zero – as one letter cant be wrong – atleast two are to be wrong
16
In 30 mins- Thief – 20km, Owner – 0km
In 1hr – Thief – 40km, Owner – 25km
In 1.5hrs – Thief – 60km, Owner – 50km
In 2hrs – Thief – 80km, Owner – 75km
In 2.5hrs – Thief – 100km, Owner – 100km
Answer: In 2.5 hours the owner would have overtaken the Theif
2hr 24min
friendly environment, supportive colleagues and place where i’ll have to work.
1:2
Time =9 seconds
Speed =(60×518) m/sec=(503)m/sec
∴ Speed =DistanceTime
Length of the train (Distance) = (Speed × Time) = (503×9) m=150 m.
P1/P2 = C1*T1/ C2*T2
where P1 = Partner 1’s Profit.
C1 = Partner 1’s Capital.
T1 = Time period for which Partner 1 contributed his capital.
P2 = Partner 2’s Profit.
C2 = Partner 2’s Capital.
T2 = Time period for which Partner 2 contributed his capital
6000/3000= 20,000*6/x*12
x=5000
The work done by A in 8 days is = 8/ 12 = 2/3
Means A alone completes 2/3 part of work.
Remaining work which is (1–2/3) = 1/3 is completed by B in 8–2 = 6 days
So the complete work done by B in 6/(1/3) = 18 days.
B alone can complete the work in 18 days.
Dear friends i ofcourse confused like you when i come
across to solve this problem. Really friends its very
simple if you understand the question clearly. First thing
is what is mean by “as many as” means its called ‘idiom and
phrase’ in english and it means “the same number of”. now
read the question “how many pairs of letters in STAINLESS
which has same number of letters between them in the word
as they have in english alphabet”.
In the alphabetical order, A-Z can be numbered as 1-26.
In A(INL)E which is same as in the alphabetical order A
(BCD)E. In both the cases E is in the Fourth position. so
we got one pair.
And in ST, there are no letters between them in the word
stainless. In alphabetical orer from A-Z also there is no
letters between them..so we got the second pair…
In STAINLESS it has two pairs ST and AE
1600 years contain 0 odd day.
300 years contain 1 odd day.
94 years = (23 leap years + 71 ordinary years)
= (46 + 71) odd days
= 117 odd days, i.e., 5 odd days
Days from 1st January 1995 to 28th February 1995
= (31 + 28) days = 59 days
= (8 weeks + 3 days) = 3 odd days
∴ Total number of odd days
= (0 + 1 + 5 + 3) = 9 odd days i.e., 2 odd days.
So, the required day is Tuesday.
10000
one-legged =5% of 10000=500
remaining=10000-500=9500
barefooted=9500/2=4750
remaining people= 9500-4750=4750
hence required number of shoes= 4750*2+500*1=100005% of 10000 = 500 one legged
9500 / 2 = 4750 bare foot
minium no of shoes = 4750*2 + 500*1 = 10000
24 days