Exercise 4.4 - Chapter 4 Life Mathematics 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.4$
Question 1.
Fill in the blanks
(i) A can finish a job in 3 days whereas $\mathrm{B}$ finishes it in 6 days. The time taken to complete the job working together is __days.
Answer:
2 days
(ii) If 5 persons can do 5 jobs in 5 days, then 50 persons can do 50 jobs in __ days. Answer: 5
(iii) $A$ can do a work in 24 days. If $A$ and $B$ together can finish the work in 6 days, then $B$ alone can finish the work in ___days.
Answer:
8
(iv) A alone can do a piece of work in 35 days. If $\mathrm{B}$ is $40 \%$ more efficient than $\mathrm{A}$, then $\mathrm{B}$ will finish the work in ___ days.
Answer:
25
(v) A alone can do a work in 10 days and $\mathrm{B}$ alone in 15 days. They undertook the work for $₹ 200000$. The amount that A will get is ___.
Answer:
$₹ 1,20,000$

Question $2 .$
210 men working 12 hours a day can finish ajob in 18 days. How many men are required to finish the job in 20 days working 14 hours a day?
Answer:
Let the required number of men be $x$.

More working hours $\Rightarrow$ less men required.
$\therefore$ It is inverse proportion.
$\therefore$ Multiplying factor is $\frac{12}{14}$
Also more number of days $\Rightarrow$ less men
$\therefore$ It is an inverse proportion.
$\therefore$ Multiplying factor is $\frac{18}{20}$
$\therefore \mathrm{x}=210 \times \frac{12}{14} \times \frac{18}{20}$

$x=162 \text { men }$
162 men are required.
 

Question $3 .$
A cement factory makes 7000 cement bags in 12 days with the help of 36 machines. How many bags can be made in 18 days using 24 machines?
Answer:
Let he required number of cement bags be $x$.

Number of days more $\Rightarrow$ More cement bags.
$\therefore$ It is direct variation.
$\therefore$ The multiplying factor $=\frac{18}{12}$
Number of machines more $\Rightarrow$ More cement bags.
$\therefore$ It is direct variation.
$\therefore$ The multiplying factor $=\frac{24}{36}$
$\therefore x=7000 \times \frac{18}{12} \times \frac{24}{36}$

$x=7000$ cement bags
7000 cement bags can be made
 

Question $4 .$
A soap factory produces 9600 soaps in 6 days working 15 hours a day. In how many days will it produce 14400 soaps working 3 more hours a day?
Answer:
Let the required number of days be $x$.

To produce more soaps more days required.
$\therefore$ It is direct proportion.
$\therefore$ Multiplying factor $=\frac{14400}{9600}$
If more hours spend, less days required.
$\therefore$ It is indirect proportion
$\therefore$ Multiplying factor $=\frac{15}{18}$
$\therefore \mathrm{x}=6 \times \frac{14400}{9600} \times \frac{15}{18}$

$x=\frac{15}{2}$ $\frac{15}{2}$ days will be needed.
 

Question $5 .$
If 6 container lorries can transport 135 tonnes of goods in 5 days, how many more lorries
are required to transport 180 tonnes of goods in 4 days?
Answer:
Let the number of lorries required more $=\mathrm{x}$.

As the goods are more $\Rightarrow$ More lorries are needed to transport.
$\therefore$ It is direct proportion.
$\therefore$ Multiplying factor $=\frac{180}{135}$
Again if more days $\Rightarrow$ less number of lorries enough.
$\therefore$ It is direct proportion.
$\therefore$ Multiplying factor $=\frac{5}{4}$
$\therefore 6+\mathrm{x}=6 \times \frac{180}{135} \times \frac{5}{4}$

$\begin{aligned}
&6+x=10 \\
&x=10-6 \\
&x=4
\end{aligned}$
$\therefore 4$ more lorries are required.
 

Ouestion 6.
A can do a piece of work in 12 hours, $B$ and $C$ can do it 3 hours whereas $A$ and $C$ can do it in 6 hours. How long will B alone take to do the same work?
Answer:
Time taken by A to complete the work $=12 \mathrm{hrs}$.
$\therefore$ A's $1 \mathrm{hr}$ work $=\frac{1}{12}-$ (1)
$(\mathrm{B}+\mathrm{C})$ complete the work in $3 \mathrm{hrs}$.
$\therefore(\mathrm{B}+\mathrm{C})$ 's 1 hour work $=\frac{1}{3}-$ (2)
(1) $+(2) \Rightarrow$
$\therefore(\mathrm{A}+\mathrm{B}+\mathrm{C})$ 's 1 hour work $=\frac{1}{12}+\frac{1}{3}=\frac{1+4}{12}=\frac{5}{12}$
Now $(A+C)$ complete the work in 6 hrs.
$\therefore(\mathrm{A} \mid \mathrm{C})$ 's 1 hour work $-\frac{1}{6}$
$\therefore$ B's 1 hour work $=(A+B+C)$ 's 1 hour work $-(A+C)$ 's 1 hr work
$=\frac{5}{12}-\frac{1}{6}=\frac{5-2}{12}=\frac{3}{12}=\frac{1}{4}$
$\therefore \mathrm{B}$ alone take 4 days to complete the work.

