Exercise 4.5 - Chapter 4 Life Mathematics 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.5$
Question $1 .$
A fruit vendor bought some mangoes of which $10 \%$ were rotten. He sold $33 \frac{1}{3} \%$ of the rest. Find the total number of mangoes bought by him initially, if he still has 240 mangoes with him.
Answer:
Let the number of mangoes bought by fruit seller initially be $x$.
Given that $10 \%$ or mangoes were rotten
$\therefore$ Number of rotten mangoes $=\frac{10}{100} \times \mathrm{x}$
Number of gond mangoes $=x-n$ n. of rotten mangoes
$=x-\frac{10}{100} x=\frac{100 x-10 x}{100}=\frac{90}{100} x \quad \ldots \ldots . . \text { (1) }$
Number of mangoes sold $=33 \frac{1}{3} \%$ of good mangoes $=\frac{100}{3} \%$
$\therefore$ Mangoes sold $=\frac{100}{3} \times \frac{90}{100} \times \times \frac{1}{100}=\frac{30}{100} x \quad \ldots \ldots$ (2)
Number of mangoes remaining $=$ No. of good mangoes - No. of mangoes sold From (1) and (2)

$\therefore \frac{90}{100} x-\frac{30}{100} x=240 \Rightarrow \frac{90 x-30 x}{100}=240$
$\begin{aligned}
\therefore \frac{60 x}{100} &=240 \\
\therefore x &=\frac{240 \times 100}{601} \\
x &=400
\end{aligned}$
$\therefore$ Intially he had 400 mangoes
 

Question 2.
A student gets $31 \%$ marks in an examination but fails by 12 marks. If the pass percentage is $35 \%$, find the maximum marks of the examination.
Answer:
Let the maximum marks in the exam be ' $x$ '
Pass percentage is given as $35 \%$
$\therefore$ Pass mark $=\frac{35}{100} \times x=\frac{35}{100} x$
Student gets $31 \%$ marks $=\frac{31}{100} \times x=\frac{31}{100} x$
But student fails by 12 marks $\rightarrow$ meaning his mark is 12 less than pass mark.
$\therefore \frac{31}{100} x=\frac{35}{100} x-12$
$\begin{aligned}
\therefore \frac{35}{100} x-\frac{31}{100} x &=12 \\
\therefore \frac{35 x-31 x}{100} &=12 \Rightarrow \frac{4 x}{100}=12
\end{aligned}$

$\therefore x=\frac{12 \times 100}{4}=300$
Maimum mark is 300
 

Question $3 .$
Sultana bought the following things from a general store. Calculate the total bill amount paid by her.
(i) Medicines costing $₹ 800$ with GST at $5 \%$
Answer:
Formula for bill amount is cost $\left(1+\frac{\mathrm{GST} \%}{100}\right)$
Medicine: bill amount is $800\left(1+\frac{5}{100}\right)=800 \times \frac{105}{100}=840$
(ii) Cosmetics coosting $₹ 650$ with GST at $12 \%$

Answer:
Formula for bill amount is cost $\left(1+\frac{\text { GST\% }}{100}\right)$
Cosmetics: Bill amount is $550\left(1+\frac{12}{100}\right)=650 \times \frac{112}{100}=728$
(iii) Cereals costing ₹ 900 with GST at $0 \%$
Answer:
Formula for bill amount is cost $\left(1+\frac{\mathrm{GST} \%}{100}\right)$
Cereals: Bill amount is $900\left(1+\frac{0}{100}\right)=900$
(iv) Sunglass costing ₹ 1750 with GST at $18 \%$

Answer:
Formula for bill amount is cost $\left(1+\frac{\mathrm{GST} \%}{100}\right)$
Sunglass: Bill amount is $1750\left(1+\frac{18}{100}\right)=1750 \times \frac{118}{100}=2065$
(v) Air Conditioner costing ₹ 28500 with GST at $28 \%$
Answer:
Formula for bill amount is cost $\left(1+\frac{\mathrm{GST} \%}{100}\right)$
Air Conditioner: Bill amount is $28500\left(1+\frac{28}{100}\right)=28500 \times \frac{128}{100}=36480$
$\therefore$ Total Bill amount $=840+728+900+2065+36480$
$=₹ 41,013$ (total bill amount)
 

