Quantitative Aptitude प्रश्न और उत्तर का अभ्यास करें
8प्र: It takes one minute to fill 2/7 th of a vessel. What is the time taken in minutes to fill the whole of the vessel ? 1884 05b5cc6d8e4d2b4197774e7fd
5b5cc6d8e4d2b4197774e7fd- 11/7 minfalse
- 27/2 mintrue
- 35/7 minfalse
- 47/5 minfalse
- उत्तर देखेंउत्तर छिपाएं
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उत्तर : 2. "7/2 min"
व्याख्या :
Answer: B) 7/2 min Explanation: Time taken to fill 2/7 = 1Then to fill full 1 = ?? = 1/(2/7) = 7/2 minutes.
प्र: Find the Odd One Out? 3, 4, 14, 48, 576, 27648 1884 05b5cc6b0e4d2b4197774d2c6
5b5cc6b0e4d2b4197774d2c6- 14false
- 214true
- 348false
- 427648false
- उत्तर देखेंउत्तर छिपाएं
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उत्तर : 2. "14"
व्याख्या :
Answer: B) 14 Explanation: Given Number series is : 3 4 14 48 576 27648 It follows a pattern that, 3 4 4 x 3 = 12 but not 14 12 x 4 = 48 48 x 12 = 576 576 x 48 = 27648 Hence, the odd one in the given Number Series is 14.
प्र: On a school’s Annual day sweets were to be equally distributed amongst 112 children. But on that particular day, 32 children were absent. Thus the remaining children got 6 extra sweets. How many sweets was each child originally supposed to get ? 1884 05b5cc741e4d2b4197774f9a4
5b5cc741e4d2b4197774f9a4- 114false
- 216false
- 312false
- 415true
- उत्तर देखेंउत्तर छिपाएं
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उत्तर : 4. "15"
व्याख्या :
Answer: D) 15 Explanation: Let 'K' be the total number of sweets.Given total number of students = 112If sweets are distributed among 112 children,Let number of sweets each student gets = 'L' => K/112 = L ....(1)But on that day students absent = 32 => remaining = 112 - 32 = 80Then, each student gets '6' sweets extra. => K/80 = L + 6 ....(2)from (1) K = 112L substitute in (2), we get112L = 80L + 48032L = 480L = 15 Therefore, 15 sweets were each student originally supposed to get.
प्र: H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 and 16 x 81 x 5 x 11 x 49 is : 1883 05b5cc6e7e4d2b4197774eda6
5b5cc6e7e4d2b4197774eda6- 1360false
- 2180true
- 390false
- 4120false
- उत्तर देखेंउत्तर छिपाएं
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उत्तर : 2. "180"
व्याख्या :
Answer: B) 180 Explanation: 4 x 27 x 3125 = 22 × 33 × 55 ; 8 x 9 x 25 x 7 = 23 × 32 × 52 × 7 16 x 81 x 5 x 11 x 49 = 24 × 34 × 5 × 11 × 72 H.C.F = 22 × 32 × 5 = 180.
प्र: 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? 1881 05b5cc698e4d2b4197774c61d
5b5cc698e4d2b4197774c61d- 136 daystrue
- 230 daysfalse
- 324 daysfalse
- 422 daysfalse
- उत्तर देखेंउत्तर छिपाएं
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उत्तर : 1. "36 days"
व्याख्या :
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.
प्र: In a certain Business, the profit is 220% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit ? 1881 05b5cc767e4d2b4197774fe45
5b5cc767e4d2b4197774fe45- 161%true
- 275%false
- 355%false
- 481%false
- उत्तर देखेंउत्तर छिपाएं
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उत्तर : 1. "61%"
व्याख्या :
Answer: A) 61% Explanation: Let C.P.= Rs. 100. Then, Profit = Rs.220, S.P. = Rs.320. New C.P. = 125% of Rs. 100 = Rs. 125 New S.P. = Rs.320. Profit = Rs. (320 - 125) = Rs. 195 Required percentage = 195320×100== 60.9 =~ 61%
प्र: Find the odd one out of the series ? 35, 48, 54, 63, 81, 105 1880 05b5cc6dae4d2b4197774e8c3
5b5cc6dae4d2b4197774e8c3- 163false
- 248false
- 335true
- 481false
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उत्तर : 3. "35"
व्याख्या :
Answer: C) 35 Explanation: 48, 54, 63, 81 and 105 are divisible by 3 but not 35.
प्र: Mr. Karthik drives to work at an average speed of 48 km/hr. Time taken to cover the first 60% of the distance is 20 minutes more than the time taken to cover the remaining distance. Then how far is his office ? 1879 05b5cc777e4d2b419777500b6
5b5cc777e4d2b419777500b6- 140 kmfalse
- 250 kmfalse
- 370 kmfalse
- 480 kmtrue
- उत्तर देखेंउत्तर छिपाएं
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उत्तर : 4. "80 km"
व्याख्या :
Answer: D) 80 km Explanation: Let the total distance be 'x' km. Time taken to cover remaining 40% of x distance is t1=40×x100×48 But given time taken to cover first 60% of x distance is t2=t1+2060hrs t2=60×x100×48 ∴60×x100×48= 40×x100×48+2060 x=80 km.

