Few more examples for direct proportion are as follows:
(i) Length of the cloth purchased and its total cost.
(ii) Number of months and total salary.
(iii) Number of hours of production and the amount of the commodity produced

\({z\over z+15}={4\over 9}\)
By cross-multiplication, we have
9z = 4(z + 15) \(\Rightarrow\) 9z = 4z + 60
Transposing 4z to LHS, we have
9z - 4z = 60
\(5z=60\Rightarrow z={60\over 5}=12\)
\(\therefore \) z = 12.

53

70 cm

We have, 43 = 40 + 3
\(\therefore\) 43² = (40 + 3)² = (40 + 3)(40 + 3)
= 40(40 + 3) + 3(40 + 3)
= 40\(\times\)40 +  40\(\times\)3 + 3\(\times\)40 + 3\(\times\)3
= 1600 + 120 + 120 + 9 = 1849

∵ \(\frac{6^{n}}{6^{-2}}=6^{3}\)
∴ 6ⁿ⁺²=6³ ⇒ n+2=3 ⇒ n=3-2
⇒ n=1  [∵ bases of two equal numbers are same, so exponents will be same]

(a) From the given graph, it is clear that the forecast temperature was the same as the actual temperature on Tuesday, Friday and Sunday. (This is indicated by the point at which both graphs meet).
(b) From the given graph, we can say that the maximum forecast temperature during the week was 35°C.
(c) From the given graph, we can say that the minimum actual temperature during the week was 15°C.

Since the maximum difference between temperatures is 7.5^o C.
Hence, the actual temperature differed the most from the forecast temperature on Thursday.

12

We have, \({4\over7}+(-{4\over9})+{3\over7}+({13\over9})\)
\(=({4\over7}+{3\over7})+(-{4\over9}+{-13\over9})={7\over7}+(-{4\over9}-{13\over9}) \)
\(\\ =1+(-{17\over9})=1-{17\over9}={9-17\over9}=-{8\over9}\)

\(\frac { { x }^{ 2 } }{ 25 } -\frac { 1 }{ 4 } \)

95°

Total annual income = Rs 1200000
∴ 5% of the total income
=1200000ｘ\(\frac{5}{100}\)=Rs 60000
Thus, he donates Rs 60000 to an orphanage

Let us draw a rough sketch of the required quadrilateral RAIL.

Steps of construction
Step I Draw AI = 4.5 cm and draw angle \(\angle \)lAX =105° and \(\angle \)AIY =110°.
Step II Draw an arc of 6 cm with centre A on AX and get the point R. So, AR = 6 cm.
Step III Now, make an angle of 60° on R.
Step IV Let it intersect IY at L.

Thus, quadrilateral RAIL is the required quadrilateral.

In a hexahedron there are 6 faces

From the above pictograph, we observe that
(i) 250 cars were produced in the month of July.
(ii) The maximum number of cars were produced in the month of September.

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.

We have
A rational number between \(\frac { 1 }{ 2 } \)
= \(\left[ \frac { 1 }{ 2 } +(-2) \right] \div 2=\left[ \frac { 1-4 }{ 2 } \right] \div 2\)
= \(\left[ \frac { -3 }{ 2 } \right] \times \frac { 1 }{ 2 } =\frac { -3 }{ 4 } \)
A rational number between \(\frac { 1 }{ 2 } \) and \(\left( \frac { -3 }{ 4 } \right) \)
= \(\left[ \frac { 1 }{ 2 } +\left( \frac { -3 }{ 4 } \right) \right] \div 2\)
\(\left[ \frac { 2-3 }{ 4 } \right] \times \frac { 1 }{ 2 } =\frac { -1 }{ 4 } \times \frac { 1 }{ 2 } =\frac { -1 }{ 8 } \)
A rational number between \(\left( \frac { -3 }{ 4 } \right) \) and (-2)
= \(\left[ \left( \frac { -3 }{ 4 } \right) +(-2) \right] \div 2=\left[ \frac { (-3)+(-8) }{ 4 } \right] \times \frac { 1 }{ 2 } \)
= \(\frac { -11 }{ 4 } \times \frac { 1 }{ 2 } =\frac { -11 }{ 8 } \)
Thus, the three rational numbers
\(\left( \frac { -3 }{ 4 } \right) ,\left( \frac { -1 }{ 8 } \right) \) and \(\left( \frac { -11 }{ 8 } \right) \) are between \(\frac { 1 }{ 2 } \) and (-2)

(a) The complete table is shown below:

(b) Graph between rows and number of white tiles are shown as below:

Graph between rows and number of purple tiles are shown below:

(c) No one graph is linear.

