Question 13 Marks
Show that the following numbers are in continued proportion: $48, 60, 75$
AnswerThe given numbers $48, 60, 75$ are in Continued proportion if $48 : 60 :: 60 : 75.$
Now, product of means $= 60 × 60 = 3600$ And,
product of extremes $= 48 × 75 = 3600$
$\therefore$ Product of means = Product of extremes
So, $48 : 60 :: 60 : 75$
Hence, the numbers $48, 60, 75$ are in continued proportion.
View full question & answer→Question 23 Marks
Find the ratio of the price of a pencil to that of a ball pen, if pencil cost Rs. 96 per score and ball pens cost $Rs. 50.40$ per dozen.
AnswerPrice of $20$ pencils $= Rs. 96 (1$ score $= 20$ pencils)
Price of $1$ pencil $= Rs. (96 ÷ 20) = Rs. 4.80$
Price of $12$ ball pens $= Rs. 50.40 (1$ dozen $= 12)$
Price of $1$ ball pen $= Rs. (50.40 ÷ 12) = Rs. 4.20.$
Ratio of the price of a pencil to that of a ball pen $= Rs. 4.80 : Rs. 4.20 = 480$
paise : $420$ paise $= 480 : 420 = 48 : 42$
View full question & answer→Question 33 Marks
Prove that $(5 : 6) > (3 : 4).$
AnswerWe can write: $(5:6)=\frac{5}6{}\text{ and }(3 : 4)=\frac{3}{4}$
By making their denominators same: (Taking the $L.C.M$ of $6$ and $4,$ which is 24) Consider, $5 : 6$
$\frac{5\times4}{6\times4}=\frac{20}{24}$ And
$\frac{3\times6}{4\times6}=\frac{18}{24}$ As $20>18$
Clearly, $(5:6)>(3:4)$
View full question & answer→Question 43 Marks
Determine if the following ratios form a proportion: $25\ cm : 1m$ and $Rs. 40 : Rs. 160$
Answer$25\ cm : 1m$ and $Rs. 40 : Rs. 160$
$=\frac{25\text{cm}}{1000\text{cm}}=\frac{1}{4}$
$=\frac{\text{Rs. }40}{\text{Rs. }160}=\frac{1}{4}$
$\because\frac{3}{5}=\frac{3}{5}$ (Ratios are equal)
$\therefore 39$ litres : $65$ litres and $6$ bottles : $10$ Bottles are in proportion
View full question & answer→Question 53 Marks
40 men can finish a piece of work in $26$ days. How many men will be needed to finish it in $16$ days?
AnswerNumber of men needed to finish a piece of work in $26$ days $= 40$
Number of men needed to finish it in $1$ day $= 26 \times 40 = 1040$ (less days means more men)
Number of men needed to finish it in $16$ days $=\frac{1040}{16}=65$
View full question & answer→Question 63 Marks
Two numbers are in the ratio $11 : 12$ and their sum is $460$, find the numbers.
