Check here the Latest detailed Answers and Solutions as per the latest guidelines of Exercise 3.10 - Chapter 3 Algebra 8th Maths Guide Samacheer Kalvi Solutions - [2024-2025]

Exercise 3.10 - Chapter 3 Algebra 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.10$
Question $1 .$
The sum of three numbers is 58 . The second number is three times of two-fifth of the first number and the third number is 6 less than the first number. Find the three numbers.
Answer:
Here what we know
$a+b+c=58$ (sum of three numbers is 58 )
Let the first number be $b$ ' $x$ '
$\mathrm{b}=\mathrm{a}+3$ (the second number is three times of of the first $\frac{2}{5}$ number)
$b=3 \times \frac{2}{5} x \frac{6}{5} \mathrm{x}$
Third number $=x-6$
Sum of the numbers is given as 58 .
$\therefore \mathrm{x}+\frac{6}{5} \mathrm{x}+(\mathrm{x}-6)=58$
Multiplying by 5 throughout, we get
$\begin{aligned}
&5 \times x+6 x+5 \times(x-6)=58 \times 5 \\
&5 x+6 x+5 x-30=290 \\
&\therefore 16 x=290+30 \\
&\therefore 16 x=320 \\
&\therefore x=\frac{320}{16} \\
&x=20 \\
&1^{\text {st }} \text { number }=20 \\
&2^{\text {nd }} \text { number }=3 \times \frac{2}{8} \times 20=24 \\
&3^{\text {rd }} \text { number }=24-16=14
\end{aligned}$

Question $2 .$
In triangle $\mathrm{ABC}$, the measure of $\angle \mathrm{B}$ is two-third of the measure of $\angle \mathrm{A}$. The measure of
$\angle \mathrm{C}$ is 200 more than the measure of $\angle \mathrm{A}$. Find the measures of the three angles.
Answer:
Let angle $\angle \mathrm{A}$ be $\mathrm{a}^{\circ}$
Given that $\angle \mathrm{B}=\frac{2}{3} \times \angle \mathrm{A}=\frac{2}{3} \mathrm{a}$
\& given $\angle C=\angle A+20=a+20$
Since $\mathrm{A}, \mathrm{B} \& \mathrm{C}$ are angles of a triangle, they add up to $180^{\circ}$ ( $\Delta$ property)
$\therefore \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$
$\Rightarrow \mathrm{a}+\frac{2}{3} \mathrm{a}+\mathrm{a}+20=180^{\circ}$
$\frac{3 a+2 a+3 a}{3}+20=180^{\circ}$
$\frac{8 a}{3}=180-20=160$
$\therefore \mathrm{a}=\frac{160 \times 3}{8}=60^{\circ}$

$\begin{aligned}
&\angle \mathrm{B}=\frac{2}{3} \times \angle \mathrm{A}=\frac{2}{\not \3} \times 60=40^{\circ} \\
&\angle \mathrm{C}=80^{\circ}
\end{aligned}$

Question $3 .$
Two equal sides of an isosceles triangle are $5 y-2$ and $4 y+9$ units. The third side is $2 y$ $+5$ units. Find ' $y$ ' and the perimeter of the triangle.
Answer:
Given that $5 y-2 \& 4 y+9$ are the equal sides of an isosceles triangle.
$\therefore$ The 2 sides are equal
$\Rightarrow \quad 5 y-2=4 y+9$ $\therefore 5 y-4 y=9+2$ (by transposing)
$\therefore \mathrm{y}=11$
$\therefore 1^{\text {st }}$ side $=5 \mathrm{y}-2=5 \times 11-2=55-2=53$
$2^{\text {nd }}$ side $=53$.
$3^{\text {rd }}$ side $=2 \mathrm{y}+5=2 \times 11+5=22+5=27$
Perimeter is the sum of all 3 sides
$\therefore \mathrm{P}=53+53+27=133$ units

Question $4 .$
In the given figure, angle $\mathrm{XOZ}$ and angle $\mathrm{ZOY}$ form a linear pair. Find the value of $\mathrm{x}$.

