## DAV Class 7 Maths Chapter 13 Worksheet 4 Solutions

The DAV Class 7 Maths Book Solutions and DAV Class 7 Maths Chapter 13 Worksheet 4 Solutions of Symmetry offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 13 WS 4 Solutions

Question 1.
How many lines of symmetry will a regular heptagon (seven sided polygon) have?
Answer: Regular heptagon (Seven sided polygon) has 7 lines of symmetry as it has 7 sides.

Question 2.
How many lines of symmetry will a regular octagon (eight sided polygon) have?
Octagon has 8 sides. Therefore, it will have eight lines of symmetry.

Question 3.
Draw a square and draw its lines of symmetry.

2 lines of symmetry are through its two pairs of opposite comers.
2 lines of symmetry are the perpendicular bisectors of opposite pairs of parallel sides.
∴ Total number of lines of symmetry of a square is 4.

## DAV Class 7 Maths Chapter 13 Worksheet 3 Solutions

The DAV Class 7 Maths Book Solutions and DAV Class 7 Maths Chapter 13 Worksheet 3 Solutions of Symmetry offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 13 WS 3 Solutions

Question 1.
How many lines of symmetry will a square have? Explain by paper folding.

Take a square paper sheet and fold it 4 times along the dotted lines l1 l2, l3 and l4. Hence, the square has 4 lines of symmetry.

Question 2.
Give reasons to show how many lines of symmetry will a square have.
A square has 4 lines of symmetry as it has 4 equal sides and 4 equal lines.

Question 3.
How many lines of symmetry will a general quadrilateral have?
General quadrilateral has no line of symmetry.

Question 4.
How many lines of symmetry will a kite have? Draw the lines of symmetry of a kite.
A kite has only one line of symmetry which will be the bisector of ∠A and ∠C of the kite quadrilateral ABCD in which AB = AD and BC = DC

Question 5.
How many lines of symmetry will an isosceles trapezium have? Draw these lines. (An isosceles trapezium is one whose non-parallel sides are equal).

Isosceles trapezium ABCD in which AD = BC and AB ∥ DC has only one line of symmetry ‘l’ which is the perpendicular bisector of AB and DC.

## DAV Class 7 Maths Chapter 13 Worksheet 2 Solutions

The DAV Class 7 Maths Book Solutions and DAV Class 7 Maths Chapter 13 Worksheet 2 Solutions of Symmetry offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 13 WS 2 Solutions

Question 1.
Find the line of symmetry of any 3 line segments on different sheets of paper.

Draw the perpendicular bisectors of the line segment which is the line of symmetry of the line segment. In above figures;
l1 is the line of symmetry of $$\overline{\mathrm{AB}}$$
l2 is the line of symmetry of $$\overline{\mathrm{PQ}}$$
and l3 is the line of symmetry of $$\overline{\mathrm{MN}}$$.

Question 2.
Check if the dotted lines are the line of symmetry of the given line segments.

(i) The dotted line is not the perpendicular bisector of PQ, so it is not a line of symmetry.
(ii) The dotted line is the perpendicular bisector of XY, so it is a line of symmetry.

Question 3.
Lines l and m are the lines of symmetry of the line segments XY and YZ respectively. If XA = 4 cm and YZ = 6 cm, find AY, YB, XZ.

l is the line of symmetry of XY.
∴ XA = AY
⇒ 4 = AY
⇒ AY = 4 cm

m is the line of symmetry of YZ, YZ = 6 cm
∴ YB = BZ = $$\frac{6}{2}$$ = 3 cm
XZ = 2XA + YZ = 2(4) + 6 = 8 + 6 = 14 cm
Hence, AY = 4 cm, YB = 3 cm and XZ = 14 cm

Question 4.
Find the line of symmetry of the following angles on different sheets of paper.
(a) 60°
(b) 150°
(c) 45°

(a) OP is the bisector of ∠AOB, so it is the line of symmetry of ∠AOB where ∠AOB = 60°
(b) SQ is the bisector of ∠RST, so it is the line of symmetry of ∠RST, where ∠RST = 150°
(c) NL is the bisector of ∠MNO, so it is the line of symmetry of ∠MNO, where ∠MNO = 45°.

Question 5.
If dotted lines represent the lines of symmetry of the given angles, find x in each case.

In fig. (z) The dotted line is the bisector of the angle
x = $$\frac{80^{\circ}}{2}$$ = 40°
In figure (ii) the dotted line is the bisector of the angle
x = 2 × 55° = 110°

Question 6.
The dotted line represents the line of symmetry of ΔABC. If ∠ABC = 40° and OB = 4.5 cm, find.

(a) ∠BAO
(b) Measure of OC. Give reasons.
(a) In ΔABC, OA is the line of symmetry 8
∴ ΔABC is an isosceles triangle ∠B = ∠C
∠A + ∠B + ∠C = 180°
∠A + 40° + 40° = 180°
∴ ∠A = 180° – 80° = 100°
AO is the line of symmetry
AO is the bisector of ∠A
∴ ∠BAO = $$\frac{100^{\circ}}{2}$$ = 50°

(b) OA is the line of symmetry of BC
OA is the perpendicular bisector of BC BO = OC = 4.5 cm
Hence, ∠BAO = 50° and OC = 4.5 cm

Question 7.
Construct an equilateral ΔABC of side 4 cm. Find all its line of symmetry.

Draw ΔABC in which AB = BC = AC = 4 cm.
Draw the bisectors of ∠A, ∠B and ∠C
Al, Bm and Cn are the line of symmetry of an equilateral ΔABC.

Question 8.
Will a scalene triangle have any line of symmetry?
No, scalene triangle has no line of symmetry.

Question 9.
Under what circumstances will a right angled triangle have a line of symmetry? Give reasons.
A right angled triangle will have a line of symmetry if it is an isosceles right triangle. Because the triangle cannot be an equilateral triangle and scalene triangle has no line of symmetry.

Question 10.
Can a right angled triangle have more than one line of symmetry under any circumstances?
No, right angled triangle will have only one line of symmetry only if it is isosceles triangle.

## DAV Class 7 Maths Chapter 13 Worksheet 1 Solutions

The DAV Class 7 Maths Book Solutions and DAV Class 7 Maths Chapter 13 Worksheet 1 Solutions of Symmetry offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 13 WS 1 Solutions

Question 1.
Make any two ink-dot designs.
Take a piece of paper and put some ink drops on it. Now, fold the paper and press two parts against each other. You will get ink-drt design.

