Help Please Water Cement ratio Prob

Can somebody help me with this problem, I am not getting it right.

Given: A concrete mix contains cement, moist sand, and moist coarse aggregate in the
following proportion by weight: 1:1.8:2.6. The following specifications are given:
Cement specific gravity = 3.15
SSD sand (m.c. = 0.5%) specific gravity = 2.70
SSD coarse aggregate (m.c. = 0.7%) specific gravity = 2.60
Added water 5.8 gal per sack cement
Air 3% (by volume)
The aggregates used for mixing the concrete had the following properties:
Wet sand: moisture content = 6%
Wet coarse aggregate: moisture content = 4%
Find: The water content of the concrete is most nearly:
(A) 6.1 gal/sack
(B) 7.1 gal/sack
© 7.9 gal/sack
(D) 8.5 gal/sack

1.8x94=169.2 lb ( sand with .5% moisture)

When wet sand (mc=6%), 5.5/100.5 x 169.2= 9.26lb

Coarse: 2.6x94= 244.4 lb ( coarse with .7% moisture)

When wet coarse (mc= 4%), 3.3/100.7 x 244.4= 8.01lb

Total extra water= 9.26+8.01= 17.27 lb=2.07 gal

Total water= 5.8 + 2.07= 7.87 gal/sack

Thank you very much

jcvv98 said:

1.8x94=169.2 lb ( sand with .5% moisture)

When wet sand (mc=6%), 5.5/100.5 x 169.2= 9.26lb

Coarse: 2.6x94= 244.4 lb ( coarse with .7% moisture)

When wet coarse (mc= 4%), 3.3/100.7 x 244.4= 8.01lb

Total extra water= 9.26+8.01= 17.27 lb=2.07 gal

Total water= 5.8 + 2.07= 7.87 gal/sack

Click to expand...

Where does 100.5 and 100.7 in the equations come from? would you mind explain it? Thank you.

JessieZ said:

jcvv98 said:

1.8x94=169.2 lb ( sand with .5% moisture)

When wet sand (mc=6%), 5.5/100.5 x 169.2= 9.26lb

Coarse: 2.6x94= 244.4 lb ( coarse with .7% moisture)

When wet coarse (mc= 4%), 3.3/100.7 x 244.4= 8.01lb

Total extra water= 9.26+8.01= 17.27 lb=2.07 gal

Total water= 5.8 + 2.07= 7.87 gal/sack

Click to expand...

Where does 100.5 and 100.7 in the equations come from? would you mind explain it? Thank you.

Click to expand...

JessieZ said:

jcvv98 said:

1.8x94=169.2 lb ( sand with .5% moisture)

When wet sand (mc=6%), 5.5/100.5 x 169.2= 9.26lb

Coarse: 2.6x94= 244.4 lb ( coarse with .7% moisture)

When wet coarse (mc= 4%), 3.3/100.7 x 244.4= 8.01lb

Total extra water= 9.26+8.01= 17.27 lb=2.07 gal

Total water= 5.8 + 2.07= 7.87 gal/sack

Click to expand...

Where does 100.5 and 100.7 in the equations come from? would you mind explain it? Thank you.

Click to expand...

JessieZ said:

jcvv98 said:

1.8x94=169.2 lb ( sand with .5% moisture)

When wet sand (mc=6%), 5.5/100.5 x 169.2= 9.26lb

Coarse: 2.6x94= 244.4 lb ( coarse with .7% moisture)

When wet coarse (mc= 4%), 3.3/100.7 x 244.4= 8.01lb

Total extra water= 9.26+8.01= 17.27 lb=2.07 gal

Total water= 5.8 + 2.07= 7.87 gal/sack

Click to expand...

Where does 100.5 and 100.7 in the equations come from? would you mind explain it? Thank you.

Click to expand...

It is the moisture content

When wet sand (m.c.=6%) is used, it contains free water, therefore, it is equal to 5.5/100.5. It is like getting the 10% of total wt. of W=127 lb. A common mistake is to calculate wt. of water = 10% of 127=12.7lb. Wrong. Since the water content is 10%, distribute the total wt. in 100:10 proportions (solids:water). Therefore, wt. of solids is 100/110 x 127=115.5 lb and the wt. of water is 11..5 lb. Hope this helps.