\(-{17\over 36}\)

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.

34500000 - 345 \(\times\) 100000
= 3.45 \(\times\) 100 \(\times\) 100000
= 3.45 \(\times\) 10² \(\times\) 10⁵
= 3.45 \(\times\) 10²⁺⁵
= 3.45 \(\times\) 10⁷
Thus, 34500000 = 3.45 \(\times\) 10⁷

Step 1: Draw D ACD using SSS construction
Step 2: With D as centre, draw an arc of radius 7 cm. (B is somewhere on this arc)
Step 3: With C as centre, draw an arc of radius 4.5 cm (B is somewhere on this arc also)
Step 4: Since B lies on both the arcs, B is the point intersection of the two arcs. Mark B and complete ABCD. ABCD is the required quadrilateral.

At least 4 planes can form to enclose a solid. Tetrahedron is the simple polyhedron. Following figure represents a simplest solid, called tetrahedron.

A tetrahedron has:
4 triangular faces, i.e., F = 4
4 vertices, i.e., V = 4
6 edges, i.e., E = 6
Now, substituting the values of F, V and E in Euler's formula, i.e.
F +V = E + 2
We have
4 +4 = 6 + 2
⇒ 8 = 8
which is true.
Thus, Euler's formula is verified for a tetrahedron.

The line graph for the area of a square as per the given table.

No, it is not a linear graph because when we join the points, then there is not a straight line formed.

p² -l0p + 25 = (p)² -l0p + (5f = p² - 2(5)p + 5²
This expression is of the form a 2 - 2ab + b² .
On comparing, we get a = p and b = 5
Since, a² - 2ab + b² = (a - b)²
\(\therefore\) p² -10p+25=(p-5)² or (p-5)p-5²

Given, 21y 5 is a multiple of 9, i.e. 21y5 is divisible by 9.
We know that a number is divisible by 9 if the sum of its digits is divisible by 9.
So, sum of digits of 21y 5 = 2 + 1+ y + 5 = 8 + y
Therefore, (8 +y) should be 0, 9, 18,27, ... , etc.
Since x is a digit.
\(\therefore\) 8 + y = 9  \(\Rightarrow\) y = 9 - 8 \(\Rightarrow\) y=1
Hence, the value of y is 1.

When x₁ = 50, y₁ = 5, then x₁y₁ = 50 x 5 = 250
When x₂ = 40, y₂ = 6, then x₂y₂ = 40 x 6 = 240
When x₃ = 30, y₃ = 7, then x₃y₃ = 30 x 7 = 210
When x₄ = 20, y₄ = 8, then x₄y₄ = 20 x 8 = 160
It is clear that 250 t:- 240 t:- 210 t:-160
i.e. x₁y₁ ≠ x₂y₂ ≠ x₃y₃ ≠ x₄y₄
So, x and y are not in inverse proportion.

The least number divisible by each one of 8, 15 and 20 is their L.C.M.
The L.C.M. of 8, 15 and 20 is
2\(\times\)2\(\times\)2\(\times\)3\(\times\)5 = 120
Now prime factorisation of 120 is
120 =\(\underline { 2\times 2 } \)\(\times\)2\(\times\)3\(\times\)5

The prime factors 2, 3 and 5 are not in pairs. Therefore, 120 is not a perfect square
In order to get a perfect square, each factor of 120 must be paired. So, we need to make pairs of 2, 3 and 5. Therefore 120 should be multiplied by 2\(\times\)3\(\times\)5; i.e. 30.
Hence, the required smallest square number is 120\(\times\)30 = 3600

Volume of water = 160 m³
Area of rectangular field = 800 m²
Let h be the height of water level in the field.
Now, volume of water = Volume of cuboid formed on the field by water
⇒ 160= Area of base x Height
⇒ 160 = 800\(\times\)h
∴ h = \(\frac { 160 }{ 800 } \)=0.2 m
Hence, the required height is 0.2 m.

34

Cost price of 12 lemons = Rs 48
or CP of 1 lemon =\(\frac{48}{12}\)=Rs 4
SP of 10 lemons = Rs 48
∴ SP of 1 lemon =Rs \(\frac{48}{10}\)=Rs 4.8
Since SP > CP, there are profit.
Now, Profit %=\(\frac{profit}{CP}\times 100=\frac{SP-CP}{CP}\times 100 \)
=\((\frac{4.8-4.0}{4.0})\times 100=\frac{0.8}{4}\times 100\)
= 0.8ｘ25 = 20%

Five rational numbers less than -4 are
\({-9\over2},{-10\over2}(=-5),{-11\over2},{-12\over2}(=-6),{-13\over2}\)