Question 2.
Make any two symmetrical paper cut designs.

Question 3.
Mark the corresponding holes for the given line of symmetry in the following figures.

Question 4.
Write three letters of English alphabet each having the following number of lines of symmetry,
(i) one line of symmetry.

(ii) two lines of symmetry.

Question 5.
Write five letters of English alphabet having no line of symmetry.
F, G, N, P, Q, J, K, S, Z

Question 6.
Draw the line of symmetry for the following figures.

### DAV Class 8 Maths Chapter 13 Worksheet 1 Notes

A figure is said to be symmetrical about a line if it is folded about that line, the two part cover completely each other.

Example:

The above figures are symmetrical about the line l.

The line of symmetry of a line segment is perpendicular bisector of the line.

A line has infinitely many lines of symmetry.

The line of symmetry of an angle is the bisector of the angle itself.

A regular polygon of n sides has i lines of symmetry.
There are infinite number of lines of symmetry of a circle. Infact each diameter of a circle its line of symmetry.

## DAV Class 8 Maths Chapter 5 Worksheet 3 Solutions

The DAV Class 8 Maths Book Solutions Pdf and DAV Class 8 Maths Chapter 5 Worksheet 3 Solutions of Profit, Loss and Discount offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 5 WS 3 Solutions

Question 1.
Rehana purchased a dress for ₹ 5400 including 8% VAT. Find the price of the dress before VAT was added.
Solution:
Let the price of the dress before the VAT added be ₹ x
∴ VAT = $$\frac{8}{100}$$ × x
= $$\frac{2 x}{25}$$
∴ Price of the dress including VAT = x + $$\frac{2 x}{25}$$
= ₹ $$\frac{27 x}{25}$$
∴ $$\frac{27 x}{25}$$ = 5400
⇒ x = $$\frac{5400 \times 25}{27}$$
⇒ x = 200 × 25
⇒ x = 5000
Hence, the required price is ₹ 5000.

Question 2.
Find the amount to be paid for the following items if 5% VAT is charged on each item.

(i) A bottle of hair styling gel at ₹ 185
Solution:
VAT on a bottle of hair style gel = $$\frac{5}{100}$$ × 185 = ₹ 9.25
∴ Price of the bottle = ₹ 185 + ₹ 9.25 = ₹ 194.25

(ii) A laptop at ₹ 55,000
Solution:
VAT on the Laptop = $$\frac{5}{100}$$ × 55,000 = ₹ 2750
∴ Price of the Laptop = ₹ 55000 + ₹ 2750 = ₹ 57750

(iii) A mobile at ₹ 36,200.
Solution:
VAT on mobile = $$\frac{5}{100}$$ × 36200 = ₹ 1810
∴ Price of the mobile = ₹ 36200 + ₹ 1810 = ₹ 38010.

Question 3.
Raman purchased a music system at ₹ 48,000 including VAT. If the cost price of the music system was 7 40,000, how much VAT (in %) has he paid?
Solution:
Let the VAT = x%
C.P. = ₹ 40,000
⇒ ₹ 400x
∴ Cost of music system including VAT = ₹ 40,000 + 400x
∴ 40000 + 400x = 48000
⇒ 400x = 48,000 – 40000
⇒ 400x = 8000
⇒ x = $$\frac{8000}{400}$$ = 20
Hence, the VAT = 20%.

### DAV Class 8 Maths Chapter 5 Value Based Questions

Question 1.
Mohit went to a shop and bought a shirt of ₹ 1,650 including 10% VAT. The shopkeeper did not give receipt to him.

(i) Find the amount of VAT not paid by the shopkeeper to the Government.
Solution:
Let the C.P. of the shirt be x such that
x + 10% of x = ₹ 1650
⇒ x + $$\frac{10}{100}$$ × x = 1650
⇒ $$\frac{110 x}{100}$$ = 1650
⇒ x = $$\frac{1650 \times 100}{110}$$ = ₹ 1500
∴ Amount of VAT not paid by the shopkeeper to the Government = 1650 – 1500 = ₹ 150

(ii) Is tax evasion correct? Will you tolerate this?
Solution:
No, never.

(iii) Why it is important to pay tax?
Solution:
It is important to pay tax as the tax paid by us is utilised to develop infrastructure of the country.

## DAV Class 8 Science Chapter 14 Notes – Reproduction in Animals

These DAV Class 8 Science Notes and DAV Class 8 Science Chapter 14 Notes – Reproduction in Animals act as excellent revision resources, particularly in preparation for board exams.

## Reproduction in Animals Class 8 DAV Notes

→ Reproduction: The process by which a living being produces its offspring is called reproduction. Reproduction in animals is of two types:

A. Asexual Reproduction: When a single parent is involved and no gamete formation takes place, the reproduction is called asexual reproduction. Asexual reproduction takes place by following methods:

→ Binary Fission: This method is observed in unicellular organisms. During binary fission, the parent cell divides into two daughter cells. Examples: Bacteria and Amoeba.

→ Multiple Fission: A repeated division of the parent cell takes place which results in the production of a number of small individuals, e.g. plasmodium.

→ Budding: In this method, a small protuberance is formed on the body of the parent. This is called bud. The bud grows to a certain size and then gets detached from the parent’s body. This finally develops into a new individual, e.g. hydra and yeast.

B. Sexual Reproduction: In sexual reproduction, two parents are involved. Two types of gametes, i.e. male and female, is formed by the respective parents. The gametes fuse through a process; called fertilisation.

→ Gametes: Specialised cells formed by male and female parents which take part in sexual reproduction. The male gamete is called sperm and female gamete is called ova; in multicellular animals.

→ Fertilisation: Fusion of male and female gametes is called fertlisation.

→ Zygote: The product of fusion of male and female gametes is called zygote.

→ Hermaphrodite Animal: When both the sexes are present on the same individual, the animal is called hermaphrodite.

→ Types of Fertilisation

→ External Fertilisation: When fertilisation takes place outside the animal’s body, it is called external fertlisation. Example: Frogs, fishes.

→ Internal Fertilisation: When fertilisation takes place inside the animal’s body, it is called internal fertilisation. Examples: humans, crocodiles, birds, etc.

→ External Development: When the embryo develops outside the animal’s body, it is called external development. External development is the norm in case of external fertilisation. In case of internal fertilisation, external development is seen in some animals; which lay eggs.

→ Internal Development: When the embryo develops inside the animal’s body, it is called internal development, e.g. Mammals.

