Question 13 Marks
If the cost of $93 \ m$ of a certain kind of plastic sheet is Rs $1395$, then what would it cost to buy $105$ m of such plastic sheet?
Answer
|
Length of plastic sheet (in metre)
|
$93$ |
$105$ |
|
Cost(in Rs)
|
$1395$ |
$x$
|
Let the cost of the plastic sheet per metre be Rs $x.$
If more sheets are bought, the cost will also be more.
Therefore, it is a direct variation.
We get:
$93 : 105 = 1395 : x$
$\Rightarrow\frac{93}{105}=\frac{1395}{\text{x}}$
Applying cross multiplication, we get:
$\text{x}=\frac{105\times1395}{93}$
$=1575$
Thus, the required cost will be $Rs 1, 575.$ View full question & answer→Question 23 Marks
In $15$ days, the earth picks up $1.2 \times 10^8 kg$ of dust from the atmosphere. In how many days it will pick up $4.8 \times 10^8 kg$ of dust?
AnswerLet $x$ be the number of days taken by the earth to pick up $4.8 \times 10^8 kg$ of dust.
Since the amount of dust picked up by the earth and the number of days are in direct variation, we have:
$\Rightarrow \frac{15}{x}=\frac{1.2 \times 10^8}{4.8 \times 10^8}$
$\Rightarrow x=15 \times \frac{4.8}{1.2}$
$\Rightarrow x=60$
Thus, the required number of days will be $60 .$
View full question & answer→Question 33 Marks
The amount of extension in an elastic string varies directly as the weight hung on it. If a weight of $150\ gm$ produces an extension of $2.9\ cm$, then what weight would produce an extension of $17.4\ cm?$
AnswerLet x gm be the weight that would produce an extension of $17.4\ cm.$
|
weight (in gm)
|
$150$ |
$x$ |
|
Length (in cm)
|
$2.9$ |
$17.4$ |
Since the amount of extension in an elastic string and the weight hung on it are in direct variation,
we have: $\frac{150}{\text{x}}=\frac{2.9}{17.4}$
$\Rightarrow 17.4\times150 =2.9\times\text{x}$
$\Rightarrow\text{x}=\frac{17.4\times150}{2.9}$
$=\frac{2610}{2.9}$
$= 900$
Thus, the required weight will be $900\ gm.$ View full question & answer→Question 43 Marks
$18$ men can reap a field in $35$ days. For reaping the same field in $15$ days, how many men are required?
AnswerLet x be the number of cows that can graze the field in $10$ days.
|
Number of days
|
$35$ |
$15$
|
|
Number of men
|
$18$ |
$x$
|
Since the number of days and the number of men required ro reap the field are in inverse variation, we have: $35\times18 =15\times\text{x}$ $\Rightarrow\text{x}=\frac{35\times18}{15}$ $=42$ Thus, the required number of men is $42.$ View full question & answer→Question 53 Marks
The cost of $97$ metre of cloth is $Rs 242.50$. What length of this can be purchased for $Rs 302.50?$
AnswerLet x metre be the length of the cloth that can be purchased for $Rs 302.50.$
|
Length (in m)
|
$97$ |
$x$ |
|
Cost (in Rs.)
|
$242.50$ |
$302.50$ |
Since the pile of the cardboards and its thickness are in direct variation,
we have: $\frac{97}{\text{x}}=\frac{242.50}{302.50}$
$\Rightarrow 97\times302.50 =\text{x}\times242.50$
$\Rightarrow\text{x}=\frac{97\times302.50}{242.50}$
$=\frac{29342.50}{242.50}$
$= 121$ Thus, the required length will be $121$ metre. View full question & answer→Question 63 Marks
A person has money to buy $25$ cycles worth Rs $500$ each. How many cycles he will be able to buy if each cycle is costing Rs $125$ more?
AnswerLet x be the number of cycles bought if each cycle costs Rs $125$ more.
|
Cost of a cycle (in Rs.)
|
$500$
|
$625$
|
|
Number of cycles
|
$25$
|
$x$
|
It is in inverse variation.
