There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors
Originally posted by Sameer Jamal:
[QB]
3. Find the values of each of the alphabets.
N O O N
S O O N
+ M O O N
----------
J U N E
Assuming each letter is different, I think this may be one answer.
E2
J9
M0
N4
O1
S5
U3
_ __ _ __ _ __ _<br />SCJP 1.4 (95%) | SCWCD 1.4 (79%)<br />Artificial intelligence is no match for natural stupidity.
There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors
Originally posted by Sameer Jamal:
5. Complete the series: 5, 20, 24, 6, 2, 8, ?
catch me if you can
_ __ _ __ _ __ _<br />SCJP 1.4 (95%) | SCWCD 1.4 (79%)<br />Artificial intelligence is no match for natural stupidity.
There is a safe with a 5 digit number as the key. The 4th digit is 4 greater than second digit, while 3rd digit is 3 less than 2nd digit. The 1st digit is thrice the last digit. There are 3 pairs whose sum is 11. Find the number.
_ __ _ __ _ __ _<br />SCJP 1.4 (95%) | SCWCD 1.4 (79%)<br />Artificial intelligence is no match for natural stupidity.
Originally posted by Deb Sadhukhan:
HS Thomas,
Are you trying to say plane will naver be able to overtake ship, even though it's speed is 10 times the speed of the ship?
Originally posted by Jignesh Malavia:
#4 (Probably not the best approach. Too long..)
Let D = number of steps
Let S = constant speed of escalator (that is, it goes down at the rate of S steps/second)
Let sa = speed of A (sa steps/second) when escalator is stationary.
Let sb = speed of B (sb steps/second) when escalator is stationary.
If the escalator is moving at speed S, and A is walking at speed sa, then the effective speed of A is S+sa
If the escalator is moving at speed S, and B is running at speed sb, then the effective speed of B is S+sb
Let T = time required by escalator to cover D steps.
Let ta = time required by A to cover D steps when escalator is stationary.
Let ta50 = time required by A to cover 50 steps when escalator is stationary.
Let taeff = effective time required by A to cover 50 steps when escalator is moving.
Let tb = time required by B to cover D steps when escalator is stationary.
Let tb90 = time required by B to cover 90 steps when escalator is stationary.
Let tbeff = effective time required by B to cover 90 steps when escalator is moving.
Therefore,
D = S*T
D = sa*ta
D = sb*tb
50 = sa*ta50 (for A walking on stationary escalator) --- 1
90 = sb*tb90 (for B running on stationary escalator) --- 2
D = (S+sa)*taeff (for A walking on moving escalator) --- 3
D = (S+sb)*tbeff (for B running on moving escalator) --- 4
For A, the time required to reach the bottom while escalator is moving is same as the time required to walk 50 steps.
Thus, ta50 == taeff.
or 50/sa = D/(S+sa) --- [from 1 and 3]
or D = ((S+sa)*50)/sa --- 5
For B, the time required to reach the bottom while escalator is moving is same as the time required to run 90 steps.
Thus, tb90 == tbeff.
90/sa = D/(S+sb) --- [from 2 and 4]
or D = ((S+sb)*90)/sb --- 6
Also, B coveres all 90 steps when A has covered only 10 steps == 1/5th the total steps for A. Thus, effective speed of B is five times effective speed of A
(S+sb) = 5*(S+sa)
Replacing S+sb with 5*(S+sa) in 6 gives
D = 5*(S+sb)*90/sb --- 7
Combine 5 and 7
(S+sa)*50/sa == 450(S+sa)/sb
thus sb/sa = 9
B goes 9 times faster then A
Solving for eq 5 and 6 using sb=9*sa gives S = sa.
Using S = sa in 5 gives D = 100.
Answer: 100 steps are visible when the escalator is not operating.
A walks at the same speed as the escalator and thus reaches in half the normal time = A covers 50 steps + escalator covers 50 steps.
B runs at the 9 times the speed of escalator and thus reaches in 1/10th the normal time = B covers 90 stpes + escalator covers 10 steps.
MH
Originally posted by Capablanca Kepler:
Train problem:both have same speed?
Originally posted by Sameer Jamal:
other short approach can be
Suppose A takes T time to reach bottom (or 50 steps) when escalater is moving
For B to take 90 steps it will require T/5 time
total no of steps in escalater= (90T-50T/5)/T-T/5
=80T/4T/5=100 STEPS
MH
Wow, I haven't see any train in India, so made this trip on Taxi. Where is train station in Bangalore.Originally posted by Capablanca Kepler:
Let v1 be velocity of train from Bangalore to Mysore.
Let v2 be the velocity of train from Mysore to Bangalore.
Let two trains meet at distance x from Bangalore and y from Mysore.
x/v1 = y/v2---(1)
also v1 = y and v2= x/4--(2)
Substituting x and y in (1),
v1 = 2v2.
Velocity of one train is twice than another.
[ April 06, 2004: Message edited by: Capablanca Kepler ]
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