Exercise 3.14 - Chapter 3 Algebra 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.14$
Solve by any one of the methods
Question 1.
The sum of a two digit number and the number formed by interchanging the digits is 110 . If 10 is subtracted from the first number, the new number is 4 more than 5 times the sums of the digits of the first number. Find the first number.
Solution:
Let the two digit number be $x y$
Its place value $=10 x+y$
After interchanging the digits the number will be $y \mathrm{x}$
Its place value $=10 y+x$
Their sum $=10 x+y+10 y+x=110$
$11 x+11 y=110$
$x+y=10 \ldots \ldots . \ldots . . .(1)$
If 10 is subtracted from the first number, the new number is $10 x+y-10$
The sums of the digits of the first number is $x+y$
Its 4 more than 5 times is $=5(x+y)+4$
$\begin{aligned}
&\therefore 10 x+y-10=5 x+5 y+4 \\
&10 x+y-5 x-5 y=4+10 \\
&5 x-4 y=14 \ldots \ldots \ldots(2)
\end{aligned}$

Substitute $\mathrm{y}=4$ in (1)
$\begin{aligned}
&x+4=10 \\
&x=10-4 \\
&x=6
\end{aligned}$
The first number is 64
 

Question $2 .$
The sum of the numerator and denominator of a fraction is 12 . If the denominator is increased by 3 , the fraction becomes $\frac{1}{2}$. Find the fraction.
Solution:

Let the numerator be $x$ Denominator be $y$
$\begin{aligned}
&x+y=12 \\
&\frac{x}{y+3}=\frac{1}{2} \\
&2 x=y+3 \\
&2 x-y=3
\end{aligned}$

Substitute $y=7$ in (1)
$\begin{aligned}
&x+7=12 \\
&x=12-7 \\
&x=5
\end{aligned}$
$\therefore$ The fraction is $\frac{x}{y}=\frac{5}{7}$
 

Question $3 .$
$\mathrm{ABCD}$ is a cyclic quadrilateral such that $\angle \mathrm{A}=(4 \mathrm{y}+20)^{\circ}, \angle \mathrm{B}=(3 \mathrm{y}-5)^{\circ}, \angle \mathrm{C}=(4 \mathrm{x})^{\circ}$ and $\angle \mathrm{D}=(7 \mathrm{x}$ $+5)^{\circ}$. Find the four angles.
Solution:
In a cyclic quadrilateral the sum of the four angles is $360^{\circ}$ and
the sum of the opposite angles is $180^{\circ}$.
$\angle \mathrm{A}+\angle \mathrm{C}=180^{\circ}$
$4 \mathrm{y}+20+4 \mathrm{x}=180^{\circ}$
$4 x+4 y=180-20$
$\begin{aligned}
&x+y=\frac{160}{4} \\
&x+y=40 \ldots \ldots \ldots \ldots(1)
\end{aligned}$
$\begin{aligned}
&\angle B+\angle D=180^{\circ} \\
&3 y-5+7 x+5=180 \\
&7 x+3 y=180 \ldots \ldots . \ldots .
\end{aligned}$

Substitute $y=25^{\circ}$ in (1)
$\begin{aligned}
&x+25=40 \\
&x=40-25 \\
&x=15^{\circ} \\
&\angle A=(4 y+20)^{\circ}=(4 \times 25+20)=100+20=120^{\circ} \\
&\angle B=(3 y-5)^{\circ}=\left(3 \times 25^{\circ}-5\right)=75^{\circ}-5^{\circ}=70^{\circ}
\end{aligned}$
$\begin{aligned}
&\angle \mathrm{C}=(4 \mathrm{x})^{\circ}=4 \times 15^{\circ}=60^{\circ} \\
&\angle \mathrm{D}=(7 \mathrm{x}+5)^{\circ}=(7 \times 15+5)=105^{\circ}+5^{\circ}=110^{\circ} \\
&\therefore \angle \mathrm{A}=120^{\circ}, \angle \mathrm{B}=70^{\circ}, \angle \mathrm{C}=60^{\circ}, \angle \mathrm{D}=110^{\circ}
\end{aligned}$

