HCL Placement Papers Bull Placement 13 ­ HCL HCL Placement Papers Question No. : 1 A computer can perform 30 identical tasks in six hours. At that rate, what is the minimum number of computers that should be assigned to complete 80 of the tasks within 3 hours? A﴿ 12 B﴿ 7 C﴿ 6 D﴿ 16 Explanation:­ 30 tasks in 6 hr. means speed of one computer is 5 task per hour. So in 3 hours, one computer will do 15 tasks. For 80 tasks to be done in 3 hours, no. of computers reqd. = 80/15 = 5.33. So, min no of computers needed are 6. Question No. : 2 Construction of a road was entrusted to a civil engineer. He was to finish the work in 124 days for which he employed 120 workmen. Two­thirds of the work was completed in 64 days. How many workmen can be reduced now without affecting the completion of the work on time? A﴿ 80 B﴿ 64 Explanation:­ We have, M D W 120 64 2/3 x 60 1/3 So here we can apply the formula ­ (M 1 D 1 H 1 /W 1 ) = (M 2 D 2 H 2 /W 2 ) 120 × 64 × 1/3 = x × 60 × 2/3 => x = 64. So workmen to be reduced is 120 ­ 64 = 56. Hence option "c" . Question No. : 3 Anu can complete a work in 10 days. Manu is 25% more efficient than Anu and Sonu is 60% more efficient than Manu. Working together, how long would they take to finish the job ? Explanation:­ Let Anu’s speed is x. Manu’s speed is 1.25x. Sonu’s speed is 2x. Combined speed is 4.25x. At speed x it takes 10 days. At speed 4.25x it will take 10/4.25 days = 40/17 days. Hence 1st option is the answer. Question No. : 4 A tradesman gives 4% discount on the marked price and 1 article free with every 15 articles bought, and still gains 35%. By what percent the marked price is more than the cost price? HCL Placement Papers A﴿ 40% B﴿ 39% C﴿ 20% Explanation:­ Let marked price be 100. After discount price is 96. He charges for only 15 articles on selling 16. So selling price becomes 96 X (15/16) = 90. CP= SP / (1.35) = 200/3. Mark Up is (MP­CP / CP) X 100 = (100 ­ 200/3)/ 200/3 = 50%. A﴿ B﴿ C﴿ 56 D﴿ 24 C﴿ D﴿ 50% D﴿ HCL Placement Papers Question No. : 5 HCL Placement Papers Even after reducing the marked price of a pen by Rs. 32, a shopkeeper makes a profit of 15%. If the cost price be Rs. 320, what percentage of profit does he make if he sells the pen at the marked price? A﴿ 25% B﴿ 20% C﴿ 10% D﴿ 30% Explanation:­ Here CP =320 and profit = 15%. => SP= 320 x 1.15 = Rs 368. Since price has benn reduced by Rs 32, so MP = 368 + 32 = Rs 400. If he sells it at Rs 400, the profit will be Rs 80 and profit % = (80/320)×100 = 25 % Question No. : 6 If 15 men or 24 women or 36 boys can do a piece of work in 12 days, working 8 hours a day, how many men must be associated with 12 women and 6 boys to do another piece of work 2.25 times in 30 days working 6 hours a day ? A﴿ 9 B﴿ 7 C﴿ 5 D﴿ 8 Explanation:­ Since men, women and children have different efficiencies, therefore we have to equate efficiencies Here 24W = 15M => 12W = 7.5M and 36B = 15M => 6B = 2.5M. Now let M men are required. So total men = M + 7.5 + 2.5 = M + 10. Hence we have M D H W 15 12 8 1 M + 10 30 6 2.25 Now using , (M 1 A﴿ 15 : 17 B﴿ 7 : 17 ) => 15 × 12 × 8 × 2.25 = (M +10) × 30 × 6 => M = 8. Hence, correct answer is D option. Question No. : 7 Two vessels contain spirit of 0.5 and 0.75 concentrations. If two litres from the first vessel and three litres from the second vessel are mixed, then what will be the ratio of the spirit and the water in the resultant solution ? Explanation:­ Spirit in 1 D 1 H 1 /W 1 ) = (M 2 D 2 H 2 /W 2 C﴿ 13 : 7 D﴿ 17 : 15 A﴿ HCL Placement Papers nd vessel = 0.75×3 = 2.25 lt. Total spirit = 