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Correct answer is

(A) proportion

(i) 2 × 9 = 3 × 6 

18 = 18 

It is a proportion 

(ii) 1 × 15 ≠ 3 × 4 

It is not a proportion 

(iii) 3 × 15 ≠ 5 × 6 

∴not a proportion 

(iv) 7 × 18 = 9 × 14

126 = 126 

It is a proportion

Correct answer is

(C) \(\frac{a + c}{b + d}\)

Correct answer is

(D) \(\frac{a}{b} < \frac{c}{d}\)

Sum of the ratios 3 + 4 + 5 = 12 

∴\(\frac{3}{12}\) × 1800 = 450 

\(\frac{4}{12}\) x 1800 = 600 

\(\frac{4}{12}\) x 1800 = 750

(a) 6

Let a + b = 6k, b + c = 7k and c + a = 8k. 

Then 2(a + b + c) = 6k + 7k + 8k = 21k 

⇒ 2 × 14 = 21k ⇒ k = \(\frac{2\times14}{21}=\frac43\)

Also a + b + c = \(\frac{21k}{2}\) and a + b = 6k

⇒ c = \(\frac{21k}{2}\) - 6k = \(\frac{9}{2}k\) = \(\frac{9}{2}\times\frac43=6\)

Correct answer is

(A) capitals

Given that the sum is divided in the ratio = 3: 5

Sum divided = ₹ 1,60,000 

Sum of the terms of the ratio 3 + 5 = 8 

∴ Smaller share = 3/8 × 1,60.000 

= ₹ 60,000

(a) Rs 500

C’s share = \(\frac{5}{12}\) Rs 6000  = Rs 2500 

B’s share = \(\frac{4}{12}\) Rs 6000  = Rs 2000 

∴ C gets Rs 500 more than B.

Let the radius of the circle is r. 

∴ circumference = 2\(\pi\)r and area = \(\pi\)\(r^{2}\) 

Ratio of circumference to the area of circle

 = \(\frac{circumference}{area}\) 

= \(\frac{2\pi{r}}{\pi{r^2}}\) 

= \(\frac{2}{r}\) 

= 2 : r

∴ The ratio of circumference of circle with radius r to its area is 2 : r.

Given total amount to be divided = 5500

Sum of the terms of the ratio = 2 + 3 = 5

Raman’s share of money = (2/5) × 5500

= 2 × 1100

= Rs. 2200

And Aman’s share of money = (3/5) × 5500

= 3 × 1100

= Rs. 3300

Given that the ratio of income and savings is 7: 2

Let the savings be 2x

2x = 500

So, x = 250

Therefore,

Income = 7x

Income = 7 × 250 = 1750

Expenditure = Income – savings

= 1750 – 500

= Rs.1250

(D) 1 : 3

We know that, number of sides in a square = 4 and number of edges in a cube = 12

So, ratio of sides to edges = 4/12

Divide both numerator and denominator by 4.

= 1/3

Therefore, ratio of sides to edges = 1 : 3

Length of side of square = 7 cm 

∴ Diagonal of square = √2 x side

= √2 x 7 

= 7 √2 cm 

Ratio of diagonal of a square to its side

= \(\frac{diagonal}{side}\)

= \(\frac{7\sqrt{2}}{7}\) 

= \(\frac{\sqrt{2}}{1}\) 

= \({\sqrt{2}}{}\) : 1
∴ The ratio of diagonal of a aquare to its side is √2 : 1.

We know angle made by hours hand in 1 hour (60 minutes) = 30° 

Let angle made by hours hand 1 minute = x° 

Then 60 : 1 = 30°: x° 

If the ratios are equal, 

The product of means = The product of the extremes 

⇒ 60 × x = 1 × 30

⇒ 60x/60 = 30/60 = 1/2

⇒ x = 1/2 (or) 0.5°

∴ Angle made, by hours hand in 1 minute = 1/2° ° (or) 0.5°

We know angle made by minutes hand in 15 m = 90°

Let angle made by minutes hand in 1 m = x° 

Then 15 : 1 = 90°: x° 

If the ratios are equal, The product of means = The product of extremes 

⇒ 15 × x = 1 × 90°

⇒ 15x/5 = 90°/15

⇒ x = 6° 

∴ Angle made by minutes hand in 1 minute = 6°

Direct proportion. (As the number of people increases, food required for them also increases)

From the composition table it is obvious that 

a * 0 = a ∀ a ∈ {0, 1/2, 3,4, 5} 

∴ 0 is the identity element 

Also ∀ a ∈ {0, 1, 2, 3, 4, 5} ∃ (6 – a) ∈ {0,1,2, 3,4,5} such that 

a * (6 – a) = (6 – a) * a = 0, a ≠ 0 

hence 6 – a is the inverse of a . 

