Exercise 3.7 - Chapter 3 Algebra 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.7$
Question 1.
Fill in the blanks:
(i) The solution of the equation $a x+b=0$ is
Answer:
$-\frac{b}{a}$
Hint:
$a x+b=0$
$a x=-b$
$\therefore \mathrm{x}=-\frac{b}{a}$
(ii) If $a$ and $b$ are positive integers then the solution of the equation $a x=b$ has to be always
Answer:
Positive
Hint:
Since a $\&$ b are positive integers, $b$
The solution to the equation $\mathrm{ax}=\mathrm{b}$ is $\mathrm{x}=\frac{b}{a}$ is also positive.
(iii) One-sixth of a number when subtracted from the number itself gives 25 . The number is
Answer:
30
Hint:
Let the number be $\mathrm{x}$.

As per question, when one sixth of number is subtracted from itself it gives 25
$\begin{aligned}
x-\frac{x}{6} &=25 \\
\therefore \frac{6 x-x}{6} &=25 \\
\therefore \frac{5 x}{6} &=25 \\
\therefore x &=\frac{25 \times 6}{5}=5 \times 6=30
\end{aligned}$
(iv) If the angles of a triangle are in the ratio $2: 3: 4$ then the difference between the greatest and the smallest angle is
Answer:
$40^{\circ}$
Hint:
Given angles are in the ratio 2:3:4
Let the angles be $2 x, 3 x$ \& $4 x$
Since sum of the angles of a triangle is $180^{\circ}$,
We get $2 x+3 x+4 x=180$
$\begin{aligned}
&\therefore 9 \mathrm{x}=180 \\
&\therefore \mathrm{x}=\frac{180}{9}=20^{\circ}
\end{aligned}$
$\therefore$ The angles are $2 \mathrm{x}=2 \times 20=40^{\circ}$
$\begin{aligned}
&3 x=3 \times 20=60^{\circ} \\
&4 x=4 \times 20=80^{\circ}
\end{aligned}$
$\therefore$ Difference between greatest \& smallest angle is
$80^{\circ}-40^{\circ}=40^{\circ}$

(v) In an equation $a+b=23$. The value of a is 14 then the value of $b$ is
Answer:
$\mathrm{b}=9$
Hint:
Given equation is $a+b=23, a=14$
$\begin{aligned}
&14+\mathrm{b}=23 \\
&\therefore \mathrm{b}=23-14=9 \\
&\mathrm{~b}=9
\end{aligned}$

Question $2 .$
Say True or False
(i) "Sum of a number and two times that number is 48 " can be written as $y+2 y=$ 48
Answer:
True
Hint:
Let the number be ' $y$ '
$\therefore$ Sum of number \& two times that number is 48
Can be written as $y+2 y=48-$ True
(ii) $5(3 x+2)=3(5 x-7)$ is a linear equation in one variable.
Answer:
True
Hint:
$5(3 x+2)=3(5 x-7)$ is a linear equation in one variable $-$ ' $x$ ' - True

(iii) $x=25$ is the solution of one third of a number is less than 10 the original number.
Answer:
False
Hint:
One third of number is 10 less than original number.
Let number be ' $x$ '. Therefore let us frame the equation
$\begin{aligned}
&\frac{x}{3}=\mathrm{x}-10 \\
&\therefore \mathrm{x}=3 \mathrm{x}-30 \\
&3 \mathrm{x}-\mathrm{x}=30 \\
&2 \mathrm{x}=30 \\
&\mathrm{x}=15 \text { is the solution }
\end{aligned}$

Question $3 .$
One number is seven times another. If their difference is 18 , find the numbers.
Answer:
Let the numbers be $\mathrm{x}$ \& $\mathrm{y}$
Given that one number is 7 times the other \& that the difference is 18 .
Let $x=7 y$
also, $x-y=18$ (given)
Substituting for $x$ in the above
We get $7 y-y=18$

$\begin{aligned}
&\therefore 6 y=18 \\
&y=\frac{18}{6}=3 \\
&\therefore x=7 y=7 \times 3=21
\end{aligned}$
The number are 3 \& 21
 

Question $4 .$
The sum of three consecutive odd numbers is 75 . Which is the largest among them?
Answer:
Given sum of three consecutive odd numbers is 75
Odd numbers are $1,3,5,7,9,11,13$
$\therefore$ The difference between 2 consecutive odd numbers is always 2 . or in other words, if one odd number is $x$, the next odd number would be $x+2$ and the next number would be $x+2+2 x+4$
i.e $x+4$
Since sum of 3 consecutive odd nos is 75
$\begin{aligned}
&\therefore \mathrm{x}+\mathrm{x}+2+\mathrm{x}+4=75 \\
&\therefore 3 \mathrm{x}+6=75 \Rightarrow 3 \mathrm{x}=75 \\
&\therefore 3 \mathrm{x}=69 \\
&\mathrm{x}=\frac{69}{3}=23
\end{aligned}$
$\therefore 3 \mathrm{x}+6=75 \Rightarrow 3 \mathrm{x}=75-6$
$\therefore$ The odd numbers are $23,23+2,23+4$
i.e $23,25,27$
$\therefore$ Largest number is 27 .
 

