We have 

This shows that 22² < 525.
Next perfect square is 23² = 529
Hence, the number to be added is 23²- 525 = 529 - 525 = 4
Therefore, the perfect square so obtained is 525 + 4 = 529
Hence, \(\sqrt { 529 } \)= 23

Resolving 288 in to prime factors
we have
i.e., 288 = 2 x 2 x 2 x 2 x 2 x 3 x 3
Grouping the factors in triples, we get
288 = [2 x 2 x 2] x 2 x 2 x 3 x 3

We observe that if 288 is multiplied by (2 x 3),then its prime factors will exist in triples.
Thus, the required smallest number by which 288 be multiplied to make it a perfect cube is (2 x 3), i.e., 6.

Since, the sum of the measures of interior angles of a quadrilateral is 360°.

Also, 115° + 70° + 60° = 245°
\(\therefore\)245° + LABC = 360°
\(\Rightarrow\)\(\angle \)ABC = 360° - 245° = 115°
Now, x = ext. \(\angle\)BCD = 180° - \(\angle\)BCD
= 180° - 115° = 65°
Similarly, y = 180° - 70° = 110°
z = 180° - 60° = 120°
w = 180° - 115° = 65°
\(\therefore\)x +y + z + w = 65° + 110°+ 120° + 65° = 360°

Steps of construction:

I. Draw AB = 6.3 em.
II. At B, draw \(\overrightarrow { BX } \) , such that \(\angle \)ABX = 90°
III. From \(\overrightarrow { BX } \), cut off BC = 6.3 cm.
IV. With centre C and radius = 6.3 cm, draw an arc.
V. With centre A and radius = 6.3 cm, draw another arc to intersect the previous arc at D.
VI. Join DA and CD.
Thus, ABCD is the required square.

Since \({ \left( \frac { 1 }{ 4 } \right) }^{ -2 }={ \left( \frac { 4 }{ 1 } \right) }^{ 2 }={ (4) }^{ 2 }=16\)
\({ \left( \frac { 1 }{ 3 } \right) }^{ -3 }={ (3) }^{ 3 }=3\times 3\times 3=27\)
\({ \left( \frac { 1 }{ 2 } \right) }^{ -4 }={ (2) }^{ 4 }=2\times 2\times 2\times 2=16\)
\(\therefore \quad { \left( \frac { 1 }{ 4 } \right) }^{ -2 }+{ \left( \frac { 1 }{ 3 } \right) }^{ -3 }+{ \left( \frac { 1 }{ 2 } \right) }^{ -4 }=16+27+16=59\)

5km/h

We have \(\frac { 1 }{ 36 } { a }^{ 2 }{ b }^{ 2 }-\frac { 16 }{ 49 } { b }^{ 2 }{ c }^{ 2 }\)
= \({ b }^{ 2 }\left( \frac { { a }^{ 2 } }{ 36 } -\frac { 16{ c }^{ 2 } }{ 49 } \right) ={ b }^{ 2 }\left( \frac { { a }^{ 2 } }{ 6\times 6 } -\frac { 4\times 4 }{ 7\times 7 } { c }^{ 2 } \right) \)
= \({ b }^{ 2 }\{ \left( \frac { a }{ 6 } \right) ^{ 2 }-\left( \frac { 4c }{ 7 } \right) ^{ 2 }\} \)
= \({ b }^{ 2 }\left( \frac { a }{ 6 } +\frac { 4c }{ 7 } \right) \left( \frac { a }{ 6 } -\frac { 4c }{ 7 } \right) \)
[by using identity, (a² - b²) = (a + b) (a - b)]

The line graph of an annual gross profit of a company for a period of five years are given below:

\(\begin{matrix}\quad 4\quad A \\ +\quad 9\quad 8 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad C\quad B\quad 3 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have three letters A, B, and C whose values are to be found.
Studying the addition in the one's column, we have A + 8 and we get 3 from this, i.e. a number whose one's digit is 3, for this A has to be 5.
\(\because\) A + 8 = 5 + 8 = 13
Now, for sum in ten's column, we have
1 + 4 + 9 = CB \(\Rightarrow\) 14 = CB
Here, B = 4 and C = 1
Therefore, the puzzle is solved as shown below:
\(\begin{matrix}\quad \quad 4\quad 5 \\ +\quad 9\quad 8 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad 14\quad 3 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 5, B = 4 and C = 1

See the figure given below to understand the solution:

