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32°F is equal to 273.15 K.

(i) 212 °F. This is the boiling point of water; others correspond to the freezing point of water.

(ii) 32 °F. This is the freezing point of water others correspond to the boiling point of water.

(1) ₹ 30, 25 paise 

(2) 34 kg 200 gm 

(3) 50 1 470 ml 

(4) 50 km 250 m 

(5) 37 m 25 cm 

(6) 56 cm 

(7) 26 km 75 m

(8) 193 kg

45 paise + 55 paise

100 paise = 1 ₹

∴ 100 rupees

We cannot subtract 8 mm from 2 mm.

So, convert 1 cm = 10 mm

∴ 20 cm 4 mm

840 ml + 250 ml

= 1090 ml 

= 11 + 90 ml 

∴ 45 l 90 ml

We cannot subtract 60 paise from 50 paise.

So convert 1 ₹ into 100 paise.

₹ 6, 90 paise

∴ ₹ 6, 90 paise

We cannot subtract 60 cm from 30 cm.

So, convert 1 m = 100 cm

∴ 2 m 70 cm

5 mm + 9 mm 

= 14 mm 14 mm 

= 1 cm 4 mm 

∴ 14 cm 4 mm

(1 – b), 

(2 – d), 

(3 – a), 

(4 – c)

(1 – b), 

(2 – d), 

(3 – a), 

(4 – e), 

(5 – c)

1. (False) 

Correct Statement: One square metre is the area enclosed inside a square of side 1 metre.

2. True

3. (False) 

Correct statement: Density of water is 1000 kg/m³

4. True.

5. (False) 

Correct statement: The lightness or heaviness of a body is due to density.

6. True.

7. True.

8. (False) 

Correct statement: The area of a figure is the region covered by the boundary of the figure.

9. True.

10. True.

1. Archimedes 

2. 10,00,000 or 10⁶ 

3. 13,600 kg/m³ 

4. 1.496×10¹¹ m 

5. graph sheet

Given length of the arc = 16 cm

Perimeter of the arc = 30 cm

l + 2r = 30

16 + 2 r = 30

2 r = 30 – 16

2 r = 14

r = 14/2

r = 7 cm

Radius of the sector = 7 cm

Length of the arc of the sector l = 50 mm

Radius r = 14 mm

Area of the sector = lr/2 sq. units 

= (50×14)/2 mm² 

= 50 × 7 mm² 

= 350 mm²

Area of the sector = 350 mm²

Euler’s formula is given by F + V- E = 2

(i) V = 6, E = 14

By Euler’s formula

= F + 6 – 14 = 2

F = 2 + 14 – 6

F = 10

(ii) F = 8, E = 10

By Euler’s formula

= 8 + V – 10 = 2

V = 2 – 8 + 10

V = 4

(iii) F = 20, V = 10

By Euler’s formula

= 20 + 10 – E = 2

30 – E = 2

E = 30 – 2

E = 28

Tabulating the required unknowns

Side of the square = 30 cm

∴ Radius of the sector design = 30 cm

Given design in the design of a circular quadrant.

Area of the quadrant = 1/4 πr² sq. units

= 1/4 × 3.14 × 30 × 30 cm²

= 3.14 × 15 × 15 cm²

∴ Area of the sector design = 706.5 cm² (approximately)

(a) Width of the door that Guna fixed = 3 feet.

When the door is open the radius of the sector = 3 feet

Angle covered = 120°

∴ Area required to open the door = 120°/360° × πr² 

= 120°/360° × π × 3 × 3 

= 37π feet²

(b) Width of the double doors that Nathan fixed = 112 feet.

Angle described to open = 120°

Area required to open = 2 × Area of the sector

= 2 × 120°/360° × π × 3/2 × 3/2 feets² = 3π/2 feet²

= 12 (3π) feet²

∴ The double door requires the minimum area.

Let A be the position of the wall AC be the gate in initial position and AB be position when it is moved 90°.

Now the arc length BC gives the distance moved by the wheel.

Length of the arc

= θ°/360° × 2πr units

= 90°/360° × 2 × 3.14 × 6 feets

= 3.14 × 3 feets

= 9.42 feets

∴ Distance moved by the wheel = 9.42 feets.

