6 minutes tab B was closed .At the moment a leak developed which can empty the full tank in 70 minutes. What is the total time taken for the completely full ? 4932 05b5cc73de4d2b4197774f952
5b5cc73de4d2b4197774f952- 124.315 minutesfalse
- 226.166 minutestrue
- 322.154 minutesfalse
- 424 minutesfalse
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Answer : 2. "26.166 minutes"
Explanation :
Answer: B) 26.166 minutes Explanation: Upto first 5 minutes I, J and K will fill => 5[(1/20)+(1/30)+(1/40)] = 65/120 For next 6 minutes, J and K will fill => 6[(1/30)+(1/40)] = 42/120 So tank filled upto first 11 minutes = (65/120) + (42/120) = 107/120 So remaining tank = 13/120 Now at the moment filling with C and leakage @ 1/60 per minute= (1/40) - (1/70) = 3/280.So time taken to fill remaining 13/120 tank =(13/120) /(3/280) = 91/6 minutes Hence total time taken to completely fill the tank = 5 + 6 + 91/6 = 26.16 minutes.
Q: A can do a work in 9 days, B can do a work in 7 days, C can do a work in 5 days. A works on the first day, B works on the second day and C on the third day respectively that is they work on alternate days. When will they finish the work ? 1582 05b5cc73de4d2b4197774f94d
5b5cc73de4d2b4197774f94d- 1[7 + (215/345)] daysfalse
- 2[6 + (11/215)] daysfalse
- 3[6 + (261/315)] daystrue
- 4[5 + (112/351)] daysfalse
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Answer : 3. "[6 + (261/315)] days"
Explanation :
Answer: C) [6 + (261/315)] days Explanation: After day 1, A finishes 1/9 of the work. After day 2, B finishes 1/7 more of the total work. (1/9) + (1/7) is finished. After day 3, C finishes 1/5 more of total work. Total finished is 143/315. So, after day 6, total work finished is 286/315. Now remaining work = 29 /315 On day 7, A will work again Work will be completed on day 7 when A is working. He must finish 29/315 of total remaining work. Since he takes 9 days to finish the total task, he will need 261/315 of the day. Total days required is 6 + (261/315) days.