It is clear that the length of the largest pole is AF, which is forming a hypotenuse of right angle ΔACF.
Thus, AC =\(\sqrt { { AB }^{ 2 }+BC^{ 2 } } =\sqrt { 12^{ 2 }+4^{ 2 } } =\sqrt { 160 } \)m
In ΔACF ,
AF =\(\sqrt { { AC }^{ 2 }+CF^{ 2 } } =\sqrt { 160+3^{ 2 } } =\sqrt { 169 } \)=13 m
Alternate Method
Length of the largest pole = Length of diagonal of the cube
\(\sqrt { { l }^{ 2 }+b^{ 2 }+h^{ 2 } } =\sqrt { { 12 }^{ 2 }+{ 4 }^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 169 } \)

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

We know that in a pack of well-shuffled cards. total cards= 52
Number of ace of club = 1
The probability of getting an ace of club = \(\frac{1}{52}\)

Given, cost price of 80 articles = Rs 2400
and profit per cent = 16%
∴ Profit = 16% of cost price of BO articles
= Rs \((\frac{16}{100}\times 2400)= \)Rs (16ｘ24)= Rs 384
Now, selling price of BO articles = Cost price of BO articles + Profit
= Rs(2400 + 384) = Rs 2784
ஃ Selling price of 1 article = \(\frac{2784}{80}\) =  Rs 34.80
Hence, selling price of one article is Rs 34.80.

The numerical factor of a term is called its numerical coefficient or simply coefficient. In given expressions, terms and corresponding coefficients are given in the following table

We have \(\frac { x }{ 2 } -\frac { 1 }{ 5 } =\frac { x }{ 3 } +\frac { 1 }{ 4 } \)
The denominators on both sides are 2,5,3 and 4. Their LCM is 60.
Multiplying both sides of the given equation by 60, then we get 60 \(\left( \frac { x }{ 2 } -\frac { 1 }{ 5 } \right) =60\left( \frac { x }{ 3 } +\frac { 1 }{ 4 } \right) \)
\(60\times \frac { x }{ 2 } -60\times \frac { 1 }{ 5 } =60\times \frac { x }{ 3 } +60\times \frac { 1 }{ 4 } \)
30x - 12 = 20x +15
30x - 20x = 15 + 12
10x = 27
x = \(\frac { 27 }{ 10 } \)
Hence x =\(\frac { 27 }{ 10 } \) is the solution of the given equation.

1.6 \(\times\) 10^-3

5 (2x+ 1)(3x+ 5) ÷ (2x+ 1)
=\(\frac{5(2x+1)(3x+5)}{(2x+1)}=5(3x+5)\)

Not direct

Yes, he stops between 10 am to 10: 30 am.

The given number is 294.
Sum of digits = 2 +9 + 4 = 15
Now, 15 \(\div\) 9 = 1 and remainder = 6
which is not divisible by 9.
Hence, 294 is not divisible by 9.

36

Given, dimensions of cuboid are 60 cm\(\times\)54 cm\(\times\)30 cm
i.e.length= 60 cm, breadth = 54 cm and height = 30 cm
∴ Volume of the cuboid = l \(\times\)b \(\times\)h = 60\(\times\)54 \(\times\)30
 = (60\(\times\)54\(\times\)30)cm³
Also,givenside of a small cube = 6 cm
∴ Volume of one cube = a³ = (6)³ = (6\(\times\)6 \(\times\)6) cm³
Here,small cubes are placed in cuboid, required number of smallcubes
=\(\frac { Volume\ of\ cuboid }{ Volume\ of\ one\ cube } =\frac { 60\times 54\times 30 }{ 6\times 6\times 6 } \)  = 450

Using the following pattern for finding the square of the given number, we get
           1² = 1
          11²=121
       111² = 12321
      1111² = 1234321
    11111² = 123454321
  111111² = 12345654321
 1111111² = 1234567654321
1111111² = 1234567654321

Rs 3200

(2pq + 3q²)\(\times\)(3pq - 2q²)
= 2pq(3pq - 2q²) + 3q²(3pq - 2q²)
= (2pq\(\times\)3pq) - (2pq\(\times\)2q²) + (3q²\(\times\)3pq)-(3q²\(\times\)2q²)
= 6p²q² -4pq³ + 9pq³ -6q⁴
= 6p²q² + (-4 + 9)pq³ - 6q⁴
= 6p²q² + 5pq³ - 6q⁴

\(A={\frac{-15}{8}},B={-14\over8},c={-11\over8},D={-8\over8}E={-7\over8}\)

Yes, to construct a unique parallelogram whose two adjacent sides are given, we need the angle included between them, because if angle is not given, then we cannot find a unique parallelogram. So, we cannot construct the parallelogram with the given information.