→ Oviparous Animals: Animals which lay eggs are called oviparous animals, e.g. reptiles, and birds.

→ Viviparous Animals: Animals which give birth to young ones are called viviparous animals, e.g. most of the mammals.

Human Reproductive System

→ Male Reproductive System:
The male reproductive system is composed of a pair of testes, two sperm ducts, urethra and a penis. Testes are oval in shape and are present outside the body in a skin pouch; called scrotum.

The sperm is produced by the testes. From the testis, the sperms pass through the sperm ducts to the urethra. The penis is a muscular organ which provides a passage to urine and semen.

Structure of a Sperm: The sperm is composed of three parts, viz. a head, middle piece and a tail. A structure are the tip of the head produces an enzyme which helps the sperm to penetrate the ovum.

→ Female Reproductive System:
The female reproductive system is composed of a uterus, a pair of ovaries and a pair of oviducts. Uterus is a muscular organ which is in the shape of a pouch. The foetus develops inside the uterus. The uterus opens on the outside through an opening called vagina.

→ Fertilisation: The sperms enter through the vagina and travel to the oviduct. Fusion of male and female gametes takes place in the oviduct. The zygote; formed after the fertilisation is single-celled structure.

→ Development of Embryo: The zygote undergoes repeated cell division and turns into a ball like cluster of cells. This is called embryo. The embryo gets implanted in the wall of the uterus. The embryo develops further and a stage comes when it begins to resemble a human being. At this stage, it is called a foetus.

→ Egg Formation in Hens: When the zygote begins to travel down to the uterus; in a hen; many protective layers are formed around it. One of the protective layers forms the egg shell. The shell is made of a form of calcium carbonate; which is called calcite. The egg is then laid by the hen. The embryo inside the egg takes about 21 days to develop into a chick. The hen sits on the eggs to provide warmth. This process is called incubation. The chick breaks open the shell and comes out of the egg. This process is called hatching.

→ Metamorphosis: In some animals, the young ones do not look like the adults. The young one undergoes a series of change to turn into an adult like animal. This process is called metamorphosis. This type of development is called indirect development. Indirect development can be seen in frogs in which a tadpole develops into a frog.

→ Gamete : Specialised cells which take part in sexual development.

→ Zygote : The product of fertilisation.

→ Embryo : The multicellular stage after the zygote.

→ Foetus : When an embryo begins to resemble an dufl animal, it is called a foetus.

→ Gestation : The perioa from pregnancy o cld rth is called gestatior.

→ Metamorphosis : A series ot changes ira which a young one changes into adult-like animal.

## DAV Class 8 Maths Chapter 5 Worksheet 2 Solutions

The DAV Class 8 Maths Book Solutions Pdf and DAV Class 8 Maths Chapter 5 Worksheet 2 Solutions of Profit, Loss and Discount offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 5 WS 2 Solutions

Question 1.
The marked price of a pant is ₹ 1250 and the shopkeeper allows a discount of 8% on it. Find the discount and the selling price of the pant.
Solution:
M.P. = ₹ 1250
Discount = 8% of M.P.
∴ Discount offered = $$\frac{8}{100}$$ × 1250 = ₹ 100
∴ Selling price = M.P. – Discount
= ₹ 1250 – ₹ 100 = ₹ 1150
Hence, the selling price of the pant = ₹ 1150
and the discount = ₹ 100.

Question 2.
The marked price of a water cooler is ₹ 5400. The shopkeeper offers an off season discount of 20% on it. Find its selling price.
Solution:
M.P. = ₹ 5400
Discount = 20% of M.P.
= $$\frac{20}{100}$$ × 5400 = ₹ 1080
∴ Selling price = M.P. – Discount
= ₹ 5400 – ₹ 1080 = ₹ 4320
Hence, the selling price = ₹ 4320.

Question 3.
An almirah of marked price ₹ 4000 is sold for ₹ 3700 after allowing certain discount. Find the rate of discount.
Solution:
M.P. = ₹ 4000
and S.P. = ₹ 3700
∴ Discount = M.P. – S.P.
= ₹ 4000 – ₹ 3700 = ₹ 300
∴ Rate of discount = $$\frac{\text { Discount }}{\text { M.P. }}$$ × 100
= $$\frac{300}{4000}$$ × 100
= $$\frac{15}{2}$$ % or 7$$\frac{1}{2}$$ %
Hence, the required rate of discount is 7$$\frac{1}{2}$$%.

Question 4.
Find the rate of discount being given on a ceiling fan whose selling price is ₹ 1175 after allowing a discount of ₹ 75 on its marked price.
Solution:
S.P. = ₹ 1175
and Discount = ₹ 75
∴ M.P. = S.P. + Discount
= ₹ 1175 + ₹ 75 = ₹ 1250
Discount = $$\frac{\text { Discount }}{\text { M.P. }}$$ × 100
= $$\frac{75}{1250}$$ × 100 = 6%
Hence, the required rate of discount = 6%.

Question 5.
Find the marked price of a washing machine which is sold at ₹ 8400 after allowing a discount of 16%.
Solution:
S.P. = ₹ 8400
and Discount = 16%
Let the M.P. be ₹ x.
∴ Discount = $$\frac{16}{100}$$ × x
= ₹ $$\frac{4x}{25}$$
∴ S.P. = M.P. – Discount
= x – $$\frac{4x}{25}$$
= ₹ $$\frac{21 x}{25}$$
∴ $$\frac{21 x}{25}$$ = 8400
⇒ x = 8400 × $$\frac{25}{21}$$
⇒ x = 400 × 25 = 10000
Hence, the marked price = ₹ 10,000.

Question 6.
A dinner set was bought for ₹ 2464 after getting a discount of 12% on its marked price. Find the marked price of the dinner set.
Solution:
C.P. = ₹ 2464
Discount = 12% of M.P.
Let the M.P. be ₹ x
∴ Discount = $$\frac{12}{100}$$ × x
= ₹ $$\frac{3 x}{25}$$
∴ C.P. = M.P. – Discount
= x – $$\frac{3 x}{25}$$
= ₹ $$\frac{22 x}{25}$$
∴ $$\frac{22 x}{25}$$ = 2464
⇒ x = $$\frac{2464 \times 25}{22}$$
⇒ x = 112 × 25 = 2800
Hence the marked price is ₹ 2800.

