Bhagya

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Let side of cube = x cm
Volume of cube = 64 cm³
[volume of cube = (side)³]
x³ = 64
x = \(\sqrt[3]{4 \times 4 \times 4}\)
x = 4 cm
∴ side of cube = 4 cm.

When cubes are joined end to end and cuboid is formed
whose Length = 2x cm = 2(4) = 8 cm
Width = x cm = 4 cm
Height = x cm = 4 cm

Surface area of cuboid = 2[LB + Bh + hL]
= 2 [8 × 4 + 4 × 4 + 4 × 8]
= 2 [32 + 16 + 32]
= 2 [80]
∴ Surface area of cuboid = 160 cm².

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere ¡s 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:

Diameter of hemisphere = Diameter of cylinder
= 14 cm
2R = 14 cm
Radius of hemisphere (R) = 7 cm
Total height of vessel = 13 cm
∴ Height of cylinder = (13 – 7) = 6 cm
Inner surface area of vessel = inner surface area of cylinder + Inner surface area of hemisphere
= 2πRH + 2πR²
= 2πR [H + R]
= 2 × \(\frac{22}{7}\) × 7(16 + 7)
= 44 × 13 = 572 cm²
Hence, Inner surface area of vessel = 572 cm²

Question 3.
A toy is ¡n the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Radius of cone = Radius of hemisphere (R) = 3.5 cm
Total height of toy = 15.5 cm
∴ Height of cone (H) = (15.5 – 3.5) = 12 cm

Slant height of cone = \(\sqrt{\mathrm{R}^{2}+\mathrm{H}^{2}}\)

= \(\sqrt{(3.5)^{2}+(12)^{2}}\)

= \(\sqrt{12.25+144}\) = \(\sqrt{156.25}\)
Slain height conk (l) = 12.5 cm
Total surface area of toy = Surface area of cone + Surface area of hemisphere
= πRL + 2πR²
= πR[L + 2R]
= \(\frac{22}{7}\) × 3.5 [12.5 + 2 (3.5)1 cm²
= \(\frac{22}{7}\) × 3.5 [19.5] cm²
= \(\frac{1501.5}{7}\) = 214.5 cm²
∴ Total surface area of toy = 214.5 cm²

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have?
Find the surface area of the solid.
Solution:
Side of cubical box = 7 cm
Diameter of hemisphere = Side of cubical box = 7 cm
2R = 7
R = \(\frac{7}{2}\) cm

Surface area of solid = (surface area of the cube) – (area of base of hemisphere) + curved surface area of hemisphere)
= 6l² – πR² + 2πR²
= 6l² + πR²
= 6(7)² + \(\frac{22}{7}\) \(\frac{7}{2}\)²
= [6 × 49 + 11 × \(\frac{7}{2}\)]cm²
= 294 + 38.5 = 332.5 cm²

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of (he hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Let each side of cube = a
∴ Diameter of hemisphere = Side of cube
2R = a
R = \(\frac{a}{2}\)

Surface area of remaining solid = Total surface area of euboid – Area of the top of cube + Inner curved Surface area of hemisphere
= 6 (side)² – πR² + 2πR²
= 6(a)² + πR²
= 6(a)² + π \(\frac{a}{2}\)²
= 6a² + π \(\frac{a^{2}}{4}\)
= a² 6 + \(\frac{\pi}{4}\) cm²

Question 6.
A medicine capsule is ¡n the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:
Diameter of capsule = Diameter of hemisphere = Diameter of cylinder = 5 mm
∴ 2R = 5 mm
R = \(\frac{5}{2}\) mm

Length of entire capsule = 14 mm
Height of cylinderical part = (14 – \(\frac{5}{2}\) – \(\frac{5}{2}\)) mm
= (14 – 5) mm
H = 9 mm
Surface area of capsule = Surface area of cylinder + 2 Surface area of hemisphere
= 2πRH + 2 (2πR²)
= 2πRH + 4πR²
= 2πR [H + 2R]
= 2 × \(\frac{22}{7} \times \frac{5}{2}\left[9+2\left(\frac{5}{2}\right)\right]\)
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) [9 + 5]
= \(\frac{22}{7}\) × 5 × 14
= 220 mm²
Hence, Surface area of capsule = 220 mm².

