CPhill

64% of the fruits at a stall are oranges. The rest are apples and pears in the ratio 4:5. There are 880 more oranges than pears. How many fruits are there at the stall altogether?

This is similar to the other problem.....the apples and pears comprise the other 36%...and if they're in the ratio of 4:5  .....  there must be  16% apples and 20% pears.....

Again, let T be the total....and we know that

.64T = .20T + 880    subtract .20T   from both sides

.44T  = 880      divide both sides by .44

T = 880/.44  = 2000 total fruits

Check  .....there are .64 x 2000 = 1280 oranges

And there are .20  x 2000 = 400  pears

And (1280 - 400)  = 880

20% of the spectators in a match are women. 10% are girls, 30% are boys and the remaining 3600 spectators are men. Find the total number of spectators at the match?

(20 +  10 + 30)%  =  60% are not men....so 40% are.....and there are 3600 men....so....

Let T be the total number of spectators

And we have.....

.40T = 3600   divide both sides by ,40

T = 3600/.40  = 9000  = the total number of spectators

51% of the fish in a pond are swordtails. The rest are mollies and guppies in the ratio 5:2. There are 222 more swordtails than guppies. How many fish are there in the pond altogether?(please help me with this within 5 minutes)

Thr mollies and guppies comprise the other 49% of the fish.....if they are in the ratio of 5:2, then 35% of the total are mollies and 14% are guppies.

Let T be the total number of fish.  Then.........

51% T   =  14%T + 222      or

.51T = .14T + 222     subtract .14T from both sides

.37T = 222   divide both sides by ,37

T = 222/.37 = 600 fish total

Check .....    51% are swordtails = 306

14% are guppies = 84

And 306 - 84 = 222

(19 + 23 + 28)%   = 70%  of the fruits are something other than mangoes.

Then, mangoes must comprise the other 30%.......let N be the total numer of fruits....so.....

180  = 30%  of N    .....  or   ......

180  = .30N      divide both sides by  .30

180/.30 = N   = 600 total fruits

1-(5/6)^x>=9/10  rearrange as

1 - 9/10  ≥ (5/6)^x    

1/10 ≥ (5/6)^x   or

(5/6)^x ≤ 1/10      take the log of both sides

log(5/6)^x ≤ log ( 1/10)  and we can write

x log(5/6) ≤ log ( 1/10)    divide both sides by log(5/6).....note....since this  is negative, we must reverse the inequality sign

x ≥ log (1/10) / log (5/6)

x ≥  12.629  (approx)

Here's another way to approach this just using some logic.....[although I have no quibble with Melody's method]

Note that , x cannot be 1 because the fraction would = 0 .  Also, if x were less than 1, the numerator would be positive and denominator negative. Tuis, the fraction would be negative.....no good, either.

And x cannot equal 2, because then, the fraction is undefined.

Likewise, if x were greater than 2, the numerator would be negative and the denominator positive, resulting in the same situation as before - a negative fraction.

Thus, x must lie between 1 and 2, because this makes numerator and denominator negative [i.e., the fraction positive and > 0]

Saina is deciding to buy a car. She can choose between a diesel car and a petrol car. 
The diesel car costs Rs. 7310 than the petrol car. However, it can travel 15.9 kilometres per litre, whereas the petrol car can only travel 11.3 kilometres per litre.
Also diesel costs Rs. 55.65 per litre, and petrol costs Rs. 58.76 per litre.
If she buys a diesel car, how many kilometres must she drive before the savings in fuel covers the extra price of the diesel car?

.................................................................................................................................................

The diesel car costs  55.65/15.9 = 3.5R/km to operate ..... and the petrol car costs  58.76/ 11.3 = 5.2R/km to operate

So  (5.2 - 3.5)x  = 7310    where x is the answer we're looking for

1.7x = 7310    divide both sides by 1.7

x =  4300km

We have....

19 =  (1/2)(17)t  + (1/2)(-1.5)t^2   simplify  by multiplying through by 2

38 = 17t - (1.5)t^2      rearrange

1.5t^2 - 17t + 38 = 0    multiply through by 2 again

3t^2 - 34t + 76  = 0      this won't factor.....using the quadratic formula and the onsite solve, we have

$${\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{34}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,\small\textbf+\,}}{\mathtt{76}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{61}}}}{\mathtt{\,-\,}}{\mathtt{17}}\right)}{{\mathtt{3}}}}\\
{\mathtt{t}} = {\frac{\left({\sqrt{{\mathtt{61}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{17}}\right)}{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{3.063\: \!250\: \!108\: \!031\: \!115\: \!2}}\\
{\mathtt{t}} = {\mathtt{8.270\: \!083\: \!225\: \!302\: \!218\: \!1}}\\
\end{array} \right\}$$

Bala is setting up a factory to manufacture tennis b***s. If he buys a machine that costs Rs. 59728, then he can manufacture a tennis ball for Rs. 3.7. On the other hand, if he buys a machine that costs Rs. 44500, he can manufacture a tennis ball for Rs. 8.4.
How many tennis ball should he make so that his total cost is the same no matter which machine he buys?

Let x  be the number of teniis b***s to make.....then, equalizing the total costs, we have

59728 + 3.7x   =  44500 + 8.4x       simplify

4.7x  =  15228      divide both sides by 4.7

x = 3240  b***s should be manufactured to equalize the cost

The corners of a square with side 12 is cut away to form an octagon with all sides equal. What is the length of a side of the octagon so formed?

Let S be the side of the octagon......then.....before the corners are to be cut from the square, an isosceles right triangle is formed at each of these vertexes. The hypotenuse of this triangle = S.....and  the other two sides are S/√2 each. 

Then, before we make the cuts, the side of the square is composed of lengths of S/√2,  S and S/√2 ...or,  put another way.....

S + 2S/√2  = 12   simplify

S + √2S  = 12      factor

S(1 + √2)  = 12

S = 12/ [ 1 + √2 ]

See the following illustration :

 GE, EF = S =  12 / [1 + √2]   ....     CE and FD  = S/√2