Question 12 Marks
The weight of $72$ books is $9\ kg.$ What is the weight of $40$ such books$?$
Answer$\because$ Weight of $72$ books $= 9\ kg.$
$\therefore$ Weight of $1$ book $=\frac{9}{72} \mathrm{kg}=\frac{1}{8} \mathrm{kg}$
$\therefore$ Weight of $40$ book $= \frac {1} {8} \times 40\ kg = 5\ kg.$
Hence, the weight of $40$ such books is $5\ kg.$
View full question & answer→Question 22 Marks
The cost of $4$ dozen bananas is $₹ 180.$ How many bananas can be purchased for $₹ 90?$
AnswerHere, $4$ dozen bananas $= (12 \times 4) = 48$ bananas
In $₹ 180$ we get $= 48$ bananas
$\therefore$ In $₹ 1 $ we get $= \frac {48}{180}$ bananas
$\therefore$ In $₹ 90$ we get $= \frac {48 \times 90}{180}$ bananas $= 24$ bananas
View full question & answer→Question 32 Marks
Shaina pays $₹ 15,000$ as rent for $3$ months. How much does she have to pay for a whole year, if the rent per month remains the same?
AnswerIn $3$ months Shaina pays as rent $= ₹ 15000$
$\therefore$ In $1$ month she pays as rent $= ₹ \frac {15000}{3}$
$\therefore$ In $12$ months she pays as rent $= ₹ \frac {15000 \times 12}{3} = ₹ 60,000$
View full question & answer→Question 42 Marks
The temperature dropped $15$ degree in the last $30$ days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days$?$
AnswerDrop in temperature in $30$ days $= 15$ degrees
$\therefore$ Drop in temperature in $1$ day $=\frac{15}{30}$ degree.
$\therefore$ Drop in temperature in $10$ days $=\frac{15}{30}$ $\times 10$ degrees $= 5$ degrees.
Hence, the temperature will drop $5$ degrees in the next ten days.
View full question & answer→Question 52 Marks
If it has rained $276\ mm$ in the last $3$ days, how many cm of rain will fall in one full week $(7$ days$)?$ Assume that the rain continues to fall at the same rate.
Answer$\because$ Rainfall in $3$ days $= 276\ mm$
$\therefore$ Rainfall in $1$ day $=\frac{276}{3}\ mm = 92\ mm$
$\therefore$ Rainfall in $7$ days $= 92\ mm \times 7 = 644\ mm.$
Hence, $644\ mm$ of rain will fall in one full week$(7$ days$).$
View full question & answer→Question 62 Marks
Ekta earns $₹ 3000$ in $10$ days. How much will she earn in $30$ days$?$
AnswerGiven: Amount of money earned in $10$ days $= ₹ 3000$
So, Amount of money earned in $1$ day $= \frac {3000}{10} = ₹ 300$
Amount of money earned in $30$ days $= 30 \times 300 = ₹ 9000$
Hence Ekta will earn $9000 ₹$ in $30$ days
View full question & answer→Question 72 Marks
If the cost of $7 m$ of cloth is $₹ 1470,$ find the cost of $5 m$ of cloth.
AnswerGiven,
Cost of $7 m$ cloth $= ₹ 1470$
So
Cost of $1 m$ cloth $= \frac {1470}{7} = ₹ 210$
Cost of $5 m$ cloth $= 210 \times 5 = ₹ 1050$
Hence the cost of $5 m$ cloth is $₹ 1050$
View full question & answer→Question 82 Marks
There are $102$ teachers in a school of $3300$ students. Find ratio of the number of teachers to the number of students.
AnswerRatio of the number of teachers to the number of students
$=\frac{102}{3300}=\frac{102 \div 6}{3300 \div 6}[\therefore H.C.F(102, 3300) = 6]$
$=\frac{17}{550} = 17 : 550$
View full question & answer→Question 92 Marks
In a year, Seema earns $₹ 1,50,000$ and saves $₹ 50,000.$ Find the ratio of money that she saves to the money she spends.
AnswerGiven that:
Money saved by Seema $= Rs. 50000$
Money spent by Seema $= Rs. 100000$
Hence,
The ratio of money saved to money spent $= \frac{50000}{100000}$
$=\frac{1}{2}$
View full question & answer→Question 102 Marks
Find the ratio of $500\ ml$ to $2$ litre.
