We have, (- 21) x (- 30) = 21 x 30 = 630

We have,\(\frac{9}{5}\times \frac{3}{5}=\frac{9\times 3}{5\times 5}=\frac{27}{25}\)

B = - 8, C = - 7, I = -1, J = 0,K = + 1

Given, a = 75 and b = 84
\(\therefore\) LHS = a - (-b) = 75 - (-84) = 75 + 84 =159
and RHS = a + b = 75 + 84 = 159
Hence, LHS = RHS

On arranging the data in ascending order, we get 128,132,135,139,141,143,146,149,150,151
Mean heigher of the girls\(=\frac { Sum\ of\ height\ of\ all\ girls }{ number\ of\ girls } \)
= \(\frac { 128+132+135+139+141+143+146+149+150+151 }{ 10 } \)
= \(\frac { 1410 }{ 10 } \)
=141.4
Hence, the mean height of the girls is 141.4 cm

In the given figure \(\angle \)1 and \(\angle \)2 are not adjacent angles because \(\angle \)1 is a part of \(\angle \)2, i.e. the other arms of the angles marked \(\angle \)1 and \(\angle \)2 are not on the opposite of the common arm or \(\angle \)1 and \(\angle \)2 do not fulfill the condition that there are no common interior point.

We have, \(\frac{7}{10}+\frac{2}{5}+\frac{3}{2}\)
\(\therefore \frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7+4+15}{10}\)
=\(\frac{26}{10}=\frac{13}{5}\)
[dividing numerator and denominator by 2]

We have, -2 and 5
In this case, firstly we go to (-2) and then move (5) step to the right of (-3).
Thus, we reached to (3), i.e. (-2) + (5) = 3

Given, the product of two decimal numbers is 2.2144
One of the decimal number = 0.64
Let the other decimal number be x.
So, according to the question.
\(0.64 \times x=2.2144\)
\(x=\frac{2.2144}{0.64} \Rightarrow x=\frac{22144}{10000}\times \frac{100}{64}\)
x=\(\frac{22144}{100}\times \frac{1}{64}=\frac{346}{100}=3.46\)
x = 3.46
Hence, the other number is 3.46.
Check = 3.46\(\times\)0.64= 346\(\times\)64= 22144
Place decimal point after 4 digits starting from the
extreme right of 22144 and 64.
So, 3.46x 0.64= 2.2144

We know that, the sum of all three angles in a triangle is equal to 180°.
So, x + 55° + 90° =180°
\(\Rightarrow\) x + 145° =180°
\(\Rightarrow\) x = 180° -145°
\(\Rightarrow\) x = 35°

For a pair of integers, whose difference is -19.
= -12-7 = -19

We have, l II m and t is a transversal.
\(\therefore \quad \angle x=\angle 1\) [ alternate interior angles]

Now, \(\angle \)1+110°=180°    [by linear pair]
 \(\therefore\) \(\angle \)1=180°-110°=70°  \(\Rightarrow\) \(\angle \)x = 70° 
Hence, the required value of x is 70°.​​​​​​

\(\angle A=90^0\) [Hint Use the concept of is isosceles triangle]

Number of students getting marks from 50-69
= Number of students getting marks from 50-59 + Number of students getting marks from 60-69
= 7+11=18

Here, the sea level is at 0 m and the plane is 5000m above the sea level.
\(\therefore\)  Distance between plane and the sea level = 5000 m
Also, the submarine is floating 1200 m below the sea level.
\(\therefore\) Distance between the submarine and the sea level = 1200 m
Hence, the vertical distance between the plane and the submarine = Distance between the plane and the sea level + Distance between the sea level and submarine
= 5000 + 1200 = 6200 m

Average number of points scared by A = 12.5 [as calculated in part (i)]
Average number of points scared by B = 4.5 [as calculated in part (iii)]
\(=\frac { sum\ of\ points\ in\ games }{ number\ of\ games } \)
= \(\frac { 8+11+13 }{ 3 } =\frac { 32 }{ 3 } \) =10.67
\(\because\)  12.5 > 10.67 > 4.5
Hence, A is the best performer.

Given, distance between A and B = 7.5 km
Distance between Band C = 12.7 km
∴ Distance travelled by Dinesh = AB + BC
= 7.5+ 12.7
= 20.2km
Distance between A and D = 9.3 km
Distance between D and C = 11.8 km
∴ Distance travelled by Ayub = AD + DC
= 9.3+ 11.8
= 21.1 km
Hence, it is clear that 21.1>20.2.
∴ Ayub travelled more distance i.e. 21.1 km.
Now, difference between distances, they both travelled
= (21.1- 20.2) km
=0.9km
Hence, Ayub travelled 0.9 km more than Dinesh.

