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A sum is sufficient to pay either George age for 15 days or marks wage to 10dayshow long together?
Read Solution (Total 11)
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- George one day work is (1/15)
Marks one day work is (1/10)
G+M=(1/15)+(1/10)=(1/6)
6 is the answer - 10 years agoHelpfull: Yes(14) No(2)
- i dont think its a tcs question
- 10 years agoHelpfull: Yes(9) No(8)
- A certain sum of money is sufficient to pay either George wages for 15 days or Mark wages for 10 days .For how long will it be sufficient if both George and Mark work together ?
a)5 b)6 c)9 d)8
This is the proper question? - 8 years agoHelpfull: Yes(9) No(0)
- george work 1/15
marks work 1/10
work together so (1/15)+(1/10)=(1/6)
then 6 will be the answer - 10 years agoHelpfull: Yes(4) No(1)
- Please write question clearly
- 10 years agoHelpfull: Yes(1) No(0)
- Ans=6days
Total work= Geroge and marks is 60 work. i.e LCM of Geroge and Marks
now George does 2Work/day.
marks does 3work /day
if both work G+M=5work/day
so total 30 work=30/5=6 days.. - 7 years agoHelpfull: Yes(1) No(0)
- Correct Question:-
A certain sum of money is sufficient to pay either George's wages for 15 days or Mark's wages for 10 days. For how long will it suffice if both George and Mark work together? - 7 years agoHelpfull: Yes(0) No(0)
- consider sum is lcm(15,10) =30rs
George one day 30/15=2rs
marks one day 30/10=3rs
total one day pay=2+3=5rs
30/5=6 its sufficient for 6days - 6 years agoHelpfull: Yes(0) No(0)
- george 1 day work is=1/15
mark 1 day work is=1/10
together=(1/15)+(1/10)
=5/30
=6
because 6 is the answer - 6 years agoHelpfull: Yes(0) No(0)
- This question is repeated in TCS Ninja 2018, :) !.
- 6 years agoHelpfull: Yes(0) No(0)
- f(2018)=f(2x1009)=f(504+1)=f(2x252+1)=f(252)+1
f(252)=f(2X126)=f(126)=f(2X63)=f(63)=f(2X31+1)=f(31)+1
f(31)=f(2X15+1)=f(15)+1
f(15)=f(2X7+1)=f(7)+1
f(7)=f(2X3+1)=f(3)+1
f(3)=f(2X1+1)=f(2)+1
f(2)=f(2X1)=f(1)
f(1)=f(2X0+1)=f(0)+1=1
So we have, f(1)=1
f(2)=1
f(3)=1+1
f(7)=2+1
f(15)=3+1
f(31)=4+1
f(252)=5+1
f(2018)=6+1=7 Ans - 5 years agoHelpfull: Yes(0) No(2)
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