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From the question it is given that, if p squares of each side 1mm makes a square of side 1cm.

We know that, area of square of side 1mm = side × side

= 1 × 1

= 1 mm²

Then, area of square of side 1cm = side × side

= 1 × 1

= 1 cm²

So, area of square of side 1mm = area of square of side 1cm

P × 1 mm² = 1 cm²

Pmm² = (10mm)²

Therefore, Pmm² = 100 mm²

(d) 1000000

We know that, 1 m = 100 cm

1 cm = 10 mm

So, 1 m = 10 × 100

1 m = 1000 mm

Then, 1 m² = 1000 × 1000

1 m² = 1000000

Correct answer is

(A) 126

It is given that x: 35:: 48: 60

But, product of extremes = product of means

∴ x × 60 = 35 × 48

= 60x = 1680

= x = 1680/60

= x = 28

Hence, x = 28

It is given that 8: x:: 16: 35

But, product of extremes = product of means

∴ 8 × 35 = x × 16

= 280 = 16x

= x = 280/16

= x = 17.5

Hence, x = 17.5

It is given that 2: 9:: x: 27

But, product of extremes = product of means

∴ 2 × 27 = 9 × x

= 54 = 9x

= x = 54/9

= x = 6

Hence, x = 6

Let the number be x,

Then the two term becomes,

= (9 + x): (16 + x) = 2: 3

= (9 + x) / (16 + x) = (2/3)

By cross multiplication,

= 3 × (9 + x) = 2 × (16 + x)

= 27 + 3x = 32 + 2x

Transposing 27 to RHS and 2x to LHS

= 3x – 2x = 32 – 27

= x = 5

Hence, the number be added to each of the ratio 9:16 to make the ratio 2: 3 is 5

Four numbers a, b, c, d are not in proportion, if a: b ≠ c: d

Let,

a = 36, b = 49, c = 6, d = 7,

Then,

Product of extremes = (36 × 7) = 252

Product of means = (49 × 6) = 294

∴ Product of extremes ≠ product of means

Hence, 30, 40, 45, 60 are not in proportion.

Let x must be subtracted from each of 7 : 4 then \(\frac{7 - x}{4 - x} = \frac{5}{2}\)

⇒ 14 – 2x = 20 – 5x 

5x – 2x = 20 – 14 

3x = 6 

⇒ x= 2.

Let x must be subtracted from each of 8:7 then

\(\frac{8 - x}{7 - x} = \frac{4}{3}\)

24 – 3x = 28 – 4x 

4x – 3x = 28 – 24 

x = 4

Let the number be x,

Then the two term becomes,

= (17 – x): (33 – x) = 7: 15

= (17 – x) / (33 – x) = (7/15)

By cross multiplication,

= 15 × (17 – x) = 7 × (33 – x)

= 255 – 15x = 231 – 7x

Transposing -15x to RHS and 231 to LHS

= 255 – 231 = 15x -7x

= 8x = 24

= x = (24/8)

= x = 3

Hence, the number be subtracted from each term of the ratio 17:33 to make the ratio 7: 15 is 3

Four numbers a, b, c, d are said to be in proportion, if a: b = c: d we write a: b:: c: d.

Let,

a = 30, b = 40, c = 45, d = 60,

Then,

Product of extremes = (30 × 60) = 1800

Product of means = (40 × 45) = 1800

∴ Product of extremes = product of means

Hence, 30, 40, 45, 60 are in proportion.

