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If the price of sugar falls by 75/2 %, a person can buy 33

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If the price of sugar falls by 75/2 %, a person can buy 33 [#permalink] New post 14 Oct 2005, 06:00
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If the price of sugar falls by 75/2 %, a person can buy 33 kgs more for $560. If the price had been risen by 75/2 %, how much sugar would he have bought for the same sum ?

a)36
b)40
c)42
d)45
e)46

Two cars A and B start from Boston and New York respectively simultaneously and travel towards each other at constant speeds. After meeting at a point between Boston and New York the two cars A and B proceed to their respective destinations of New York and Boston. Car A reaches New York 40 minutes after the two cars have met and Car B reaches Boston 90 minutes after they have met. How long did Car A take to cover the distance between Boston and New York?

A. 1 hour
B. 1 hour 10 minutes
C. 2 hours 30 minutes
D. 1 hour 40 minutes
E. 2 hours 10 minutes


Need smiple explanation pls for the above 2.
thks
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Re: PS : price & cars [#permalink] New post 14 Oct 2005, 08:00
[quote="sushom101"]If the price of sugar falls by 75/2 %, a person can buy 33 kgs more for $560. If the price had been risen by 75/2 %, how much sugar would he have bought for the same sum ?

a)36
b)40
c)42
d)45
e)46

if x is original price per bag and y is original number of bags he can buy. x * y = 560. After rebate .625 * x * (y+33) = 560. solve for x and y.
Then 560/(1.375*x) is answer.
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 [#permalink] New post 14 Oct 2005, 08:21
Is question B complete? Looks like one more information is required to calculate the solution.
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 [#permalink] New post 14 Oct 2005, 08:33
Second question:

The formula first (commit it):
if two trains starts simulateneously, constant speed, opposite direction:

After they meet:

sqrt (time taken for A to reach / time taken by B to reach) = avg speed of B / avg speed of A

from the question, therefore, we can deduce:
speed of B/speed of A = sqrt(40/90) = 2/3

let A took t time to complete journey, we can say B took t + 50 mins(remember after they meet, B took 90 mins while A took 40 mins)

therefore,
3 * t = 2 * (t + 50) or
t = 100 mins

A took an hour and 40 mins to complete the journey
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Re: PS : price & cars [#permalink] New post 14 Oct 2005, 08:52
the solution for second question is :
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 [#permalink] New post 14 Oct 2005, 09:04
At= time that car A travels until it meets B
Bt= time that car B travels until it meets A

1. At=90B min
2. Bt=40A min

To solve for t, we must substitute equation 2 into equation 1

t=60

60+40=1hr 40min

Choose D.
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 [#permalink] New post 15 Oct 2005, 03:50
thanks all,

can i get a similar ...simpler method for question 1 pls. :?: :?:
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 [#permalink] New post 15 Oct 2005, 10:48
sushom101 wrote:
thanks all,

can i get a similar ...simpler method for question 1 pls. :?: :?:


Let me explain in fractions. Hope you will find this easier:

x = actual price per kg
y = number of kgs

We have: x * y = 560

Price decreased by 75/2% = x - 75/200x = 5/8x. With this we could buy an addl 33 kgs for the same amount. Hence 5/8x * (y+33) = 560

Solving above, we will have x = 112/11 (Note: there is a unknown y in the above eqn ... but when y is multiplied by x, we could replace with 560).

Now, lets assume that x is increased by 75/2% = x + 75/200x = 275/200x = 11/8x = 11/8 * 112/11 = 112/8 = 14

Subt this in x * y = 560
y = 560 / 14 = 40

Answer is B
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 [#permalink] New post 15 Oct 2005, 17:23
sushom101 wrote:
thanks all,

SIMPLE Answers for 1 & 2

can i get a similar ...simpler method for question 1 pls. :?: :?:



Ask and you shall receive

Question 1.


discount is 75/2% a fancy way of saying 37.5% = change to simple fraction = 3/8 thus the discount is 5/8 which is 62.5%

with a 37.5% discount you can 33 more kgs for $560
price x volume = $560

write down the equation 5/8P(V+33) = $560

P*V = 560 thus, V = 560/P substitute

5/8P(560/P+33) = 560

350 + (33)5/8P = 560


(33)*5/8P = 210

P = 210 x 8/(3*11*5)

P = (14*8)/11

So what will V be when you increase the Price by 37.5%?

1.375 = 11/8

11/8*P = 11/8 x (14)(8)/11 = 14


Thus 14V=560

V=560/14=40



*remember don't make any unnecessary calculations



Second problem

These rate problems require you to find a common element between the 2 objects. In this particular problem, Time (the time it took to meet) is the common element.
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  [#permalink] 15 Oct 2005, 17:23
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