Question 13 Marks
A truck covers a distance of $510\ km$ in $34$ litres of diesel. How much distance would it cover in $20$ litres of diesel$?$
AnswerLet the required distance be $x\ km.$ Then,
We have:
| Quantity of diesal(in litres) |
$34$ |
$20$ |
| Distance(in km) |
$510$ |
$x$ |
Clearly, the less the quantity of diesel consumed, the less is the distance covered.
So, this is a case of direct proportion.
Now, $\frac{34}{510}=\frac{20}{\text{x}}$
$\Rightarrow\frac{1}{15}=\frac{20}{\text{x}}$
$\Rightarrow\text{x}\times1=20\times15=300$
Therefore, the required distance is $300\ km.$ View full question & answer→Question 23 Marks
A car takes $5$ hours to reach a destination by travelling at the speed of $60\ km/hr.$ How long will it take when the car travels at the speed of $75\ km/hr?$
AnswerLet $x\ h$ be the required time taken. Then, We have:
| Speed(in km/h) |
$60$ |
$75$ |
| Time(in h) |
$5$ |
$x$ |
Clearly, the higher the speed, the lesser will be the time taken. So, it is a case of inverse proportion. Now, $60\times5=75\times\text{x}$ $\Rightarrow\text{x}=\frac{60\times5}{75}$ $\text{x}=4$ Therefore, the car will reach its destination in $4\ h$ if it travels at a speed of $75\ km/h.$ View full question & answer→Question 33 Marks
$30$ men can finish a piece of work in $28$ days. How many days will be taken by $21$ men to finish it$?$
AnswerLet $x$ be the number of days taken by $21$ men to finish the piece of work.
| No. of men |
$30$ |
$21$ |
| No. of days |
$28$ |
$x$ |
More men will take less time to complete the work. So, this is a case of inverse proportion.
Now, $30 \times 28 = 21 \times x$
$\Rightarrow\text{x}=\frac{30\times28}{21}$
$\Rightarrow\text{x} = 40$
$\therefore 21$ men will take 40 days to finish the piece of work. View full question & answer→Question 43 Marks
A taxi charges a fare of $Rs. 2550$ for a journey of $150\ km$. How much would it charge for a journey of $124\ km?$
AnswerLet the charge for a journey of $124\ km$ be $Rs. x$
| Price(in Rs.) |
$2550$ |
$x$ |
| Distance(in km) |
$150$ |
$124$ |
More is the distance travelled, more will be the price.
So, it is a case of direct proportion.
$\therefore\frac{2550}{150}=\frac{\text{x}}{124}$
$\Rightarrow\text{x}=\frac{2550\times124}{150}=2108$
Thus, the taxi charges $Rs. 2,108$ for the distance of $124\ km.$ View full question & answer→Question 53 Marks
If $x$ and $y$ vary inversely and $x = 15$ when $y = 6$, find y when $x = 9$.
Answer
| $x$ |
$15$ |
$9$ |
| $y$ |
$6$ |
$y_1$ |
$x$ and $y$ vary inversely.
i.e., $xy$= constant
Now, $15\times6=9\times\text{y}_1$
$\Rightarrow\text{y}_1=\frac{15\times6}{9}$
$\text{y}_1=10$
$\therefore$ value of $\text{y}=10,$
when$\text{x}=9$ View full question & answer→Question 63 Marks
If the thickness of a pile of $12$ cardboards is $65\ mm,$ find the thickness of a pile of $312$ such cardboards.
AnswerLet $x\ mm$ be the required thickness. Then, We have:
| Thickness of cardboard (in mm) |
$65$ |
$x$ |
| No. of cardboards |
$12$ |
$312$ |
Clearly, when the number of cardboard is more, the thickness will also be more.
