Quantitative Aptitude Practice Question and Answer

Q: When I was married 10 years ago my wife is the 6th member of the family. Today my father died and a baby born to me.The average age of my family during my marriage is same as today. What is the age of Father when he died ? 5055 0

  • 1
    50 yrs
    Correct
    Wrong
  • 2
    60 yrs
    Correct
    Wrong
  • 3
    70 yrs
    Correct
    Wrong
  • 4
    65 yrs
    Correct
    Wrong
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Answer : 2. "60 yrs"
Explanation :

Answer: B) 60 yrs Explanation: Let the Father be x years when he died Average Age 10 years ago be A Total Age 10 years ago = 6*A Total Age after 10 years(Just before father's Death) = 6A + 6*10 = 6A + 60 Father Died and Baby was born => the Total number of people in the family is Same (6) Baby born today so age of baby = 0 (6A +60 - x)/6 = 6A/6 => A + 10 -(x/6) = A => x/6 = 10 => x = 60 Therefore we can conclude that the father was 60 years old when he died.

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Answer : 4. "Rs. 15000"
Explanation :

Answer: D) Rs. 15000 Explanation: It won't requires that much solving. Here given the increase in weekly collection by Rs. 1,05,000 That is this increase denotes for all 7 days. => Gross Collection increase per day = Rs. 1,05,0007 = Rs. 15,000 Hence, the gross collection increase per day = Rs. 15,000.

Q: The product of a number and its multiplicative inverse is 4949 0

  • 1
    1
    Correct
    Wrong
  • 2
    0
    Correct
    Wrong
  • 3
    -1
    Correct
    Wrong
  • 4
    Infinity
    Correct
    Wrong
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Answer : 1. "1"
Explanation :

Answer: A) 1 Explanation: The mutiplicative inverse of a number is nothing but a reciprocal of a number. Now, the product of a number and its multiplicative inverse is always equal to 1.   For example : Let the number be 15 Multiplicative inverse of 15 = 1/15 The product of a number and its multiplicative inverse is = 15 x 1/15 = 1.

Q: Ratan can complete a work in \(5 {1 \over 2}\) days while Rupesh can complete it in \(10 {1 \over 2}\) days. In how many days can they both complete it working on it together? 4946 2

  • 1
    \(3 {39 \over 64 }\)
    Correct
    Wrong
  • 2
    \(4 {39 \over 64 }\)
    Correct
    Wrong
  • 3
    \(5 {39 \over 64 }\)
    Correct
    Wrong
  • 4
    \(6 {39 \over 64 }\)
    Correct
    Wrong
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Answer : 1. "\(3 {39 \over 64 }\) "

Q: In a class of 39 students the ratio of boys and girls is 2 : 1. Radhika ranks 15th among all the students from top and 8th among girls from bottom. How many boys are there below Radhika ? 4923 0

  • 1
    9
    Correct
    Wrong
  • 2
    17
    Correct
    Wrong
  • 3
    11
    Correct
    Wrong
  • 4
    15
    Correct
    Wrong
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Answer : 2. "17"
Explanation :

Answer: B) 17 Explanation: Total student= 39, boys : girls = 2:1, hence no of boys= 2/3(39)= 26num of girls= 1/3(39) = 13 , since radhika ranked 15 among 39 student from topand ranked 8 among girls from bottom , this means 7 girls are below radhika, rest 13-9= 5 girls are above her, now from 14 toper 5 are girls hence 14-5=9 boys.Hence num of boys below radhika are 26-9=17

Q: Ramesh and Dinesh can complete a work in 15 and 20 days respectively. Ramesh started working but after 5 days Dinesh replaced Ramesh. Find the share of Dinesh in Rs 15000 4915 0

  • 1
    Rs 5000
    Correct
    Wrong
  • 2
    Rs 9000
    Correct
    Wrong
  • 3
    Rs 10000
    Correct
    Wrong
  • 4
    Rs 8000
    Correct
    Wrong
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Answer : 3. "Rs 10000"

Q: On selling oranges 5 for Rs.1. One gets 20% profit. How many orange were purchased for Rs.1. 4903 0

  • 1
    3
    Correct
    Wrong
  • 2
    7
    Correct
    Wrong
  • 3
    4
    Correct
    Wrong
  • 4
    6
    Correct
    Wrong
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Answer : 4. "6"

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Answer : 2. "6 mile/hr"
Explanation :

Answer: B) 6 mile/hr Explanation: Let the speed of river and boat be 'r' m/min and 'b' m/min.so relative speed in upstream (b-r)m/min and in downstream (b+r)m/min.Now in upstream distnce covered in 5 min is 5(b-r)milesso total distnce covered => 1 + 5(b-r)miles in upstreamIn downstream distance covered in 5min is 5(b+r)miles Now 1 + 5(b-r) = 5(b+r)1+5b-5r = 5b+5r1 = 10rr = 1/10 mile/min => 6 mile/hr.

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