Quantitative Aptitude प्रश्न और उत्तर का अभ्यास करें

प्र: 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  ? 1643 0

  • 1
    1/7 min
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  • 2
    7/2 min
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  • 3
    5/7 min
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  • 4
    7/5 min
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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: 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 ? 1642 0

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    40 km
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  • 2
    50 km
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  • 3
    70 km
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  • 4
    80 km
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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.

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उत्तर : 4. "8:3"
व्याख्या :

Answer: D) 8:3 Explanation: Let the speed of the boat upstream be p kmph and that of downstream be q kmph   Time for upstream = 8 hrs 48 min = 845hrs   Time for downstream = 4 hrs   Distance in both the cases is same.   => p x  845= q x 4   => 44p/5 = 4q   => q = 11p/5    Now, the required ratio of Speed of boat : Speed of water current   = q+p2:q-p2   => (11p/5 + p)/2  : (11p/5 - p)/2   => 8 : 3

प्र: A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions made by the larger wheel is ? 1640 0

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    11
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  • 2
    7
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  • 3
    9
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  • 4
    13
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उत्तर : 3. "9"
व्याख्या :

Answer: C) 9 Explanation: Here the number of cogs is inveresly proportional to number of revolutions => More cogs less revolution=> 6 : 14 :: x : 21=> (21x6)/14 = 9

प्र: How many numbers up to 101 and 300 are divisible by 11  ? 1639 0

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    18
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  • 2
    20
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  • 3
    19
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  • 4
    17
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उत्तर : 1. "18"
व्याख्या :

Answer: A) 18 Explanation: (300 – 101)/11 = 199/11 = 18 1/1118 Numbers.

प्र: 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 ? 1635 1

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    14.28 % profit
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  • 2
    24.18 % profit
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  • 3
    14.28 % loss
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  • 4
    24.18 % loss
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उत्तर : 1. "14.28 % profit"
व्याख्या :

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.

प्र: There are 41 students in a class, number of girls is one more than number of guys. we need to form a team of four students. all four in the team cannot be from same gender. number of girls and guys in the team should NOT be equal. How many ways can such a team be made ? 1633 0

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    49450
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  • 2
    50540
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  • 3
    46587
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  • 4
    52487
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उत्तर : 2. "50540"
व्याख्या :

Answer: B) 50540 Explanation: Given G + B= 41 and B = G-1 Hence, G = 21 and B = 20Now we have 2 options, 1G and 3M (or) 3G and 1M(2G and 2M or 0G and 4M or 4G and oM are not allowed),   Total : C121×C320+C321×C120= 50540 ways.

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उत्तर : 2. "100"
व्याख्या :

Answer: B) 100 Explanation: Let the Efficiency of pavan be E(P) Let the Efficiency of sravan be E(S) Here Work W = LCM(25,20) = 100 Now, E(P+2S) = 100/25 = 4 ....(1) E(2P+S) = 100/20 = 5 ....(2) Hence, from (1) & (2) we get E(S) = 1 => Number of days Savan alone work to complete the work  = 100/1 = 100 days.

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