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of 500 per (Note that the base of the tent will not be
covered with canvas.)
Solution:
Diameter of cone = Diameter of cylinder
2R = 4
R = 2 m

Radius of cone = Radius of cylinder
Height of cylinder (H) = 2.1 in
Slant height of cone (L) = 2.8 m
Curved surface area of tent = Curved surface of cylinder + Curved surface of conical part
= 2πRH + πRL
= πR [2H + L]
= \(\frac{22}{7}\) × 2[2(2.1) + 2.8]
= \(\frac{22}{7}\) × 2[4.2 + 2.8]
= \(\frac{22}{7}\) × 2 × 7
= 44 m²
∴ Curved surface area of tent = 44 m²
Cost of 1m² canvas = ₹ 500
Cost of 44 m² canvas = 44 × 500 = ₹ 22000
Hence, Total cost of canvas = ₹ 22000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Solution:
Diameter of cylinder (D) = 1.4 cm = Diameter of cone
∴ Radius of cylinder = Radius of cone (R) = 0.7 cm
Height of cylinder (H) = 2.4 cm

As we know, L² = R² + H² + (2.4)²
L = \(\sqrt{(0.7)^{2}+(2.4)^{2}}\)
= \(\sqrt{0.49+5.76}\) = \(\sqrt{6.25}\)
L = 2.5 cm
Total surface area of remaining solid = curved surface area of cylinder + Area of base of cylinder + Surface area of cone
= 2πRH + πR² + πRL
= πR [2R +R + L]
= \(\frac{22}{7}\) × 0.7 [2(2.4) + 0.7 + 2.5]

=\(\frac{22}{7}\) × \(\frac{7}{10}\) [4.8 + 3.2]

= \(\frac{22}{10}\) [8]

= \(\frac{176}{10}\) = 17.6 cm²

Hence, Total surface area remaining solid to nearest cm² = 18 cm².

Question 9.
A wooden article was made by scooping out a hemisphere form each end of a solid cylinder, as shown in Fig.

If the height of the cylinder is 10 cm, and its base ¡s of radius 3.5 cm, find the total surface area of the article.
Solution:
Height of cylinder (H) = 10 cm
Radius of cylinder = Radius of hemisphere (R) = 3.5 cm

Surface area of article = curved surface area of cylinder + 2 curved surface area of hemisphere
= 2πRH + 2 (2πR²)
= 2πR [H + 2R]
= 2 × \(\frac{22}{7}\) × 3.5 [10 + 2(3.5)]
= \(\frac{44}{7}\) × \(\frac{35}{10}\) [10 + 7]
= 44 × \(\frac{5}{10}\) × 17
= 44 × \(\frac{1}{2}\) × 17
= 22 × 17 = 374 cm²
Hence, total surface area of article = 374 cm².

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O. Join A to O. Show that:
(i) OB = OC
(ii) AO bisects ∠ A
Answer:

In ∆ ABC, AB = AC
∴ ∠ ABC = ∠ ACB (Theorem 7.2)
∴ \(\frac{1}{2}\) ∠ ABC = \(\frac{1}{2}\) ∠ ACB
∴ ∠ OBC = ∠ OCB (BO bisects ∠ ABC and CO bisects ∠ ACB)
Now, in ∆ OBC, ∠ OBC = ∠ OCB
∴ OB = OC (Theorem 7.3)
Similarly, ∠ ABC = ∠ ACB gives
∴ \(\frac{1}{2}\) ∠ ABC = \(\frac{1}{2}\) ∠ ACB
∴ ∠ ABO = ∠ ACO
Now, in ∆ ABO and ∆ ACO,
AB = AC (Given)
∠ ABO = ∠ ACO
and OB = OC
∴ ∆ ABO ≅ ∆ ACO (SAS rule)
∴ ∠ BAO = ∠ CAO (CPCT)
But, ∠ BAO + ∠ CAO = ∠ BAC (Adjacent angles)
∴ ∠ BAO = ∠ CAO = \(\frac{1}{2}\) ∠ BAC
Thus, AO bisects ∠ A.

