that's why have asked @deedeedudu bhai
@sowmyanarayanan said:explanation please
d1=2a
d2=2r
Total distance=d1+d2=2(a+r)
Now
2(a+r)/a=t
2(a+r)/r=t+3
2(a+r)/r=2(a+r)/a+3
multiply both sides to get
3/2=(a^2-r^2)/ar
3/2=(1-(r/a)^2)/r/a
Let r/a=x
Quadratic eqn solve karne par
x=1/2
d2=2r
Total distance=d1+d2=2(a+r)
Now
2(a+r)/a=t
2(a+r)/r=t+3
2(a+r)/r=2(a+r)/a+3
multiply both sides to get
3/2=(a^2-r^2)/ar
3/2=(1-(r/a)^2)/r/a
Let r/a=x
Quadratic eqn solve karne par
x=1/2
@deedeedudu said:d1=2ad2=2rTotal distance=d1+d2=2(a+r)Now2(a+r)/a=t2(a+r)/r=t+32(a+r)/r=2(a+r)/a+3multiply both sides to get 3/2=(a^2-r^2)/ar3/2=(1-(r/a)^2)/r/aLet r/a=xQuadratic eqn solve karne parx=1/2
is there any way we can use the formula S1/S2=sqrt(t2/t1)
@bhatkushal said:Q)At 7:00 AM,Ravish started from A towards B and Amit started from B towards A.At 9:00 AM they crossed each other and continued towards respective destination.if time taken by Ravish is 3 hours more than time taken by Amit. Find ratio of Ravish speed to Amit's speed?Could someone explain it using V1/V2=sqrt(t1/t2)?
1/2 ??
i wont do this qs. via ur method. apply simple logic of TSD
@bhatkushal said:is there any way we can use the formula S1/S2=sqrt(t2/t1)
I guess not coz time given is for the total journey that too relative time.
If exact values were given there may hav been a possibility
If exact values were given there may hav been a possibility
@bhatkushal said:Q)At 7:00 AM,Ravish started from A towards B and Amit started from B towards A.At 9:00 AM they crossed each other and continued towards respective destination.if time taken by Ravish is 3 hours more than time taken by Amit. Find ratio of Ravish speed to Amit's speed?Could someone explain it using V1/V2=sqrt(t1/t2)?
my question to u.. what r t1 and t2 here??
@bhatkushal said:is there any way we can use the formula S1/S2=sqrt(t2/t1)
Doing it your way would also require forming 2 equations and solving a quadratic equation.
It's one and same thing I suppose.
@bhatkushal said:Q)At 7:00 AM,Ravish started from A towards B and Amit started from B towards A.At 9:00 AM they crossed each other and continued towards respective destination.if time taken by Ravish is 3 hours more than time taken by Amit. Find ratio of Ravish speed to Amit's speed?Could someone explain it using V1/V2=sqrt(t1/t2)?
Let ravish speed =a
and Amit speed =b
=>2a+2b=d
Also, d/a=d/b+3
Solving both the equations:
2(b/a)=2(a/b)+3
2(b/a)^2-3(b/a)-2=0
b/a=2
a/b=ravish speed/amit speed=1/2
@bullseyes refer my post in the previous page. i have posted the formula and the respective variables
@bhatkushal said:is there any way we can use the formula S1/S2=sqrt(t2/t1)
While deriving formula
case is both reach destination at the same time
eq will be like this T/K - TK = 0 K=sqrt(T) where K is the ratio of speed and T is time taken to meet.
Here T/K- TK = 3 and T is 2
2/k+2k=3
(2k-1)(k+2)=0 l k=1/2
case is both reach destination at the same time
eq will be like this T/K - TK = 0 K=sqrt(T) where K is the ratio of speed and T is time taken to meet.
Here T/K- TK = 3 and T is 2
2/k+2k=3
(2k-1)(k+2)=0 l k=1/2
@bhatkushal said:is there any way we can use the formula S1/S2=sqrt(t2/t1)
yes u can..
to make it more clear. t2 here refers to the time taken by S2 to reach his destination AFTER meeting.. and t1 here refers to the time taken by S1.
let time taken by Amit(S2) after meeting is x hrs. then by Ravish(S1) is x+3 hrs.
now according to ur formula.
S1/S2 = sqrt(x/x+3) --------> 1
let they meet at point C. hence distance AC = 2*S1 ( applied speed = dist/time)
now time taken by Ravish to travel distance AC = x hrs
hence
(2*S1)/S2 = x hrs.
from eqns 1
2*sqrt(x/x+3) = x
solve for x here.. it comes out to be 1
putt x as 1 in eqn 1..
S1/S2 = 1/2
i hope this clear all ur concepts abt tht formula
@SAHILVANSIL said:11. In a bombing attack there is 50% chance that anyone bomb will strike the target. Two direct hits are required to destroy the target. How many bombs must be dropped to give a 99% chance or better of completely destroying the target.
say in every attempt 's' is chance of success and 'f' is chance of failure. So for n attempts binary probability distribution is given by
(f+s)^n = 1
=>C(n,0)f^n + C(n,1)*{f^(n-1)}*{s^1}.......................+C(n,n)*{s^n} = 1
the first two terms represent the case when there is less than 2 successful hit. So we want it to be less than 1%.
