One-fifth of the attendees at an international banquet were French men. If the number of French women at the banquet was 2/3 more than the number of French men and no other French citizens were present at the banquet, what was the share of non-French people at the banquet?
In the international banquet- French guy=1/5 French women=2/3×1/5 +1/5 =2/15+1/5 =5/15 Remaining non-French=1-(1/5+5/15) =1-8/15 =7/15.
`(601/49)xx(399/81)` ÷ `(29/201)` = ?
30.05% of 250.05 – 15.15% of 99.99 × 2.02 = ?
12.5% of 6400 + (17 × 25) = ?% of 2200+ 125
(? + 6.063.03 ) ÷ 10.08 + 21.89 × 6.97 = 1979.97 ÷ 10.96
In a cyclic quadrilateral ABCD, angle A = 70 degrees and angle B = 100 degrees. What is the measure of angle C?
? = 650.24 + 1124.97 – 14.992
? 2.03 + 121.96 – 55.98 + 129.78% of 59.98 = 399.98
185.92 ÷ 5.98 - (4.002 2 + 114.03 of 5.03 ÷ 18.99 of 6.04 = 5.01 of 2.99 + ? ÷ 12.02
√49 + 6.66% of 1725 + 22² = ?² - √361
(71.78 × 59.88 ÷ 17.92 ÷ 120.22 × 161.72)% of 32 = ?