AnswerLet the numbers be $11x$ and $12x.$
Then, $11x + 12x = 460$
$\Rightarrow23\text{x}=460$
$\Rightarrow\frac{23\text{x}}{23}=\frac{460}{23}$ (Dividing both sides by $23)$
$\Rightarrow\text{x}=20.$
$\therefore$ Required numbers are ($11 \times 20$) and $12 \times 20$,that is $220$ and $240$
View full question & answer→Question 73 Marks
Divide $Rs. 3450$ among $A, B$ and $C$ in the ratio $3 : 5 : 7.$
AnswerTotal amount $= Rs. 3450$
Ratio in $A, B$ and $C$ shares $= 3 : 5 : 7$
Sum of share $= 3 + 5 + 7 = 15$
$\therefore\text{A's share}=\text{Rs. }\frac{3450\times3}{15}$
$=\text{Rs. }230\times3=\text{Rs. }690$
$\text{B's share}=\text{Rs.}\frac{3450\times5}{15}$
$=\text{Rs. }230\times5=\text{Rs. }1150$
$\text{C's }\text{share}=\text{Rs. }\frac{3450\times7}{15}$
$=\text{Rs. }230\times7=\text{Rs. }1610$
View full question & answer→Question 83 Marks
Find the value of $x$ in the following proportions: $x : 92 :: 87 : 116$
AnswerWe have Product of means $= 85 \times 57$
Product of extremes $= 51 \times x$
$\therefore 51\times\text{x}=85\times57$
$\Rightarrow\text{x}=\frac{85\times57}{51}=95$
View full question & answer→Question 93 Marks
Express the following ratios in the simplest form: $777 : 1147$
AnswerThe given ratio $=777:1147=\frac{777}{1147}$ To express it in simplest form,
we divide the numerator and denominator by the $HCF$ of $777$ and $1147.$
$=\frac{777\div37}{1147\div37}=\frac{21}{31}$
$(\text{HCF of 777 and 1147 is 37})$
$=21:31$ View full question & answer→Question 103 Marks
Mr Sahai and his wife are both school teachers and earn $Rs. 16800$ and $Rs. 10500$ per month respectively. Find the ratio of: Mr Sahai’s income to his wife’s income.
AnswerEarning of Sahai $= Rs. 16800$
And of his wife $= Rs. 10500$
Then total income $= Rs. 16800 + 10500 = Rs. 27300$
Ratio in sahai's income and his wife $16800 : 10500$
$=\frac{16800}{10500}=\frac{16800\div2100}{10500\div2100}$
$(\text{HCF}$ of $16800$ and $105000$ is $2100)$
$=\frac{8}{5}=8:5$
View full question & answer→Question 113 Marks
Rohit earns $Rs. 15300$ and saves $Rs. 1224$ per month. Find the ratio of: His income and savings.
AnswerRohit monthly earnings $= Rs. 15300$
And his savings $= Rs. 1224$
So, his expenditure $= Rs. 15300 - 1224= = Rs. 14076$
Now, Ratio in his income and savings $=15300:1224=\frac{15300}{1224}$
$=\frac{15300\div612}{1224\div612}$ (Dividing by $HCF$ of $15300$ and $1224)$
$=\frac{25}{2}=25:2$ View full question & answer→Question 123 Marks
Write $(T)$ for true and $(F)$ for false in case of the following: $36 : 45 :: 80 : 100$
AnswerTrue Solution: We have, $36:45=\frac{36}{45}=\frac{4}{5}$ And $80:100=\frac{80}{100}=\frac{4}{5}$ $\therefore36:45=80:100$ So, given statemant is true
View full question & answer→Question 133 Marks
Write the following ratios in the simplest form: $Rs. 6.30 : Rs. 16.80$
AnswerThe given ratio $= Rs. 6.30 : Rs. 16.80$
$=\frac{\text{Rs. }6.30}{\text{Rs. }16.80}$
$=\frac{630}{1680}$ To express it in simplest form,
we divide the numerator and denominator by the $HCF$ of $630$ and $1680$.
Now, $HCF$ of $630$ and $1680 = 210$
$\therefore\frac{630}{1680}=\frac{630\div210}{1680\div210}=\frac{3}{8}$
$\therefore\text{Required ratio}= 3:8$ View full question & answer→Question 143 Marks
Mr Sahai and his wife are both school teachers and earn $Rs. 16800$ and $Rs. 10500$ per month respectively. Find the ratio of:
Mr Sahai’s income to the total income of the two.
AnswerEarning of Sahai $= Rs. 16800$
and of his wife $= Rs. 10500$
Then total income $= Rs. 16800 + 10500$
$= Rs. 27300$
Sahai's income and total of their income
$=16800:27300=\frac{16800}{27300}$
$=\frac{16800\div2100}{27300\div2100}=\frac{8}{13}=8:13$
View full question & answer→Question 153 Marks
Express the following ratios in the simplest form:
$480 : 384$
AnswerThe given ratio $=480:384=\frac{480}{384}$
To express it in simplest form, we divide the numerator and denominator by the $HCF$ of $480$ and $384$.
