Concrete mix w/c ratio solution

Determine density of cement in the mix:
(6.28 sacks/cy) x (94 lb/sack) - 590 lb/cy

Determine densities of aggregates in the mix:
Fines: (590 lb/cy) x 2.25 = 1328 lb/cy
Coarse Aggregates: (590 lb/cy) x 3.25 = 1919 lb/cy

Determine volumes of aggregates, cement, and water in the mix:
Aggregates: (1328+1919)lb/cy / (2.65 x 62.4 pcf) = 19.6 cf/cy
Cement: (590 lb/cy)lb/cy / (3.15 x 62.4 pcf) = 3.00 cf/cy
Water: 1 cy - 19.6 cf - 3.00 cf = 4.37 cf = 32.7 gal

Calculate water/cement ratio in gallons of water per sack in one cubic yard:
w/c = 32.7 gal / 6.28 sacks = 5.2

The answer is (D) 5.2

Concrete mix w/c ratio question

From NCEES PE Civil Sample Questions & Solutions, 2008.

A highway project requires a concrete mix. On a weight basis, the design mix has the proportions 1:2.25:3.25. Cement content was specified at 6.28 sacks/cy. The aggregates are SSD and have specific gravities of 2.65 for both the fine and the coarse aggregate. The specific gravity of the cement is 3.15.

The water/cement ratio (gal/sack) of the concrete mix is most nearly:
(A) 3.6
(B) 4.2
(C) 4.7
(D) 5.2

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