Question 2.
In ∆ ABC, AD is the perpendicular bisector of BC (see the given figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.

Answer:
In ∆ ABC, AD is the perpendicular bisector of BC.
∴ BD = CD and ∠ ADB = ∠ ADC = 90°
In ∆ ADB and ∆ ADC,
AD = AD (Common)
∠ ADB = ∠ ADC (Right angles)
and BD = CD
∴ ∆ ADB ≅ ∆ ADC (SAS rule)
∴ AB = AC (CPCT)
Now, in ∆ ABC, AB = AC.
Hence, ∆ ABC is an isosceles triangle in which AB = AC.

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.

Answer:
In ∆ ABC, AC = AB
∴ ∠ ABC = ∠ ACB
∴ ∠ FBC = ∠ ECB
Now, in ∆ FBC and ∆ ECB,
∠ FBC = ∠ ECB
∠ BFC = ∠ CEB (Right angles)
BC = CB (Common)
∴ ∆ FBC ≅ ∆ ECB (AAS rule)
∴ CF = BE (CPCT)
Thus, the altitudes CF and BE on equal sides AB and AC respectively of ∆ ABC are equal.

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that
(i) ∆ ABE ≅ ∆ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.

Answer:
In ∆ ABE and ∆ ACF,
∠ AEB = ∠ AFC (Right angles)
∠ A = ∠ A (Common)
BE = CF (Given)
∴ ∆ ABE ≅ ∆ ACF (AAS rule)
∴ AB = AC (CPCT)
Thus, ∆ ABC is an isosceles triangle.

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ ABD = ∠ ACD.

Answer:
∠ ABC and ∠ DBC are adjacent angles.
∴ ∠ ABC + ∠ DBC = ∠ ABD ………… (1)
∠ ACB and ∠ DCB are adjacent angles.
∴ ∠ ACB + ∠ DCB = ∠ ACD ………….. (2)
In ∆ ABC, AB = AC.
∴ ∠ ABC = ∠ ACB (Theorem 7.2)
In ∆ DBC, DB = DC.
∴ ∠ DBC = ∠ DCB (Theorem 7.2)
∴ ∠ ABC + ∠ DBC = ∠ ACB + ∠ DCB
∴ ∠ ABD = ∠ ACD [From (1) and (2))

Question 6.
∆ ABC is an isosceles triangle in which AB = AC. side BA is produced to D such that AD = AB (see the given figure). Show that ∠ BCD is a right angle.

Answer:
AB = AC and AD = AB
∴ AC = AD
In ∆ ABC, AB = AC
∴ ∠ ACB = ∠ ABC (Theorem 7.2) ……………… (1)
In A ADC, AC = AD
∴ ∠ ACD = ∠ ADC (Theorem 7.2) ……………… (2)
Adding (1) and (2),
∠ ACB + ∠ ACD = ∠ ABC + ∠ ADC
∴ ∠ BCD = ∠ DBC + ∠ BDC
(Adjacent angles and A lies on BD)
In ∆ BCD,
∠ DBC + ∠ BDC + ∠ BCD = 180°
∴ ∠ BCD + ∠ BCD = 180° (from (3)]
∴ 2 ∠ BCD = 180°
∴ ∠ BCD = 90°
Thus, ∠ BCD is a right angle.

Question 7.
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠ B and ∠ C.
Answer:

In ∆ ABC, AB = AC
∴ Z C = Z B (Theorem 7.2)
In ∆ ABC,
∠ A + ∠ B + ∠ C = 180°
∴ 90° + ∠ B + ∠ B = 180° (Given and ∠ C = ∠ B)
∴ 2 ∠ B = 90°
∴ ∠ B = 45°
∴ ∠ C = 45°

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Answer:

∆ ABC is an equilateral triangle.
∴ AB = BC = AC
In ∆ ABC, AB = BC
∴ ∠ C = ∠ A (Theorem 7.2)
In ∆ ABC, AB = AC
∴ ∠ C = ∠ B (Theorem 7.2)
Hence, ∠ A = ∠B = ∠ C.
Now, in ∆ ABC, ∠ A + ∠ B + Z C = 180°
∴ ∠ A = ∠ B = ∠ C = \(\frac{180^{\circ}}{3}\) = 60°
Thus, the angles of ah equilateral triangle are 60° each.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

1. Construct ΔABC, given AB = 6 cm, ∠A = 30° and ∠B = 75°.
Solution:
Given. One side of ΔABC as AB = 6 cm, m∠A = 30° and m∠B = 75°.
To construct: A triangle with one side and these two angles.
Steps of Construction :
Step 1. We first draw a rough sketch of ΔABC and indicate the measures of side and two angles.

