CBSE Class 10

CBSE Sample Papers for Class 10 Maths Paper 4 is part of CBSE Sample Papers for Class 10 Maths . Here we have given CBSE Sample Papers for Class 10 Maths Paper 4. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Sample Papers for Class 10 Maths Paper 4

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 4 of Solved CBSE Sample Papers for Class 10 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

GENERAL INSTRUCTIONS:

• All questions are compulsory.
• This question paper consists of 30 questions divided into four sections A, B, C and D.
• Section A comprises of 6 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 8 questions 1 of 4 marks each.
• There is no overall choice. However, internal choice has been provided in one question of 2 marks, 1 three questions of 3 marks each and two questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• In question of construction, drawings shall be neat and exactly as per the given measurements.
• Use of calculators is not permitted. However, you may ask for mathematical tables.

SECTION A

Question numbers 1 to 6 carry 1 mark each.

Question 1.
The decimal expansion of the rational number \(\frac { 11 }{ { 2 }^{ 3 }\times { 5 }^{ 2 } } \) will terminate after how many places of decimal?

Question 2.
Is x = -2 a solution of the equation x² – 2x + 8 = 0?

Question 3.
In the adjoining figure, P and Q are points on the sides of AB and AC respectively of ∆ABC such that AP = 3.5 cm, PB = 7 cm, QC = 6 cm. Prove that PQ || BC.a

Question 4.
If tan θ = cot (30° + θ), find the value of θ.

Question 5.
A cylinder and a cone are of same base radius and c the ratio of volume of cylinder to that of the cone.

Question 6.
Two coins are tossed simultaneously. Find the probability of getting exactly one head.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
Complete the following factor tree and find the numbers x, y and z.

Question 8.
If α and β are zeroes of the polynomial f(x) = x² – x – k such that α – β = 9, find k.

Question 9.
The coordinates of the vertices of ∆ABC are A(4, 1), B(-3, 2) and C(0, k). Given that the area of ∆ABC is 12 unit², find the value of k. .

Question 10.
Without using trigonometric tables, evaluate the following:

Question 11.
Prove that

Question 12.
Area of a sector of a circle of radius 36 cm is 54π cm². Find the length of the corresponding arc of the sector.

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
The HCF of 65 and 117 is expressible in the form 65m – 117. Find the value of m. Also find the LCM of 65 and 117 using prime factorization

Question 14.
Solve the following pair of linear equations: px + qy = p – q;qx – py = p + q.
OR
In the given figure, ABCDE is a pentagon with BE || CD and BC || ED. BC is perpendicular to CD. If the perimeter of pentagon ABCDE is 21 cm, find the values of x and y.

Question 15.
Find the roots of the following equation

Question 16.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work for 30 days?
OR
Jaspal Singh repays his total loan of Rs 118000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, what amount will be paid by him in the 30th instalment? What amount of loan does he still to pay after the 30th instalment?

Question 17.
Prove that : sin θ(1 + tan θ) + cos θ(1 + cot θ) = sec θ + cosec θ.

Question 18.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².
OR
In the adjoining figure, DB ⊥ BC, DE ⊥ AB
and AC ⊥ BC. Prove that \(\frac { BE }{ DE } =\frac { AC }{ BC } \)

Question 19.
In the adjoining figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 4 cm and 3 cm respectively. If area of ∆ABC = 21 cm², then find the lengths of sides AB and AC.

Question 20.
In the adjoining figure, APB and CQD are semicircles of diameter 7 cm each, while ARC and BSD are semicircles of diameter 14 cm each. Find the perimeter of the shaded region.

Question 21.
Two different dice are rolled together. Find the probability of getting:
(i) the sum of numbers on two dice as 5.
(ii) even numbers on both dice.
OR
The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is \(\frac { 1 }{ 4 }\). The probability of selecting a blue ball at random from the same jar is \(\frac { 1 }{ 3 }\). If the jar contains 10 orange balls, find the total number of balls in the jar.

Question 22.
Find the value of p for the following distribution whose mean is 10:

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
If pth, qth and rth terms of an AP are a, b and c respectively, then show that (a – b) r + (b – c) p + (c – a) q = 0.

Question 24.
An icecream seller sells his icecreams in two ways:
(A) In a cone of r = 5 cm, h – 8 cm with hemispherical top.
(B) In a cup in shape of cylinder with r – 5 cm, h – 8 cm.
He charges the same price both but prefers to sell his icecream in a cone.
(a) Find the volume of the cone and the cup.
(b) Which out of the two has more capacity?
(c) By choosing a cone, which value is not followed by the icecream seller?

Question 25.
Solve the following pair of linear equations graphically:
x + 3y = 6; 2x – 3y = 12
Also find the area of the triangle formed by the lines representing the given equations with y-axis.

