Terms: 1,+x,+x²  Co-efficients:  1,1,1 

34x³y³z³ ÷ 51xy²z³=\(\frac{2}{3}x^{2}y\)

(i) In the rectangle OABC, the coordinates of O is (0, 0), A is (2, 0), B is (2, 3) and C is (0, 3)
(ii) In the parallelogram PQRS, the coordinates of Pis (4, 3), Qis (6, 1), R is (6, 5) and S is (4, 7)

(iii) In the ΔKLM, the coordinates of K is (10, 5), Lis (7, 7) and Mis (10, 8).

It is the case of direct proportion.
Let number of pencils be x and cost of pencil be Rs.y
\(x\propto y\)
x₁ =10, y₁ = Rs.90, x₂ =19,y₂ =?
Using formula,
\(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \Rightarrow \frac { 10 }{ 90 } =\frac { 19 }{ { y }_{ 2 } } \)
y₂ x 10 = 90 \(\times\)19
\(\ { \therefore y }_{ 2 }=\frac { 90\times 19 }{ 10 } =171\)
So, cost of 19 pencils is Rs.171.

Yes

48

Given, pair of numbers 1000² and 1001².
Non-square numbers between 1000² and 1001² = 2n
= 2 \(\times\) 1000 = 2000 [\(\because\) n = 1000]

There are infinitely many rational numbers greater than -2.
Five of them are \({-3\over2},{-1},{-1\over2},0,{1\over2}\)

No, since, using angle sum property of a quadrilateral PLAN, we see that
\(\angle\) P+ \(\angle\)L + \(\angle\)A + \(\angle\)N=75° + 150°+ 140° + \(\angle\)N
= 365°+ \(\angle\)N, which is greater than 360°.
But sum of all the interior angles of a quadrilateral must be equal to 360°. So, construction of quadrilateral PLAN is not possible.

We have 1.6 =\(\frac { y }{ 1.5 } \) \(\Rightarrow\) 1.6 x 1. 5 = \(\frac { y }{ 1.5 } \) x 1.5 [multiplying both sides by 1.5]
1.6 x 1.5 = y \(\Rightarrow\) 2.40 = y
y = 2.4, Which is the required solution

m-\(\frac{m-1}{2}\)=1-\(\frac{m-2}{2}\)
We have m-\(\frac{m-1}{2}\)=1-\(\frac{m-2}{2}\)
It is a linear equation since it involves linear expressions only.
\(\Rightarrow\) m-\(\frac{m}{2}+\frac{1}{2}\)=1-\(\frac{m}{3}+\frac{1}{3}\)
\(\Rightarrow\) m-\(\frac{m}{2}+\frac{m}{3}\)=1+\(\frac{2}{3}+\frac{1}{2}\)
Transposing -\(\frac{m}{3}\)to LHS  and \(\frac{1}{2}\)to RHS 
\(\Rightarrow\) \(\frac { 6m-3m+2m }{ 6 } =\frac { 6+4-3 }{ 6 } \)
Taking LCM
\(\Rightarrow\) \(\frac { 5m }{ 6 } =\frac { 7 }{ 6 } \)
\(\Rightarrow\) m = \(\frac { 7 }{ 6 } \times \frac { 6 }{ 5 } =\frac { 7 }{ 5 } \)
Multiplying both sides by \(\frac{6}{5}\)
This is the required solution.

(a) This a line graph.
(b) It represents the number of people, who visited the store at a particular time.
(c) The busiest time of day is 1 pm at a shop.
(d) When, it opens less than 5 people enter the shop.
(e) There are 20 people in the shop at 1 : 30 prn.

Total number of students = 6250
Number of students forming a square = 6250 - 9
= 6241
Thus, 6241 students form a big square which has number of rows equal to the number of students in each row.
Let the number of students in each row be x, then the number of rows will be x.
Therefore, x x x = 6241 \(\Rightarrow\) x = \(\sqrt { 6241 } \) = 79
Hence, there are 79 students in each row of the square formed.
The value depicted here is that exercise is essential for good health. It makes our body fit and strong.

\((a) \because {1\over2}={1\over2}\times{11\over11}={11\over22} and {5\over11}={5\over11}\times{2\over2}={10\over22}\)
So, paper recycled is less than \({1\over2}\)
\((b) Similarly ,{5\over8}is \ greater \ than {1\over2}(={4\over8})\)
\(\\ Also, {2\over5}={2\times2\over5\times2}={4\over10}<{1\over2}(={5\over10})\)
\(\\ and {3\over4}>{1\over2}(={2\over4}) \)
So, the quantity of paper and glass recycled is less than \({1\over2}\)
(c) Quantity of cans = \({5\over8}(={10\over16})\) is more than half of the quantity of aluminium cans
\(={5\over8}\times {1\over2}={5\over16}\)
(d) Scrap> Aluminium cans> Paper> Glass
(e) Recycling is the process, which reduce our  wastage and create the same matter from the used part of the matter and hence, it is very useful