Step 2. Draw a ray AB of length 6 cm.

Step 3. At A; draw a ray AX making an angle 30° with AB.

Step 4. With the help of compass. At B; draw a ray BY making an angle of 75° with AB.

Step 5. Both rays AX and BY intersect, at a point. So the point of intersection of the two rays is C Then. ΔABC is now obtained.

2. Construct an isosceles ΔABC such that AB = 5.3 cm and each base angle = 45°.
Solution:
Given : Isosceles ΔABC with AB = 5.3 cm each base angle = 45°.
To Construct: A triangle with one side and two base angles.
Steps of construction :
Step 1. Draw a rough sketch of ΔABC with given measures

Step 2. Draw a line segment AB = 5.3 cm.

Step 3. Taking A as centre with the help of compass. Draw a ray AX making an angle 45° with AB.

Step 4. With the help of compass and taking B as a centre. Draw a ray BY making an angle 45° with the line segment AB.

Step 5. Rays AX and BY intersect, at a point say C, then ABC is the required triangle.

3. Construct ΔXYZ if XY = 4 cm, ∠X = 45° and ∠Z = 60°.
[Hint : ∠Y = 180° – 45° – 60° = 75°]
Solution:
One side of ΔXYZ as
XY = 4 cm,
∠X = 45°
and ∠Z = 60°.
As we know that by angle-sum property of a triangle; sum of all three angles of a triangle is equal to 180°.
∴ ∠X + ∠Y + ∠Z = 180°
⇒ 45° + ∠Y + 60° = 180°
⇒ 105° + ∠Y = 180°
⇒ ∠Y = 75°.
Now it will be easy to construct triangle with side
XY = 4 cm,
∠X = 45°
and ∠Y = 75°.
Steps of Construction :
Step 1. We first draw a rough sketch of ΔXYZ and indicate the measure of side and two angles.

Step 2. Draw a ray XY of length 4 cm.

Step 3. At X draw a ray XA making an angle of 45° with XY.

Step 4. At Y; draw a ray YB making an angle of 75° with XY.

Step 5. Z has to lie on both rays XA and YB. So, the point of intersection of two rays is Z.
ΔXYZ is now obtained.

4. Examine whether you can construct ΔPQR such that ∠P = 100°, ∠Q = 90° and PQ = 4.3 cm If not possible given reason.
Solution:
No, we cannot construct given ΔPQR.
Reason :
As we know that by angle sum property of a triangle; sum of all three angles a triangle is equal to 180°. But in given question sum of two angles;
m∠P + m∠Q
= 100° + 90°
= 190°
The sum of these two angles should be less than 180°. So triangle with given measures cannot be constructed as it violates the angle sum property of a triangle.

5. Question (i).
In which of the following cases a unique triangle can be drawn ?
(a) BC = 5 cm, ∠B = 90° and ∠C = 100°
(b) AB = 4 cm, BC = 7 cm and CA = 2 cm
(c) XY = 5 cm, ∠X = 45°, ∠Y = 60°
(d) An isosceles triangle with length of each equal side equal to 5 cm.
Answer:
(c) XY = 5 cm, ∠X = 45°, ∠Y = 60°

Question (ii).
A triangle can be constructed by taking two of its angles as.
(a) 110°, 40°
(b) 70°, 115°
(c) 135°, 45°
(d) 90°, 90°
Answer:
(a) 110°, 40°

Punjab State Board PSEB 7th Class English Book Solutions English Grammar Tense Exercise Questions and Answers, Notes.