Question $5 .$
The length of a rectangle is $\frac{1}{3}$ rd of its breadth. If its perimeter is $64 \mathrm{~m}$, then find the length and breadth of the rectangle.
Answer:
Let length \& breadth of rectangle be ' $l$ ' and ' $b$ ' respectively
Given that length is $\frac{1}{3}$ of breadth,

Also given that perimeter is $64 \mathrm{~m}$
Perimeter $=2 \times(1+b)$
$2 \times 1+2 \times b=64$
Substituting for value of $b$ from (1), we get
$21+2(31)=64$
$\therefore 21+6 l=64$
$81=64$
$\therefore 1=\frac{64}{8}=8 \mathrm{~m}$
$\mathrm{b}=31=3 \times 8=24 \mathrm{~m}$
Ienglh $1=8 \mathrm{~m}$ in \& breadth $\mathrm{b}=24 \mathrm{~m}$

Question $6 .$
A total of 90 currency notes, consisting only of $₹ 5$ and $₹ 10$ denominations, amount to $₹ 500$. Find the number of notes in each denomination.
Answer:

Let the number of ₹ 5 notes be ' $x$ ' And number of ₹ 10 notes be ' $y$ ' Total numbers of notes is $x+y=90$ (given) The total value of the notes is 500 rupees. Value of one ₹ 5 rupee note is 5
Value of $x$ ₹ 5 rupee notes is $5 \times x=5 x$
$\therefore$ Value of $\mathrm{y} ₹ 10$ rupee flotes is $10 \times y=10 \mathrm{y}$
$\therefore$ The total value is $5 \mathrm{x}+10 \mathrm{y}$ which is 500
$\therefore$ we have 2 equations:
$x+y=90 \ldots \text {... (1) }$
$5 x+10 y=500 \ldots .$ (2)
Multiplying both sides of $(1)$ by 5 , we get
$\begin{aligned}
&5 \times x+5 \times y=90 \times 5 \\
&5 x+5 y=450
\end{aligned}$
Subtracting (3) from (2), we get

$\therefore y=\frac{50}{5}=10$
Substitute $y=10$ in equation (1)
$x+y=90 \Rightarrow x+10=90 \Rightarrow x=90-10 \Rightarrow x=80$
There are $₹ 5$ denominations are 80 numbers and $₹ 10$ denominations are 10 numbers
 

Question $7 .$
At present, Thenmozhi's age is 5 years more than that of Murali's age. Five years ago, the ratio of Then mozhi's age to Murali's age was $3: 2$. Find their present ages.
Answer:
Let present ages of Thenmozhi \& Murali be ' $t$ ' \& ' $m$ '
Given that at present
Then mozhi's age is 5 years more than Murali
$\therefore \mathrm{t}=\mathrm{m}+5 \ldots . .$ (1)
5 years ago, Thenmozhi's age would be $t-5$
$\&$ Murali's age would be $m-5$
Ratio of their ages is given as $3: 2$

$\therefore \frac{t-5}{m-5}=\frac{3}{2}[\therefore$ By cross multiplication $]$ $2(\mathrm{t}-5)=3(\mathrm{~m}-5)$
$2 \times \mathrm{t}-2 \times 5=3 \times \mathrm{m}-3 \times 5 \Rightarrow 2 \mathrm{t}-10=3 \mathrm{~m}-15$
Substituting for t from (1)
$2(m+5)-10=3 m-15$
$\begin{aligned}
&2 m+1 v-1 /=3 m-15 \\
&2 m=3 m-15 \\
&3 m-2 m=15 \\
&m=15 \\
&t-m+5-15+5-20
\end{aligned}$
$\therefore$ Present ages of Thenmozhi \& Murali are $20 \& 15$

