We have, 1.3 \(\times\) 10 = 13

Given, length of the rectangle = 5.7 cm
Breadth of the rectangle = 3 cm
Area of the rectangle = Length X Breadth
=5.7x3=17.1 cm²
Hence, the area of the rectangle is 17.1 cm ².

\(\frac{2}{5}\)

We have, (- 36) \(\div\) (- 4) = 36 \(\div\) 4 = 36/4 = 9

We have, 64 \(\div\) (-16) = -(64 \(\div\) 4) = -64/16 = -4

We have, 80 \(\div\) (- 5) = - (80 \(\div\) 5) = -80/5 = -16

(-15) x (-23) = 15 x 23
Taking LHS = (-15) x (-23) = +345         [  \(\because\) (-a) x (-b) = ab ]
Taking RHS = 15 x 23 = 345                    [  \(\because\) a x b = ab ]
\(\because\)  LHS = RHS
So, property of multiplication of integer is verify.

We have, (-2) x (-4) x (-2)
First integer (x) = - 2                                                                [given]
Second integer (y) = - 4                                                           [given]
Third integer (z) = - 2                                                               [given]
\(\therefore\) Product (xyz) = (-2) x (-4) x (-2) = -(2 x 4 x 2) = -16          [ \(\because\) (-a) x (-b) x (-c) = - (a x b x c) ]

(-21) - (30) = (-21) + (-30) = -51     [\(\because\) (- )(-) = (+), higher integer sign is (-)]

(-6) - (-16) = -6 + (16)= + 10 or 10    [\(\because\) (-)( -) = (+), difference = 10, higher integer sign is '+']

Given, algebraic expression
2x² + 3y² + 5x² - 9y² - 3xy + 5xy
Here, 2x², 5x² are like terms and 3y², 9y² are
like terms. Also, - 3xy and 5xy are like terms.

Given, algebraic expression
5x² - 6xy + 6x² +1
Here, 5x² and 6x² are like terms

Given, 6xy and 3xy
Here, 6xy and 3xy is like pair.

We have, 2.03 or 2.30
First, we compare the digits on the left of the decimal point starting from the leftmost digit. Here, both the digits to the left at the decimal point are same. i.e. 2. So, we compare the digits on the right of the decimal point starting from the tenth place.
Here, 0 < 3
So, 2.03 < 2.30
Hence, 2.30 is greater.

We have, 1.37 or 1.49
First, we compare the digits on the left of the decimal point starting from leftmost digit. Here, both digits to the left of the decimal point are same. i.e. 1.
So,we compare the digits on the right of the decimal point starting from the tenth place.
Here, 3 < 4
So, 1.37 < 1.49
Hence, 1.49 is greater.

We have, 7 or 0.7
First, we compare the digits on the left of the decimal point starting from the leftmost digit.
Here, 7> 0
[∴  7 has no decimal point, so we can write 7 as 7.0]
So, 7.0> 0.7

We have, 0.7 or 0.5
First, we compare the digits on the left of the decimal point i.e. 0 is same. So, we compare the digits on the right of the decimal point starting from the tenth place.
Here, 7> 5 So, 0.7> 0.5
Hence, 0.7 is greater.

Number line with all numbers is shown below:

We have, the temperature of Lahulspiti = - 8° C
and temperature of Srinagar = - 2° C
\(\therefore\) The temperature difference between Lahulspiti and Srinagar
= Temperature of Srinagar - Temperature of Lahulspiti
= (-2) - (-8)= - 2 +8 = 6°C

Number line with all numbers is shown below:

We know that a place whose temperature is highest, will be the hottest place and the place whose temperature is lowest, will be the coldest place.
Here, the temperature of hottest place (Bangalore) = 22° C and temperature of coldest place (Lahulspiti) = - 8° C
\(\therefore\) Temperature difference between the hottest and coldest places = Temperature of Bangalore - Temperature of Lahulspiti.
= 22 - (- 8) = 22 + 8 = 30
Hence, the temperature difference between the hottest and the coldest places is 30°C.

Number line with all numbers is shown below:

From the above number line, the temperature of the indicated places are as follow:

Area = \(\frac { 1\times 1 }{ 2 } =\frac { 1 }{ 2 } \)cm²

As per the given information,
Profit on Monday = Rs 47
Loss on Tuesday = Rs 12
Loss on Wednesday = Rs 8
\(\therefore\) Net profit = Total profit - Total loss
Now, total profit = 47
and total loss =12 + 8 = 20
\(\therefore\) Net profit = 47 - 20 = Rs 27

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

Let O be the point of the level of the sea.
Difference between these two points A and B = Height between sea level and point A + Height between point B and sea level.
= AO + OB = 1800 + 700
= 2500 m

As per the given information, initial position of motor was 380 m.
During the ride, change in elevation was 540 m, -268 m, 116 m, -152 m, 490 m, -844 m and 94 m.
Net change in position
= 540 + (-268) + (116) + (-152) + (490) + (-844) + 94
= -24m
Initial position was 380 m. So, at the end of the ride the position would be = 380 + (-24) = 356 m.

For two integers, which are greater than -10, but their sum is smaller than -9.
So, first integer = - 6 and second integer = - 5
 \(\therefore\) Sum of integers = -6 + (-5) = -11

For two integers which are smaller than -5 but their difference is -5.
So, first integer = -11 and second integer = -6
Their difference = -11 - (-6) = -11 + 6 = - 5

For a pair of integers whose product is -36 and whose difference = 15.
So, first integer = -3 and second integer = 12
Their product = (-3) \(\times\) 12 = -(3 \(\times\) 12)= -36
and the difference between these two integer is 15.

