-5

Sum of 5x² - 6x + 4 and -x²r - 2x + 3
= (5x² - 6x + 4) + (-4x² - 2x + 3)
= 5x² - 6x + 4 - 4x² - 2x + 3
= (5x² - 4x²) + (-6x - 2x) + (4 + 3)
= (5 - 4)x² + (-6 - 2)x + (7)
= (I)x² + (-8x) + 7
=x²-8x+7
Now subtract x² - 8x + 7 from 0, we get 
∴  0 - [x² - 8x + 7] = 0 - x² + 8x - 7
= -x² + 8x - 7

To finish the work in 10 months, number of men required = 480
To finish the work in 1 month, number of men required = 480 x 10
To finish the work in 8 months, number of men required = \(480\times10\over8\)
\(={60\times10\over 1}=600\)
∴ Number of extra men required
= 600 - 480 = 120
Thus, to complete the work in 8 months, 120 extra men are to be employed.

40°, 60°, 80°

ΔPQR and ΔLMN:
We have 


Therefore, using ASA congruence rule, the two triangles are congruent.
Now, R \(\leftrightarrow \) L,Q \(\leftrightarrow \) N and P\(\leftrightarrow \) M
\(\therefore \)ΔPQR ≅ ΔMNL

Area of circle = \(\pi\)r²
\(\Rightarrow\) \(\pi\)r² = 38.5 cm²
\(\Rightarrow\)\(\frac { 22 }{ 7 } \) r² = 38.5
\(\Rightarrow\) r² = \(\frac { 38.5\times 7 }{ 22 } \)
\(\Rightarrow\) r² = 12.25
\(\Rightarrow\) r = 3.5cm
Circumference = 2\(\pi \)r
= 2 x \(\frac { 22 }{ 7 } \)x 3.5
= 22 cm

In the given figure \(\angle AOB\) and \(\angle COB\) are the angles of linear pair.
So, \(\angle AOB+\angle BOC=180^0\)
(3x + 10°) + (2x - 30°) = 180°
\(\Rightarrow\) 3x + 10° + 2x - 30° = 180°
\(\Rightarrow\) 5x - 20° = 180°
\(\Rightarrow\) 5x = 180° + 20°
\(\Rightarrow\) 5x = 200°
\(\Rightarrow x=\frac{200^0}{5}\)
Thus, x=40⁰
Now, \(\angle AOB=3x+10^0\)
= 3 (40°) + 10°
= 120° + 10°
= 130°.​​​​​​​

Since,
\(\frac { 2x-1 }{ 3 } -\frac { 6x-2 }{ 5 } =\frac { 1 }{ 3 } \)
\(\therefore \quad \frac { 5(2x-1) }{ 3\times 5 } -\frac { 3(6x-2) }{ 3\times 5 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \quad \frac { 10x-5 }{ 15 } -\frac { (18x-6) }{ 15 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \quad \frac { 10x-5-18x-6 }{ 15 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \quad \frac { -18x+1 }{ 15 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \) -18x+1=\(\frac { 15 }{ 3 } =5\)
\(\Rightarrow\) -18x = 5 - 1 = 4
\(\Rightarrow \quad x=\frac { 4 }{ -18 } \)
Thus, x=\(\frac { -2 }{ 9 } \)

Steps of Construction:
(a) Draw a line segment BC = 4·5cm.
(b) Construct \(\angle CBX\) = 60°​​​​​​​ at B.

(c) Construct \(\angle BCY\) = 40°​​​​​​​ at C.
(d) Let the rays \(\overline { BX } \) and \(\overline { CY } \)intersect at A.
Thus, ΔABC is required triangle.

We have,
n = Number of observations = 40, Mean = 160
\(\therefore Mean=\frac{Sum\ of\ the\ observations}{Number\ of\ observations}\)
\(\Rightarrow 160=\frac{Sum\ of\ the\ observations}{40}\)
\(\Rightarrow 160\times40\)  = Sum of the observations.
Thus, incorrect sum of observations = 160 x 40
Now,
Correct sum of the observations = Incorrect sum of the observations - Incorrect observation + correct observation
\(\Rightarrow\) Correct sum of the observations \(=160\times40-125+165\)
\(\Rightarrow\) Correct sum of the observations
= 6400 + 40 = 6440
\(\therefore\) correct mean
\(=\frac{Correct\ sum\ of\ the\ observations}{Number\ of\ observations}\)
\(=\frac{6440}{40}=161\)

$ 3200

Total scores of team A = - 30 + 5 + 0
= -30 + 5 = -25
Total scores of team B = 20 + 0 + (-10) = 20 -10
= 10
Hence, score of team B was more than team A.
Yes, we can add integer in any order.

Given, a=137, b=92
Taking LHS,
a + (-b) =137 + (-92) = 45
Now, taking RHS,
a - b = 137 - 92 = 45
Hence, a+(-b) = (a-b)

1972

Cricket is most popular sport

(i) Arrange the given data (in kg) in ascending order,
we get 32,35,36,37,38,38,38,40,42,43,43,43,45,47,50
Now, arrange it in tabular form

Here, 38 and 43 occur more frequently i. e. 3 times.
\(\therefore\)  Mode = 38 and 43
The value of the middle observation = 40
\(\therefore\) Median = 40
(ii) Yes,there are two mode i.e. 38 and 43.

From the given figure, it is clear that the directions East and West are opposite to each other.
According to the question,

If moving towards East is a positive integer, then moving towards West is represented by a negative integer.
Now, distance moved towards West = - 30 km 
and distance moved towards East = 20 km
\(\therefore\) Rita's final position from A = 20 + (- 30) = 20 - 30
= -10 km (i.e. West)

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

We have 7, 3, - 1, - 5, ___, ___, ___
Difference between first two term = Second term - First term = 3 - 7 = - 4
Similarly, difference between another two terms = (-1) - 3 = -1- 3 = - 4
Thus, the difference between two consecutive term is - 4.
Now, add this difference from first to the last term
i.e. 7 + (-4)
= 7 - 4 = 3; 3 + (-4) = 3 - 4 = -1; -1 + (-4)
= - 5; -5+ (-4)= - 5 - 4= -9; -9 + (-4)
= -9 - 4= -13 and -13+(-4)= -13-4= -17
Hence, the pattern is 7, 3, -1, - 5, - 9, - 13,- 17.

The frequency table is shwn below.

From the above frequency table,
Highest number = 9
Lowest number = 1
\(\therefore\)  A range of data = Highest number - Lowest number
     =9-1=8
Hence, the range of the data is 8.