\(\frac{37}{15}\)

Let us make Ceasar Cipher table first. Here, we have to set to + 4 table. For that, we have to start letter e to set as A, f as B … likewise d as Z. Now, the + 4 Ceasar Cipher table looks like

The given plain text is
fvieo mr gshiw ger fi xvmgoc
To crack this secret code, follow the steps given below.
Step 1: Using Ceasar Cipher table, let us first match the most repeated letters. This will help us to progress faster.
fvieo mr gshiw ger fi xvmgoc

Step 2: Then, let us find remaining letters to complete the code.

Thus, the secret sentence is, BREAK IN CODES CAN BE TRICKY

2x = 9 − 5
2x = 4
x = \(\frac { 4 }{ 2 } \)
x = 2

The ladder, wall and the ground form a right triangle with the ladder as the hypotenuse. From the figure, by Pythagoras theorem,
20² = 16² + x²
⇒ 400 = 256 + x²
⇒ x² = 400 − 256 = 144 = 12²
⇒ x = 12 feet
Therefore, the base (foot) of the ladder is 12 feet away from the wall.

Take a = 5, b = 12 and c = 13
Now, a² + b² = 5² +12² = 25 +144 = 169 = 13² = c²
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.

Let the number be x.
x-\(\frac { 25 }{ 100 } x\) = 120
\(\frac { 100x-25x }{ 100 } \) = 120
\(\frac { 75x }{ 100 } \) = 120
x = \(\frac { 120\times 100 }{ 75 } \)
x = 160

[\(\because\) using a²-2ab+b² = (a-b)²]
4a²-4a+1 = (2a)²-2\(\times\)(2a)\(\times\)1+1²
= (2a-1)²
= (2a-1)(2a-1)

In \(\triangle \)APB and \(\triangle \)DPC
\(\angle \)A =\(\angle \)D = 90° [given]
\(\angle \)APB =\(\angle \)DPC [Vertically opposite angles]
\(\angle \)ABP =\(\angle \)DCP [Remaining angle]
 ∴ \(\triangle \)APB ~ \(\triangle \)DPC [AAA criteria]
 \( \frac {AP }{ DP } =\frac { BP }{ PC }\) [Corresponding sides are proportional]
AP x PC = BP x DP

Comparing (x + 4)³ with (a + b)³, we get a = x, b = 4
We know(a + b)³ = a³ + 3a²b + 3ab² + b³
(x + 4)³ = (x)³ + 3(x)² + 3(x)(4)² + (4)³
= (x)⁵ + 3x²(4)+  3(x)(16) + 64
(x+4)³ = x³ + 12x² + 48x + 64

\(-5.8=\frac { -58 }{ 10 } =\frac { -29 }{ 5 } =-5\frac { 4 }{ 5 } \)

Now, \(\frac { PQ }{ PR } =\frac { 20 }{ 15 } =\frac { 4 }{ 3 } \)
\(\frac { PR }{ PS } =\frac { 15 }{ 11.25 } =\frac { 4 }{ 3 } \)
Also, \(\frac { QR }{ RS } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)
We find \(\frac { PQ }{ PR } =\frac { PR }{ PS } =\frac { QR }{ RS } \)
That is, their corresponding sides are proportional.
∴ By SSS Similarity, ΔPQR ～ ΔPRS

Since, the sectors are cut from an equilateral triangle, the central angle of each of them is 60°.
∴ Area of the shape formed, \(A=3\times \left( \frac { { \theta }^{ 0 } }{ { 360 }^{ 0 } } \times { \pi r }^{ 2 } \right) \)
\(=3\times \frac { { 60 }^{ 0 } }{ { 360 }^{ 0 } } \times \pi \times 5\times 5\)
\(=3\times \frac { 1 }{ 6 } \times \pi \times 5\times 5\)
= 12.5 π sq.cm

18

90

89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765

The first term -xy can be factorized as: (-1) × x × y and
The second term -ay can be factorized as : (-1) × a × y
The common factor for both the terms is -y
Taking out the common factor we get,
- xy - ay = -y(x + a)

