Question 12 Marks
Find the value of $x,$ if: $45\%$ of marks $x$ is $405.$
AnswerGiven, $45\%$ of $Rs. x\ kg$ is $Rs. 280.$
$\therefore\frac{45}{100}\times\text{x}=405$
$\Rightarrow\text{x}=\frac{405\times100}{45}$
$\Rightarrow\text{x}= 900\text{ marks.}$
View full question & answer→Question 22 Marks
Find the value of $x,$ if:
$8\%$ of $Rs.x$ is ? $100.$
AnswerGiven, $8\%$ of $Rs. x$ is $Rs. 100.$
$\therefore\frac{8}{100}\times\text{x}=100$
$\Rightarrow\text{x}=\frac{100\times100}{8}$
$\Rightarrow\text{x}=\text{Rs. }1250$
View full question & answer→Question 32 Marks
What sum of money lent out at $16\%$ per annum simple interest, so that it would produce $Rs. 9600$ as interest in $2$ year$?$
AnswerHere, $I = 9600, T = 2$ year and $R = 16\%$
$\because\text{I}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$\therefore\text{P}=\frac{\text{I}\times100}{\text{R}\times\text{T}}=\frac{9600\times100}{16\times2}$
$\Rightarrow\text{P}=\text{Rs. }30000$
Hence, the sum/ principal is $Rs. 30000.$
View full question & answer→Question 42 Marks
Harish bought a gas-chullah for $Rs. 900$ and later sold it to Archana at a profit of $5\%.$ Archana used it for a period of two years and later sold it to Babita at a loss of $20\%.$ For how much did Babita get it$?$
AnswerIt is given that, Harish bought the chullah for $Rs. 900$ and sold it to Archana at a profit of $5\%.$
$\therefore$ Cost price of chullah for Archana $= 900 + 5\%$ of $900$
$=900+\frac{5}{100}\times900$
$=900+45=945$
Now Aechana soid it to Babita at a loss of $20\%.$
$\therefore$ Cost price of chullah for Babita $= 945 - 20\%$ of $Rs. 945$
$=945-\frac{20}{100}\times945$
$=945-189=\text{Rs. }756$
Hence, Babitagot chullah at $Rs. 756.$
View full question & answer→Question 52 Marks
$150$ students are studying English, Maths or both. $62$ per cent of students study English and $68$ percent are studying Maths. How many students are studying both$?$
AnswerTotal students $= 150$
Student who study English $= 62\%$ of $150$
$=\frac{62}{100}\times150=93$
Student who study Maths $= 68\%$ of $150$
$=\frac{68}{100}\times150=102$
Total students studying English or Maths $= 93 + 102 = 195$
$\therefore$ Student who study English and Maths both $= 195 - 150 = 45$
View full question & answer→Question 62 Marks
In how many years, will the simple interest on a certain sum be $4.05$ times the principal at $13.5\%$ per annum$?$
AnswerLet principal $= P$
$R = 13.5\%$
$I = 4.05$ times principal $= 4.05 \times P$
$T = ?$
We know that, $\text{I}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$\Rightarrow4.05\times\text{P}=\frac{\text{P}\times13.5\times\text{T}}{100}$
$\Rightarrow\text{T}=\frac{405\text{P}}{\text{P}\times13.5}$
$\Rightarrow\frac{405\times10}{135}=30\text{ year}$
View full question & answer→Question 72 Marks
Life Science Application:A $DNA$ model was built using the scale $2\ cm : 0.0000001\ mm.$ If the model of the $DNA$ chain is $17\ cm$ long, what is the length of the actual chain$?$
AnswerLet the length of the actual chain be $x\ mm.$
Therefore.$\frac{2\text{cm}}{0.0000001\text{mm}}=\frac{17\text{cm}}{\text{x mm}}$
$\Rightarrow\text{x}=\frac{17\times0.0000001}{2}$
$=8.5\times0.0000001=0.00000058\text{mm}$
View full question & answer→Question 82 Marks
Dhruvika invested money for a period from May $2006$ to April 2008 at the rate of $12\%$ per annum. If interest received by her is $Rs. 1620,$ then find the money invested.
