2⁸

Steps of construction:
I. Draw a line segment BC = 4.5 cm. 
II. Construct \(\angle\)CBX = 60° at B.
III. From BX, cut off line segment BA = 5.5 cm,
IV. Join AC
Thus, ABC is the required triangle.

Points of intersection are A, B and C.
Measure of LA = 60°
Measure of LB = 60°
Measure of LC = 60°

Sum of the observations (heights)
= [142 + 150 + 146 + 154 + 148] cm =740 cm
Number of observations = 5
\(\therefore \)Mean height = \(\frac { Sum\ of\ observations }{ Number\ of\ observations } \)
\(\frac { 140 }{ 5 } \) cm = 148 cm
Thus, the required mean height = 148 cm

1

Faces = 5, edges = 9, vertices = 6

Given, base (b)= 6 cm and height (h)= 4 cm
\(\therefore\) Area of a triangle = \(\frac{1}{2}\) x b x h = \(\frac{1}{2}\) x 6 x 4 = 12 cm²
and perimeter of a triangle = Sum of the length of all three sides
                                      = Base + 2 x Equal sides
                                      = 6 + 2 x 5 [.: two sides are equal]
                                      = 6 + 10 = 16 cm.

The actual length of an iron rod = 19.33 cm
measured length = 19.34 cm
Error = Measured value - Actual value
= (19.34 -19.33) cm + .01cm
Kamal measured = 19.25 cm
Error = (19.25 -19.33) cm - 0.08 cm
Tabish measured = 19.27 cm
Error = (19.27 -19.33) cm = - 0.06 cm

\(\angle QRS=140^0\)

Original price of the car = Rs 350000
Increased price of the car = Rs 370000
Then. increase in price = Rs (370000 - 350000) =Rs 20000
\(\therefore\)   Percentage of price increase \(=\left( { Increase\ in\ price\over Original\ price\times100 } \right)\%\)
\(=\left( {20000\over350000} \times100 \right)\%\)
\(={40\over7}\%=5{5\over7}\%\)
Hence, the percentage of price increase was\(5{5\over 7}\%\)

In ΔABC and ΔFED,
ㄥB = ㄥE       [each 90° given]
\(ㄥ\)A = ㄥF        [given]
∴ ㄥC = ㄥD
[if two corresponding angles of two triangles are equal than their third corresponding angles are equal)
Also, BC = ED           [given]
It implies that ΔABC and ΔFED are congruent by ASA congruent criterion.
[∵ ㄥB = ㄥE, ㄥC = ㄥD and BC = ED)

For a negative integer and a positive integer whose difference is +2,
First integer = 1,
and second integer = -1
\(\therefore\) 1 - (-1) = 1 + 1 = 2

By observing the above pattern, we find that denominators are multiples of 3, so we will progress in this way 
\(\frac { -2 }{ 6 } =\frac { -1\times 2 }{ 3\times 2 } ,\frac { -3 }{ 9 } =\frac { -1\times 3 }{ 3\times 3 } \)
so,\(\\ \frac { -1\times 4 }{ 3\times 4 } =\frac { -4 }{ 12 } ,\frac { -1\times 5 }{ 3\times 5 } =\frac { -5 }{ 15 } \)

Let the number be x.
Three-fourths of the number = \(\frac{3}{4}x\)
According to the question,
On adding 3 to it, we get 2l.
i.e.   \(\frac{3}{4}x+3=21\)
which is the required equation.
Now, to solve this equation, transposing (+3) from LHS to RHS, we get
       \(\frac{3}{4}x=21-3\quad \Rightarrow \frac{3}{4}x=18\)
On multiplying both sides by 4, we get
       \(\frac{3}{4}x\times 4=18\times 4\Rightarrow 3x=72\)
Again, dividing both sides by 3, we get
       \(\frac{3x}{3}=\frac{72}{3}\quad \Rightarrow x=24\)
Hence, the required number is 24.

1x10⁴+2x10³+3x10²+4x10¹+5x10⁰

Area of a tile = 25 x 20 cm² = 500 cm²
Area of 2400 tiles = 2400 x 500 cm²
= 1200000 cm²
= \(\frac { 1200000 }{ 10000 } \)m²
 [\(\therefore\) 10000 cm² = 1 m²]
= 120 m²
Let the height of the wall be h metres then,
Area of the wall = 4h m²
Since 2400 tiles completely cover the wall.
\(\therefore\) Area of the wall = Area of 2400 tiles
\(\Rightarrow\) 4h = 120
\(\Rightarrow\) \(\frac { 4h }{ 4 } \)= \(\frac { 120 }{ 4 } \)
\(\Rightarrow\) h = 30
[Dividing both sides by 4]
Hence the height of the wall is 30 metre.