Now, $HCF$ of $480$ and $384 = 96$
$\therefore\frac{480}{384}=\frac{480\div96}{384\div96}=\frac{5}{4}$
$\therefore\text{Required ratio}=5:4$ View full question & answer→Question 163 Marks
Determine if the following ratios form a proportion: $200\ mL : 2.5L$ and $Rs. 4 : Rs. 50$
Answer$200mL : 2.5L$ and $Rs. 4 : Rs. 50$
$\frac{200\text{mL}}{2.5\text{L}}=\frac{200}{2.5\times1000}$
$=\frac{200\times10}{25\times1000}=\frac{2}{25}$ and
$\frac{\text{Rs. }4}{\text{Rs. }50}=\frac{2}{25}$ (Ratios are equal)
$200mL : 2.5L$ and $Rs. 4 : Rs. 50$ are in proportion.
View full question & answer→Question 173 Marks
Express the following ratios in the simplest form: $85 : 561$
AnswerThe given ratio is $85:561=\frac{85}{561}$
To express in it simplest form, we divided the numerator and denominator by the $HCF$ of $85$ and $561.$
By the $HCF$ of $85$ and $561.$
Now, $HCF$ of $85$ and $561 = 17$
$\therefore\frac{85}{561}=\frac{85\div17}{561\div17}=\frac{5}{33}$
$\therefore \text{Required ratio}=5.33$ View full question & answer→Question 183 Marks
Divide $Rs. 1575$ between Kamal and Madhu in the ratio $7 : 2.$
AnswerTotal amount $= Rs. 1575$
Ratio in Kamal and Madhu’s share $= 7 : 2$
Sum of ratios $= 7 + 2 = 9$
$\therefore$ kamal's share $=\text{Rs. }1225$
$\text{Rs. }175\times7=\text{Rs. }1225$
Madhu's share $=\text{Rs.}\frac{1575\times2}{9}$
$=\text{Rs. }175\times2=\text{Rs. }350$
View full question & answer→Question 193 Marks
Express the following ratios in the simplest form:
$186 : 403$
AnswerThe given ratio $=186:403=\frac{186}{403}$
To express it in simplest form, we divide the numerator and denominator by the $HCF$ of $186$ and $403$.
Now, $HCF$ of 186 and $403 = 31$
$\therefore\frac{186}{403}=\frac{186\div31}{403\div31}=\frac{6}{13}$
$\therefore\text{Required ratio}=6:13$ View full question & answer→Question 203 Marks
Write the following ratios in the simplest form: $48$ min : $2$ hours $40$ min.
AnswerThe given ratio $= 48 min $
$: 2 hours 40 min. = 48 min $
$: (2 \times 60 + 40) min = 48 min$
$: (120 + 40) min = 48 min $
$: 160 min = 48 $
$: 160 = 3 $
$:10$
View full question & answer→Question 213 Marks
The ratio of the number of male and female workers in a textile mill is $5 : 3$. If there are $115$ male workers, what is the number of female workers in the mill?
AnswerLet the number of male and female workers in the mill be $5x$ and $3x$ respectively.