Question 7.
The marked price of a computer is ₹ 22,000. After allowing a 10% discount, a dealer still makes a profit of 20% . Find the cost price of a computer.
Solution:
M.P. = ₹ 22000
Discount = 10% of M.P.
= $$\frac{10}{100}$$ × 22000 = ₹ 2200

∴ S.P. = M.P. – Discount
= ₹ 22000 – ₹ 2200 = ₹ 19800
Let C.P. be ₹ x.
Profit = 20% of C.P.
= $$\frac{20}{100}$$ × x = ₹ $$\frac{x}{5}$$

∴ S.P. = C.P. + Profit
= x + ₹ $$\frac{x}{5}$$ = ₹ $$\frac{6 x}{5}$$
∴ $$\frac{6 x}{5}$$ = 19800
⇒ x = 19800 × $$\frac{5}{6}$$
⇒ x = 3300 × 5 = ₹ 16500
Hence, the cost price is ₹ 16500.

Question 8.
The marked price of a double bed is ₹ 9575. A shopkeeper allows a discount of 12% on its marked price and still gains 10%. Find the cost price of the double bed.
Solution:
M.P. = ₹ 9575
Discount = 12% of M.P.
= $$\frac{12}{100}$$ × 9575 = ₹ 1149
∴ S.P. = M.P. – Discount
= ₹ 9575 – ₹ 1149 = ₹ 8426
Gain = 10%
∴ S.P. = C.P. (1 + $$\frac{\text { Gain }}{100}$$)
⇒ 8426 = C.P. (1 + $$\frac{10}{100}$$)
⇒ 8426 = C.P. ($$\frac{11}{10}$$)
⇒ C.P. = $$\frac{8426 \times 10}{11}$$
= 766 × 10 = ₹ 7660
Hence the cost price of the double bed = ₹ 7660.

Question 9.
A dealer buys a bicycle for ₹ 1,250 and marks it at 40% above its cost price. If he allows 8% discount, find
(i) Selling price of the bicycle
(ii) Profit percentage
Solution:
(i) C.P. of bicycle = ₹ 1250
M.P. of bicycle = C.P. + 40% of C.P.
= 1250 + $$\frac{40}{100}$$ × 1250
= 1250 + 500 = ₹ 1750

S.P. = M.P. (1 – $$\frac{\text { Discount% }}{100}$$)
= 1750 (1 – $$\frac{8}{100}$$)
= 1750 $$\left(\frac{100-8}{100}\right)$$
= 1750 $$\left(\frac{92}{100}\right)$$
= $$\frac{161000}{100}$$ = ₹ 1610

(ii) Profit = S.P. – C.P.
= 1610 – 1250 = ₹ 360
Profit = $$\frac{\text { Profit }}{\text { C.P. }}$$ × 100
= $$\frac{360 \times 100}{1250}$$
= $$\frac{144}{5}$$ = 28.8%.

Question 10.
Priti allows 8% discount on the marked price of the suits and still makes a profit of 15%. If her gain over the sale of a suit is ₹ 156, find the marked price of the suit.
Solution:
Let M.P. be ₹ x.
∴ Discount = $$\frac{8}{100}$$ × M.P.
= $$\frac{8}{100}$$ × x
= $$\frac{2 x}{25}$$

∴ S.P. = M.P. – Discount
= x – $$\frac{2 x}{25}$$
= ₹ $$\frac{23 x}{25}$$

∴ S.P. = C.P. (1 + $$\frac{\text { Gain }}{100}$$)
⇒ $$\frac{23 x}{25}$$ = C.P. (1 + $$\frac{15}{100}$$)
∴ $$\frac{23 x}{25}$$ = C.P. × $$\frac{23}{20}$$
⇒ C.P. = $$\frac{23 x}{25} \times \frac{20}{23}=₹ \frac{4 x}{5}$$

∴ Gain = S.P. – C.P.
= $$\frac{23 x}{25}-\frac{4 x}{5}=₹ \frac{3 x}{25}$$
∴ $$\frac{3 x}{25}$$ = 156
⇒ x = $$\frac{156 \times 25}{3}$$
⇒ 52 × 25 = ₹ 1300
Hence the marked price is ₹ 1300.

## DAV Class 8 Maths Chapter 5 Worksheet 1 Solutions

The DAV Class 8 Maths Book Solutions Pdf and DAV Class 8 Maths Chapter 5 Worksheet 1 Solutions of Profit, Loss and Discount offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 5 WS 1 Solutions

Question 1.
By selling a bedsheet for ₹ 640, a shopkeeper earns a profit of 28%. How much did it cost the shopkeeper?
Solution:
S.P. = ₹ 640
Profit = 28%
C.P. = ?
S.P. = C.P. (1 + $$\frac{\text { Profit }}{100}$$)
⇒ 640 = C.P. (1 + $$\frac{28}{100}$$)
⇒ 640 = C.P. × $$\frac{128}{100}$$
⇒ C.P. = $$\frac{640 \times 100}{128}$$ = ₹ 500
Hence, the required cost price is ₹ 500.

Question 2.
Rajan purchased 250 packets of blades at the rate of ₹ 8 per packet. He sold 70% of the packets at the rate of ₹ 11 per packet and the remaining packets at the rate of ₹ 9 per packet. Find the gain percent.
Solution:
Cost price of 250 packets = 250 × 8 = ₹ 2000
70 % of 250, i.e. 175 packets are sold for ₹ 11 per packet
∴ S.P. of 175 packets = 11 × 175 = ₹ 1925
The remaining packets are 250 – 175 = 75
∴ S.P. of 75 packets = 75 × 9 = ₹ 675
∴ Total S.P. = ₹ 1925 + ₹ 675 = ₹ 2600
∴ Gain = S.P. – C.P. = 2600 – 2000 = ₹ 600
∴ Gain % = $$\frac{\text { Gain }}{\text { C.P. }}$$ × 100
= $$\frac{600}{2000}$$ × 100 = 30%
Hence, the gain % is 30%.

Question 3.
Ankit sold two jeans for ₹ 990 each. On one, he gains 10% and on the other he lost 10%. Find his gain or loss per cent in the whole transaction.
Solution:
S.P. of two jeans = 2 × 990 = ₹ 1980
Now for 1st jeans,
S.P. = C.P. (1 + $$\frac{\text { Gain }}{100}$$)
⇒ 990 = C.P. (1 + $$\frac{10}{100}$$)
⇒ 990 = C.P. × $$\frac{110}{100}$$
⇒ C.P. = $$\frac{990 \times 100}{110}$$ = ₹ 900

For 2nd jeans,
S.P. = C.P. (1 – $$\frac{\text { Loss }}{100}$$)
990 = C.P. (1 – $$\frac{10}{100}$$)
990 = C.P. × $$\frac{90}{100}$$
C.P. = $$\frac{990 \times 100}{90}$$ = 1100.
∴ Total C.P. = ₹ 900 + ₹ 1100 = ₹ 2000
Now C.P. > S.P.
∴ Loss = ₹ 2000 – ₹ 1980 = ₹ 20
∴ Loss % = $$\frac{\text { Loss }}{\text { C.P. }}$$ × 100
= $$\frac{20}{2000}$$ × 100 = 1 %.
Hence the loss = 1 %.