Question 26.
If the centroid of ∆ABC, in which A(a, b), B(b, c), C(c, a) is at the origin, then calculate the value of (a³ + b³ + c³).
OR
The three vertices of a parallelogram ABCD are A(3, -4), B(-1, -3) and C(-6, 2). Find the coordinates of vertex D and find the area of parallelogram ABCD.

Question 27.
From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the cars. (Use √3 = 1.73)
OR
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Question 28.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
OR
Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O’P are tangents to the two circles. Find the length of the common chord PQ.

Question 29.
Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ∆ABC in which PQ = 8 cm. Also justify the construction.

Question 30.
Size of agriculture holding in a survey of 200 families is given in the following table:

Compute the median and mode size of holding.

Answers

Answer 1.

Hence, the decimal expansion of given rational number will terminate after 3 places of decimal.

Answer 2.
(-2)² – 2 x (-2) + 8 = 4 + 4 + 8 = 16≠0
∴ x = -2 is not a solution of the equation x² – 2x + 8 = 0.

Answer 3.

∴ By the converse of B.P.T., PQ || BC.

Answer 4.
tan θ = cot (30° + θ)
= tan [90° – (30° + θ)]
= tan (60° – θ)
θ = 60° – θ => 2θ = 60° => θ = 30°.

Answer 5.
Let r be the base radius of cylinder and cone and h be their heights.

Hence, the ratio of volume of cylinder to that of cone = 3:1.

Answer 6.
When two coins are tossed simultaneously, then the outcomes are HH, HT, TH, TT.
So total number of outcomes = 4.
The outcomes favourable to the event ‘getting exactly one head’ are HT and TH.
∴The number of outcomes favourable to the event = 2.
∴ P(exactly one head) = \(\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)

Answer 7.
x = 3381 x 2 = 6762
y = 161 x 7 = 1127
and z = \(\frac { 161 }{ 7 }\) = 23.

Answer 8.
Given f(x) = x² – x – k

Answer 9.
Area of ∆ABC = \(\frac { 1 }{ 2 }\)| 14(2 – k) – 3(k – 1) + 0(1 – 2) |
=> 12 = \(\frac { 1 }{ 2 }\)|8 – 4k – 3k + 3|
=> 24 = |11 – 7k|
=> 11 – 7k = ±24
=> 11 – 7k = 24 or 11 – 7k = -24
=> 7k = -13 or 7k = 35
=> k = \(-\frac { 13 }{ 7 }\) or k = 5
Hence, k = \(-\frac { 13 }{ 7 }\) or k = 5

Answer 10.

Answer 11.

Answer 12.
Let the central angle (in degrees) be θ, then

Hence, the length of the corresponding arc of the sector = 3π cm.

Answer 13.
By Euclid’s division algorithm, we have
117 = 65 x 1 + 52; 65 = 52 x 1 + 13; 52 = 13 x 4 + 0
∴ HCF of 65 and 117 = 13
∴ 65m – 117 = 13 => 65m = 130 => m = 2.
Prime factorisation of 65 and 117 are as follows:
65 = 5 x 13 and 117 = 3 x 3 x 13
∴ LCM of 65 and 117 = 3 x 3 x 5 x 13 = 585.

Answer 14.
Given equations are
px + qy = p – q …(i)
qx – py = p + q …(ii)
Multiplying (i) by p and (ii) by q, we get
p²x + pqy = p² – pq …(iii)
and q²x – pqy – pq + q² …(iv)
On adding (iii) and (iv), we have

Putting x = 1 in equation (i), we have
p x 1 + qy = p – q => qy = -q => y = -1
∴ x = 1 ,y = -1.
OR
Given BE || CD and BC || ED, also BC ⊥ CD
=> BCDE is a rectangle
CD = BE
=> x + y = 5
Again, given perimeter of pentagon ABCDE = 21 cm
=> AB + BC + CD + DE + EA = 21 cm
=> 3 + x – y + x + y + BC + 3 = 21cm
=> 2x + x – y = 21 – 6
=> 3x – y = 15
Adding (i) and (ii), we get
4x = 20 => x = \(\frac { 20 }{ 4 }\) => x = 5
Putting x = 5 in (i), we get
5 + y = 5 =>y = 0
Hence, x = 5, y = 0.

Answer 15.
Given

(x + 4) (x – 7) = -30 => x² – 7x + 4x – 28 = -30
=> x² – 3x + 2 = 0 => x² – 2x – x + 2 = 0
=> x(x – 2) – 1(x – 2) = 0 => (x – 2) (x – 1) = 0
=> x – 2 = 0 or x – 1 = 0 => x = 2 or x = 1.
Hence, the roots of the given equation are 2 and 1.