Then, $5x = 115$
$\Rightarrow\frac{5\text{x}}{5}=\frac{115}{5}$ (Dividing both sides by $5)$
$\Rightarrow \text{x} = 23$
Number of female workers in the mill $=3\text{x}$ $=3\times23=69$
View full question & answer→Question 223 Marks
Find the value of $x$ in the following proportions: $55 : 11 :: x : 6$
AnswerWe have $55 : 11 :: x : 6$
Product of means $= 11 \times x = 11x$
Product of extremes $= 55 \times 6 = 330$
$\therefore11\text{x}=330$
$\Rightarrow\frac{11\text{x}}{11}=\frac{330}{11}$ (Dividing both sides by $11)$
$\Rightarrow\text{x}=30$
View full question & answer→Question 233 Marks
Express the following ratios in the simplest form:
$36 : 90$
Answer$36 : 90$$=\frac{36}{90}$
$=\frac{36\div18}{90\div18}$
$(\text{HCF of 36, 90 = 18})$
$=\frac{2}{5}=2:5$
View full question & answer→Question 243 Marks
The boys and the girls in a school are in the ratio $9 : 5$. If the total strength of the school is 448. find the number of girls.
AnswerLet the number of boys and girls in the school be $9x$ and $5x$ respectively.
According to the question, $9\text{x}+5\text{x}=44$
$\Rightarrow14\text{x}=448$
$\Rightarrow\frac{14\text{x}}{14}=\frac{448}{14}$ (Dividing both sides by $14)$
$\Rightarrow\text{x}=32$
Number of girls $=5\text{x}$
$=5\times32$
$=160$
View full question & answer→Question 253 Marks
Write $(T)$ for true and $(F)$ for false in case of the following: $30$ bags $: 18$ bags $:: Rs. 450 : Rs. 270$
AnswerWe have, $30$ bags $: 18$ bags $=30:18=\frac{30}{18}=\frac{5}{3}$
$\text{And Rs. }450 : \text{Rs. } 270$
$\therefore 30\text{ bags}:18\text{ Rs. }270$
$=450:270=\frac{450}{270}=\frac{5}{3}$
$\therefore30\text{bags}:18\text{ bags}=\text{Rs. }450:270$
So, the given statement is true.
View full question & answer→Question 263 Marks
If $9, x, x, 49$ are in proportion, find the value of $x$. Hint.
$x^2=(9 \times 49)=(32 \times 72)=(3 \times 7)^2=(21)^3$
AnswerIt is given that $9, x, x, 49$ are in proportion, that is, $9 : x :: x : 49$
$\therefore$ Product of means = Product of extremes
$\text{x}\times\text{x}=9\times49$
$\Rightarrow\text{x}^2=9\times49$
$\Rightarrow\text{x}=\sqrt{9\times49}$
$\Rightarrow\text{x}=\sqrt{3\times3\times7\times7}$
$\Rightarrow\text{x}=3\times7=21$
$\therefore\text{x}=21$
View full question & answer→Question 273 Marks
Divide $Rs. 1400$ among $A, B$ and $C$ in the ratio $2 : 3 : 5$
Answer$A : B : C = 2 : 3 : 5$
Sum of ratio $= 2 + 3 + 5$
Total money $= Rs. 1400$
Then share of $A = 2 \times Rs. 1400 = Rs. 2800 = Rs. 280$
Share of $B = 3 \times Rs. 1400 = Rs. 4200 = Rs. 420$
Share of $C =\frac{5}{10}\times\text{Rs. }1400$
$=\text{Rs. }\frac{7000}{10}=\text{Rs. }700$
View full question & answer→Question 283 Marks
Show that the following numbers are in continued proportion:
$36, 90, 225$
AnswerThe given numbers $36, 90, 225$ are in
Continued proportion of $36 : 90 :: 90 : 225$
Now, product of means $= 90 \times 90 = 8100$
And, product of extremes $= 36 \times 225 = 8100$
$\therefore$ Product of means = Product of extremes
So, $36 : 90 :: 90 : 225$
Hence, the numbers $36, 90, 225$ are in continued proportion.