So, it is a case of direct proportion. Now, $\frac{65}{12}=\frac{\text{x}}{312}$
$\Rightarrow\text{x}=\frac{65\times312}{12}$ $\text{x}=1690$
Therefore, the thickness of the pile of $312$ cardboards is $1690\ mm$. View full question & answer→Question 73 Marks
The railway fare for $61\ km$ is $Rs. 183$. Find the fare for $53\ km$.
AnswerLet $Rs. x$ be the railway fare for a journey of distance $53\ km$.
| Distance (in km) |
$61$ |
$53$ |
| Railway fare (in rupees) |
$183$ |
$xx$ |
The lesser the distance, the lesser will be the fare.
So, it is a case of direct proportion.
Now, $\frac{61}{183}=\frac{53}{\text{x}}$
$\Rightarrow\text{x}=\frac{53\times183}{61}$
$\Rightarrow\text{x}=159$
The railway fare for a journey of distance $53\ km$ is $Rs.\ 159$. View full question & answer→Question 83 Marks
If $35$ men can reap a field in $8$ days, in how many days can $20$ men reap the same field?
AnswerLet $x$ be the required number of days. Then, We have:
| No. of days |
$8$ |
$x$ |
| No. of men |
$35$ |
$20$ |
Clearly, less men will take more days to reap the field.
So, it is a case of inverse proportion.
Now, $8\times35=\text{x}\times20$
$\Rightarrow\frac{8\times35}{20}=\text{x}$
$14=\text{x}$
Therefore, $20$ men can reap the same field in $14$ days. View full question & answer→Question 93 Marks
A school has $9$ periods a day each of $40$ minutes duration. How long would each period be, if the school has $8$ periods a day, assuming the number of school hours to be the same?
AnswerLet $x$ min be the duration of each period when the school has $8$ periods a day.
| No. of periods |
$9$ |
$8$ |
| Time( in min) |
$40$ |
$x$ |
Clearly, if the number of periods reduces, the duration of each period will increase.
So, it is a case of inverse proportion.
Now, $9\times40=8\times\text{x}$
$\Rightarrow\text{x}=\frac{9\times40}{8}$ $\text{x}=45$
Therefore, the duration of each period will be $45$min if there were eight periods a day. View full question & answer→Question 103 Marks
A farmer has enough food to feed $28$ animals in his cattle for $9$ days. How long would the food last, if there were $8$ more animals in his cattle?
AnswerLet $x$ be the required number of days. Then, We have:
| No. of days |
$9$ |
$x$ |
| No. of animals |
$28$ |
$36$ |
Clearly, more number of animals will take less number of days to finish the food.
So, it is a case of inverse proportion.
Now, $9\times28=\text{x}\times36$
$\Rightarrow\text{x}=\frac{9\times28}{36}$
$\text{x}=7$ Therefore, the food will last for $7$ days. View full question & answer→Question 113 Marks
In $8$ days, the earth picks up $(6.4 × 10^7)$kg of dust from the atmosphere. How much dust will it pick up in $15$ days?
AnswerLet $x$ kg be the required amount of dust. Then, We have:
| No. of days |
$8$ |
$15$ |
| Dust(in kg) |
$6.4 × 10^7$ |
$x$ |
Clearly, more amount of dust will be collected in more numbers of days.
So, this is a case of direct proportion.
Now, $\frac{8}{6.4\times10^7}=\frac{15}{\text{x}}$
$\Rightarrow\text{x}=\frac{15\times6.4\times10^7}{8} $
$\text{x}=12\times10^7$
Therfore, $12,00,00,000\ kg$ of the dust will be picked up in $15$ days. View full question & answer→Question 123 Marks
In a hostel, $75$ students had food provision for $24$ days. If $15$ students leave the hostel, for how many days would the food provision last?
AnswerLet $x$ be the required number of days. Then, We have:
|
No. of students
|
$9$
|
$60$
|
|
No. of days
|
$24$
|
$x$
|
Clearly, less number of students take more days to finish the food.
So, it is a case of inverse proportion.