Using standard identity: (a + b)² = a² + 2ab + b²
Here, a = 7x and b = 5
(7x + 5)² = (7x)² + [2(7x)(5)] + 5²
49x²-70x+25

\(\frac{4}{3}\)

Given that ∠B = 2/3 x ∠A = 2/3a & given ∠C = ∠A + 20 = a + 20
Since A, B & C are angles of a triangle, they add up to 180° (∆ property)
∠A + ∠B + ∠C = 180°
a + 2/3 a + a + 20 = 180°
(3a+2a+3a)/3 + 20 = 180°
8a/3 = 180 – 20 = 160°
\(\frac{160 \times 3}{8} = 60^o\)
\( \angle B = \frac{2}{3}\times \angle A = \frac{2}{\not 3} \times \not 60 = 40^o \)
∠C = 80°
60°, 40°, 80°

(1) x = -7

(ii) y = 6

a) (3, 2)
b) (- 2, 0)
c) (- 2, - 2) 
d) (- 2, 2)
e) (-1, -1)
f) (3, -1)
g) (0, 2)
h) (2, 0)
i) (- 3, 3) 
j) (0, -2)

Given:
AI = 6 cm,
IM = 5 cm,
AM = 9 cm,
MS = 6.5 cm.
Rough Diagram

Steps:
1. Draw aline segment AI = 6 cm.
2. With A and I as centres draw arcs of radius
9 cm and 5 cm respectively and let them cut at M.
3. Join AM and IM.
4. Draw MX parallel to AI.
5. With M as centre, draw an arc of radius6.5 cm cutting MX at S.
6. |oin AS. AIMS is the required trapezium
Calculation of Area:
Area of the trapezium AIMS
\(=1 / 2 \times h \times(a+b) \)
\(=1 / 2 \times 4.8 \times(6+6.5) \)
\(=1 / 2 \times 4.8 \times 12.5=30 \mathrm{sq} . \mathrm{cm} . \)
2. Given:
CU = 7cm; o
CE = 6cm; TE = 5cm.
Rough Diagram 

Steps: 
1. Draw a line segment CU = 7cm
2. Constract an angle o at c
3. With C as centre, draw an arc of radius 6 cm cutting CX at E.
4. Draw EY parallel to CU.
5. With E as centre, draw an arc of radius 5 cm cutting EY at T.
6. Join TLI CUTE is the required trapezium.
Calculation of Area:
Area of the trapezium CUTE
\(=1 / 2 \times \mathrm{h} \times(\mathrm{a}+\mathrm{b}) \)
\(=1 / 2 \times 6 \times(5+7) \)
\(=1 / 2 \times 6 \times 12=36 \text { sq. } \mathrm{cm} . \)
3. Given:
AR = 7 cm; RM = 6.5 cm
ZRAY = 100^o; ZARM = 60^o
Rough Diagram
​​​​​
Steps: -
1. Draw a line segment AR 7 cm.
2. Construct an angle o at R. S.
3. With R as centre draw an arc of radius 6.5 cm cutting RX at M.
4. Draw MZ parallel to AR.
5. Construct an angle ZRAY = 100^o A cutting MZ at Y.
6. ARMY is the required trapezium
Calculation of Area:
Area of the trapezium ARMY
\(=1 / 2 \times h \times(a+b) \)
\(=1 / 2 \times 5.6 \times(7+7) \)
\(=1 / 2 \times 5.6 \times 14=39.2 \text { sq. cm } \)
4. Given:
CI = 7 cm ; IT = 5.5 cm
TY = 4 cm ; YC = 6cm.
Rough Diagram

Steps:
1..Draw a line segment CI = 7 cm.
2. Mark the point A on CI such that CA = 4 cm.
3. With A and I as centres, draw arcs of radii 6 cm and S.S cm. Let them cut at T. join AT and IT.
4. With C and T as centres, draw arcs of radii 6 cm and 4 cm respectively. Let them cut at Y. Join TY and CY.
5. CITY is the required trapezium.
Calculation of Area:
Area of the trapezium CITY
\(=1 / 2 \times h \times(a+b) \)
\(=1 / 2 \times 6 \times(4+4) \)
\(=1 / 2 \times 6 \times 8=24 \text { sq. } \mathrm{cm} . \)