Question 4.
Nidhi purchased two sarees for ₹ 2150 each. She sold one saree at a loss of 8% and the other at a gain. If she had a gain of ₹ 1230 on the whole transaction, find the selling price of the second saree.
Solution:
C.P. for one saree = ₹ 2150
Loss = 8%
∴ S.P. = C.P. [1 – $$\frac{\text { Loss }}{100}$$]
= 2150 [1 – $$\frac{8}{100}$$]
= 2150 × $$\frac{92}{100}$$ = 1978

Let gain on the other saree be ₹ x.
∴ S.P. for other saree = (₹ 2150 + ₹ x)
∴ Total S.P. for 2 sarees = ₹ 1978 + ₹ 2150 + ₹ x = ₹ (4128 + x)
Total C.P. for 2 sarees = 2 × 2150 = ₹ 4300
∴ Total gain on the whole transaction = S.P. – C.P.
= ₹ (4128 + x) – ₹ 4300 = ₹ (x – 172)
∴ x – 172 = 1230
⇒ x = 1230 + 172 = 1402
∴ S.P. for second saree = C.P. + Gain
= ₹ 2150 + ₹ 1402 = ₹ 3552.

Question 5.
By selling 35 greeting cards, a shopkeeper loses an amount equal to the selling price of 5 greeting cards. Find his loss per cent.
Solution:
Let S.P. for one greeting card be ₹ x.
∴ S.P. for 35 greeting cards = ₹ 35x
and Loss = ₹ 5x
∴ C.P. = S.P. + Loss
= ₹ 35x + ₹ 5x = ₹ 40x
∴ Loss % = $$\frac{\text { Loss }}{\text { C.P. }}$$ x 100
= $$\frac{5 x}{40 x}$$ x 100
= $$\frac{25}{2}$$
= 12 $$\frac{1}{2}$$%
Hence, the loss = 12$$\frac{1}{2}$$%.

Question 6.
A man bought bananas at the rate of 10 for ₹ 15 and sold at the rate of one dozen bananas for ₹ 15. Find his gain or loss per cent.
Solution:
C.P. for 10 bananas = ₹ 15
∴ C.P. for 1 banana = ₹ $$\frac{15}{10}$$ = ₹ 1.50
S.P. for 12 bananas = ₹ 15
∴ S.P. for 1 banana = ₹ $$\frac{15}{12}$$ = ₹ 1.25
Here C.P. > S.P.
∴ Loss = C.P. – S.P
= 1.50 – 1.25 = ₹ 0.25
∴ Loss percent = $$\frac{\text { Loss } \times 100}{\text { C.P. }}$$
= $$\frac{0.25 \times 100}{1.50}$$
= $$\frac{50}{3}$$
= $$16 \frac{2}{3}$$ %
Hence, the loss per cent is 16 $$\frac{2}{3}$$ %.

### DAV Class 8 Maths Chapter 5 Worksheet 1 Notes

Cost price (C.P) :
The price at which an article is purchased or manufactured is called its cost price (C.P.).

Selling price (S.P):
The price at which an article is sold is called its selling price (S.P.).

Profit = S.P. – C.P.
Profit % = $$\frac{\text { Profit }}{\text { C.P. }}$$ × 100

Loss = C.P. – S.P.
Loss % = $$\frac{\text { Loss }}{\text { C.P. }}$$ × 100
[Note: Profit and Loss are always calculated on the cost price of the article]

Direct formulae

(i) S.P. = C.P. (1 + $$\frac{\text { Profit }}{100}$$)
(ii) S.P. = C.P. (1 – $$\frac{\text { Loss }}{100}$$)

Market Price:
The price written on the article or tagged with a card is called its Market value.

Discount:
It is a certain percentage of rebate on the market price offered by the shopkeeper.
Discount = M.P. – S.P.

Discount % = $$\frac{\text { Discount }}{\text { M.P. }}$$ × 100

It is a new method of realising tax by the Government at every sale/purchase right from the manufacturer to the retailer. It is not in addition to the existing sales tax, but is the replacement of Sales Tax.

Example 1.
Find the profit or loss per cent, if
(i) C.P. = ₹ 55 and S.P. = ₹ 72.60
(ii) C.P. = ₹ 112, overheads = ₹ 14 and S.F. = ₹ 49.50.
Solution:
(i) Here S.P. > C.P.
∴ Profit = S.P. – C.P.
= 72.60 – 55
= ₹ 17.60

∴ Profit % = $$\frac{\text { Profit }}{\text { C.P. }}$$ × 100
=$$\frac{17.60}{55}$$ × 100 = 32%.

(ii) C.P. = ₹ 112 and overheads = ₹ 14
∴ Net C.P. = 112 + 14 = 126
S.P. = ₹ 49.50
Here C.P. > S.P.
∴ Loss = C.P. – S.P.
= ₹ 126 – ₹ 94.50
= ₹ 31.50
∴ Loss % = $$\frac{\text { Loss } \times 100}{\text { C.P. }}$$
= $$\frac{31.50 \times 100}{126}$$
= 25%.

Example 2.
Find C.P. when a bicycle is sold for ₹ 1485 at a profit of 8%.
Solution:
Here S.P. = ₹ 1485;
Profit = 8%;
C.P. = ?
S.P. = C.P. [1 + $$\frac{\text { Profit }}{100}$$]
⇒ 1485 = C.P. [1 + $$\frac{8}{100}$$]
⇒ 1485 = C.P. × $$\frac{108}{100}$$
⇒ C.P. = $$\frac{1485 \times 100}{108}$$ = ₹ 1375.

