Yes ! it is a case of direction variation as
\(\frac { 100 }{ 10 } =\frac { 200 }{ 20 } =\frac { 300 }{ 30 } =\frac { 500 }{ 50 } =\frac { 1000 }{ 100 } \)
=10(constant)

983² – 17² = (983 + 17) (983 – 17)
[Here a = 983, b =17, a² – b² = (a + b) (a – b)]
Therefore, 983² – 17² = 1000 x 966 = 966000

By prime factorisation, we have

     

 2(m - 2n)(2p -q)

The given number is 14567. Unit's digit of this number is an odd number. Hence, this number is not divisible by 2.

99856

(i) If the number of workers increases, then time to complete the job would decrease. So, it is the case of inverse proportion.
(ii) For longer distance, more time would be required. So, it is not the case of inverse proportion.
(iii) For more area of cultivated land, more crops would-he harvested. So, area of cultivated land and the crop harvested are not in inverse proportion.
(iv) If speed of a vehicle is more, then time to cover a fixed journey would be less. So, it is a case of inverse proportion.
(v) For more population, less area per person would be there. So, it is a case of inverse proportion

We have, \((-3)^{4}\times (\frac{5}{3})^{4}\)
=\((-1\times 3)^{4}\times (\frac{5}{3})^{4}\)    [∵ -a =-1хa]
=\((-1)^{4}\times3^{4}\times \frac{5^{4}}{3^{4}}\)
[∵ \((a\times b)^{m}=a^{m}\times b^{m}, (\frac{a}{b})^{m}=\frac{a^{m}}{b^{m}}\)]
= 1 x 5⁴ = (5)⁴     [∵ (-1)⁴=1]
which is the required form.

Let the cost price of the article be Rs x.
Then, x + 40% of x = 1120000
⇒ x + \(\frac{40}{100}\)ｘx =1 12000
⇒ \(\frac{100x+40x}{100}=112000\)
⇒ 140x = 112000ｘ100
∴ x = \(\frac{112000\times 100}{140}=80000\)
Hence, the cost price is Rs 80000.

The multiplicative inverse of  \({-11\over13}\) is \({-13\over11}\)

Given d₁ = 7.5 cm and d₂ = 12cm
We know that ,
Area of rhombus =\(\frac { 1 }{ 2 } \times \) d₁\(\times\)d₂
=\(\frac { 1 }{ 2 } \times \) 7.5\(\times\) 12 = 7.5\(\times\) 6 = 45m²
Hence area of a rhombus is 45 m²

In the trapezium ABCD, we have
AB II CD
Also,sum of interior angles B and C is 180°
∴ (x + 20)° + (x - 30)° =180°
⇒ 2x⁰ -10⁰ =180°
⇒ 2x⁰ = 190°
⇒ x⁰ = 95°

In a pack of well-shuffled cards, there are
number of black face cards = 6; number of red jack = 2;
number of red card of ace = 2; number of black kings = 2 ;
number of ordinary card = 52
whereas, in a pack, there are total 52 cards.
\(\therefore\) Probability of getting
a red jack = \(\frac{2}{52}=\frac{1}{26}\)

The front view, side view and top view of a nut are as follow:

In case of a quadrilateral, it is necessary to have atleast the knowledge of five parts to be able to construct it uniquely. We shall need data about specified parts of the quadrilateral as follows:
(a) When four sides and one diagonal are given.
(b) When two diagonals and three sides are given.
(c) When two adjacent sides and three angles are given.
(d) When three sides and two included angles are given.
(e) When some special properties are given.

