Good 1 
was more of logic and less of quants.
@gautam22 said:Using the digits 1 up to 9, two numbers must be made. The product of these two numbers should be as large as possible. All digits must be used exactly once. Which are the requested two numbers. ?
OK got it. I think :P
Let the two numbers be a and b. They have to start with 9 and 8. So a = 9..., b = 8....
For second digit, I want the largest possible thing to multiply with 9, so will put 7 after 8 and 6 after 9. So a = 96... b = 87...
For 3rd digit, again want largest to multiply with 96 so put 5 after 87....a = 964... b= 875...
For 4th digit, same logic, a = 9642... and b = 8753...
Now one number has to be 5 digits, adding the 1 on b makes sense so that it multiplies with the large number. Hence a = 9642 b = 87531
Does this make sense?
regards
scrabbler
Let the two numbers be a and b. They have to start with 9 and 8. So a = 9..., b = 8....
For second digit, I want the largest possible thing to multiply with 9, so will put 7 after 8 and 6 after 9. So a = 96... b = 87...
For 3rd digit, again want largest to multiply with 96 so put 5 after 87....a = 964... b= 875...
For 4th digit, same logic, a = 9642... and b = 8753...
Now one number has to be 5 digits, adding the 1 on b makes sense so that it multiplies with the large number. Hence a = 9642 b = 87531
Does this make sense?
regards
scrabbler
@gautam22 said:Using the digits 1 up to 9, two numbers must be made. The product of these two numbers should be as large as possible. All digits must be used exactly once. Which are the requested two numbers. ?
Your numbers are
9642 87531
9642 87531
@scrabbler said:OK got it. I think Let the two numbers be a and b. They have to start with 9 and 8. So a = 9..., b = 8....For second digit, I want the largest possible thing to multiply with 9, so will put 7 after 8 and 6 after 9. So a = 96... b = 87...For 3rd digit, again want largest to multiply with 96 so put 5 after 87....a = 964... b= 875...For 4th digit, same logic, a = 9642... and b = 8753...Now one number has to be 5 digits, adding the 1 on b makes sense so that it multiplies with the large number. Hence a = 9642 b = 87531Does this make sense?regardsscrabbler
Hey scrabbler 😁 long time
A trains meets with a accident and travels 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it travelled at its regular speed?
@veertamizhan said:A trains meets with a accident and travels 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it travelled at its regular speed?
48min?
@veertamizhan said:@19rsb yes, how did you arrive at this answer bro?
let he takes T min with his normal speed(from accident place to destination)
after accident his speed decreased by 4/7........implies he will tale 7/4 of the normal time which he usually takes to travel the given distance
hence,(7T/4) - T =36
=>T=48min
after accident his speed decreased by 4/7........implies he will tale 7/4 of the normal time which he usually takes to travel the given distance
hence,(7T/4) - T =36
=>T=48min
@veertamizhan said:A trains meets with a accident and travels 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it travelled at its regular speed?
48 mins
d/(4v/7)-d/v=36...(i)
regular time d/v=t
from (i)... d/v=t=36*4/3=48 mins
@veertamizhan said:A trains meets with a accident and travels 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it travelled at its regular speed?
48 mins...
X = total distance
S= speed
a = distance where accident occurs
Case 1) Total time = a/s + (x-a)/(4/7)s ----- eq 1
Case 2) Total time= x/s ------eq 2
Now eq 1 + 36 = eq 2
we get a/s = 48 min..
@veertamizhan said:A trains meets with a accident and travels 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it travelled at its regular speed?
7/4*x/v-x/v=3/5
3/4*x/v=3/5
x/v=4/5*60=48 min
a right angled triangle with longest side 80,construct 3 paths from the corner point opposite to the longest side such that these three paths divide the longest side into four equal segments.determine the sum of the lengths of three paths.
@gautam22 said:An (infinitely small) ball starting out in the middle of a 5 pointed star table (outer 5 points - 10m radius..... inner 5 points - 5m radius) has a starting angle of a random value from 0 to 360 degrees. The ball is now set loose and travels around the table.On average, how many sides will have been hit once the ball has travelled 1000m ?
@gautam22 said:A sphere 5.5 metres in diameter is filled with 1m diameter hemi-spheres.(1) What is the theoretical maximum amount of hemi-spheres that can be crammed into the big sphere given that the following condition is met:Each hemi-sphere's flat side (which I'll now refer to as its 'disc') has a central point (indicated by the white point shown in the hemisphere diagram to the right). The point must not 'see' another hemisphere's disc. (2) By cramming them as efficiently as possible, a relatively small volume will be left. How large is this volume?
