Quantitative Aptitude प्रश्न और उत्तर का अभ्यास करें

प्र: At a certain rate of interest the compound interest of 3 years and simple interest of 5 years for certain sum of money is respectively Rs. 1513.2 and Rs. 2400. Find the common rate of interest per annum ? 2828 0

  • 1
    5%
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  • 2
    6%
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  • 3
    4%
    सही
    गलत
  • 4
    3%
    सही
    गलत
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उत्तर : 1. "5%"
व्याख्या :

Answer: A) 5% Explanation: Given compound interest for 3 years = Rs. 1513.2   and simple interest for 5 years = Rs. 2400   Now, we know that  C.I = P1+R100n - 1   => 1513.2 = P1+R1003 - 1 ...........(A)   And S.I = PTR/100   => 2400 = P5R/100 ..................(B)   By solving (A) & (B), we get   R = 5%.

प्र: A vertical toy 18 cm long casts a shadow 8 cm long on the ground. At the same time a pole casts a shadow 48 m. long on the ground. Then find the height of the pole ? 3860 0

  • 1
    1080 cm
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  • 2
    180 m
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  • 3
    108 m
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  • 4
    118 cm
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उत्तर : 3. "108 m"
व्याख्या :

Answer: C) 108 m Explanation: We know the rule that,   At particular time for all object , ratio of height and shadow are same.   Let the height of the pole be 'H'   Then  188=H48   => H = 108 m.

प्र: Raju sold an article by giving a discount of 8% for Rs. 17,940 and earn profit of 19.6 %. If he did not give discount then how much profit percentage he gets ? 1535 0

  • 1
    27%
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  • 2
    30%
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  • 3
    32%
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  • 4
    25%
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उत्तर : 2. "30%"
व्याख्या :

Answer: B) 30% Explanation: Let the M.R.P of an article be 100 %   Selling price of article =Rs. 17940   If he give a discount of 8 % then S.P = 92 % 92 % = 17490 100 % = 19500   Now C.P of the article is    17490 x 100/119.6 = 15000   If he did not give the discount, then    Required profit percentage =

प्र: The length of two superfast trains are 140 mts and 160 mts respectively. If they run at the speed of 60 km/hr and 80 km/hr respectively in opposite direction, find the time in which they will cross each other ? 1492 0

  • 1
    7.71 sec
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  • 2
    10.48 sec
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  • 3
    9.36 sec
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  • 4
    8.45 sec
    सही
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उत्तर : 1. "7.71 sec"
व्याख्या :

Answer: A) 7.71 sec Explanation: Given L1 = 140 m L2 = 160 m S1 = 60 km/hr S2 = 80 km/hr From the question we get, S1 + S2 = (L1 + L2) / T => (60 + 80) 5/18 m/s = 140 + 160/T => T = 54/7 = 7.71 sec

प्र: The average of marks in 3 subjects is 224. The first subject marks is twice the second and the second subject marks is twice the third. Find the second subject marks ? 1337 0

  • 1
    384
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  • 2
    96
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  • 3
    192
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  • 4
    206
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उत्तर : 3. "192"
व्याख्या :

Answer: C) 192 Explanation: Let the third subject marks be 'x'=> Second subject marks = 2x=> Third subject marks = 4xGiven avg = 224x + 2x + 4x = 224 x 3=> 7x = 224 x 3=> x = 96Hence, Second subject marks = 2x = 2 x 96 = 192.

प्र: A person can row 8 1/2 km an hour in still water and he finds that it takes him twice as long to row up as to row down the river. The speed of the stream ? 917 0

  • 1
    1.78 kmph
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  • 2
    2.35 kmph
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  • 3
    2.83 kmph
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  • 4
    3.15 kmph
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उत्तर : 3. "2.83 kmph"
व्याख्या :

Answer: C) 2.83 kmph Explanation: Given speed of the person = 8 1/2 = 17/2 kmphLet the speed of the stream = x kmphspeed of upstream = 17/2 - xspeed of downstream = 17/2 + xBut given that,2(17/2 - x) = 17/2 + x=> 3x = 17/2=> x = 2.83 kmph.

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उत्तर : 4. "45 kg"
व्याख्या :

Answer: D) 45 kg Explanation: Let the ratio of initial quantity of oils be 'x' => 4x, 5x & 8x.Let k be the quantity of third variety of oil in the final mixture.Let the ratio of initial quantity of oils be 'y'From given details,4x + 5 = 5y ..... (1)5x + 10 = 7y .....(2)8x + k = 9y ......(3) By solving (1) & (2), we getx = 5 & y = 5 From (3) => k = 5 Therefore, quantity of third variety of oil was 9y = 9(5) = 45kg.

प्र: A contractor undertake to finish a certain work in 124 days and employed 120 men. After 64 days, he found that he had already done 2/3 of the work. How many men can be discharged so that the work may finish in time ? 1274 0

  • 1
    64
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  • 2
    62
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  • 3
    58
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  • 4
    56
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उत्तर : 4. "56"
व्याख्या :

Answer: D) 56 Explanation: Days remaining 124 – 64 = 60 days Remaining work = 1 - 2/3 = 1/3 Let men required men for working remaining days be 'm' So men required = (120 x 64)/2 = (m x 60)/1 => m = 64 Men discharge = 120 – 64 = 56 men.

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