View full question & answer→Question 293 Marks
The ratio of income to savings of a family is $11 : 2$. Find the expenditure if the savings is $Rs. 1520.$
AnswerRatio in income and savings of a family $= 11 : 2$
But Total savings $= Rs. 1520$
Let income $= x 11 : 2 = x : 1520$
$\Rightarrow\text{x}=\frac{11\times1520}{2}=11\times760$
$=\text{Rs. }8360$ Expenditure = total income - savings $=\text{Rs. }8360-1520$
$=\text{Rs. }6840$
View full question & answer→Question 303 Marks
In an army camp, there were provisions for $425$ men for $30$ days. How long did the provisions last for $375$ men?
AnswerNumber of days for which provisions last for $425$ men $= 30$ days
Number of days for which provisions last for $1$ men $= 30 \times 425 = 12750$ days. (less men means more days)
Number of days for which provisions last for $375$ men $=\frac{12750}{375}=34\text{ days}$
Hence, provisions will last for $34$ days for $375$ men.
View full question & answer→Question 313 Marks
The ratio of zinc and copper in an alloy is $7 : 9$ If the weight of copper in the alloy is $12.6\ kg$ find the weight of zinc in it.
AnswerRatio of zinc and copper in an alloy is $7 : 9$
Let the weight of zinc and copper in it be $(7x)$ and $(9x)$, respectvely.
Now, the weight of a copper $= 12.6kg$ (given)
$\therefore9\text{x}=12.6$
$\Rightarrow\text{x}=\frac{12.6}{9}=1.4$
$\therefore$ Weight of zinc $= 7x = 7 \times 1.4 = 9.8\ kg$
View full question & answer→Question 323 Marks
From each of the given pairs, find which ratio is larger: $(3 : 4)$ or $(9 : 16)$
Answer$(3 : 4)$ or $(9 : 16) LCM$ of $4, 16 = 16$
$\therefore\frac{3}{4}=\frac{3\times4}{4\times4}=\frac{12}{16}$
We see that $\frac{12}{16}>\frac{9}{16}\text{ or }\frac{3}{4}>\frac{9}{16}$
$\therefore 3 : 4$ the greater ratio
View full question & answer→Question 333 Marks
From the given pairs, find which ratio is larger: $(5 : 12)$ or $(17 : 30)$
Answer$(5 : 12)$ or $(17 : 30)$
$\frac{5}{12},\frac{17}{30}$
$LCM$ of $12$ and $30 = 60$
$\therefore\frac{5}{12}=\frac{5\times5}{12\times5}=\frac{25}{60}$
$\text{and }\frac{17}{30}=\frac{17\times2}{30\times2}=\frac{34}{60}$
We see that $\frac{34}{60}>\frac{25}{60}\text{or }\frac{17}{30}>\frac{5}{12}$
$\therefore17:30$ is greater ratio
View full question & answer→Question 343 Marks
A 35cm line segment is divided into two parts in the ratio $4 : 3$. Find the length of each part.
AnswerLength of line segment $= 35\ cm$
Ratio $= 4 : 3$
Sum of ratio $= 4 + 3 = 7$
$\therefore\text{First part}=\frac{35\times4}{7}=20\text{cm}$
$\text{Second part }= \frac{35\times3}{7}=15\text{cm}$
View full question & answer→Question 353 Marks
Find the value of x in the following proportions: $51 : 85 :: 57 : x$
AnswerWe have $51 : 85 :: 57 : x$
Product of means $= 85 \times 57$
Product of extremes $= 51 \times x$
$\therefore 51\times\text{x}=85\times57$
$\Rightarrow\text{x}=\frac{85\times57}{51}=95$
View full question & answer→Question 363 Marks
From the given pairs, find which ratio is larger: $(1 : 2)$ or $(13 : 27)$
Answer(1 : 2) or (13 : 27)$\frac{1}{2},\frac{13}{27}$
$\text{LCM of 2 and 27 = 54}$
$\therefore\frac{1}{2}=\frac{1\times27}{2\times27}=\frac{27}{54}$
$\text{and }\frac{13}{27}=\frac{13\times2}{27\times2}=\frac{26}{54}$
We see that $\frac{27}{54}>\frac{26}{54}\text{ or }\frac{1}{2}>\frac{13}{27}$
$\therefore1:2$ is greater ratio
View full question & answer→Question 373 Marks
The $1st, 3rd$ and $4th$ terms of a proportion are $12, 8$ and $14$ respectively. Find the 2nd term.