Given that three numbers are in the ratio 2: 3: 5 and sum of them is 800

Therefore sum of the terms of the ratio = 2 + 3 + 5 = 10

First number = (2/10) × 800

= 2 × 80

= 160

Second number = (3/10) × 800

= 3 × 80

= 240

Third number = (5/10) × 800

= 5 × 80

= 400

The three numbers are 160, 240 and 400

Let x must be added to each of 4:5 then

\(\frac{7 - x}{4 - x}\) = \(\frac{5}{2}\)

x = 3

Let the numbers are 7x and 11x,

Then,

= (7x + 7): (11x + 7) = 2: 3

= (7x + 7)/ (11x + 7) = (2/3)

By cross multiplication,

= 3 × (7x + 7) = 2 × (11x + 7)

= 21x + 21 = 22x + 14

Transposing 21x to RHS and 14 to LHS

= 21 – 14 = 22x – 21

= x = 7

Hence the numbers are 7x = 7 × 7 = 49

11x = 11 × 7 =77

Let the number to be added is x

Then (7 + x) + (13 + x) = (2/3)

(7 + x) 3 = 2 (13 + x)

21 + 3x = 26 + 2x

3x – 2x = 26 – 21

x = 5

Hence the required number is 5.

Let the required numbers be 7x and 11x

If 7 is added to each of them then

(7x + 7)/ (11x + 7) = (2/3)

3 (7x + 7) = 2 (11x + 7)

21x + 21 = 22x + 14

22x – 21x = 21 – 14

x = 21 – 14 = 7

Thus the numbers are 7x = 7 (7) =49

And 11x = 11 (7) = 77

Correct answer is

(B) 32 : 7

a)Jumman’s aunt was referred to as T. 

b)The speaker was an unwanted person in Jumman’s house.

c)The speaker demanded a monthly allowance so that she could set up a separate kitchen.

a) Algu said it. 

b) Algu. 

c) He was not ready to go against the person.

a) Sahu, the cart driver of the village said this. 

b) This was said to Algu. 

c) He brought ruin to his owner by death.

1. The phono-receptors of the body receive the sound waves and transfer the nerve impulses to the auditory areas of the brain. 

2. The interpretation, of the sound is a combined effort of sensory and association areas [auditory areas] of temporal lobes of the brain. 

3. This is how we are able to hear the chirping of the birds and recognize the sound of the bird.

(1) (a) ↔ (ii)

(b) ↔ (iii) and

(c) ↔ (i)

(2) (a) ↔ (iii)

(b) ↔ (i) 

(c) ↔ (iv), 

(d) ↔ (ii)

(3) (a) ↔ (iv)

(b) ↔ (iii)

(c) ↔ (ii),

(d) ↔ (i)

(4) (a) ↔ (i) 

(b) ↔ (iii)

(c) ↔ (iv), 

(d) ↔ (ii)

(5) (a) ↔ (iii) 

(b) ↔ (i) 

(c) ↔ (ii)

(i) Yes. It can be observed that trapezium ABCD and triangle PCD have a common base CD and these are lying between the same parallel lines AB and CD.

(ii) No. It can be observed that parallelogram PQRS and trapezium MNRS have a common base RS. However, their vertices, (i.e., opposite to the common base) P, Q of parallelogram and M, N of trapezium, are not lying on the same line.

(iii) Yes. It can be observed that parallelogram PQRS and triangle TQR have a common base QR and they are lying between the same parallel lines PS and QR.

(iv) No. It can be observed that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and BC. However, these do not have any common base.

(v) Yes. It can be observed that parallelogram ABCD and parallelogram APQD have a common base AD and these are lying between the same parallel lines AD and BQ.

(vi) No. It can be observed that parallelogram PBCS and PQRS are lying on the same base PS. However, these do not lie between the same parallel lines.

(i) ΔPCD and trapezium ABCD or on the same base CD and between the same parallels AB and DC. 

(ii) Parallelogram ABCD and APQD are on the same base AD and between the same parallels AD and BQ. 

(iii) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC but they are not on the same base. 

(iv) ΔQRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS 

(v) Parallelogram PQRS and trapezium SMNR on the same base SR but they are not between the same parallels. 

(vi) Parallelograms PQRS, AQRD, BCQR and between the same parallels also, parallelograms PQRS, BPSC and APSD are between the same parallels. 

(i) Triangle APB and trapezium ABCD are on the common base AB and between the same parallels AB and DC. 

