Yes ; \((-4) \times [(-8) \times (-5)] = [(-4) \times (-8)] \times (-5) = 160 \)

\((-4)^{-3} \times(5)^{-3} \times(-5)^{-3}=[(-4) \times 5 \times(-5)]^{-3}=[100]^{-3}=\frac{1}{100^{3}}\)

22 km

19mn² (mn - 3)

Let side of a square be a cm then its perimeter be 4a cm.
If we denote perimeter = P, then P = 4a
where, 4 is constant
∴ \(P\propto a\)
Hence, side of a square is directly proportional to its perimeter.

Here, the remainder = 0. So, N is an even number, i.e. its one's digit is even. Therefore, the one's digit must be 0, 2, 4, 6 or 8.

4913

Not perfect square

1050 m²

\(\therefore\) Total number of outcomes = 5
\(\therefore\) Required probability = \(\frac{2}{5}\)
Hence, option (a) is correct.

5.1 x 5.2= (5+ 0.1) (5+ 0.2)
= (5)² + (0.1 + 0.2)(5) + (0.1) (0.2)
(x + a)(x + b) = x² + (a + b)x + ab
= 25 + 1.5+ 0.02 = 26.52

Rs 46656

35 km

Number of sides of first polygon = 4
Number of sides of second polyqon = 12

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.

64cm³

Let the length of the diagonal BD be x cm
\(\because\) Area of a quadrilateral  \(=\frac{1}{2}\times(diagonal)\times\) (Sum of the length of perpendiculars on the diagonal from the opposite vertices)
\(\therefore\) Area of the quadrilateral ABCD
\(=\frac{1}{2}\times BD\times(AP+CQ)\)
\(=\frac{1}{2}\times x\ cm\times(4.5\ cm+3.5\ cm)\)
\(=\frac{1}{2}\times x\times 8\ cm^2\)
Since area of the quadrilateral ABCD = 32 cm²
\(\therefore \frac{1}{2}x\times 8=32\Rightarrow x=\frac{32\times2}{8}\ cm=8\ cm\)
Thus, the required length of the diagonal BD = 8 cm.

5.463x10³

Suppose the man was absent on x days. Then,
he worked for (28 - x) days.
Thus, he will get the amount as per the given
condition for the February month
(28 - x) x 500 - x x 100 = 9200
\(\Rightarrow\) 28 x 500 - 500x -100x= 9200
\(\Rightarrow\) 600x = 9200 -14000 \(\Rightarrow\) -600 x = -4800
\(\Rightarrow\) x = -4800 x \(\left( -\frac { 1 }{ 600 } \right) =8\)
So, the man works for (28 - 8) i.e. 20 days

Number of observations = 4x²+9 

\(\begin{matrix} \quad \quad A\ B \\ \quad \times \ \ \ \ \ 3 \\ \_ \_ \_ \_ \_ \_ \_\_ \\ \quad C\ A\ B \\ \_\_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have three letters A, Band C whose values all to be found.
Since unit's digit of 3 x B is B. So, it must be B = 0 or B = 5
When B = 0, then
\(\begin{matrix} \quad \quad A\ 0 \\ \quad \times \quad 3 \\ \_ \_ \_ \_ \_ \_ \_ \\ C\ A\ 0 \\ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
and when B = 5, then
\(\begin{matrix} \quad \ A\ 5 \\ \quad \times \ 3 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ C\quad A\quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Now, unit's digit of 3 x A is A. So, A must be 0 or 5 but A cannot be 0 because if A = 0, then AB becomes the one-digit number. So, A must be 5 and multiplication is either 50 x 3 or 55 x 3.
Here, the second possibility fails, since 55 x 3 = 165 but the first possibility is correct.
\(\therefore\)  50 x 3 = 150
Therefore, the puzzle is solved as shown below:
\(\begin{matrix} \quad \quad 5\quad 0 \\ \quad \times \quad 3 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ 1\quad 5\quad 0 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 5, B = 0 and C = 1

\(-3{1\over3}\)