Answer$2$ litre $= 2 \times 1000\ ml = 2000\ ml$
$\therefore$ Ratio of $500\ ml$ to $2$ litre
$=\frac{500}{2000}=\frac{500 \div 500}{2000 \div 500}[\therefore H.C.F. (500, 2000) = 500]$
$=\frac{1}{4} = 1 : 4$
View full question & answer→Question 112 Marks
Find the ratio of $55$ paise to $Rs.1$
Answer$Rs.1 = 100$ paise.
$\therefore$ Ratio of $55$ paise to $Rs.1$
$\frac{55}{100}=\frac{55 \div 5}{100 \div 5} [ \therefore H.C.F. (55, 100) = 5]$
$\frac{11}{20} = 11 : 20$
View full question & answer→Question 122 Marks
Find the ratio of $40 \ cm$ to $1.5 m.$
Answer$1.5 m = 1.5 \times 100 \ cm = 150 \ cm.$
$\therefore$ Ratio of $40 \ cm$ to $1.5 m$
$=\frac{40}{150}=\frac{40 \div 10}{150 \div 10}[\therefore H.C.F. (40, 150) = 10]$
$= =\frac{4}{15} = 4 : 15$
View full question & answer→Question 132 Marks
Find the ratio of $30$ minutes to $1.5$ hours.
Answer$1.5$ hours $= 1.5 × 60$ minutes $= 90$ minutes.
$\therefore$ Ratio of $30$ minutes to $1.5$ hours
$=\frac{30}{90}=\frac{30 \div 30}{90 \div 30}[H.C.F. (30, 90) = 30]$
$=\frac{1}{3} = 1 : 3$
View full question & answer→Question 142 Marks
Find the ratio of $33\ km$ to $121\ km.$
AnswerRatio of $33\ km$ to $121\ km$
$=\frac{33}{121}=\frac{33 \div 11}{121 \div 11}[H.C.F. (33, 121) = 11]$
$=\frac{3}{11} = 3 : 11$
View full question & answer→Question 152 Marks
Find the ratio of $98$ to $63.$
AnswerRatio of $98$ to $63$
$=\frac{98}{63}=\frac{98 \div 7}{63 \div 7}[H.C.F. (98, 63) = 7]$
$=\frac{14}{9} = 14 : 9$
View full question & answer→Question 162 Marks
Find the ratio of $30$ minutes to $45$ minutes.
AnswerRatio of $30$ minutes to $45$ minutes
$=\frac{30}{45}=\frac{30 \div 15}{45 \div 15}[H.C.F. (30, 45) = 15]$
$=\frac{2}{3} = 2 : 3$
View full question & answer→Question 172 Marks
Find the ratio of $81$ to $108.$
AnswerRatio of $81$ to $108$
$=\frac{81}{108}=\frac{81 \div 27}{108 \div 27} [ \therefore H.C.F.(81, 108) = 27]$
$=\frac{3}{4} = 3 : 4$
View full question & answer→Question 182 Marks
Distance travelled by Hamid and Akhtar in an hour are $9\ km$ and $12\ km$ respectively. Find the ratio of speed of Hamid to the speed of Akhtar.
AnswerRatio of speed of Hamid to the speed of Akhtar
$=\frac{9}{12}=\frac{9 \div 3}{12 \div 3}[\therefore H.C.F. (9, 12) = 3]$
$=\frac{3}{4} = 3 : 4$
View full question & answer→Question 192 Marks
Present age of father is $42$ years and his son is $14$ years. Find the ratio of present age of father to the present age of son.
AnswerRatio of present age of father to the present age of son
$=\frac{42}{14}=\frac{42 \div 14}{14 \div 14} [ \therefore H.C.F. (42, 14) = 14]$
$ =\frac{3}{1} = 3 : 1$
View full question & answer→Question 202 Marks
Divide $20$ pens between Sheela and Sangeeta in the ratio $3 : 2.$
AnswerTotal number of pens $= 20$
Ratio $= 3 : 2$
Sum of the ratio $= 3 + 2 = 5$
$\therefore$ Sheela's share $=\frac{3}{5} \times 20 = 12$
and Sangeeta's share $= \frac{2}{5} \times 20 = 8.$
View full question & answer→Question 212 Marks
Out of $1800$ students in a school, $750$ opted basketball, $800$ opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of number of students who opted basketball to the number of students who opted table tennis.