Question $7 .$
$\mathrm{A}$ and $\mathrm{B}$ can do a piece of work in 12 days, while $\mathrm{B}$ and $\mathrm{C}$ can do it in 15 days whereas $\mathrm{A}$ and $C$ can do it in 20 days. How long would each take to do the same work?
Answer:
$(A+B)$ complete the work in 12 days.
$\therefore(\mathrm{A}+\mathrm{B})$ 's 1 day work $=\frac{1}{12}-(1)$
$(\mathrm{B}+\mathrm{C})$ complete the work in 15 days
$\therefore(\mathrm{B}+\mathrm{C})$ 's 1 day work $=\frac{1}{15}-(2)$
$(A+C)$ complete the work in 20 days
$\therefore(\mathrm{A}+\mathrm{C})$ 's 1 day work $=\frac{1}{20}-$ (3)
Now $(1)+(2)+(3)=$
$[(\mathrm{A}+\mathrm{B})+(\mathrm{B}+\mathrm{C})+(\mathrm{A}+\mathrm{C})]$ 's 1 day work $=\frac{1}{12}+\frac{1}{15}+\frac{1}{20}$
$(2 \mathrm{~A}+2 \mathrm{~B}+2 \mathrm{C})$ 's 1 day work $=\frac{5}{60}+\frac{4}{60}+\frac{3}{60}$
$2(\mathrm{~A}+\mathrm{B}+\mathrm{C})$ 's 1 day work $=\frac{5+4+3}{60}$

$\begin{aligned}
&\mathrm{LCM}=5 \times 4 \times 3=60 \\
&(\mathrm{~A}+\mathrm{B}+\mathrm{C}) \text { 's } 1 \text { day work }=\frac{12}{60 \times 2} \\
&(\mathrm{~A}+\mathrm{B}+\mathrm{C}) \text { 's } 1 \text { day work }=\frac{1}{10} \\
&\text { Now A's I day's work }=(\mathrm{A}+\mathrm{B}+\mathrm{C}) \text { 's } 1 \text { day work }-(\mathrm{B}+\mathrm{C}) \text { 's } 1 \text { day work } \\
&=\frac{1}{10}-\frac{1}{15}=\frac{3}{30}-\frac{2}{30}=\frac{1}{30} \\
&\therefore \mathrm{A} \text { takes } 30 \text { days to complete the work. } \\
&\mathrm{B} \text { 's } 1 \text { day work }-(\mathrm{A}+\mathrm{B}+\mathrm{C}) \text { 's } 1 \text { day's work }-(\mathrm{A}+\mathrm{C}) \text { 's } 1 \text { day's work } \\
&=\frac{1}{10}-\frac{1}{20}=\frac{6}{60}-\frac{3}{60} \\
&=\frac{6-3}{60}=\frac{3}{60}=\frac{1}{20} \\
&\mathrm{~B} \text { takes } 20 \text { days to complete the work. } \\
&\text { C's } 1 \text { day work }(\mathrm{A} \mid \mathrm{B} \text { । } \mathrm{C}) \text { 's I day work }-(\mathrm{A} \mid \mathrm{B}) \text { 's I day work } \\
&\therefore \mathrm{C} \text { takes } 60 \text { days to complete the work. }
\end{aligned}$

Question $8 .$
Carpenter A takes 15 minutes to fit the parts of a chair while Carpenter B takes 3 minutes more than A to do the same work. Working together, how long will it take for them to fit the parts for 22 chairs?
Answer:
Time taken by A to fit a chair $=15$ minutes
Time taken by $B=3$ minutes more than $A$
$=15+3=118$ minutes
$\therefore$ As 1 minute work $=\frac{1}{15}$
B's 1 minute work $=\frac{1}{18}$
$(\mathrm{A}+\mathrm{B})$ 's 1 minutes work $=\frac{1}{15}+\frac{1}{18}$
$\frac{12}{180}+\frac{10}{180}=\frac{22}{180}=\frac{11}{90}$
$\therefore$ Time taken by $(\mathrm{A}+\mathrm{B})$ to fit a chair
$\mathrm{LCM}=3 \times 5 \times 6=180$
$=\frac{1}{\frac{11}{90}}=\frac{90}{11} \mathrm{~m}$
$\therefore$ Time taken by $(\mathrm{A}+\mathrm{B})$ to fit a chair
$=\frac{90}{11} \times 22=180$ minutes
$=\frac{180}{60}=3$ hours

Question $9 .$
A can do a work In 45 days. He works at it for 15 days and then, B alone finishes the remaining work in 24 days. Find the time taken to complete $80 \%$ of the work, if they work together.
Answer:
A completes the work in 45 days.
$\therefore$ A's 1 day work $=\frac{1}{45}$
A's 15 days work $=\frac{1 / 3}{45}=\frac{1}{3}$
Remaining work $=1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}$
B finishes $\frac{2}{3}$ rd work in 24 days
$\begin{aligned}
\text { B's 1 day work } &=\frac{\frac{2}{3}}{24} \\
&=\frac{2}{3 \times 24}=\frac{1}{36}
\end{aligned}$
$(A+B)$ 's 1 day work $=\frac{1}{45}+\frac{1}{36}$
$=\frac{4+5}{180}$

$=\frac{9}{180}=\frac{1}{20}
$Let $x$ days required
$\therefore \frac{x}{20}=\frac{80}{100} \Rightarrow x=\frac{80}{190} \times 20=16 \text { days. }$

Question $10 .$
$\mathrm{A}$ is thrice as fast as $\mathrm{B}$. If $\mathrm{B}$ can do a piece of work in 24 days then, find the number of days they will take to complete the work together.
Answer:
If $B$ does the work in 3 days, $A$ will do it in I day.
$B$ complete the work in 24 days.
$\therefore$ A complete the same work in $\frac{24}{3}=8$ days.
$\therefore(\mathrm{A}+\mathrm{B})$ complete the work in $\frac{a b}{a+b}$ days
$\begin{aligned}
&=\frac{24 \times 8}{24+8} \text { days } \\
&=\frac{24 \times 8}{32} \text { days }=6 \text { days }
\end{aligned}$

They together complete the work in 6 days.