Consider following sums of fractions:
A. 16/19 + 1/4
B. 11/14 + 1/5
C. 19/21 + 1/3
Choose the correct ascending order of above sums of fractions from the following options:
B < A < C
To compare the three sums, convert each pair of fractions to a single fraction or decimal so they can be directly compared.
Sum A: 16/19 + 1/4 = (16 × 4 + 1 × 19)/(19 × 4) = (64 + 19)/76 = 83/76 ≈ 1.092.
Sum B: 11/14 + 1/5 = (11 × 5 + 1 × 14)/(14 × 5) = (55 + 14)/70 = 69/70 ≈ 0.986.
Sum C: 19/21 + 1/3 = (19 × 3 + 1 × 21)/(21 × 3) = (57 + 21)/63 = 78/63 = 26/21 ≈ 1.238.
Comparing the decimal values: B ≈ 0.986 is the smallest, A ≈ 1.092 is the middle value, and C ≈ 1.238 is the largest.
Therefore, the correct ascending order of the sums is B < A < C.
Similar Questions
If numerator of a fraction is increased by 25% and the denominator is decreased by 15%, the fraction becomes 15/17. Find the value of the original fraction?
Important Questions from Fractions
Which fraction among the following is the least ?
\(\frac{5}{11}, \frac{7}{12}, \frac{8}{13}, \frac{9}{17}\)
Find the value of the following expression:
\(\frac{{3 \div 1 \times 2 + 5 - 2}}{{3 \times 3 - 2}}\)
Simplify the expression 441 ÷ \(\left[270 \div \frac{3}{7}+\left(17\div \frac{1}{3}\right)-\left(8\frac{1}{2}-\frac{5}{2}\right)\right]\)
If the sum of two positive numbers is 65 and the square root of their product is 26, then the sum of their reciprocals is:
The value of \(9 \div [\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}\div(\frac{3}{4}-\frac{1}{3})\;of\;\frac{2}{9}]\) is:
