In the given data Rs. 800 occurs thrice. Hence the mode is Rs. 800

In the cyclic quadrilateral \(\angle\)A + \(\angle\)C = 180⁰

6y = 180⁰
\(y=\cfrac { { 180 }^{ 0 } }{ 6 } ={ 30 }^{ 0 }\)
\(\angle\)B+\(\angle\)D = 6\(\times\)-4+7\(\times\)+2
13\(\times\)-2 = 180⁰
13\(\times\) = 180 + 2 = 182⁰
x = \(\cfrac{182}{13}={14}^{0}\)
\(\therefore\) \(\angle\)A = 2(30) + 4⁰ = 64⁰
\(\angle\)B = 6(14) - 4⁰ = 84 - 4 = 80⁰
\(\angle\)C = 4(30) - 4 = 120 - 4 = 116⁰
\(\angle\)D = 7(14) + 2 = 98 + 2 = 100⁰

(2x)²+(3y)²+(5z)²+2(2x)(3y)+2(3y)+(5z)+2x3yx5z
(2x+3y+5z)²
∵ (a+b+1)² = a²+b²+c²+2ab+2bc+2ca

(a+b+c) (a²+b²+c²-ab-bc-ca) = a³+b³+c³-3abc
If a+b+c = 0, then a³+b³+c³ = 3abc
ஃ 7-10+3 = 0
⇒ 7³-10³+3³ = 3\(\times\)7\(\times\)-10\(\times\)3
= 9\(\times\)-70 = -630

(x+4) (x+5) (x+6) = x³+(4 + 5 + 6)x²+(4\(\times\)5 + 5\(\times\)6 + 6\(\times\)4)x + 4\(\times\)5\(\times\)6
= x³+15x²+ (20 + 30 + 24)x + 120
(x+4) (x+5) (x+6) = x³+15x²+74x+120

(4000000)³ ÷ (0.00002)⁴
\( =\left(4.0 \times 10^{6}\right)^{3} \div\left(2.0 \times 10^{-5}\right) 4 \)
\(=(4.0)^{3} \times\left(10^{6}\right)^{3} \div(2.0)^{4} \times\left(10^{-5}\right)^{4} \)
\(=\frac{64.0 \times 10^{18}}{16.0 \times 10^{-20}} \)
\(=4 \times 10^{18} \times 10^{+20} \)
\(=4.0 \times 10^{38} \)

\( =1.598 \times 10^{17}-4.58 \times 10^{15} \\ =\left(1.598-4.58 \times 10^{-2}\right) 10^{17} \\ =(1.598-0.0458) \times 10^{17} \\ =1.5522 \times 10^{17} \)

\((\sqrt[3]{49})^{5}=\left(\sqrt[3]{7^{2}}\right)^{5}=7^{\frac{2 \times 5}{3}}=7^{\frac{10}{3}}\)

(i) (P∪Q)∪R
(P∪Q) = {1, 2, 5, 7, 9} U {2, 3, 5, 9, 11} = {1,2,3,5,7,9,11}
(P∪Q)∪R = {1,2,3,5,7,9,11} U {3,4,5,7,9} = {1,2,3,4,5,7,9,11}
(ii) (P⋂Q)⋂S
(P⋂Q) = {1,2,5,7,9} ⋂ {2,3,5,9,11} = {2,5,9}
(P⋂Q)⋂S = {2,5,9} ⋂ {2,3,4,5,8} = {2,5}
(iii) (Q⋂S)⋂R
(Q⋂S) = {2,3,5,9,11} ⋂ {2,3,4,5,8} = {2,3,5}
(Q⋂S)⋂R = {2,3,5} ⋂ {3,4,5,7,9} = {3,5}

Given, A = {2, 3, 4, 5},B = {2, 3, 5,7} and C = {1, 3, 5}
Now, \(B\cup C\) = {1,2, 3,5,7}
\(A\cup (B\cup C)\) ={1, 2, 3, 4, 5, 7}        .....(1)
Then, \(A\cup B\) = {2, 3, 4,5,7}
\((A\cup B)\cup C\) = {1,2, 3, 4, 5, 7}    ...(2)
From (1) and (2), it is verified that \(A\cup (B\cup C)\) = \((A\cup B)\cup C\).

