dstubbornkanav Saysthe highest power of 360 which divides 520! is ???pla explain how to solve dis type of problems?
is the answer 1 ??
the procedure for such questions is to find out the highest power of 360 in 520!. for this divide 520 successively with 360.
divide 520 with 360 implies quotient is 1'n remainder is 160. further division is not possible. hence the highest power is 1
is the answer 1 ??
the procedure for such questions is to find out the highest power of 360 in 520!. for this divide 520 successively with 360.
divide 520 with 360 implies quotient is 1'n remainder is 160. further division is not possible. hence the highest power is 1
The answer is 128.
360=2^3*3^2*5
So we find the highest powers of 2,3 and 5 in 520! . Once we find the powers, we have to divide the highest power of 2 by 3 and highest power of 3 by 2. Then choose the least among the powers thus obtained.
overlooked expressing 360 as the product of prime factors
Please post answer for this one. I am getting 424. Is it right?
for 1,5,9,...660 which is in A.P. (ignoring the last term), there are 165 terms. For 2,9,16..660 which is A.P. there are exactly 95 terms. But 9,37,58...are repeating,which we need to subtract as they are added twice, which are 24 terms. =>165+95-24=236.
Therefore, total burgers with tomato filling is 660-236= 424.
A bartender stole champagne from a bottle that contained 50% of spirit and he replaced what he had stolen with champagne having 20% spirit .The bottle then contained only 25 % spirit .How much of the bottle did he steal ????
(a) 80%
(b) 83.33%
(c) 85.71%
(d) 88.88%
plz post the approach as well..
A bartender stole champagne from a bottle that contained 50% of spirit and he replaced what he had stolen with champagne having 20% spirit .The bottle then contained only 25 % spirit .How much of the bottle did he steal ????
(a) 80%
(b) 83.33%
(c) 85.71%
(d) 88.88%
plz post the approach as well..
Determine the first term of GP, the sum of whose 1st term and 3rd term is 40 and the sum of 2nd and 4th term is 80
ans is 8 plss help me
a + ar^2 = 40 ==> a(1+r^2) = 40
ar + ar^3 = 80 ==> ar(1+r^2) = 80
Comparing line 1 and 2 --> r = 2
GP is 8,16,32,64
first term is 8 ..
hi all friends, please help me out with couple of question from HCF and LCM :
Q1.) A forester wants to plant 44 apple trees, 66 banana tress and 110 mango trees in equal rows (in terms of number of trees).Also,he wants to make distinct rows of trees(i.e. only one type of tree in one row).the number of rows (minimum) will be:
a.)2 b.) 3 c.) 10 d.) 11
Ans: c.) 10 How???:|
Q2.) 4 bells toll together at 9:00 A.M..They toll after 7,8,11,12 seconds respectively .How many times will they toll together again in next 3 hours?
a.)3 b.)4 c.)5 d.)6
Ans: c.) 7 how???
Thank you,
chandan
Q1) hcf (44,66,110) = 22
44 trees - 2 rows
66 - 3
110 - 5
total 10 rows
Q2) lcm (7,8,11,12) = 1848
after every 1848 secs it will toll together .. in 3 hours there are 10800 secs ..
so it will toll together 5 times in the next 3 hours ..
hi fellow puys...i cant solve a TSD question(review test 2 q no.6)...please help me by posting the approach...thanks in advance
Every day asha's husband meets her at railway station at 6:00pm and drives her to their residence.One day she left early from the office and reached railway station by 5:00pm.She started walking towards her home,met her husband coming from their residence on the way and they reached home 10 mins earlier than usual time.For how long did she walk.
(a) 1 hr (b) 50 mins (c) half hour (d) 55mins
I can eliminate options (a) and (c) but cant finish it...pls post approach
cheers:cheers:
hi fellow puys...i cant solve a TSD question(review test 2 q no.6)...please help me by posting the approach...thanks in advance
Every day asha's husband meets her at railway station at 6:00pm and drives her to their residence.One day she left early from the office and reached railway station by 5:00pm.She started walking towards her home,met her husband coming from their residence on the way and they reached home 10 mins earlier than usual time.For how long did she walk.
(a) 1 hr (b) 50 mins (c) half hour (d) 55mins
I can eliminate options (a) and (c) but cant finish it...pls post approach
cheers:cheers:
it is 50 mins...
let the speed of husband be N speed of asha be C and the house to station distance be D and distance children walked be x
Husband left his home at 6-D/N hrs to reach station at 4 pm
ow on saturday---> 5+x/C = 6-D/N+(D-X)/N ---> x*(N+C)/NC = 1
and 5+X/C + (D-X)/N + 1/5 = 6+D/N ----> X*(1/C-1/N) = 2/3
Solvin then we get N = 5C
so X/C comes out to be 5/6 hrs or 50 minutes
it is 50 mins...I think it's 55min.I used trial and error...
let the speed of husband be N speed of asha be C and the house to station distance be D and distance children walked be x
Husband left his home at 6-D/N hrs to reach station at 4 pm
ow on saturday---> 5+x/C = 6-D/N+(D-X)/N ---> x*(N+C)/NC = 1
and 5+X/C + (D-X)/N + 1/5 = 6+D/N ----> X*(1/C-1/N) = 2/3
Solvin then we get N = 5C
so X/C comes out to be 5/6 hrs or 50 minutes
five pipes fill a cistern in 5 hours and 2 pipes empty it in 4 hours if all the pipes are open together find the time when the cistern will overflow
I need the solution for this someone plastic help me
priyamvada_89 Saysfive pipes fill a cistern in 5 hours and 2 pipes empty it in 4 hours if all the pipes are open together find the time when the cistern will overflow
Sorry it was each fill individually in 4 hours and empty individually in 5 hrs
priyamvada_89 Saysfive pipes fill a cistern in 5 hours and 2 pipes empty it in 4 hours if all the pipes are open together find the time when the cistern will overflow
priyamvada_89 SaysI need the solution for this someone plastic help me
priyamvada_89 SaysSorry it was each fill individually in 4 hours and empty individually in 5 hrs