AnswerNumber of students opting table tennis $= 1800 - (750 + 800) = 1800 - 1550 = 250$
$\therefore$ Ratio of number of students who opting basketball to number of students opting table tennis
$=\frac{750}{250}=\frac{750 \div 250}{250 \div 250} [ \therefore H.C.F(750, 250) = 250]$
$ =\frac{3}{1} = 3 : 1$
View full question & answer→Question 222 Marks
Out of $1800$ students in a school, $750$ opted basketball, $800$ opted cricket and remaining opted table tennis. If a student can opted only one game, find the ratio of the number of students who opted basketball to the total number of students.
AnswerRatio of number of students who opted basketball to the total number of students.
$=\frac{750}{1800}=\frac{750 \div 150}{1800 \div 150} [ \therefore H.C.F of (750, 1800) = 150]$
$=\frac{5}{12} = 5 : 12$
View full question & answer→Question 232 Marks
Out of $1800$ students in a school, $750$ opted basketball, $800$ opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of number of students who opted cricket to the number of students opting basketball.
AnswerRatio of number of students who opted cricket to the number of students opting basketball.
$=\frac{800}{750}=\frac{800 \div 50}{750 \div 50} [H.C.F of (800, 750) = 50]$
$=\frac{16}{15} = 16 : 15$
View full question & answer→Question 242 Marks
In a college out of $4320$ students, $2300$ are girls. Find the ratio of number of boys to the number of girls.
AnswerNumber of boys $= 4320 - 2300 = 2020$
$\therefore$ Ratio of number of boys to the number of girls
$=\frac{2020}{2300}=\frac{2020 \div 20}{2300 \div 20}[ \therefore H.C.F.(2020, 2300) = 20]$
$=\frac{101}{115}= 101 : 115$
View full question & answer→Question 252 Marks
In a college out of $4320$ students, $2300$ are girls. Find the ratio of number of girls to the total number of students.
AnswerRatio of number of girls to the total number of students
$=\frac{2300}{4320}=\frac{2300 \div 20}{4320 \div 20} [ \therefore H.C.F.(2300, 4320) = 20]$
$ =\frac{115}{216} = 115 : 216$
View full question & answer→Question 262 Marks
Are $30, 40$ and $45, 60$ in proportion$?$
AnswerSince, Ratio of $30$ to $40 = \frac{30}{40} = 3 : 4.$
and Ratio of $45$ to $60 = \frac{45}{60} = 3 : 4.$
Such that $30 : 40 = 45 : 60$
Therefore, $30, 40, 45, 60$ are in proportion.
View full question & answer→Question 272 Marks
Are the ratios $25g : 30g$ and $40 \ kg : 48 \ kg$ in proportion$?$
AnswerWe know that
$25 g : 30 g = \frac{25}{30} = 5 : 6$
and $40 \ kg : 48 \ kg = \frac{40}{48} = 5 : 6$
So, $25 : 30 = 40 : 48.$
Therefore, the ratios $25 g : 30 g$ and $40 \ kg : 48 \ kg$ are in proportion,
i.e. $25 : 30 :: 40 : 48$
The middle terms in this are $30, 40$ and the extreme terms are $25, 48.$
View full question & answer→Question 282 Marks
Divide $₹ 60$ in the ratio $1 : 2$ between Kriti and Kiran.
AnswerThe two parts are $1$ and $2.$
Sum of the parts $= 1 + 2 = 3.$
This means if there are $₹ 3,$ Kriti will get $₹ 1$ and Kiran will get $₹ 2.$
Or, we can say that Kriti gets $1$ part and Kiran gets $2$ parts out of every $3$ parts.