Therefore, we get: $500\times25 =625\times\text{x}$
$\Rightarrow\text{x}=\frac{500\times25}{625}$
$=20$
$\therefore$ The required number of cycles is $20.$ View full question & answer→Question 73 Marks
The amount of extension in an elastic spring varies directly with the weight hung on it. If a weight of $250\ gm$ produces an extension of $3.5\ cm$, find the extension produced by the weight of $700\ gm.$
AnswerLet x cm be the extension produced by the weight of $700\ gm.$
|
weight (in gm)
|
$250$ |
$700$ |
|
Length (in cm)
|
$3.5$ |
$x$ |
Since the amount of extension in an elastic spring varies and the weight hung on it is in direct variation,
we have: $\frac{250}{700}=\frac{3.5}{\text{x}}$
$\Rightarrow \text{x}\times250 =3.5\times700$
$\Rightarrow\text{x}=\frac{3.5\times700}{250}$
$=\frac{2450}{250}$
$= 9.8$
Thus, the required extension will be $9.8\ cm.$ View full question & answer→Question 83 Marks
If the thickness of a pile of $12$ cardboards is $35$ mm, find the thickness of a pile of $294$ cardboards.
AnswerLet $x \ cm$ be the thickness of a pile of $294$ cardboards.
|
Thickness (in cm)
|
$3.5$ |
$x$ |
|
Cardboard
|
$12$ |
$294$ |
Since the pile of the cardboards and its thickness are in direct variation,
we have: $\frac{3.5}{\text{x}}=\frac{12}{204}$
$\Rightarrow 3.5\times294 =\text{x}\times12$
$\Rightarrow\text{x}=\frac{3.5\times294}{12}$
$=\frac{1029}{12}$
$= 85.75\text{cm}$
Thus, the thickness of $294$ cardboards will be $85.75\ cm$ (or $857.5\ mm).$ View full question & answer→Question 93 Marks
$55$ cows can graze a field in $16$ days. How many cows will graze the same field in $10$ days?
AnswerLet $x$ be the number of cows that can graze the field in $10$ days.
|
Number of days
|
$16$
|
$10$
|
|
Number of cows
|
$55$
|
$x$
|
Since the number of cows and the number of days taken by them to graze the field are in inverse variation,
we have: $16\times55 =10\times\text{x}$
$\Rightarrow\text{x}=\frac{16\times55}{10}$
$=88$
$\therefore$ The required number of cows is $88.$ View full question & answer→Question 103 Marks
A worker is paid Rs $210$ for $6$ days work. If his total income of the month is Rs $875$, for how many days did he work?
AnswerLet $x$ be the number of days for which the worker is paid Rs.$875.$
|
Income (in Rs.)
|
$210$ |
$875$ |
|
Number of days
|
$6$ |
$x$ |
Since the income of the worker and the number of working days are in direct variation, we have:
$\frac{210}{875}=\frac{6}{\text{x}}$
$\Rightarrow 210\times\text{x} =875\times6$
$\Rightarrow\text{x}=\frac{875\times6}{210}$
$=\frac{5250}{210}$
$= 25$
Thus, the required number of days is $25.$ View full question & answer→Question 113 Marks
If $x$ and $y$ vary inversely as: $x = 5$ when $y = 15$, find x when $y = 12$
AnswerSince $x$ and $y$ vary inversely, we have:$xy = k$
For $x = 5$ and $y = 15$, we have:
$5 \times 15 = k$
$\Rightarrow k = 75$
For $y = 12,$ we have:
$12x = 75$
$\Rightarrow\text{x}=\frac{75}{12}$
$=\frac{25}{4}$
$\therefore\text{x}=\frac{25}{4}$
View full question & answer→Question 123 Marks
Explain the concept of direct variation.
AnswerWhen two variables are connected to each other in such a way that if we increase the value of one variable, the value of other variable also increases and vice−versa. Similarly, if we decrease the value of one variable, the value of other variable also decreases and vice−versa.Therefore, if the ratio between two variables remains constant, it is said to be in direct variation.
View full question & answer→Question 133 Marks
In $10$ days, the earth picks up $2.6 \times 10^8$ pounds of dust from the atmosphere. How much dust will it pick up in $45$ days?
AnswerLet the amount of dust picked up by the earth in $45$ days be $x$ pounds.
Since the amount of dust picked up by the earth and the number of days are in direct variation, we have:
Ratio of the dust picked up by the earth in pounds $=$ ratio of the number of days taken.
$\Rightarrow \frac{10}{45}=\frac{2.6 \times 10^8}{x}$
$\Rightarrow x=10=45 \times 2.6 \times 10^8$
$\Rightarrow x=\frac{45 \times 2.6 \times 10^8}{10}$
$=\frac{117 \times 10^8}{10}$
$=11.7 \times 10^8$
Thus, $11.7 \times 10^8$ pounds of dust will be picked up by the earth in $45$ days.