23

In ΔPQR we have mㄥQ = 30°, mㄥR = 60° and QR = 4.7 cm.
Here, two angles and included side are given, so ΔPQR can be constructed.
Step V Rays QX and RY intersect at P.
Step I Firstly, draw a rough sketch of triangle with given measures marked on it.

Step II Draw a line segment QR = 4.7 cm.

Step III At point Q, draw a ray QX making an angle of 30° with QR.

Step IV At point R, draw a ray RY making an angle of 600 with QR

Step V Rays QX and RY intersect at P.

Thus, ΔPQR is the required triangle.

Selling price of each radio = Rs 5000
Let the cost price of each radio be Rs x.
10% profit on one radio
i.e.\({110\over 100}\times x=5000\)
\(x={5000\over 110}\times100={50000\over11}=Rs4545.45\\ 10\% \)

Let we add A to x to find y,
A+x=y\(\Rightarrow \) \(A+\left( \frac { -4 }{ 9 } \right) =\left( \frac { 5 }{ 12 } \right) \)
\(A=\frac { 5 }{ 12 } -\left( \frac { -4 }{ 9 } \right) =\frac { 5 }{ 12 } +\frac { 4 }{ 9 } \) 
\(=\frac { 5\times 3+4\times 4 }{ 36 } =\frac { 15+16 }{ 36 } =\frac { 31 }{ 36 } \)

x = 34°, y=56°

As per the given information, we can draw the given diagram.

Let A be the point above the sea level and B be the point below the sea level.
\(\therefore\) Vertical distance between points A and B = Distance between point A and sea level + Distance between point B and sea level.
= AO + OB = 8848 + 10911 = 19759 m

169.86 m²

(a) The above double bar graph compares the sale of brands A and 8during the months of January to June.
(b) We can clearly see from the double bar graph that sales for brand A reduced in the month of March compared to that of February.
(c) Sales of brand A in June = Rs. 57 lakh
and sales of brand 8 in June = Rs. 54 lakh
Difference in sales = 57 - 54 = Rs. 3 lakh
(d) Average sales of brand 8
=\(\frac { total\ sales\ of\ brand\ b\ in\ sin\ months\ from\ january\ to\ june }{ 6 } \)
= \(\frac { 36+38+43+35+45+54 }{ 6 } \)
= \(\frac { 251 }{ 6 } \) Rs.41.83 lakh
(e) We can clearly see from the double bar graph that sales of brand 8is less than sales of brand A in the month of April and June
(f) Sales of brand A in January = 31
and sales of brand 8 in January = 36
Required ratio = \(\frac { 31 }{ 36 } \)or 31:36

2x+3y=220°

As per the given information in the question, the total number of students in the class = 60
Let x be the number of boys in the class.
So, the number of girls in the class = \(\frac{x}{3}\)
\(\therefore x+\frac{x}{3}=60\Rightarrow \frac{3x+x}{3}=60\Rightarrow \frac{4x}{3}=60\)
4x = 60 x 3  \(\Rightarrow\) 4x = 180 \(\Rightarrow x=\frac{180}{4}=45\)
Hence, the number of boys in the class is 45 and number of girls in the class is \(\frac{45}{3}=15\)

We know that in a triangle, an exterior angle is equal to the sum of two interior opposite angles. Therefore,
(i) Sum of interior opposite angles = Exterior angle
=> x+50°=115°​​​​​​​=>x=115°​​​​​​​-50°​​​​​​​=65°​​​​​​​
Hence, the value of the unknown interior angle x is 65°.
(ii) The value of the unknown interior angle x is 30°.
(iii) The value of the unknown interior angle x is 35°.
(iv) The value of the unknown interior angle x is 60°.
(v) The value of the unknown interior angle x is 50°.
(vi) The value of the unknown interior angle x is 40°.

We may draw AC = 5 cm and \(\angle\)C = 30°. CA is one arm of \(\angle\)C. Point B should be lying on the other arm of \(\angle\)C. But we observe that point B cannot be located uniquely. So, the given data is not sufficient for construction of \(\triangle\)ABC.

6s + 24 = 0
\(\Rightarrow\) 6s = - 24
Thus, s = -4

44875 \(\times\) 99- (- 44875) = 44875 \(\times\) 99- 44875 \(\times\) - 1
= 44875 \(\times\) [99- (-1)]
[using distributively over addition]
= 44875 \(\times\) (99 + 1)
= 44875 \(\times\) 100
= 4487500.