Given, 212 x 5 is a multiple of 3.
So, sum of all its digits 2 + 1 + 2+ x + 5 must be divisible by 3. \(\Rightarrow\) 10 + x is a multiple of 3.
\(\Rightarrow\)10 + x = 0,3,6,9,12, ...
But x is a digit of number 212x5.
\(\therefore\) x can take values 0, 1, 2, 3, ... , 9.
\(\Rightarrow\)10 + x can take value 12, 15, 18.
So, 10 + x = 12 or 15 or 18
\(\Rightarrow\)10 + x = 12 or 10 + x = 15 or 10 + x = 18
\(\Rightarrow\)x = 2, x = 5, x = 8                                            .....................(i)
Similarly, 212 x 5 is a multiple of 11
Sum of the digits at even places is x + 1.
Sum of the digits at odd places = 5 + 2 + 2 = 9
\(\therefore\)Difference = 9 - x -1 = 8 - x
Now, 212 x 5 will be a multiple of 11.
So, 8 - x is a multiple of 11.
\(\Rightarrow\)8 - x = 0 or 11 or 22 \(\Rightarrow\) x = 8 or 3 or 14
But x is a digit of the number 212x5.
So, x = 0, 1,2 ... 9 \(\Rightarrow\) x = 8                                  ...............(ii)
From Eqs. (i) and (ii), we get x = 8 and the number is 21285

736000

Given, three numbers are in the ratio 1 : 2 : 3.
Let the numbers be x, 2x and 3x.
If the sum of their cubes is 4500, then
(x)³ + (2x)³ + (3x)³ = 4500
x³ + 8x³ + 27x³ = 4500
36x³ = 4500
⇒ x³ = \(\frac { 4500 }{ 36 } \)
⇒ x³ = 125
⇒ x = \(\sqrt [ 3 ]{ 125 } \)
⇒ x = \(\sqrt [ 3 ]{ 5\times 5\times 5 } \)
⇒ x = 5
So, the numbers are 5 x 1 ,5 x 2 ,5 x 3, i.e. 5,10 and 15.

Let length of the ship and mast be x cm and y cm, respectively.
Now, we can make a table as shown below:

Here, as the height of the mast decreases, so the length of the ship also decreases in the same ratio. So, this is a case of direct proportion.
Here, x₁ = 28m y₁ = 12m y₂ = 9cm,x₂ = ?
Now, by using the relation\(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \)
\(\frac { 28m }{ 12m } =\frac { { x }_{ 2 }cm }{ 9cm } \Rightarrow { x }_{ 2 }\times 12=28\times 9\)
\({ x }_{ 2 }=\frac { 28\times 9 }{ 12 } =21\Rightarrow { x }_{ 2 }=21cm\)
Hence, the length of model ship is 21 cm.

Let box B's weight be x kg.
Since, box A weights \(2\frac { 1 }{ 2 } \) kg more than box Band  box C weights \(10\frac { 1 }{ 4 } \) kg more than box B.
Weight of box A = \(\left( x+2\frac { 1 }{ 2 } \right) kg=\left( x+\frac { 5 }{ 2 } \right) kg\)
Weight of box B = \(\left( x+10\frac { 1 }{ 4 } \right) kg=\left( x+\frac { 41 }{ 4 } \right) kg\)
Total weight of all the boxes
\(\left( x+\frac { 5 }{ 2 } +x+x+\frac { 41 }{ 4 } \right) kg\)
According to the question,
Total weight = \(48\frac { 3 }{ 4 } kg=\frac { 195 }{ 4 } kg\)
\(\therefore\)  \(x+\frac { 5 }{ 2 } +x+x+\frac { 41 }{ 4 } =\frac { 195 }{ 4 } \)
\(\Rightarrow\) 4x + 10+ 4x + 4x + 41 = 195  [multiplying both sides by 4]
\(\Rightarrow\) 12x + 51 = 195 \(\Rightarrow\) 12x = 144 \(\Rightarrow\) x = 12
\(\therefore\) weight box
A = \(\left( 12+\frac { 5 }{ 2 } \right) kg=\frac { 29 }{ 2 } kg=14\frac { 1 }{ 2 } kg\)