View full question & answer→Question 143 Marks
Complite the following tables given that $x$ varies directly as $y.$
|
$x$
|
$3$
|
$5$
|
$7$
|
$9$
|
|
$y$
|
...
|
$20$
|
$28$
|
...
|
Answersolution Here, $x$ and $y$ vary directly. $\therefore x = ky x =5$ and $y =20$
i.e., $5= k \times 20 $
$\Rightarrow k =\frac{5}{20}=\frac{1}{4}$
For $x =3$ and $k =\frac{1}{4}$,
we have: $\Rightarrow 3=\frac{1}{4} \times y$
$\Rightarrow y=4 \times 3=12$ For $x=9$ and $k=\frac{1}{4}$,
we have: $x=k y $
$\Rightarrow 9=\frac{1}{4} \times y $
$\Rightarrow y=9 \times 4=36$
View full question & answer→Question 153 Marks
If $36$ men can do a piece of work in $25$ days, in how many days will $15$ men do it?
AnswerLet $x$ be the number of days in which $15$ men can do a piece of work.
|
Number of men
|
$36$
|
$15$
|
|
Number of days
|
$25$
|
$x$
|
Since the number of men hired and the number of days taken to do a piece of work are in inverse variation,
we have: $36\times25 = \text{x}\times15$
$\Rightarrow\text{x}=\frac{36\times25}{15}$
$=\frac{900}{15}$
$=60$ Thus, the required number of days is $60.$ View full question & answer→Question 163 Marks
A work force of $50$ men with a contractor can finish a piece of work in $5$ months. In how many months the same work can be completed by $125$ men?
AnswerLet $x$ be the number days required to complete a piece of work by $125$ men.
|
Number of men
|
$50$ |
$125$
|
| Months |
$5$
|
$x$
|
Since the number of men engaged and the number of days taken to do a piece of work are in inverse variation, we have:
$50\times5 = 125\text{x}$
$\Rightarrow\text{x}=\frac{50\times5}{125}$
$=2$
Thus, the required number of months is $2.$ View full question & answer→Question 173 Marks
In a hostel of $50$ girls, there are food provisions for $40$ days. If $30$ more girls join the hostel, how long will these provisions last?
AnswerLet $x$ be the number of days with food provisions for $80$ (i.e., $50 + 30)$ girls.
|
Number of girls
|
$50$ |
$80$ |
| Number of days |
$40$ |
$x$ |
Since the number of girls and the number of days with food provisions are in inverse variation, we have: $50\times40 = 80\text{x}$
$\Rightarrow\text{x}=\frac{50\times40}{80}$
$=\frac{2000}{80}$
$=25$ Thus, the required number of days is $25.$ View full question & answer→Question 183 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 number of men required to dig a trench of $27$ metre.
|
Number of men
|
$11$ |
$x$ |
|
Length (in m)
|
$\frac{27}{4}$ |
$27$ |
Since the length of the trench and the number of men are in direct variation, we have:
$\frac{11}{\text{x}}=\frac{\frac{27}{4}}{27}$
$\Rightarrow 11\times27 =\text{x}\times\frac{27}{4}$
$\Rightarrow\text{x}=\frac{11\times27\times4}{27}$
$= 44$
Thus, $44$ men will be required to dig a trench of $27\ m.$ View full question & answer→Question 193 Marks
Three spraying machines working together can finish painting a house in $60$ minutes. How long will it take for $5$ machines of the same capacity to do the same job?
AnswerLet the time taken by $5$ spraying machines to finish a painting job be $x$ minutes.
|
Number of Machines
|
$3$ |
$5$
|
|
Time (in minutes)
|
$60$
|
$x$ |
Since the number of spraying machines and the time taken by them to finish a painting job are in inverse variation, we have:
$3\times60 =5\times\text{x}$
$\Rightarrow800=5\text{x}$
$\Rightarrow\text{x}=\frac{180}{5}$
$=36$
Thus, the required time will be $36$ minutes. View full question & answer→Question 203 Marks
$1200$ men can finish a stock of food in $35$ days. How many more men should join them so that the same stock may last for $25$ days?
Answer
|
Number of men
|
$1200$ |
$x$ |
| Days |
$35$ |
$25$ |
Let $x$ be the number of additional men required to finish the stock in $25$ days.