Given U = {−2,−1,0,1,2,3,4,5,6,7,8,9,10},
A = {−1,1,3,5,7,9} and B = {−2,1,4,7,10}
Law (i) \(\left( A\cup B \right)' =A'\cap B'\)
Now, AUB = {-2,-1,1,3,4,5,7,9,10}
\(\left( A\cup B \right) '\)= {0,2,6,8}   ............... (1)
Then, A'= {-2,0,2,4,6,8,10} and B' = {-1,0,2,3,5,6,8,9}
\(A'\cap B'\) = {0,2,6,8}  ........... (2)
From (1) and (2), it is verified that \(\left( A\cup B \right) =A'\cap B'\)
Law (ii) \(\left( A\cap B \right) '=A'\cup B'\)
Now, \(A\cap B\) = {1,7}
\((A\cap B)'\) = {-2,-1,0,2,3,4,5,6,8,9,10} ........ (3)
Then \(A'\cup B'\) = { -2, -1,0,2,3,4,5,6,8,9,10} ......... (4)
From (3) and (4), it is verified that \(\left( A\cap B \right) '=A'\cup B'\)

From (1) and (2), it is verified that \(\left( A\cup B \right) '\)=\(A'\cap B'\)

Data in ascending order 27, 27, 28, 28, 29, 31, 32, 33, 33, 33, 34, 35, 63 
Number of terms = 13
If n is odd rhen Median = \(\left[\frac{\mathrm{N}+1}{2}\right]^{t h}\) observation
\(=\left(\frac{13+1}{2}\right)^{t h}\)observation
= 7^th observation
7^th observation = 32

Let us arrange the marks in the two subjects in ascending order.

Since the number of students is 12, the marks of the middle-most student would be the mean mark of 6^th and 7^th students.
Therefore, Median mark in Mathematics \(={44+50\over2}=47\)
Median mark in Science \(={42+42\over2}=42\)
Here the median mark in Mathematics is greater than the median mark in Science. Therefore, the level of achievement of the students is higher in Mathematics than Science.

Let, p(x) = x³+13x²+ 32x + 20
Sum of all the coefficients = 1 + 13 + 32 + 20 = 66 ≠ 0
Hence, (x -1) is not a factor.
Sum of coefficients of even powers and constant term = 13 + 20 = 33
Sum of coefficients of odd powers = 1 + 32 = 33
Hence, (x +1) is a factor of p (x)
Now we use synthetic division to find the other factors

If 2x-3y−4z = 0 then 8x³− 27y³− 64z³=?
If x+y+z = 0 then x³+y³+z³ = 3xyz
8x³− 27y³− 64z³ = (2x)³+(−3y)³+(−4z)³
= 3\(\times\)2\(\times\)−3y\(\times\)−4z = 72xyz

Let p(x) = 2x³- 6x²+ mx + 4
By factor theorem, (x - 2) is a factor of p(x) , if p(2) = 0
[To find the zero of x-2;
put x-2 = 0
we get x = 2
p(2) = 0
2(2)³- 6(2)²+ m(2)+ 4 = 0
2(8)- 6(4)+2m + 4 = 0
- 4+2m = 0
m = 2

\(\frac { \sqrt { 7 } -2 }{ \sqrt { 7 } +2 } =a\sqrt{7}+b\)
L.H.S = \(\frac{\sqrt{7}-2\times\sqrt{7}-2}{\sqrt{7}+2\times\sqrt{7}-2}=\frac{(\sqrt{7}-2)^2}{\sqrt{7}^2-2^2}=\frac{\sqrt{7}^2-2\sqrt{7}\times2+2^2}{7-4}\)
=\(\frac{7-4\sqrt{7}+4}{3}=\frac{11-4\sqrt{7}}{3}=\frac{11}{3}-\frac{4\sqrt{7}}{3}\)
\(\frac{-4\sqrt{7}}{3}+\frac{11}{3}=a\sqrt{7}+b\)
ஃa √7=\(\frac{-4\sqrt{7}}{3}\)
a = \(\frac{-4}{3}\)
b = \(\frac{11}{3}\)