For a pair of integers whose product is -12 and there lies seven integers between them.
First integer = - 2 and second integer = 6
Their product = - 2 \(\times\) 6 = -12
Total number of integers lying between -2 to 6 = 7    [excluding the given integers].

Given, marks for every correct answer = + 3
Marks for every incorrect answer = - 2
(i) Total marks scored by Radhika = 20
Marks scored for correct answer = 12 \(\times\) 3 = 36
\(\therefore\) Marks given for incorrect answers
= Total marks - Marks given for correct answers
= 20 -36= -16
\(\therefore\) Number of incorrect answers
= \(\frac { Total\ Marks\ of\ incorrect\ answers }{ Marks\ of\ every\ incorrect\ answer } \)
= \(\frac { \left( -16 \right) }{ \left( -2 \right) } =\frac { -16 }{ -2 } \)= 8
Hence, Radhika attempted 8 questions incorrectly
(ii) Marks obtained by Mohini = - 5
Marks obtained for 7 correct answers = 7 \(\times\) 3 = 21
\(\therefore\)Marks obtained for incorrect answers
= - 5 - 21 = - 26
\(\therefore\)Number of incorrect answers
= \(\frac { Total~ Marks \ obtained \ for \ incorrect \ answers }{ Marks \ of \ every \ incorrect \ answer } \)
= \(\frac { -26 }{ -2 } \)
= 13

Given, temperature at 12 noon = + 10°C
Rate of change in temperature = - 2°C/h
Total time from 12 noon to mid-night = 12 h
Change in temperature in 12 h = 12°C \(\times\) (- 2) = - 24°C
Temperature at mid-night = 10 + (- 24) = 10 - 24= -14° C
Now, temperature difference between 10°C and - 8°C
= 10 - (- 8) = 10 + 8 = 18° C = 18/2 = 9
So, temperature change of 18° C will take place in 9 h from 12 noon
Thus, the temperature 8° C below 0° (- 8° C) would be at 9 pm.

(i) We know that, 3 \(\div\) 1= 3, so 3 \(\div\) (- 1)= - 3
On comparing with a \(\div\) b = - 3, we get a = 3 and b = -1
Hence, the required pair of integers is (3, -1).
(ii) We know that, 12 \(\div\) 4 = 3, so 12 \(\div\)(- 4) = - 3
On comparing with a \(\div\) b = - 3, we get a = 12 and b = -4
Hence, the required pair of integers is (12, - 4).
(iii) We know that, 9 \(\div\) 3 = 3, so 9 \(\div\) (- 3) = - 3
On comparing with a \(\div\) b = - 3, we get a = 9 and b = -3
Hence, the required pair of integers is (9, - 3).
(iv) We know that,12 \(\div\) 4 = 3, so (-12) \(\div\) 4 = -3
On comparing with a \(\div\) b = - 3, we get a = -12 and b=4
Hence, the required pair of integers is (-12, 4).
(v) We know that, 15 \(\div\) 5 = 3, so 15 + (- 5) = -3
On comparing with a \(\div\) b = - 3, we get a = 15 and b = -5
Hence, the required pair of integers is (15, - 5).

(a) Since profit of Rs 8 is earned on a bag of white cement and a loss of Rs 5 on a bag of grey cement.
Number of white cement bags sold = 3000
Number of grey cement bags sold = 5000
\(\therefore\) Profit = Total number of white cement bags X Profit per bag
= 3000 \(\times\) 8 = Rs 24000
Loss =Total number of grey cement bags X Loss per bag
= 5000 \(\times\) 5 = Rs 25000
\(\because\) Loss > Profit, so Loss = Rs 25000 - Rs 24000 = Rs 1000
(b) Let x be the number of white cement bags sold,
According to the question,
Number of white cement bags X Profit per bag = Number of grey cement bags X Loss per bag
\(\Rightarrow\) x \(\times\) 8 = 6400 \(\times\) 5 \(\Rightarrow\) x = 6400 \(\times\) 5 / 8
\(\Rightarrow\) x = 800 \(\times\) 5 = 4000
Hence, 4000 bags of white cement must be sold.

Given, total number of questions = 10
Marks for a correct answer = 5
Marks for an incorrect answer = (- 2)
Marks for not attempted the question = 0
(i) Mohan's score = (Correct questions \(\times\) Marks) + (Incorrect questions \(\times\) Marks)
= 4 \(\times\) (5) + 6 \(\times\) (- 2)
= 20 + (-12) = 20 -12 = 8
(ii) Reshmas score = 5 \(\times\) 5 + 5 \(\times\) (- 2)
=25 + (-10) = 15
(iii) Heena's score = 2 \(\times\) 5 + 5 \(\times\) (- 2) + 3 \(\times\) 0
=10 + (-10) + 0 = 0

(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°

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

Arranging the observations (marks) in ascending order:
2, 2, 5, 5, 5, 8, 8, 8, 8, 13, 13, 13, 13, 16, 16, 16, 17, 17, 18,18,18,18,18,19,20
Sum of observations
\(2\times2=4\\5\times3=15\\8\times4=32\\13\times4=52\\16\times3=48\\17\times2=34\\18\times5=90\\19\times1=19\\20\times1=20\\\quad\quad\quad\quad\_\_\_\_\\\ Total=314\\ \quad\quad\quad\quad\_\_\_\_\)Number of students = 25
\(\therefore\) Mean marks \(=\frac{Sum\ of\ the\ observations}{Total\ number\ of\ students}\)
\(=\frac{314}{25}=12.5\)
So, the number of students who scored marks more than mean marks is
4 + 3 + 2 + 5 + 1 + 1 = 16.