it takes six days for three women and two men working together to complete a work. Three men would do the same work five days sooner than nine women. How many times does the output of a man exceeds that of a women?
a. 3 times b. 4 times c. 5 times d. 6 times e. 7 times
Again, solving by options.
Randomly picking up option (b) = 4 times: For the first statement, 3w+2m can complete the work in 6 days.
Since 1m = 4w, number of workers = 3w+2*4w = 11w So, the number of units of work = 11w*6 (since all 11 women work 6 days) = 77 units
Now, work done by 3m = 3*4w should be equal to 77 units Since, 77 is not divisible by 12, this is unlikely to be the answer.
Checking for option (d) = 6 times:
3w+2m = 3w+12w = 15w Work done in 6 days = 15w*6 = 90 units
If 3m have to do 90 units, i.e. 18w can do 90 units in 90/18 = 5 days Similarly, 9w can do 90 units in 90/9 = 10 days, which satisfies the criteria of "Three men would do the same work five days sooner than nine women"
Since it satisfies both the criteria, this has to be the answer and you can stop checking other options 😁 Answer is 6 times 😃 Why break head over algebraic equations when you can do without them??
20. gives remainder and {.} denotes fractional part of that. The fractional part is of the form (0.bx). The value of x could be OP: a. 2 b. 4 c. 6 d. 8 e. None of these remainder is 4 so 4/50=0.08
25. The expression (333^555) + (555^333) is divisible by Op: a. 2 b. 3 c. 37 d. 11 e. All of these For this its divisible by 2,3,37 but not 11... Question shd not divisible by...
39. Find the smallest no. n such that n! is divisible by 990. OP: a. 3 b. 5 c. 11 d. 12 e. None of these (My ans. 11 ; Given ans. None of these) 990=11x9x5x2, so yeah it shd be 11
64. Find the 28383rd term of series 123456789101112........ OP: a. 3 b. 4 c. 9 d. 7 (Gettin 7 bt not correct) This problem posted many times... Here u can get the lucid explanation of avinav bhai http://www.pagalguy.com/forum/quantitative-questions-and-answers/23813-quant-by-arun-sharma-167.html#post1636356
77. (12^33) * (34^23) * (2^47) Find the sum of odd factors. Please explain the approach for this....I waste quite a time to solve these type of ques. Ans: ((3^34-1)/2) * ((17^24-1)/16) well to calculate sum of all odd factors u shd not consider powers of 2...
Find last two digits of follwing nos.
1. 101*102*103*197*198*199 : Getting 64 (even tried calci.); given ans 54 yeah another wrong answer in Arunsharma
2. 65*29*37*63*71*87*62 : Any shortcuts for solving que 65*29*37*63*71*87*62/100 divide by 5 13*29*37*63*71*87*62/20 =>13x9x-3x3x-9x7x2 =>78 Find the remainder when
1. 7^99 divides 2400 (getting no clue whatsoever) is it 343 solved using chinese theorem
The remainder comes out to be 90 (-2 for division of 20).So the last two digits should be 90. But ans given is 10. For ur ref. it is que. no. 83 (page no. 47)
dint know abt chinese theorem.....will read and see if I get ans for last one...
1. two trains A and B start simultaneously in opposite direction from two points A and B and arrive at their destination 9 hr and 4 hr respectively after their meetin each other. At what rate does second train B travel if the first train travels at 8o km/hr.
a. 60 kmph b. 100 kmph c. 120 kmph d. 80kmph e. none
2. Sambhu beats kali by 30 metres or 10sec. how much time was taken by sambhu to complete a race of 1200 metres.
a. 6 min 30 s b. 3 min 15 sec c. 12 min 10 s d. 2 min 5 s e. none
1. two trains A and B start simultaneously in opposite direction from two points A and B and arrive at their destination 9 hr and 4 hr respectively after their meetin each other. At what rate does second train B travel if the first train travels at 8o km/hr.