Yes, two situation are as follows:
Distance covered by a car
(i) Distance = Speed (s) ｘ Time (t)
[If the speed of car is increased by 2 units, i.e. (s + 2)and time is decreased by 1, i.e. (t -1), then distance  
= (s + 2) ｘ (t -1)
(ii) \(SimpleInterest=\frac { Principal\times Rate\times Time }{ 100 } \)
\(=SI=\frac { P\times R\times T }{ 100 } \)
PｘRｘT = 51ｘ100
\(\Rightarrow P=\frac { SI\times 100 }{ R\times T } \)
\(\Rightarrow R=\frac { SI\times 100 }{ P\times T } \quad \Rightarrow T=\frac { SI\times 100 }{ P\times R } \)

From the given table, we observe that the sum of angles
(interior angles) of a polygon having m sides is (m - 2) x 180°.
(a) Here, number of sides of a polygon (m) = 7
∴ Sum of the angles of a polygon of 7 sides
= (7 - 2) x 180° = 5 x 180° = 900°
(b) Here, number of sides of a polygon (m) = 8 .
∴ Sum of the angles of a polygon of 8 sides
= (8 - 2) x 180° = 6 x 180° = 1080°
(c) Here, number of sides of a polygon (m) = 10
∴ Sum of the angles of a polygon of 10 sides
= (10 - 2) x 180° = 8 Xx 180° = 1440°
(d) Here, number of sides of a polygon (m) = n
∴ Sum of the angles of a polygon of n sides
= (n-2) x 180°.

The area of this triangle and the area of the trapezium WXYZare same because this triangle has been formed by adding the triangle ZYA(cut out from trapezium WXYZ) to the remaining part WXAZof the trapezium WXYZ such that Y coincides with X and Z coincides with B.
So, we get the expression for the area of trapezium by using the expression for the area of triangle.

\(-\frac{3}{8}+\frac{1}{7}=\frac{-13}{56}\)
\(\frac{1}{7}+(\frac{-3}{8}) = \frac{-13}{56}\)
So, \(-\frac{3}{8}+\frac{1}{7}\)=\(\frac{1}{7}+(\frac{-3}{8})\).

\(\frac { 1 }{ n } \)

100

Here,
Number of faces (F) = 20
Number of vertices (V) = 12
Let the number of edges be E.
∴ Using Euler's formula, we have
F + V = E + 2
⇒ 20 + 12 = E + 2
⇒ 32 = E + 2
⇒ E = 32 - 2 = 30
Thus, the required number of edges = 30.

Given, (5, 7)
In this point, x-coordinate (abscissa) is 5, i.e.
(x = 5).

(7x - 6y)

No

Given number is 4489.
Here, number of digits, n = 4 [even]
\(\therefore\) Number of digits in the square root of 4489
\(=\frac { n }{ 2 } =\frac { 4 }{ 2 } =2\)
Hence, the number of digits in the square root of 4489 is 2 digits.

3

We have,3.02 x 10^-6 = \(\frac{3.02}{10{6}}=\frac{3.02}{1000000}\)
= 0.00000302     \([\because a^{-m}=\frac{1}{a^{m}}]\)

6

No, as we know that the sum of measures of angles of a quadrilateral is 360°.
Here, \(\angle\)A + \(\angle\)B + \(\angle\)C =120° + 105° + 160°
= 385° > 360°
So, it is not possible to have such a quadrilateral.

we have,\(\left( -\frac { 10 }{ 3 } p{ q }^{ 3 } \right) \times \left( \frac { 6 }{ 5 } { p }^{ 3 }q \right) \)
\(=\left\{ \left( -\frac { 10 }{ 3 } \right) \times \frac { 6 }{ 5 } \right\} \times \left( p\times { p }^{ 3 } \right) \times \left( { q }^{ 3 }\times q \right) \)
\(=-\frac { 10 }{ 3 } \times \frac { 6 }{ 5 } \times \left( { p }^{ 4 } \right) \times \left( { q }^{ 4 } \right) =-4{ p }^{ 4 }{ q }^{ 4 }\)

x= 80°, y = 70°, z = 30°

768000

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.

Rs 1260;7%

x = 72

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. 

(a) Subjects and marks obtained (out of 10) by Sonia in two tests.
(b) In Maths, she scored the highest in test I.
(c) In English and Hindi, she scored the least in test II.
(d) In Hindi and Maths, she scored the same marks in both tests.
(e) She scored 6 marks in English in test II.
(f) Same performance in both tests.
(g) Test I, in maths, she scored full marks i.e. 10 marks.

