A student adds \(\tfrac35\) unit of a solution, then adds 0.4 unit of the same solution, and finally removes \(\tfrac1{10}\) unit. What is the net change in the quantity of the solution?
increase of 0.9 unit
Net change: \(\tfrac35+0.4-\tfrac{1}{10} = 0.6+0.4-0.1 = 0.9\).
Hence, the net change in the quantity of the solution is an increase of 0.9 unit.
Simplify the given expression.
$9 \times 0.9 \times 0.09 \times 0.009 \times \frac{1}{0.3} \times \frac{1}{0.03} \times \frac{1}{0.003}$
Arrange the following numbers in their increasing order.
(1) $-0.96$
(2) $0.83$
(3) $0.24$
(4) $-0.64$
(5) $0.58$
The value of \(\frac{1}{4} + \frac{{[{{(20.35)}^2} - {{(8.35)}^2}] \times 0.0175}}{{{{(1.05)}^2} + (1.05)(27.65)}}\) is:
The value of \(0.4\overline 6 + 0.7\overline {23} - 0.3\overline 9 \times 0.\overline 7 \) is:
The value of \(\frac{48.3\times[(4.95)^2+4.95\times13.25]}{[(12.55)^2-(5.65)^2]\times19.8} \) is:
Find the value of (1.6) 3 - (0.9) 3 - (0.7) 3.
What is the value of x, if \(5\left( {1 - \frac{x}{5}} \right) - (5 - x) - \frac{1}{{200}}{\rm{of (20 - x) = 0}}{\rm{.08}}\) ?