24 cm

\(\triangle\) AFI is a right angled triangle
(AI)² = (AF)² + (FI)²
25² = (AF)² + 15²
625 = (AF)² + 225
(AF)² = 625 - 225
= 400
\(\mathrm{AF}=\sqrt{400}\)
AF = 20ft
\(\triangle\) FRI is a right angled triangle.
(FI)² + (FR)² = (RI)²
(FR)² + 15² = 17²
(FR)² + 225 = 289
(FR)² = 289 - 225
= 64
\(A R=\sqrt{64}\)
= 8ft
AR = AF+FR
= 20 + 8
= 28 ft
AR = 28 ft

From the figure AC is to be found.
By using Pythagoras theorem,
AC² = AB² + BC² = 282 + 212 = 784 + 441 = 1225 = 352
∴ AC = 35 km

(i) From the figure,
x² = 9² + 40²
= 81 + 1600
= 1681
\(x=\sqrt{1681}\)
x = 41
(ii) From the figure
34² = 30² + y²
1156 = 900 + y²
y = 1156 - 900
y² = 256
\(y=\sqrt{256}\)
y = 16
(iii) From the figure
39² = 36² + z²
1521 = 1296 + Z²
z² = 1521- 1296
z² = 225
\(z=\sqrt{225}\)
z = 15

Let x be the initial number of mangoes.
Nurnber of Rotten mangoes = 10%
Number of good mangoes
\(=90 \% \text { of } x=\frac{90}{100} x\)
Percentage of sold mangoes \(=33 \frac{1}{3} \%\)
Percentage of unsold mangoes \(=100-33 \frac{1}{3} \%\)
\(=66 \frac{2}{3} \%\)
\( \frac{66 \frac{2}{3}}{100} \times \frac{90}{100} x =240 \)
\(\frac{200}{3 \times 100} \times \frac{90}{100} x =240 \)
\(\frac{2}{3} \times \frac{9}{10} \times x =240 \)
\(=\frac{240 \times 3 \times 10}{2 \times 9} =400 \)
The vendor bought 400 mangoes initially.

Let the numerator be x and the denominator be y.
The original fraction = x/y
Given the numerator of a fraction is increased by 50%
\( \therefore \mathrm{Nr} =x+50 \% \text { of } x \)
\(=x+\frac{50}{100} x=x+\frac{1}{2} x \\ \mathrm{Nr} =\frac{3}{2} x \)
Also given the denominator is decreased by 20 %
\( \text { Dr } =y-20 \% \text { of } y \)
\(=y-\frac{20}{100} y=y-\frac{1}{5} y=\frac{4}{5} y \)
By given data
\( \frac{\frac{3}{2} x}{\frac{4}{5} y}=\frac{3}{5} \Rightarrow \frac{3}{2} x \times \frac{5}{4 y}=\frac{3}{5} \)
\(\frac{x}{y}=\frac{8}{25} \)
The original fraction is 8/25

[ using (a - b)²=a²-2ab+b²]
(2x-3y)² = (2x)²-2(2x)(3y)+(3y)²
= 4x²-12xy+9y²

\(\left( \frac { -7 }{ 18 } \times \frac { 15 }{ -7 } \right) -\left( 1\times \frac { 1 }{ 4 } \right) +\left( \frac { 1 }{ 2 } \times \frac { 1 }{ 4 } \right) =\left( \frac { -7\times 15 }{ 18\times -7 } \right) \left( \frac { 1\times 1 }{ 1\times 4 } \right) +\left( \frac { 1\times 15 }{ 18\times 1 } \right) -\left( \frac { 1 }{ 4 } \right) +\left( \frac { 1 }{ 8 } \right) \)
\(\\ \\ =\frac { 5 }{ 6 } -\frac { 1 }{ 4 } +\frac { 1 }{ 8 } =\frac { (5\times 4)-(1\times 6)+(1\times 3) }{ 24 } \)
\(=\frac { 20-6+3 }{ 24 } =\frac { 14+3 }{ 24 } =\frac { 17 }{ 24 } \)

