[5 marks sum]

Question 15 Marks

The sum of three numbers, whose ratios are $3 \frac{1}{3}: 4 \frac{1}{5}: 6 \frac{1}{8}$ is 4917 . Find the numbers.

Answer

Sum of three numbers $=4917$
Ratio between them $=3 \frac{1}{3}: 4 \frac{1}{5}: 6 \frac{1}{8}$
$ =3 \frac{1}{3}: 4 \frac{1}{5}: 6 \frac{1}{8}=\frac{10}{3}: \frac{21}{5}: \frac{49}{8} $
$=\mathrm{LCM}$ of denominators 3, 5, and 8 is 120.
$ =\frac{10.120}{3}: \frac{21.120}{5}: \frac{49.120}{8} $
$ =10.40: 21.24: 49.15 $
Multiplication of the above Numbers, we will get this,
$ =400: 504: 735 $
Sum of ratio's $=400+504+735=1639$
$\therefore$ First number $=\frac{400}{1639}$ of $4917=1200$
Second number $=\frac{504}{1639}$ of $4917=1512$
and third number $=\frac{735}{1639}$ of $4917=2205$
Therefore, the first, second, and third number is 1200, 1512, and 2205 respectively.

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Question 25 Marks

A bag contains ₹ 1,600 in the form of ₹ 10 and ₹ 20 notes. If the ratio between the numbers of ₹ 10 and ₹ 20 notes is 2 : 3; find the total number of notes in all.

Answer

Total amount in the bag $=1600$
It contains notes in the denomination of ₹ 10 and 20
Ratio between the number of ₹ 10 and 20 notes is $=2: 3$
Let number of ₹ 10 note $=x$
and number of ₹ 20 notes $=y$
According to condition,
$ 10 x+20 y=1600..........(i) $
and $\mathrm{x}=\frac{2}{3} \mathrm{y}$..........(Ii)
Now, substitute the value of $x$ in eq (i)
$ 10 \times \frac{2}{3} y+20 y=1600 $
$\Rightarrow \frac{20}{3} \mathrm{y}+20 y=1600$
$\Rightarrow \frac{20+60}{3} y=1600$
$\Rightarrow \frac{80}{3} \mathrm{y}=1600$
$\Rightarrow \mathrm{y}=\frac{1600 \times 3}{80}$
$\therefore y=60$
Now, substitute the value of $y$ in eq (ii), we get
$ x=\frac{2}{3} \times 60=40 $
Total nuber of notes in all $=x+y$
$ =60+40=100 \text { notes } $

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