22/7 and 3.14 are rational numbers n has non-terminating and non -repeating decimal expansion. So it is not a rational number. It is an irrational number.

If r = 2r₁ 

⇒ Area of the circle = πr² = π(2r₁)² = π4r₁² = 4πr₁²

Area = 4 × old area.

(i) 3

(ii) 4

(iii) 5

(iv) 2

(v) 1

Here θ = 90°; radius r = 7cm

Length of the arc = θ°/360° × 2πr units

= 90°/360° × 2 × 22/7 × 7 = 11 cm

∴ Length of the arc = 11 cm

Least number of planes = 4, the solid is tetrahedron.

For a triangular prism

Faces = 5, Edges = 9, Vertices = 6

By Euler’s formula 

F + V – E 

= 5 + 6 – 9 

= 11 – 9 

= 2

A pyramid has faces = 5, Vertices = 5, Edges = 8

By Euler’s formula 

F + V – E 

= 5 + 5 – 8 

= 10 – 8 

= 2

(1) 1 km 250 m 

(2) 2 m 50 cm 

(3) 3050 ml 

(4) 250 paise

181 km – 95 km = 86 km

∴ The distance between Amravati and Jalgaon is 86 km.

249 km – 154 km = 95 km

∴ The distance between Bhusaval and Nagpur is 95 km

We cannot subtract 960 m from 225 m. So, convert 1 km = 1000 m

∴ 22 km 265 m

We cannot subtract 65 cm from 15 cm. 

So, convert l 

m = 100 cm 

∴ 10 m 50 cm

740 m + 950 m

= 1690 m 1690 m

= km 690 m

∴ 29 km 690 m

650 gm + 770 gm 

= 1420 gm 

= 1 kg 420 gm 

∴ 55 kg 420 gm

We cannot subtract 750 ml from 200 ml.

1 l = 1000 ml

∴ 7 l 450 ml

We cannot subtract 900 gin from 880 gm. 

So, convert 1 kg = 1000 gm 

∴ 20 kg 980 gm

50 cm + 75 cm

= 125 cm

= 1 m 25 cm

∴ 48 m 25 cm

We cannot subtract 470 gm from 150 gm.

So, convert I kg = 1000 gm

∴ 8 kg 680 gm

450 ml + 800 ml

= 1250 ml

= 11 + 250 ml

∴ Ajay donated 49 l 250 ml milk

The speed of the train is 90 kmph. 

That is, it travels 90 km in one hour. It travels 90 more km in the second hour.

In the next half an hour, 90 ÷ 2 = 45 km

The total distance travelled is 90 + 90 + 45 = 225 km.

30 kmph means

In 60 minutes 30 km and 30 minutes 15 km and 15 minutes \(\frac{15}{2}=\frac{15\times5}{2\times5}=\frac{75}{10}\) = 7.5 km

∴ In 45 minutes 15 km + 7.5 km = 22.5 km

Hour and quarter = 1 + \(\frac{1}{4}\) hours

= 40 km + \(\frac{1}{4}\) x 40 km

= 40 km + 10 km

= 50 km

∴ Motorcycle will travel in a hour and a quarter 50 km

750 m cannot be subtracted from 575 m.

So, convert 1 km = 1000 m.

∴ 825 m remained to be cleaned

\(\frac{1}{2}+\frac{1}{2}=\frac{1+1}{2}=\frac{2}{2}\) = 1 litre

22 + 22 = ₹44

That is, 1 litre cost ₹ 44

∴ 7 litres costs 44 x 7 = ₹ 308

∴ 7 litres costs ₹ 308

(a) directly proportional to the number of atoms

(b) Both assertion and reason are true but reason is not the correct explanation of the assertion 

Correct explanation :

In SI system the units are precisely defined and have the same value everywhere.

(c) Assertion is true, but reason is false 

Correct explanation : 

Least count is the minimum measurement that can be measured accurately by an instrument.

(a) Both assertion and reason are true and reason is the correct explanation of the assertion

(a) Both assertion and reason are true and reason is the correct explanation of the assertion

32°F. is the 'Lower fixed point' of the Fahrenheit scale.

The value of every measurement contains some uncertainty. These uncertainties are called as ‘Errors’.