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True

We have.4 = {1,2, 3}, 5= {3,4} andC= {4,5,6}

AxB= {(1, 3), (1,4), (2, 3), (2,4), (3, 3), (3,4)}

And A x C = {(1,4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3,4), (3, 5), (3, 6)}

(A x B)∪(A xC)= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3,3), (3,4), (3, 5), (3,6)}

Answer is (B) Isosceles

Circumference of Circle = Circumference of circle with radius 19 cm + Circumference of circle with radius 9 cm

2πr = 2π(19) + 2π(9)

2πr = 38π + 18 π

r = 28

Radius of the circle is 28 cm.

Let the radii of the two circles be r₁ and r₂.

And, the circumferences of the two circles be C₁ and C₂.

We know that, circumference of circle (C) = 2πr

Given,

Sum of radii of two circle i.e., r₁ + r₂ = 140 cm … (i)

Difference of their circumference,

C₁ – C₂ = 88 cm

2πr₁ – 2πr₂ = 88 cm

2(22/7)(r₁ – r₂) = 88 cm

(r₁ – r₂₎ = 14 cm

r_{1 =} r₂ + 14….. (ii) 

Substituting the value of r₁ in equation (i), we have,

r₂ + r₂ + 14 = 140

2r₂ = 140 – 14

2r₂ = 126

r₂ = 63 cm

Substituting the value of r₂ in equation (ii), we have,

r₁ = 63 + 14 = 77 cm

Therefore,

Diameter of circle 1 = 2r₁ = 2 x 77 = 154 cm

Diameter of circle 2 = 2r₂ = 2 × 63 = 126 cm

Given diameter of the wheel = 140 cm

Desired speed of the bus = 66 km/hr

Distance covered by the wheel in 1 revolution = Circumference of the wheel 

Circumference of the wheel = πd

C = 22/7 × 140

C = 440 cm

Now, the desired speed of the bus = 66 km/hr = 66 × 1000 × (100/60) = 110000 cm/min

Number of revolution per minute = 110000/440

= 250

Therefore, the bus must make 250 revolutions per minute to keep the speed at 66 km/hr.

Given the diameter of the wheel = 60 cm

Distance covered by the wheel in 1 revolution = Circumference of the wheel

Distance covered by the wheel in 1 revolution = πd

= 22/7 × 60 cm

Distance covered by the wheel in 140 revolutions = 22/7 × 60 × 140

= 26400 cm

Thus, the wheel covers 26400 cm in 1 minute. Then,

Speed = 26400/100 × 60 m/hr

= 264 × 60 m/hr

= 264 × 60/1000 km/hr

= 15.84 km/hr

The speed with which the boy is cycling is 15.84 km/hr.

The perimeter of a circle is called the circumference of the circle.

π = Circumference of a circle/ Diameter

Form this we get circumference of a circle is the product of π and its diameter.

C = πd

= 2πr ( since d = 2r).

Total distance covered = 1 km = 100000 cm 

Distance covered by circular wheel in 1 revolution = circumference of circle 

Circumference of circle = 2πr 

Total no. of revolution = 450 

= 2πr × 450 = 100000

r = \(\frac{10000\times7}{450\times2\times22}\) = 35.35 cm

Given total distance covered by bicycle in 5000 revolutions = 11 km = 11000 m

Therefore distance covered in 1 revolution = 11000/5000 = 2.2 m = 11/5

Distance covered in 1 revolution = Circumference of the wheel

C = πd

11/5 = (22/7) d

d = (11 × 7)/ (5 × 22)

d = 77/110

d = 0.7 m

Therefore diameter of wheel = 0.7 m = 70 cm

Let r₁, h₁ and r₂, h₂ be the radii and height of the two cylinders respectively. Then

\(πr^2_1h_1 = πr^2_2h_2 ⇒\frac{h_1}{h_2}=\frac{r^2_2}{r^2_1}=\big(\frac{r_2}{r_1}\big)^2=\bigg(\frac{\frac32}{\frac22}\bigg)^2=\frac94.\)

(d) 13860 cm³

Given, 2πr = 66 ⇒ r = \(\frac{66\times7}{2\times22}= \frac{21}{2}\) cm

∴ V = πr²h = \(\frac{22}{7}\times\frac{22}{7}\times\frac{22}{7}\times40\) = 13860 cm³.