Now, $75\times24=60\times\text{x}$
$\Rightarrow\text{x}=\frac{75\times24}{60}$
$\text{x}=30$ Therefore, the food will now last for $30$ days. View full question & answer→Question 133 Marks
A garrison of $900$ men had provisions for $42$ days. However, a reinforcement of $500$ men arrived. For how many days will the food last now?
AnswerLet $x$ be the required number of days. Then, We have:
| No. of men |
$900$ |
$400$ |
| No. of days |
$42$ |
$x$ |
Clearly, more men will take less number of days to finish the food.
So, it is a case of inverse proportion.
Now, $900\times42=1400\times\text{x}$
$\Rightarrow\text{x}=\frac{900\times42}{1400}$
$\text{x}=27$
Therefore, the food will now last for $27$ days. View full question & answer→Question 143 Marks
The cost of $140$ tennis balls is $Rs.\ 4900$. Find the cost of $2$ dozen such balls.
AnswerLet $Rs\ x$ be the cost of $24$ tennis balls.
| No. of balls |
$140$ |
$24$ |
| Cost of balls |
$4900$ |
$xx$ |
More tennis balls will cost more.
Now, $\frac{140}{4900}=\frac{24}{\text{x}}$
$\Rightarrow\text{x}=\frac{24\times4900}{140}$
$\Rightarrow\text{x} = 840$
$\therefore$ The cost of $2$ dozen tennis balls is $Rs.\ 840$. View full question & answer→Question 153 Marks
$350$ boxes can be placed in $25$ cartons. How many boxes can be placed in $16$ cartons?
AnswerLet $x$ be the required number of boxes.
| No. of Boxes |
$350$ |
$xx$ |
| No. of Cartons |
$25$ |
$16$ |
Less number of boxes will require less number of cartons.
So, it is a case of direct proportion.
Now, $\frac{350}{25}=\frac{\text{x}}{16}$
$\Rightarrow\text{x}=\frac{350\times16}{25}$
$\Rightarrow\text{x} = 224$
$\therefore$ $224$ boxes can be placed in $16$ cartoons. View full question & answer→Question 163 Marks
$8$ taps of the same size fill a tank in $27$ minutes. If two taps go out of order, how long would the remaining taps take to fill the tank?
AnswerLet $x$ min be the required number of time. Then, We have:
| No. of taps |
$8$ |
$6$ |
| Time(in min) |
$27$ |
$x$ |
Clearly, less number of taps will take more time to fill the tank.
So, it is a case of inverse proportion.
Now, $8\times27=6\times\text{x}$
$\Rightarrow\text{x}=\frac{8\times27}{6}$
$\text{x}=36$ Therefore, it will take $36$min to fill the tank. View full question & answer→Question 173 Marks
A garrison of $200$ men had provisions for $45$ days. After $15$ days, $40$ more men join the garrison. Find the number of days for which the remaining food will last.
AnswerClearly, the remaining food is sufficient for $200$ men for $(45 - 15)$, i.e., $30$ days.
Total number of men $= 200 + 40 = 240$ Let the remaining food last for $x$ days.
| No. of men |
$200$ |
$240$ |
| No. of days |
$30$ |
$xx$ |
Clearly, more men will take less number of days to finish the food.
So, it is a case of inverse proportion.
Now, $200 \times 30 = 240 \times x$
$\Rightarrow\text{x}=\frac{200\times30}{240}$
$\Rightarrow\text{x} = 25$
$\therefore$ The remaining food will last for $25$ days. View full question & answer→Question 183 Marks
The cost of $15$ metres of a cloth is $Rs.\ 981$. What length of this cloth can be purchased for $Rs.\ 1308$?
AnswerLet the length of cloth be $x\ m$. Then, We have:
| Length of cloth(in $m.$) |
$15$ |
$x$ |
| Cost of cloth(in $Rs.$ ) |
$981$ |
$1308$ |
Clearly, more length of cloth can be bought by more amount of money. So, this is a case of direct proportion.