Example 3.
If oranges are bought at 11 for ₹ 30 and sold 10 for 31. Find the loss or gain per cent.
Solution:
C.P. for 11 oranges = ₹ 30
∴ C.P. for 1 orange = ₹ $$\frac{30}{11}$$
SP. for 10 oranges = ₹ 31
S.P. for 1 orange = ₹ $$\frac{31}{10}$$
Here S.P. > C.P.
∴ Gain = S.P. – C.P.
= ₹ $$\frac{31}{10}$$ – ₹ $$\frac{30}{11}$$
= ₹ $$\frac{41}{110}$$
∴ Gain % = $$\frac{\text { Gain }}{\text { C.P. }}$$ × 100
= $$\frac{\frac{41}{110}}{\frac{30}{11}}$$ × 100
= $$\frac{41}{3} \%=13 \frac{2}{3} \%$$.

Example 4.
By selling a book for ₹ 115.20, a man loses 10%. At what price should he il it to gain 5%?
Solution:
S.P. = ₹ 115.20;
Loss = 10%
C.P. =?
∴ S.P. = C.P. (1 – $$\frac{\text { Loss }}{100}$$)
115.20 = C.P. (1 – $$\frac{10}{100}$$)
115.20 = C.P. × $$\frac{90}{100}$$
C.P. = $$\frac{115.20 \times 100}{90}$$ = ₹ 128
Now C.P. = 128;
Gain = 5%;
S.P. = ?
∴ S.P. = C.P. (1 + $$\frac{\text { Gain }}{100}$$)
= 128 (1 + $$\frac{5}{100}$$)
= $$\frac{128 \times 105}{100}$$
= ₹ 134.40

Example 5.
Find the rate of discount being given on a sweater whose price has been slahed down from ₹ 975 to ₹ 760.50.
Solution:
Discount = ₹ 975 – ₹ 760.50 = ₹ 214.50
Rate of discount = $$\frac{\text { Discount }}{\text { M.P. }}$$ × 100
= $$\frac{214.50}{975}$$ × 100 = 22%
Hence, the rate of discount given = 22%.

Example 6.
A shopkeeper fixed the market price of an item 35% above its cost price. Find the discount per cent so a to gain 8%.
Solution:
Let C.P. be ₹ x.
∴ M.P. = x + $$\frac{35}{100}$$ x
= ₹ $$\frac{27}{20}$$ x

S.P. = C.P. (1 + $$\frac{\text { Gain }}{100}$$)
= x (1 + $$\frac{8}{100}$$)
= ₹ $$\frac{27}{25}$$ x

Discount = M.P. – S.P.
= $$\frac{27}{20}$$ x – $$\frac{27}{25}$$ x
= ₹ $$\frac{27 x}{100}$$

Discount % = $$\frac{\text { Discount }}{\text { M.P. }}$$ × 100
= $$\frac{\frac{27 x}{100}}{\frac{27 x}{20}}$$ × 100
= 20%
Hence, the discount is 20%.

Example 7.
A shopkeeper bought a TV at a discount of 30% of the list price of ₹ 24000. He offers a discount of 10% of the listed price to his customer. If the VAT is 10%, find:
(i) The amount paid by the customer.
(ii) The VAT to be paid by the shopkeeper.
Solution:
List price = ₹ 24000
Discount = 30%
Selling price = 24000 – $$\frac{30}{100}$$ × 24000
= ₹ 24000 – ₹ 7200
= ₹ 16800

Tax = $$\frac{10}{100}$$ × 16800 = ₹ 1680

New S.P. to customer = 24000 – 24000 × $$\frac{10}{100}$$
= ₹ 24000 – ₹ 2400 = ₹ 21600

(i) Amount paid by customer = ₹ 21600 + ₹ 2160 = ₹ 23760
(ii) Total VAT to be paid by the shopkeeper = ₹ 2160 – ₹ 1680 = ₹ 480.

## DAV Class 8 Maths Chapter 4 Brain Teasers Solutions

The DAV Class 8 Maths Book Solutions and DAV Class 8 Maths Chapter 4 Brain Teasers Solutions of Direct and Inverse Variation offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 4 Brain Teasers Solutions

Question 1A.
Tick (✓) the correct option.

(i) Both x and y are in direct proportion, then $$\frac{1}{x}$$ and $$\frac{1}{y}$$ are
(a) in direct proportion
(b) in inverse proportion
(c) neither in direct nor in inverse proportion
(d) sometimes in direct and sometimes in inverse proportion
Solution:
(a) in direct proportion

(ii) If two quantities x and y vary inversely with each other, then
(a) $$\frac{x}{y}$$ remains constant
(b) (x – y) remains constant
(c) (x + y) remains constant
(d) (xy) remains constant
Solution:
(d) (xy) remains constant

(iii) Both p and q vary directly with each other. When p is 10 and q is 15, which of the following is not a possible pair of corresponding values of p and q?
(a) 15 and 20
(b) 20 and 30
(c) 2 and 3
(d) 5 and 7.5
Solution:
(a) 15 and 20

(iv) If x and y vary directly with each other and x variation is 3?
(a) 21
(b) $$\frac{1}{8}$$
(c) 8
(d) 27
Solution:
(c) 8
$$\frac{x}{y}$$ = 3
⇒ $$\frac{24}{y}$$ = 3
⇒ y = $$\frac{24}{3}$$
⇒ y = 8.

(v) Which of the quantities vary inversely with each other?
(a) Distance travelled and cab fare.
(b) Area of a land and its cost.
(c) Number of workers and amount of work done.
(d) Number of workers and time taken to finish a job.
Solution:
(d) Number of workers and time taken to finish a job.

Question 1B.

(i) l varies directly as m and l is equal to 5, when m = $$\frac{2}{3}$$. Find l when m = $$\frac{16}{3}$$.
(ii) A bowler throws a cricket ball at a speed of 36 km/hr. How long does this ball take to travel a distance of 20 metres to reach the batsman?
(iii) Sweets from a packet were distributed among 50 children and each of them received four sweets. If it is distributed among 40 children, how many sweets will each child get?
(iv) If y is directly proportional to $$\frac{1}{x}$$ and x = 2 when y = 20, what is the value of x when y = 1.25?
(v) If l is inversely proportional to √m and l = 6 when m = 4. What is the value of m when l = 4?
Solution:
(i) Let the required value of l be x.

As this is a case of direct variation.

∴ $$\frac{5}{2 / 3}=\frac{x}{16 / 3}$$
⇒ $$\frac{2}{3} x=5 \times \frac{16}{3}$$
⇒ x = 5 × $$\frac{16}{3}$$ × $$\frac{3}{2}$$ = 40

(ii) Speed = 36 km/hr = $$\frac{36 \times 1000}{60 \times 60}$$ m/s
= 10 m/s
Time taken = $$\frac{\text { Distance }}{\text { Speed }}=\frac{20}{10}$$
= 2 seconds

(iii) Let the number of sweets be x.