Here, 324 -1 = 323,323 - 3 = 320
320 - 5 = 315,315 -7 = 308
308 - 9 = 299, 299 - 11 = 288
288 -13 = 275,275 -15 = 260
260 - 17 = 243, 243 - 19 = 224
224 - 21 = 203,203 - 23 =180
180 - 25 =155, 155 - 27 =128
128 - 29 = 99, 99 - 31 = 68
68 - 33 = 35, 35 - 35 = 0
So, to get 0, we use 18 steps.
Hence, square root of 324 is 18.

3

Given, {5² + (12²)^1/2}³
Now, (5)² = 25
(12²)^1/2=(12) ^{2 x 1/2} =(12)¹ =12
So, (25 + 12)³ =(37)³ = 50653

We know that, the dimensions of cuboid, l=70 m, b = 35 m, h = 40 m,
diameter of cylinder is 35 m and cost of painting is Rs. 2 per m²
Area of cylindrical to be painted
\(\frac { 1 }{ 2 } \)\(\times\)Total surface area =\(\frac { 1 }{ 2 } \) [2\(\pi\) r(r+h)] 
= \(\frac { 1 }{ 2 } \left[ 2\times \frac { 22 }{ 7 } \times \frac { 35 }{ 2 } \left( \frac { 35 }{ 2 } +70 \right) \right] \)=4812.5 m²
Area of cuboid to be painted = Area of three walls = lh+2bh
= 70\(\times\)40 + 2\(\times\)40\(\times\)35 = 2800 + 2800
 = 5600 m²
Total area to be painted = 4812.5 + 5600=10412.5 m²
∵ Cost of painting per m² = Rs. 2
∴ Cost of painting 10412.5 ² = Rs. 10412.5\(\times\)2 = Rs. 20825
Verification Verify your answer by adopting some other plan i.e. here in this problem instead of taking area in two steps, let us find in one step.
Area to be painted= Area of three walls + Area of cylindrical part
 =  2bh+lh +\(\frac { 1 }{ 2 } \)[2\(\pi\)rh+2\(\pi\)r²]
= h[2b+l]+ [\(\pi\)R(r+h)]
= 40 [2\(\times\)35+70]+ \(\frac { 22 }{ 7 } \frac { 35 }{ 2 } \left( \frac { 35 }{ 2 } +70 \right) \)
= 40 [140] + 55\(\times\)87.5
= 5600+ 4812.5=10412.5m²
∴  Final cost (similar as in previous method) = Rs. 20825

(p-3)(p²+2-q)

Rs \(275\over9\)

(i)(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

22

We have, if 
    2 B
+  A B  
    8 A   
In column II, we see that sum of 2 and A is equal to 8, A cannot be a 2-digit number. Clearly, A =6.
\(\because\)  2 + 6 = 8
Now, we study the column I.
Since, B + B = 6, \(\Rightarrow\) 2B = 6 \(\Rightarrow\) B =3
Hence, A = 6 and B =3.
\(\therefore\)    23
+  63   
    86  

(a) P (Suhail winning) = \(\frac{1}{7}\)
(b) P (Suhail winning now) = \(\frac{4}{15}\)
                         [\(\therefore\) there are 4 face cards]
    P (Reshma winning) = \(\frac{4}{15}\)
                       [\(\therefore\) there are four 4 numbers card]

72°, 108°

Let PQRS be the required rectangle, where PQ = 4 cm and PR = 6 cm. Draw a rough sketch of rectangle PQRS

Steps of construction
Step I Draw a line PQ = 4 cm.
Step II Make line QX such that \(\angle\) PQX = 90°
Step III With P as a centre draw an arc of length 6 cm, which intersects ray QX at R.
Step IV Draw another ray PY such that \(\angle\)QPY = 90°.
Step V With R as a centre and radius 4 cm draw an arc, which intersect line PY 5. Then, PORS is the required rhombus.

28 x 7 = 196 = 14²

9x³, - 8x³, 6x³.

10^-10. 10¹⁰ = 10^{-10+ 10} = 100 = 1.

x² +x +\(\frac{1}{4}\)=x² + 2(X)\((\frac{1}{2})\)+\((\frac{1}{2})^2\)
=\((x+\frac{1}{2})^2\)

Amount of zinc = 400 \(\times \frac { 45 }{ 100 } \)= 180 g.

9z - 15 = 9 - 3z \(\Rightarrow\) 9z + 3z = 9 + 15
\(\Rightarrow\) 12z = 24 \(\Rightarrow\) z =\(\frac{24}{12}\)=2

Frequency of class 290-325 = Frequency of class 465-500 = 5.

2 \(\times\)2 \(\times\) 2 = 8.

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