Answer 16.
200, 250, 300, … is in AP.
whose first term = 200 and common difference = 50.
Let the number of days for which work is delayed be n, then

=> n[200 + 25 (n- 1)] = 27750
=> 25n² + 175n = 27750
=> n² + 7n – 1110 = 0
=> n² + 37n – 30n – 1110 = 0
=> n(n + 37) – 30(n + 37) = 0
=> (n + 37) (n – 30) = 0
=> n = -37 or n = 30
But n cannot be negative
∴ n = 30
Hence, the work was delayed by 30 days.
OR
Jaspal Singh starts repaying loan with first instalment of Rs 1000 and increases the instalment every month by Rs 100, so the numbers involved in instalments are 1000, 1100, 1200, …
These numbers form an AP with a = 1000 and d = 100.
30th term = a + 29d = 1000 + 29 x 100 = 3900.
Hence, his 30th instalment = Rs 3900
Sum of 30 terms = \(\frac { 30 }{ 2 }\) (100 + 3900)
= 15 x 4900 = 73500.
∴ The total amount of loan repaid = Rs 73500.
∴ Amount of loan still to be paid = Amount of total loan – amount of loan repaid
= Rs 118000 – Rs 73500 = Rs 44500.

Answer 17.

Answer 18.

In ∆ABC, ∠C = 90°
∴ AB² = AC² + BC²
In AECD, ∠ECD = 90°
∴ DE² = CD² + EC²
In ∆AEC, ∠ACE = 90°
∴ AE² = AC² + EC²
In ∆BCD, ∠BCD = 90°
∴ BD² = BC² + CD²
Adding (iii) and (iv), we get
AE² + BD² = (AC² + BC²) + (CD² + EC²)
= AB² + DE²
OR

Given, DB ⊥ BC, AC ⊥ BC
=> ∠DBC = ∠ACB = 90°
∴ AC || DB(sum of interior ∠s on the same side of BC is 180°)
∠ABD = ∠BAC (alt ∠s)
Now, in ABED and AACB,
∠EBD = ∠ABD = ∠BAC
∠C = ∠E (each = 90°)
∴ ∆BED ~ ∆ACB (AA similarity)

Answer 19.
Since tangents drawn from an external point to a circle are equal.
∴BF = BD = 4 cm (given)
and CE = CD = 3 cm (given)
Let AF = AE = x cm

Now, area (∆ABC) = area (∆AOB) + area (∆BOC) + area (∆AOC)
21 = \(\frac { 1 }{ 2 }\)AB x OF + \(\frac { 1 }{ 2 }\)BC x OD + \(\frac { 1 }{ 2 }\)AC x OE
=> 21 = \(\frac { 1 }{ 2 }\)(AF + BF) x 2 + \(\frac { 1 }{ 2 }\)(BD + CD) x 2 + \(\frac { 1 }{ 2 }\)(AE + CE) x 2
=> 21 = (x + 4) + (4 + 3) + (x + 3)
=> 21 = 2x + 14 => 2x = 7 => x = \(\frac { 7 }{ 2 }\) =>x = 3.5
AB = x + 4 = (3.5 + 4) cm = 7.5 cm
and AC = x + 3 = (3.5 + 3) cm = 6.5 cm.

Answer 20.
Given BC = R = 7 cm
and AE = CF = \(\frac { 7 }{ 2 }\) cm = r (say)
The perimeter of the shaded region
= perimeter of semicircle APB + perimeter of semicircle BSD + perimeter of semicircle DQC + perimeter of semicircle CRA

Answer 21.
Total possible outcomes = 6 x 6 = 36.
(i) The possible outcomes favourable to the event ‘sum of numbers on two dice is 5’ are (2, 3), (3, 2), (1, 4), (4, 1) i.e. 4 in number.
∴ Required probability = \(\frac { 4 }{ 36 } =\frac { 1 }{ 9 } \)
(ii) The possible outcomes favourable to the event ‘even numbers on both dice’ are (2, 2), (2, 4), – (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) i.e. 9 in number.
∴ Required probability = \(\frac { 9 }{ 36 } =\frac { 1 }{ 4 } \)
We know that the sum of probabilities of all elementary events = 1
=> P(red ball) + P(blue ball) + P(orange ball) = 1

=> x = 24.
Hence, the total number of balls in the jar = 24.

Answer 22.

Answer 23.
Let A be the first term and D be the common difference of the AP.
Given,
a = pth term => a = A + (p – 1)D ….(i)
b = qth term => b = A +(q – 1)D ….(ii)
c = rth term => c = A + (r – 1)D ….(iii)
Subtracting (ii) from (i), we get
a – b = (p – q) D ….(iv)
Similarly, b – c = (q – r) D ….(v)
and c – a = (r – p) D ….(vi)
∴ (a – b) r + (b – c) p + (c – a) q = [(p – q) r + (q – r) p + (r – p)q] D
= (pr – qr + pq + qr – pq) D
= 0 x D = 0.

