Question 13 Marks
A truck requires $108$ litres of diesel for covering a distance of $594\ km.$ How much diesel will be required by the truck to cover a distance of $1650 \ km$?
Answer$\because$ Diesel required for covering a distance of $594\ km = 108$ litres.
$\therefore$ Diesel required for covering a distance of $1\ km =$ $\frac{108}{594}$ litre.
$\therefore$ Diesel required for covering a distance of $1650 \ km$
=$\frac{108}{594} \times$ $1650$ litres $= 300$ litres.
Hence, $300$ litres of diesel will be required by the truck to cover a distance of $1650 \ km.$
View full question & answer→Question 23 Marks
Cost of $5 \ kg$ of wheat is $₹ 91.50$. What quantity of wheat can be purchased in $₹ 183$?
AnswerAs per the question,
Wheat purchased in $₹ 91.50 = 5 \ kg$
Therefore,
Wheat purchased in $₹ 1 =$ $\frac{5}{91.50}$ $kg$
Hence,
Wheat purchased in $₹ 183 =$ $\frac{5}{91.50} \times$$183$
= $\frac{1}{18.3} \times$ $183$
$= 10 \ kg$
View full question & answer→Question 33 Marks
Cost of $5 \ kg$ of wheat is $₹ 91.50$. What will be the cost of $8 \ kg$ of wheat?
AnswerGiven that:
Cost of $5 \ kg$ wheat $= ₹ 91.50$
Therefore,
Cost of $1 \ kg$ wheat = $\frac{91.50}{5}$ $= 18.3$
$= ₹ 18.3$
Hence,
Cost of $8 \ kg$ wheat $= 18.3$ $\times$ $8$
$= ₹ 146.40$
View full question & answer→Question 43 Marks
Anish made $42$ runs in $6$ overs and Anup made $63$ runs in $7$ overs. Who made more runs per over?
AnswerFor Anish
$\because$ Runs made in 6 overs $= 42$
$\therefore $ Runs made per over = $\frac {42} {6}$ $= 7$
For Anup
$\because$ Runs made in $7$ overs $= 63$
$\therefore $ Runs made per over = $\frac {63} {7}$ = 9
So, Anup made more runs per over.
View full question & answer→Question 53 Marks
Raju purchases $10$ pens for $₹150$ and Manish buys $7$ pens $₹ 84$. Can you say who got the pens cheaper?
AnswerFor Raju
$\because$ Cost of $10$ pens $= ₹ 150$
$\therefore $ Cost of $1$ pen = ₹$\frac{150}{10}$ $= ₹15$
For Manish
$\because$ Cost of $7$ pens $= ₹ 84$
$\therefore $ Cost of $1$ pen = ₹$\frac {84} {7}$ $= ₹12$
So, Manish got the pens cheaper.
View full question & answer→Question 63 Marks
Determine if the ratio form a proportion. Also, write the middle term and extreme term where the ratio form a proportion:
$200\ ml:$ $2.5$ litre and $₹4 : ₹50.$
Answer$2.5\ litre = 2.5$ $\times$ $1000\ ml = 2500\ ml$
$\therefore$ $200\ ml : 2.5\ litre = 200\ ml : 2500\ ml$
$=\frac{200}{2500}=\frac{200 \div 100}{2500 \div 100}$ $[$$\therefore$$H.C.F.(200, 2500) = 100]$
$=\frac{2}{25}$ $= 2 : 25$
$₹4 : ₹ 50$ $=\frac{4}{50}=\frac{4 \div 2}{50 \div 2}$ $[$$\therefore$$H.C.F.(4, 50) = 2]$
$=\frac{2}{25}$ $= 2 : 25$
Since the two ratios are equal, therefore, the given ratios are in proportion.
Middle terms are $2.5$ $litres$ and $₹4$.
Extreme terms are $200\ ml$ and $₹50.$
View full question & answer→Question 73 Marks
Determine if the ratios form a proportion. Also, write the middle term and extreme term where the ratio form a proportion: $2\ kg : 80\ kg$ and $25g : 625g.$
Answer$2\ kg : 80\ kg$ $=\frac{2}{80}=\frac{2 \div 2}{80 \div 2}$[$\therefore$$H.C.F.(2, 80) = 2]$
$=\frac{1}{40}$ $= 1 : 40$
$25g : 625g$ $=\frac{25}{625}=\frac{25 \div 25}{625 \div 25}$[$\therefore$$H.C.F.(25, 625) = 25]$
$=\frac{1}{25}$ $= 1 : 25$
Since the two ratios are not equal, therefore, the given ratios are not in proportion.
View full question & answer→Question 83 Marks
Determine if the ratio form a proportion. Also, write the middle form and extreme term where the ratio form a proportion: $39$ litres : $65$ litres and $6$ bottles : $10$ bottles.
