Depth drilled in one day = –22 feet
Depth of water = –110 feet
Number of days required = –110 ÷ –22 = 5
Hence the workers wil reach resource in 5 days.

Distance from A to D = 7 m
Distance from A to D (via B) = 4 + 5 = 9 m
Distance from A to D (via C) = 2 + 6 = 8 m
Therefore, the route directly from A to D requires minimum cable of length 7 m.

Let the number of correct answers be x
Thus the number of wrong answers = 60 − x
Then, 4x − 1 (60 − x) = 130
4x − 60 + x = 130
4x + x − 60 + 60 = 130 + 60 [Add 60 on both sides]
5x + 0 = 190
5x = 190
Divide by 5 on both the sides,
\(\frac { 5x }{ 5 } =\frac { 190 }{ 5 } \)=38
x = 38
Hence, the number of correct answer is 38.

The numbers are natural and consecutive. Let the numbers be x and x + 1.
Given that,
x + (x + 1) = 75
2x + 1 = 75
2x + 1 − 1 = 75 − 1 [Subtract 1 on both sides]
2x + 0 = 74
\(\frac {2x }{2 } =\frac { 74 }{ 2 } \) [Divide 2 at both sides]
x = 37 and x + 1 = 38
Therefore, the required numbers are 37 and 38.

Consider, [3 x (–4) + 6],3 x  [(–4) + 6] = 3 x 2 = 6
Consider, [3 x(–4)] + [3 x 6],[3 x(–4)] + [3 x 6] = –12 + 18 = 6
Therefore, [3 x [(–4) + 6] and [3 x (–4)] + 3 x 6 are equal.
It is the distributive property of multiplication over addition.

In the first case (–2) and (3) are grouped together and in the second case (3) and (–4) are grouped together
L.H.S = [(–2) 3] x (–4)
= (–6) x (–4) = 24
R.H.S = (–2)  x [3 x (–4)]
= (–2) x (–12) = 24
Therefore, L.H.S. = R.H.S.
[(–2) x  3] x (–4) = (–2) x [3 x (–4)]
Hence it is proved.

(–7) x (+8) = (–56)
Hence, –56 is an integer.
Therefore, (–7) x (+8) is closed under multiplicaton.

Profit of 1 kg mangoes = Rs. 15
Profit of 5 kg mangoes = Rs. 15 x 5
= Rs. 75
Loss of 1 kg apples = Rs. 30
Loss of 3 kg apples = 30 x 3
= Rs. 90
Loss = Rs. 90 − Rs. 75
= Rs. 15

1.

Therefore, there are nine tetrominoes in a tile.
2. Given, the cost of a tile is Rs. 52
There are six tiles in the front elevation
Therefore, the total cost = 6 x 52 = Rs. 312

Let, the length of the required side be ‘x’ cm.
Then, area of the trapezium = \(\cfrac { 1 }{ 2 } \times h\left( a+b \right) sq.units\)
= \(\cfrac { 1 }{ 2 } \times 16(25+x)\)
= 200 + 8x
But, the area of the trapezium = 352 sq. cm (given)
Therefore, 200 + 8x = 352
\(\Rightarrow\) 8x = 352 – 200
\(\Rightarrow\) 8x = 152
 \(\Rightarrow\) \(x=\cfrac { 152 }{ 8 } \)
x = 19
Therefore, the length of the other side is 19 cm.

Given, height (h) = 9 cm
Parallel sides are (a) = 23 cm and (b) = 12 cm
Area of the trapezium = \(\cfrac { 1 }{ 2 } \times h(a+b)\)
= \(\cfrac { 1 }{ 2 } \times 9\left( 23+12 \right) \)
= \(\cfrac { 1 }{ 2 } \times 9(35)\)
= 157.5 sq. cm
Therefore, Area of the trapezium is 157.5 sq. cm.