Answer 24.
Volume of type A = Volume of cone + Volume of hemisphere

Answer 25.
The given equations can be written as

Select the coordinate axes. Take 1 cm = 1 unit on both axes.
Plot the points (0, 2) and (3, 1) on the graph paper and join these points by a straight line. Plot the points (3, -2) and (6, 0) on the same graph paper and join these points by a straight line.
The given lines intersect at the point P(6, 0). Therefore, the solution of the given pair of linear equations is x = 6, y = 0.
The area bounded by the given lines and y-axis has been shaded.

From the figure AB = 6 units and OP = 6 units
The area of shaded region = area of ∆ABP
= \(\frac { 1 }{ 2 }\) x AB x OP = \(\frac { 1 }{ 2 }\) x 6 x 6
= 18 sq. units

Answer 26.
Given vertices of ∆ABC as A (a, b), B (b, c), C (c, a) and the centroid of ∆ABC is G(0, 0).
As we know that centroid of A whose vertices are (x1 y1), (x2, y2) and (x3, y3)

Answer 27.
Let AB be a tower 100 m high. P and Q be the position of two cars.
Angles of depressions are shown in adjoining figure.
∴∠APB = 90° and ∠AQB = 45°.
In ∆APB, ∠ABP = 90°
30° = \(\frac { AB }{ PB }\)

Answer 28.
Given: P is an external point to a circle with centre O. PA and PB are two tangents drawn from P to the circle, A and B being points of contact.
To prove: PA = PB
Construction: Join OA, OB and OP.
Proof: Since the tangent at any point of a circle is perpendicular to the radius through the point of contact.
∠OAP = 90° and ∠OBP = 90°
Now, in ∆OAP and ∆OBP,
OA = OB (radii of same circle)
OP = OP (common)
∠OAP = ∠OBP (each 90°)

Answer 29.
As ∆PQR ~ ∆ABC, so the side PQ of ∆PQR is the corresponding side AB of ∆ABC.
Given AB = 6 cm and we need PQ to be 8 cm, so \(\frac { PQ }{ AB } =\frac { 8 }{ 6 } =\frac { 4 }{ 3 } \)
Thus, we are required to construct ∆PQR similar to ∆ABC such that sides of ∆PQR are \(\frac { 4 }{ 3 }\) times of the corresponding sides of ∆ABC.
Steps of construction:
1. Draw AB = 6 cm.
2. With A as centre and radius 6 cm, draw an arc.
3. With B as centre and radius 5 cm, draw an arc to meet the previous arc at C.
4. Join AC and BC, then ABC is an isosceles triangle with AB = AC = 6 cm and BC = 5 cm.
5. Take points A and P same.
6. Draw any ray AX making an acute angle with AB on the side opposite to the vertex C.

7. Locate 4 (the greater of 4 and 3) points A1, A2, A3 and A4 on AX such that AA1 = A1A2 = A2A3 = A3A4.
8. Join A3B.
9. Through A4, draw a line parallel to A3B (making an angle equal to ∠AA3B) to intersect the extended line segment AB at Q.
10. Through Q, draw a line parallel to BC (making an angle equal to ∠ABC) to intersect the extended line segment AC at R.
Then AQR i.e. PQR is the required triangle.
Justification:

Answer 30.
Construct the cumulative frequency distribution table as under:

We hope the CBSE Sample Papers for Class 10 Maths Paper 4 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 4, drop a comment below and we will get back to you at the earliest.

Value Based Questions in Science for Class 10 Chapter 15 Our Environment

These Solutions are part of Value Based Questions in Science for Class 10. Here we have given Value Based Questions in Science for Class 10 Chapter 15 Our Environment

Question 1.
Why have the railway vendors been switching over from re-usable glasses to plastic cups, then clay cups and now paper cups while we stress greater role for renewable articles.
Answer:
There is no denying the fact that reusable articles are ecofriendly. However, they are also liable to get contaminated, especially when railway vendors do not have time and resources to sterilise them. Therefore, railways switched over to disposable articles. Reusable glasses were replaced with disposable plastic cups. However, plastic cups are not biodegradable. Their disposal posed a big problem. They could not be burnt as burning produced toxic gases. The dumping place became sites of breeding flies, mosquitoes and others. The practice was, therefore, discontinued and replaced by clay cups or kulhads. However, making lakhs of kulhads daily removed top fertile soil from several hundred acres of land. The clay cups also required dumping places. Ultimately, disposable paper cups have come to stay with the vendors. The waste cups are recyclable as well as can be burnt without causing problem of dumping.