3.25 lt. Total water = 5­3.25= 1.75 lt. Ratio is 3.25/1.75 = 13:7 Hence , option C Question No. : 8 A man takes 6 hours 30 min. in going by a cycle and coming back by scooter. He would have lost 2 hours 10 min. by going on cycle both ways. How long would it take him to go by scooter both ways? Explanation:­ Extra time when going by cycle is 2h 10m. total time when going by cycle will be 8h 40m. So one way cycle would take 4h 20m. It takes 2h 10m extra than scooter so scooter takes 2h 10m for one way. Both way scooter will take 4h 20m. Hence answer is 3rd option. st vessel = 0.5×2= 1 lt. Spirit in 2 B﴿ 5 hours C﴿ D﴿ HCL Placement Papers Question No. : 9 HCL Placement Papers The priest told the devotee, "The temple bell is rung at regular intervals of 45 minutes. The last bell was rung 5 minutes ago. The next bell is due to be rung at 7.45 a.m." At what time did the priest give this information to the devotee ? A﴿ 7.00 a.m B﴿ 7.05 a.m C﴿ 7.40 a.m D﴿ 6.55 a.m Explanation:­ Next bell will be rung at 7.45 so, previous bell was rung at 7.00. This was 5 min ago so. Time is now 7.00 +.05 = 7.05am Question No. : 10 A can hit a target 4 times in 5 shots, B hits 3 times in 4 shots, and C hits twice in 3 shots. They fire together. Find the probability that at least two shots hit the target. A﴿ 13/30 B﴿ 5/6 C﴿ 11/40 D﴿ None of these Explanation:­ P =4/5, Q=3/4, R=2/3 Probability that all of them to hit the target is 4/5x3/4x2/3=2/5 Probability that two of them hit the target is P(P& Q)=4/5x3/4x1/3=1/5( 1/3 means that R does not hit, ) ,prob (P&R)=4/5x2/3x1/4=2/15 Prob (Q&R)=3/4x2/3x1/5=1/10 or probability of at least 2 is 2/5+1/5+2/15+1/10=5/6 Question No. : 11 The odds that A speaks the truth are 3 : 2 and the odds that ‘B’ speaks the truth are 5 : 3. In what percent of cases are they likely to agree each other on an identical point? A﴿ 47.5% B﴿ 37.5% C﴿ 63.5% Explanation:­ Question No. : 12 Explanation:­ A﴿ 200 m B﴿ 300 m C﴿ 270 m D﴿ 160 m D﴿ None of these HCL Placement Papers A﴿ 10 litres B﴿ 20 litres HCL Placement Papers D﴿ A﴿ 46 B﴿ 4 C﴿ 2 47 A﴿ Question No. : 13 There are two identical vessels X and Y. Y is filled with water to the brim and X is empty. There are two pails A and B, such that B can hold half as much water as A. One operation is said to be executed when water is transferred from Y to X using A once and water is transferred to Y from X using B once. If A can hold a litre of water and it takes 40 operations to equate the water level in X and Y, what is the total volume of water in the system? Explanation:­ Question No. : 14 In a class with a certain number of students if one student weighing 50 kg is added then the average weight of the class increases by 1 kg. If one more student weighing 50 kg is added then the average weight of the class increases by 1.5 kg over the original average. What is the original average weight ﴾in kg﴿ of the class? Explanation:­ Let x be the original average and n be the number of students, From the first increase in the average, we get nx + 50 = (n + 1) (x + 1) => 50 – x = n + 1.....(1) From the second increase in the average, we get nx + 100 = (n + 2)(x + 1.5) =>100 – 2x = 1.5(n+2)........(2) Solving (1) and (2), we get the value of x = 47. Question No. : 15 A portion of a 30 m long tree is broken by a tornado and the top strikes the ground making an angle of 30° with the ground level. The height of the point where the tree is broken is equal to Explanation:­ B﴿ 10 m C﴿ C﴿ 40 litres D﴿ D﴿ 60 m HCL Placement Papers @ ​
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