Since the number of men and the time taken to finish a stock are in inverse variation, we have:
$1200\times35 = 25\text{x}$
$\Rightarrow\text{x}=\frac{1200\times35}{25}$
$=1680$
$\therefore$ Required number of men $= 1680 - 1200 = 480$
Thus, an additional $480$ men should join the existing $1200$ men to finish the stock in $25$ days. View full question & answer→Question 213 Marks
Complite the following tables given that $x$ varies directly as $y.$
|
$x$
|
$6$
|
$8$
|
$10$
|
$...$
|
$20$
|
|
$y$
|
$15$
|
$20$
|
$...$
|
$40$
|
$...$
|
AnswerHere, $x$ and $y$ vary directly.
$\therefore\text{x}=\text{ky}$
$x = 6$ and $y = 15$
i.e., $6 = k × 15$
$\Rightarrow\text{k} =\frac{6}{15}=0.4$
For $x = 10$ and $k = 0.4$, we have:
$\Rightarrow\text{y}=\frac{10}{0.4}=25$
For $y = 40$ and $k = 0.4$, we have:
$x = 0.4 × 40 = 16$
For $x = 20$ and $k = 0.4$, we have:
$\Rightarrow\text{y}=\frac{20}{0.4}=50$
View full question & answer→Question 223 Marks
If $x$ and $y$ vary inversely as:
$x = 3$ when $y = 8$, find $y$ when $x = 4$
AnswerSince $x$ and $y$ vary inversely, we have:
$xy = k$
For $x = 3$ and $y = 8$, we have:
$3 \times 8 = k$
$\Rightarrow k = 24$
For $x = 4$, we have:
$4y = 24$
$\Rightarrow\text{y}=\frac{24}{4}$
$=6$
$\therefore\text{y}=6$
View full question & answer→Question 233 Marks
Complite the following tables given that $x$ varies directly as $y.$
|
$x$
|
$2.5$
|
$...$
|
$...$
|
$15$
|
|
$y$
|
$5$
|
$8$
|
$12$
|
$...$
|
AnswerHere, $x$ and $y$ vary directly.
$\therefore\text{x}=\text{ky}$
$x = 2.5$ and $y = 5$
i.e., $2.5 = k \times 5$
$\Rightarrow\text{k} =\frac{2.5}{5}=0.5$
For $y = 8$ and $k = 0.5$, we have:
$x = ky$
$\Rightarrow x = 8 \times 0.5 = 4$
For $y = 12$ and $k = 0.5$, we have:
$x = ky$
$\Rightarrow x = 12 \times 0.5 = 6$
For $x = 15$ and $k = 0.5$, we have:
$x = ky$
$\Rightarrow 15 = 0.5 \times y$
$\Rightarrow\text{y}=\frac{15}{0.5}=30$
View full question & answer→Question 243 Marks
A woker is paid Rs. $200$ for $8$ days work. If he works for $20$ days, how much will he get?
AnswerLet $x$ be the number of days for which the worker is paid Rs. $875$
|
Income (in Rs.)
|
$200$ |
$x$ |
|
Number of days
|
$8$ |
$20$ |
Since the income and the number of working days are in direct variation,
we have: $\frac{200}{\text{x}}=\frac{8}{20}$
$\Rightarrow 200\times20 =8\text{x}$
$\Rightarrow\text{x}=\frac{200\times20}{8}$
$=\frac{4000}{8}$
$= 500$ Thus, the worker will get Rs. $500$ for working $20$ days. View full question & answer→Question 253 Marks
A group of $3$ friends staying together, consume $54\ kg$ of wheat every month. Some more friends join this group and they find that the same amount of wheat lasts for $18$ days. How many new members are there in this group now?
AnswerLet x be the number of new members in the group.
|
Number of members
|
$3$ |
$x$
|
|
Number of days
|
$30$ |
$18$ |
Since more members can finish the wheat in less number of days, it is a case of inverse variation.
Therefore, we get: $3\times30 =\text{x}\times18$
$\Rightarrow90=18\text{x}$
$\Rightarrow\text{x}=\frac{90}{18}$
Thus, the number of new members in the group $= 5 - 3 = 2.$ View full question & answer→Question 263 Marks
A work-force of $420$ men with a contractor can finish a certain piece of work in $9$ months. How many extra men must he employ to complete the job in $7$ months?
AnswerLet $x$ be the extra number of men employed to complete the job in $7$ months.
|
Number of men
|
$420$ |
$x$ |
| Months |
$9$ |
$7$ |
Since the number of men hired and the time required to finish the piece of work are in inverse variation,
we have: $420\times9 = 7\text{x}$ $\Rightarrow\text{x}=\frac{420\times9}{7}$ $=540$
Thus, the number of extra men required to complete the job in $7$ months $= 540 - 420 = 120$ View full question & answer→