PSEB 7th Class English Grammar Tense

Study the Chart

Exercise 1

I. Fill in the blanks with the right tense form of the verb given in the brackets:

1. Our school ……………………… (begin) with a prayer everyday.
2. Ahmed ……………………… (keep awake) till midnight these days.
3. What…………….(make) you do so ?
4. This box ………………….. (contain) a gift for him.
5. This road is closed. They …………………….. (repair) it.
6. The clerk ……………. (type) the letter still, he never …………… (finish) the work in time.
7. Don’t make a noise. An important meeting ……………………. (go on) here. 8. She
8. ………… (say) a prayer to God regularly before going to bed.
9. The peon ……………….. (ring) the bell now.
10. We ……………. (feel) uneasy on a very hot day.
Hints:
1. begins
2. keeps awake
3. makes
4. contains
5. are repairing
6. is typing, finishes
7. is going on
8. says
9. is ringing
10. feel.

Exercise 2

Fill in the blanks with the right tense form of the verb given in the brackets: (Use Present Perfect Tense)

1. The water level …………………… (go up) because of rains.
2. The doctor ……………………. (examine) the patient. He is improving now.
3. I ……………………… (finish) my work. I am going home now.
4. They ………………….. (leave) the place.
5. Sheela …………………. (learn) her lesson.
6. Father ………………………. (not come) home for lunch yet.
7. ………………… you ………………. (finish) your work ? Can you come with me now?
8. I ……………………. (study) the problem. It is easy to solve.
9. ………….. he ………. (take charge) of his new assignment ?
10. The train …………………………… (leave) the station. The platform looks deserted.
Hints:
1. has gone up
2. has examined
3. have finished
4. have left
5. has learnt
6. has not come
7. Have, finished
8. have studied
9. Has, taken charge
10. has left.

Exercise 3

Fill in the blanks with the right tense form (Past Continuous or Simple Past) of the verbs given in brackets:

1. She ………………… (look) for a book shop, when I ………………….. (meet) her.
2. The policeman …………………… (arrest) the thief, when I …………………… (see) him.
3. I……………………………… (go) to the cattle-shed, when I …………………….. (hear) someone quarrelling.
4. The sarpanch ………… (take) the horse out of the stable, when I ……. (call) him.
5. Yesterday as I …………. (walk) along the street. I ……………………… (meet) my friend.
6. In January 1948 Gandhiji ……………………… (stay) in Delhi. He was shot while he ……………. (come out) of the prayer meeting.
7. While I …………………. (watch) the T.V., the lights ………….. (go off).
Hints:
1. was looking, met
2. was arresting, saw
3. was going, heard
4. was taking, called
5. was walking, met
6. was staying, was coming out
7. was watching, went off.

Exercise 4

Fill in the blanks using the verbs in the brackets according to the tense form indicated:

1. A group of officials ……….. (go) to Delhi tomorrow. (Simple Future)
2. We …………………… (visit) Kashmir next January. (Future Continuous)
3. What …………………….. you ……………(do) tomorrow evening ? (Future Continuous)
4. I …….. (go) with my brother. I am sure I ……………… (have) a very nice time on this (Simple Future)
5. I want to give this book to Jaspreet …………………… you ……………………… (go) to give this book to her ? (Simple Future)
Hints:
1. will go
2. shall be visiting
3. will, be doing
4. shall go, will have
5. will, go.

Exercise 5

Fill in the blanks to express Future Perfect aspects of the verbs given in the brackets.

1. The police …………………… (arrest) the thief by tomorrow.
2. Davinder. ………………… (tell) me all before you talk to him.
3. He ………………….. (return) before you arrive.
4. Gobind ……………………. (write) the story before you return.
5. We …………………… (enjoy) our holidays for about a month before he arrives.
Hints:
1. will have arrested
2. will have told
3. will have returned
4. will have written
5. shall have enjoyed.

Exercise 6
(Miscellaneous)

I. Underline the verb and write in the space given whether the sentence is in the Present, Past or Future Perfect tense. One has been done for you.