Let the number of steps moved down to be represented by negative integers and number of steps moved up to be represented by positive integers. Given that, the total number of steps to be moved down to reach water level= 9
(i) Number of steps moved down in one jump = - 3
and number of steps moved up in one jump = 2

From the above figure, monkey will reach the water level in 11 jumps.
Alternate Method:
The monkey is sitting on the topmost step i.e. the first step. While going down, the monkey jumps 3 steps down and then jumps back 2 steps up. To reach the water level, he has to jump as follows:
-3 + 2 -3 + 2 -3 + 2 -3 + 2 -3 + 2 -3 = -8
Hence, he takes 11 jumps to reach the water level.
ii) In one jump number of steps moved up = 4
In one jump, number of steps moved down = - 2

Distance covered in single jump = 4 - 2 =2
Now, total number of steps for top = 9
From the above figure, he will reach back the top step in 5 jumps.
Alternate Method
Afterdrinkingwaters,he has to jump as followsto go back:
4 -2 + 4 - 2 + 4 = 8
Hence, he takes 5 jumps to reach back the top
(iii) (a) We have,
-3 + 2 -3 + 2 - 3 + 2 - 3 + 2 - 3 + 2 -3 = -8
(b) 4 - 2 + 4 - 2 + 4 = 8
According to the question, the sum 8 in part (b) represents going up 8 steps.

(i) Given, AB = \(\frac{5}{2}\)cm
BE = 2 \(\frac{3}{4}\)cm \(=\frac{2\times4+3}{4}=\frac{11}{4}\)cm
and AE = 3\(\frac{3}{5}\)cm\(=\frac{3\times5+3}{5}=\frac{18}{5}\)cm
Now, perimeter of △ABE = Sum of all sides of a triangle = AB + BE + AE = \(\frac{5}{2}\)+\(\frac{11}{4}\)+\(\frac{18}{5}\)

∴ LCM of 2, 4, 5 = 2 x 2 x 5 = 20
ஃ \(\frac{5}{2}\)+\(\frac{11}{4}\)+\(\frac{18}{5}\)
\(=\frac{5\times10+11\times5+18\times 4}{20}\)
=\(\frac{50+55+72}{20}=\frac{177}{20}\) and or \(8\frac{17}{20}\) cm
(ii) Given, length of rectangle BCDE,
I = BE = CD = \(2-\frac{3}{4}\) cm \(=\frac{2\times 4+3}{4}=\frac{11}{4}\)cm
and breadth of rectangle, b = BC = DE =\(\frac{7}{6}\)cm
∴ Now, perimeter of rectangle BCDE = 2(l + b)
\(=2(\frac{11}{4}+\frac{7}{6})=2(\frac{11\times 3+7\times 2}{12})\)    [∵ LCM of 4 and 6 = 12]
=\(2(\frac{33+14}{12})=\frac{47}{6}\)cm
Now, perimeter of MBC =\(\frac{177}{20}\)cm
and perimeter of rectangle BCDE =\(\frac{47}{6}\)cm

LCM of 20 and 6 = 2 x 2 x 3 x 5 = 60
On converting both fractions into like fractions, we get
\(\frac{177}{20}\)=\(\frac{177\times 3}{20\times 3}=\frac{531}{60}​​\)cm
and \(\frac{47}{6}=\frac{47\times 6}{47\times10}=\frac{470}{60}​​\)
Since, 531>470   [numerators of fractions]
So, \(\frac{531}{60}>\frac{470}{60} i.e., \frac{177}{20}>\frac{47}{6}\)
Hence, the perimeter of triangle is greater than the perimeter of rectangle.

Given, x + 7 = 16  \(\Rightarrow\) x = 16 - 7 = 9
So, x + 7 = 16 has solution 9.

Mode = maximum occuring observation = 67  (occurs 2 times)

Since, arithmetic mean of 3,9,4,x 9,7,6 is 6
Arithmetic mean = \(\frac { sum\ of\ observations }{ Number\ of\ observations } \)
6 = \(\frac { 3+9+4+x+9+7+6 }{ 7 } \)
6 x 7 = 38 + x \(\Rightarrow\) 42 = 38 + x
x = 42-38 = 4
x = 4

(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°