8

(2²⁰ \(\div\) 2¹⁵) x 2³ = (2^20-15)x2³
=2⁵ x2³ =2⁵⁺³ =2⁸

Given, radius (r) = 7 cm
\(\therefore\) Circumference of a circle = 2\(\pi\)r
                                         = 2 x \(\frac{22}{7}\) x 7 = 44 cm.

Given 4x³ + 3x² + 6
The expression 4x³ + 3x² + 6 is having three terms, i.e. 4x³, 3x² and + 6.
So, it is a trinomial.

For more rational numbers, firstly we make both denominators same. Since LCM of 7 and 8 is 56.
Now , \(\frac { -5\times 8 }{ 7\times 8 } =\frac { -40 }{ 56 } ,\frac { -3 }{ 8 } \times \frac { 7 }{ 7 } =\frac { -21 }{ 56 } \Rightarrow \) -40<-39<-38<....<-21
\(\therefore \) \(\frac { -40 }{ 56 } <\frac { -39 }{ 56 } <\frac { -38 }{ 56 } <\frac { -37 }{ 56 } <\frac { -36 }{ 56 } <\frac { -35 }{ 56 } <.....<\frac { -21 }{ 56 } \)
Hence, Five rational numbers between \(\frac { - 5 }{ 7 } \) and \(\)\(\frac { - 3 }{ 8 } \) are \( \frac { - 39 }{ 56 } ,\frac { - 38 }{ 56 } ,\frac { - 37 }{ 56 } ,\frac { - 36 }{ 56 } ,\frac { - 35 }{ 56 }\).

The multiple lines of symmetry of each figure are shown below:
This figure has 2 lines of symmetry which are shown by dotted lines.

Let the number be x.
\(\therefore\) 75% of \( x =15 \Rightarrow {75\over100}\times x =15\)
\(\Rightarrow x=15\times {100\over75} \Rightarrow x=20\)
Hence, the number is 20.

In ΔABC and ΔPQR, we have
AB = 3.5 cm and PQ = 3.5 cm, so AB = PQ
AC = 3.5 cm and PR = 3.5 cm, so AC = PR
LA=60° and LP=60°, so ㄥA = ㄥP
Hence, two sides and an angle in ΔABC and ΔPOR are equal. Therefore, the two triangles are congruent.
∴ ΔABC ≅ ΔPOR

We know that in a triai.gle, the exterior angle is equal to the sum of two interior opposite angles.
By exterior angle property of a triangle,
Sum of interior opposite angles = Exterior angle
\(\Rightarrow\) x + 50° = 120° \(\Rightarrow\) = 120° - 50° = 70°
Since, the sum of all the angles of triangle is 180^0.
Now, by angle sum property of a triangle,
\(\Rightarrow\) x+ y+500=180° \(\Rightarrow\) 70°+ y+500=180°
\(\Rightarrow\) y +120° = 180° \(\Rightarrow\) Y = 180° -120° = 60°
Hence, the value of the unknown x and yare 70° and 60°, respectively.

Given, angle 48°
Complement of 48° = 90° - 48° = 42°

First five even numbers = 2, 4, 6, 8,10
Sum of first five even numbers
=2+4+6+8+10=3
\(Mean=\frac { sum\ of\ the\ numbers }{ number\ of\ observations } =\frac { 30 }{ 5 } =6\)
Hence, mean of first five even numbers is 6.

We have, \(\frac{4}{5}\times \frac{12}{7}\times \frac{4\times12 }{5\times 7}=\frac{48}{35}=1\frac{13}{35}\)

\(\frac{-6}{-25}=\frac{-6\times-1}{-25\times-1}=\frac{6}{25}\)

\(3\frac{1}{3}\div10=\frac{10}{3}\div10=\frac{10}{3}\times\frac{1}{10}=\frac{10\times1}{3\times10}=\frac{1}{3}\)

(i) x + y = 1200                                  ...(1)
The exterior angle of a triangle is equal to the sum of its two interior opposite angles
x + y + y = 1800
Base angles opposite to the equal sides of an isosceles triangle are equal and the sum of the measures of the three angles of a triangle is 180⁰
\(\Rightarrow\)x + 2y = 1800                                  ... (2)
Subtracting equation (1) from equation (2),
y = 60°
Put y = 60° in equation (1),
x + 60° = 120°
\(\Rightarrow\)x = 120° - 60°
\(\Rightarrow\)x = 60°

Let ABCD be the rectangular park of sides 40 m and 25 m, and the shaded region represents the path 2.5 m wide.
Now, PQ = (40 + 2.5 + 2.5) m = 45 m
PS = (25 + 2.5 + 2.5) m = 30 m
\(\therefore\)Area of rectangle ABCD = I x b = 40 m x 25 m
= 1000 m2
Area of rectangle PQRS = 45 m x 30 m = 1350 m²
So, Area of the path = [Area of rectangle PQRS] - [Area of rectangle ABCD]
= 1350 m² - 1000 m² = 350 m²