More Resources

• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Hospital waste is not allowed to be mixed with municipal trash and garbage. Neither it is dumped separately. How is it disposed off and why ?
Answer:
Hospital waste is highly contaminated and can be a source of several diseases wherever it is dumped or disposed off. It also contains infected, cancerous and other waste tissues that can be source of several diseases. Therefore, hospital wastes are collected in separate bags and taken to incinerators for disposal. In incinerators the articles are burnt at 850°C under aerobic conditions. Smoke and grit are precipitated to avoid air pollution. Only ash and unburnt metallic materials are left. They are disposed off in land fills.

Question 3.
Most Australians develop skin problems as they grow old. Why ?
Answer:
Australia lies near Antarctica in the zone where ozone hole appears every year. More high energy ultraviolet rays (UV – B) reach the earth in the area of ozone hole. UV radiations cause skin cancers, mutations and cataracts besides reducing immunity. Therefore, most Australians come to have skin problems as they grow old.

Question 4.
Mahi’s father is a manufacturer who purchases raw materials from only one firm and sells the finished product to another single firm. While studying the difference between food chain and food web Mahi noted that his father is following a faulty practice which must be stopped. He also tried to convince his father about it.
What is the fault in the business of Mahi’s father and how has Mahi got this idea ?
Answer:
In a food chain, members of one trophic level are eaten by members of higher trophic level. A disease or scarcity due to drought that kills members of one trophic level, affects the population of both lower trophic level (very high number, glut) as well as higher trophic level (starvation and death). By anology, Mahi found that any problem in the supplier firm or purchaser firm will ruin his father’s business. Therefore, like a food web, there should be a few alternative sources of raw material supply as well as sale of finished products.
You plan to organise a compaign on “Harmful effects of smoking on human health” in your neighbourhood areas and guide them.

Question 5.
(a) List any three reasons that you will give to convince the people about harmful effects of smoking on human health.
(b) List any three values that are indicated with such approach. (CCE 2014)
Answer:
(a)

1. Smoking causes cancer of oral cavity and lungs, bronchitis, emphysema, gastric ulcers and heart disease.
2. Tobacco smoke is equally harmful to nonsmokers who live along with the smoker. They develop respiratory problems and CO toxicity.
3. Smoking is a wasteful, rather harmful drain on the resources of the family.
4. Smoke is irritant to nonsmokers.

(b) Values:

1. To protect the residents from the harmful effect of smoking .
2. It will educate your classmates and others in the compaign not to indulge in smoking despite peer pressure.
3. Participants in the compaign will be able to spread the message to their families and acquaintances. Ecoclub of your school is organising a debate on the topic “Nature’s fury unleashed by human greed (Uttarakhand disaster) “.

Question 6.
(a) List three arguments that you will use to convince the audience that humans are responsible for this disaster.
(b) List three values that will be inculcated with this debate. (CCE 2014)
Answer:
(a) Nature’s Fury: Humans are responsible for excessive landslides and floods in the hilly areas due to following reasons :

1. Building dams to store water and generate electricity. However, the geology of hills is usually unstable.
   Seepage from dams make the hills prone to excessive landslides and earthquakes.
2. Hills attract a large number of tourists. Roads are build up and a great deal of construction activity is carried out along the roads to accomodate tourists.
3. Trees are cut off for construction activity and roads. Debris of construction sites is often thrown into gorges and channels near the roads and buildings.
4. A lot of garbage and other wastes are generated by tourists and the locals. However, there is no system for their proper disposal.

(b) Values:

1. It is important to know the features of our environment and surroundings.
2. No human activity should disturb the balance of nature.
3. Trees are life line against disturbances in nature. They should not be felled unless protective measures have been taken.
4. There should be proper and scientific method for disposal of wastes.
   You have been asked to talk on “Ozone layer and its protection” in the school assembly on ‘Envrionment Day’.

Question 7.
(a) Why should ozone layer be protected to save the environment ?
(b) List any two ways that you would stress in your talk to bring in awareness amongst your fellow friends that would also help in protection of ozone layer as well as envronment. (CBSE Delhi 2017)
Answer:
Ozone layer is present in the stratosphere roughtly 23-25 km above the equator and 10-16 km above the poles. It protects the earth by filtering out the harmful UV radiations,. However, due to release of ozone depleting substances (ODS), ozone layer has thinned out by 8% over the equator. A big hole appears every year in August-September over Antarctica. This has increased the level of UV-B radiations reaching the earth by 15-20%. These radiations are causing increased number of skin cancers, cataracts and reduced immunity in human beings. There is also increased incidence of blinding of animals, death of young ones, reduced photosynthesis, higher number of mutations and damage to articles.
The remedy lies in banning the use of ozone depleting substances like halons and chlorofluorocarbons (CFCs).
All the classmates and the public in general should be made aware of the consequences of depletion of ozone layer, causes of depletion as well other aspects of degradation of environment. This could be done by

1. Arranging poster making competition as well as exhibition highlighting the effects of ozone layer depletion,
2. Conducting street plays about the harmful effects of our tampering with envronment and how to remedy the situation.

