[prakharc]

Originally Posted by shilpakokkanti

2)In a Km race,R beats S by one minute and S beats T by 30 sec.If R beats T by 250m in the same race,find te time taken by R torun the race.

ans is 270sec.

Please help me by giving the solutions.

1000/s - 1000/r= 60
1000/t-1000/s=30 => 1000/t-1000/r=90 ---- (i)

T covers 250 m in 90 secs => t= 250/90=25/9

Putting in (i)
9/25-9/100 = 1/r = 27/100

So time taken by r= 1000/r= 270 secs

[prateek7563]

Side AC of the right triangle ABC right angled at A, is divided into 8 equal parts by drawing 7 lines segments parallel to BC meeting AB at 7 different points. What is the sum of all the 7 lines drawn parallel to BC, if the length of BC is 10 cm? Answer is 35

[arya5139]

Originally Posted by shilpakokkanti

Hi guys,,

2)In a Km race,R beats S by one minute and S beats T by 30 sec.If R beats T by 250m in the same race,find te time taken by R torun the race.

ans is 270sec.

Please help me by giving the solutions.

Let t be the time (I n minutes) taken by R to complete the race=>Time taken by T to complete the race = t + 1 + 30/60 = t + 3/2 =(2t + 3)/2

Since R beats T by 250 meters

=> t*1000/t - t*1000/(2t + 3)/2 = 250
=>1000 - 2000t/(2t + 3) =250
750 =2000t/(2t + 3) => 3 =8t/(2t + 3) => 6t + 9 =8t => t = 4.5 minutes
=> t = 4.5*60 sec = 270 sec

[ashish.88]

Originally Posted by shilpakokkanti

Hi guys,,

1)R and S are competing a race 1000m race.R gives saleem a head stat of 200m,but his speed is twice that of S.When R reaches the 600m mark,R reduces his speed by half but still beats S by 20 sec.What is S's speed?

ans is 5m/s.


2)In a Km race,R beats S by one minute and S beats T by 30 sec.If R beats T by 250m in the same race,find te time taken by R torun the race.

ans is 270sec.

Please help me by giving the solutions.

first one has been answered so I will answer th second one

1/s-1/r=1..(1)

1/t-1/s=0.5..(2)

hence 1/t-1/r=1.5..(3)

also 1/r=0.75/t

t=3r/4

sub in (3)
r=1/4.5

total time taken is 1/r ie 4.5 mins or 270 seconds

[prakharc]

Originally Posted by prateek7563

Side AC of the right triangle ABC right angled at A, is divided into 8 equal parts by drawing 7 lines segments parallel to BC meeting AB at 7 different points. What is the sum of all the 7 lines drawn parallel to BC, if the length of BC is 10 cm? Answer is 35

AB= 8, BC= 6

AB is divided into equal segments of h=1
BC is divided into equal segments of b= 3/4

General h= n ; general b = 3n/4 for nth parellel line

l= sqrt(n^2 + 9n^2/16) = 5n/4

Sum of lines = 5n(n+1)/8 (for n=7) = 35 cms

[crazy_mn2002]

Originally Posted by arya5139

consider 2^101%(2^100 + 1) = (2^100 + 2^100)%(2^100 + 1)
= (2^100 +1 +2^100 +1 -2)%(2^1001)= -2

=>2^101 can be substituted by -2
=>given number =(2^101)^4 + (2^101)^3 + (2^101)^2 + 2^101 +1
=> remainder = -2^4 + -2^3 + -2^2 + -2^1 +1
=16-8+4-2+1=11

hope it clears..

how did u replace 2^101 by -2... after the step : (2^100+1+2^100+1-2)/2^100+1 , i think it will be 1+1+(-2/2^100+1)... but how did u replace it by -2. please explain.

[simrat_sandy]

Originally Posted by akhilmisra

21p^2 +pq =45pq -15q^2
21p^2+15q^2 =44pq

21(p/q) + 15(q/p) = 44

let p/q=x
21x +15/x=44
21x^2 -44x + 15=0
solving the quadratic eqation we get
x=3/7 or 5/3
p/q = 3/7 or 5/3

pls confirm....
cheers...

i got the same thing

[akhilmisra]

Originally Posted by prateek7563

may be the answer is wrong

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....

[prateek7563]

Originally Posted by prakharc

AB= 8, BC= 6

AB is divided into equal segments of h=1
BC is divided into equal segments of b= 3/4

General h= n ; general b = 3n/4 for nth parellel line

l= sqrt(n^2 + 9n^2/16) = 5n/4

Sum of lines = 5n(n+1)/8 (for n=7) = 35 cms

thanks, i got it.

[knight27]

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..

The ans is -41.........
X and Y will be 7 and 23 respectively.......

Pls confirm