Both the problem have been picked up from the link below.
And after reading the details on this link, I think both these problems can be avoided for the time being on the forum and we can limit ourselves to the problems and concepts related to B-School enterance exams.
As far as I know there is no simple and easily comprehendable solution to the above problems, such problems are best solved using computer similations.
ATDH.
@sujamait said:a right angled triangle with longest side 80,construct 3 paths from the corner point opposite to the longest side such that these three paths divide the longest side into four equal segments.determine the sum of the lengths of three paths.
getting very weird answer :O
Let triangle be ABC with Right angle at B
it's a Pythagoras triplet 48^2+64^2=80^2
so, AB=48, BC=64
Three dividing point be D, E, F
mid path will be of length 40. (circumradius)
Now, Cos AEB=(40^2+40^2-48^2)/2*40*40
and for triangle DEB, Cos AEB=(20^2+40^2-BD^2)/2*20*40
=> (40^2+40^2-48^2)/2*40*40=(20^2+40^2-BD^2)/2*20*40
=> BD^2=1552
Similarly, Cos BEC=(40^2+40^2-64^2)/2*40*40
and for triangle BEF, Cos BEC=(20^2+40^2-BF^2)/2*20*40
=> (40^2+40^2-64^2)/2*40*40=(20^2+40^2-BF^2)/2*20*40
=> (40^2+40^2-64^2)/2*40*40=(20^2+40^2-BF^2)/2*20*40
=> BF^2=2448
==> length of three paths=40+sqrt(1552)+sqrt(2448)
If it is length of squares of paths then answer would be 1600+1552+2448=5600
@sujamait said:a right angled triangle with longest side 80,construct 3 paths from the corner point opposite to the longest side such that these three paths divide the longest side into four equal segments.determine the sum of the lengths of three paths.
Here, the length of the middle path will be half the longest side..i.e the hypotenuse
Hence, l(middle path) = 40m
Length of other 2 paths = root (40^2+20^2)
Sum of the lengths of the 3 paths = 40+2root(40^2+20^2)
Please correct me if i am wrong
@soumitrabengeri said:Here, the length of the middle path will be half the longest side..i.e the hypotenuseHence, l(middle path) = 40mLength of other 2 paths = root (40^2+20^2)Sum of the lengths of the 3 paths = 40+2root(40^2+20^2)Please correct me if i am wrong
I guess Both the cases are possible. (the one with Pythagoras and one with yours).
So, questioner should be asking for squares of length of paths. (which is 5600 in each case) :lookaround:
@sujamait said:a right angled triangle with longest side 80,construct 3 paths from the corner point opposite to the longest side such that these three paths divide the longest side into four equal segments.determine the sum of the lengths of three paths.
Assuming triangle to be an isosceles triangle...
The middle path will be equal to the Circum Radius of the triangle. and hence will be equal to R.
let three paths be p1, p2 and p3
here, p2=40= 80/2= circumradius
p1= p3 can easily be found out by pythagoras theorem,
p1^2= 40^2+20^2
=> p1=p3= 20rt(5)
Hence the required sum= p1+p2+p3
=> 40+ 20rt(5) + 20rt(5)= 40(1+rt(5))
correct me if wrng.
A rope is cut into 3 pieces of unequal lengths. Find the length of shortest piece.
1) Length of 2 longest pieces is 12 cm
2) Length of 2 shorter pieces is 11 cm.
a) can be answered by a alone
b) can be answered by b alone
c) can be answered by using both
d) can be answered by either
e) cannot be answered
@bullseyes said:A rope is cut into 3 pieces of unequal lengths. Find the length of shortest piece.1) Length of 2 longest pieces is 12 cm2) Length of 2 shorter pieces is 11 cm. a) can be answered by a aloneb) can be answered by b alonec) can be answered by using bothd) can be answered by eithere) cannot be answered
e) cannot be answered
@bullseyes said:A rope is cut into 3 pieces of unequal lengths. Find the length of shortest piece.1) Length of 2 longest pieces is 12 cm2) Length of 2 shorter pieces is 11 cm. a) can be answered by a aloneb) can be answered by b alonec) can be answered by using bothd) can be answered by eithere) cannot be answered
cannot be answered?