Given that, 

Height = 15 cm 

Circumference = \(2\pi r\)

r = \(\frac{44\times7}{2\times22}\)

= 7 cm 

We know that, 

Volume of cylinder = \(2\pi r\)

= \(\frac{22}{7}\times7\times7\times15\)

= 2310 cm³ 

Hence, option B is correct

Solution:

False

If 0< r< 2, then numerical value of circumference is greater than numerical value of area of circle and if r > 2, area is greater than circumference.

True

Explanation:

The distance travelled by a circular wheel of radius r m in one revolution is equal to the circumference of the circle = 2πr

No. of revolutions completed in 2πr m distance = 1

No. of revolutions completed in 1 m distance = (1/2πr)

No. of revolutions completed in s m distance = (1/2πr) × s = s/2πr

Thus, the statement “in covering a distance s metres, a circular wheel of radius r metres makes s/2πr revolutions” is true.

Given,

Diameter of roller = 84 cm

So radius of roller =\(\cfrac{84}2\) = 42 cm = .42 m

Length of roller = 1.5 m

Area covered by roller in one revolution = 2πrh = 2 \(\times\cfrac{22}7\times \)4.2 \(\times\) 1.5 =  3.96 m²

Area covered by it in 100 revolution = 100 × 3.96 = 396 m²

Cost of levelling 1m² area = Rs. .50

∴cost of levelling 396m² = .50 × 396 = Rs. 198

Volume of a right circular cylinder = (Area of base) × height = πr² x h =  πr²h

Volume of a hollow circular cylinder = πR²h −πr²h = πh(R+r) (R–r)

Radius = \(\frac{\text{Circumference}}{2π}\) = \(\frac{6π}{2π}\) = 3 cm

⇒ Height = 6 cm.

∴ Volume of the cylinder = πr² =π x (3)² × 6 cm³ = 54π cm³

= \(\frac{54π}{1000}\) litres = 0.54π litres.

2r = 84cm

∴ r = 84/2cm = 42cm

h = 120cm

Area of the playground leveled in one complete revolution = 2πrh

= 2 × 22/7 × 42 × 120 × 31680cm²

∴  Area of the playground = 31680 × 500cm²

= (31680 x 500)/(100 x 100)m² = 1584m²

∴ Cost of leveling @ Rs 25 per square metre = Rs 1584 × 25 = 39600.

Solution:

True

The distance covered in one revolution is 2πr i.e., its circumference.

Solution:

False

Because the distance travelled by the wheel in one revolution is equal to its circumference

i.e., πd.

i.e., π(2r) = 2 πr = Circumference of wheel  [∵d = 2r]

We have,

Diameter of roller = 84 cm

Radius of roller = 84/2 = 42cm

Length of roller = 120 cm

Curved surface area of roller = 2πrh

= 2 × 22/7 × 42 × 120

= 31680 cm²

It takes 500 complete revolutions to level the playground.

∴ Area of playground = 500 × 31680

= 15840000 cm²

= 1584 m²

Area covered by the roller in 50 revolutions = 50 × Curved surface area of the roller

= 50 x 2 x \(\frac{22}{7}\) x 35 x 200 cm²

= 2200000 cm² = 220 m².

Diameter of the roller = 84 cm 

Thus radius = 84/2 = 42 cm

= 42/100 = 0.42m

Length of the roller =120 cm 

= 120/100 = 1.2m

It takes 500 complete revolutions to roll over the play ground. 

