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A specialized type of sand consists of 40% mineral X by volu [#permalink]

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02 Oct 2011, 14:28

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A specialized type of sand consists of 40% mineral X by volume and 60% mineral Y by volume. If mineral X weighs 2.5 grams per cubic centimeter and mineral Y weighs 3 grams per cubic centimeter, how many grams does a cubic meter of specialized sand combination weigh? (1 meter = 100 centimeters)

A. 5,500,000 B. 2,800,000 C. 55,000 D. 28,000 E. 280

A specialized type of sand consists of 40% mineral X by volume and 60% mineral Y by volume. If mineral X weighs 2.5 grams per cubic centimeter and mineral Y weighs 3 grams per cubic centimeter, how many grams does a cubic meter of specialized sand combination weigh? (1 meter = 100 centimeters)

a) 5,500,000 b) 2,800,000 c) 55,000 d) 28,000 e) 280

Good luck...cheers.

Let me share what goes through my mind here to arrive at the answer orally: Volume of mineral X is 40% and that of mineral Y is 60% i.e. they are in the ratio 2:3. So if we have 2 \(cm^3\) of X, we have 3 \(cm^3\) of Y. Weight of X is 2.5 gms/\(cm^3\) so weight of 2 \(cm^3\) of X is 5 gm Weight of Y is 3 gms/\(cm^3\) so weight of 3 \(cm^3\) of Y is 9 gm therefore, weight of 5 \(cm^3\) of mixture is 14 gm So weight of 10 \(cm^3\) of mixture is 28 gm Since 1 \(m^3\) = 1,000,000 \(cm^3\), weight of 1,000,000 \(cm^3\) of mixture must be 2,800,000 gms
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Let the volume be 1 m^3 = 1m * 1m * 1m = 100cm * 100cm * 100cm = 1,000,000 cm^3 By volume 40% is X = 400,000 cm^3 60% is Y = 600,000 cm^3

By weight, In 1 cm^3, X is 2.5 gms In 400,000 cm^3 , X = 2.5 * 400,000 = 1,000,000 gms In 1 cm^3, Y is 3 gms In 600,000 cm^3, Y = 3 * 600,000 = 1,800,000 gms Total gms in 1 m^3 = 1,000,000 + 1,800,000 = 2,800,000

OA B.
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My dad once said to me: Son, nothing succeeds like success.

A specialized type of sand consists of 40% mineral X by volume and 60% mineral Y by volume. If mineral X weighs 2.5 grams per cubic centimeter and mineral Y weighs 3 grams per cubic centimeter, how many grams does a cubic meter of specialized sand combination weigh? (1 meter = 100 centimeters)

a) 5,500,000 b) 2,800,000 c) 55,000 d) 28,000 e) 280

Good luck...cheers.

1 cubic cm = 0.4 x 2.5 + 0.6 x 3 = 2.8g 1000000 cubic cm = 1 cubic m = 2.8 x 10^6 (B)
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hey karishma, can this question be done using your method of weighted averages? I'm trying to do the question using weighted averages but am not able to. Your help would be appreciated.

Yes, you can.

Weight of X is 2.5 gms/\(cm^3\) Weight of Y is 3 gms/\(cm^3\) Weight of a mixture of X and Y such that they are in the ratio 40%:60% i.e. 2:3 will be Wavg.

Note: If you notice carefully, you don't need to do this calculation. The Wavg will lie between 2.5 and 3. Also, Wavg will be in units of gms/\(cm^3\) so you will need to multiply it by 1,000,000 to convert to gms/\(m^3\). Hence, only suitable answer will be 2.8*1,000,000
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A specialized type of sand consists of 40% mineral X by volume and 60% mineral Y by volume. If mineral X weighs 2.5 grams per cubic centimeter and mineral Y weighs 3 grams per cubic centimeter, how many grams does a cubic meter of specialized sand combination weigh? (1 meter = 100 centimeters)

a) 5,500,000 b) 2,800,000 c) 55,000 d) 28,000 e) 280

OA to follow

Good luck...cheers.