It is a case of indirect variation.
Thus, 50 × 4 = 40 × x
∴ x = $$\frac{50 \times 4}{40}$$
= 5 sweets

(iv) Let the required value of x be a.

It is a case of direct variation.
∴ $$\frac{\frac{1}{2}}{20}=\frac{\frac{1}{a}}{1.25}$$
⇒ $$\frac{1}{2}$$ × 1.25
= $$\frac{1}{a}$$ × 20
a = $$\frac{2 \times 20}{1.25}$$
= $$\frac{40}{1.25}$$ = 32.

(v) Let the required value of m be x.

It is a case of indirect proportional.
As it is given that
l ∝ $$\frac{1}{\sqrt{m}}$$
⇒ l√m = k
Now, 6 × √4 = 4 × √x
√x = $$\frac{6 \times \sqrt{4}}{4}$$
x = $$\left(\frac{6 \times \sqrt{4}}{4}\right)^2$$
= $$\frac{36 \times 4}{4 \times 4}$$ = 9.

Question 2.
Determine from the values of x and y given below whether they vary directly, inversely or in neither of these ways.

(i)

(ii)

(iii)

Solution:
(i) $$\frac{x}{y}=\frac{12}{9} \neq \frac{3}{36} \neq \frac{27}{4} \neq \frac{6}{18} \neq \frac{2}{54}$$
So, they are not directly proportional.
For inversely proportional,
x1y1 = x2y2
So, 12 × 9 = 3 × 36
= 27 × 4
= 6 × 18
= 2 × 54
= 108
Hence, they vary inversely.

(ii) $$\frac{x}{y}=\frac{4}{2} \neq \frac{20}{8} \neq \frac{30}{10} \neq \frac{45}{15} \neq \frac{60}{20} \neq \frac{80}{25}$$
So, they do not vary directly.
For inversely proportional,
x1y1 = x2y2
So, 4 × 2 ≠ 20 × 8 ≠ 30 × 10 ≠ 40 × 15 ≠ 60 × 20 = 80 × 25
Hence, they are neither directly nor inversely proportional.

(iii) $$\frac{x}{y}=\frac{2}{10}=\frac{9}{45}=\frac{7}{35}=\frac{4}{20}=\frac{3}{15}=\frac{9}{45}$$
Hence, they vary directly.

Question 3.
If x and y vary inversely and x = 25, when y = 3. Find y, when x = 15.
Solution:
Here x1 = 25, y1 = 3
x2 = 15 and y2 = y
If they vary inversely, then
x1y1 = x2y2
⇒ 25 × 3 = 15 × y
⇒ y = $$\frac{25 \times 3}{15}$$
⇒ y = 5.
Hence, the value of y is 5.

Question 4.
If x and y vary inversely and y = 45, find x when constant of variation = 9.
Solution:
If x andy vary inversely, then
x1y1 = x2y2 = K (constant variation)
∴ x × 45 = 9
⇒ x = $$\frac{9}{45}=\frac{1}{5}$$
Hence, the value of x is $$\frac{1}{5}$$.

Question 5.
Oranges cost 54 for 3 dozens in the super market. What is the cost of 8 oranges?
Solution:
Let the required cost be x.

[∵ 3 dozen = 3 × 12 = 36]
As there is a direct variation,
$$\frac{36}{54}=\frac{8}{x}$$
⇒ 36 × x = 8 × 54
⇒ x = $$\frac{8 \times 54}{36}$$
⇒ x = 12
Hence, the required cost is ₹ 12.

Question 6.
A car travels 60 km in 1 hr 30 min. How long will it take to cover a distance of 100 km at the same speed?
Solution:
Let the required time be t hours.

[∵ 1 hr 30 min = 1.5 hours]
As there is a direct variation,
$$\frac{60}{1.5}=\frac{100}{t}$$
⇒ 60 × t = 100 × 1.5
⇒ t = $$\frac{100 \times 1.5}{60}$$
⇒ t = 2.5 hours = 2 hours 30 min.
Hence, the required time is 2 hours 30 min.

Question 7.
The extension of an elastic spring is found to vary directly with the weight suspended from it. If a weight of 75 kg produces an extension of 1.4 cm, calculate the weight that would produce an extension of 9.8 cm?
Solution:
Let the required weight be W kg.

As there is a direct variation,
$$\frac{75}{1.4}=\frac{W}{9.8}$$
⇒ 1.4 × W = 75 × 9.8
⇒ W = $$\frac{75 \times 9.8}{1.4}$$
⇒ W = 525 kg
Hence, the required weight is 525 kg.

Question 8.
In 25 days, the earth picks up 6 × 108 pounds of dust from the atmosphere. How much dust will it pick up in 15 days?
Solution:
Let the required dust to be pick up be x pounds.

As there is direct variation,
$$\frac{6 \times 10^8}{25}=\frac{x}{15}$$
⇒ x × 25 = 15 × 6 × 108
⇒ x = $$\frac{15 \times 6 \times 10^8}{25}$$
⇒ x = 3.6 × 108
Hence, the required amount of dust in 15 days = 3.6 × 108 pounds.

### DAV Class 8 Maths Chapter 4 HOTS

Question 1.
While driving his car at a speed of 50 km/hr, Ramit covers a distance from home to his office in 1 hour 30 minutes. One day he was 15 minutes late from his home. In order to reach office at time, what should be the speed of the car?
Solution:
Also, find the total distance covered by Ramit daily.

It is a case of inverse variation.
∴ 50 × $$\frac{3}{2}$$ = x × $$\frac{5}{4}$$
⇒ x = 50 × $$\frac{3}{2} \times \frac{4}{5}$$
= 60 km/hr
Now, speed = 50 km/hr
and time = $$\frac{3}{2}$$ hr
distance covered = 50 × $$\frac{3}{2}$$ = 75 km
Total distance covered daily = 75 km + 75 km = 150 km

Question 1.
The cost of 8 mineral water bottles is ₹ 120. Find the cost of such 18 bottles?
Solution:
Let the required cost be ₹ x.

As there is a direct variation,
$$\frac{8}{120}=\frac{18}{x}$$
⇒ 8 × x = 18 × 20
⇒ x = $$\frac{18 \times 120}{8}$$
⇒ x = 270
Hence, the number of required bottles are 270.

