Let's say that s is the total number of students, b is the number of boys, g is the total number of girls, n is the number of non-blonde girls, and e is the number if blonde girls. We know that s = b + g, b = g, g = n + e, e = g/3, and n = 10. Substituting for b in the first equation gives us s = g + g = 2g Then we substitute for n and e in the third equation and solve for g: g = g/3 + 10 g - g/3 = 10 g - (1/3)g = 10 (2/3)g = 10 g = 10 x (3/2) = 15 Finally, solve for s: s = 2g = 2 x 15 = 30
goldilocks and the 3 bears
3 wishes from the genie in the bottle
B/G = 4/3 so 3B = 4G B+G = 21 Multiply by 3: 3B+3G = 63 But 3B = 4G so 4G+3G = 63 ie 7G = 63 so that G = 9 So B+G=21 means B = 12 3 more boys so B' = 12+3 = 15 1 more girl so G' = 9+1 = 10 So new ratio = B'/G' = 15/10 = 3/2 or 3 to 2
This is a Ditloid 147 maximum break in a game of snooker.
The B. G. was born on September 3, 1980.
3 b with g is 3 bears with goldilocks
3 strikes you're out at the old ball game
Suppose there are B boys and G girls. Then Sophie has B brothers and (G-1) sisters - remember to leave Sophie out. So B = (G-1) + 5 ie B = G + 4 Also Joe has (B-1) brothers and G sisters So (B-1) = 2*G Substituting the value of B from Sophie's equation into this one gives (S+4 - 1) = 2*G ie G+3 = 2*G Subtract G from both sides: 3 = G and then B = G+4 = 7 So number of children in the family = B+G = 7+3 = 10.
The B. G. was born on September 3, 1980.
B. G. S. Ellett has written: 'The capacity of British Standard road gullies'
The Notorious B. I. G. was born on May 21, 1972.
Intro: g-g-a-a-a-a-a-a c-c-g-g-a-a-a Verse: g-g-g-g g-g-g-c g-g-g-g-g-g-g-g-g-a-b-a Refrain/Bridge: g-b-d-e-d-b-b-b-b-a g-b-d-e-d-b-a Chorus: b-b-a-g-g-g-g-g-g-a-b-a b-b-a-g-g-g-g-g-g-a-b-a b-b-a-g-g-g-g-g-g-b-a-a b-b-a-g-a-b b-b-b-b-b-a-g-g-g-g-g-g-a-b-a b-b-a-g-g-g-g-g-g-a-b-a b-b-a-g-g-g-g-g-g-b-a-a b-b-a-g-a-b b-b-b-b-b-a-g-a-b b-b-b-b-b-a-g
Most Medals:Apolo Anton Ohno: 8 (G, S, B) (2,2,4)Most Golds:Bonnie Blair: 5 (G, S, B) (5,0,1)Eric Heiden: 5 (G, S, B) (5,0,0)
3 Bears in Goldilocks
The answer is NOT a)a) I(g) + e → I-(g)b) I2(g) → 2I(g)c) I(g) → I+(g) + ed) Na(g) + I(g) → NaI(s)e) Na(s) + 1/2I2(s) → NaI(s)
G. B. Bogatov has written: 'Televidenie na zemle i v kosmose G. B. Bogatov. --' -- subject(s): Television