Abhishek starts to paint a fence on one day. On the second day, two more friend of Abhishek join him. On the third day 3 more friends of him join him and so on. If the fence is completely painted this way in exactly 20 days, then find the number of days in which 10 girls painting together can paint the fence completely, given that every girl can paint twice as fast as Abhishek and his friends(Boys)?(Assume that the friends of Abhishek are all boys).
Answer: D The number of boys-days => 1/2[1+ (1+2)+(1+2+3)+.,.................+(1+2+3+....+20)]= 1440 => But , each boy =1/2 girls=> 770 girl-days. 10 girls will take 770/10 = 77 days.
Q. No. 44:
A man takes 20 days to reach the point B under normal circumstances. But, due to the increasingly hostile weather conditions the distance they travel every day reduces by 20%. In how many days would the man reach the point B, taking into consideration weather conditions?
Answer: D Let the total distance to be covered be d On each day under normal weather conditions they travel 'd/20' of the distance. On 1st day they would travel = d/20 On 2nd day they would travel = 0.8 * d/20 On 3rd day they would travel = 0.8* (0.8*d/20) Let he will reach the point B in nth day => d/20 + 0.8d/20 + 0.82d/20+...................+0.8nd/20 = d => d/20* (-(0.8)n+1)/(-0.8+1) = d => 1- (0.8)n = 4(0.8)n = -3 Since (0.8)n >0, thus it is never equal to -3 The man will never reach the point B.
Q. No. 45:
A work is done by 30 workers not all of them have the same capacity to work. Every day exactly 2 workers, do the work with no pair of workers working together twice. Even after all possible pairs have worked once, all the workers together works for two more days to finish the work. Find the number of days in which all the workers together will finish the whole work?
Answer: A 30 workers work in pairs, with no same pair of workers working together. Each worker will be working with other 29 which means each workers will work for 29 days in pair. Let the time taken by each worker be W1, W2, W3..........W30 respectively According to Question {work done when the workers work in pairs}+{work done when all the woorkers work together for two days}= 1 29[1/W1 + 1/W2 + 1/W3. ....1/W30] + 2[1/W1+ 1/W2+ 1/W3+....1/W30] =1 => [1/W1 + 1/W2 + 1/W3 + .............1/W30] = 1/31. If all the workers work together they will finish the whole work in 31 days.
Q. No. 46:
There are four varieties of pipes Pipe A, Pipe B, Pipe C and Pipe D. Each pipe can be either an inlet pipe or an outlet pipe but cannot be both. there are 5 tanks of equal volume. Tank P is filled by Pipe A and Pipe B Tank Q is filled by Pipe A and Pipe C Tank R is filled by Pipe A and Pipe D Tank S is filled by Pipe B and Pipe C Tank T is filled by Pipe C and Pipe D Time taken for the first 3 tanks(P, Q and R) to get filled are in the ratio 1:2:4 and the time taken for the S and T tanks to be filled are in the ratio 7:10. Find the outlet pipes among the 4 varieties.
Answer: D Let A,B,C and D do a,b,c and d with of work in an hour. Let A and B fill the tank in 1 hour. Then A and C would fill the tank in 2 hours while A and D in 4 hours. a+b =1 ..........(i) a+c=1/2.........(ii) a+d = 1/4.......(iii) Let B and C take 7k hours while c and D take 10k hours to fill the tank => b+c = 1/7k..............(iv) c+d= 1/10k...................(v) a = {(i)+(ii)-(iv)}/(ii) = {(ii) + (iii) - (v)}/(ii) a = {1+1/2-1/7k}/2 = {1/2+1/4-1/10k}/2............(vi) => k= 4/70 On substituting value of k in equation(vi) we get a < 0 => a<0, b>0, c>0 and d>0 Hence only A is the only outlet pipe.
Q. No. 47:
A contract is to be completed in 50 days and 105 men were set to work, each working 8 hour a day. After 25 days, 2/5th of the work is finished. How many additional men be employed so that the work may be completed on time, each man now working 9 hours a day ?
Answer: C Since 2/5th of the work is completed in 25 days. remaining 3/5th of the work is to be completed in 25 days. Let, x men work for 25 days to complete 3/5th of the work. MDH/W = M1 D1 H1 /W1 25*105*8*5/2 = x*25*9*5/3 => x= 140. Additional men required = 140-105 = 35.
Q. No. 48:
A can build up a structure in 8 days and B can break it is 3 days. A has worked for 4 days and then B joined to work with A for another 2 days only. In how many days will A alone build up the remaining part of the structure ?
Answer: D A can build the structure in 8 days. Fraction of structure built in a day by A = 1/8 similarly, fraction of structure broken by B in a day = 1/3. Amount of work dome by A in 4 days = 4/8 = 1/2. Now, both A and B together for 2 days. So, fraction of structure built in 2 days = 2(1/8 - 1/3) = -5/12 Fraction of structure still to be built = 1/2 + 5/12 = 11/12. If A takes x days to build up the remaining structure, then x/8 = 11/12 => x= 22/3 days.