Question
✓
Answer
1. (b): Perimeter of farmland $X =6 a^2+2 a^2 b-2 a b-4 b^2$.
Length of farmland $X =3 a^2+a^2 b-2 a b$.
Now, $2(l+b)=$ perimeter [for a rectangle]
$
\begin{aligned}
\Rightarrow \quad b & =\left(\frac{1}{2} \times \text { perimeter }\right)-l=\frac{1}{2}\left(6 a^2+2 a^2 b-2 a b-4 b^2\right)-\left(3 a^2+a^2 b-2 a b\right) \\
& =\left(3 a^2+a^2 b-a b-2 b^2\right)-\left(3 a^2+a^2 b-2 a b\right) \\
& =\left(3 a^2+a^2 b-a b-2 b^2-3 a^2-a^2 b+2 a b\right)=a b-2 b^2
\end{aligned}
$
2. (a): Length of the combined farmland ( $X + Y$ )
$
\begin{array}{l}
=\left(3 a^2+a^2 b-2 a b\right)+\left(a^2+a b\right) \\
=3 a^2+a^2 b-2 a b+a^2+a b=4 a^2+a^2 b-a b=4 a^2-a b+a^2 b .
\end{array}
$
3. $\begin{array}{l}\text { (d): Perimeter of combined farmland (X }+ Y ) \\ \qquad \begin{aligned} & =2 \text { (length }+ \text { breadth })=2\left\{\left(4 a^2-a b+a^2 b\right)+\left(a b-2 b^2\right)\right\} \\ & =2\left\{4 a^2-a b+a^2 b+a b-2 b^2\right\}=2\left(4 a^2+a^2 b-2 b^2\right)=8 a^2+2 a^2 b-4 b^2\end{aligned}\end{array}$
4. (d): Area of farmland $Y =($ length of farmland Y $) \times($ breadth of farmland Y $)$
$\begin{array}{l}=\left(a^2+a b\right)\left(a b-2 b^2\right)=a^3 b-2 a^2 b^2+a^2 b^2-2 a b^3 \\ =a^3 b-a^2 b^2-2 a b^3 .\end{array}$
Length of farmland $X =3 a^2+a^2 b-2 a b$.
Now, $2(l+b)=$ perimeter [for a rectangle]
$
\begin{aligned}
\Rightarrow \quad b & =\left(\frac{1}{2} \times \text { perimeter }\right)-l=\frac{1}{2}\left(6 a^2+2 a^2 b-2 a b-4 b^2\right)-\left(3 a^2+a^2 b-2 a b\right) \\
& =\left(3 a^2+a^2 b-a b-2 b^2\right)-\left(3 a^2+a^2 b-2 a b\right) \\
& =\left(3 a^2+a^2 b-a b-2 b^2-3 a^2-a^2 b+2 a b\right)=a b-2 b^2
\end{aligned}
$
2. (a): Length of the combined farmland ( $X + Y$ )
$
\begin{array}{l}
=\left(3 a^2+a^2 b-2 a b\right)+\left(a^2+a b\right) \\
=3 a^2+a^2 b-2 a b+a^2+a b=4 a^2+a^2 b-a b=4 a^2-a b+a^2 b .
\end{array}
$
3. $\begin{array}{l}\text { (d): Perimeter of combined farmland (X }+ Y ) \\ \qquad \begin{aligned} & =2 \text { (length }+ \text { breadth })=2\left\{\left(4 a^2-a b+a^2 b\right)+\left(a b-2 b^2\right)\right\} \\ & =2\left\{4 a^2-a b+a^2 b+a b-2 b^2\right\}=2\left(4 a^2+a^2 b-2 b^2\right)=8 a^2+2 a^2 b-4 b^2\end{aligned}\end{array}$
4. (d): Area of farmland $Y =($ length of farmland Y $) \times($ breadth of farmland Y $)$
$\begin{array}{l}=\left(a^2+a b\right)\left(a b-2 b^2\right)=a^3 b-2 a^2 b^2+a^2 b^2-2 a b^3 \\ =a^3 b-a^2 b^2-2 a b^3 .\end{array}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Earth and Moon. The moon is earth’s only natural satellite. The mass of the moon is $\frac{1}{100}$and its radius $\frac{1}{4}$times that of earth. As a result, the gravitational attraction on the moon is about $\frac{1}{6}$when compared to earth.
Based on the above information, answer the following questions:
Q.1. The earth’s mass is approximately 6,000,000,000, 000,000,000,000,000 kg.
What is the approximate mass of the moon?
Q.2. The radius of earth is 6371 km. What is the radius of the moon?
Q.3. If an object weighs $5 \frac{3}{5}$ kg on earth, how much would it weigh on the moon?
→Based on the above information, answer the following questions:
Q.1. The earth’s mass is approximately 6,000,000,000, 000,000,000,000,000 kg.