Hope given Value Based Questions in Science for Class 10 Chapter 15 Our Environment are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

HOTS Questions for Class 10 Science Chapter 15 Our Environment

These Solutions are part of HOTS Questions for Class 10 Science. Here we have given HOTS Questions for Class 10 Science Chapter 15 Our Environment

Question 1.
Study the figure. What does it depict ?

Answer:
It is representation of movement of energy and matter in the biosphere.
Energy flow is unidirectional while matter is repeatedly recycled.

More Resources

• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Which trophic level eats nothing and which one is not eaten.
Answer:
Eats Nothing: Producers.
Not Eaten: Top carnivores.

Question 3.
What is the reason that a food chain consists of only 3-5 steps ? (CCE 2011)
Answer:
As per 10% law of Lindeman (1942), the energy available decreases by 90% with the rise of trophic level. 2000 J of energy available at the producer or T₁ level will provide only 2 J of energy to second order carnivores (T₄).
Therefore, an ecosystem cannot have food chains of several steps.

Question 4.
Describe how decomposers facilitate recycling of matter in order to maintain balance in the ecosystem. (CBSE Foreign 2010)
Answer:
Decomposers act on organic or biodegradable wastes by secreting digestive enzymes over them. Organic waste is broken down into soluble simpler substances. Decomposers pick up the simple organic substances for their own use leaving the inorganic substances. The phenomenon is called mineralisation. Minerals released from decaying organic matter become available to plants for reuse. Decomposers, therefore, help in recycling of minerals and maintain the balance in the ecosystem.

Hope given HOTS Questions for Class 10 Science Chapter 15 Our Environment are helpful to complete your science homework.

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NCERT Exemplar Solutions for Class 10 Science Chapter 9 Heredity and Evolution

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 9 Heredity and Evolution

NCERT Exemplar Solutions for Class 10 Science Chapter 9 Short Answer Questions

Question 1.
How is a sex of new born determined in humans ?
Answer:
Natural selection is preferential survival and differential reproduction of individuals of a population which possess variations that provide better adaptability to the existing environment. Peppered Moth of England has two forms, light grey and dark grey. Prior to industrial revolution, tree trunks in the forests around.
Manchester were light greyish green due to presence of lichens. Most of the Peppered Moth found in the area were light coloured with dark spots which could not be spotted easily by predator birds. During 1845 to 1890, air pollution killed the lichen flora. The deposition of soot changed the colour of the tree trunks to blackish one. Peppered Moth of the area also exhibited switch over to melanic or blackish form.
It provided better survival value against dark background. The light grey form largely disappeared as it could be easily spotted by predator birds and devoured.

More Resources

• NCERT Exemplar Solutions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Does genetic combination of mothers play a significant role in determining the sex of a new born ?
Answer:
No. Mothers have no role in determining the sex of the new born. Mothers are homogametic, that is, they produce only one type of ova (22 + X). Fathers are heterogametic, that is, they produce two types of sperms, gynosperms (22 + X ) and androsperms (22 + Y). If gynosperm (22 + X) fertilises the ovum (22 + X), the sex of new born will be female (44 + 2X). If androsperm (22 + Y) fuses with ovum (22 + X), the new born will be boy (44 + XY).

Question 3.
Mention three important features of fossils which help in the study of evolution.
Answer:
Fossils are remains or impressions of past organisms that are found in the rocks. Fossils of lower strata belong to early periods while those of upper strata are of later periods. Arranging the fossils stratumwise will indicate the occurrence of different forms of life at different times. It is found that the early fossils generally belong to simple organisms. Complexity and elaboration increased gradually with evolution. Evolution has never been linear or straight. A number of variants or branches appeared, some of which were more complex while others were less complex.

1. Fossils indicate the path of evolution of different groups.
2. They can indicate the phylogeny of some organisms, e.g, Horse, Elephant.
3. Some fossils have characteristics intermediate between two groups,
   e.g., toothed bird Archaeopteryx. They indicate how one group has evolved from another.

Question 4.
Why do all the gametes formed in human females have an X-chromosome ?
Answer:
Human females are homogametic. Their generic constitution is 44 + XX The two sex chromosomes are similar. Their ova which are produced after meiosis carry a gametic constitution of 22 + X No other combination is possible.

Question 5.
In human beings the satistical possibility of getting either a male or female child is 50 : 50. Give suitable explanation.
Answer:
Human females (44 + XX) are homogametic, that is, they produce only one type of ova (22 + X). Human males are heterogametic. They produce two types of sperms (22 + X and 22 + Y) in equal proportion, that is, 50 : 50 ratio. The chance of male or female child is also 50 : 50, as there is equal chance of androsperm (22 + Y) or gynosperm (22 + X) fertilizing an ovum.

