[arya5139]

[QUOTE=shilpakokkanti;1586354]1)In a race of 500 m,L beats M by40 sec and beats N by 12 5mm.If M and N run a 500m race,M beats N by 40 sec.Find the time taken by M to run the race?

280 is the ans.

Also find the speed of N??
ans is 1.5625 m/s.


Similar to one u posted earlier...

let t be the time taken(in seconds) by L to complete the race
=>time taken by M = t+40 and time taken by N to complete the race = t+40+40 = t+80 sec

since L beats N by 125 m

t*500/t - t*500/(t+80) = 125 => 375 =500t/(t + 80)
15 (t+80) = 20t => 5t = 15*80 => t = 240 sec

=> time taken by M to complete the race = t + 40 =240 + 40 =280 sec

spedd of N = 500/(240 +80) = 500/320 = 25/16 = 1.5625 m/s

[Mani_23]

Originally Posted by akhilmisra

Regarding the ladder ques (IIFT )

(24 - x/2)^2 + (7 + x)^2 =25^2
solving we get
x=8
According to me the answer will be 8 as given in the soln above,
hope this helps....

(24 - x/2)^2..i guess here dere;s a problem?

if u have got 24..it means its the height from the top of the ladder to the foot of the wall..

and if ladder has been brought down to the half of the distance..dat means its 12

[prateek7563]

problem in the language of the question

[$U!\!$H|!\!E]

Originally Posted by simrat_sandy

the numerator is divisible by 16+8+4+2=30
and the denominator is divisble by 3
hence nr is a multiple of 30 and dr is multiple of 3
so i think remainder should be 0??
is this logic correct?

1+(2)^101+(4)^101+(8 )^101+(16)^101

is equivalent to

1 + (2)^101 + (2)^(2*101) + (2)^(3*101) + (2)^(4*101)

If we consider 2^101 = a then the problem becomes what is the reminder of
1+a+a^2+a^3+a^4 when it's divided by a+1

According to Reminder theorem:
f(a=-1) = 1+(-1)+(-1)^2+(-1)^3+(-1)^4 = 1

so Reminder is 1

[kinscool]

Originally Posted by effort

LCM of two no.s x and y is 161.find the value of 4x-3y, if y>x,x>1..

is teh ans -41?

[prakharc]

Originally Posted by kinscool

Originally Posted by effort
LCM of two no.s x and y is 161.find the value of 4x-3y, if y>x,x>1..

is teh ans -41?

Why only -41 ??

LCM is 161 = 7*23

so possible values of x,y = (7,161) , (7,23) , (23,161)

4x-3y= -455 , -41 , -391

So any one of these 3 can be the solution

[prateek7563]

Originally Posted by prakharc

Why only -41 ??

LCM is 161 = 7*23

so possible values of x,y = (7,161) , (7,23) , (23,161)

4x-3y= -455 , -41 , -391

So any one of these 3 can be the solution

since no option was given , we solved for 'only' value

[shekhar255]

Originally Posted by prakharc

Why only -41 ??

LCM is 161 = 7*23

so possible values of x,y = (7,161) , (7,23) , (23,161)

4x-3y= -455 , -41 , -391

So any one of these 3 can be the solution

i hav d same doubt...

[shilpakokkanti]

If the hands of a clock coincide every 80 minutes and the hands of another clock coincide evry 65 minutes..What is the app time difference between the two clocks in exactly 24hrs time interval as shown by a correct clock?

[$U!\!$H|!\!E]

What would be the total number of factors of the expression
62^3 -54^3-8^3?