We have \(\frac { 2x }{ 3 } =18\Rightarrow \frac { 2x }{ 3 } \times 3=18\times 3\)[multiplying both side by 3]
\(\Rightarrow\)   2x =18 x 3
\(\Rightarrow\) 2x = 54
\(\Rightarrow\) \(\frac { 2x }{ 2 } =\frac { 54 }{ 2 } \)     [dividing both sides by 2]
\(\Rightarrow\) x = \(\frac { 54 }{ 2 } \), 27 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.

x = 10

13

1.SQUARE
2.PYTHAGOREAN
3.PERFECT SQUARE
4.TRIPLET
5.SQUARE ROOT
6.FACTOR

Given, number of observations = (x + 5)
Sum of the observations =(x⁴- 625)
\(Mean\ of\ the\ observations=\frac { sum\ of\ the\ observations }{ number\ of\ observations } =\frac { \left( { x }^{ 4 }-625 \right) }{ \left( x+5 \right) } \)
We have to factorise (x⁴ - 625)
= (x²)² _ 25 x 25
= (x²)² -(25)²
= (x² + 25) (x²- 25)
[by using identity, (a² - b²) = (a + b)(a - b)where, a = x² ,b = 25]
= (x² + 25)(x² - 5 \(\times\) 5)
= (x² + 25) {(x)² -(5)²}
= (x² + 25)(x + 5)(x - 5)
\(\frac { \left( { x }^{ 4 }-625 \right) }{ \left( x+5 \right) } =\frac { \left( { x }^{ 2 }+25 \right) \left( x+5 \right) \left( x-5 \right) }{ \left( x+5 \right) } \)
= (x² + 25)(x - 5)
Hence, the mean of the observations is (x² + 25) (x - 5).

(a) From the given graph, the car travel in 2 h is 80km.
(b) 5 h taken by car to reach R.
(c) 2 h taken by car to cover 80 km.

12 days

The fraction of substance remains after 7 half- lives =3^-7=\(\frac{1}{3^{7}}\)
ஃ \(3^{-x}=\frac{1}{243}=\frac{1}{3\times3\times3\times3\times3}=\frac{1}{3^{5}}\)
⇒ \(3^{-x}=3^{-5}\)
⇒ \((3)^{-x}=(3)^{-5}\)
⇒ -x=-5  [∵ bases are same]]
Hence, after 5 half-lives, the fraction will be \(\frac{1}{243}\) of the original.

Area of figure A, for radius 8 cm
 =\(\pi\)r²h = \(\frac { 22 }{ 7 } \times \)8\(\times\)8=201.14 cm²
Since, there are two pipes.
∴ Area = 2\(\times\)201.14 = 402.28 cm²
Now, area of figure B, for radius 15 cm
 =\(\pi\)r² =\(\frac { 22 }{ 7 } \times \) 15\(\times\)15 = 707.14 cm²
Hence, pipe B fill the swimming pool faster.

P = 9, Y = 1 and Z = 8

\(\frac{1}{2}\)

Given,51²-49²=100p
2601-2401=100p
200 =100p
\(p=\frac { 200 }{ 100 } =2\)
So, it is true
Alternate Method
51²- 49² =100p
(51- 49)(51 + 49) =100p
2 x100 =100p
200 =100p
\(p=\frac { 200 }{ 100 } =2\)

113

54°, 144°, 72°, 90°

Here, P = Rs 10000, T = 1 year
R = 8% p.a. compounded half yearly.
\(\therefore\) R = 8% p.a. = 4% per half yearly
T = 1 year \(\rightarrow\) n = 2 x 1 = 2
Now, amount = P\((1+{R\over 100})^n\)
= Rs 10,000\((1+{4\over 100})^2\)= Rs 10,000 \(({26\over25})^2\)
= Rs 10 000 x \({26\over25}\) x \({26\over25}\)
= Rs16 x 26 x 26 = Rs10816
CI = Rs 10816 - Rs 10000 = Rs 816

We know that, the length PQ and QR of a rectangle PQRS. Also, we know that in a rectangle opposite sides have equal length and each of the angle is 90°. 
Thus, we have PQ=RS and QR=PS and \(\angle \)PQR = 90°.
Steps of construction
Step I Draw PQ.
Step II Makes \(\angle \)PQX=90°.
Step III Cut QR from QX
Step IV From P cut-off an arc equal to QR.
Step V From R, cut-off an arc equal to PQ.
Step VI Mark S as the intersection point of both the arcs. Join PS and RS to get the required rectangle PQRS.
 