Answer:
Since $\angle \mathrm{XOZ} \& \angle \mathrm{ZOY}$ form a linear pair, by property, we have their sum to be $180^{\circ}$
$\begin{aligned}
&\therefore \angle \mathrm{XOZ}+\angle \mathrm{ZOY} 180^{\circ} \\
&\therefore 3 \mathrm{x}-2+5 \mathrm{x}+6=180^{\circ} \\
&8 \mathrm{x}+4=180=8 \mathrm{x}=180-4 \\
&\therefore 8 \mathrm{x}=76 \Rightarrow \mathrm{x}=\frac{176}{8} \Rightarrow \mathrm{x}=22^{\circ} \\
&\mathrm{XOZ}=3 \mathrm{x}-2=3 \times 22-2=66-2=64^{\circ} \\
&\mathrm{YOZ}=5 \mathrm{x}+6=5 \times 22+6 \\
&=110+6=116
\end{aligned}$

Question $5 .$
Draw a graph for the following data:

Answer:

Challenging problems
Question $6 .$
Three consecutive integers, when taken in increasing order and multiplied by 2,3 and 4 respectIvely, total up to 74 . Find the three numbers.
Answer:
Let the 3 consecutive integers be ' $x$ ', ' $x+1$ ' $\&$ ' $x+2$ '
Given that when multiplied by 2,3 \& 4 respectively \& added up, we get 74
i.e $2 \times x+3 \times(x+1)+4 \times(x+2)=74$
Simplifying the equation, we get
$2 x+3 x+3+4 x+8=74$ $9 x+11=63$ $9 x=63 \Rightarrow x=\frac{63}{9}=7$ First numberi $=7$
Second numbers $=x+1 \Rightarrow 7+1=8$
Third numbërs $=x+2 \Rightarrow 7+2=9$
$\therefore$ The numbers are $7,8 \& 9$

Question 7.
331 students went on a field trip. Six buses were filled to capacity and 7 students had to travel in a van. How many students were there in each bus?
Answer:
Let the number of students in each bus be ' $x$ '
$\therefore$ number of students in 6 buses $=6 \times x=6 x$
A part from 6 buses, 7 students went in van
A total number of students is 331
$\therefore 6 \mathrm{x}+7=331$
$\therefore 6 x=331-7=324$
$\therefore x=\frac{324}{6}=54$
$\therefore$ There are 54 students in each bus.

Question 8.
A mobile vendor has 22 items, some which are pencils and others are ball pens. On a particular day, he is able to sell the pencils and ball pens. Pencils are sold for $₹ 15$ each and ball pens are sold at $₹ 20$ each. If the total sale amount with the vendor is ₹ 380 , how many pencils did he sell?
Answer:
Let vendor have ' $p$ ' number of pencils \& ' $b$ ' number of ball pens
Given that total number of items is 22
$\therefore \mathrm{p}+\mathrm{b}=22$
Pencils are sold for ₹ 15 each \& ball pens for $₹ 20$ each
total sale amount $=15 \times \mathrm{p}+20 \times \mathrm{b}$
$=15 p+20 b$ which is given to be 380 .
$\therefore 15 \mathrm{p}+20 \mathrm{~b}=380$
Dividing by 5 throughout,
$\frac{15 p}{5}+\frac{20 b}{5}=\frac{380}{5} \Rightarrow-3 \mathrm{p}+4 \mathrm{~b}=76$
Multiplying equation (1) by 3 we get
$\begin{aligned}
&3 \times \mathrm{p}+3 \times \mathrm{b}=22 \times 3 \\
&\Rightarrow 3 \mathrm{p}+3 \mathrm{~b}=66
\end{aligned}$
Equation (2) - (3) gives

$\begin{aligned}
&\therefore \mathrm{b}=10 \\
&\therefore \mathrm{p}=12
\end{aligned}$
He sold 12 pencils

Question $9 .$
Draw the graph of the lines $y=x, y=2 x, y=3 x$ and $y=5 x$ on the same graph sheet. Is there anything special that you find in these graphs?
Answer:
(i) $y=x$
(ii) $\mathrm{y}=2 \mathrm{x}$,
(iii) $y=3 x$
(iv) $y=5 x$

(i) $\mathrm{y}=\mathrm{x}$
When $x=1, y=1$
$\begin{aligned}
&x=2, y=2 \\
&x=3, y=2
\end{aligned}$
(ii) $\mathrm{y}=2 \mathrm{x}$
$\begin{aligned}
&\text { When } x=1, y=2 \\
&x=2, y=4 \\
&x=3, y=6
\end{aligned}$
(iii) $\mathrm{y}=3 \mathrm{x}$
When $x=1, y=3$
$\begin{aligned}
&x=2, y=6 \\
&x=3, y=9
\end{aligned}$
(i) $y=5 x$
When $x=1, y=5$
$\begin{aligned}
&x=2, y=10 \\
&x=3, y=15
\end{aligned}$
When we plot the above points \& join the points to form line, we notice that the lines become progressively steeper. In other words, the slope keeps increasing.