Therefore, Kriti's share $= \frac{1}{3} \times 60 = ₹ 20$
and, Kiran's share $ = \frac{2}{3} \times 60 = ₹ 40.$
View full question & answer→Question 292 Marks
Give two equivalent ratios of $6 : 4.$
AnswerThe given ratio is $6 : 4 = \frac{6}{4}=\frac{6 \times 2}{4 \times 2}=\frac{12}{8}$
Therefore, $12 : 8$ is an equivalent ratio of $6 : 4$
Similarly, the ratio $6 : 4 = \frac{6}{4}=\frac{3 \times 2}{2 \times 2}=\frac{3}{2}$
So, $3:2$ is another equivalent ratio of $6 : 4.$
View full question & answer→Question 302 Marks
There are $45$ persons working in an office. If the number of females is $25$ and the remaining are males, find the ratio of:
$a.\ $The number of females to number of males.
$b.\ $The number of males to number of females.
AnswerWe are given that:
Number of females $= 25$
Total number of workers $= 45$
Number of males $= 45 – 25 = 20$
Therefore, the ratio of number of females to the number of males $= 25 : 20 = 5 : 4$ And the ratio of number of males to the number of females $= 20 : 25 = 4 : 5.$
$($Notice that there is a difference between the two ratios $5 : 4$ and $4 : 5)$
View full question & answer→Question 312 Marks
Find the ratio of $90 \ cm$ to $1.5 m.$
AnswerThe two quantities are not in the same units.
Therefore, we need to bring them into the same units.
$1.5 m = 1.5 \times 100 \ cm = 150 \ cm.$
Therefore, the required ratio is $90 : 150$
$= \frac{90}{150}=\frac{90 \times 30}{150 \times 30}=\frac{3}{5}$
Hence, the required ratio is $3 : 5$
View full question & answer→Question 322 Marks
Cost of $105$ envelopes is $₹ 350.$ How many envelopes can be purchased for $₹ 100?$
AnswerIt is given that:
In $₹ 350,$ the number of envelopes that can be purchased $= 105$
Therefore, in $₹ 1,$ number of envelopes that can be purchased$ = \frac{105}{350}$
Therefore, in $₹ 100,$ the number of envelopes that can be purchased $= \frac{105}{350} \times 100 = 30$
Thus, $30$ envelopes can be purchased for $₹ 100.$
View full question & answer→Question 332 Marks
If the cost of a dozen soaps is $₹ 153.60,$ what will be the cost of $15$ such soaps$?$
AnswerWe know that $1$ dozen $= 12$ units
Since, cost of $12$ soaps $= ₹ 153.60$
Therefore, cost of $1$ soap $= \frac{153.60}{12} = ₹ 12.80$
Hence, cost of $15$ soaps $= ₹ 12.80 \times 15 = ₹ 192$
Thus, cost of $15$ soaps is $₹ 192.$
View full question & answer→Question 342 Marks
A motorbike travels $220 \ km$ in $5$ litres of petrol. How much distance will it cover in $1.5$ litres of petrol$?$
AnswerIn the question we are given that
In $5$ liters of petrol, a motorbike can travel $220 \ km.$
Therefore, in $1$ litre of petrol, motorbike travels $= \frac{220}{5} \ km$
Therefore, in $1.5$ litres, motorbike travels $= \frac{220}{5}\times 1.5 \ km$
$= \frac{220}{5} \times \frac{15}{10} \ km = 66 \ km.$
Thus, the motorbike can travel $66 \ km$ in $1.5$ litres of petrol.
View full question & answer→Question 352 Marks
If the cost of $6$ cans of juice is $₹ 210,$ then what will be the cost of $4$ cans of juice$?$
AnswerCost of 6 cans of juice $= ₹ 210$
Therefore, cost of one can of juice $= \frac{210}{6} = ₹ 35$
And hence, cost of $4$ cans of juice $= ₹ 35 \times 4 = ₹ 140.$
Thus, cost of $4$ cans of juice is $₹ 140.$
View full question & answer→Question 362 Marks
Length and breadth of a rectangular field are $50 m$ and $15 m$ respectively. Find the ratio of the length to the breadth of the field.
AnswerHere, we are given that:
Length of the rectangular field $= 50 m$
Breadth of the rectangular field $= 15 m$
The ratio of the length to the breadth is $50 : 15$
The ratio can be written as $\frac{50}{15}=\frac{50 \div 5}{15 \div 5}=\frac{10}{3} = 10 : 3$
Thus, the required ratio is $10 : 3$
View full question & answer→