Now, $\frac{15}{981}=\frac{\text{x}}{1308}$
$\Rightarrow\text{x}=\frac{15\times1308}{981}$
$\text{x}=20$
Therfore, $20m$ of cloth can be bought for $Rs.\ 1,308$ View full question & answer→Question 193 Marks
$11$ men can dig $6\frac{3}{4}$ metre-long trench in one day. How many men should be employed for digging $27$-metre-long trench of the same type in one day?
AnswerLet $x$ be the required number of men. Now, $6\frac{3}{4}\text{m}=\frac{27}{4}\text{m}$ Then, we have:
| Number of men |
$11$ |
$x$ |
| Length of trench (in metres) |
$\frac{27}{4}$ |
$27$ |
Clearly, the longer the trench, the greater will be the number of men required.
So, it is a case of direct proportion.
Now, $\frac{11}{\frac{27}{4}}=\frac{\text{x}}{27}$
$\Rightarrow\frac{\text{11}\times4}{27}=\frac{\text{x}}{27}$
$\text{x}=44$
Therefore, $44$ men should be employed to dig a trench of length $27\ m$. View full question & answer→Question 203 Marks
Ravi walks at the uniform rate of $5$km/hr. What distance would he cover in $2$ hours $24$ minutes?
AnswerLet $x\ km$ be the required distance covered by Ravi in $2$h $24$min.
Then, We have: $1$h = $60$min i.e., $2$h $24$min $= (120 + 24)$min $= 144$min
| Distance covered (in km) |
$5$ |
$x$ |
| Time(in min) |
$60$ |
$144$ |
Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.
Now, $\frac{5}{60}=\frac{\text{x}}{144}$
$\Rightarrow\text{x}=\frac{5\times144}{60} $
$\text{x}=12$
Therfore, the distance covered by Ravi in $2$h $24$min is $12$km. View full question & answer→Question 213 Marks
$12$ men can dig a pond in $8$ days. How many men can dig it in $6$ days?
AnswerLet x be the required number of men. Then, We have:
| No. of days |
$8$ |
$6$ |
| No. of men |
$12$ |
$x$ |
Clearly, more men will require less number of days to dig the period.
So, it is a case of inverse proportion.
Now, $8\times12=\text{x}\times6$
$\Rightarrow\frac{8\times12}{6}=\text{x}$ $16=\text{x}$
Therefore, $16$ men can dig the pond in $14$ days. View full question & answer→Question 223 Marks
If $x$ and $y$ vary inversely and $x = 18$ when $y = 8$, find $x$ when $y = 16.$
Answer
| $x$ |
$18$ |
$x_1$ |
| $y$ |
$8$ |
$16$ |
$x$ and $y$ vary inversely.
i.e., $xy =$ constant$$
Now, $18\times8=\text{x}_1\times16$
$\Rightarrow\frac{18\times8}{16}=\text{x}_1$
$9=\text{x}_1$
$\therefore$ value of $\text{x}=9$ View full question & answer→Question 233 Marks
If $9\ kg$ of sugar costs $Rs.\ 238.50$, how much sugar can be bought for $Rs.\ 371$?
AnswerLet the quantity of sugar bought for $Rs. 371$ be $x\ kg$.
| Quantity(in kg) |
$18$ |
$x$ |
| Price(in Rs.) |
$630$ |
$455$ |
The price increases as the quantity increases.
Thus, this is a case of direct proportion.
$\therefore\frac{9}{238.50}=\frac{\text{x}}{371}$
$\Rightarrow\text{x}=\frac{9\times371}{238.50}=14$
Thus, the quantity of sugar bought for $Rs.\ 371$ is $14\ kg$. View full question & answer→Question 243 Marks
$7$ teps of the same size fill a tank in $1$ hour $36$ minutes. How long will $8$ taps of the same size take to fill the tank?
