6 men and 8 women finish a job in 10 days . In how many days can 5 men and 9 women do the same job?
Please solve it .
Certain quantities of 2 distinct varieties of rice V1 and V2 are mixed and d mixture is sold at rs.X per kg ,giving 25% profit.If rice of variety V1 is sold at Rs.X per kg,a loss of (100/11)% would be incurred.If the ratio of the quantities of rice of variety V2 and variety V1 in d mixture is 8:3,then what is the %age of profit obtained when rice of variety V2 is sold at rs.X per kg?
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plzz explain dis sum..i m nt gettin it..:)
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plzz explain dis sum..i m nt gettin it..:)
@vbhvgupta said:6 boys and 8 women finish a job in 6 days and 14 boys and 10 women finish the same job in 4 days. In how many days working together 1 boy and 1 woman can finish the work?
6/B+8/W=1/6
14/B+10/W=1/4
solve to get B and W then 1/[1/B+1/W ]
14/B+10/W=1/4
solve to get B and W then 1/[1/B+1/W ]
@vbhvgupta said:6 boys and 8 women finish a job in 6 days and 14 boys and 10 women finish the same job in 4 days. In how many days working together 1 boy and 1 woman can finish the work?
B = 2W/5
so 6B+8W = 12W/5+8W = 52W/5
so 1 women does 5/(52*6) work in one day
and 1 B does 2/(52*6)
so 7/(52*6) is done in one day
total number of days = 7n/(52*6) = 1
44 4/7 days
@meenu05 said:Please solve it .
on solving it comes out to be 3[ log y/log10-1/3(logy)/log10]= 8( log 10/log y)
put log y=x
and solve get the +v value of x and again solve for y i.e. log y=+ve value of x
put log y=x
and solve get the +v value of x and again solve for y i.e. log y=+ve value of x
@PURITAN said:@viewpt me too looking for the xplanation. ?? dont have the solution
may be..but we are getting 260 only .
@chandrakant.k said:B = 2W/5so 6B+8W = 12W/5+8W = 52W/5so 1 women does 5/(52*6) work in one dayand 1 B does 10/(52*6) so 15/(52*6) is done in one daytotal number of days = 15n/(52*6) = 120.8 days21??
44 4/7
@meenu05
Question: log17 6^log7[log 20 25 *log 25 20]
Pl note first no is base. did I get the question correctly?
If yes, Im getting the ans as '0'.
Question: log17 6^log7[log 20 25 *log 25 20]
Pl note first no is base. did I get the question correctly?
If yes, Im getting the ans as '0'.
@chandrakant.k said:B = 2W/5so 6B+8W = 12W/5+8W = 52W/5so 1 women does 5/(52*6) work in one dayand 1 B does 10/(52*6) so 15/(52*6) is done in one daytotal number of days = 15n/(52*6) = 120.8 days21??
bro where did u derive/assume B=2W/5 from ???
@vbhvgupta said:44 4/7
yeah calculation mistake.. please check..
@DeAdLy said:bro where did u derive/assume B=2W/5 from ???
@VJ12 said:@chandrakant.kB=2W/5 kaise mila bhai?
see the given data is
6B+8W)*6 = 1 (i have taken 1 unit as the work)
14B+10W)*4 = 1
equate these 2 as it is the same work you will get that equation
1. What is the number of ending zeros in 1^1 * 2^2 * 3^3........ 99^99 * 100^100.
a. 1050
b. 1300
c. 1250
d. None of these.
a. 1050
b. 1300
c. 1250
d. None of these.
@TootaHuaDil said:1. What is the number of ending zeros in 1^1 * 2^2 * 3^3........ 99^99 * 100^100.a. 1050b. 1300c. 1250d. None of these.
1300
OA??
@chandrakant.k said:yeah calculation mistake.. please check.. see the given data is6B+8W)*6 = 1 (i have taken 1 unit as the work)14B+10W)*4 = 1equate these 2 as it is the same work you will get that equation
My bad 
....let see u all 2omoro....feeling too sleepy now....
@TootaHuaDil said:1. What is the number of ending zeros in 1^1 * 2^2 * 3^3........ 99^99 * 100^100.a. 1050b. 1300c. 1250d. None of these.
1300?
take only power of 5s
5^5 *10^10.........100^100
(5 + 10 + 15 + 20.........+100) + (25 + 50 + 75 + 100)
1050 + 250 = 1300