AnswerGiven, $I = Rs. 1620$ and $R = 12\%$
Time $=$ from my $2006$ to April $2008 = 2$ year
$\because\text{I}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$\therefore\text{P}=\frac{\text{I}\times100}{\text{R}\times\text{T}}=\frac{1620\times100}{12\times2}$
$\Rightarrow\text{P}=\text{Rs. }6750$
Hence, the invested money is $Rs. 6750.$
View full question & answer→Question 92 Marks
$45\%$ of the population of a town are men and $40\%$ are women. What is the percentage of children$?$
AnswerPercentage of men in town $= 45\%$
Percentage of women in town $= 40\%$
So, percentage of children in town $= 100 - 45 - 40 = 100 - 85 = 15\%$
Hence, $15\%$ of the population of a town are children.
View full question & answer→Question 102 Marks
Find the value of $x,$ if: $35\%$ of $Rs.x$ is? $280.$
AnswerGiven, $35\%$ of $Rs. x\ kg$ is $Rs. 280.$
$\therefore\frac{35}{100}\times\text{x}=280$
$\Rightarrow\text{x}=\frac{280\times100}{35}$
$\Rightarrow\text{x}=\text{Rs. }800$
View full question & answer→Question 112 Marks
What’s the Error? An analysis showed that $0.06\%$ of the T-shirts made by one company were defective. A student says this is $6$ out of every $100.$ What is the student’s error?
AnswerAccording to the analysis,Defective T-Shirt made bt one company $= 0.06\%$
$=\frac{0.06}{100}=\frac{6}{100}$
But according to the student, defrctive T-shirts $= 6$ out every
$100 = \frac{6}{100}$
Hence, student's error is that, the defective T-shirts are 6 out of every $10000 ($not $100).$
View full question & answer→Question 122 Marks
Geography Application: Earth’s total land is about $148428950 ~km^2$. The land area of Asia is about 30% of this total. What is the approximate land area of Asia to the nearest square kilometre?
AnswerTotal land area of Earth = $148428950~km^2$
$\therefore \text { Land area Asia }=30 \% \text { of land area of Earth }$
$=\frac{30}{100} \times 148428950=3 \times 14842895$
$=44528685 \mathrm{~km}^2$
View full question & answer→Question 132 Marks
Find the value of $x,$ if: $32\%$ of $x\ kg$ is $400\ kg.$
AnswerGiven, $32\%$ of $x\ kg$ is $400\ kg.$
$\therefore\frac{32}{100}\times\text{x}=400$
$\Rightarrow\text{x}=\frac{400\times100}{32}$
$\Rightarrow\text{x}=1250\text{kg}$
View full question & answer→Question 142 Marks
What per cent of $Rs. 80$ is $Rs. 100?$
AnswerLet $x\%$ of $Rs. 80$ be $Rs. 100.$
Then, $\frac{\text{x}}{100}\times80=100$
$\Rightarrow \text{x}=\frac{100\times10}{8}$
$\therefore\text{x}=125\%$
Hence, $125\%$ of $Rs. 80$ is $Rs. 100$
View full question & answer→Question 152 Marks
What per cent of $1\ km$ is $1000\ m?$
AnswerLet $x\%$ of $1\ km$ be $1000\ m.$
Then, $\frac{\text{x}}{100}\times1\text{km}=1000\text{m}$
$[\therefore 1\ km = 100\ m]$
$\Rightarrow\frac{\text{x}}{100}\times1000\text{m}=1000\text{m}$
$\Rightarrow\text{x}\times10=1000$
$\therefore\text{x}=100\%$
Hence, $100\%$ of $1\ km$ is $30\ m.$
View full question & answer→Question 162 Marks
What per cent of $1$ day is $1\ min?$
AnswerLet $x\%$ of $1\ day$ be $1\ min.$
Then, $\frac{\text{x}}{100}\times1\text{day}=1\text{min}$
$[\therefore 1\ day = 24\ h$ and $1h = 60\ min ]$
$\Rightarrow\frac{\text{x}}{100}\times24\text{h}=1\text{min}$
$\Rightarrow\frac{\text{x}}{100}\times1440\text{min}=1\text{min}$
$\Rightarrow\text{x}=\frac{100}{1440}=\frac{10}{144}$
$\therefore\text{x}=0.069\%$
Hence, $0.069\%$ of $1\ day$ is $1\ min.$
View full question & answer→Question 172 Marks
$800\ kg$ of mortar consists of $55\%$ sand, $33\%$ cement and rest lime. What is the mass of lime in mortar$?$
AnswerWe have,
Percentage of sand in mortar $= 55\%$
Percentage of cement in mortar $= 33\%$
So, percentage of lime in mortar $= 100 - 55 - 33 = 100 - 88 = 12\%$
$\therefore$ Weight of mortar $= 800kg$
$\therefore$ Mass of lime in mortar $= 12%$ of $800\ kg$
$=\frac{12}{100}\times800=12\times8=96\text{kg}$
Hence, weight of lime in mortar is $96\ kg.$
View full question & answer→Question 182 Marks
The strength of a school is $2000.$ If $40\%$ of the students are girls, then how many boys are there in the school$?$
AnswerAs per the given information in the question,
The strength of school $= 2000$
Percentage of girls in school $= 40\%$
Percentage of boys in school $= 100 - 40 = 60\%$
Number of boys in school $= 60\%$ of $2000$
$=\frac{60}{100}\times2000=60\times20=1200$
Hence, number of boys in school is $1200.$
View full question & answer→Question 192 Marks
What’s the Error? A clothing store charges $Rs. 1024$ for $4$ T-shirts. A student says that the unit price is $ Rs. 25.6$ per T-shirt. What is the error? What is the correct unit price?