Question 8 .
A number consists of two digits whose sum is 9 . If 27 is subtracted from the original number, its djgit.s are interchanged. Find the original number.
Answer:
Let the units/digit of a number be ' $u$ ' \& tens digit of the number be ' $t$ '
Given that sum of it's digits is 9
$\therefore \mathrm{t}+\mathrm{u}=9 \ldots . . .$ (1)
If 27 is subtracted from original number, the digits are interchanged
The number is written as $10 \mathrm{t}+\mathrm{u}$
[Understand: Suppose a 2 digit number is 21
it can be written as $2 \times 10+1$
$\therefore 32=3 \times 10+2$
$\begin{aligned}
&45=4 \times 10+5 \\
&t u=t \times 10+u=1 \text { ot }+\mathrm{u}]
\end{aligned}$
Given that when 27 is subtracted, digits interchange
$10 \mathrm{t}+\mathrm{u}-27=10 \mathrm{u}+\mathrm{t}$ (number with interchanged digits)
$\therefore$ By transposition \& bringing like variables together
$\begin{aligned}
&10 \mathrm{t}-\mathrm{t}+\mathrm{u}-10 \mathrm{u}=27 \\
&\therefore 9 \mathrm{t}-9 \mathrm{u}=27
\end{aligned}$
Dividing by ' 9 ' throughout, we get $\frac{9 t}{9}-\frac{9 u}{9}=\frac{27}{9} \Rightarrow \mathrm{t}-\mathrm{u}=3 \ldots \ldots .(2)$
Solving (1) \& (2):

$t=\frac{12}{2}=6 \therefore u=3$
$t=6$ substitute in (1)
$\mathrm{t}+\mathrm{u}=9 \Rightarrow 6+\mathrm{u}=9 \Rightarrow \mathrm{u}=9-6=3$
Hence the number is 63 .

Question $9 .$
The denominator of a fraction exceeds Its numerator by 8 . If the numerator is increased by 17 and the denominator is decreased by 1 , we get $\frac{3}{2}$. Find the original fraction.
Answer:
Let the numerator \& denominator be ' $n$ ' \& ' $d$ '
Given that denominator exceeds numerator by 8 $\therefore \mathrm{d}=\mathrm{n}+8 \ldots \ldots$ (1)
If numerator increased by 17 \& denominator decreased by 1 , it becomes $(\mathrm{n}+17) \&(\mathrm{~d}-1)$, fraction is $\frac{3}{2}$.
i.e $\frac{n+17}{d-1}=\frac{3}{2}$ by cross multiplying, we get $\frac{n+17}{d-1}=\frac{3}{2}$
$\begin{aligned}
&2(\mathrm{n}+17)=3(\mathrm{~d}-1) \\
&2 \mathrm{n}+2 \times 17=3 \mathrm{~d}-3 \\
&2 n+34=3 d-3
\end{aligned}$
$\begin{aligned}
&\therefore 34+3=3 \mathrm{~d}-2 \mathrm{n} \\
&\therefore 3 \mathrm{~d}-2 \mathrm{n}=37 \ldots \ldots . .(2)
\end{aligned}$

Substituting eqn. (1) in (2), we get,
$\begin{aligned}
&3 \times(\mathrm{n}+8)-2 \mathrm{n}=37 \\
&3 \mathrm{n}+3 \times 8-2 \mathrm{n}=37 \\
&\therefore 3 n+24-2 n=37 \\
&\therefore \mathrm{n}=37-24=13 \\
&\mathrm{~d}=\mathrm{n}+8=13+8=21
\end{aligned}$
The fraction is $\frac{n}{d}=\frac{13}{21}$
 

Question $10 .$
If a train runs at $60 \mathrm{~km} / \mathrm{hr}$ it reaches its destination late by 15 minutes. But, if it runs at $85 \mathrm{kmph}$ it is late by only 4 minutes. Find the distance to be covered by the train.
Answer:
Let the distance to be covered by train be 'd'
Using the formula, time take $(t)=\frac{\text { Distance }}{\text { Speed }}$
Case 1:
If speed $=60 \mathrm{~km} / \mathrm{h}$
The time taken is 15 minutes more than usual $\left(\mathrm{t}+\frac{15}{60}\right)$
Let usual time taken be ' $t$ ' hrs.
Caution: Since speed is given in $\mathrm{km} / \mathrm{hr}$, we should take care to maintain all units such as time should be in hour and distance should be in kin.
Given that in case 1 , it takes $15 \mathrm{~min}$. more
$15 \mathrm{~m}=\frac{15}{60} \mathrm{hr}=\frac{1}{4} \mathrm{hr}$.
$\therefore$ Substituting in formula,
$\frac{\text { Distance }}{\text { Speed }}=$ time
$\therefore \frac{d}{60}=t+\frac{1}{4}$