In triangles \(\triangle \)RPQ and \(\triangle \) RTS, we have
\(\angle \)RPQ=\(\angle \)RTS [∵ given]
\(\angle \)PRQ =\(\angle \)TRS [∵ common]
\(\angle \)PQR =\(\angle \)RST l∵ Remaining angle]
\(\triangle \)RPQ ~ \(\triangle \)RTS [∵ By AAA similarity]

3x²(x⁴ - 7x³+ 2) 
= 3x⁶-21x⁵ + 6x²
Highest power is 6 in x⁶

The lower triangle cannot be constructed as the sum of two sides 5 + 3 = 8 < 10 cm. So this quadrilateral cannot be constructed.

1-(iv), 2-(iii), 3-(i), 4-(ii)

\(1\div \frac { 1 }{ 2 } =\frac { 1 }{ 1 } \times \frac { 2 }{ 1 } =\frac { 1\times 2 }{ 1\times 1 } =2\)

\(\frac { 3 }{ 4 } \times \frac { 5 }{ 7 } =\frac { 3\times 5 }{ 4\times 7 } =\frac { 15 }{ 28 } \)

By Euler's formula F + V - E = 10 + 15 - 20
= 25 - 20 = 5 ≠ 2
No, it cannot have F = 10, E = 20 and V = 15

Verifying Euler's formula
F + V - E = 12 + 6
= 18 - 16 = 2
Yes, the polyhedron can have F = 12, E = 16 and V = 6

(i) Cube - dice, building blocks, jewel box.
(ii) Cuboid - books, bricks, containers.
(iii) Cylinder - candles, electric tube, water pipe.
(iv) Cone - Funnel, cap, ice cream cone
(v) Sphere - ball, beads, Lemmon.

\(\frac { 5 }{ -4 } =-\frac { 5 }{ 4 } =-1\frac { 1 }{ 4 } \)
-1 \(\frac{1}{4}\) lies between -1 and -2. The unit part between -1 and -2 is divided into four equal parts and the first part is taken.

(10 x 5) + (9 x 4)

(i) in proportion
(ii) shape
(iii) equal
(iv) congruent
(v) similar

All integers are rational numbers. 
∴ Rational numbers less than -2 are -10, -15, -20, -25,-30.

(4n)² − 2n  +3 = 4n²− 2n + 3 (false)
(4n)² − 2n  +3 = 16n²− 2n + 3

Given in \(\triangle\) AIM
P is the mid point of MI and
N is the mid point of AI
\(\therefore \mathrm{PN} \| \mathrm{AM} \)
\(Now \)
\(\angle \mathrm{TMA}=\angle \mathrm{IAM}=\angle \mathrm{PNI} \)
\(\angle \mathrm{TAM}=\angle \mathrm{IMA}=\angle \mathrm{IPN} \)
\(\angle \mathrm{ATM}=\angle \mathrm{MIA}=\angle \mathrm{PIN} \)
\( \therefore \triangle \mathrm{PIN} \sim \triangle \mathrm{ATM} \)
Hence proved

Shanthi wears a chudithar in 5 different ways.
She also wears a frock in 4 different ways.
So she wears either a chudithar or a frock in 5 + 4 different ways.
That is 9 different ways.