Question 6.
A very small population of a species faces a greater threat of extinction than a larger population. Provide a suitable genetic explanation.
Answer:
A small population is always at a risk of degeneration and extinction due to

1. Excessive inbreeding that brings about inbreeding depression or degeneration,
2. Fewer recombinations and variations which are otherwise essential for maintaining vitality and vigour of the species.
3. Lesser adaptability to changes in the environment,
4. Increased threat to survival due to poaching, habitat destruction and environmental change.

Question 7.
What are homologous structures ? Give an example. Is it necessary that homologous structures always have a common ancestor ?
Answer:
Homologous structures or organs are those structures which have similar origin, similar development, similar internal structure and similar basic plan but show different external form and function, e.g., forelimbs of amphibians, reptiles, mammals and birds. Homologous structures are always have a common ancestry because otherwise there cannot be any similarity in basic plan, internal structure, development or origin. Modifications have occurred in them due to varied adaptations.

Question 8.
Does the occurrence of diversity of animals on earth suggest their diverse ancestry also ? Discuss this point in the light of evolution.
Answer:
Diversity of animals does not mean that they have diverse ancestry. Animals can be grouped into distinct lineages (e.g., mammals, birds, reptiles, annelids). Many of the lineages further show some similarities in basic traits indicating a common ancestry, e.g.,
vertebrates. Therefore, animals having a common ancestor in the remote past have successively developed newer and newer traits forming various groups of animals.

Question 9.
Give the pair of contrasting traits of the following characters in Pea plant and mention which is dominant and recessive

1. Yellow seed
2. Round seed.

Answer:

1. Seed Colour: Yellow (dominant), green (recessive).
2. Seed Shape: Round (dominant), wrinkled (recessive).

Question 10.
Why did Mendel choose Pea plant for his experiments ?
Answer:
Mendel chose Garden Pea or Edible Pea (Pisum sativum) as his experimental material because of

1. Easy availability of detectable contrasting traits of several characters,
2. Flower structure normally allows self fertilization but cross fertilization can also be carried out easily.
3. Annual nature of the plant,
4. Formation of a large number of seeds by each plant,
5. Requirement of little care except at the time of cross breeding.

Question 11.
A woman has only daughters. Analyse the situation genetically and provide a suitable, explanation.
Answer:
A woman produces only one type of ova (22 + X) while her husband produces two types of sperms, gynosperms (22 + X) and androsperms (22 + Y) in equal proportion. It is a chance that each time the woman conceived, only the gynosperm fertilised the egg so that only daughters were born.

NCERT Exemplar Solutions for Class 10 Science Chapter 9 Long Answer Questions

Question 12.
Does geographical isolation of individuals of a species lead to formation of a new species ? Provide a suitable explanation.
Answer:
Yes. Geographical isolation of a population will lead to genetic drift as there will be no gene flow between it and the parent species. Inbreeding in small population will result in fixation of certain alleles and elimination of others. There will be change in gene frequency. Mutations will produce new alleles and hence a new gene pool. Accumulation of new alleles and hence new variations over several generations will ultimately lead to the formation of new species.

Question 13.
Bacteria have a simpler body plan when compared with human beings. Does it mean that human beings are more evolved than bacteria. Provide a suitable explanation.
Answer:
Both bacteria and human beings perform all the activities of life and live comfortably in their environments. They, therefore, seem
to be equally evolved. However, human beings have a far more complex organisation and differentiation which are absent in bacteria. Since complex organisation and differentiation develop only through evolution, humans are far more evolved than bacteria.

Question 14.
All the human races like Africans, Asians, Europeans, Americans and others might have evolved from a common ancestor. Provide a few evidences in support of this view.
Answer:
All the human races have evolved from a common ancestor because they possess

1. Common body plan
2. Common structure
3. Similar physiology
4. Similar metabolism
5. Similar chromosome number
6. Common genes or genetic blue print
7. Free interbreeding.

Question 15.
Differentiate between inherited and acquired characters. Give one example of each type.
Answer:

Question 16.
Give reasons why acquired characters are not inherited.
Answer:
Many of the variations have no immediate benefit to the species. They function as preadaptations which can be beneficial under certain environmental conditions like heat tolerance variation if the temperature of the area rises.

Question 17.
Evolution has exhibited greater stability of molecular structures when compared with morphological structures. Comment on the statement and justify your opinion.
Answer:
Life is a highly organised system of biochemicals and their reactions. It is expressed inside the cells. They are the same right from bacteria to human beings. However, the method of procurement of the raw materials for forming bio-chemicals is not the same in various organisms. It has helped in avoiding cut throat competition amongst the living beings. They developed different morphological features and modes of obtaining nutrients. As a result a huge diversity has appeared in the living world.

Question 18.
In the following crosses, write the characteristics of the progency

Answer:
(a) All Round Yellow
(b) Round Yellow, Round Green, Wrinkled Yellow and Wrinkled Green in the ratio of 9 : 3 : 3 : 1.
(c) All Wrinkled Green (et) All Round Yellow (Rr Yy).

