In a bag containgn only white and black balls. Half the number of white balls is equal to one third of black?

Question

balls. Twicw the total number of balls exceeds three times the number of black balls by 4. How many black and white balls in the bag.
2. While covering a distance of 30km salman takes 2 hours more than Amir. If salman doubles his speed, he would take 1 hour less than Amir. Find their rate of walking.

Answer 1

1/2 w = 1/3 b

2(b+w) = 4+ 3b

After simplifying and solving simultaneously,

w = 8 and b = 12 for a total of 20 balls

Answer 2

W = number of white balls
B = number of black balls
W/2 = B/3 implies W = 2B/3
2(W+B) = 3B + 4
2((2B/3) + B) = 3B + 4
Solve for B:
4B/3 + 2B = 3B + 4
4B + 6B = 9B + 12
B = 12; this implies W = 8
Total number of balls in bag = W + B = 20

If Salman doubles his speed, he cuts his travel time by 3 hours.
Therfore, he originally took 6 hours to walk 30 km, for an original speed of 5 km/hour.
Amir takes 2 hours less than Salman's original 6 hours, or 4 hours, to walk 30 km, for a speed of 7.5 km/hour.

Answer 3

We have two variables (W=white, B=black), so we need two equations. The problem gives the info to create the 2 equations:
[Half the number of white balls is equal to one third of black]
1/2W=1/3B or 1/2W-1/3B=0
[Twice the total number of balls exceeds three times the number of black balls by 4]
2(B+W)=3B+4
Now comes the algebra. Start with one of the equations solving by one of the variables:
2(B+W)=3B+4
2W=B+4
W=1/2B+2 [this is the value of W. Throw this value into the W of the other equation to get B]
1/2(1/2B+2)-1/3B=0
Now solving for B:
1/4B+1-1/3B=0
1/4B-1/3B= -1
-1/12B=-1 [1/4 minus 1/3 is negative 1/12]
1/12B=1
B=12
Now that we have the value of B we can find the real value of W using the equation for W we found above:
W=1/2B+2
W=1/2(12)+2
W=6+2
W=8

Answer 4

let's call the balls ,spheres (s) there are black(b) and white (w) so our first eqn is
1 s=b+w
then we have
2 (1/2)w=(1/3)b which becomes
3 w=(2/3)b
the original (1) becomes
4 s=b+(2/3)b=(5/3)b
finally we're given
5. 2s=3b+4
substituting what we already discovered
2((5/3)b)=3b+4
(10/3)b-3b=4
(1/3)b=4
b=12
now go back to 3.
w=(2/3)b=(2/3)(12)=8
finally fill these findings into 1.
s=12+8=20

Answer 5

1.let the black balls be 'x' and the white balls be 'y'
theequations are 1/2y=1/3x
i.e.3y-2x=0........(1) and 2(x+y)=3x+4=>-x+2y=4..........(2)
(2)*2 -(1)=-y=-8=>y=8 and substituting in (1) x=12
so the white balls=8 and the black balls=12

2.let the speeds of salman and amir be 'x' and 'y' respectively
the equations are 30/x=30/y +2=>(30+2y)/y=30/x
30y=30x+2xy =>1/x-1/y=1/15..........(1)
30/2x=30/y -1 =>30/2x=30-y/y 60x-2xy=30y=>60x-30y=2xy
2/y-1/x=2/30.(2)
adding 1/y=15/2 therefore y=7.5 and substituting x=5
therefore the speeds are salman 5 kmph and amir 7.5 kmph

Answer 6

W/2 = B/3 { *6=} 3W = 2B => W = B*2/3
- -- - - - -


2*(B + W) = 3B + 4
- - - - - - -
2B + 2W = 3B + 4
2B + 2*(B*2/3) = 3B +4

{mult by 3}

6B + 4B = 9B + 12
10B = 9B +12
B = 12
B/3 = 4 = W/2
W = 8

B + W = 12 + 8 =20

sheeeesh .....
velocity = v //distance = d // time = t

v = d/t
---
vs = ds/ts and va = da/ta (da=ds = 30km)

ts = ta + 2 (ii)

{"While covering a distance of 30km salman takes 2 hours more than Amir"}

call "second time" walking "vs2 & ts2"

ts2 = ds/(vs2) ((but vs2 = 2*vs)
sooo ... ts2 = ts/2

(ts2 = ) ... ts/2 = ta-1 => ts = 2ta-2
but ....
ts = ta + 2 (ii)
so we see that
2ta - 2 = ta +2 { subtract "ta" from both sides}
ta - 2 = 2 {add "2" to both sides}

ta = 4 ......
ts = 6

if ta = 4, then he walks 30km/4hour = 7.5 km/hr
if ts = 6, then he walks 30km/6hour = 5 km/hr

sorry for the clumsy notation, but the 'math'
is sound and the 'answer' is right