If we make this parallelograms into a rectangle, we get the breadth and height of the rectangle 8 cm and 6 cm respectively
 Area of the shaded part
= Area of the rectangle = bh
= 8 x 5 = 48 cm²

Length of the arc of the sector l = 50 mm
Radius r = 14mm
Area of the sector = \(\frac { lr }{ 2 } \) sq. units
= \(\frac { 50\times 14 }{ 2 } \) mm² = 50 x 7 mm² = 350 mm²
Area of the sector = 350mm²

Here, we have to find HCF of 270, 384 and 588
STEP 1: First find the HCF of any two of the given numbers (follow the same step 1, 2 and 3 of the above example). Here, find HCF of (384, 588) first.
STEP 2: The HCF of the first two numbers which is 12 becomes the divisor and the third number 270 becomes the dividend.
STEP 3: Repeat this division process till the remainder becomes zero. The last divisor is the HCF. Here, 6 is the last divisor.
Hence, HCF of 270, 384 and 588 is 6. Therefore, we needs 6 containers so that each of them contains (270 ÷ 6 = 45) 45 ginger chocolates, (384 ÷ 6 = 64) 64 milk chocolates and (588 ÷ 6 = 98) 98 coconut chocolates.

Let the two digit number be xy (i.e., ten’s digit is x, ones digit is y)
Its value can be expressed as 10 x + y.
Given, x + y = 8 which gives y = 8 − x
Therefore its value is 10 x + y
= 10x + 8 − x
= 9x + 8.
The new number is yx with value is 10y + x
= 10(8 − x) + x
= 80 – 9x
Given, when 18 is added to the given number (xy) gives new number (yx)
(9x + 8) + 18 = 80 – 9x
This simplifies to 9x + 9x = 80 – 8 – 18
18x = 54
x = 3 ⇒ y = 8 – 3 = 5
The two digit number is xy = 35

Given:
DE = 7 cm, EA = 6.5 cm ㄥEDN = 100⁰ and ㄥDEA = 70⁰ and \(\overset { \_ \_ }{ DE } \) || \(\overset { \_ \_ }{ NA} \)


Steps:
1. Draw a line segment DE = 7cm.
2. Construct an angle ㄥDEX = 70⁰ at E.
3. With E as centre draw an arc of radius 6.5cm cutting EX at A.
4. Draw AY parallel to DE.
5. Construct an angle ㄥEDZ = 100⁰ at D cutting AY at N.
6. DEAN is the required trapezium.
Calculation of area:
Area of the trapezium DEAN = \(\frac { 1 }{ 2 } \) x h x (a+b) sq. units
= \(\frac { 1 }{ 2 } \) x 6.1 x (7 + 5.8) = 39.04 sq. units

Let the C.P of one strawberry box be Rs.  x.
Then C.P of 20 strawberry boxes = 20 x and
S.P of 20 strawberry boxes = C.P of 16 strawberry boxes = 16 x
Thus, S.P < C.P, hence there is a loss.
Loss = C.P –S.P = 20 x − 16 x = 4 x
∴ Loss % =\(\left( \frac { Loss }{ C.P } \times 100 \right) \)%
=\(\left( \frac { 4x }{ 20x } \times 100 \right) \)%
= 20 %

\(\frac { QT }{ PR } =\frac { QR }{ QS } \\ \frac { QT }{ QR } =\frac { PR }{ QS } \) [given] ...(1)
\(\angle \)1 = \(\angle \)2 
PR=PQ ...(2)  [∵ Sides opposite to equal angles are equal]
From (1) and (2)
\(\frac { QT }{ QR } =\frac { PQ }{ QS } \\ \frac { PQ }{ QT } =\frac { QS }{ QR } \)
\(\angle \)PQS = \(\angle \)TQR =\(\angle \)Q
:. By SAS criteria \(\triangle \)PQS ~\(\triangle \)TQR.

Area of quadrant AOBCA = \(\frac14\)πr²
= \(\frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times 3.5\times 3.5\)cm²
= \(\frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \)cm² = \(\frac { 27 }{ 8 } \)cm²
= 9.625 cm²
Area of ΔAOD = \(\frac { 1 }{ 2 } \) x b x h sq. units
= \(\frac { 1 }{ 2 } \)x 3.5 x 2 cm²
= 3.5 cm²
Area of the shaded portion = Area of the quadrant - Area of triangle
= 9.625 - 3.5 cm²
= 6.125 cm²

Total number of pictures = 100x²y³
Each page contains = 4xy pictures

= 25xy² pages

This is one of the solutions. Try for more solutions