Solving the given expression,
Suggested Test Series
Value of0.97×0.97×0.97+0.03×0.03×0.030.97×1.94−1.94×0.03+0.03×0.06 \frac{0.97 \times 0.97 \times 0.97 + 0.03 \times 0.03 \times 0.03}{0.97 \times 1.94 - 1.94 \times 0.03 + 0.03 \times 0.06}0.97×1.94−1.94×0.03+0.03×0.060.97×0.97×0.97+0.03×0.03×0.03 is equal to:
The value of (0.29‾÷0.32‾)×6.60.95‾÷1.05‾×1.2‾\frac{(0.\overline{29} \div 0.\overline{32}) \times 6.6}{0.\overline{95} \div 1.\overline{05} \times 1.\overline{2}}0.95÷1.05×1.2(0.29÷0.32)×6.6 is
4110−[212−{56−(25+310−415)}]4\frac{1}{10}-\left[2\frac{1}{2}-\left\{\frac{5}{6}-\left(\frac{2}{5}+\frac{3}{10}-\frac{4}{15}\right)\right\}\right]4101−[221−{65−(52+103−154)}]on simplification gives
Find the value of [(36÷9)×{568+141×(6−5)}]\left[\left(36 \div 9\right) \times \left\{\frac{56}{8} + \frac{14}{1} \times (6 - 5)\right\}\right][(36÷9)×{856+114×(6−5)}]