Thus 500 × L.S.A. of the roller

= Area of the play ground 

∴ Area of the play ground = 500 × 2πrh

= 500 × 2 × 22/7 × 0.42 × 1.2 = 1584 m²

Correct Answer is: (c, d)

मानव को दूर तथा पास की वस्तुएँ पूर्णत: देखते के लिए नेत्र सुनियोजित करते पड़ते है | इस प्रकार मानव के अभिनेत्र लेंस की वह क्षमता जिससे वह अपनी फोकस दुरी कोण सुनियोजित कर लेता है , समाजंन क्षमता कहलाती है |

प्रतिबिंब दूरी सदैव एक जैसी रहती है | इसका कारण है कि वस्तु की दुरी मानव नेत्र के लेंस की फोकस दुरी इस प्रकार समायोजित हो जाती है जिससे प्रतिबिंब दृष्टि पटल पर ही बने |

(b)    समंजन

वायुमंडलीय अपवर्तन के कारण सूर्योदय से पहले ही और सूर्यास्त के बाद तक हमे दरअसल सूर्य का अभासी प्रतिबिम्ब दिखाई देता रहता है। इसलिए सूर्योदय से 2 मीनट पहले ही और सूर्यास्त के 2 मीनट बाद तक हमे सूर्य दिखाई देता हैं।

प्रकाश के अवयवी वर्णो में विभाजन को प्रकाश का विक्षेपण कहते हैं ।

(d) 41

Let the two digit number be 10x + y. 

According to the question,

10x + y = 8(x + y) + 1  \(\Rightarrow\) 2x – 7y = 1 ….......(i)

and 10x + y = 13(x – y) + 2 \(\Rightarrow\) 3x – 14y = –2 ...…(ii)

Now we can get x and y.

(c) 14

Suppose I have x, one-rupee coins and y, 20 – paise coins.

x × 1 + y × 0.2 = 14.40 \(\Rightarrow\) x + 0.2y = 14.4 …(i) 

After shopping, I had y one-rupee coins and x 20-paise coins.

Also, x × 0.2 + y × 1 = \(\frac{1}{3}\) x 14.4

\(\Rightarrow\) 0.2x + y = 4.8 ........(ii)

[Note: To solve the equations of the form ax + by = c and bx + ay = d, where a \(\ne\) b, we can use the following method also.]

Adding eqn (i) and (ii), we get

1.2 x + 1.2 y = 19.2 \(\Rightarrow\)x + y = \(\frac{19.2}{1.2}\)= 16 .........(iii)

and subtracting eqn (ii) from eqn (i), we get 

– 0.8x + 0.8y = –9.6 \(\Rightarrow\) x – y = 12 ..........(iv)

Now adding (iii) and (iv), we get

2x = 28 \(\Rightarrow\) x = 14.

In case of direct variation both the quantities vary in the ratio i.e if one quantity is doubled then other will also get double. If one quantity gets half the other will also get half.

Let the number of men required be ‘x’ men

Total men = 420

We know k = xy

420 × 9 = x × 7

x = (420×9)/7

= 540

So, 540 – 420 = 120 Men

∴ 120 extra men are required to finish the job in 7 months.

Let number of men be ‘x’ men

We know k = xy

35 × 18 = 15 × x

x = (35×18)/15

= 42

∴ 42 men are required to reap the same field in 15 days.

Given Salman’s investment = ₹ 75000 

Salman’s period in business = 1 year = 12 months 

Deepak’s investment = ₹ 80000 

Deepak’s period in business = 7 months 

The ratio of investments of Salman to 

Deepak = 75000 : 80000 = 15 : 16 

The ratio of periods of business of 

Salman to Deepak = 12 : 7 

The profit should be distributed on the basis of compound ratio of 15 : 16 and 12 : 7 is 

15 × 12 : 16 × 7 = 180 : 112 

= 45 : 28 

Therefore, compound ratio = 45 : 28 

Profit = ₹ 73,000 

Total parts = 45 + 28 = 73 

∴ Salman’s profit = 73000 × 45/73 = ₹ 45000 

Deepak’s profit = 73000 × 28/73 

= ₹ 28,000

Let us consider the speed of the car as ‘x’ km/hr.

We know k = xy

10 × 48 = 8 × x

x = (10×48)/8

= 60

Increased speed = 60 – 48 = 12km/hr.

∴ The speed should be increased by 12 km/hr. to cover the same distance.