What is the approximate mass of the moon?
Q.2. The radius of earth is 6371 km. What is the radius of the moon?
Q.3. If an object weighs $5 \frac{3}{5}$ kg on earth, how much would it weigh on the moon?
Given below is a piece of cardboard.
$1.$ What is its area$?$
$2.$ Jatin placed another cardboard of same size along the $12\ cm$ long edge.
What is the perimeter of the combined shape$?$
→$1.$ What is its area$?$
$2.$ Jatin placed another cardboard of same size along the $12\ cm$ long edge.
What is the perimeter of the combined shape$?$
A mathematical operation of rational numbers is shown on the number line.
$1.$ Which operation is it$?$
$A.$ Addition
$B.$ Subtraction
$C.$ Multiplication
$D.$ Division
$2.$ Which number will come in place of $x ?$
$3.$ Solve: $\frac{7}{10}, \frac{5}{10}$
$A. \frac{5}{7}$
$B. \frac{2}{10}$
$C.\frac{7}{5}$
$D. \frac{35}{135}$
→$1.$ Which operation is it$?$
$A.$ Addition
$B.$ Subtraction
$C.$ Multiplication
$D.$ Division
$2.$ Which number will come in place of $x ?$
$3.$ Solve: $\frac{7}{10}, \frac{5}{10}$
$A. \frac{5}{7}$
$B. \frac{2}{10}$
$C.\frac{7}{5}$
$D. \frac{35}{135}$
Sunder is a fruit seller. Today he bought plums to sell. He purchased them at the rate of 9 for 100 from the wholesale market.
Q.1. How many plums should he sell for ₹40 if he wishes to gain 20%?
(a) 3$\quad$ (b) 4$\quad$(c) 5$\quad$(d) 8
Q.2. He found that another fruit seller Kishan was selling plums at the rate of 6 for ₹90. Find the gain per cent of Kishan. (Assume that the wholesale rate is the same for all fruit sellers.
(a) 15%$\quad$(b) 20%$\quad$(c) 25%$\quad$(d) 35%
Q.3. Sunder decided to sell his plums at the rate of 7 for₹ 100. Find his gain per cent.
(a) $15 \frac{5}{9} \%$$\quad$(b) $17 \frac{7}{9} \%$$\quad$(c) $21 \frac{1}{9} \%$$\quad$(d) $22 \frac{2}{9} \%$
Q.4. In a box of 180 plums, he found that 28 were rotten and he had to throw them away. He sold the remaining plums at the rate of 8 for 100. Find his gain or loss per cent on this box.
(a) 2% gain$\quad$(b) 5% gain$\quad$(c) 2% loss$\quad$(d) 5% loss
→Q.1. How many plums should he sell for ₹40 if he wishes to gain 20%?
(a) 3$\quad$ (b) 4$\quad$(c) 5$\quad$(d) 8
Q.2. He found that another fruit seller Kishan was selling plums at the rate of 6 for ₹90. Find the gain per cent of Kishan. (Assume that the wholesale rate is the same for all fruit sellers.
(a) 15%$\quad$(b) 20%$\quad$(c) 25%$\quad$(d) 35%
Q.3. Sunder decided to sell his plums at the rate of 7 for₹ 100. Find his gain per cent.
(a) $15 \frac{5}{9} \%$$\quad$(b) $17 \frac{7}{9} \%$$\quad$(c) $21 \frac{1}{9} \%$$\quad$(d) $22 \frac{2}{9} \%$
Q.4. In a box of 180 plums, he found that 28 were rotten and he had to throw them away. He sold the remaining plums at the rate of 8 for 100. Find his gain or loss per cent on this box.
(a) 2% gain$\quad$(b) 5% gain$\quad$(c) 2% loss$\quad$(d) 5% loss
In an equilateral triangle $ABC,$ the length of $AC = 10\ cm$ and altitude $AD = 6\ cm.$ $P$ is a point on $AB.$
The length of $BP = 4x – 1.$ The length of $PA = 3x+ 4$
$1.$ What is the length of $BP?$
$A. 1\ cm$
$B. 3\ cm$
$C. 5\ cm$
$D. 10\ cm$
$2.$ What is the length of the median on $BC?$
$A. 3\ cm$
$B. 5\ cm$
$C. 6\ cm$
$D. 10\ cm$
→The length of $BP = 4x – 1.$ The length of $PA = 3x+ 4$
$1.$ What is the length of $BP?$
$A. 1\ cm$
$B. 3\ cm$
$C. 5\ cm$
$D. 10\ cm$
$2.$ What is the length of the median on $BC?$
$A. 3\ cm$
$B. 5\ cm$
$C. 6\ cm$
$D. 10\ cm$
Jacob and Mariya participated in Archery.
Jacob’s scores for ive shots are given below.