1. The rain had stopped before I arrived. (Past Perfect Tense)
2. I have lived in Bhatinda since childhood. (Present Perfect Tense)
3. We shall have reached home before they arrive. (Future Perfect Tense)
4. The children shall have eaten something by the time I reach home, (Future Perfect Tense)
5. She has not finished writing the book. (Present Perfect Tense)
6. The watchman had run away before the owner reached. (Past Perfect Tense)
7. The children have learnt the song. (Present Perfect Tense)
8. Naaz has posted the letter. (Present Perfect Tense)
9. My uncle has given me a room in the new house. (Present Perfect Tense)
10. They will have left for Patiala by night. (Future Perfect Tense)

II. Complete the following letter by using the verb in the bracket in present perfect/ past tense. One has been done for you:

Dear sir
I wrote (write) to you some time ago-asking about conditions of entry to your competition. You replied (reply) enclosing an entry form which I filled up (fill up) and sent (send) without delay. I have heard (hear) nothing from you and am beginning to wonder if my application has gone (go) astray. Please check if you have received (receive) it or not. In case you have not got (not get) it. Kindly inform me.

Thanking you
Yours faithfully
Amarjit Singh

III. You have planned to undertake a railway journey in your summer holidays. In about ten sentences describe your forth coming trip using simple future tense.

I will go to Amritsar during summer holidays. I will go by train. I will catch the train from Ludhiana. I will go to the railway station. I will buy a ticket and go to the platform to board the train. The train will reach Amritsar within three hours. There, I will stay in a hotel. I will visit Sri Harmandar Sahib. I will have a holy dip in the tank there. I will also visit the Durgiana Mandir. I will not miss to see the Jallianwala Bagh before I come back.

IV. Your house is flooded due to heavy rains. You saved yourself by sitting on the roof top for almost three days and nights. Using simple and continuous past tense, write your experience.

I found water all around me. It was rising continually. I, with my family, took shelter on the roof of my house. We took with us all the eatables available in home. There was a large carpet at the roof. At night we slept on it. But the rain did not stop for three days. All our eatables were consumed. We grew weaker and weaker. We prayed to God for help. Then one day, it stopped raining. We came down, took buckets and threw the water out of the house. We thanked God for saving our lives.

V. Supply for the blanks the future perfect tense of the verb given in the brackets:

1. Our maid will have broken all the cups (break).
2. He …………..by that time. (return)
3. The sun …………… when we reach home. (set)
4. We …………. all the cakes by evening. (eat)
5. She ……………for the family by night. (cook)
6. He …………. his pledge. (keep)
7. Each child …………… a new pull over. (bought)
8. The shoeshine …………… my shoes. (polish)
9. The commander ………….. the army to march. (order)
10. I……………. the job by sunset. (complete)
11. She …………. to speak English by the year end. (learn)
12. You ………….. tea before we reach there. (drink)
13. Meeta ………….. the beauty contest. (win)
14. Mini …………… to India by early September. (fly)
Hints:
2. will have returned
3. will have set
4. shall have eaten
5. will have cooked
6. will have kept
7. will have bought
8. will have polished
9. will have ordered
10. shall have completed
11. will have learnt
12. will have drunk
13. will have won
14. will have flown.

VI. Given below is a complaint letter. Fill in the blanks with the correct form of the verb given in the brackets. One has been done.

The Secretary
DIV/4901
Vasant Kunj
New Delhi.
Dear Sir

I regret to bring (bring/brought) to your notice that Bihari Lal, the sweeper is not doing (is not doing / has not done) his duty well. He sweeps (sweeps / sweep) the road only once a day. He leaves (leave /leaves) the garbage on the road or throws (threw / throws) them all around. As a result the area is filthy. I have requested (am requesting /have requested) him many times but he refused (refuses / refused) to obey. It seems (seem / seems) he does not care (do not care/does not care).

Yours sincerely
Anil Sharma

VII. Rewrite the following sentences after changing the verbs into the present or past continuous tense:

1. Sudha lies on the bed.
2. Raja plays with his brother.
3. The servant rang the bell.
4. The children scream.
5. The sun sets in the west.
6. They go out for a picnic.
7. She likes the game.
8. I eat my food.
Answer:
1. Sudha is lying on the bed.
2. Raja is playing with his brother.
3. The servant was ringing the bell.
4. The children are screaming.
5. The sun is setting in the west.
6. They are going out for a picnic.
7. She is liking the game.
8. I am eating my food.