\({7y+4\over y+2}={-4\over 3}\)
By cross-multiplication, we have
3 x (7y + 4) = - 4 x (y + 2)
or 21y + 12 = - 4y - 8
Transposing 12 to RHS and (-4y) to LHS,
we have
21y + 4y = -8 - 12
or 25y = -20 or y = \({-20\over 25}={-4\over 5}\)
(Dividing both sides by 25)
\(\therefore y={-4\over 5}\)

A cube has 6 equal surfaces.
Let side of the cube be x.
\(\therefore\)Area of a surface =x²
\(\Rightarrow\) Area of the given cube = 6x²
\(\therefore\)6x²= 486 \(\Rightarrow x^2={486\over6}=81\)
\(\Rightarrow x =\sqrt{81}=9\)
Thus, the volume of the given cube = 729 cm³ .

A = 5, B = 3

We have to divide x³ y³ + x²y³ - xy⁴+ xy by xy.
Method I Cancellation method
We will divide each term of polynomial by given monomial.
\(\frac{x_3y_3+x_2y_3-xy_4+xy}{xy}=\frac{x^{3}y^{3}}{xy}+\frac{x^{2}y^{3}}{xy}-\frac{xy^{4}}{xy}+\frac{xy}{xy}=x^{2}y^{2}+xy^{2}-y^{3}+1\)
Method II Common factor method
We will express each term of polynomial in factor form, then we will separate common factors.
\(x^{3}y^{3}+x^{2}y^{3}-xy^{4}+xy=x\times x\times x\times y\times y\times y\times+x \times x \times y \times y \times y -x\times y \times y \times y \times y+x \times y\)
= \(xy(x \times x \times y \times y+x \times y \times y-y \times y\times y+1 )\)
Then, \((x^{3}y^{3}+x^{2}y^{3}-xy^{4}+xy)\div xy=\frac{xy(x\times x \times y \times y+x \times y \times y - y \times y \times y+1)}{xy}\)
=\((x^{2}\times y^{2}+x \times y^{2}-y^{3}+1)=(x^{2}y^{2}+xy^{2}-y^{3}+1)\)

(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.

We have, h = 2 m and b = 45 m 
Since, in 60 min length of water flowing = 3 km
∴ 1 min, the length of water flowing  =\(\frac { 3\times 1000 }{ 60m } \)=50 m
∴ Amount of the water flowing= 1\(\times\)b\(\times\)h = 50\(\times\)45\(\times\)2 = 4500 m³

Given number is 121.
Now, we subtract successive odd numbers starting from 1 as follows:
121 - 1 = 120,
117 - 5 = 112,
120 - 3 = 117,
112 - 7 = 105,
105 - 9 = 96, 96 -11 = 85,
85 -13 = 72, 72 -15 = 57,
57-17 = 40, 40 - 19 = 21, 21 - 21 = 0
We observe that the number 121 reduced to zero after subtracting first 11 odd numbers. So, 121 is a perfect square.
\(\therefore \ \sqrt { 121 } =11\)
Hence, the square root of 121 is 11.

Let the parking charges be x and time be y.
If x = Rs. 60 and y = 4 h, then
\(\frac { x }{ y } =\frac { 60 }{ 4 } =\frac { 60\div 4 }{ 4\div 4 } =\frac { 15 }{ 1 } \) [HCF of 60 and 4 = 4]
If x = Rs.100 and y = 8 h, then
\(\frac { x }{ y } =\frac { 100 }{ 8 } =\frac { 100\div 4 }{ 8\div 4 } =\frac { 25 }{ 2 } \) [HCF of 100 and 8 = 4]
If x = Rs.140 and y = 12 h, then
\(\frac { x }{ y } =\frac { 140 }{ 12 } =\frac { 140\div 4 }{ 12\div 4 } =\frac { 35 }{ 3 } \) [HCF of 140 and 12 = 4]
If x = 180 and y = 24 h, then
\(\frac { x }{ y } =\frac { 180 }{ 24 } =\frac { 180\div 12 }{ 24\div 12 } =\frac { 15 }{ 2 } \) [HCF of 180 and 24 = 12]
Since,all the values of ratio \(\frac { x }{ y } \) are not same, so the parking charges are not in direct proportion to the parking time.

