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Question

1) the cost of manufacturing the car is made of three items-cost of materials, labour and overheads. in 1992, the cost of these items was in the ratio of 5:4:3. in 1993, the cost of materials rose by 16%, the cost of labour rose by 10%, cost of overheads reduced by 8 %. find the increase % in th price of the car.

2) X litres if milk and water contains x% milk. how many litres of milk should be added to mixture so that it contains 3x% milk.

3) a motorist reduces his distance covered annual(in km) by x% when the price of petrol is increased by y%, find the increase percent in his annual petrol bill.

P.S. All these problems are related to PERCENTAGE

Jun Agruda · Accepted Answer

= 100%(0.16[5/12] + 0.1[4/12] + 0.08[3/12])
= 100%(1/15 + 1/30 + 1/50)
= 100%([10 + 5 + 3]/150)
= 100%(3/25)
= 300/25% or 12%

Answer (1): 12%
------------
x/100 + y = 3x/100
x + 100y = 3x
100y = 2x
y = 2x/100 or x/50

Answer (2) = x/50 liters
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Answer (3): Infinitely many solutions therefore there is no definite solution to this problem.

? · Answer

1)
Our ratio is 5:4:3.  If we add those together, we get 12.
Now let's increase each by the listed percentage.
5 times 1.16 becomes 5.8
4 times 1.10 becomes 4.4
3 times 1.08 becomes 3.24
Adding all these together, we get 13.44
(13.44 - 12)/12 = 1.44/12 = 0.12
So our total cost increased by 12%


2)
You used two different capitalizations of X and x, so I'm treating them as two different variables.

Let Y = the amount of milk we add
The amount of milk we have now is x% of X = xX/100
The amount of milk we will end up with is 3x% of (X+Y) = 3x(X+Y)/100
So we have:
xX/100 + Y = 3x(X+Y)/100
Multiply both sides by 100:
xX + 100Y = 3xX + 3xY
Y(100 - 3x) = 2xX
Y = 2xX / (100 - 3x)

So we need to add 2xX/(100-3x) litres of milk to triple the concentation.
(if your X and x were supposed to be the same variable, then the numerator becomes 2x^2)


3)
Percentage change = (new value - old value) / (old value) Ã 100%
Let P = old petrol bill
(1 - x/100)(1 + y/100)P = new petrol bill
So we have:
[(1 - x/100)(1 + y/100)P - P]/P Ã 100%
= [(1 + y/100 - x/100 - xy/10,000)P - P]/P Ã 100%
=[(1 + y/100 - x/100 - xy/10,000) - 1] Ã 100%
=(y/100 - x/100 - xy/10,000) Ã 100%
=(y - x - xy/100)%

Deb · Answer

I find all of you have missed the point that overhead is reduced by 8%.

Material  5    becomes 5.8
Labour    4   becomes  4.4
Overhead 3  becomes   2.76

Therefore, 
Price  12 becomes 12.96
That is increase of 8%.

? · Answer

4. extracts of solutions given for  you to fill in all the details ... hope you are smart enough !

1.  cost A : 5+4+3 =12
     cost B  : 5.8+ 4.4+ 3.24 = 13.44 , so % rise in car price is 12% 

2. milk is X*x/100, water is X - X*x/100 
 add y milk , so Xx/100 + y milk in total  X +y quantity.
 then  100 *(X*x/100 +y)/(X+y) = 3*x  gives
y = 2X*x/(100-3*x).....................................................* is multiplication

3. bill = price * km = p*k

 new bill = p*(y/100+1) * k*(1- x/100)

new bill and old bill...... difference is p*k*((y/100+1) * (1- x/100)-1)

so % change is 100**((y/100+1) * (1- x/100)-1)

ranjankar · Answer

5+ 4 + 3 = 12

5 X 1.16 = 5.8
4 X 1.1 =   4.4
3 X 0.92 = 2.76
Total = 12.96

0.96 / 12 X 100 = 8% ANSWER

Vaibhav · Answer

first
12%

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Can you solve any of these problems or all these problems? are you smart enough?

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