30 cm

9,10,11

108⁰

(a) Length of the cube
= Width of the cube
= Height of the cube
= l
(b) Area of each face = l x l  = l²
(c) Total surface area of this cube = 6l²
(d) Area of each face = l²
Yes ; we can say that the total
surface area of a cube of side l is 6l²

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.

Case of Simple Interest
[P = Rs 1000, r = 8% p.a.]

\(\because\) In each case the ratio \(\frac { SI }{ T } \) is the same.
\(\therefore\) The simple interest changes in direct proportion with time period.
Case of compound Interest [P = Rs 1000, r = 8% p.a.]

\(\because \ \frac { CI }{ T } \) is not the same in each case.
\(\therefore\)  The compound interest does not change in direct proportion with time period. 

The value depicted here is that if we make a plan on expenditure, then we can save some money,
 

Number of chocolates Abhinav bought = (x + 4)
Cost of one chocolate = Rs.(x + 4)
Total amount paid by him = Number of chocolates \(\times\) Cost of one chocolate
= Rs. ( x + 4) x (x + 4) = Rs.(x + 4)²
= Rs.{(x² + (4)² + 2 \(\times\) x \(\times\) 4))
[using identity, (a + b)² = (a² + b² + 2ab)]
= Rs.(x² + 4\(\times\)4+ 8x)
= Rs.(x² + 16 + 8x)
It is given that, if x = 10 then,
the amount paid by him
= Rs. {(10)² + 16 + 8 \(\times\)10}
= Rs.(10 x10 + 16+ 8 \(\times\)10)
= Rs.(100+16+ 80)=Rs.196
(ii) The value depicted here is that Abhinav is helpful and friendly by nature to his friend.

We have, a 3-digit number 2a3, when added to the number 326, gives a 3-digit number 5b9.
We can write 2a3 and 5b9 in generalised form as.
2a³ = 2 x 100 + a x10 + 3 = 200 + 10a + 3
5b9 = 5 x 100 + b x 10 + 9 = 500 + 10b + 9
Now, it is given that
2a3 + 326 = 5b9
200 + 10a + 3 + 326 = 500 + 10b+ 9
10a + (200 + 3 + 326) = 10b + (500 + 9)
10a + (200 + 329) = 10b+ 509
10a + 529 = 10b + 509
529 - 509 = 10b -10a
20 = 10 x (b-a)
b - a = \({20 \over 10}\)
b - a = 2                        .........(i)
Now, it is given that 5b9 is divisible by 9.
So, sum of digits of 5b9 will be divisible by 9, i.e
5 + b + 9 = Multiple of 9
14 + b =18 (nearest multiple of 9)
b = 18 - 14 = 4
Put b = 4 in Eq. (i), we get
b - a = 2 \(\Rightarrow\)4-a = 2
4 - 2 = a \(\Rightarrow\) 2 = a
\(\Rightarrow\) a = 2
Hence, a = 2, b = 4 and b - a = 2

Let side of two squares be a₁ and a₂·
Then, perimeter of first square = 4a₁
According to the question,
 Perimeter = 40
\(\therefore\) 4a₁ = 40 \(\Rightarrow\) a₁ = 10
Similarly, second square's perimeter = 4a₂
But according to the question,
Perimeter = 96
\(\therefore\) 4a₂ = 96 \(\Rightarrow\) a₂ = 24
Let a be the side of another square.
Now, sum of the areas of first and second square area of
= Area of (first square) + (Area of second square)
=\({ a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }\)
Area =(10)² + (24)² =100 + 576
\(\therefore\) Area = 676
\(\Rightarrow\)a² = 676 \(\Rightarrow\) a = \(\sqrt { 676 } \)
Then, a=26 m
Another square's perimeter = 4a = 4 x 26 = 104 m
So, perimeter of another square is 104 m.

729 cm³

2

RHS =(9982)² -(18)²
= (9982 + 18) (9982 - 18)
[∴ a²- b² = (a + b) (a - b)]
= 10000 x 9964
where, LHS = (10000) x y
On comparing LHS and RHS, we get
10000y=10000x9964
\(y=\frac { 10000\times 9964 }{ 10000 } =9964\)

Central angle for oceans = \((\frac{Value\quad of\quad the\quad component}{Sum\quad of\quad the\quad components}\times 360)^o\)
For pacific ocean = 40% = \(\frac{40}{100}\times 360^o\) = 144^o
For atlantic ocean = 30% = \(\frac{30}{100}\times 360^o\) = 108^o
For Indian ocean = 20% = \(\frac{20}{100}\times 360^o\) = 72^o
For others ocean = 10% = \(\frac{10}{100}\times 360^o\) = 36^o
On the basis of above data we can draw the following pie chart:

(a) The quadrilateral park have kite shape, since PA = AR and PK = RK
(b) In a kite, the diagonals are perpendicular to each other, so \(\angle \)AOR = 90°.
(c) The park is having a well structured as kite shape with a running track for the runner.

