| Work done =$ \dfrac{5}{8} $ |
| Balance work =$ \left(1 -\dfrac{5}{8} \right) $=$ \dfrac{3}{8} $ |
Let the required number of days be $ x $.
| Then,$ \dfrac{5}{8} $:$ \dfrac{3}{8} $= :: 10 : $ x $ $\Leftrightarrow$ $ \dfrac{5}{8} \times x $ =$ \dfrac{3}{8} \times$ 10 |
| $\Rightarrow x $ =$ \left(\dfrac{3}{8} \times 10 \times\dfrac{8}{5} \right) $ |
$\Rightarrow x $ = 6.
Let the required weight be $ x $ kg.
Less weight, Less cost (Direct Proportion)
$\therefore$ 250 : 200 :: 60 : $ x $ $\Leftrightarrow$ 250 x $ x $ = (200 x 60)
| $\Rightarrow x $=$ \dfrac{(200 \times 60)}{250} $ |
$\Rightarrow x $ = 48.
Let the required number of revolutions made by larger wheel be $ x $.
Then, More cogs, Less revolutions (Indirect Proportion)
$\therefore$ 14 : 6 :: 21 : $ x $ $\Leftrightarrow$ 14 x $ x $ = 6 x 21
| $\Rightarrow x $ =$ \dfrac{6 \times 21}{14} $ |
$\Rightarrow x $ = 9.
Let the height of the building $ x $ metres.
Less lengthy shadow, Less in the height (Direct Proportion)
$\therefore$ 40.25 : 28.75 :: 17.5 : $ x $ $\Leftrightarrow$ 40.25 x $ x $ = 28.75 x 17.5
| $ x $ =$ \dfrac{28.75 \times 17.5}{40.25} $ |
$\Rightarrow x $ = 12.5
There is a meal for 200 children. 150 children have taken the meal.
Remaining meal is to be catered to 50 children.
Now, 200 children $\equiv$ 120 men.
| 50 children $\equiv \left(\dfrac{120}{200} \times 50\right) $= 30 men. |
Le the required time be $ x $ seconds.
More metres, More time (Direct Proportion)
$\therefore$ 0.128 : 25 :: 1 : $ x $ $\Leftrightarrow$ 0.128$ x $ = 25 x 1
| $ x $ =$ \dfrac{25}{0.128} $=$ \dfrac{25 \times 1000}{128} $ |
$\Rightarrow x $ = 195.31.
$\therefore$ Required time = 195 sec (approximately).
Let the required number of days be $ x $.
Less men, More days (Indirect Proportion)
$\therefore$ 27 : 36 :: 18 : $ x $ $\Leftrightarrow$ 27 x $ x $ = 36 x 18
| $\Rightarrow x $ =$ \dfrac{36 \times 18}{27} $ |
$\Rightarrow x $ = 24.
Let the number of revolutions made by the larger wheel be $x$
More cogs, less revolutions (Indirect proportion)
Hence we can write as
(cogs) 6 : 14 :: $x$ : 21
$\Rightarrow 6 \times 21 = 14 \times x\\$
$\Rightarrow 6 \times 3 = 2 \times x\\$
$\Rightarrow 3\times 3 = x\\$
$\Rightarrow x = 9$
Let the number of hours be x.
12 : 30 :: 20 : x
$\dfrac{20}{30}$ = $\dfrac{20}{x}$
x = $\dfrac{20 \times 30}{12}$ = 50
Let the whole ration now lasts for x days.
Equating the consumption on both sides, we get (90 × 70) = (90 × 20) + (100 × x) ⇒ x = 45 days.