Follow BODMAS rule, Bars, remove brackets in (), {},[] order, Of, Division, Multiplication, Addition, Subtraction respectively.
32 Ã· 8 $\displaystyle \times$ 24 + 16 => 4 $\displaystyle \times$ 24 + 16 = 96 + 16 = 112.
65 X 65 Ã· 13 + [ 4 + 18 Ã· 3] => 65 X 65 Ã·13 +[ 4 + 6] => 65 X 5 + 20 = 325 + 10 = 335
60 $\displaystyle \times \frac{3}{4}$ = 45, 60 $\displaystyle \times \frac{8}{5}$ = 96
45 â€“ 96 + ? = 12 => 96 + 12 = 108 â€“ 45 = 63
[63 + 45 = 108 â€“ 96 = 12]
Let total number of land x acre
Elder son = $\displaystyle x \times \frac{1}{3}$ = $\displaystyle \frac{1}{3}x$
Remaining = $\displaystyle x \ â€“ \ \frac{1}{3}$ $\displaystyle x \ = \ \frac{2}{3} x $
Second son = $\displaystyle \frac{1}{2}$ of $\displaystyle \frac{2}{3} x $ = $\displaystyle \frac{1}{3} x$
Remaining = $\displaystyle \frac{1}{3} x$ = 24 acre [12 + 12 = 24, younger son + Daughter]
$\displaystyle \frac{1}{3} x$ = 24, then x = 24 $\displaystyle \times \frac{3}{1}$ = 72
Quantity of milk : water = $\displaystyle \frac{4}{5}$ : $\displaystyle \frac{1}{5}$
First replacement = $\displaystyle \frac{1}{5}$ of $\displaystyle \frac{4}{5}$ = $\displaystyle \frac{4}{25}$
Remaining milk = $\displaystyle \frac{4}{5}$ â€“ $\displaystyle \frac{4}{25}$ = $\displaystyle \frac{16}{25}$
2^{nd} replacement = $\displaystyle \frac{1}{8}$ th of $\displaystyle \frac{16}{25}$ = $\displaystyle \frac{16}{200}$ or $\displaystyle \frac{2}{25}$
Remaining = $\displaystyle \frac{16}{25}$ â€“ $\displaystyle \frac{2}{25}$ = $\displaystyle \frac{14}{25}$
Quantity of milk = $\displaystyle \frac{14}{25}$
Water = 1 â€“ $\displaystyle \frac{14}{25}$ = $\displaystyle \frac{11}{25}$ th part
[Or]
Let the container contains 100 lr of milk and water
Then milk = $\displaystyle \frac{4}{5}$ = 80lr
1^{st} repl = $\displaystyle \frac{1}{5}$ th of 80 = 16 lr
Remaining = 80 â€“ 16 = 64
2^{nd} repl = $\displaystyle \frac{1}{8}$ th of 64 = 8 lr
Remaining milk = 64 â€“ 8 = 56 lr
Water = 100 â€“ 56 = 44 lr, if expressed in a fraction $\displaystyle \frac{44}{100}$ = $\displaystyle \frac{11}{25}$
Follow reverse process
48 + 12 = 60 â€“ 10 = 50 Ã· 5 = 10 $\displaystyle \times$ 4 = 40
Number of marbles on 4^{th} day = 20
4^{th} = 20 = 1 $\displaystyle \frac{1}{3}$ = $\displaystyle \frac{4}{3}$ = 20 => 20 $\displaystyle \times \frac{3}{4}$ = 15
3^{rd} = $\displaystyle \frac{1}{2}$ = 15 = 15 $\displaystyle \times \frac{2}{1}$ = 30
2^{nd} = 30 = 1 + $\displaystyle \frac{1}{4}$ = 30 = $\displaystyle \frac{5}{4}$ = $\displaystyle \frac{4}{5} \times $ 30 = 24
1^{st} day = 1 - $\displaystyle \frac{1}{5}$ = $\displaystyle \frac{4}{5}$ = 24 = 24 $\displaystyle \times \frac{5}{4}$ = 30
In the beginning he had 30 marbles
As every day the worms increases by 3 times
$\displaystyle \frac{1}{3} \ \times $ 3 = 1 [full]
Therefore on 11th day
160 is actually in 4^{th} hour
We have to calculate for the first 3 hrs
On third hour $\displaystyle \frac{160}{2}$ = 80 km/h
Second hr $\displaystyle \frac{80}{2}$ = 40km/hr
First hour $\displaystyle \frac{40}{2}$ = 20km/hr
Total distance covered = 80 + 40 + 20 = 140 km
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