Question $4 .$
P's income is $25 \%$ more than that of Q. By what percentage is Q's income less than P's?
Answer:
Let Q's income be 100 .
$\mathrm{P}$ 's income is $25 \%$ more than that of Q
$\therefore$ P's income $=100+\frac{25}{100} \times 100=125$
Q's income is 25 less than that of P
In percentage terms, Q's income is less than P's with respect to P's income is
$\frac{\mathrm{P}-\mathrm{Q}}{\mathrm{P}} \times 100=\frac{125-100}{125} \times 100=\frac{25}{125} \times 100=20 \%$

Question $5 .$
Vaidegi sold two sarees for $₹ 2200$ each. On one she gains $10 \%$ and on the other she loses $12 \%$. Find her total gain or loss percentage in the sale of the sarees.
Answer:
Saree 1:
The selling price is $₹ 2200$, let cost price be $\mathrm{CP}_{1}$, gain is $10 \%$
Cost price? Using the formula
$\mathrm{SP}=\mathrm{CP}_{1}\left(1+\frac{\text { gain } \%}{10}\right)$ \& substituting the values
$2200=\mathrm{CP}_{1}\left(1+\frac{10}{100}\right) \therefore \mathrm{CP}_{1}=220 \emptyset \times \frac{100}{11 \emptyset}=2000$
Saree 2:
The selling price is 2200 , let cost price be $\mathrm{CP}_{2}$, loss is given as $12 \%$. We need to find $\mathrm{CP}_{2}$

$\begin{aligned}
&\text { using the formula as before, }\\
&\mathrm{SP}=\mathrm{CP}_{\overline{2}}\left(1-\frac{\operatorname{loss} \%}{100}\right) \& \text { substituting the values }\\
&2200=\mathrm{CP}_{2}\left(1-\frac{12}{100}\right)=\mathrm{CP}_{2} \times\left(\frac{100-12}{100}\right)\\
&=\mathrm{CP}_{2} \times \frac{88}{100}\\
&\therefore \mathrm{CP}_{2}=\frac{2200 \times 100}{88}=2500\\
&\therefore \text { Cost price of both together is } \mathrm{CP}_{1}+\mathrm{CP}_{2}\\
&=2000+2500=4500 \ldots \ldots . . \text { (1) }\\
&\text { Selling price of both together is } 2 \times 2200=4400 \ldots \ldots \text { (2) }\\
&\text { Since net selling price is less than net cost price, there is a loss. }\\
&\text { Loss } \%=\frac{\text { loss }}{\text { cost price }} \times 100\\
&\text { Loss }=\text { Net cost price }-\text { Net selling price }
\end{aligned}$

$(1)-(2)=4500-4400=100$ 100100202
$\begin{aligned}
&\therefore \text { loss } \%=\frac{100}{4500} \times 100=\frac{100}{45}=\frac{20}{9}=2 \frac{2}{9} \% \\
&=2 \frac{2}{9} \% \text { loss }
\end{aligned}$

Question $6 .$
If 32 men working 12 hours a day can do a work in 15 days, then how many men working 10 hours a day can do double that work in 24 days?
Answer:

$\begin{aligned}
&\text { Let } \\
&\mathrm{P}_{1}=32 \\
&\mathrm{P}_{2}=\mathrm{x} \\
&\mathrm{H}_{1}=12 \\
&\mathrm{H}_{2}=10 \\
&\mathrm{D}_{1}=15 \\
&\mathrm{D}_{2}=24 \\
&\mathrm{~W}_{1}=1 \\
&\mathrm{~W}_{2}=1 \\
&\mathrm{U}_{\text {sing formula }} \frac{\mathrm{P}_{1} \times \mathrm{D}_{1} \times \mathrm{H}_{1}}{\mathrm{~W}_{1}}=\frac{\mathrm{P}_{2} \times \mathrm{D}_{2} \times \mathrm{H}_{2}}{\mathrm{~W}_{2}} \\
&\frac{32 \times 15 \times 12}{1}=\frac{x \times 24 \times 10}{1} \\
&
\end{aligned}$

$x=24$ persons
To complete the same work 24 men needed.
To complete double the work $24 \times 2=48$ men are required.
 

Question $7 .$
Amutha can weave a saree in 18 days. Anjali is twice as good a weaver as Amutha. If both of them weave together, then in how many days can they complete weaving the saree?
Answer:
Amutha can weave a saree in 18 days.
Anjali is twice as good as Amutha.
ie. If Amutha weave for 2 days, Anjali do the same work in 1 day.
If Anjali weave the saree she will take $\frac{18}{2}=9$ days.
Hence time taken by them together $a b$
$\begin{aligned}
&=\frac{a b}{a+b} \text { days } \\
&=\frac{18 \times 9}{18+9}=\frac{18 \times 9}{27}=6 \text { days }
\end{aligned}$

In 6 clays they complete weaving the saree.
 