(i) Since the angles are linear pair, \(\angle\)ACD + \(\angle\)BCD = 180°
123° + \(\angle\)BCD = 180°
Subtracting 123° on both sides
123° + \(\angle\)BCD – 123° = 180° – 123°
\(\angle\)BCD = 57°
(ii) Since the angles are linear pair, \(\angle\)LNO + \(\angle\)MNO = 180°
46° + \(\angle\)MNO = 180°
Subtracting 46° on both sides
46° + \(\angle\)MNO – 46° = 180° – 46°
\(\angle\)MNO = 134°

We know that \(\angle\)POR = \(\angle\)POQ + \(\angle\)QOR
62° = 23° + \(\angle\)QOR
Subtracting 23° on both sides
62° – 23° = 23° + \(\angle\)QOR – 23°
\(\angle\)QOR = 39°

Time taken to cover 90 km = 2hrs 30mins
= 150 minutes [\(\begin{matrix} 1hour=60minutes \\ 2hour=120mnutes \end{matrix}\)]
Time taken to cover 1 km = \(\frac { 150 }{ 90 } \) x 210 minutes
= 350 minutes
= 5 hours 50 minutes
Thus, the time taken to travel 210 km is 5 hours 50 minutes.

Let x be the required number of charts.
As weight increases, the number of charts also increases. So the quantities are in direct proportion.
Hence \(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \)

\(\frac { 15 }{ 50 } =\frac { x }{ 2500 } \)
15 x 2500 = x \(\times\) 50
x \(\times\) 50 = 15 x 2500
x = \(\frac { 15\times 2500 }{ 50 } \) = 750
Therefore, 750 charts will weigh 2\(\frac {1}{2}\) kilogram.

i) 4x + 7y = 4 (3) + 7 (2) = 12 + 14 = 26
ii) 3x + 2y – 5 = 3 (3) + 2 (2) – 5 = 9 + 4 – 5 = 8
iii) x – y = 3 – 2 = 1

Let the height of the parallelogram = h cm
Then the base of the parallelogram = 3h cm
Area of the parallelogram = 192 sq. cm
b x h = 192
3h x h = 192
3h² = 192
h² = 64
h x h = 8 x 8
h = 8 cm
base = 3h = 3 x 8 = 24 cm
Therefore, base of the parallelogram is 24 cm and height is 8 cm.

Given: Area = 368 sq. cm , base b = 23 cm
Area of the parallelogram = 368 sq. cm
b × h = 368
23 × h = 368
\(h=\cfrac { 368 }{ 23 } \)
Thus, the height of the parallelogram = 16 cm.

(i) From the Fig.2.4
Base of a parallelogram (b) = 15 cm, Height of a parallelogram (h) = 4 cm
Area of a parallelogram = b x h sq.units. Therefore, Area = 15 × 4 = 60 sq. cm.
Thus, area of the parallelogram is 60 sq. cm.
Perimeter of the parallelogram = sum of the length of the four sides.
= (15 + 5 +15 + 5) = 40 cm.
(ii) From the Fig.2.5
Base of a parallelogram (b) = 9 m, Height of a parallelogram (h) = 16 cm.
Area of a parallelogram = b × h sq.units Therefore, Area = 9 × 16 = 144 sq. cm
Thus, area of the parallelogram is 144 sq. cm.
Perimeter of the parallelogram = sum of the length of the four sides.
= (18 + 9 +18 + 9) = 54 cm.

(i) Closure Property
(ii) Commutative Property
(iii) Closure Property
(iv) Commutative Property
(v) Associative Property
(vi) Additive Identity

The amount Sita has Rs. 225
The amount spent for stationery on credit = Rs. 400
The due amount to be paid = Rs. 225 – Rs. 400 = – Rs.175
Therefore, Sita has to pay Rs. 175

(i) (–70) + (–12) = –70 – 12 = – 82
(ii) 103 + 39 = 142

Let us add the intergers using number line
(i) 10 and –15

On the number line we first start at zero facing positive direction and move 10 steps forward, reaching 10. Then we move 15 steps backward to represent –15 and reach at –5. Thus, we get 10 + (–15) = –5.
(ii) –7 and –9

On the number line we first start at zero facing positive direction and move 7 steps backward, reaching –7. Then we move 9 steps backward to represent –9 and reach at –16. Thus, we get (–7) + (–9) = –16.

(i) When we add, 120 + 51 = 171 ; 51 + 120 = 171
In both the cases we get same answer. This means that integers can be added in any order. Hence, addition of integers is commutative.
(ii) (-5) + [(-4) + (-3)] and [(-5) + (-4)] + (-3)
In (–5) + [(–4) + (–3)], (–4) and (–3) are added first and their result is then added with(–5).

(–5) + [(–4) + (–3)] = –12 
Whereas in [(-5) + (-4) + (-3), (–4) and (–3) are added first and then the result is added with (–5)

In both the cases, we get −12
That is (–5) + [(–4) + (–3)] = (–5) + [(–4) + (–3)]
So, addition is associative.