A specialized type of sand consists of 40% mineral X by volu

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.

2/3 = (3 - Wavg)/(Wavg - 2.5)
Wavg = 2.8 gms/\(cm^3\) = 2,800,000 gms/\(m^3\)

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

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

\(0.6*3*(100)^3+0.4*2.5*(100)^3\)

\(=10^6(0.6*3+0.4*2.5)=10^6(1.8+1)=2.8*10^6=2800000 \hspace{2} grams\)

Ans: "B"

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)

Well done! OA is posted...and is in fact B.

I thought I'd get some of you on the conversion

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.

"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 (https://gmatclub.com/forum/after-multipl ... l#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.

The more exposed you are to high level questions, the more intuitive they will get with time.

I am going to solve only the mixture part of this question. I used Karishma's method described here https://gmatclub.com/forum/john-and-ingrid-pay-30-and-40-tax-annually-respectively-107983.html#p855286

40/60=2/3

(3/5)*(3-2.5)+2.5=2.8

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

(0.40x)(2.5g/1cm3)(1,000,000cm3/1m3) + (0.60)(3g/1cm3)(1,000,000cm3/1m3)

= 2,800,000

B.

I think this is simple example not 700 level question.

1 cubic meter = \(100^3\) cubic centimeter
= 1,000,000 cubic centimeter

out of that 40% = In 400,000 cubic centimeter is mineral X
So Weight of Mineral X = 400,000 x 2.5 = 1,000,000 gm

60% = In 600,000 cubic centimeter is mineral Y
So Weight of Mineral Y = 600,000 x 3 = 1,800,000 gm

Total = 1,000,000 + 1,800,000 = 2,800,000

B is the answer

KEY is to read the question carefully.