Yes,we can guess without prime factorisation. Firstly, we separate1331 into two groups (i.e. one's, tens and hundred's place digit and remaining digit).
So, we separate the given number 1331 into two groups as 331 and 1.
Take first group number 331, whose one's digit is 1.
∴ Unit's digit of cube root of 1 = 1
and take second group number 1, whose one's digit is 1.
∴ Unit's digit of cube root of 1= 1
Thus, the cube root of 1331 is 11.
(i) We have, 4913
Here, unit's digit of 4913 = 3
∴ Unit's digit of its cube root = 7
After striking three digits from the right most side of 4913, we get the number 4.
As 1³ = 1 and 2³ = 8
So, 1³ < 4 < 2³
Therefore, the ten's digit of cube root of 4913 is 1.
\(\sqrt [ 3 ]{ 4913 } \)= 17
(ii) We have, 12167
Here, unit's digit of 12167 = 7
:. Unit's digit of its cube = 3
After striking three digits from the rightmost side of
12167, we get the number 12.
As 2³ = 8 and 3³ = 27
So, 2³ < 12< 3³
Therefore, the ten's digit of the cube root of 12167 is 2.
∴   \(\sqrt [ 3 ]{ 12167 } \)=23
(iii) We have, 32768
Unit's digit of 32768 = 8
Unit's digit of its cube root = 2
After striking three digits from the rightmost side of
32768, we get the number 32.
As 3³ = 27 and 4³ = 64
So, 3³ < 32 < 4³
Therefore, the ten's digit of the cube root of 32768 is 3.
∴  \(\sqrt [ 3 ]{ 32768 } \) = 32

We know that, in a rhombus, all sides are of equal length.
Here, BE=4.5 cm
Firstly, draw a rough sketch of rhombus BEST, which helps us in deciding steps of construction.
So, BE = ES = ST = BT = 4.5 cm
Steps of construction
Step IDraw BE = 4.5 cm.
Step II With B as centre and radius 4.5 cm, draw an arc.
Step III With E as centre at E and radius 6 cm, draw another arc which intersects the arc in Step II at T
Step IV With E as centre and radius 4.5 cm, draw an arc on the side opposite to B with reference to ET
Step V With T as centre and radius 4.5 cm, draw another arc which intersects the arc drawn in step IV at S.
Step VI Join ES, ST, TB and TE.

Thus, BEST is the required rhombus. 

100 = 10²

\(\frac { 500\times 5\times T }{ 100 } =100\)
\(\Rightarrow T=\frac { 100\times 100 }{ 500\times 5 } =4\).

x2 - 9 = (x)2 - (3)2 = (x - 3) (x + 3).

200 + 200 = 400

(x - 3y)(3x + 5)

Suppose, 28 men will do the piece of work in x days. The given information can be exhibited in the following tabular form:

Clearly, less is the number of men, more will be the number of days to finish the work, therefore it is the case of inverse proportion.
Ratio of number of men = Inverse ratio of number of days
52: 28 = x : 35
\(\Rightarrow \frac { 52 }{ 28 } =\frac { x }{ 35 } \Rightarrow 52\times 35=28\times x\Rightarrow x=\frac { 52\times 35 }{ 28 } =65\)
Hence, 28 men will do the work in 65 days.

Edge of a small cube = 30 cm
Surface area of each small cube = 6\(\times\)(Edge)²
=  6\(\times\)30\(\times\)30 = 5400 cm²
Edge of the cubical box = 3 m = 300 cm
Surface area of cubical box = 6\(\times\)(Edge)²
= 6\(\times\)(300)² = 6\(\times\)300\(\times\)300 = 540000 cm²
∴ Number of small cubes accomodated in the cubical box of edge 3 m or 300 cm 
= \(\frac { Surface\quad area\quad of\quad the\quad cubical\quad box }{ Surface\quad area\quad of\quad each\quad small\quad cubes } \)
= \(\frac { 54000 }{ 5400 } =100\)
Hence, 100 small cubes can be accomodated ina cubical box of 3 m edge.

(21a²bc) + (-45abc²) + (18a²bc) + (21abc²)
= 21a²bc - 45abc² + 18a²bc + 21abc²
= 21a²bc + 18a²bc - 45abc² + 21abc²
= 39a²bc - 24abc²

Let the required number be x.
Then, according to the question
x³ = x ...(1) and x²≠x ...(2)
From (1), x³ -x = 0
⇒ x(x² - 1) = 0
⇒ x = 0, ± 1
⇒ x = 0,1,-1
If x = 0, then x² = x.
∴ x = is inadmissible
If x = 1 then x² = x.
∴ x = 1 is inadmissible
If x = - 1, then x² = (- 1)2 = 1 ≠ x(= - 1)
Hence, the required number is - 1.

Let the number of plates on the table be y and the number of puris be x.
Then, according to the first condition of the problem
4(y - 1) = x ...(1)
and, according to the second condition of the
problem
3y + 1= x    ... (2)
From (1) and (2), we have
4(y - 1) = 3y + 1
⇒ 4y - 4 = 3y + 1
⇒ 4y - 3y = 1 + 4
⇒ y=5
Put y = 5 in (1), we get
x = 4(5) - 4 = 20 - 4 = 16
Hence, the number of plates and number of puris on the table are 5 and 16 respectively.

 We observe about T and A that they increase together and is \(T\over A\) same every time.
Yes; The angle turned by the minute hand is directly proportional to the time that has passed.
On checking, we find that
T₂ : T₃ = A₁ : A₃ = 2: 3
and T₃ : T₄ = A₃ : A₄ = 3 : 4

The highest observation = 62
The lowest observation = 17
One of the class intervals = 40-48
∴ Class size = Upper class limit - Lower class limit
= 48 - 40 = 8
∴ The appropriate classes can be:
16-24, 24-32, 32-40, 40-48, 48-56, 56-64
Thus, the frequency distribution table for the above data can be shown using the Tally marks

Groups [Class intervals]	Tally marks	Frequency
16-24	||	7
24-32	||||	4
32-40	|	11
40-48	||	2
48-56	||||	9
56-64	||	7
Total		40