When x = 20,y= 40, then
\(\frac { x }{ y } =\frac { 20 }{ 40 } =\frac { 20\div 20 }{ 40\div 20 } \) [.: HCF of 20 and 40 = 20]
\(\Rightarrow \frac { x }{ y } =\frac { 1 }{ 2 } \)
When x = 17,y = 34, then
\(\frac { x }{ y } =\frac { 17 }{ 34 } =\frac { 17\div 17 }{ 34\div 17 } =\frac { 1 }{ 2 } \) [HCF of17 and 34 = 17]
When x = 14,y = 28, then
\(\frac { x }{ y } =\frac { 14 }{ 28 } =\frac { 14\div 14 }{ 28\div 14 } =\frac { 1 }{ 2 } \) [HCF of 14 and 28 = 14​​​​​​​]
When x = 11,y = 22, then​​​​​​​
\(\frac { x }{ y } =\frac { 11 }{ 22 } =\frac { 11\div 11 }{ 22\div 11 } =\frac { 1 }{ 2 } \) [HCF of 11 and 22=11]​​​​​​​
When x = 8,y = 16,then​​​​​​​
\(\frac { x }{ y } =\frac { 8 }{ 16 } =\frac { 8\div 8 }{ 16\div 8 } =\frac { 1 }{ 2 } \) [HCF of 8 and 16=8]
When x = 5,y = 10,then​​​​​​​
\(\frac { x }{ y } =\frac { 5 }{ 10 } =\frac { 5\div 5 }{ 10\div 5 } =\frac { 1 }{ 2 } \) [HCFof 5and10=5]
When x = 2,y = 4, then​​​​​​​
\(\frac { x }{ y } =\frac { 2 }{ 4 } =\frac { 2\div 2 }{ 4\div 2 } =\frac { 1 }{ 2 } \) [HCF of 2 and 4 = 2]
From above,we can say that the values of​​​​​​​ \(\frac { x }{ y } \) is same​​​​​​​,in everycaseit is​​​​​​​ \(\frac { 1 }{ 2 } \) So, these values of x and y are directly proportional​​​​​.

Given,\(\ x+\frac { 1 }{ x } =5\)
\(\ \left( x+\frac { 1 }{ x } \right) ^{ 2 }=25\)
\(\left( x+\frac { 1 }{ x } \right) ^{ 2 }={ x }^{ 2 }+2\times x\times \frac { 1 }{ x } +\left( \frac { 1 }{ x } \right) ^{ 2 }\quad \)
(a+ b)² =a² + 2ab + b². Here, a = x and \(b=\frac { 1 }{ x } \)
\(={ x }^{ 2 }+2+\left( \frac { 1 }{ { x }^{ 2 } } \right) ={ x }^{ 2 }+\left( \frac { 1 }{ { x }^{ 2 } } \right) +2\)
\(\left( x+\frac { 1 }{ x } \right) ^{ 2 }=25\)
\(\quad { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2=25\)
\({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =25-2=23\)

30

\(\frac { 401 }{ 607 } \)

Let the regular polygon have n sides.
∴ Exterior angle =\(\frac { { 360 }^{ 0 } }{ n } \)
and interior angle = \({ 180 }^{ 0 }-\frac { { 360 }^{ 0 } }{ n } =\frac { 180^{ 0 }n-{ 360 }^{ 0 } }{ n } \)
According to the question,
\(\frac { { 360 }^{ 0 } }{ n } ;\frac { { 180 }^{ 0 }n-{ 360 }^{ 0 } }{ n } \)=1:3
⇒ \(\frac { { 360 }^{ 0 } }{ n } \times \frac { n }{ { 180 }^{ 0 }n-{ 360 }^{ 0 } } =\frac { 1 }{ 3 } \)
⇒ 360⁰ x 3 =180⁰n-360⁰
⇒ 1080⁰+360⁰=180⁰n
⇒ \(\frac { { 1440 }^{ 0 } }{ 180^{ 0 } } \)=n  ⇒ n=8
∴ Exterior angle = \(\frac { { 360 }^{ 0 } }{ n } =\frac { 360^{ 0 } }{ 8 } \)=45⁰
and interior angle = 180⁰-45⁰=135⁰