Question 2.
Ramesh deposited a sum of ₹ 12000 in a nationalised Bank at a certain rate of interest for 2 years and earns an interest of ₹ 900. How much interest would he earn for a deposit of ₹ 15000 for the same period and at the same rate of interest?
Solution:
Let the required amount of interest be ₹ x.

As there is a direct variation,
$$\frac{12000}{900}=\frac{15000}{x}$$
⇒ 12000x = 15000 × 900
⇒ x = $$\frac{15000 \times 900}{12000}$$
⇒ x = 1125
Hence, the required amount of interest is ₹ 1125.

Question 3.
In a picnic there is a provision for 400 students for 20 days. If 80 more students attend the picnic, for how many days will the provision be last?
Solution:
Let the number of days be x

[∵ Total number of students = 400 + 80 = 480]
As there is an inverse variation,
x1y1 = x2y2
⇒ 400 × 20 = 480 × x
⇒ x = $$\frac{400 \times 20}{480}$$
⇒ x = $$\frac{50}{3}$$ days = 16$$\frac{2}{3}$$ days
Hence, the required number of days = 16$$\frac{2}{3}$$.

Question 4.
An electric pole, 20 m high, casts a shadow of 15 m. Find the height of the tree that casts a shadow of 24 m under the similar conditions.
Solution:
Let the height of the tree required be x m.

As there is a direct variation,
∴ $$\frac{x_1}{y_1}=\frac{x_2}{y_2}$$
⇒ $$\frac{20}{15}=\frac{x}{24}$$
⇒ 15 × x = 20 × 24
⇒ x = $$\frac{20 \times 24}{15}$$
⇒ x = 32
Hence, the required height of tree is 32 m.

Question 5.
A loaded truck travels 14 km in 25 minutes. If the speed of the truck remains the same, how far can it travel in 5 hours?
Let the required distance travelled be x km.

[∵ 5 hours = 5 × 60 = 300 minutes]
$$\frac{x_1}{y_1}=\frac{x_2}{y_2}$$
⇒ $$\frac{14}{25}=\frac{x}{300}$$
⇒ 25 × x = 14 × 300
⇒ x = $$\frac{14 \times 300}{25}$$
⇒ x = 168.
Hence, the required distance is 168 km.

Question 6.
A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?
Solution:
Let the required number of machines be x.

[Total number of animals = 20 + 10 = 30]
As there is an inverse variation,
x1y1 = x2y2
⇒ 42 × 63 = x × 54
⇒ x = $$\frac{42 \times 63}{54}$$
⇒ x = 49
Hence, the required number of machines is 49.

Question 7.
A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Solution:

[Total number of animals = 20 + 10 = 30]
As there is an inverse variation,
x1y1 = x2y2
⇒ 20 × 6 = 30 × x
⇒ x = $$\frac{20 \times 6}{30}$$
⇒ x = 4
Hence, the number of days required is 4 days.

Question 8.
A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?
Solution:
Let the required number of boxes be x.

As there is an inverse variation,
x1y1 = x2y2
⇒ 25 × 12 = x × 20
⇒ x = $$\frac{25 \times 12}{20}$$
⇒ x = 15
Hence, the required number of boxes be 15.

Question 9.
A car travels 60 km in 1 hr 30 min. How long will it take to cover a distance of 100 km at the same speed?
Solution:
Let the required time be x hours.

$$\frac{60}{1.5}=\frac{100}{x}$$
⇒ 60 × x = 10 × 1.5
⇒ x = $$\frac{100 \times 1.5}{60}$$
⇒ x = $$\frac{5}{2}$$ = 2 hours 30 min.
Hence, the required time is 2 hours 30 min.

Question 10.
Ramesh can finish his work in 25 days, working 8 hrs a day. If he wants to finish the same work in 20 days, how many hours should he work in a day?
Solution:
Let the number of hours required be x.

As there is an inverse variation,
x1y1 = x2y2
⇒ 25 × 8 = 20 × x
⇒ x = $$\frac{25 \times 8}{20}$$
⇒ x = 10
Hence, the required time is 10 hours.

Question 11.
A garrison of 120 men has provisions for 30 days. At the end of 5 days, 5 more men joined them. How many days can they sustain on the remaining provision?
Solution:
Let the number of days be x.

[∵ Remaining days = 30 – 5 = 25]
[∵ Total men = 120 + 5 = 125]
As there is an inverse variation,
∴ x1y1 = x2y2
⇒ 120 × 25 = 125 × x
⇒ x = $$\frac{120 \times 25}{125}$$
⇒ x = 24
Hence, the required days are 24.

## DAV Class 7 Maths Chapter 14 Brain Teasers Solutions

The DAV Class 7 Maths Book Solutions and DAV Class 7 Maths Chapter 14 Brain Teasers Solutions of Visualising Solids offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 14 Brain Teasers Solutions

Question 1.
A. Tick (✓) the correct option:

represents a net of—
(a) cone
(b) cube
(c) cubiod
(d) cylinder
(d) cylinder

It represents a net of cylinder.
Hence, (d) is the correct option.

(ii) How many cubes are there in the given figure?

(a) 13
(b) 14
(c) 15
(d) 16
(c) 15

15 cubes are there in the figure.
Hence, (c) is the correction option.

(iii) A tetrahedron has ___________ vertices.
(a) 2
(b) 3
(c) 4
(d) 5
(c) 4

A tetrahedron has four vertices A, B, C and D.
Hence, (c) is the correct

(iv) I have one square and four triangles. I am a
(a) cube
(b) pyramid
(c) prism
(d) cubiod
(b) pyramid

A pyramid has one square and four triangles as shown in the given figure.
Hence, (b) is the correctc option.

can be folded into a:
(a) cone
(b) sphere
(c) cylinder
(d) circle
(a) cone

It can be folded into a cone.
Hence, (a) is the correct option.

(i) Draw a net for a tetrahedron.
Nets for a tetrahedron

(ii) Can this be a net for a die? Why or why not?

No, When the net is folded, it will look like the figure given alongside, in which sum of opposite face digits is not equal to seven (7).

(iii) Complete the net for making a cube.

(iv) Two cubes of edge 2 cm each are placed side to side to form a cuboid. Draw an oblique sketch for the resulting figure.

(v) What is the difference between a cube and a cuboid?
All faces of a cube are squares and all faces of a cuboid are rectangles.

## DAV Class 7 Maths Chapter 14 HOTS

Question 1.
Which figure will be opposite to the triangle when the net is folded into a cube?