If a and b are positive numbers and 100/a =95/b,
prove that:
(i) a > b
\(\frac{100}{a}=\frac{95}{b}\\ 100b=95a\\ 95a=100b\\ a=\frac{100b}{95}\\ a=\frac{95b}{95}+\frac{5b}{95}\\ a=b+more\\ a>b \)
i) prove \(\frac{105}{a}<\frac{100}{b}\)
\(\frac{100}{a}=\frac{95}{b}\\ \frac{100}{a}\times 1.05=\frac{95}{b}\times 1.05\\ \frac{105}{a}=\frac{99.75}{b}\\ \frac{105}{a}=\frac{100}{b}-\frac{0.25}{b}\\ \frac{105}{a}=\frac{100}{b}-\;\;a\; bit\\ so\\ \frac{105}{a}<\frac{100}{b} \)
If a and b are positive numbers and 100/a =95/b,
prove that:
(i) a > b
(ii) 105/a <100/b
If a and b are positive numbers and 100/a =95/b,
prove that:
(i) a > b
\(\frac{100}{a}=\frac{95}{b}\\ 100b=95a\\ 95a=100b\\ a=\frac{100b}{95}\\ a=\frac{95b}{95}+\frac{5b}{95}\\ a=b+more\\ a>b \)
i) prove \(\frac{105}{a}<\frac{100}{b}\)
\(\frac{100}{a}=\frac{95}{b}\\ \frac{100}{a}\times 1.05=\frac{95}{b}\times 1.05\\ \frac{105}{a}=\frac{99.75}{b}\\ \frac{105}{a}=\frac{100}{b}-\frac{0.25}{b}\\ \frac{105}{a}=\frac{100}{b}-\;\;a\; bit\\ so\\ \frac{105}{a}<\frac{100}{b} \)
If a and b are positive numbers and 100/a =95/b,
prove that:
(i) a > b
(ii) 105/a <100/b
(i) if 100/a = 95/b, this implies that 95a = 100b, which implies that a = (100/95)b......Then a must be greater than b since we have to multiply b by a quantity > 1 to get a
(ii) 105/a < 100/b..... Cross-multiplying........
105b < 100a........but, by definition, a =(100/95)b....so....
105b < 100(100/95)b .......divide both sibes by 100 →
(105/100)b < (100/95)b ........reduce the fractions →
(21/20)b < (20/19)b ........divide both sides by b →
(21/20) < (20/19) → cross-multiply, again
19*21 < 20*20 →
399 < 400......and since the left side is less than the right side......then the left side of the original inequality is also less than the right