We have, \(\frac{3^{-5}\times 10^{-5} \times 125}{5^{-7}\times 6^{-5}}\)
=\(​​\frac{(3)^{-5}\times (2\times5)^{5}\times 125}{(5)^{-7}\times(2\times3)^{-5}}\)         \([\because (a\times b)^{m}=a^{m}\times b^{m}]\)
=\(\frac{(5)^{-5}\times (5)^{3}}{(5)^{-7}}\)   [\(\because 125=5\times5\times5=5^{3}\)]
=\(\frac{5^{7}\times 5^{3}}{5^{5}}\)    [\(\because a^{-m}=\frac{1}{a^{m}}\)]
=\(\frac{5^{7+3}}{5^{5}}\)  [\(\because a^{m}\times a^{n}=a^{m+n}\)]
=\(\frac{5^{10}}{5{5}}=5^{10-5}=5^{5}\)     [\(\because \frac{a^{m}}{a^{n}}=a^{m-n}\)]

We have,
\({3\over7}\times{28\over15}\div{14\over5}={3\over7}\times{28\over15}\times{5\over14} \)
\(\\{3\over7}\times({28\over15}\times{5\over14})={3\over7}\times{2\over3}={2\over7}\)

313.3%

The probability of getting a comics = \(\frac{Number\quad of\quad events\quad of\quad getting\quad a\quad comics}{Total\quad number\quad of\quad events}=\frac{3}{8}\)

We have, -7 pq²r³ x (-26 p³ qr²)
= (-7) x (- 26) x p x q² x r³ x p³ x q x r²
=182p⁴ q³ r⁵

Number of faces is equal to 14

We know that in a ·rectangle, opposite sides are parallel and D equal and each angle is of 90°. Firstly, we draw a rough sketch of rectangle PQRS, which helps us in deciding steps of construction.

Steps of construction
Step I Draw PQ = 5 cm.
Step II At Q, draw a ray QX making \(\angle \)PQX = 90°.
Step III Cut QR = 4 cm from ray QX.
Step IV At P, draw a ray PY making \(\angle \)QPY = 90°.
Step V Cut PS = 4 cm from PY.
Step VI Join SR.

Thus, PQRS is the required rectangle.

Given,PQRS is a trapezium in which
ㄥQ = 130°, ㄥR = 90° and \(\bar { SP } ||\bar { RQ } \)
Since, \(\bar { SP } ||\bar { RQ } \)  and PQ is a transversal, so
ㄥP + ㄥQ = 180°
[∵ interior angles on the same side]
⇒ ㄥP + 130° =180°
⇒ ㄥP = 180° - 130° = 50°
Similarly, ㄥS + ㄥR = 180° [∵ SR is a transversal]
⇒  ㄥS + 90⁰=180°
⇒ ㄥS = 180° - 90° = 90°
Hence, the measure of ㄥP = 50° and ㄥS = 90° .
We may find ㄥP by one more method, which is given below: We know that, the sum of all the angles of a quadrilateral is 180°.
∴ ㄥP + ㄥQ + ㄥR + ㄥS = 360°
⇒ ㄥP +130° +90° +90° = 360°
⇒ ㄥP +310° = 360°
⇒ ㄥP = 360° - 310° = 50°
Hence, the measure of ㄥP is 50°.

Curved surface = 2 x \(\frac { 22 }{ 7 } \times 14\times 5\)
= 440 cm²

\(n = 3,{n\over2}=2\)

8-7=1

7x², 5x, - 5.

14a²b = 2 x 7 x a x a x b
35a⁴b² = 5 x 7 x a x a x a x a x b x b.

Loss = 10000 - 8000 = 2000
∴ Loss% = \(\frac { 2000 }{ 10000 } \times \) 100% = 20%.

1296 = 2\(\times\) 2\(\times\) 2 \(\times\) 2 \(\times\) 3 \(\times\) 3 \(\times\) 3 \(\times\) 3
= 23 \(\times\) 2\(\times\) 33 \(\times\) 3.

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.

 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