Quantitative Aptitude Practice Question and Answer
8Q: 8 couples (husband and wife) attend a dance show "Nach Baliye' in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is : 3200 05b5cc7d0e4d2b41977751218
5b5cc7d0e4d2b41977751218- 18/39false
- 215/39true
- 312/13false
- 4None of thesefalse
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Answer : 2. "15/39"
Explanation :
Answer: B) 15/39 Explanation: P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize) = 1-16C1× 14C1×12C1×10C116C4=1539
Q: Find the odd one out of the following options. 3198 05b5cc6d6e4d2b4197774e679
5b5cc6d6e4d2b4197774e679- 1Sapphirefalse
- 2Rubyfalse
- 3Topazfalse
- 4Granitetrue
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Answer : 4. "Granite"
Explanation :
Answer: D) Granite Explanation: Except Granite all other options belong to the group of precious stones.
Q: A watch gains 5 seconds in 3 minutes and was set right at 8 AM. What time will it show at 10 PM on the same day ? 3197 05b5cc6e7e4d2b4197774edd4
5b5cc6e7e4d2b4197774edd4- 110 : 27 : 41 AMfalse
- 28 : 51 : 04 AMfalse
- 39 : 45 : 15 PMfalse
- 410 : 23 : 20 PMtrue
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Answer : 4. "10 : 23 : 20 PM"
Explanation :
Answer: D) 10 : 23 : 20 PM Explanation: The watch gains 5 seconds in 3 minutes = 100 seconds in 1 hour. From 8 AM to 10 PM on the same day, time passed is 14 hours. In 14 hours, the watch would have gained 1400 seconds or 23 minutes 20 seconds. So, when the correct time is 10 PM, the watch would show 10 : 23 : 20 PM
Q: A man spend Rs. 810 in buying trouser at Rs. 70 each and shirt at Rs. 30 each. What will be the ratio of trouser and shirt when the maximum number of trouser is purchased ? 3192 05b5cc731e4d2b4197774f7b5
5b5cc731e4d2b4197774f7b5- 11:3false
- 22:1false
- 33:2true
- 42:3false
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Answer : 3. "3:2"
Explanation :
Answer: C) 3:2 Explanation: Let us assume S as number of shirts and T as number of trousersGiven that each trouser cost = Rs.70 and that of shirt = Rs.30Therefore, 70 T + 30 S = 810=> 7T + 3S = 81......(1)T = ( 81 - 3S )/7 We need to find the least value of S which will make (81 - 3S) divisible by 7 to get maximum value of TSimplifying by taking 3 as common factor i.e, 3(27-S) / 7In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7 Hence T = (81-3xS)/7 = (81-3x6)/7 = 63/7 = 9 Hence for S, put T in eq(1), we getS = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.The ratio of T:S = 9:6 = 3:2.
Q: The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. If its present value is Rs. 8,748, its purchase price was ? 3167 05b5cc75ce4d2b4197774fc2f
5b5cc75ce4d2b4197774fc2f- 1Rs.14,000false
- 2Rs.12,000true
- 3Rs. 12,800false
- 4Rs. 15,000false
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Answer : 2. "Rs.12,000"
Explanation :
Answer: B) Rs.12,000 Explanation: We have P = 8748, R = -10 and n = 3Therefore, the purchase price of the machine= P1+R100n= 87481-101003= 8748 x (100/90) x (100/90) x (100/90)= 12,000. Therefore, the purchase price of the machine was Rs. 12,000.
Q: In a race of 1000 m, A can beat by 100 m, in a race of 800m, B can beat C by 100m. By how many meters will A beat C in a race of 600 m ? 3157 05b5cc6e5e4d2b4197774ed42
5b5cc6e5e4d2b4197774ed42- 1127.5 mtrue
- 2254 mfalse
- 3184 mfalse
- 4212 mfalse
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Answer : 1. "127.5 m"
Explanation :
Answer: A) 127.5 m Explanation: When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m. When B runs 900 m, distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 m. In a race of 1000 m, A beats C by (1000 - 787.5) = 212.5 m to C. In a race of 600 m, the number of meters by which A beats C = (600 x 212.5)/1000 = 127.5 m.
Q: What is the next number in the given Number Series? 5, 24, 94, 279, ? 3153 05b5cc6b9e4d2b4197774d73e
5b5cc6b9e4d2b4197774d73e- 1587false
- 2554true
- 3489false
- 4499false
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Answer : 2. "554"
Explanation :
Answer: B) 554 Explanation: The given number series follows a pattern that 5 x 5 – 1 = 24 24 x 4 – 2 = 94 94 x 3 – 3 = 279 279 x 2 – 4 = 554
Q: In a G - 20 meeting there were total 20 people representing their own country. All the representative sat around a circular table. Find the number of ways in which we can arrange them around a circular table so that there is exactly one person between two representatives namely Manmohan and Musharraf. 3108 05b5cc7d1e4d2b4197775128b
5b5cc7d1e4d2b4197775128b- 12 x (17!)false
- 22 x (18!)true
- 3(3!) x (18!)false
- 4(17!)false
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Answer : 2. "2 x (18!)"
Explanation :
Answer: B) 2 x (18!) Explanation: A person can be chosen out of 18 people in 18 ways to be seated between Musharraf and Manmohan. Now consider Musharraf, Manmohan, and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Musharraf and Manmohan can also be arranged in 2 ways. Required number of permutations = 18 x (17!) x 2 = 2 x 18!