So, 

Common base = AB 

Parallel lines: AB and DC 

(ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ. 

Common base = AD 

Parallel lines: AD and BQ 

(iii) Consider, parallelogram ABCD and ΔPQR, lies between the same parallels AD and BC. But not sharing common base. 

(iv) ΔQRT and parallelogram PQRS are on the same base QR and lies between same parallels QR and PS. 

Common base = QR 

Parallel lines: QR and PS 

(v) Parallelograms PQRS and trapezium SMNR share common base SR, but not between the same parallels. 

(vi) Parallelograms: PQRS, AQRD, BCQR are between the same parallels. Also, 

Parallelograms: PQRS, BPSC, APSD are between the same parallels.

In parallelogram ABCD, AB = 16 cm, AE = 8 cm and CF = 10 cm 

Since, opposite sides of a parallelogram are equal, then 

AB = CD = 16 cm 

We know, Area of parallelogram = Base x Corresponding height 

Area of parallelogram ABCD: 

CD x AE = AD x CF 

16 x 18 = AD x 10 

AD = 12.8 

Measure of AD = 12.8 cm

Area of a parallelogram ABCD: 

From figure: 

AD × CF = CD × AE 

6 x 10 = CD x 8 

CD = 7.5 

Since, opposite sides of a parallelogram are equal. 

=> AB = DC = 7.5 cm

We have AB = 16 cm, AE = 8 cm CF = 10 cm. 

We know that are of parallelogram = Base × Height [Base = CD, height = AE] 

ABCD = CD × AE = 16 × 8 = 128 cm² 

 Again, Area of parallelogram = Base × Height = AD × CF [Base = AD, height = CF] 

128 = AD × 10 

⇒ AD =128/10 =12.8 cm

ABCD is a parallelogram. 

AE ⊥ DC and CF ⊥ AD. 

AB = 16 Cm, AE = 8 cm, 

CF = 10 cm, then AD = ? 

(i) Area of ABCD = base × altitude

= DC × AE ( ∵ AB=DC) 

= 16 × 8 

= 128 sq. cm. 

Whether ABCD is quadrilateral, 

ADCB is a quadrilateral. 

∴ Area of ADCB = 128 sq. cm. 

Base = AD = ? 

Altitude, CF= 100 cm. 

Base × Altitude = Area 

AD × CF = 128 

AD × 10 = 128 

∴ AD = \(\frac{128}{10}\)

∴ AD = 12.8 cm.

Two prime numbers are said to be twin primes, if they differ each other by 2. Twin primes less than 50 are (3, 5); (5, 7); (11,13); (17, 19); (29, 31) and (41, 43).

Solution:

i) Let us assume the larger number is x and smaller number is y. Then we have following equations, as per question:

x = y + 26 and x = 3y

Substituting the value of x from second equation in the first equation, we get;

x = y + 26

Or, 3y = y + 26

Or, 2y = 26

Or, y = 13

Substituting the value of y in second equation, we get;

x = 3y

Or, x = 13 x 3 = 39

Hence, x = 39 and y = 13

ii)  Let us assume the larger angle is x and smaller angle is y. Then we have following equations;

x = y + 18 and x + y = 180o

Substituting the value of x from first equation in second equation, we get;

x + y = 180o

Or, y + 18 + y = 180o

Or, 2y = 180o - 18o = 162o

Or, y = 81o

Substituting the value of y in first equation, we get;

x = y + 18

Or, x = 81o + 18o = 99o

Hence, x = 99o and y = 81o

iii)  Let cost of each bat  =  Rs x

Cost of each ball  = Rs y

Given that coach of a cricket team buys 7 bats and 6 balls for Rs 3800.

