Quantitative Aptitude Practice Question and Answer

Q: 5 years ago Sushma was 5 times as old as her Son. 5 years hence her age will be 8 less than three times the corresponding age of her Son. Find their ages ? 4401 0

  • 1
    24 and 13 years
    Correct
    Wrong
  • 2
    48 and 24 years
    Correct
    Wrong
  • 3
    35 and 11 years
    Correct
    Wrong
  • 4
    33 and 15 years
    Correct
    Wrong
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Answer : 3. "35 and 11 years"
Explanation :

Answer: C) 35 and 11 years Explanation: Let the age of sushma be x and the age of her son is yThen five years before x-5=5(y-5) ...(1)Five years hence x+5 = 3(y+5)-8 .....(2) By soving (1) & (2), we get5y - 15 = 3y + 7y = 11 => x = 35 Therefore, the age of Sushma = 35 and her son = 11.

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Answer : 3. "11 बजे सुबह"

Q: In an election only two candidates contested. A candidate secured 70% of the valid votes and won by a majority of 172 votes. Find the total number of valid votes  ? 4391 0

  • 1
    446
    Correct
    Wrong
  • 2
    415
    Correct
    Wrong
  • 3
    404
    Correct
    Wrong
  • 4
    430
    Correct
    Wrong
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Answer : 4. "430"
Explanation :

Answer: D) 430 Explanation: Let the total number of valid votes be 'V'. 70% of V = 70/100 x V = 7V/10 Number of votes secured by the other candidate = V - 7V/100 = 3V/10 Given, 7V/10 - 3V/10 = 172 => 4V/10 = 172 => 4V = 1720 => V = 430.

Q: If there are 150 questions in a 3 hr examination. Among these questions 50 are type A problems, which requires twice as much as time be spent on the rest of the type B problems. How many minutes should be spent on type A problems ? 4376 0

  • 1
    75 min
    Correct
    Wrong
  • 2
    82 min
    Correct
    Wrong
  • 3
    90 min
    Correct
    Wrong
  • 4
    101 min
    Correct
    Wrong
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Answer : 3. "90 min"
Explanation :

Answer: C) 90 min Explanation: Let X = Time taken for each of Type B Problems(100 Problems)And 2X = Time taken for each of Type A problems(50 problems) Total time period = 3hrs = 3 x 60min = 180 minutes 100X + 50(2X) = 180100X + 100X = 180200X = 180 min X = 180/200X = 0.90 min By convertiing into seconds, X = 0.90 x 60 secondsX = 54 sec So, time taken for Part A problems is = 54 x 2 x 50 = 5400 seconds= 5400/60sec = 90 minutes.

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Answer : 1. "\({40\over3}\)"

Q: 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 ? 4347 0

  • 1
    1080 cm
    Correct
    Wrong
  • 2
    180 m
    Correct
    Wrong
  • 3
    108 m
    Correct
    Wrong
  • 4
    118 cm
    Correct
    Wrong
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Answer : 3. "108 m"
Explanation :

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.

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Answer : 2. "30 min"
Explanation :

Answer: B) 30 min Explanation: Distance travelled by Ramu = 45 x 4 = 180 km Somu travelled the same distance in 6 hours. His speed = 180/6 = 30 km/hr Hence in the conditional case, Ramu's speed = 45 - 9 = 36 km/hr and Somu's speed = 30 + 10 = 40km/hr. Therefore travel time of Ramu and Somu would be 5 hours and 4.5 hours respectively. Hence difference in the time taken = 0.5 hours = 30 minutes.

Q: Ten years ago, Karan was thrice as old as Shiva was but 10 years hence, he will be only twice as old. Find Shiva’s present age ? 4314 0

  • 1
    70 years
    Correct
    Wrong
  • 2
    30 years
    Correct
    Wrong
  • 3
    60 years
    Correct
    Wrong
  • 4
    40 years
    Correct
    Wrong
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Answer : 2. "30 years"
Explanation :

Answer: B) 30 years Explanation: Let Karan’s present age be x years and Shiva’s present age be y years.Then, according to the first condition,x - 10 = 3(y - 10) or x - 3y = - 20 ..(1) Now, Karan’s age after 10 years = (x + 10) yearsShiva’s age after 10 years = (y + 10)(x + 10) = 2 (y + 10) or x - 2y = 10 ..(2) Solving (1) and (2), we get x = 70 and y = 30Karan’s age = 70 years and Shiva’s age = 30 years.

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