[#38] 2 Min. Challenge : Sand

Question

1. Time yourself
2. Solve as fast as you can
 3. Write your solution here, and your time please 

In what proportion must sand be mixed with cement to gain 25%, by selling it at the cost price? (Assume sand is free of cost.)

a. 1 : 5 
b. 1 : 3 
c. 1 : 4 
d. 1 : 2
e. 1 : 6

kpadma · Answer

Ans: 1:4 

(1-X) * 1.25 = 1 
Solve for X
X : 1-X = 1:4

Geethu · Answer

B) 1:3 

I am not sure whether this answer is correct. I took less than 2 mins

I assumed 100$ for 100 pounds of cement. To get 25% gain only 75 pounds of cement has to be taken. So, remaining 25 pounds will be sand.

25:75 => 1:3

mdfrahim · Answer

agreed with geethu. 1:3 (20 seconds)
if the ratio is 1:3 then in 100 Kg cement we will have...
Sand : 1/(1+3) * 100 = 25
Cement : 3/(1+3) 8 100 = 75

Praetorian · Answer

c. 1: 4 is the correct answer. only kpadma got this right.

Let cost price of 1 lb cement = $ 1
and cost price of 1 lb sand = 0
As the mixture is sold at cost price, cost price of mixture = 1/1.25 = 4/5

Ratio of quantity of sand to quantity of cement to gain 25%

= 1/5 : 4/5 = 1/4

Makky07 · Answer

Kpadma,

Can you please explain this:  (1-X) * 1.25 = 1  


Thanks

kpadma · Answer

Assume 1 unit of mixture, that has X frational units of sand and 
the rest , 1-X, cement.

Again, assume that you buy 1 units of cements at 1$ and sell the
mixture for a 1$ , thus making 25% profit.

(1-X) is the cement in 1 unit of mixture, sold for 1$ , thus making 25%
profit.

So, (1-X) * 1 = cost to procure the cement.
 1.25 = profit+cost
 1*1 = sale price

(1-X) * 1 * 1.25 = 1*1
(1-X) 1.25 = 1

Makky07 · Answer

Thank you very much. That was a good explanation.

[#38] 2 Min. Challenge : Sand

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