Question $4 .$
On selling a T.V. at $5 \%$ gain and a fridge at $10 \%$ gain, a shopkeeper gains ₹ 2000 . But if he sells the T.V. at $10 \%$ gain and the fridge at $5 \%$ loss, he gains $₹ 1500$ on the transaction. Find the actual price of the T.V. and the fridge.
Solution:
Let the actual price of a T.V. $=x$
Let the actual price of a Fridge $=y$
$\begin{aligned}
&\frac{5}{100} x+\frac{10}{100} y=2000 \\
&\frac{5}{100}(x+2 y)=2000 \\
&x+2 y=2000 \times \frac{100}{5} \\
&x+2 y=40000 \ldots . . (1)
\end{aligned}$

$\begin{aligned}
&\frac{10}{100} x-\frac{5}{100} y=1500 \\
&2 x-y=1500 \times \frac{100}{5} \\
&2 x-y=30000 \ldots \ldots (2)
\end{aligned}$

Substitute $y=10000$ in (1)
$\begin{aligned}
&x+2(10000)=40000 \\
&x+20000=40000 \\
&x=40000-20000 \\
&x=20000
\end{aligned}$
$\therefore$ Actual price of T.V. $=₹ 20000$
Actual price of Fridge $=₹ 10000$
 

Question $5 .$
Two numbers are in the ratio $5: 6$. If 8 is subtracted from each of the numbers, the ratio becomes 4 :5 . Find the numbers.
Solution:
Let the two numbers be $x, y$
$\begin{aligned}
&\frac{x}{y}=\frac{5}{6} \\
&\Rightarrow 6 \mathrm{x}=5 \mathrm{y} \\
&6 \mathrm{x}-5 \mathrm{y}=0 \ldots \ldots \ldots \ldots(1) \\
&\frac{x-8}{y-8}=\frac{4}{5}
\end{aligned}$
$\begin{aligned}
&\Rightarrow 5(x-8)=4(y-8) \\
&5 x-40=4 y-32 \\
&5 x-4 y=40-32 \\
&5 x-4 y=8 \ldots \ldots \ldots . \ldots(2)
\end{aligned}$

Substitute $y=48$ in (1)
$\begin{aligned}
&6 x-5(48)=0 \\
&6 x-240=0 \\
&6 x=240 \\
&x=\frac{240}{6}=40 \\
&\frac{x}{y}=\frac{40}{48}=\frac{5}{6}
\end{aligned}$
$\therefore$ The numbers are in the ratio $5: 6$

Question $6 .$
4 Indians and 4 Chinese can do a piece of work in 3 days. While 2 Indians and 5 Chinese can finish it in 4 days. How long would it take for 1 Indian to do it? How long would it take for 1 Chinese to do it?
Solution:
Let for one Indian the rate of working be $\frac{1}{x}$
Let for one Chinese the rate of working be $\frac{1}{y}$
$\begin{array}{r}
\therefore \frac{4}{x}+\frac{4}{y}=\frac{1}{3} \\
\frac{2}{x}+\frac{5}{y}=\frac{1}{4}
\end{array}$
Put $\frac{1}{x}=a, \frac{1}{y}=b$
$\begin{array}{r}
4 a+4 b-\frac{1}{3}=0 \\
2 a+5 b=\frac{1}{4} \\
2 a+5 b-\frac{1}{4}=0
\end{array}$
For cross multiplication method, we write the co-efficients as

$\frac{a}{-1+\frac{5}{3}}=\frac{b}{-\frac{2}{3}+1}=\frac{1}{20-8}$
$\frac{3 a}{2}=3 b=\frac{1}{12}$
$\therefore \frac{3 a}{2}=\frac{1}{12}$
$3 b=\frac{1}{12}$
$a=\frac{1}{12} \times \frac{2}{3}=\frac{1}{18} \quad b=\frac{1}{36}$
$\therefore 1$ Chinese can do the same piece of work $=\frac{1}{y}=\mathrm{b}=\frac{1}{36}=36$ days 1 Indian can do the piece of work $=\frac{1}{x}=\mathrm{a}=\frac{1}{18}=18$ days