-13

125

Since, each of the above three positions fits into the original.
∴ It has a rotational symmetry of order 3.

In \(\triangle \)PLO and \(\triangle \)PMO, we have
\(\angle PLO=\angle PMO=\)90°[Given]
\(\overline { OP } =\overline { OP } \)  [Hypotenuse]
PL=PM      [Given]
\(\therefore \)Using RHS congruency, we get 
\(\triangle \)PLO \(\cong \)\(\triangle \)PMO

\(\because \)  Large number of data is given
\(\therefore \)Putting them in a tabular form, we have

Heights (in cm)	Tally marks	Number of students (Frequency)
160		5
161	\(\parallel \)	2
162		5
163		10
164	\(||||\)	4
165	\(|||\)	3
166		0
167		0
168	\(|\)	1

\(\because \)The frequency of 163 cm is maximum, i.e. 10.
\(\therefore \)163 em occurring the maximum number of times.
Thus, the mode of the given data is 163 cm.

Fraction =\(\frac { 5 }{ 9 } \)
5 is added to both numerator and denominator.
New fraction =\(\frac { 5+5 }{ 9+5 } =\frac { 10 }{ 14 } \)
By comparing fraction \(\frac { 5 }{ 9 } ,\frac { 10 }{ 14 } \)
   [by cross multiplication]
5\(\times\)14 < 9\(\times\)10
70 <  90
So, new fraction is increased in value as compared to the given fraction

A = 3x² + 2x and B = 3x + 1
A - B = 3x² + 2x - 3x - 1
= 3x² -x-1

Perimeter of square = 4 x side
\(\Rightarrow\) 4 x side = 200
\(\Rightarrow\) side \(={200\over 4}\) = 50 m
\(\Rightarrow\) Area of park = (side)²
= (50)² = 50 x 50
= 2500 m².

118-[121 ÷ (11 \(\times\) 11)-(-4)-{3- \(\overline { 9-2 } \)}]
= 118-[121 ÷ (11 \(\times\) 11)-(-4)-{3-7}]
= 118- [121 ÷ 121- (- 4) - {3- 7}]
= 118- [1- (- 4) - {- 4}]
= 118- [1 + 4 + 4]
= 118 - 9
= 109.

Let us place two cubes with 2 cm edge side by side as shown in the diagram. Its length only increases and becomes (2 + 2) cm. i.e., 4 cm. But breadth and height remain same as 2 cm.

We have,
[32 + 2 \(\times\) 17 + ( - 6) ] ÷ 15
= [32 + 34 + (- 6)] ÷ 15
= (66-6) ÷ 15
= 60 ÷ 15
=\(\frac{60}{15}\)
=4

We have, -91 ÷ 13 = \(\frac{-91}{13}=-\frac{91}{13}\)
=-7

Since, b - 2 = 1
⇒ b = 1 + 2
⇒ b = 3

6

(i) ➝ Side, (ii) ➝ Top, (iii) ➝ Front

We have, (- 75) \(\div\) 5 = - (75 \(\div\) 5) = -75/5 = -15

We have, (-1) x (- 2) x (- 3) x 4 = - (1 x 2 x 3 x 4)
= -24

We have, 6 x (-19) = - ( 6 x 19 ) = -114

A pair of integers, whose sum is - 7, can be - 2 and - 5.
\(\because\)  Sum = -2 + (-5) = -2 -5 = -7

18 - (-24)
​​​​​​​Forsubtraction 18 - (-24) = 18 + 24                      [\(\because\) (-)(-) = (+)]
= 42                                             [\(\because\) higher integer sign is (+)]
Verification First integer (x) = 18           [given]
Second integer (y) = (-24)             [given]
For closure property x - y = 18 - (-24) = 18+ 24 = 42 (integer) i.e. subtraction is closed.

9 + (-64)
​​​​​​​For addition 9 + (-64) = 9 - 64                 [\(\because\) (+)(-) = (-)]
= - 55                                 [\(\because\) higher integer sign is (-)]
Verification First integer (x) = 9          [given]
Second integer (y) = -64          [given]
For closure property x + y = 9 + (-64) = 9 - 64 = - 55 (integer) i.e. addition is closed

Given, a = 28 and b = 11
\(\therefore\) LHS = a - (-b) = 28 - (-11) = 28+11 = 39
and RHS = a + b = 28 + 11= 39
Hence, LHS = RHS

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

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