Since usually it takes ' $t$ ' hr, but when running at $60 \mathrm{k}$, it kes $15 \mathrm{~min}\left(\frac{1}{4} \mathrm{hr}\right)$ extra.
Multiplying by 60 on both sides
$\mathrm{d}=60 \times \mathrm{t}+60 \times \frac{1}{4}=6 \mathrm{ot}+15 \ldots \ldots \text { (1) }$
Case 2:
Speed is given as $85 \mathrm{~km} / \mathrm{h}$
Time taken is only $4 \mathrm{~min}\left(\frac{4}{60} \mathrm{hr}\right)$ more than usual time
$\therefore$ time taken $=\left(\mathrm{t}+\frac{1}{15}\right) \mathrm{hr} .\left(\frac{4}{60}=\frac{1}{15}\right)$
Using the formula,
$\frac{\text { Distance }}{\text { Speed }}=$ time
$\therefore \frac{d}{85}=t+\frac{1}{15}$
Multiplying by 85 on both sides
$\begin{aligned}
&\frac{d}{85} \times 85=85 \times \mathrm{t}+85 \times \frac{1}{15} \\
&\therefore \mathrm{d}=85 \mathrm{t}+\frac{17}{3} \ldots . .(2)
\end{aligned}$
From (1) \& (2), we will solve for ' $r$ '
Equating \& eliminating 'd' we get
$60 t+15=85 t+\frac{17}{3}$
$\therefore$ By transposing, we get
$\begin{aligned}
15-\frac{17}{3} &=85 t-60 \mathrm{t} \\
\frac{45-17}{3} &=25 \mathrm{t} \\
\therefore 25 t &=\frac{28}{3}
\end{aligned}$

$\therefore t=\frac{28}{3 \times 25}=\frac{28}{75} \mathrm{hr}\left(\frac{28}{75} \times 60=22.4 \mathrm{~min}\right)$
Substituting this value of ' $t$ ' in eqn. (1), we get $d=60 t+15$
$\begin{aligned}
&=60 \times \frac{28}{75}+15=\frac{1680}{75}+15=22.4+15 \\
&=37.4 \mathrm{~km}
\end{aligned}$

Question 11 .
Sum of a number and its half is 30 then the number is
(a) 15
(b) 20
(c) 25
(d) 40
Answer:
(b) 20
Hint:
Let number be ' $x$ '
$\therefore$ half of number is $\frac{x}{2}$
Sum of number and it's half is given by

$\mathrm{x}+\frac{x}{2}=30$ [Multiplying by 2 on both sides]
$\begin{aligned}
&2 x+x=30 \times 2 \\
&3 x=60 \\
&x=\frac{60}{3}=20
\end{aligned}$

Question $12 .$
The exterior angle of a triangle is 1200 and one of its interior opposite angle $58^{\circ}$, then the other opposite interior angle is
(a) $62^{\circ}$
(b) $72^{\circ}$
(c) $78^{\circ}$
(d) $68^{\circ}$
Answer:
(a) $62^{\circ}$
As per property of A. exterior angle is equals to sum of interior opposite angles Let the other interior angle to be found be ' $\mathrm{x}$ '
$\therefore \mathrm{x}+58=120^{\circ}$
$\therefore \mathrm{x}=120-58=62^{\circ}$

Question $13 .$
What sum of money will earn 500 as simple interest in 1 year at $5 \%$ per annum?
(a) 50000
(b) 30000
(c) 10000
(d) 5000
Answer:
(c) 10000
Hint:
Let sum of money be $\mathrm{P}$ '
Time period (n) is given as l yr.
Ratc of simple interest (r) is given as $5 \%$ p.a
$\therefore$ As per formula for simple interest.
$\begin{aligned}
&\mathrm{S} . \mathrm{I}=\frac{\mathrm{P} \times r \times n}{100}=\frac{\mathrm{P} \times 5 \times 1}{100}=500 \\
&\therefore \mathrm{P} \times 5 \times \mathrm{n}=500 \times 100 \\
&\therefore \mathrm{p}=\frac{500 \times 100}{5}=100 \times 100=10,000
\end{aligned}$

Question $14 .$
The product of LCM and HCF of two numbers is 24 . If one of the number is 6 , then the other number is
(a) 6
(b) 2
(c) 4
(d) 8
Answer:
(C) 4
Hint:
Product of LCM \& HCF of 2 numbers is always product of the numbers. [this is property]
Product of LCM \& HCF is given as 24
$\therefore$ Product of the 2 nos. is 24

Given one number is 6 . Let other number be ' $x$ '
$\begin{aligned}
&\therefore 6 \times \mathrm{x}=24 \\
&\mathrm{x}=\frac{24}{6}=4
\end{aligned}$
 

Question $15 .$
The largest number of the three consecutie numbers is $x+1$, then the smallest number is
(A) $x$
(B) $x+1$
(C) $x+2$
(D) $x-1$
Answer:
(D) $x-1$
Hint:
The 3 consecutive numbers are: $x-1, x, x+1$