Question $8 .$
$P$ and $Q$ can do a piece of work in 12 days and 15 days respectively. P started the work alone and then after 3 days, Q joined him till the work was completed. How long did the work last?
Answer:
P can do a piece of work in 12 days.
$\therefore$ P's 1 day work $=\frac{1}{12}$
P's 3 day's work $=3 \times \frac{1}{12}=\frac{3}{12}$
Q can do a piece of work in 15 days.
$\therefore$ Q's 1 day work $=\frac{1}{15}$
Remaining work after 3 days $=1-\frac{3}{12}=\frac{9}{12}$
$(P+Q) ' s ~ 1$ day work $=\frac{1}{12}+\frac{1}{15}=\frac{5}{60}+\frac{4}{60}=\frac{9}{60}$
$\therefore$ Number of days required to finish the remaining work 9

$\therefore$ Number of days required to finish the remaining work 9
$=\frac{\text { Remaining work }}{(\mathrm{P}+\mathrm{Q}) \text { 's 1 day work }}=\frac{\frac{9}{\frac{12}{9}}}{\frac{60}{6}}=\frac{9}{12} \times \frac{60}{9}=5$
Remaining work lasts for 5 days
Total work lasts for $3+5=8$ days.

Question $9 .$
If the numerator of a fraction is increased by $50 \%$ and the denominator is decreased by $20 \%$, then it becomes $\frac{3}{5}$. Find the original fraction.
Answer:
Original fraction $=\frac{x}{y}$
numerator increased by $50 \%$
$\therefore$ Numerator $=\frac{150}{100} x$
Denominator decreased by $20 \%$
$\therefore$ Denominator $=\frac{80}{100} y$
Hence

$\begin{gathered}
\frac{\frac{156}{106} x}{\frac{88}{108} y}=\frac{3}{5} \\
\frac{\frac{15}{10} x}{\frac{8}{10} y}=\frac{3}{5} \\
\frac{15 x}{19} \times \frac{18}{8 x}=\frac{3}{5} \\
\frac{x}{y}=\frac{3}{5} \times \frac{8}{18}=\frac{8}{25}
\end{gathered}$

Question $10 .$
Gopi sold a laptop at $12 \%$ gain. If it had been sold for ₹ 1200 more, the gain would have been $20 \%$. Find the cost price of the laptop.

Answer:
Let the cost price of the laptop be ' $x$
Gain $=12 \%$
$\therefore$ Selling price $(\mathrm{SP})=\mathrm{CP}\left(1+\frac{\text { gain } \%}{100}\right)$
Substituting in formula $=x\left(1+\frac{12}{100}\right)=x\left(\frac{100+12}{100}\right)=\frac{112}{100} x$
If the selling price was 1200 more
i.e $\frac{112}{100} x+1200$, the gain is $20 \%$
$\begin{aligned}
\text { i.e new selling price } &=x\left(1+\frac{20}{100}\right) \\
&=\frac{112}{100} x+1200=x\left(100+\frac{20}{100}\right)=\frac{120}{100} x \\
\therefore 1200 &=\frac{120}{100} x-\frac{112}{100} x=\frac{8}{100} x
\end{aligned}$

$\therefore x=\frac{1200 \times 100}{8}=15000$
Cost price of the laptop is $₹ 15,000 /-$
 

Question 11.
A shopkeeper gives two successive discounts on an article whose marked price is ₹ 180 and selling price is $₹ 108$. Find the first discount percentage if the second discount is $25 \%$.
Answer:
Marked price is given as ₹ 180
Let 1 discount be $\mathrm{d}_{1} \%=$ ? (to find)
2 nd discount be $\mathrm{d}_{2} \%=25 \%$
Selling price is 108 (given)
Price after $1^{\text {st }}$ discount $=180\left(1-\frac{d_{1}}{100}\right)=\mathrm{P} \ldots \ldots$ (1)
Price after $2^{\text {nd }}$ discount $=P_{1}\left(1-\frac{d_{2}}{100}\right)=108$
Substituting for $P_{1}$ from (1), we get

