Quantitative Aptitude Practice Question and Answer

Q: If a girl cycles at 10 kmph, then she arrives at a certain place at 1 p.m. If she cycles at 15 kmph, she will arrive at the same place at 11 a.m. At what speed must she cycle to get there at noon? 1901 0

  • 1
    14 kmph
    Correct
    Wrong
  • 2
    13 kmph
    Correct
    Wrong
  • 3
    12 kmph
    Correct
    Wrong
  • 4
    11 kmph
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 3. "12 kmph"
Explanation :

Answer: C) 12 kmph Explanation: The distance is constant in this case. Let the time taken for travel with a speed of 10 kmph be 't'. Now the speed of 15 kmph is 3/2 times the speed of 10 kmph. Therefore, time taken with the speed of 15 kmph will be 2t/3 (speed is inversely proportional to time) Extra time taken = t - 2t/3 = t/3 => 1pm - 11am = 2hrs => t/3 = 2h => t = 6 hrs. Now, Distance = speed x time = 10 x 6 = 60 kms Time he takes to reach at noon = 6 - 1 = 5 hrs Now, Speed = 60/5 = 12 kmph.

Q: What is the sum of all 3 digits number that can be formed using digits 0,1,2,3,4,5 with no repitition ? 1901 0

  • 1
    28450
    Correct
    Wrong
  • 2
    26340
    Correct
    Wrong
  • 3
    32640
    Correct
    Wrong
  • 4
    36450
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 3. "32640"
Explanation :

Answer: C) 32640 Explanation: We know that zero can't be in hundreds place. But let's assume that our number could start with zero.   The formula to find sum of all numbers in a permutation is   111 x no of ways numbers can be formed for a number at given position x sum of all given digits   No of 1 s depends on number of digits   So,the answer us   111 x 20 x (0+1+2+3+4+5) = 33300   We got 20 as follows. If we have 0 in units place we can form a number in 4*5 ways. This is for all numbers. So we have substituted 20 in formula.   Now, this is not the final answer because we have included 0 in hundreds place. so we have to remove the sum of all numbers that starts with 0.   This is nothing but the sum of all 2 digits numbers formed by 1 2 3 4 5. Because 0 at first place makes it a 2 digit number.   So the sum for this is 11 x 4 x (1+2+3+4+5). =660   Hope u understood why we use 4. Each number can be formed in 4x1 ways   So, the final answer is 33300-660 = 32640

Q: There are two numbers such that the sum of twice the first number and thrice the second number is 300 and the sum of thrice the first number and twice the second number is 265. Which is the largest number ? 1901 0

  • 1
    24
    Correct
    Wrong
  • 2
    39
    Correct
    Wrong
  • 3
    85
    Correct
    Wrong
  • 4
    74
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 4. "74"
Explanation :

Answer: D) 74 Explanation:

Q: The number of ways that 7 teachers and 6 students can sit around a table so that no two students are together is 1899 0

  • 1
    7! x 7!
    Correct
    Wrong
  • 2
    7! x 6!
    Correct
    Wrong
  • 3
    6! x 6!
    Correct
    Wrong
  • 4
    7! x 5!
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 2. "7! x 6!"
Explanation :

Answer: B) 7! x 6! Explanation: The students should sit in between two teachers. There are 7 gaps in between teachers when they sit in a roundtable. This can be done in 7P6ways. 7 teachers can sit in (7-1)! ways.    Required no.of ways is = 7P6.6! = 7!.6!

Q: A laborer is engaged for 30 days on the condition that he receives Rs.25 for each day he works and is fined Rs.7.50 for each day is absent. He gets Rs.425 in all. For how many days was he absent ? 1897 0

  • 1
    10 days
    Correct
    Wrong
  • 2
    14 days
    Correct
    Wrong
  • 3
    11 days
    Correct
    Wrong
  • 4
    9 days
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 1. "10 days"
Explanation :

Answer: A) 10 days Explanation: For 30 days he receives 30 x 25 = 750                Actually he gets 425 Therefore, Fine is 750 - 425 =  325   => No of days absent = 325/(25+7.50)= 10

Q: If x = 2 + 223 + 213 then find the value of x3-6x2+6x ? 1894 0

  • 1
    1
    Correct
    Wrong
  • 2
    2
    Correct
    Wrong
  • 3
    3
    Correct
    Wrong
  • 4
    4
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 2. "2"
Explanation :

Answer: B) 2 Explanation: Given that x = 2+ 223 + 213   So x - 2 = 223 + 213   Now cubing on both sides, we get     => x3-8-6x2+12x = 4 + 2 + 6(x-2)   Therefore, x3-6x2+6x = -6 + 8 = 2

Q: A woman fills a bucket in 6 minutes. 1845 buckets have to be filled from 8 am. to 9:30 am. How many woman employees should be employed for this task? 1892 0

  • 1
    111
    Correct
    Wrong
  • 2
    117
    Correct
    Wrong
  • 3
    123
    Correct
    Wrong
  • 4
    139
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 3. "123"
Explanation :

Answer: C) 123 Explanation: We know that, M1 × D1W1 = M2 × D2W2 Here M1 = 1, D1 = 6 min, W1 = 1 and M2 = M, D2 = 90 min, W2 = 1845   1 × 61 = M × 901845  => M = 123

Q: 12 men complete a work in 9 days. After they have worked for 6 days, 4 more men join them. How many days will they take to complete the remaining work ? 1892 0

  • 1
    2 days
    Correct
    Wrong
  • 2
    2.5 days
    Correct
    Wrong
  • 3
    2.25 days
    Correct
    Wrong
  • 4
    3 days
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 3. "2.25 days"
Explanation :

Answer: C) 2.25 days Explanation: 1 man's 1 day work = 1/108 12 men's 6 day's work = 1/9 x 6 = 2/3 Remaining work = 1 - 2/3 = 1/3 16 men's 1 day work = 1/108 x 16 = 4/274/27 work is done by them in 1 day. 1/3 work is done by them in 27/4 x 1/3 = 9/4 days.

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully