Quantitative Aptitude Practice Question and Answer
8Q: The Simple interest accrued on an amount of Rs. 2,500 at the end of six years is Rs.1875. What would be the simple interest accrued on an amount of Rs. 6875 at the same rate and Same period? 2309 05b5cc7c7e4d2b41977750d78
5b5cc7c7e4d2b41977750d78- 1Rs. 4,556.5false
- 2Rs. 5,025.25false
- 3Rs.5,245.5false
- 4None of thesetrue
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Answer : 4. "None of these"
Explanation :
Answer: D) None of these Explanation:
Q: A man covers half of his journey by train at 60 km/hr, half of the remaining by bus at 30 km/hr and the rest by cycle at 10 km/hr. Find his average speed during the entire journey? 2307 05b5cc6ade4d2b4197774d12f
5b5cc6ade4d2b4197774d12f- 132 kmphfalse
- 220 kmphfalse
- 318 kmphfalse
- 424 kmphtrue
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Answer : 4. "24 kmph"
Explanation :
Answer: D) 24 kmph Explanation: Let the total distance of the journey of a man = d kms Now, the average speed of the entire journey = dd120 + d120 + d40 = 24 kmph.
Q: A single pipe of diameter x has to be replaced by six pipes of diameters 12 cm each. The pipes are used to covey some liquid in a laboratory. If the speed/flow of the liquid is maintained the same then the value of x is ? 2307 05b5cc73ae4d2b4197774f902
5b5cc73ae4d2b4197774f902- 114.69 cmfalse
- 229.39 cmtrue
- 318.65 cmfalse
- 422.21 cmfalse
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Answer : 2. "29.39 cm"
Explanation :
Answer: B) 29.39 cm Explanation: Volume of water flowing through 1 pipe of diameter x = Volume discharged by 6 pipes of diameter 12 cms. As speed is same, area of cross sections should be same. Area of bigger pipe of diameter x = Total area of 6 smaller pipes of diameter 12 i.e πR2 = 6πR12 Here R = R and R1 = 12/2 = 6 ⇒R2 = 6×6×6 R = 14.696 => D = X = 14.696 * 2 = 29.3938 cm.
Q: A class has 8 football players. A 5-member team and a captain will be selected out of these 8 players. How many different selections can be made ? 2304 05b5cc6d0e4d2b4197774e38a
5b5cc6d0e4d2b4197774e38a- 1210false
- 2168true
- 31260false
- 410!/6!false
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Answer : 2. "168"
Explanation :
Answer: B) 168 Explanation: we can select the 5 member team out of the 8 in 8C5 ways = 56 ways. The captain can be selected from amongst the remaining 3 players in 3 ways. Therefore, total ways the selection of 5 players and a captain can be made = 56x3 = 168 ways. (or) Alternatively, A team of 6 members has to be selected from the 8 players. This can be done in 8C6 or 28 ways. Now, the captain can be selected from these 6 players in 6 ways. Therefore, total ways the selection can be made is 28x6 = 168 ways.
Q: A watch which gains uniformly is 2 minutes low at noon on Tuesday and is 4 min 48 sec fast at 2 p.m. on the following Tuesday. When was it correct ? 2302 05b5cc761e4d2b4197774fd0c
5b5cc761e4d2b4197774fd0c- 112 p.m. on Wednesdayfalse
- 22 p.m. on Thursdaytrue
- 33 p.m. on Thursdayfalse
- 42 p.m. on Wednesdayfalse
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Answer : 2. "2 p.m. on Thursday"
Explanation :
Answer: B) 2 p.m. on Thursday Explanation: Time from 12 p.m. on Tuesday to 2 p.m. on the following Tuesday = 7 days 2 hours.= 170 hours.The watch gains = (2 + 4 x 4/5) min= 34/5 min. in 170 hrs.Now, 34/5 min are gained in 170 hrs.Then, 2 min are gained in (170 x 5/34 x 2) hrs.Watch is correct after 2 days 2 hrs after 12 p.m. on Tuesday, i.e., it will be correct at 2 p.m. on Thursday.
Q: Find the greatest number that will divide 964, 1238 and 1400 leaving remainder of 41,31 and 51 respectively ? 2298 05b5cc745e4d2b4197774fa54
5b5cc745e4d2b4197774fa54- 164false
- 269false
- 371true
- 458false
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Answer : 3. "71"
Explanation :
Answer: C) 71 Explanation: To find the greatest number which divides the numbers 964, 1238 and 1400 leaving the remainders 41, 31 and 51 is nothing but the HCF of (964 - 41), (1238 - 31), (1400 - 51). Therefore, HCF of 923, 1207 and 1349 is 71.
Q: P, Q, and R can do a job in 12 days together. If their efficiency of working be in the ratio 3 : 8 : 5, Find in what time Q can complete the same work alone? 2294 05b5cc698e4d2b4197774c61d
5b5cc698e4d2b4197774c61d- 136 daystrue
- 230 daysfalse
- 324 daysfalse
- 422 daysfalse
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Answer : 1. "36 days"
Explanation :
Answer: A) 36 days Explanation: Given the ratio of efficiencies of P, Q & R are 3 : 8 : 5 Let the efficiencies of P, Q & R be 3x, 8x and 5x respectively They can do work for 12 days. => Total work = 12 x 16x = 192x Now, the required time taken by Q to complete the job alone = 162x8x = 24 days.
Q: A seller uses 840 gm in place of 1 kg to sell his goods. Find his actual profit/loss % When he sells his article on 4% loss on cost price ? 2293 15b5cc74ae4d2b4197774fa72
5b5cc74ae4d2b4197774fa72- 114.28 % profittrue
- 224.18 % profitfalse
- 314.28 % lossfalse
- 424.18 % lossfalse
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Answer : 1. "14.28 % profit"
Explanation :
Answer: A) 14.28 % profit Explanation: Let 1kg of Rs. 100 then 840gm is of Rs. 84.Now (label on can 1kg but contains 840kg ) so for customer it is of Rs. 100 and further gives 4% discount [he sells his article on 4% loss on cost price.] So now S.P = Rs. 96But actually it contains 840 gm so C.P for shopkeeper = Rs. 84S.P = Rs. 96C.P = Rs. 84Profit% = {(S.P-C.P)/C.P}x100{(96-84)/84} x 100 = 14.28571429% PROFIT.