Answer$39$ litre $: 65$ litre $=\frac{39}{65}=\frac{39 \div 13}{65 \div 13}$[$\therefore$$H.C.F.(39, 65) = 13]$
$=\frac{3}{5}$$= 3 : 5$
$6$ bottle $: 10$ bottle $=\frac{6}{10}=\frac{6 \div 2}{10 \div 2}$ [$\therefore$$H.C.F.(6, 10) = 2]$
$=\frac{3}{5}$ $= 3 : 5$
Since the two ratios are equal, therefore, the given ratios are in proportion.
Middle terms are $65$ litres and $6$ bottles. Extreme terms are $39$ litres and $10$ bottles.
View full question & answer→Question 93 Marks
Determine if the following ratios form a proportion. Also, write the middle forms and extreme terms where the ratios form a proportion. $25\ cm : 1m$ and $₹40 : ₹160.$
Answer$\because$ $1m = 100\ cm$
$\therefore$ $25\ cm : 1m = 25\ cm : 100\ cm$
$=\frac{25}{100}=\frac{25 \div 25}{100 \div 25}$ $[$$\therefore$$H.C.F.(25, 100) = 25]$
$=\frac{1}{4}$ $= 1 : 4$
$₹40 : ₹160$$=\frac{40}{160}=\frac{40 \div 40}{160 \div 40}$ $[$$\therefore$$H.C.F.(40, 160) = 40]$
$=\frac{1}{4}$ $= 1 : 4$
Since the two ratios are equal, therefore, the given ratios are in proportion. Middle terms are $1m$ and $₹40.$
Extreme terms are $25 \ cm$ and $₹160.$
View full question & answer→Question 103 Marks
$45 \ km : 60 \ km$ $= 12$ hours $: 15$ hours
View full question & answer→Question 113 Marks
$32 m : 64 m = 6\ sec : 12\ sec$
View full question & answer→Question 123 Marks
$99\ kg : 45\ kg = ₹44 : ₹20$
View full question & answer→Question 133 Marks
$7.5$ litre : $15$ litre $= 5\ kg : 10\ kg.$
View full question & answer→Question 143 Marks
$40$ persons $: 200$ persons $= ₹15 : ₹75$
View full question & answer→Question 153 Marks
Determine if the $33, 44, 75, 100$ are in proportion.
AnswerWe know that, a proportion happens when two ratios are forced to be equal to each other.
Thus,
$\frac{33}{44}=\frac{3}{4}$ And $\frac{75}{100}=\frac{3}{4}$
Therefore,
$33 : 44 = 75 : 100$
Hence, the given numbers are in proportion.
View full question & answer→Question 163 Marks
Determine if the $4, 6, 8, 12$ are in proportion.
Answer$4 : 6$ $=\frac{4}{6}=\frac{4 \div 2}{6 \div 2}$[$\therefore$$H.C.F.(4, 6) = 2]$
$=\frac{2}{3}$ $= 2 : 3$
$8 : 12$ $=\frac{8}{12}=\frac{8 \div 4}{12 \div 4}$[$\therefore$$H.C.F. (8, 12) = 4]$
$=\frac{2}{3}$ $= 2 : 3$
$\because$ $4 : 6 = 8 : 12$
$\therefore$ $4, 6, 8, 12$ are in proportion.
View full question & answer→Question 173 Marks
Determine if the $32, 48, 70, 210$ are in proportion.
Answer$32 : 48$ $=\frac{32}{48}=\frac{32 \div 16}{48 \div 16}$[$\therefore$$H.C.F.(32, 48) = 16]$
$=\frac{2}{3}$ $= 2 : 3$
$70 : 210$ $=\frac{70}{210}=\frac{70 \div 70}{210 \div 70}$[$\therefore$$H.C.F. (70, 210) = 70]$
$\frac {1 } {3}$ $= 1 : 3$
$\because$ $2 : 3$ $\neq$ $1 : 3$
$\therefore$ $32, 48, 70, 210$ are not in proportion.
View full question & answer→Question 183 Marks
Determine if the $24, 28, 36, 48$ are in proportion.
Answer$24 : 28$ $=\frac{24}{28}=\frac{24 \div 4}{28 \div 4}$[$\therefore$$H.C.F. (24, 28) = 4]$
$=\frac{6}{7}$ $= 6 : 7$
$36 : 48$ $=\frac{36}{48}=\frac{36 \div 12}{48 \div 12}$[$\therefore$$H.C.F. (36, 48) = 12]$
$=\frac{3}{4}$ $= 3 : 4$
$\because$ $6 : 7$ $\neq$ $3 : 4$
$\therefore$ $24, 28, 36, 48$ are not in proportion.
