Quantitative Aptitude Practice Question and Answer
8Q: Find the number which multiplied by 15 is increased by 56 ? 1902 05b5cc6e2e4d2b4197774ec43
5b5cc6e2e4d2b4197774ec43- 16false
- 24true
- 33false
- 412false
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Answer : 2. "4"
Explanation :
Answer: B) 4 Explanation: Let the number be x. Then, 15x - x = 56 x = 4
Q: The letters of the word PROMISE are to be arranged so that three vowels should not come together. Find the number of ways of arrangements? 1902 05b5cc6ade4d2b4197774d0f9
5b5cc6ade4d2b4197774d0f9- 14320true
- 24694false
- 34957false
- 44871false
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Answer : 1. "4320"
Explanation :
Answer: A) 4320 Explanation: Given Word is PROMISE. Number of letters in the word PROMISE = 7 Number of ways 7 letters can be arranged = 7! ways Number of Vowels in word PROMISE = 3 (O, I, E) Number of ways the vowels can be arranged that 3 Vowels come together = 5! x 3! ways Now, the number of ways of arrangements so that three vowels should not come together = 7! - (5! x 3!) ways = 5040 - 720 = 4320.
Q: If a student walks at the rate of 4 mts/min from his home, he is 4 minutes late for school, if he walks at the rate of 6 mts/min he reaches half an hour earlier. How far is his school from his home ? 1900 05b5cc764e4d2b4197774fd80
5b5cc764e4d2b4197774fd80- 1100 mtsfalse
- 296 mtsfalse
- 398 mtstrue
- 4120 mtsfalse
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Answer : 3. "98 mts"
Explanation :
Answer: C) 98 mts Explanation: Let the distance between home and school is 'x'.Let actual time to reach be 't'.Thus, x/4 = t + 4 ---- (1)and x/6 = t - 30 -----(2)Solving equation (1) and (2)=> x = 98 mts.
Q: How many seconds in 10 years? 1900 05b5cc695e4d2b4197774c534
5b5cc695e4d2b4197774c534- 131523500 secfalse
- 2315360000 sectrue
- 3315423000 secfalse
- 4315354000 secfalse
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Answer : 2. "315360000 sec"
Explanation :
Answer: B) 315360000 sec Explanation: We know that, 1 year = 365 days 1 day = 24 hours 1 hour = 60 minutes 1 minute = 60 seconds. Then, 1 year = 365 x 24 x 60 x 60 seconds. = 8760 x 3600 1 year = 31536000 seconds. Hence, 10 years = 31536000 x 10 = 315360000 seconds.
Q: Find the odd man out in the following number series? 71, 88, 113, 150, 203, 277 1900 05b5cc678e4d2b4197774c158
5b5cc678e4d2b4197774c158- 1277true
- 2203false
- 3150false
- 4113false
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Answer : 1. "277"
Explanation :
Answer: A) 277 Explanation: The given number series is 71 88 113 150 203 277 +17 +25 +37 +53 +74 +8 +12 +16 +21 +4 +4 +5 The series should be 71 88 112 150 203 276 i.e, the differences of the difference should be +4. Hence, the odd man in the given series is 277.
Q: The ratio of investments of two partners A and B is 7:5 and the ratio of their profits is 7:10. If A invested the money for 5 months, find for how much time did B invest the money ? 1897 05b5cc6dfe4d2b4197774eaf3
5b5cc6dfe4d2b4197774eaf3- 111 monthsfalse
- 29 monthsfalse
- 37 monthsfalse
- 410 monthstrue
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Answer : 4. "10 months"
Explanation :
Answer: D) 10 months Explanation: 7x5: 5xk = 7:10k = 10 months
Q: 5/9 of the part of the population in a village are females. If 30 % of the females are married. The percentage of unmarried males in the total males is ? 1896 05b5cc72ce4d2b4197774f7a1
5b5cc72ce4d2b4197774f7a1- 162.5 %true
- 2125 %false
- 384.32 %false
- 446.87 %false
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Answer : 1. "62.5 %"
Explanation :
Answer: A) 62.5 % Explanation: Let total population = p number of females = 5p/9number of males =(p-5p/9) = 4p/9married females = 30% of 5p/9 = 30x5p/100x9 = p/6married males = p/6unmarried males =(4p/9-p/6) = 5p/18Percentage of unmarried males in the total males = {(5p/18)/4p/9}x100 = (5p/18)x(9/4p) = 125/2 % = 62.5%
Q: 14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together. 1895 05b5cc6b6e4d2b4197774d5b8
5b5cc6b6e4d2b4197774d5b8- 111/379false
- 221/628false
- 324/625true
- 426/247false
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Answer : 3. "24/625"
Explanation :
Answer: C) 24/625 Explanation: Total no of ways = (14 – 1)! = 13! Number of favorable ways = (12 – 1)! = 11! So, required probability = 11!×3!13! = 39916800×66227020800 = 24625