However when a = 0, 6 – a g {0, 1, 2, 3,4, 5} 

hence 6 – a is the inverse of a when a ≠ 0.

Given: 

Set N = {1, 2, 3, 4, …} 

= Set of natural numbers 

A relation R in N × N is defined as 

(a, b) R(c, d) ⇔ ad = bc 

where a, b,c, d ∈ N ∀ (a, b), (c, d) ∈ N × N 

Here, to prove R is a equivalence relation, we have to show that R is reflexive, symmetric and transitive. 

(i) Reflexivity: 

Let (a, b) ∈ N × N 

{a, b) ∈ N × N 

⇒ a.b = ba 

(Commutative law of multiplication) 

⇒ (a, b) R(a, b) ∀ (a, b) ∈ N × N 

R is reflexive relation. 

(ii) Symmetricity: 

Let (a, b) (c, d) ∈ N × N is in this way 

(a, b) R(c, d) (a, b) R(c, d) 

⇒ ad = bc 

⇒ bc = ad 

⇒ c.b = d.a 

⇒ (c, d) R(a, b) 

So, (a, b) R(c, d) 

⇒ (c, d) R(a, b) ∀ (a, b)(c, d) ∈ N × N 

R is a symmetric relation. 

(iii) Transitivity: 

Let (a, b), (c, d), (e, f) ∈ N × N is in this way 

(a, b) R(c, d) and (c, d) R(e, f) 

(a, b) R(c, d) ⇒ ad = bc 

(c, d) R(e, f) ⇒ cf = de 

(a, b)R (c, d) and (c, d)R (e, f) 

⇒ (ad) (cf) = (bc)(de) (on multiplication) 

⇒ af = be 

⇒ (a.b) R(e, f) 

So, (a, b) and R(c, d) and (c, f) R(e, f) 

⇒ (a, b) R(e, f) ∀ (a, b), (c, d), (e, f) ∈ N × N 

R is a transitive relation. 

Hence, according to (i), (ii) and (iii), the given relation is an equivalence relation. 

Hence Proved.

Fish, crocodile, tortoise, snake etc. live in water.

Among them, only fish can live in water.

Crocodile, snake, fish, crab, tortoise, lichen etc. are some aquatic animals.

Fish can breathe in deep water. So, they can survive in water.

Tortoise and crocodile live both on land and in water.

Some animals and plants are able to survive in water due to their specific physical characteristics.

(i) Plant breeding

(ii) Better yield and disease resistance

(iii) Superior improved variety

(iv) Green revolution, M.S. Swaminathan.

Terrestrial animals cannot find any favourable condition in water, as well as they do not have the special physical characteristics by which they can exist in water.

Dog dips its face in water at the time of swimming is False.

Terrestrial animal.

Bear lives in the snowy areas.

Singhara, lotus, water lily, jalkumbhi etc. are found in water.

The Eichhornia plant (jalkumbhi) spreads very fast in the ponds and lakes etc. on its own without growing and covers the water surface of the entire pond, which disturbs the growing of other aquatic plants and animals. Due to this plant, the sunrays cannot enter the pond water and the quantity of oxygen of water decreases. So it is necessary’ to remove it from time to time.

(B) TtRr

‘Tt’ and ‘Rr’ are the two pairs of heterozygous alleles.

Fish can swim with their fins and tail is True.

Fish, tortoise, dolphins, snake, crocodile etc.

The shape of the fish is round is false.

Algae only grows in water is True.

(A) heterozygous

If organisms resemble dominant parent but are genetically different, they are heterozygous.

(b) Tortoise

Dog lives only on land.

Bath sponge is Euspongia.

Goldfuss first used the term protozoa.

The cloud was formed in space about 4-5 billion years ago

The Earth is the only planet where life is known to exist

After the creation of the Earth, it took about 80 crore years for its surface to cool down and for water bodies to be formed on it.

Animals of some species lay eggs and their young ones are bom out of an egg. While in some other species, a mother gives birth to their young ones.

The protozoa appeared first in water.

The sun and planets which revolve around the sun forms the solar system.

We need a microscope to see a unicellular organism.