View full question & answer→Question 193 Marks
Determine if the $33, 121, 9, 96$ are in proportion.
Answer$33 : 121$ $=\frac{33}{121}=\frac{33 \div 11}{121 \div 11}$[$\therefore$$H.C.F.(33, 121) = 11]$
$=\frac{3}{11}$$ = 3 : 11$
$9 : 96$ $=\frac{9}{96}=\frac{9 \div 3}{96 \div 3}$[$\therefore$$H.C.F.(9, 96) = 3]$
$=\frac{3}{32}$ $= 3 : 32$
$\because $ $3 : 11$ $\neq$ $3 : 32$
$\therefore$ $33, 121, 9, 96$ are not in proportion.
View full question & answer→Question 203 Marks
Determine if the $15, 45, 40, 120$ are in proportion.
Answer$15 : 45$ $=\frac{15}{45}=\frac{15 \div 15}{45 \div 15}$[$\therefore$$H.C.F.(15, 45) = 15]$
$=\frac{1}{3}$ $= 1 : 3$
$40 : 120$ $=\frac{40}{120}=\frac{40 \div 40}{120 \div 40}$[$\therefore$$H.C.F.(40, 120) = 40]$
$=\frac{1}{3}$ $= 1 : 3$
$\because$ $15 : 45 = 40 : 120$
$\therefore$ $15, 45, 40, 120$ are in proportion.
View full question & answer→Question 213 Marks
In a year, Seema earns $₹ 1,50,000$ and saves $₹ 50,000$. Find the ratio of money that Seema earns to the money she saves.
AnswerIn the question it is given that
Money earned by Seema $= Rs. 150000$
Also,
Money saved by Seema $= Rs. 50000$
Therefore,
Money spent by Seema $= Rs. 150000 – Rs. 50000$
$= Rs. 100000$
Now,
Ratio of money earned to money saved = $\frac{150000}{50000}$
= $\frac{3}{1}$
View full question & answer→Question 223 Marks
Present age of father is $42$ years and his son is $14$ years. Find the ratio of age of father to the age of son when father was $30$ years old.
AnswerFather was $30$ years old $42 - 30 = 12$ years before
Age of son $12$ years before $= 14 - 12 = 2$ years
$\therefore$ Ratio of age of father to the age of son when father was $30$ years old.
$=\frac{30}{2}=\frac{30 \div 2}{2 \div 2}$[$\therefore$$H.C.F.(30, 2) = 2$]
$=\frac{15}{1}$ $= 15 : 1$
View full question & answer→Question 233 Marks
Present age of father is $42$ years and his son is $14$ years. Find the ratio of age of father after $10$ years to the age of son after $10$ years.
AnswerAge of father after $10$ years $= 42 + 10 = 52$ years.
Age of son after $10$ years $= 14 + 10 = 24$ years.
$\therefore$ Ratio of age of father after 10 years to the age of son after $10$ years.
$=\frac{52}{24}=\frac{52 \div 4}{24 \div 4}$[$\therefore$$H.C.F. (52, 24) = 4$]
$=\frac{13}{6}$ $= 13 : 6$
View full question & answer→Question 243 Marks
Present age of father is $42$ years and his son is $14$ years. Find the ratio of age of the father to the age of son, when son was $12$ years old.
AnswerSon was $12$ years old $14 - 12 = 2$ years before
Age of the father $2$ years before $= 42 - 2 = 40$ years.
$\therefore$ Ratio of age of the father to the age of son, when son was $12$ years old.
$=\frac{40}{12}=\frac{40 \div 4}{12 \div 4}$ [$\therefore$$H.C.F.(40, 12) = 4]$
$=\frac{10}{3}$ $= 10 : 3$
View full question & answer→Question 253 Marks
Consider the statement: Ratio of breadth and length of a hall $2 : 5.$ Complete the following table that shows some possible breadth and lengths of the hall.
| Breadth of the hall in metres |
$10$ |
- |
$40$ |
| Length of the hall in metres |
$25$ |
$50$ |
- |
Answer$\frac{10}{25}=\frac{\square}{50} \Rightarrow 25 \times \square=10 \times 50$$\Rightarrow \square=\frac{10 \times 50}{25} \Rightarrow \square$ $= 20$
$\frac{10}{25}=\frac{40}{\square} \Rightarrow 10 \times \square=25 \times 40$$\Rightarrow \square=\frac{25 \times 40}{10} \Rightarrow \square$ $= 100$
Hence, the completed is as follows:
| Breadth of the hall in metres |
$10$ |
$20$ |
$40$ |
| Length of the hall in metres |
$25$ |
$50$ |
$100$ |
View full question & answer→Question 263 Marks
Cost of a dozen pens is $₹180$ and cost of 8 ball pens is $₹56$. Find the ratio of cost of a pen to the cost of a ball pen.