a. 60 kmph b. 100 kmph c. 120 kmph d. 80kmph e. none
taking a que from Aninav bhai approach from options I would usually solve with equations
U can very well note that speed of B is more than A.. So I will take option c 120KMPH
A travelled a distance of 80x9=720km.. so B would take 6 hours to cover the same distance... so in 6 hours A would have travelled a distance of 80x6=480km... and the same distance B would take 4 hours...Hence 120 is the right answer.. Which satisfies the condition in prob
2. Sambhu beats kali by 30 metres or 10sec. how much time was taken by sambhu to complete a race of 1200 metres.
a. 6 min 30 s b. 3 min 15 sec c. 12 min 10 s d. 2 min 5 s e. none
in 10sec sambhu would travel a distance of 30m, speed of 3m/s for 1200/3 = 400 sec => 6 min 40 sec
taking a que from Aninav bhai approach from options I would usually solve with equations
U can very well note that speed of B is more than A.. So I will take option c 120KMPH
A travelled a distance of 80x9=720km.. so B would take 6 hours to cover the same distance... so in 6 hours A would have travelled a distance of 80x6=480km... and the same distance B would take 4 hours...Hence 120 is the right answer.. Which satisfies the condition in prob
in 10sec sambhu would travel a distance of 30m, speed of 3m/s for 1200/3 = 400 sec => 6 min 40 sec
For first one of course answer is correct....But I would like to share a formula to solve such questions which can give ans directly.....I never tried 4 proof of it nor do I remember where I read it but IT WORKS....
For second que. I think question has to be how much time Kalu takes......B'coz Shambhu reaches 10 sec before Kalu and by that time Kalu is 30m behind Shambhu (or finish line).....i.e. It would take 10 sec for Kalu to reach finish line (i.e. to cover 30m)...So actually Kalu's speed is 3m/sec....(I hope I have explained my logic clearly....just draw a straight line and plot points).......So unless the length of track is given nothing can be commented abt Shambhu's speed.....
@Divya : Can you plz recheck the ques....Give page no. if u can
Since sambhu beats kalu by 30m and 10 sec... There would be 2 cases, in the first case I will assume both of them started together and completed the race where shambo finished 30m ahead of kali...
In the second case assume that sambhu gives a lead of 10 secs to kali so that both of them finish the race at same time..
Case 1 - since time is same, D/Ss=D-30/Sk Ss/Sk=1200/1170
case- 2=> 1200-10Sk/Sk=1200/Ss
solve both, 1200-10sk/1200 = 1170/1200
sk=3m/s =>Ss=40/13 m/s
so time sambho takes is 1200/(40/13) =>390 secs =>6min 30 secs
@ divya wats the OA
Since kali takes 400 secs, sambho shd take 390 secs... All those steps I did above was complete bakar, waste of time
Yeah.....but in both the cases you have assumed that that the track is 1200m long whereas in que it is not explicitly mentioned that the race in which Shambhu beats Kalu is also on the same track.....But again as there is no "can not be determined" option I think it is quite safe to assume that......Had it been in option I would have gone with it......What say????
Also where can I find the lucid explanation of Chinese theorem which you mentioned about in reply to my last query......Of course THANKS in advance...
Yeah.....but in both the cases you have assumed that that the track is 1200m long whereas in que it is not explicitly mentioned that the race in which Shambhu beats Kalu is also on the same track.....But again as there is no "can not be determined" option I think it is quite safe to assume that......Had it been in option I would have gone with it......What say????
Also where can I find the lucid explanation of Chinese theorem which you mentioned about in reply to my last query......Of course THANKS in advance...
yes if there was a cannot be determined option then that shd be the solution.. For chinese - http://www.pagalguy.com/forum/quantitative-questions-and-answers/55747-cat-2010-concepts-fundas-tips-2.html#post2196538
didnt understand the solution..plz explain why you took 15c5 + .... the ans is 4944
Detailed explanation : We want no. of subsets of not more than 5 elements to be selected from 15 elements. So the possible no. of elements in the subset could be 5, 4, 3, 2, 1 or 0. Now, a subset of 5 elements can be selected from 15 elements in 15C5 ways; that of 4 elements in 15C4 ways and so on upto 15C0...... Total no. of subsets = 15C5 + ....... + 15C0
FYI, 15C0 can be viewed as null set (It is also subset of the main set having not more than 5 elements).
seven different objects must be divided among three people. in how many ways can this be done *if one or two of them must get no objects? ans 381 *if at least one of them gets exactly one object? ans 1176