Let ABGD is a rhombus

Then, BD = 8 cm, AC =15 cm
and AB = BG = CD = DA
We know that, in a rhombus, diagonals bisect each other at right angles.
So, DO = OB and AO = OC
∴ DO =4 cm and OC = 7.5 cm
Now, we see that a /lDOC is formed such that
ㄥDOC = 90°
∴ (DC)² = (DO)² + (OC)²
[∵ in a right angled triangle, the square of the side opposite to the right angle is equal to the sum of the squares of other two sides]
(DC)² =(4)² + (7.5)² = 16+ 56.25
= 72.25 cm²
or DC = \(\sqrt { 72.25 } \) = 8.5 cm
Thus, side of the rhombus is 8.5 cm.

Here, P = Rs 8000, T = 1 year, R = 9% p.a.
Interest is compounded half yearly,
\(\therefore\)T = 1 year = 2 half years
R = 9% p.a =\(9\over2\)% half yearly
\(\therefore\)Amount = p\((1+{R\over100})^n\)
= Rs 8000 x\((1+{9\over200})^2\)
= 8000 x \({209\over200}\times{209\over200}\) =Rs \({2 \times 209 \times 209\over 10}\)
= Rs \(87362\over10\)= Rs 8736.20
CI = Rs  8736.20 - Rs 8000 =Rs 736.20

(a) The shaded (coloured) map according to the question is as follows:

(b) Intersection of road C and Nehru road is marked by X and intersection of Gandhi road and road A is marked by Yin map given in (a) part.
(c) Short street route from library to the bus depot is Library ⟶ Nehru road ⟶ Road C ⟶ Bus depot.
(d) City Park is in the East.
(e) Sr. Secondary School is in the South.

Let the number of Rs. 5 coins be x.
Then, the number of Rs. 2 coins = 3x
The total number of coins is 160 .
The number of coins of Rs. 1= 160 - (x + 3x)
= (160 - 4x)
The amount that I have from Rs.  5 coins = 5 x x = 5x
The amount that I have from Rs.  2 coins = 2 x 3x = 6x
The amount that I have from Rs.  1 coins
= 1 x (160 - 4x) = 160 - 4x
According to the question,
Total amount = 300
\(\Rightarrow\) 5x + 6x + (160 - 4x) = 300
\(\Rightarrow\) 5x +6x +160 - 4x = 300
\(\Rightarrow\) 7x + 160 = 300   [transposing 160 to RHS]
\(\Rightarrow\) 7x = 300 -160  [transposing 160 to RHS]
\(\Rightarrow\) 7x = 140
\(\Rightarrow\) x = \(\frac { 140 }{ 7 } \) = 20   [dividing both sides by 7]
Number of Rs. 5 coins = x = 20
Number of Rs. 2 coins = 3x = 3 x 20 = 60
and number of Rs.1 coins = 160 - 4x = 160 - 4 x 20
= 160 - 80 = SO
Hence, I have SO,60 and 20 coins of denomination Rs.1, Rs. 2 and Rs 5, respectively.

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

7 x 7 = 49

0 \(\rightarrow \) (0,0)

Area = (2l²m)(3lm²) = 6l³m³.

(2⁰+4^-1)x2² = \(\left( 1+\frac { 1 }{ 4 } \right) \)x4 = \(\frac { 5 }{ 4 } \)x4 = 5

\({120\over 600}={400\over ?}⇒?=2000\)

12a² b + 15ab² = 3ab(4a + 5b).

One's digit of 125 is 5.

100 - 20 = 80
∵ If C.P. is 100, then S.P. = Rs. 80
∴ If C.P. is Rs. 500, then S.P.
= \(\frac { 80 }{ 500 } \times \)500 = Rs.400.

Required measure = \({360^\circ-(110^\circ+100^\circ)\over 2}=75^\circ\)

3y + 4 = 5y - 4 \(\Rightarrow\) 5y - 3y = 4 + 4
\(\Rightarrow\) 2y = 8   y=\(\frac{8}{2}\)=4

Required number = 5 + 2 + 4 + 6 + 7 + 5 = 29.

3 \(\times\)3 \(\times\) 3 = 27.