Mariya’s scores for three shots are given below.
Mariya won the competition.
$1.$ How much she did score in her fourth and fifth shots$?$
$2.$ Team Alpha and Beta played the Paint Ball game. Each team had $6$ members and each member shot the paint ball three times.
Team Alpha hit the opponent team $12$ times. Team Beta hit the opponent team $15$ times.
Which team got more penalty points and how many penalty points did they get$?$
$3.$ In another match, each member of team Alpha got hit $3$ times.
$4$ members hit back $5$ times each and the rest hit back $2$ times each.
Which calculation shows the team’s score$?$
$A. 6 × [3 + (-10)] + 4 × (5 + 20) + 2 × (2 + 20)$
$B. 6 × [3 × (-10)] + 4 × (5 × 20) + 2 × (2 × 20)$
$C. 6 + [3 + (-10)] + 4 + (5 + 20) + 2 + (2 + 20)$
$D. 6 + [3 + (-10)] + 4 + (5 + 20) + 2 + (2 + 20)$
→Jacob’s scores for ive shots are given below.
| First shot | Second shot | Third shot | Fourth shot | Fifth shot |
| $0$ | $4$ | $8$ | $10$ | $6$ |
| First shot | Second shot | Third shot |
| $0$ | $4$ | $8$ |
$1.$ How much she did score in her fourth and fifth shots$?$
$2.$ Team Alpha and Beta played the Paint Ball game. Each team had $6$ members and each member shot the paint ball three times.
Team Alpha hit the opponent team $12$ times. Team Beta hit the opponent team $15$ times.
Which team got more penalty points and how many penalty points did they get$?$
$3.$ In another match, each member of team Alpha got hit $3$ times.
$4$ members hit back $5$ times each and the rest hit back $2$ times each.
Which calculation shows the team’s score$?$
$A. 6 × [3 + (-10)] + 4 × (5 + 20) + 2 × (2 + 20)$
$B. 6 × [3 × (-10)] + 4 × (5 × 20) + 2 × (2 × 20)$
$C. 6 + [3 + (-10)] + 4 + (5 + 20) + 2 + (2 + 20)$
$D. 6 + [3 + (-10)] + 4 + (5 + 20) + 2 + (2 + 20)$
Gaming apps allow users to download and play games in online or offline mode. A play store data shows 20 million downloads of 6 popular games. The table below shows the number of daily active users of the games
(i) Around 50% of those who downloaded the games are active users. Is the statement correct? Give reason to justify your answer.
(ii) Every fourth person who downloaded the games spends more than an hour a day playing them.
What percentage of active players play more than an hour a day?
| Game | Active users (in million) |
| KC | 2.07 |
| RJ | 1.99 |
| CM | 1.82 |
| SN | 1.56 |
| CR | 1.34 |
| HC | 1.20 |
(ii) Every fourth person who downloaded the games spends more than an hour a day playing them.
What percentage of active players play more than an hour a day?
Riya wrote an algebraic expression.
$56t^3 + 12t^2 + 6t + 16s^2 + 2s + 106$
$1.$ Which of the following terms has 6 as the coeficient$?$
$A. s$
$B. s^2$
$C. t$
$D. t^3$
$2.$ Write the factors of $56t^3.$
$3.$ What is the type of the algebraic expression written by Riya$?$
$A.$ Monomial
$B.$ Binomial
$C.$ Trinomial
$D.$ Polynomial
$4.$ Riya said, “There are two like terms in the algebraic expression.” Is Riya correct? Give reason.
$5.$ Riya added an algebraic expression to $56t^3 + 12t^2 + 6t + 16s^2 + 2s + 106.$ The resultant expression is $14t^2+ 7t + 9s.$ Which of the following algebraic expressions did she add?
$A. 56t^3+ 2t^2+ t - 16s^2+ 7s + 106$
$B. -56t^3+ 2t^2+ t - 16s^2+ 7s – 106$
$C. -56t^3 - 2t^2 - t - 16s^2 - 7s – 106$
$D. 56t^3+ 26t^2+ 11t + 16s^2+ 11s + 106$
→$56t^3 + 12t^2 + 6t + 16s^2 + 2s + 106$
$1.$ Which of the following terms has 6 as the coeficient$?$
$A. s$
$B. s^2$
$C. t$
$D. t^3$
$2.$ Write the factors of $56t^3.$
$3.$ What is the type of the algebraic expression written by Riya$?$
$A.$ Monomial
$B.$ Binomial
$C.$ Trinomial
$D.$ Polynomial
$4.$ Riya said, “There are two like terms in the algebraic expression.” Is Riya correct? Give reason.
$5.$ Riya added an algebraic expression to $56t^3 + 12t^2 + 6t + 16s^2 + 2s + 106.$ The resultant expression is $14t^2+ 7t + 9s.$ Which of the following algebraic expressions did she add?