Answer$1st$ term $=12$, third term $= 8$ and fourth term $= 14$
Let 2nd term $= x$, Then,
$12:\text{x}::9:14$
$\therefore\text{x}=\frac{12\times14}{8}$
$\Big(\text{b}=\frac{\text{ad}}{\text{c}}\Big)$
$=21$$\therefore\text{second term}=21$
View full question & answer→Question 383 Marks
Rohit earns $Rs. 15300$ and saves $Rs. 1224$ per month. Find the ratio of: His income and expenditure.
AnswerRohit monthly earnings $= Rs. 15300$
And his savings $= Rs. 1224$
So, his expenditure $= Rs. 15300 - 1224= = Rs. 14076$
Now, Ratio in his income and expenditure$=1500:14076=\frac{15300}{14076}$
$=\frac{15300\div612}{14076\div612}$
$(\text{HCF}=612)$
$=\frac{25}{23}=25:23$
View full question & answer→Question 393 Marks
Show that the following numbers are in continued proportion:
$16, 84, 441$
AnswerThe given numbers $16, 84, 441$ are in
Continued proportion if $16 : 84 :: 84 : 441.$
Now, product of means $= 84 \times 84 = 7056$
And, product of extremes $= 16 \times 441 = 7056$
Product of means = Product of extremes.
So, $16 : 845 :: 84 : 441$
Hence $16, 84, 441$ are in continued proportion.
View full question & answer→Question 403 Marks
A bus covers $128\ km$ in $2$ hours and a train covers $240\ km$ in $3$ hours. Find the ratio of their speeds.
AnswerA bus covers in $2$ hours $= 128\ km$
$128$ It will cover in $1$ hour $=\frac{128}{2}=64\text{km}$
A train cover in $3$ hours $= 240\text{km}$
It will cover in $1$ hour $=\frac{240}{3}$
$=80\text{km}$
Ratio in their speeds $= 64: 80$
$= 4 : 5$
{Dividing by $16$, the $LCM$ of $64, 80$}
View full question & answer→Question 413 Marks
Find the value of $x$ when $36 : x :: x : 16$
AnswerGiven:
$36 : x :: x : 16$
We know:
Product of means = Product of extremes
$x × x = 36 × 16$
$ \Rightarrow x^2=576 $
$ \Rightarrow x^2=24^2 $
$ \Rightarrow x=24 $
View full question & answer→Question 423 Marks
A factory produces electric bulbs. If $1$ out of every $10$ bulbs is defective and the factory produces $630$ bulbs per day, find the number of defective bulbs produced each day.
AnswerTotal bulbs produced per day $= 630$
Out of every $10$ bulbs, defective bulb $= 1$
Out of every $10$ bulbs, lighting bulbs $= 10 - 1 = 9$
$\therefore$ Ratio of defective bulbs to lighting bulbs $= 1 : 9$
Sum of the terms of the ratio $= (1 + 9) = 10$
$\therefore$ Number of defective bulbs produced each day$=\Big(\frac{1}{10}\times630\Big)=63.$
View full question & answer→Question 433 Marks
Rohit earns $Rs. 15300$ and saves $Rs. 1224$ per month. Find the ratio of: His expenditure and savings.