1 m^3= (100)^3 cm^3

So, 3 gram in 1 cm^3; Y=3*(100)^3 gram in (100)^3 cm^3 OR 1 m^3

Likewise, 2.5 gram in 1 cm^3; X=2.5*(100)^3 gram in (100)^3 cm^3 OR 1 m^3

A specialized type of sand consists of 40% mineral X by volume and 60% mineral Y by volume. If mineral X weighs 2.5 grams per cubic centimeter and mineral Y weighs 3 grams per cubic centimeter, how many grams does a cubic meter of specialized sand combination weigh? (1 meter = 100 centimeters)

a) 5,500,000 b) 2,800,000 c) 55,000 d) 28,000 e) 280

Good luck...cheers.

Let me share what goes through my mind here to arrive at the answer orally: Volume of mineral X is 40% and that of mineral Y is 60% i.e. they are in the ratio 2:3. So if we have 2 \(cm^3\) of X, we have 3 \(cm^3\) of Y. Weight of X is 2.5 gms/\(cm^3\) so weight of 2 \(cm^3\) of X is 5 gm Weight of Y is 3 gms/\(cm^3\) so weight of 3 \(cm^3\) of Y is 9 gm therefore, weight of 5 \(cm^3\) of mixture is 14 gm So weight of 10 \(cm^3\) of mixture is 28 gm Since 1 \(m^3\) = 1,000,000 \(cm^3\), weight of 1,000,000 \(cm^3\) of mixture must be 2,800,000 gms

hey karishma, can this question be done using your method of weighted averages? I'm trying to do the question using weighted averages but am not able to. Your help would be appreciated.

Re: A specialized type of sand consists of 40% mineral X by volu [#permalink]

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25 Mar 2013, 22:58

Thanks for the quick reply Karishma! Now i'm digressing from the topic, but I want your advice on the 700 level questions. No matter what topic I pick-up, i usually end up doing those questions wrong or having no idea as to how to go about them. Is there some specific stratergy or technique involved or should I just keep on practicing?

Thanks for the quick reply Karishma! Now i'm digressing from the topic, but I want your advice on the 700 level questions. No matter what topic I pick-up, i usually end up doing those questions wrong or having no idea as to how to go about them. Is there some specific stratergy or technique involved or should I just keep on practicing?

The thing about 700+ level questions is that you need to be more innovative in them. You need to think of all that you know in Quant and link it up - i.e. focusing on one topic (on which the question seems to be based) is not enough. As an example, check out this question I made for Veritas some time back.

"After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?"

What do you think it tests? By just reading the question, you cannot guess that all you need to know to solve it within one minute are your divisibility rules. Also, you need to start with whatever is definitive. Check out this link for the solution (after-multiplying-a-positive-integer-a-which-has-n-digits-147350.html#p1183033)

700+ level questions use the same fundamentals but are harder because of the way the question is worded. Practice high level questions and check out every different method available to solve them. You can use methods learned in one topic to solve questions of another if you can just identify the parallel. e.g. check out this post where I solve a permutations question using sets (method 2) http://www.veritasprep.com/blog/2012/01 ... e-couples/

The more exposed you are to high level questions, the more intuitive they will get with time.
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1. We need to find the weight of a certain mix. 2. The mix is 40% x and 60% y. The volume of the mix for which we need to find the weight is 1 cubic meter 3. 1 cubic meter of the mix contains 0.4 cu.m of x and 0.6 cu.m of y. 4. Weight of 1 cubic meter of x is 2500000g and weight of 1 cubic meter of y is 3000000g 5. So weight of 0.4 cubic meter of x is 1000000g and weight of 0.6 cubic meter of y is 1800000g 6. Total weight of 1 cubic meter of the mix is 2,800,000g
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Re: A specialized type of sand consists of 40% mineral X by volu [#permalink]

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10 Jun 2015, 23:42

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