AnswerLet $x$ be the required number of taps. Then,
We have:
$1$h = $60$min
i.e., $1$h $36$min $= (60 + 36)$min $= 96$min
| No. of taps |
$7$ |
$8$ |
| Time( in min) |
$96$ |
$x$ |
Clearly, more numbers of taps will require less time to fill the tank.
So, it is a case of inverse proportion.
Now, $7\times96=8\times\text{x}$
$\Rightarrow\text{x}=\frac{7\times96}{8}$
$\text{x}=84$
Therefore, 8 taps of the same size will take $84$min or $1$h $24$min to fill the tank. View full question & answer→Question 253 Marks
If $18$ dolls cost $Rs.\ 630$, how many dolls can be bought for $Rs.\ 455$?
AnswerLet the required number of dolls be $x$. Then,
We have:
| No of dolls |
$18$ |
$x$ |
| Cost of dolls(in rupees) |
$630$ |
$455$ |
Clearly, the less the amount of money, the less will be the number of dolls bought.
So, this is a case of direct proportion.
Now, $\frac{18}{630}=\frac{\text{x}}{455}$
$\Rightarrow\frac{1}{35}=\frac{\text{x}}{455}$
$\Rightarrow\text{x}=\frac{455}{35}$
$\text{x}=13$
Therefore, $13$ dolls can be bought for $Rs.\ 455.$ View full question & answer→Question 263 Marks
$10$ people can dig a trench in $6$ days. How many people can dig it in $4$ days?
AnswerLet $x$ people dig the trench in $4$ days.
|
No. of people
|
$10$
|
$xx$
|
|
No. of days
|
$6$
|
$4$
|
More people will take less number of days to dig the trench.
Hence, this is a case of inverse proportion.
Now, $10 \times 6 = 4$
$\Rightarrow\text{x}=\frac{60}{4}$
$\Rightarrow\text{x} = 15$
$\therefore$ $15$ people can dig the trench in $4$ days. View full question & answer→Question 273 Marks
$6$ cows can graze a field in $28$ days. How long would $14$ cows take to graze the same field?
AnswerLet $x$ be the number of days. Then, We have:
| No. of days |
$28$ |
$x$ |
| No. of cows |
$6$ |
$14$ |
Clearly, more numbers of cows will take less number of days tograze the field.
So, it is a case of inverse proportion.
Now, $28\times6=\text{x}\times14$
$\Rightarrow\frac{28\times6}{14}=\text{x}$
$\text{x}=12$
Therefore, $14$ cows will take $12$ days to graze the field. View full question & answer→Question 283 Marks
A car is travelling at the average speed of $50$km/hr. How much distance would it travel in $1$ hour $12$ minutes?
AnswerLet $x\ km$ be the required distance.
Then, We have: $1$h= $60$min i.e., $1$h $12$min $= (60 + 12)$min $= 72$min
| Distance covered (in km) |
$50$ |
$x$ |
| Time(in min) |
$60$ |
$72$ |
Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.
Now, $\frac{50}{60}=\frac{\text{x}}{72}$
$\Rightarrow\text{x}=\frac{50\times72}{60} $
$\text{x}=60$
Therfore, the distance travelled by the car in $1$h $12$min is $60$km. View full question & answer→Question 293 Marks
If $x$ and $y$ are directly proportional, find the values of $x_1, x_2$ and $y_1$ in the table given below.
| $x$ |
$3$ |
$x_1$ |
$x_2$ |
$10$ |
| $y$ |
$72$ |
$120$ |
$192$ |
$y_1$ |
AnswerSince $x$ and $y$ are directly proportional,We have:
$\frac{3}{72}=\frac{\text{x}_1}{120}=\frac{\text{x}_2}{192}=\frac{10}{\text{y}_1}$
Now, $\frac{3}{72}=\frac{\text{x}_1}{120}$
$\Rightarrow\text{x}_1=\frac{120\times3}{72}=5$
And, $\frac{3}{72}=\frac{\text{x}_2}{192}$
$\Rightarrow \ \text{x}_2=\frac{3\times192}{72}=8$
And, $\frac{3}{72}=\frac{10}{\text{y}_1}$
$\Rightarrow \ \text{y}_1=\frac{72\times10}{3}=240$
Therefore, $\text{x}_1=5,\text{x}_2=8$ and $\text{y}_1=240.$
View full question & answer→Question 303 Marks
Let Reenu type $x$ words in $8$ minutes.