$A. 56t^3+ 2t^2+ t - 16s^2+ 7s + 106$
$B. -56t^3+ 2t^2+ t - 16s^2+ 7s – 106$
$C. -56t^3 - 2t^2 - t - 16s^2 - 7s – 106$
$D. 56t^3+ 26t^2+ 11t + 16s^2+ 11s + 106$
Vibhor has a keen interest to invest in the share market. His investment consultant advised him to study the fluctuations in the values of certain shares over a period of 5 days. Vibhor found that on July 1, the price of one share of ABC company was ₹3735.
Q.1. On July 2, the price of the share increased by ₹ 147. What was the price of each share of ABC company on July 2?
(a) ₹3582$\quad$(b) ₹3872$\quad$(c) ₹3882$\quad$(d) ₹3588
Q.2. On July 3, the share opened 27 below the previous day closing and again slipped by ₹9 by the end of the day. At what price did the share close on July 3?
(a) ₹3900 $\quad$(b) ₹3864$\quad$(c)₹ 3873 $\quad$(d) ₹3846
Q.3. On July 4, the share opened 128 above the previous day closing but again went down by ₹76 by the end of the day. What was the price of each share of ABC company at the end of the day on July 4?
(a) ₹3898$\quad$(b) ₹4050$\quad$(c) ₹3642$\quad$(d) ₹3787
Q.4. On July 5, the price of the share jumped ₹291 when the market opened. It rose further by ₹66 by the end of the day. What was the price of the share at the end of the day on July 5?
(a) ₹4123$\quad$(b) ₹3673$\quad$(c) ₹3541 $\quad$(d) ₹4255
→Q.1. On July 2, the price of the share increased by ₹ 147. What was the price of each share of ABC company on July 2?
(a) ₹3582$\quad$(b) ₹3872$\quad$(c) ₹3882$\quad$(d) ₹3588
Q.2. On July 3, the share opened 27 below the previous day closing and again slipped by ₹9 by the end of the day. At what price did the share close on July 3?
(a) ₹3900 $\quad$(b) ₹3864$\quad$(c)₹ 3873 $\quad$(d) ₹3846
Q.3. On July 4, the share opened 128 above the previous day closing but again went down by ₹76 by the end of the day. What was the price of each share of ABC company at the end of the day on July 4?
(a) ₹3898$\quad$(b) ₹4050$\quad$(c) ₹3642$\quad$(d) ₹3787
Q.4. On July 5, the price of the share jumped ₹291 when the market opened. It rose further by ₹66 by the end of the day. What was the price of the share at the end of the day on July 5?
(a) ₹4123$\quad$(b) ₹3673$\quad$(c) ₹3541 $\quad$(d) ₹4255
Ramlal is the incharge of a laboratory. In a bacteria culture under observation in his laboratory,the population of bacteria doubles every hour. When Ramlal started the observation, there were100 bacteria.
Q.1. Which of the following expressions gives the bacterial population after n hours?
(a)$\frac{100}{2^n}$$\quad$(b) $\frac{100}{2^{n-1}}$$\quad$(c)$2^n \times 100$$\quad$(d) $2^{n-1} \times 100$
Q.2. The population size of the bacteria after 3 hours will bе
(a) 800$\quad$ (b) 300$\quad$(c) 600$\quad$(d) 120О
Q.3. How many bacteria will be there in the culture after 1 day?
(a) $\frac{100}{2^{12}}$$\quad$(b) $2^{24} \times 100$$\quad$(c) $2^{12} \times 100$$\quad$(d) $\frac{100}{2^{24}}$
Q.4. Ramlal observes the culture after every one hour. Find the number of hours after which the population size of the bacteria will be larger than 3000.
(a) 5 hours$\quad$(b) 6 hours$\quad$(c) 7 hours$\quad$(d) 8 hours
→Q.1. Which of the following expressions gives the bacterial population after n hours?
(a)$\frac{100}{2^n}$$\quad$(b) $\frac{100}{2^{n-1}}$$\quad$(c)$2^n \times 100$$\quad$(d) $2^{n-1} \times 100$
Q.2. The population size of the bacteria after 3 hours will bе
(a) 800$\quad$ (b) 300$\quad$(c) 600$\quad$(d) 120О
Q.3. How many bacteria will be there in the culture after 1 day?
(a) $\frac{100}{2^{12}}$$\quad$(b) $2^{24} \times 100$$\quad$(c) $2^{12} \times 100$$\quad$(d) $\frac{100}{2^{24}}$
Q.4. Ramlal observes the culture after every one hour. Find the number of hours after which the population size of the bacteria will be larger than 3000.
(a) 5 hours$\quad$(b) 6 hours$\quad$(c) 7 hours$\quad$(d) 8 hours