Answer$1$ dozen $= 12$ items
$\because$ Cost of $12$ pens $= ₹180$
$\therefore$ Cost of $1$pen = ₹$\frac{180}{12}$ $= ₹15$
$\because$ Cost of $8$ ball pens $= ₹56$
$\therefore$ Cost of $1$ ball pen = ₹$\frac{56}{8}$ $= ₹7$
$\therefore$ Ratio of cost of a pen to the cost of a ball pen = $\frac{15}{7}$ $= 15 : 7$
View full question & answer→Question 273 Marks
Ratio of distance of the school from Mary’s home to the distance of the school from John’s home is $2 : 1.$
$a.\ $Who lives nearer to the school?
$b.\ $Complete the following table which shows some possible distances that Mary and John could live from the school.
| Distance from Mary’s home to school $($in $km.)$ |
$10$ |
|
$4$ |
|
|
| Distance from John’s home to school $($in $km.)$ |
$5$ |
$4$ |
|
$3$ |
$1$ |
$c.\ $If the ratio of distance of Mary’s home to the distance of Kalam’s home from school is 1 : 2, then who lives nearer to the school?
Answer$a.\ $John lives nearer$($closer$)$ to the school $($As the ratio is $2 : 1).$
$b.\ $
| Distance from Mary’s home to school $($in $km.)$ |
$10$ |
$8$ |
$4$ |
$6$ |
$2$ |
| Distance from John’s home to school $($in $km.)$ |
$5$ |
$4$ |
$2$ |
$3$ |
$1$ |
$c.\ $Since the ratio is $1 : 2$, So, Mary lives nearer$($closer$)$ to the school. View full question & answer→Question 283 Marks
Fill in the missing numbers: $\frac{14}{21}=\frac{\square}{3}=\frac{6}{\square}$
AnswerIn order to get the first missing number, we consider the fact that $21 = 3$ $\times$ $7$. i.e.
when we divide $21$ by $7$ we get $3$.
This indicates that to get the missing number of second ratio, $14$ must also be divided by $7$.
When we divide, we have, $14$ $\div$ $7 = 2$
Hence, the second ratio is $\frac{2}{3}$
Similarly, to get third ratio we multiply both terms of second ratio by $3.$
Hence, the third ratio is $\frac{6}{9}$
Therefore, $\frac{{14}}{{21}} = \frac{{\boxed2}}{3} = \frac{6}{{\boxed9}}$
View full question & answer→Question 293 Marks
A car travels $90 \ km$ in 2$\frac{1}{2}$ hours.
$i.\ $How much time is required to cover $30 km$ with the same speed?
$ii.\ $Find the distance covered in $2$ hours with the same speed.
Answer$i.\ $In this case, time is unknown and distance is known.
Therefore, we shall proceed as follows:
$2\frac{1}{2}$ hours = $\frac{5}{2}$ hours = $\frac{5}{2} \times 60$ minutes $= 150$ minutes.
Here we are given that, $90 \ km$ is covered in $150$ minutes
Therefore, $1 \ km$ can be covered in $\frac{150}{90}$ minutes
And hence, $30 \ km$ can be covered in $\frac{150}{90} \times 30$ minutes i.e. $50$ minutes
Thus, $30 \ km$ can be covered in $50$ minutes.
$ii.\ $In this case, distance is unknown and time is known.
Therefore, we proceed as follows:
Here we are given that, Distance covered in 2$\frac{1}{2}$ hours $($i.e. $\frac{5}{2}$ hours$) = 90 \ km$
Therefore, distance covered in $1$ hour $= 90 \div \frac{5}{2} km = 90 = 90 \times \frac{2}{5} = 36 \ km$
And hence, distance covered in $2$ hours $= 36 \times 2 = 72 \ km.$
Thus, in $2$ hours, the distance covered is $72 \ km.$
View full question & answer→Question 303 Marks
Do the ratios $15 \ cm$ to $2 m$ and $10$ sec to $3$ minutes form a proportion?
AnswerWe have, $2 m = 200 \ cm,$ because $1 m = 100 cm.$
Therefore, ratio of $15 \ cm$ to $2 m = 15 : 2$ $\times$ $100$
$= 3 : 40$
Now, as $1$ min $= 60$ sec. So, $3$ min $= 180$ sec.
Therefore, Ratio of $10$ sec to $3$ min $= 10 : 3$ $\times$ $60$ ($1$ min = $60$ sec)
$= 1 : 18$
Since, $3 : 40 $$\neq$ $1 : 18.$
Therefore, the given ratios do not form a proportion.
View full question & answer→