Letcost price of Ist buffalo = Rs x
Given,gain per cent on Ist buffalo = 5%
∴ Gain amount on Ist buffalow = 5% of cost price of I st buffalo
= 5% of Rs x=Rs\((\frac{5}{100}\times x)\)=Rs \(\frac{5x}{100}\)
Selling price of Ist buffalo = Cost price of Ist buffalo + Gain amount on Ist buffalo
= Rs \((x+\frac{5x}{100})\)
= Rs \((\frac{100x+5x}{100})=Rs \frac{105x}{100}\)
Butselling price ofIst buffalo = Rs 20000
∴ \(\frac{105x}{100}\) = 2000 ⇒ 105x = 20000ｘ100
⇒ x = \(\frac{20000\times100}{105}⇒ x=\frac{400000}{21}\)
⇒ x = Rs 19047.62
Letcostprice ofIInd buffalo be Rs y.
Given,lossper cent on IInd buffalow = -10%.
Lossamount on IInd buffalo = 10% of cost price of IInd buffalo
= 10% of Rs y =Rs \((\frac{10}{100}\times y)\)=\(\frac{Rs 10 y}{100}\)
Sellingprice of IInd buffalo = Cost price of IInd buffalo - Loss amount on the buffalo
=Rs \((y-\frac{10 y}{100})=Rs (y-\frac{10 y}{100})\)
=Rs \(\frac{100y-10y}{100}=Rs \frac{90 y}{100}\)
But selling price ofIInd buffalo = Rs 20000
∴ \(\frac{90 y}{100}\)=20000 ⇒ 90y =20000ｘ100
⇒ \(y=\frac{20000\times100}{90}\)
⇒ y=\(\frac{200000}{9}\)
⇒ y=Rs 22222.22
∴  Total cost price of both buffaloes = Rs (x + y)
= Rs 19047.62 + Rs 22222.22 = Rs 41269.84
∴ Total selling price of both buffaloes = Rs 20000 + Rs 20000
= Rs 40000
∴ SP< CP
∴ Loss amount on both buffaloes
Cost price of both buffaloes - Selling price of both buffaloes
= Rs 41269.84 - Rs 40000
= Rs 1269.84
Hence, overall loss is Rs 1269.84.

Firstly, draw a rough sketch of quadrilateral MORE, which helps us in deciding steps of construction.
Steps of construction
Step I Draw MO = 6 cm.
Step II At O, draw a ray OX making \(\angle \)MOX = 105°.
Step III Cut OR = 4.5 cm on ray OX.
Step IV At M, draw a ray MY making \(\angle \)OMY =60°.
Step V At R, draw a ray RZ making \(\angle \)ORZ = 105°, which meets the ray MY at E.

PLAN is the required quadrilateral.

Here, comparison between percentage wins in ODI by 8 top cricket teams is given, so to represent the given information, we draw the double bar graph. So, draw two perpendicular axes OX and OY. Take teams on OX and percentage wins in ODI on OY Choose a suitable scale for OX and OY to draw double'bar graph.
Thus, we get the following double bar graph:

\(\because\) We have \(({4\over 5})^3={4 \times4 \times4 \over 5\times 5\times 5}\)
\(=[\because ({m\over n})^3={m^3\over n^3}]={64\over 125}\)
\(Thus\ ({4\over5})^3={64\over 125}\)

In this case, d = 5.5 cm, h1 = 2.5cm, h2 = 1.5 cm,
\( \text { Area } =\frac{1}{2} d\left(h_{1}+h_{2}\right) \)
\(=\frac{1}{2} \times 5.5 \times(2.5+1.5) \mathrm{cm}^{2} \)
\(=\frac{1}{2} \times 5.5 \times 4 \mathrm{~cm}^{2}=11 \mathrm{~cm}^{2}\)

It is given that a and b are in inverse proportion.
Hence, product of a and b will be a constant.
In case I, ab = 12 \(\times\) 30 = 360
In case II, \(5\times x=360 \Rightarrow x=\frac{360}{5}=72\)
in case III 
\(8\times y=360 \Rightarrow y=\frac{360}{8}=45\)
Hence, the required value of x = 72 and y = 45.