Total number of students = 8
Sum of the marks obtained =85 + 62 + 36 + 48 + x + 75 + 39 + 72=417+x
\(\therefore \)Mean marks =\(\frac { (417\quad +x) }{ 8 } \)
\(\Rightarrow \)\(\frac { (417\quad +x) }{ 8 } \)=60
\(\Rightarrow \)417 + x = 60 x 8 = 480
\(\Rightarrow \)x = 480 - 417 = 63
Thus, the required value of x = 63

Substituting a = 3 and b = 2 in
(a) a + b, we get
a + b = 3 + 2 = 5
(b) 7a - 4b,  we get
7a - 4b = 7 x 3 - 4 x 2
= 21 - 8 = 13
(c) a² + 2ab + b², we get
a² + 2ab + b² = 3² + 2 x 3 x 2 + 2²
= 9 + 2 x 6 + 4
= 9 + 12 + 4 = 25.
(d) a3 - b3, we get
a³ - b³ = 3³ - 2³ = 3 x 3 x 3 - 2 x 2 x 2
= 27 -8 = 19.

\(⇒\ \left(1\over 5\right)^{5+19}=\left(1\over 5\right)^{8x}\)
[∵ a^m x aⁿ = a^m+n]
\(⇒\ \left(1\over 5\right)^{24}=\left(1\over 5\right)^{8x}\)
When bases are equal, then by equating their exponents, we get
8x = 24
\(∴\ x={24\over 8}=3\)

Quantity of pure alcohol in 400 ml. of 15% = 400 \(\times \frac { 15 }{ 100 } \)
= 60 ml
Now, we add x ml of pure alcohol to the sample.
So, total pure alcohol = (60 + x) ml.
But volume of new sample = (400 + x) ml.
\(\therefore\) Percentage of pure alcohol in new sample \(=\frac { (60+x) }{ (400+x) } \times 100\)
which is equal to 32%
\(\Rightarrow \quad \frac { 60+x }{ 400+x } \times 100=32\)
\(\Rightarrow \quad \frac { 60+x }{ 400+x } =\frac { 32 }{ 100 } \)
\(\Rightarrow\) 100(60 + x) = 32 (400 + x)
\(\Rightarrow\) 100x + 6000 = 32x + 12800
\(\Rightarrow\) 100x-32x = 12800 - 6000
\(\Rightarrow\)  68x = 6800
\(\Rightarrow \ x=\frac { 6800 }{ 68 } =100\ ml\)

Let the cost of property be Rs. 100.
\(\Rightarrow \) Commission is Rs.3
\(\Rightarrow \) When commission is Rs.3, then cost of property
= Rs.100
\(\therefore \) When commission is Rs.42,660, then cost of property
=\(\frac { 100 }{ 3 } \times \)42660
=100 x 14220
=Rs.1422000
Hence, the cost of property is Rs.14,22,000.

Given pattern is
\(-\frac { 1 }{ 3 } ,\frac { 2 }{ 6 } ,\frac { 3 }{ 9 },-\frac { 4 }{ 12 } ...\)
Here, \(-\frac { 1 }{ 3 } =\frac { (-1)\times 1 }{ 3\times 1 } \)
\(-\frac { 2 }{ 6 } =\frac { (-1)\times 2 }{ 3\times 2 } \)
\(-\frac { 3 }{ 9 } =\frac { (-1)\times 3 }{ 3\times 3 } \)
and \(-\frac { 4 }{ 12 } =\frac { (-1)\times 4 }{ 3\times 4 } \)
Hence, next four numbers are
\(\frac { (-1)\times 5 }{ 3\times 5 } =-\frac { 5 }{ 15 } \)
\(\frac { (-1)\times 6 }{ 3\times 6 } =-\frac { 6 }{ 18 } \)
\(\frac { (-1)\times 7 }{ 3\times 7 } =-\frac { 7 }{ 21 } \)
\(\frac { (-1)\times 8 }{ 3\times 8 } =-\frac { 8 }{ 24 } \).

In the above figure, AB and CD are two poles whose heights are 9 m and 14m respectively.
\(\Rightarrow\) AB = EC = 9m
and BD = 13m
DE = 14 - 9
= 5m
Now in right ΔBDE, by Pythagoras
BD² = BE² + DE²
13² = BE² + 5²
\(\Rightarrow\) BE² = (13)² - (5)²
= 169-25
BE² = 144
\(\Rightarrow\) BE = \(\sqrt { 144 } \)
\(\Rightarrow\) BE = 12m.
Hence, distance between their feet = 12 m.