Long Jump Competition

Let a be the number of successful competitions and b be the failed competitions. Note that we are finding a/(a+b)

(6.5a+4.5b)/(a+b)=4.9

65a+45b=49a+49b

16a=4b

4a=b

a/b=1/4

a:b=1:4

So a/(a+b) = 1/(1+4) = 20%

The Very Basic Solution
Let N = Total no. of jumps
S = Total no. of successful jumps
Then, N - S = Total no. of unsuccessful jumps
According to question
Sum of all jumps = [ 4.9 $\times$ N ]
Sum of all successful jumps = [ 6.5 $\times$ S ]
Sum of all unsuccessful jumps = [ 4.5 $\times$ (N - S) ]
$\Rightarrow$ [ 4.9 $\times$ N ] = [ 6.5 $\times$ S ] + [ 4.5 $\times$ (N - S) ]
$\Rightarrow$ S = [ 0.2 $\times$ N ]
Thus 20% successful jumps.

Let x = percentage of succesfull competitors.
6.5(x) + 4.5(1-x) = 4.9
Solving this equation give x = 0.2
Thus, x = 20%

let total no of competitors be 100 successful of them be X ; unsuccessful are 100 -X ; X 6.5 +(100-X ) 4.5 =100 *4.9 ; 6.5 X -4.5 X =490-450 =40 ; 2 X =40 ; X=20 ;It is for a total of 100 persons so 20%

suppose the successful is x and the failed is y. 6,5x + 4,5y = 4,9(x + y) 0,4y = 1,6y x/y = 1/4 so, for the successful 1/5 100 = 20% and the failed 4/5 100 = 80%

let x=successful y=unsuccessful ave= 4.9 4.9=(6.5x)(4.5y)/(x+y) (4.9x+4.9y)100=(6.5x+4.5y)100 (49-45)y=(65-49)x 4y=16x 1y=4x 1:4=y:x therefore %=((1/(1+4))100 =20%

(6.5x/100)+((4.5)*(100-x)/100)=4.9 6.5x+450-4.5x=490 2x=40 x=20

Simply (4.9-4.5=.4/(6.5-4.5)=20%

s = sucessful; t = total

6.5s + (4.5)(t-s) = 4.9t

6.5s + 4.5t - 4.5s = 4.9t

s = (0.4/2.0)t

Let number of successful=S and Unsuccessful=U. Then, distance covered by successful and unsuccessful = 6.5S+4.5U and this must be equal to 4.9(S+U) i.e., 6.5S+4.5U = 4.9S+4.9U or, 6.5S-4.9S=4.9U-4.5U or, S=(0.4/1.6)U. Let us divide both sides by (S+U). We get S/(S+U) = 0.25U/(U+S) = 0.25U/(0.25U+U)=0.25/1.25=0.2. If we multiply it by 100 for percentage, we get 20%.

nice problem

Thanks for all......... please reshare

6.5-4.5=2 4.9-4.5=.4 .4 is 20 percent of 2

Let X were sucessful ,eqation becomes 6.6Xx +4.5(100-x) =100X4.9 By solving we get x =20,so successful are 20 % K.K.GARG,India

20% is the answer.
6.5 x + 4.5(100-x) = 4.9 100, solving this we will get x = 20,

(6.5x +4.5y)/(x + y) = 4.9 or 1.6x = 0.4y and x + y = 1 solving x = 0.2 i.e. 20%

6,5x+4,5y=4,9 , x+y=100%=1

65x+45y=49 , y = 1 - x

65x + ( 45 ( 1- x )) = 49

65x + 45 - 45x = 49

20x = 4

x = 4/20 = 1/5 = 0,2 = 20%

6.5x+4.5y=(4.9)100, x+y=100, (x, y) = (20, 80)

call the percentage of success is y, we have ((6.5y+ 4.5(1-y))/1= 4.9 => 6.5y+ 4.5-4.5y=4.9 => 2y = 0.4 => y= 0.2 or 20%

X = SUCCESSFUL AND Y= UNSUCCESSFUL

6.5X + 4.5 Y = 4.9 (X+Y)

65X + 45Y = 49X+49Y

X= 1/4 Y AND X+Y = 5/4Y

100*X/(X+Y) = (Y * 1/4 )/(Y * 5/4) = 20

we have average jumpers =4.5 succesful jumpers=6.5 unsuccesful jumpers=4.9

then we have 6.5/4.5=2.0 also we have 4.9/4.5=0.4 the total percentage=0.4/2.0=20%

all average of competitor is 4,9 m

x= total competitor who succesful y= total competitor who unseccesful

to solve it, (x(6,5) + y(4,5)) / (x+y) = 4,9

6,5 X + 4,5 Y = 4,9X + 4,9Y

1,6X = 0,4Y

Y = 0,4/2 * 100% = 20%

NOTE : 2 = 1,6+0,4

k(6.5)+l(4.5)=4.9 were k and l are fractions of sucessfull and unsucessfull competers then k+l=1 k(6.5)+(1-k)(4.5)=4.9 by solving we get k=0.2 percentage=0.2*100=20%

actually ,we can know the answer to this, by calculating the % increase or % drop in the ave. jump value of successful &unsuccessful players as, compared to ave. jump value of total players.......................................... like,,,, ave.value of successful players=6.5...................... (1) " " " unsuccessful players=4.5 ...............(2) and,,,,, " " " total players=4.9..............(3) now,,,, (1)-(3)=1.6 & (3)-(2)=0.4............... i.e., let 0.4 =x ,,,then 1.6 =4x .........and 5x=100%.....then x=20%

let x is number of successful competitors having average jump =6.5 m and y is number unsuccessful competitor having average jump=4.5 m.

So the average jump of total competitors =4.9 m will be equal to (6.5x+4.5y)/(x+y)

which gives x/y=1/4.

So the percentage of winning competitor will be=(x/(x+y))*100 %=100/5%=20%

Let x be the number of successful competitors and y be the unsuccessful competitors. Note that we are finding x/(x+y)

(6.5x+4.5y)/(x+y)=4.9

65x+45y=49x+49y

16x=4y

4x=y

x/y=1/4

x:y=1:4

Then x/(x+y) = 1/(1+4) = 20%

let x denote the succeful tries. then,

x1+x2+x3+.......+xn=6.5*(n)----------equation 1

and,let y denote the number of unsuccesful tries then, y1+y2+y3+......+ym=4.5*(m)---------equation 2

also, (x1+x2+x3+....+xn)+(y1+y2+......+ym)=4.9*(m+n)-------equation 3

now subtract the sum of equation 1 and equation2 from equation 3, we get,

m= 4 *(n)

where n are succesful total tries and m are unsuccesful total tries.

we have to find, n /(m+n)

which is equal to 20%