To find the square roots of 100, we subtract successive odd number starting from 1 as follows:
100 - 1 = 99, 99 - 3 = 96, 96 - 5 = 91, 91 - 7 = 84
84 -9 = 75, 75 -11 =64, 64 -13 = 51,51-15 =36
36-17 = 19, 19 - 19 = 0
We observe that the number 10 reduced to zero after subtracting first 10 odd numbers.
So, 100 is a perfect square.
\(\therefore \ \sqrt { 100 } =10\)
Hence, the square root of 100 is 10.

3\(\pi \) unit 

(48)²=(50-2)²
Since, (a - b)² = a² - 2ab + b²
(50-2)²=(50)²-2x50x2+(2)²
= 2500 - 200 + 4
[(a-b)²=a²-2ab+b²]
= 2504 - 200 = 2304
(ii) 181² -19² =(181-19)(181 + 19)
where, a = 50 and b = 2
= 162 x 200 = 32400
(iii) 497 x 505 =(500 - 3)(500 + 5)
= 500² + (-3 + 5) x 500 + (-3)(5)
[.: (x + a)(x + b) = x² + (a +b) x + ab]
= 250000 + 1000 -15 = 250985
(iv) 2.07 x1.93 = (2 + 0.07)(2 - 0.07)
= 2² -(0.07)²
[.: where, a = 50 and b = 2]
= 3.9951

Steps of construction
Step I Draw a line AB = 8 cm. [\(\because\)  8 cm = 800 m]
Step II Draw rays BX and AY, such that \(\angle\) ABX = 90° and \(\angle\)BAY = 90°, respectively.
Step III Cut BC = 6 cm on AX and AD = 6 cm on AY.
Step IV Join C to D, then ABCD is the required rectangular field.
Step V Join A to C.
Step VI Make altitude MD on AC.
Step VII Make altitude MN on CD and write name of the regions.
The value depict, by this type of management of area by government is that they want to maintain all social, environmental aspects in society.

Given,principal (P) = Rs 8000, time (n) = 2 yr, rate (R) = 5%
We know that,
Amount (A) = P\((1+\frac{R}{100})^{n}=8000(1+\frac{5}{100})^{2}\)
=\(8000(\frac{100+5}{100})^{2}\)
= \(8000(\frac{105}{100})^{2}=8000\times \frac{21}{20}\times \frac{21}{20}= \frac{80\times 21\times 21}{2\times2}\)
= 20ｘ21ｘ21 = Rs 8820
ஃ Compound Interest (CI) = Amount (A) - Principal (P)
= Rs(8820 - 8000) = Rs 820.
Alternate Method
Principalfor the Isr year, PI = Rs 8000, R = 5% and T = 2 yr
ஃ SI = SI at 5% per annum for Ist year
\(=\frac{8000\times 5\times 1}{100}\)=Rs 400        \([\therefore =\frac{P\times R\times T}{100}]\)
Amount at the end of Ist year = Principal + Simple interest
= Rs 8000 + Rs 400 = Rs 8400
Now, principal for the IInd year =Amount at the end of Ist year
= Rs 8400
SI₂ = SI at 5% per annum for IInd year =Rs \(\frac{8400\times 5\times 1}{100}\)
= Rs(84ｘ5) = Rs 420
Amount at the end of IInd year = Amount at the end of 1st year + SI₂
= Rs(8400 + 420) = Rs 8820
ஃ Compound Interest (CI) =Amount (A) - Principal (P)
= Rs (8820 - 8000) = Rs 820
Hence, the compound interest is Rs 820.