\(\frac { 5\sqrt { 3 } +\sqrt { 2 } }{ \sqrt { 3 } +\sqrt { 2 } } =\frac{(5\sqrt{3}+\sqrt{2})\times(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})\times(\sqrt{3}-\sqrt{2})}\)

=\(\frac{5\sqrt{3}\times\sqrt{3}+\sqrt{2}\times\sqrt{3}-5\sqrt{3}\times\sqrt{2}-\sqrt{2}^2}{\sqrt{3}^2-\sqrt{2}^2}\)
=\(\frac{5\times3+\sqrt{6}-5\sqrt{6}-2}{3-2}=\frac{13-4\sqrt{6}}{1}=13-4\sqrt{6}\)

K = {a,b,d,e,f}, L = {b,c,d,g} and M = {a,b,c,d,h}
(i) K\(\cup\)(L\(\cap \)M)
L\(\cap \) M = {b,c,d,g} \(\cap \) {a,b,c,d,h} = {b,c,d}
K\(\cup\)(L\(\cap \)M) = {a,b,d,e,f} \(\cup\) {b,c,d} = {a,b,c,d,e,f}
(ii) K∩(L\(\cup\)M)
L\(\cup\)M = {a,b,c,d,g,h}
K⋂(L \(\cup\) M) = {a,b,d,e,f) \(\cap \) {a,b,c,d,g,h} = {a,b,d}
(iii) (K \(\cup\) L) \(\cap \) (K\(\cup\)M)
K \(\cup\) L = {a,b,c,d,e,f,g}
K \(\cup\) M = {a,b,c,d,e,f,h}
(K \(\cup\) L) \(\cap \) (K \(\cup\) M) = {a,b,c,d,e,f}
(iv) (K \(\cap \) L) \(\cup\) (K \(\cap \) M)
(K \(\cap \) L) = {b,d}
(K \(\cap \) M) = {a,b,d}
(K \(\cap \) L) \(\cup\) (K \(\cap \) M) = {b,d} \(\cup\) {a,b,d} = {a,b,d}

Given: Chord AB of the outer circle cuts the inner circle at C and D.

To prove: AB − CD = 2AC
Construction: Draw OM ⊥ AB
Proof: Since, OM ⊥ AB (By construction)
Also, OM ⊥ CD
Therefore, AM = MB ... (1) (Perpendicular drawn from centre to chord bisect it)
CM = MD ... (2)
Now, AB – CD = 2AM–2CM
= 2(AM–CM) from (1) and (2)
AB – CD = 2AC

In PQR,
PQ = 5 cm,
PR = 6 cm
\(\angle \)QPR = 60°

Construction :
Step 1: Draw △ PQR with the given measurement
Step 2: Draw perpendicular bisectors of any two sides (PQ and QR) to find the mid points of PQ and QR.
Step 3: Draw medians PD and RE. Let them meet at G.
Step 4: G is the centroid of the given △PQR.

In \(\triangle \)ABC, AB = 6 cm, LB= 110°, AC = 9 cm

Construction :
Step 1: Draw \(\triangle \)ABC with AB = 6 cm, ∠B =110°, AC = 9 cm
Step 2: Draw perpendicular bisectors of any two sides (BC and AB) to find the mid points of BC and AB.
Step 3: Construct medians AD and CE. Let them meet at G.
Step 4: G is the centroid of the given \(\triangle \)ABC.

Step 1 : Draw the ΔABC with AB = 6cm, ㄥB = 65° and AC = 7cm

Step 2 : Construct the angle bisectors of any two angles (A and B) and let them meet at I.
Then I is the incentre of ΔABC. Draw perpendicular from I to any one of the side (AB) to meet AB at D

Step 3: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally

Step 4: Measure inradius 
In radius = 1.9 cm