So that 7x + 6y = 3800

6y = 3800 – 7x

Divide by 6 we get

y = (3800 – 7x) /6         … (1)

Given that she buys 3 bats and 5 balls for Rs 1750.so that

3x + 5y = 1750

Plug the value of y

3x + 5 ((3800 – 7x) /6) = 1750

Multiply by 6 we get

18 x  + 19000 – 35 x = 10500

-17x =10500 - 19000

-17x = -8500

x = - 8500 / - 17

x =  500

Plug this value in equation first we get

y = ( 3800 – 7 * 500) / 6

y = 300/6

y = 50

Hence cost of each bat = Rs 500 and cost of each balls is Rs 50

Two lines, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

If \(\frac{a_1}{a_2}\)= \(\frac{b_1}{b_2}\)= \(\frac{c_1}{c_2}\), then the lines coincide

If   \(\frac{a_1}{a_2}\)= \(\frac{b_1}{b_2}\)≠ \(\frac{c_1}{c_2}\) , then the lines are parallel

If   \(\frac{a_1}{a_2}\)≠ \(\frac{b_1}{b_2}\)≠ \(\frac{c_1}{c_2}\) , then the lines intersect

Given the linear equation 2x + 3y - 8 = 0.

An intersecting line is x + 2y – 4 =0

A parallel line is 4x + 6y – 12 = 0

A coincident line is 4x + 6y – 16 = 0

152x - 378y =  - 74

- 378x +158y = - 604

Adding (1) and (2)

⇒ -226x – 226y = -678

⇒ x + y = 3----- (3)

(1) – (2)

⇒ 530x – 530y = 530

⇒ x – y = 1 ------ (4)

Adding (3) and (4)

⇒ 2x = 4

⇒ x = 2

Thus,

y = 1

Two lines, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

If \(\frac{a_1}{a_2}\)=\(\frac{b_1}{b_2}\)=\(\frac{c_1}{c_2}\), then the lines coincide

If \(\frac{a_1}{a_2}\)= \(\frac{b_1}{b_2}\)≠\(\frac{c_1}{c_2}\), then the lines are parallel

If \(\frac{a_1}{a_2}\)≠ \(\frac{b_1}{b_2}\)≠\(\frac{c_1}{c_2}\), then the lines intersect

Given line 2x + 3y - 8 = 0

(i) intersecting line:

3x + 2y – 6 = 0

(ii) parallel line:

4x + 6y = 15

(iii) coincident line:

4x + 6y – 16 = 0

3x - 7y + 10 = 0

y - 2x - 3 = 0

Multiplying eq 1 by 2 and eq 2 by 3 and adding.

⇒ 6x – 14y + 20 + 3y – 6x – 9 = 0

⇒ - 11y = - 11

⇒ y = 1

Substituting value of y in eq 1, we get

x = - 1

 \(\frac{7x-2y}{xy}\) = 5

⇒ 7/y – 2/x = 5 ---- (1)

 \(\frac{8x+7y}{xy}\) = 15

⇒ 8/y + 7/x = 15 ---- (2)

Multiplying eq1 by 7 and eq2 by 2 and adding

⇒ 65/y = 65

⇒ y = 1

Thus,

7 – 2/x = 5

⇒ x = 1

11x + 15y + 23 = 0 [1]

7x - 2y - 20 = 0 [2]

Multiplying the 1st equation by 2 and 2nd equation by 15, we get

22x + 30y + 46 = 0 [3]

105x - 30y - 300 = 0 [4]

Adding [3] and [4]

⇒ 22x + 30y + 46 + 105x – 30y – 300 = 0

⇒ 127x = 254

⇒ x = 2

Substituting value of x in equation 1.

⇒ 11(2) + 15y + 23 = 0

⇒ 22 + 15y + 23 = 0

⇒ 15y = -45

⇒ y = -3

Botanical garden, museum and Herbarium are the device to study plant taxonomy.

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The wind wrinkles the water of the sea.

Plant taxonomy help to categories plants in different groups on the basis of similarities and dissimilarities.

Botanical gardens are also known as Outdoor laboratories for studies of botany and plants because they are used for research and investigation work.

Blotting paper or newspaper should change every 24 hours for proper drying of plant part so fungal growth can be avoided