(i) Given, ABCD is a parallelogram in which ㄥB = 100°.
We know that, the sum of any two adjacent angles of a parallelogram is 180°.
ㄥA + ㄥB = 180°
[∴ ㄥA and ㄥB are adjacent angles]
⇒ z+100⁰=180° ⇒ z=180⁰-100° =80°
Also, opposite angles of a parallelogram are of equal measure.
∴  ㄥD=ㄥB ⇒ y=100⁰
and ㄥC=ㄥA ⇒ x=z=80°
Hence, the measure of x, y and z are 80°,100° and 80°, respectively.

(ii) Let a parallelogram be ABCD in which ㄥD = 50° .
We know that, the sum of any two adjacent angles of a parallelogram is 180°.
∴ ㄥA + ㄥD=50° ⇒ x + 50⁰= 180°
⇒ x = 180°-50° = 130°
Also, opposite angles of a parallelogram are of equal measure.
∴  ㄥC=ㄥA ⇒ y=x=130⁰ and ㄥB=ㄥD=50°
Now, ㄥB + exterior ㄥB = 180°
⇒ 50° + z = 180° [by linear pair angle]
⇒ z = 180⁰- 50° = 130°
Hence, the measure of angles x, y and z are 130°, 130° and 130° , respectively.

(iii) Let a parallelogram be ABCD, in which ㄥCBO = 30°. Here, AC and BD intersect each other at O and ㄥAOD=90°.
∴ ㄥCOB = ㄥAOD=90°
[vertically opposite angles]
We know that, the sum of three angles of a triangle is 180°.
In ΔOBC,
ㄥCOB + ㄥOCB + ㄥCBO = 180°
⇒ 90° +y +30° = 180° ⇒ Y + 120° = 180⁰
 y=180⁰ - 120⁰= 60⁰
As AD II BC and AC is a transversal.
 y = z = 60° [alternate interior angles]
Hence, the measure of angles x, y and z are 90°, 60 and 60° , respectively.

(iv) Let parallelogram be ABCD, in which ㄥB = 80°
We know that, the sum of any two adjacent angles of a parallelogram is 180°.
∴  ㄥB + ㄥC = 180° ⇒ 80⁰+ ㄥC= 180°
⇒ ㄥC = 180° - 80°= 100°
Also, ㄥC + exterior ㄥC = 180° [by linear pair
:. exterior LC or z = 180° - 80°= 100°
Also, opposite angles of a parallelogram are of equal measure.
So, ㄥA = ㄥC ⇒ x = 100°
and ㄥD = ㄥB ⇒  y = 80°
Hence, the measure of angles x, y and z are 100°,80° and 100°, respectively.

(v) Let parallelogram be ABCD, in which
ㄥB = 112° and ㄥDAC = 40°.
We know that, the sum of any two adjacent angles of a parallelogram is 180°.
∴ ㄥA + ㄥB = 180°
⇒ (40° + z) + 112° = 180°
⇒ 40⁰ + z + 112° = 180° ⇒ z+152°=180⁰
⇒ z = 180⁰-152°=28⁰
Also, opposite angles of a parallelogram are of equal measure.
So, ㄥD = ㄥB ⇒ y=112°
Now, in ㄥACD, using angle sum property of a triangle,
ㄥCAD + ㄥD + ㄥDCA = 180⁰
⇒ 40⁰+y+x=180⁰
⇒ 40° +112° + x = 180°
⇒ 152° + x =180⁰ - x=180⁰-152°=28⁰
Hence, the values of x, y and z are 28⁰, 112° and 28°, respectively.