Let the number of men joined be ‘x’ men

Total men = 1200

We know k = xy

1200 × 35 = x × 25

x = (1200×35)/25

= 1680

So, 1680 – 1200 = 480 Men

∴ 480 men should join for the same stock to last for 25 days.

Here Prabhu and Suresh started a business with ₹ 1,00,000. 

Prabhu continued till the end of the year.

Suresh continued only for 3 months. 

The ratio of the contributions = 100000: 100000 = 1 : 1 

Ratio of their period of business = 12 : 3 = 4 : 1 

So, their profits should be divided on the basis of compound ratio 

Compound ratio = 1 × 4 : 1 × 1 

Profit = ₹ 20,000 

Total parts= 4 + 1 = 5 

Suresh’s profit = 20000 × 1/5 

= ₹ 4000 

Prabhus profit = 20000 – 4000 

= ₹ 16000

In a hostel of 50 girls, food is available = 40 days

For 1 girl, food provision = 50× 40 = 2000 days

Now, for (50 + 30) girls i.e. 80 girls, the food provision = 2000/80 = 200/8

⇒ 25 days

The food provision will last for 25 days, if 30 more girls join.

Let the principle be P 

then A = 2P 

T =8 years 4 months = 8y \(\frac {4}{12}\) y = 8 \(\frac13\) years

R = R% say 

We know that

I = \(\frac {PTR}{100}\) 

Here I = A – P 

= 2P – P = P 

∴ P = \(\frac {P.\frac {25}{3}\times R}{100}\)

∴ R = 100 x \(\frac {3}{25}\) = 12%

P = ₹3000 R = 9% T = 2 1/2 years = \(\frac 52\)

Interest = P x R% x T

3000 x \(\frac {9}{100} \times \frac 52\) = ₹675

Here A = ₹ 15,624 

R = 9% T = ? 

P = ₹ 12,600 

∴ I = A – P = 15,624 – 12,600 = ₹ 3024

Also I = \(\frac {PRT}{100}\)

∴ 3024 = \(\frac {12600\times9\times T}{100}\)

∴  T = \(\frac {3024 \times 100}{12600\times 9} = \frac {24}{9} = \frac {8}{3} = 2\frac 23\) years

Number of circles = 4; 

Number of squares = 6 

Number of circles: Number of squares = 4 : 6 

= 2 : 3

From the table itself we observe that as x increases y decreases. 

∴ x and y are inversely proportional 

∴ xy = 8 x 32 = 16 x 16 = 32 x 8 = 256 x 1 

= 256

Some other examples of such kind are 

(i) Cost of book and number of books 

(ii) Distance and time to travel 

(iii) Men workers and wages.

Lesser the number of periods in a day, more the duration of them. So, it is an inverse proportion. 

Let the required duration be x.

⇒ 9 × 40 = 8 × x

⇒ x = \(\frac{9\,\times\,40}{8}\) = 45 days

Let the cost of 8 quintals of soyabean be Rs x. The quantity of soyabeans and their cost are in direct proportion.

\(\therefore \frac{12}{36000} = \frac{8}{x}\)

∴ 12x = 8 × 36000 ….(Multiplying both sides by 36000x)

\(\therefore x = \frac{8 \times 36000}{12} = 24000\)

∴ The cost of 8 quintals of soyabean is Rs 24000.

Let the bunches of feed bought for Rs 1280 be x. The quantity of feed bought and their cost are in direct proportion.

∴ \(\frac{600}{15} = \frac{1280}{x}\)

∴ 600x = 1280 × 15 …. (Multiplying both sides by 15x)

∴ \(x = \frac{1280 \times 15}{600} = 32\)

∴ 32 bunches of feed can be bought for Rs 1280.

After enlarging 50,000 times the length is 5 cm. 

Without enlarging (1 – times) the length Is x cm say 

∴ 50,000 : 5 :: 1 : x 

By rule of properties 50,000 × x = 5 × 1

x =  \(\frac {5}{50000} = \frac {1}{10000}\) = 0.0001 cm

Let the number of students be x and the number of chocolates be y. As the valve of x increases also correspondingly increases. 

i.e., \(\frac{x}{y}\) = k 

∴ x and y are in direct proportion.