We have, 14pq = 2 x 7 x p x q 
and 28p²q² = 2 x 2 x 7 x p x p x p x q 
The two terms have 2,7, P and q as common factors .
So, common factor of given terms
= 2 x 7 x p x q = 14 pq 

Given, \(\frac{5^{m}\times5^{3}\times5{-2}}{5^{-5}}=5^{12}\)
⇒ \(\frac{5^{m+3-2}}{5^{-5}}=5^{12}\)
⇒ 5^m+1+5 = 5¹²
⇒ (5)^m+6 = (5)¹²
⇒ m + 6 = 12   [ ∵  bases are same]
⇒ m= 12 - 6 = 6

(a) 54
(b) 108

we have \(\frac { \left( 2+y \right) \left( 7-y \right) }{ \left( 5-y \right) \left( 4+y \right) } =1\)
By cross-multiplication, we get
(2 + y)(7 - y) =(5 - y)(4 + y)
\(\Rightarrow\)   14 - 2y + 7y - y² = 20 + 5y - 4y - y²
\(\Rightarrow\) 14 + 5y = 20 + Y   \(\Rightarrow\) 5y - y = 20-14
\(\Rightarrow\) 4y = 6   \(\Rightarrow\) y = \(\frac { 6 }{ 4 } =\frac { 3 }{ 2 } \)
Thus, solution of the given equation is \(\frac { 3 }{ 2 } \)

Cuboid

15

(3x + 2y)² =(3x)² + (2y)² + 2 x(3x)(2y)
= 9x² + 4y² + 12xy
Similarly, (3x - 2y)² = 9x² + 4y² -12xy
(3x + 2y)² + (3x - 2y)²
= (9x² + 4y² + 12xy) + (9x² + 4y² -12xy)
=18x²+8y²

Firstly, we draw a rough sketch of kite EASY, which helps us in deciding steps of construction.

We know that, a kite has two distinct consecutive pairs of sides of equal length and its both diagonals are perpendicular to each other.
So, EY = AE = 4 cm and SY = SA = 6 cm
In \(\Delta\) AYE , EY +AE= 4+4=8 =AY
Since, the sum of the lengths of any two sides of a triangle must be greater than the third side, so \(\Delta\) AYE is not possible.
Hence, construct of kite EASY with given sides is not possible.

Given, cost of one bar of soap = Rs 35
∴ Cost of two bars of soap = Rs (35 x 2) =Rs 70
Also, sale tax = 5%
Now, sale tax charged = 5% of cost of two bars of soap
= 5% of Rs 70
= Rs \((\frac{5}{100}\times 70)=\) Rs \(\frac{350}{100}\) = Rs 3.50
∴ Buyingprice of two bars of soap = Cost of two bars of soap + Sale tax
= Rs (70 + 350) = Rs 73.50
Hence, the buying price of two bars of soap is Rs 73.50.

Given, three cut-outs of congruent triangles of sides 3 cm, 4 cm and 5 cm. On arranging them, we get a trapezium from the given figure, we have

ㄥDEC = ㄥECB = 90°    [alternate angles]
∴  DEIIBC
and DE = BC =3 cm
Also, EB = DC = 5 cm
So, EBCD is a parallelogram.
∴ EBIIDC
Also, AEB is a straight line, so AEB II DC. Therefore, in trapezium ABCD, sides AB and DC are parallel and non-parallel sides AD and BC are unequal. Also, non-parallel sides of a trapezium mayor may not be equd By using the same set of triangles, we can get two more trapeziums.

\(\frac { (12+8)d }{ 2 } =40\Rightarrow d=4cm\)

2002 \(\rightarrow \) 300
2003 \(\rightarrow \) 500
500-300=200

36 = 6 x 6 = 6²

\({7\over 875}={16\over ?}⇒?={875\times16\over 7}=2000\)

6x² - 5x - 6
= 6x² - 9x + 4x - 6
= 3x(2x - 3) + 2(2x - 3)
= (2x - 3)(3x + 2).

Number of Science students
= 340 \(\times \frac { 55 }{ 100 } \) = 187.
.. Number of students of Commerce
= 340 - 187 = 153.

Required measure = \({360^\circ\over 9}=40^\circ\)

Lowest frequency = 2 ⟶ 325-360.

x -1, x, x + 1

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