(i) Total length of the fence surrounding it = Perimeter of -he park = 30m + 20m + 30m + 20m = 100 m
(ii) Land occupied by the park = Area of the park = (30 \(\times\) 20) m² = 600 m²
(iii) Area of cemented path = Area of the park - Area of the park left after cementing the path
= 600 m² - [(30 - 2) \(\times\) (20 - 2)] m² = 600 m² - (28 \(\times\) 18) m² = 600 m² - 504 m² = 96 m²
Number of cement bags = \(\frac { Area\quad of\quad the\quad path }{ Area\quad cemented\quad by\quad 1\quad bag } =\frac { 96 }{ 4 } \)=24
(iv) Area covered by the grass =Area of the park left after cementing the path - Area of rectangular beds
= 504 m² -2(1.5 2)m²= 504m²-6m²=498m²

Here, to represent the given information, we draw the bar graph. So, draw two perpendicular axes OX and OY. Take months on OX and number of watches sold on OY. Choose a suitable scale for OX and OY and drawbars of equal width with equal gaps in between.
Thus, we get the following bar graph

\(\frac{3}{8} \times 1\)=\(1 \times \frac{3}{8}=\frac{3}{8}\).

Yes, some of these are convex polygons while the others are concave polygons.

1.496 \(\times\) 10¹¹ m

Here, x₁= 2, y₁= 9 x 10⁶, x₃ = 1.2 and y₃ =?
Now, by using the relation\(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 3 } }{ { y }_{ 3 } } \)
\(\frac { 2 }{ 9\times { 10 }^{ 6 } } =\frac { 1.2 }{ { y }_{ 3 } } \Rightarrow 2\times { y }_{ 3 }=1.2\times 9\times { 10 }^{ 6 }\)
\(\Rightarrow { y }_{ 3 }=\frac { 10.8\times { 10 }^{ 6 } }{ 2 } =5.4\times { 10 }^{ 6 }\Rightarrow { y }_{ 3 }=5.4\times { 10 }^{ 6 }\)
Hence, there are 5.4 \(\times\) 10⁶ crystals of sugar in 1.2 kg of sugar.

Given mathematical statement is incorrect.
Hence, the correct statement is x(3x + 2) =3x² + 2x because when we multiply the expression enclosed in a bracket by a constant (or a variable) outside, each term of the expression has to be multiplied by the constant (or the variable).

(a) 50^oF   (b)20^oF

706

\(\frac { 50653 }{ 343 } \).

441

7040 cm²

25000

(1.5x - 4y)(1.5x + 4y + 3) - 4.5x + 12y
= 1.5x(1.5x + 4y +3) - 4y(1.5x + 4y + 3) -4.5x + 12y
= (1.5x\(\times\)1.5x) + (1.5x\(\times\)4y) + (1.5x\(\times\)3) - (4y\(\times\)1.5x) - (4y\(\times\)4y) - (4y\(\times\)3)-4.5x + 12y
= 2.25x² + 6xy + 4.5x - 6xy -16y²- 12y - 4.5x + 12y
= 2.25x²+ (6 - 6)xy + (4.5 - 4.5)x - 161 + (12 - 12)y
= 2.25x² + (0)xy + (0)x-16y²+ (o)y
= 2.25x²+ 0 + 0 - 16y²+ 0
= 2.25x²- 16y²

On the wheel, there are three sectors of red colour and five sectors of green colour.
Number of outcomes of getting a green sector on this wheel = 5
Number of outcomes of not getting a green sector on this wheel
Number of outcomes of getting a red sector on this wheel = 3.

Yes, the quadrilateral MIST can be constructed
with \(\angle\)M = 100°.
By angle sum property of a quadrilateral,
\(\angle\)M+ \(\angle\)I+ \(\angle\)S+ \(\angle\)T=360°
\(\Rightarrow\) 100°+105°+120°+ \(\angle\)T = 360°
\(\Rightarrow\) \(\angle\)T=35°
Thus, we have, MI = 3.5 cm, IS = 6.5 cm, \(\angle\)M =100°, \(\angle\)I = 105° and \(\angle\)S =120° and \(\angle\)T = 35°.

Shivaji road

a = \(\frac { -2 }{ 31 } \)

10 - 8 = 2

4 = 2 x 2 = 2²

Sum ={8 + (-17)}pq = - 9pq.

2 + 4 + 7 = 13 is not divisible by 9.

0\(\times\)0\(\times\)0=0.

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