$180\left(1-\frac{d_{1}}{100}\right)\left(1-\frac{d_{2}}{100}\right)=108 \Rightarrow$ Since $d_{2}=25$, we get
$180\left(1-\frac{d_{1}}{100}\right)\left(1-\frac{25}{100}\right)=108$
$\therefore\left(1-\frac{d_{1}}{100}\right)=\frac{108}{180} \times \frac{100}{75}=\frac{4}{5}$
$1-\frac{d_{1}}{100}=\frac{4}{5}$
$\therefore \frac{d_{1}}{100}=1-\frac{4}{5}=\frac{1}{5}$
$\therefore d_{1}=\frac{1}{5} \times 100=20 \%$
lst discount $=20 \%$
 

Question $12 .$
Find the rate of compound interest at which a principal becomes $1.69$ times itself in 2 years.
Answer:
Let principal be ' $\mathrm{P}$ '
Amount is given to be $1.69$ times principal
i.e $1.69 \mathrm{P}$
Time period is $2 \mathrm{yrs} .=(\mathrm{n})$
Rate of interest $=r=$ ?(required)
Applying the formula,

Amount = Principal $\left(1+\frac{r}{100}\right)^{n}$
Substituting, 1.69 $\mathrm{P}=\mathrm{P}\left(1+\frac{r}{100}\right)^{2}$
$\therefore\left(1+\frac{r}{100}\right)^{2}=\frac{1.69 \mathrm{P}}{\mathrm{P}}=1.69$
Taking square root on both sides, we get
$\begin{aligned}
\sqrt{1.69} &=1+\frac{r}{100} \\
\therefore 1+\frac{r}{100} &=1.3 \\
\therefore \frac{r}{100} &=1.3 \quad r=30 \%
\end{aligned}$
$\therefore$ rate of compound interest is $30 \%$
 

Question $13 .$
A small - scale company undertakes an agreement to make 540 motor pumps in 150 days and employs 40 men for the work. After 75 days, the company could make only 180 motor pumps. How many more men should the company employ so that the work is completed on time as per the agreement?
Answer:
Let the number of men to be appointed more be $\mathrm{x}$.

To produce more pumps more men required
$\therefore$ It is direct variation.
$\therefore$ The multiplying factor is $\frac{360}{180}$
More days means less employees needed.
$\therefore$ It is Indirect proportion. 75
$\therefore$ The multiplying factor is $\frac{75}{75}$
Now $40+x=40 \times \frac{360}{180} \times \frac{75}{75}$
$\begin{aligned}
&40+x=80 \\
&x=80-40 \\
&x=40
\end{aligned}$
40 more man should be employed to complete the work on time as per the agreement.
 

Question $14 .$
P alone can do $\frac{1}{2}$ of a work in 6 days and $Q$ alone can do $\frac{2}{3}$ of the same work in 4 days. in how many days will they finish $\frac{3}{4}$ of the work, working together?
Answer:
$\frac{1}{2}$ of the work is done by $P$ in 6 days

$\therefore$ Full work is done by P in $\frac{6}{\frac{1}{2}}=6 \times 2=12$ days $\frac{2}{3}$ of work done byQin4days.
$\therefore$ Full work done by $\mathrm{Q}$ in $\frac{4}{\frac{2}{3}}=4 \times \frac{3}{2}=6$ days
$(\mathrm{P}+\mathrm{Q})$ will finish the whole work in $\frac{a b}{a+b}$ days $=\frac{12 \times 6}{12+6}=\frac{12 \times 6}{18}=4$ days $(P+Q)$ will finish $\frac{-3}{4}$ of the work in $4 \times \frac{3}{4}=3$ days.
 

Question $15 .$
$\mathrm{X}$ alone can do a piece of work in 6 days and $\mathrm{Y}$ alone in 8 days. $\mathrm{X}$ and $\mathrm{Y}$ undertook the work for $₹ 48000$. With the help of $Z$, they completed the work in 3 days. How much is Z's share?
Answer:
$\mathrm{X}$ can do the work in 6 days.
X's $I$ day work $=\frac{1}{6}$

X's share for 1 day $=\frac{1}{6} \times 48000=₹ 800$
$X$ 's share for 3 days $=3 \times 800=₹ 2400$
$\mathrm{Y}$ can complete the work in 8 days.
Y's 1 day work $=\frac{1}{8}$
Y's I day share $=\frac{1}{8} \times 4800=₹ 600$
$\mathrm{Y}$ 's 3 days share $=₹ 600 \times 3=₹ 1800$
$(\mathrm{X}+\mathrm{Y})$ 's 3days share $=₹ 2400+₹ 1800=₹ 4200$
Remaining money is Z's share
$\therefore$ Z's share $=₹ 4800-₹ 4200=₹ 600$