| No. of words |
$540$ |
$x$ |
| Times taken (in min) |
$30$ |
$8$ |
Clearly, less number of words will be typed in less time.
So, it is a case of direct proportion.
Now, $\frac{540}{30}=\frac{\text{x}}{8}$
$\Rightarrow\text{x}=\frac{540-8}{30}$
$\text{x}=144$
Therefore, Reenu will type $144$ words in $8$ minutes. AnswerLet $x$ be the required number of men. Now, $6\frac{3}{4}\text{m}=\frac{27}{4}\text{m}$ Then, we have:
| Number of men |
$11$ |
$x$ |
| Length of trench (in metres) |
$\frac{27}{4}$ |
$27$ |
Clearly, the longer the trench, the greater will be the number of men required.
So, it is a case of direct proportion.
Now, $\frac{11}{\frac{27}{4}}=\frac{\text{x}}{27}$
$\Rightarrow\frac{\text{11}\times4}{27}=\frac{\text{x}}{27}$
$\text{x}=44$
Therefore, $44$ men should be employed to dig a trench of length $27\ m$. View full question & answer→Question 313 Marks
A loaded truck covers $16$km in $25$ minutes. At the same speed, how far can it travel in $5$ hours?
AnswerLet the required distance be $x\ km$.
Then, We have: $1h$ = $60$min i.e., $5$h $= 5 \times 60 = 300$min
| Distance(in km) |
$16$ |
$x$ |
| Times(in min) |
$25$ |
$300$ |
Clearly, the more the time taken, the more will be the distance covered.
So, this is a case of direct proportion.
Now, $\frac{16}{25}=\frac{\text{x}}{300}$
$\Rightarrow\text{x}=\Big(\frac{16\times300}{25}\Big)$
$\text{x}=192$
Therefore, the required distance is $192\ km$. View full question & answer→Question 323 Marks
A factory requires $42$ machines to produce a given number of articles in $56$ days. How many machines would be required to produce the same number of articles in $48$ days?
AnswerLet $x$ be the number of machine required to produce same number of articles in $48$. Then, We have:
| No. of machines |
$42$ |
$x$ |
| No. of days |
$56$ |
$48$ |
Clearly, less number of days will require more number of machines.
So, it is a case of inverse proportion.
Now, $42\times56=\text{x}\times48$
$\Rightarrow\text{x}=\frac{42\times56}{48}$
$\text{x}=49$
Therefore, $49$ machines would be required to produce the same number of articles in $48$ days. View full question & answer→Question 333 Marks
In a model of a ship, the mast is $9\ cm$ high, while the mast of the actual ship is $15\ m$ high. If the length of the ship is $35$ metres, how long is the model ship?
AnswerLet $x\ m$. be the length of the model of the ship.
Then, We have: $1m = 100\ cm$ Therefore, $15m = 1500\ cm$ $35m = 3500\ cm$
| |
Length of the mast(in cm) |
Length of the ship(in cm) |
| Actual ship |
$1500$ |
$3500$ |
| Model of the ship |
$9$ |
$x$ |
Clearly, if the length of the actual ship is more, then the length of the model ship will also be more:So, this is a case of direct proportion.
Now, $\frac{1500}{9}=\frac{3500}{\text{x}}$
$\Rightarrow\text{x}=\frac{3500\times9}{1500}$
$ \text{x}=21\text{cm}$
Therfore, the length of the model of the ship is $21\ cm$. View full question & answer→