Aptitude Time and Work Test Paper 836) A contractor undertakes to complete a work in 120 days with the help of 100 workers. After 45 days, he finds that only ¼ of the work has been completed. To complete the remaining work, how many extra workers will be required?
Answer: A Explanation: Apply formula: [Man1 * day1* hour1*]/ work1 = [Man2 * day2* hour2*]/ work2 ATQ, target = 120 days M1 = 100, M2 =? Now, apply the formula [M1 * D1 * H1]/ W1 = [M2 * D2 * H2]/ W2 Or, (100 * 45 * 3/4) = (M2 * 75 * ¼) 37) A and B individually can finish a job in 20 and 30 days respectively. They start working together, and B leaves the job 5 days before completion, now the work will be finished in how many days?
Answer: B Explanation: Note: Assume the total work = LCM of the given days Take the LCM of days = LCM of (20 and 30) = 60 A's one day work = 60/20 = 3 unit If (A+B) work, then they can finish the work in 60/5 = 12 days. But ATQ, B left the job 5 days before Now, the total work = 60+10 = 70units So, to complete the work it requires = 70/5 = 14 days. 38) A, B, and C can finish a work in 18, 27, and 36 days respectively. All three start working together. A leaves the job after 8 days, and B leaves the job 6 days before it is finished, then the work will be finished in how many days?
Answer: A Explanation: Note: Assume the total work = LCM of the given days Take the LCM of days = LCM of (18, 27, and 36) = 108 A's one day work = 108/18 = 6 units ATQ, A works only for 8 days. If (B+C) work till last, they can finish the work in 60/7= 8.57 days. But ATQ, B left the job 6 days before it is completed. Now, the total remaining work = 60+24 = 84units. (B+C)'s one day efficiency = 7 So, to complete the work it requires = 84/7 = 12 days. 39) A and B together can finish a work in 4 days. If A reduces his efficiency by 30%, and B increases his efficiency by 10 %, the work is finished in 5 days. If A works for his original efficiency and B works for 1[1/2] times of his efficiency, they together can finish the job in how many days?
Answer: A Explanation: ATQ, (A+B) can complete a work in 4 days......................... (i) The work will be the same. Total work = man* days So, equation 1 = equation 2 (A+B)*4 = (0.7 A + 1.1 B)* 5 That means the efficiency ratio of A and B = 3: 1 ATQ, the efficiency of A remains same and B's efficiency gets 1[1/2] of the original efficiency. Now, (A+B)'s one day work = 3+ 3/2 = 9/2 So, days requires = Total work/ efficiency of (A+B) 40) A hostel has sufficient food for 200 students for 50 days. After 10 days, 50 more students join the hostel. Now, the food will continue for how many days?
Answer: B Explanation: ATQ, M = 200, D = 50 Apply the formula: [M1 * D1 * H1]/ W1 = [M2 *D2 * H2]/ W2 But here W1, W2, H1, and H2 are not present so assume these to 1. So, M1 * D1 = M2 * D2 After 10 days, 200 students can eat for 40 days, but on the other hand, 50 more students join the hostel. So, 250 men can eat it for D2 days So, the food will continue for 32 days. Aptitude Time and Work Test Paper 1 Aptitude Time and Work Test Paper 2 Aptitude Time and Work Test Paper 3 Aptitude Time and Work Test Paper 4 Aptitude Time and Work Test Paper 5 Aptitude Time and Work Test Paper 6 Aptitude Time and Work Test Paper 7 Aptitude Time and Work Test Paper 9 Time and Work Concepts
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