Question 19.
Study the following cross showing self pollination in F₁. Fill in the blank and answer the question that follows :

Answer:
Rr Yy (Round Yellow ).

Question 20.
In the above question, what are combination of characters in F₂ progeny ? What are their ratios ?
Answer:
Round Yellow – 9,
Round Green – 3,
Wrinkled Yellow – 3,
Wrinkled Green -1
i.e., 9 : 3 : 3 : 1.

Question 21.
Give the basic features of the mechanism of inheritance.
Answer:
Basic Features of Inheritance

1. Unit Characters: An organism is made of a large number of characters, each of which behaves as a unit.
2. Genes: Characters are controlled by genes.
3. One Gene-One Character: A single gene generally controls one character.
4. Location of Genes: Genes are located on chromosomes.
5. Alleles: A gene may have two or more forms called alleles. They represent different traits of a character.
6. Paired Alleles: An individual possesses two alleles of every gene. The two may be similar or dissimilar.
7. Dominance: Where there are two different forms or alleles of the gene, generally one expresses its effect. It is called dominant allele. The other which does not express its effect in presence of dominant allele, is called recessive allele.
8. Segregation: The two alleles separate at the time of gamete formation. A gamete has only one allele or form of the gene.
9. Independent Assortment: The alleles of different genes located on separate chromosomes behave independent of one another.
10. Pairing: Fusion of gametes during fertilization, brings together the two forms of a gene in the zygote.

Question 22.
Give reasons for the appearance of new combination of characters in the F₂ progeny.
Answer:
Independent Assortment: The two forms of a gene separate and pair independent of the two forms of other genes during gametogenesis and fertilisation. It causes new combination of characters, e.g.;

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 9 Heredity and Evolution are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

Value Based Questions in Science for Class 10 Chapter 9 Heredity and Evolution

These Solutions are part of Value Based Questions in Science for Class 10. Here we have given Value Based Questions in Science for Class 10 Chapter 9 Heredity and Evolution

Question 1.
Theory of natural selection stresses upon struggle for existence and survival of the fittest. How is this applicable to us ?
Answer:
Struggle is natural to our existence. A toddler will struggle to get up and walk. Every body struggled to get admission in a good school. There is struggle to get selected in the school team for various sports and extracurricular activities. Every student works hard to score well in the examination. People struggle to get job after finishing the studies. There is struggle to remain fit, and so on. In the struggle for any aspect of life only the fittest are able to win and obtain what they are aspiring whether in sports team or merit in studies or a job after studies.

More Resources

• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Everybody cannot be topper or a good sportsperson. But everybody has some good quality where he or she can excel whether it is painting, gardening, singing, playing instruments, dancing, a good salesman, a good entertainer or event management. What is social impact of emigration and acclimitisation ?
Answer:
A person emigrates to another state or country in order to get better opportunities, better perks and better living. Initially the person calls his home more frequently and also visits his place as often as possible. However, slowly he mixes up in the society where he works and lives. As acclimitisation increases, his calls and visits to his home become less frequent, His childern who are brought up in a different environment are unable to adjust themselves in the family and his old circle of relations. This disinterest in visiting fathers old home and family adds to inability of the person to go home as frequendy. In the new area the children grow up in a different set up. Their attitude and perceptions are often at variance with that of the parents. Clashes may occur. Ultimately the emigrant feels quite unsatisfied, dejected but helpless.

Question 3.
Recessive traits do not express their effect in the presence of dominant traits. How is this fact useful in overcoming hereditary diseases in the families ?
Answer:
Most hereditary diseases are caused by recessive traits in the homozygous state. Homozygosity increases if marriages are performed within relatives of upto 4-6 generations. In heterozygous state the recessive traits remain suppressed due to presence of dominant traits. Therefore, for overcoming hereditary diseases, it is important that individuals should remain heterozygous. This is possible only through out crossing or marriages between unrelated individuals or not having common ancestors on either side for 4-6 generations.

Question 4.
Raghu often taunts his wife for having only daughters and no son. As a student of biology how will you convince Raghu that his wife has no role in giving birth to girls only ?
Answer:
By telling Raghu that sex of the child is determined at the time of conception. Women are homogametic, i.e., they produce only one type of ova (22 + X). Males are heterogametic. They produce two types of sperms, androsperms (22 + Y) and gynosperms (22 + X). The two types of sperms are formed in equal proportion. It is chance factor that gynosperm fuses with the ovum (22 + X and 22 + X) resulting in the female child. The same chance is possible for the second and even the third time. In any case, for the sex of the child, only the father is responsible.

Hope given Value Based Questions in Science for Class 10 Chapter 9 Heredity and Evolution are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.