Given s= 0.5 and f = 0.5
hence we want C(n,0)*(0.5)^n + C(n,1)(0.5^n-1)*(0.5)
=> (0.5^n)*(1+n)
=> (1+n)*100
smallest value of n = 11.
Hence 11 attempts give a chance of more than 99%.
ATDH.
@SAHILVANSIL said:7. The odds that A speaks the truth is 3:2 and the odds that B speaks the truth is 5:3. In what percentage of cases are they likely to contradict?
19/40*100=47.5%?
@bullseyes said:yes u can.. to make it more clear. t2 here refers to the time taken by S2 to reach his destination AFTER meeting.. and t1 here refers to the time taken by S1. let time taken by Amit(S2) after meeting is x hrs. then by Ravish(S1) is x+3 hrs.now according to ur formula. S1/S2 = sqrt(x/x+3) --------> 1 let they meet at point C. hence distance AC = 2*S1 ( applied speed = dist/time) now time taken by Ravish to travel distance AC = x hrshence (2*S1)/S2 = x hrs. from eqns 1 2*sqrt(x/x+3) = x solve for x here.. it comes out to be 1 putt x as 1 in eqn 1.. S1/S2 = 1/2 i hope this clear all ur concepts abt tht formula
awsome explaination @bullseyes
I had got the answer but this cleared my understanding of d formula.......
@SAHILVANSIL said:7. The odds that A speaks the truth is 3:2 and the odds that B speaks the truth is 5:3. In what percentage of cases are they likely to contradict?
p(A) = 3/5, p(B) = 5/8
p(A,B) = (3/5)*(5/8) = 3/8
=> p'(A,B) =1 - { 3/5 +5/8 -3/8} = 3/20
hence contradiction happens = 1- p(A,B) - p'(A,B) = 1 - 3/8 -3/20 = (40-15 -6)/40 = 21/40 = 52.5%
ATDH.
@bhatkushal said:Q)At 7:00 AM,Ravish started from A towards B and Amit started from B towards A.At 9:00 AM they crossed each other and continued towards respective destination.if time taken by Ravish is 3 hours more than time taken by Amit. Find ratio of Ravish speed to Amit's speed?Could someone explain it using V1/V2=sqrt(t1/t2)?
In this case we cannot directly apply V1/V2 = sqrt(t2/t1). because in this case we are given the difference between t1 and t2.
rather than applying final result directly it would be better if you know the fundamental equations which are used to derive the formula you are quoting
if t is time till they meet then they take t1 and t2 to complete the remaining journey then:
V1*t/V2 = t2.............(i)
V2*t/V1 = t1.............(ii)
if you divide (i) by (ii) you get the formula you are taking abt.
But in our case we need to subtract rather than divide as we have the difference of t1-t2 given.
Say speeds are R and A hence we get
2R/A = tA; 2A/R = tR
Subtracting we get (tR-tA) = 2(A/R - R/A) = 3
=> (A/R) = 2.
ATDH.
@Cat.Aspirant123 said:The average weight of 8 people decreases by 2 kgs, when two new people are added to thegroup. If the difference between the original weight and new weight is 36 kgs, find the new totalweight of the people.
260.
@Cat.Aspirant123 said:The distance between Shyam's house and Suman's house is 16 kms. Shyam left his house at 7a.m. and headed towards Suman's house at the rate of 5 km per hour. Suman left her house at 8 a.m.and headed towards Shyam's house at the rate of 6 km per hour. At what time will they meet?a) 11 a.m. b) 2 p.m. c) 10 a.m. d) 9 a.m. e) 12.00 noon
d. 9am
@audiq7 said:N is a natural number which gives remainders 1 and 2 when divided by 6 and 5, respectively. All such N €™s are written in the ascending order, side by side from left to right. What is the 99th digit from the left?(a) 2 (b) 0 (c) 1 (d) 7
d. 7?
@gautam22 said:Does the equation, x^2 + y^3 = z^4 have solutions in prime numbers? Find at least one if yes, give a nonexistence proof otherwise
y^3 = z^4 - x^2
y^3 = (z^2 - x)(z^2 + x)
y^3 can be written as 1*y^3 or y*y^2
So, possible cases are:-
i) z^2 - x = 1 and z^2 + x = y^3
x + 1 = z^2
x = (z - 1)(z + 1), so z has to be 2 for x to be prime
but then z^2 + x is not a perfect cube (so not possible)
ii) z^2 - x = y and z^2 + x = y^2
z^2 = (y + y^2)/2 = y(y + 1)/2
here y and (y + 1) are coprime, so one of y and (y + 1) will be a perfect sq and another one will be twice of another perfect square (which is not possible as y is a prime number)
So, no solution