AnswerBy unitary method, Cost of $4$ T-shirts $= Rs. 1024$
Cost of $1$ T-shirt $=\frac{1024}{4}=\text{Rs. }256$
Hence, the correct unit price is $Rs. 256.$
View full question & answer→Question 202 Marks
What per cent of $1h$ is $30$ min$?$
AnswerLet $x\%$ of $1h\ 30\min.$
Then, $\frac{\text{x}}{100}\times1\text{h}=30\text{min} [ \therefore 1h = 60\min.]$
$\Rightarrow\frac{\text{x}}{100}\times60\text{min}=30\text{min}$
$\Rightarrow\text{x}=\frac{30\times100}{60}$
$\therefore\text{x}=50\%$
Hence, $50\%$ of $1h$ is $30\min.$
View full question & answer→Question 212 Marks
A piece of cloth $5\ m$ long shrinks $10\%$ on washing. How long will the cloth be after washing$?$
AnswerLength of shrink clotg $= 10\%$ of $5m =\frac{10}{100}\times5=\frac{1}{2}\text{m}$
$\therefore$ Length of cloth wash $=5-\frac{1}{2}=\frac{9}{2}=45\text{m}$
View full question & answer→Question 222 Marks
Science Application: The king cobra can reach a Length of $558\ cm.$ This is only about $60\%$ of the length of the largest reticulated python. Find the length of the largest reticulated python.
AnswerLength of the king cobra $= 558\ cm$
According to the question,
$60\%$ of lenght of reticulated python $= 558\ cm$
$\Rightarrow\frac{60}{100}\times$ Lenght of reticulated python $= 558\ cm$
$\therefore$ Length of reticulated
$=558\times\frac{100}{60}=930\text{cm}$
View full question & answer→Question 232 Marks
What’s the Error? A student said that the ratios $\frac{3}{4}$ and $\frac{9}{16}$ were proportional. What error did the student make?
AnswerTwo ratios $a : b$ and $c : d$ are said to be proportional,
if $\frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}$ or $ad =bc.$
But in the given ration $\frac{3}{4}$ and $\frac{9}{16},3\times16\neq4\times9.$
Hence, the ration are not proportional.
To make a ration proportional to another ratio,
we just simply multiply both numerator and denominator by same number.
In our given case, student had multiply numerator by $3$ and denominator by $4,$ which is incorrect.
View full question & answer→Question 242 Marks
Physical Science Application: Unequal masses will not balance on a fulcrum, if they are at equal distance from it, one side will go up and the other side will go down. Unequal masses will balance when the following proportion is true. $\frac{\text{Mass 1}}{\text{Length 2}}=\frac{\text{Mass 2}}{\text{Length 1}}$
AnswerIt is given that, for balancing.
$\frac{\text{Mass 1}}{\text{Length 2}}=\frac{\text{Mass 2}}{\text{Length 1}}$
According to the question,
Mass $1 = 24\ kg,$
Length $1 = 3m$ and length $2 = 2m$
$\therefore\frac{24}{2}=\frac{\text{Mass 2}}{3}$ [by cross - multiplication]
$\Rightarrow\text{Mass }2=\frac{24\times3}{2}=36\text{kg}$
View full question & answer→