AnswerRohit monthly earnings $= Rs. 15300$ And his savings $= Rs. 1224$
So, his expenditure $= Rs. 15300 - 1224= = Rs. 14076$
Now, His expenditure and savings $=14076:1224=\frac{14076}{1224}$
$=\frac{14076\div612}{1224\div612}=\frac{23}{2}$
$=23:2$
View full question & answer→Question 443 Marks
Find the value of $x$ in the following proportions: $27 : x :: 63 : 84$
AnswerWe have $27 : x :: 63 : 84$
Product of means $= x \times 63$
Product of extremes $= 27 \times 84$
$\therefore\text{x}\times63=27\times84$
$\Rightarrow\text{x}=\frac{27\times84}{63}=36$
View full question & answer→Question 453 Marks
In a proportion the $1st, 2nd$ and $4th$ terms are $51, 68$ and $108$ respectively. Find the $3rd$ term.
AnswerLet the third term be $x.$
Then $51 : 68 :: x : 108$
Now, product of means $= x \times 68$ And,
product of extremes $= 51 \times 108 x \times 68 = 51 \times 108$
$\Rightarrow\text{x}=\frac{51\times108}{68}$
$\Rightarrow3\times27=81$
$\text{x}=81$ Hence the third term of the given proportion is $81$
View full question & answer→Question 463 Marks
Write the following ratios in the simplest form: $1L 35\ mL : 270\ mL$
AnswerThe given ratio
$= 1L 35mL : 270mL = (1 \times 1000 + 35)mL : 270mL = 1035mL : 270mL = 1035 : 270$
$=\frac{1035}{270}$ To express it in simplest form, we divide its numerator and denominator by the
$HCF$ of $1035$ and $270.$
Now, $HCF$ and $1035$ and $270 = 45$
$\therefore\frac{1035}{270}=\frac{1035\div45}{270\div45}=\frac{23}{6}$
$\therefore\text{Required ratio}=23:6$ View full question & answer→Question 473 Marks
The ratio of zinc and copper in an alloy is $7 : 9$. If the weight of copper in the alloy is $11.7\ kg$., find the weight of zinc in it.
AnswerIt is given that the ratio of zinc and copper in an alloy is $7 : 9.$
If the weight of zinc in the alloy is $7\ kg$ then the weight of copper in the alloy is $9\ kg.$
Now, if the wieght of copper is $9\ kg$
Then weight of zinc $= 7\ kg$
If the wieght of copper is $1\ kg$ then
Weight of zinc $=\frac{7}{9}\text{kg}$
If the wieght of copper is $11.7\ kg$ then
Weight of zinc
$=\Big(\frac{7}{9}\times11.7\Big)\text{kg}$
$=(7\times1.3)\text{kg}$
$=9.1\text{kg}$
View full question & answer→Question 483 Marks
Determine if the following ratios form a proportion: $39$ litres : $65$ litres and $6$ bottles : $10$ bottles
Answer$39$ litres : $65$ litres and $6$ bottles $: 10$ bottles$\frac{39}{65}=\frac{3}{5}$
$\text{and }\frac{6}{10}=\frac{3}{5}$
$\because\frac{3}{5}=\frac{3}{5}$ (Ratios are equal)
$\therefore 39$ litres : $65$ litres and $6$ bottles : $10$
Bottles are in proportion
View full question & answer→Question 493 Marks
The ratio of the length of a field to its width is $5 : 3.$ Find its length if the width is $42$ metres.
AnswerIt is given that the ratio of the length of a field to its width is $5 : 3.$
If the width of the field is $3$ metres then length $= 5$ metres.
If the width of the field is $1$ metres than length $=\frac{5}{3}\text{metres}$
If the width of the field is $42$ metres then length $=\frac{5}{3}\times14\text{ metres}$
$=5\times14\text{ metres}$
$=70 \text{ metres}$
View full question & answer→Question 503 Marks
Write the following ratios in the simplest form:
$3$ weeks $: 30$ days
AnswerThe given ratio $= 3$ weeks $: 30$ days
$= (3 × 7)$ days $: 30$ days
$= 21$ days $: 30$ days
$= 21 : 30$
$= 7 : 10$
View full question & answer→