Quantitative Aptitude Practice Question and Answer

Q: If A1, A2, A3, A4, ..... A10 are speakers for a meeting and A1 always speaks after, A2 then the number of ways they can speak in the meeting is 1945 0

  • 1
    9!
    Correct
    Wrong
  • 2
    9!/2
    Correct
    Wrong
  • 3
    10!
    Correct
    Wrong
  • 4
    10!/2
    Correct
    Wrong
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Answer : 4. "10!/2"
Explanation :

Answer: D) 10!/2 Explanation: As A1 speaks always after A2, they can speak only in  1st  to 9th places and    A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place    A2 can speak in 9 places the remaining     A3, A4, A5,...A10  has no restriction. So, they can speak in 9.8! ways. i.e   when A2 speaks in the first place, the number of ways they can speak is 9.8!.   When A2 speaks in second place, the number of ways they can speak is  8.8!.   When A2 speaks in third place, the number of ways they can speak is  7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!       Therefore,Total Number of ways they can  speak = (9+8+7+6+5+4+3+2+1) 8! = 92(9+1)8! = 10!/2

Q: If three eighth of a number is 141. What will be the approximately value of 32.08% of this number ? 1942 0

  • 1
    101
    Correct
    Wrong
  • 2
    112
    Correct
    Wrong
  • 3
    104
    Correct
    Wrong
  • 4
    120
    Correct
    Wrong
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Answer : 4. "120"
Explanation :

Answer: D) 120 Explanation: K x 3/8 = 141 => K = 376 376 x 32.08/100 = 120

Q: Write 0.125 as a percent %? 1942 0

  • 1
    12%
    Correct
    Wrong
  • 2
    12.5%
    Correct
    Wrong
  • 3
    13%
    Correct
    Wrong
  • 4
    13.5%
    Correct
    Wrong
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Answer : 2. "12.5%"
Explanation :

Answer: B) 12.5% Explanation: Here to convert 0.125 into % First convert to fractio Multiply and divide by 1000 => 0.125 x 10001000 = 1251000 = 18 => Now, in percentage18 x 100 = 12.5%

Q: A wholesale dealer allows a discount of 20 % on the marked price to the retailer. The retailer sells at 5% below the marked price. If the customer pays Rs.19 for an article, what profit is made by the retailer on it ? 1942 0

  • 1
    Rs. 4
    Correct
    Wrong
  • 2
    Rs. 3
    Correct
    Wrong
  • 3
    Rs. 5
    Correct
    Wrong
  • 4
    Rs. 1
    Correct
    Wrong
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Answer : 2. "Rs. 3"
Explanation :

Answer: B) Rs. 3 Explanation: Let wholesaler dealers marked price = 100%, Retailer's C.P = 80% And the retailer sells at 5% less than the marked price => S.P = 95%If S.P of 95% of the retailer costs Rs.19 to customer,so its C.P of 80% will cost 80 x 19/95 = 16 Profit made by the retailer = 19-16 =  Rs.3

Q: How many iron rods, each of length 7 mts and diameter 2 cms can be made out of 0.88 cubic metre of iron ? 1940 0

  • 1
    500
    Correct
    Wrong
  • 2
    600
    Correct
    Wrong
  • 3
    400
    Correct
    Wrong
  • 4
    300
    Correct
    Wrong
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Answer : 3. "400"
Explanation :

Answer: C) 400 Explanation: Given length of iron rod =7 mts and  and diameter of rod = 2 cms  => r = 1/100 mts Therefore,  Volume of one rod = πr2h cu. m =  227×11002×7cu. m = 115000 cu. m Given volume of iron = 0.88 cu. m  Therefore, Number of rods = 0.88×500011 = 400.

Q: There are four prime numbers written in ascending order of magnitude. The product of the first three is 385 and of the last three is 1001. Find the fourth number ? 1939 0

  • 1
    17
    Correct
    Wrong
  • 2
    19
    Correct
    Wrong
  • 3
    13
    Correct
    Wrong
  • 4
    11
    Correct
    Wrong
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Answer : 3. "13"
Explanation :

Answer: C) 13 Explanation: (385/1001) = 5/13 First Number is 5 => Fourth number is 13

Q: The probability that a number selected at random from the first 100 natural numbers is a composite number is  ? 1937 0

  • 1
    3/2
    Correct
    Wrong
  • 2
    2/3
    Correct
    Wrong
  • 3
    1/2
    Correct
    Wrong
  • 4
    34/7
    Correct
    Wrong
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Answer : 1. "3/2"
Explanation :

Answer: A) 3/2 Explanation: The number of exhaustive events = 100 C₁ = 100. We have 25 primes from 1 to 100. Number of favourable cases are 75. Required probability = 75/50 = 3/2.

Q: 640 ml of a mixture contains milk and water in ratio 6:2. How much of the water is to be added to get a new mixture containing half milk and half water ? 1936 0

  • 1
    360 ml
    Correct
    Wrong
  • 2
    320 ml
    Correct
    Wrong
  • 3
    310 ml
    Correct
    Wrong
  • 4
    330 ml
    Correct
    Wrong
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Answer : 2. "320 ml"
Explanation :

Answer: B) 320 ml Explanation: Here total parts of milk and water in the solution is 6+2 = 8 parts. 1part = 640/8 = 80   old mixture contains 6parts of milk and 2 parts of water(6:2).   To get new mixture containing half milk and half water, add 4parts of water to the old mixture then 6:(2+4) to make the ratio same. i.e, 4 x 80 = 320 ml.

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