Fractions and Decimals (.538 x .538 - .462 x .462) / (1-.924) = ? 1 11 31 21 1 11 31 21 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Given expression = [ (.538)² - (.462)² ] / .076 = [(.538 + .462)(.538 - .462)] /.076=.076 / .076 = 1
Fractions and Decimals [{(0.1)² - (0.01)²} / {0.0001 + 1 }] is equal to ? 75 50 100 25 75 50 100 25 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Given expression = (0.01 - 0.0001) / (0.00001 + 1)= (.0099 / .0001) + 1 = 99 + 1= 100
Fractions and Decimals When 0.36 is written in simplest form, the sum of the numerator and the denominator is : 12 34 23 45 12 34 23 45 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 0.36 = 36/100 = 9/25.Sum of the numerator and denominator is 9 + 25 = 34
Fractions and Decimals If √4096 =64 ,then the value of √40.96 + √0.4096 + √0.004096 + √0.00004096 is ? 71.104 7.1104 711.04 .71104 71.104 7.1104 711.04 .71104 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP √4096/100+ √4096/10000 + √4096/1000000 + √4096/100000000=√4096 / 10 + √4096/ 100 + √4096/ 1000 + √4096/ 10000=64/10 + 64/100 + 64/1000 + 64/10000=6.4 + .64 + .064 + .0064 = 7.1104
Fractions and Decimals 999.99 + 25.012 = ? - 174.469 215 56 116 14 215 56 116 14 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP ? = (999.99)/25.012 + 174.469 = (1000/25) + 174.469= (40 + 174.469) = 214.469 = 215
Fractions and Decimals The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.3 2013 2011 2014 2012 2013 2011 2014 2012 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Suppose commodity X wil cost 40 paise more than Y after Z years.Then, (4.20 + 0.40Z) - (6.30 + 0.15Z) = 0.40 0.25Z = 0.40 + 2.10 Z= X will cost 40 paise more than Y 10 years after 2001 . i.e in 2011
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