VIII. Fill in the blanks with the past perfect tense of the verb given in brackets:

1. He …………. a tiger before I reached the forest. (kill)
2. She ………….. a sweater before I bought a new one. (knit)
3. I ……………. money from my friend before I received my salary. (borrow)
4. The river ……………. its bank before the dam was built. (overflow)
5. She ……………. my book before I could check her. (steal)
6. The train …………….. before I could reach the station. (arrive)
7. I ………….. a funny story before the sad one. (hear)
8. The thief ………….. from the jail before the police arrived. (escape)
Answer:
1. He had killed a tiger before I reached the forest. (kill)
2. She had knitted a sweater before I bought a new one. (knit)
3. I had borrowed money from my friend before I received my salary. (borrow)
4. The river had overflown its bank before the dam was built. (overflow)
5. She had stolen my book before I could check her. (steal)
6. The train had arrived before I could reach the station. (arrive)
7. I had heard a funny story before the sad one. (hear)
8. The thief had escaped from the jail before the police arrived. (escape)

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3

Question 1.
Find the area of the shaded region in Fig., If PQ = 24 cm, PR =7 cm and O is the centre of the circle.

Solution:
PQ = 24 cm
PR = 7 cm
RQ is diameter of circle
∠RPQ = 90° Angle in semi circle
In ∆PQR,
QR² = RP² + PQ²
QR = \(\sqrt{(7)^{2}+(24)^{2}}=\sqrt{49+576}\)
= \(\sqrt{625}\)
QR = 25 cm
∴ Diameter of circle (QR) = 25 cm
Radius of circle (R) = \(\frac{25}{2}\) cm
Area of shaded region = Area of the semicircle – Area of ∆RPQ
= \(\frac{1}{2} \pi \mathrm{R}^{2}-\frac{1}{2} \mathrm{RP} \times \mathrm{PQ}\)

= \(\left[\frac{1}{2} \times \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2}-\frac{1}{2} \times 7 \times 24\right]\) cm²

= \(\left[\frac{6875}{28}-84\right]\)
= 245.53 – 84 = 161.53 cm²
∴ Area of shaded region = 161.53 cm².

Question 2.
Find the area of the shaded region in Fig., if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Solution:
Radius of smaller circle (r) = 7 cm
Radius of bigger circle (R) = 14 cm
Central angle ∠AOC (θ) = 40°
Area of shaded region = Area of bigger sector OAC – Area of smaller sector OBD
= \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}-\frac{\pi r^{2} \theta}{360^{\circ}}\)

= \(\frac{\pi \theta}{360^{\circ}}\) [R² – r²]

= \(\frac{22}{7} \times \frac{40}{360}\) × [14² – 7²]

= \(\frac{22}{63}\) [196 – 49]

= \(\frac{22}{63}\) × 147 = 51.33 cm²
∴ Shaded Region = 51.33 cm².

Question 3.
Find the area of the shaded region in fig., if ABCD is a square of side 14 cm and APD and BPC are semi circles.

Solution:
Side of square = 14 cm
Diameter of semicircle (AB = BC) = 14 cm
Radius of semi circle (R) = 7 cm
Area of square = (Side)²
= 14 × 14 = 196cm²
Area of a semi circles = \(\frac{1}{2}\) πR²
= \(\frac{1}{2} \times \frac{22}{7}\) × 7 × 7
= 77 cm²
Area of two semi circle = 2(77) = 154 cm²
Area of shaded region = Area of square ABCD – Area of two semi circles
= (196 – 154) = 42 cm²
Area of shaded region = 42 cm².

Question 4.
Find the area of the shaded region in fig., where a circular arc of radius 6 cm has been drawn iith vertex O of an equilateral triangle OAB of side 12 cm as centre.

[Each angle of equilateral triangle is 60°]
Area of major sector of circle = Area of circle – Area of sector
= πR² – \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7}\) × 6 × 6 – \(\frac{22}{7}\) × 6 × 6 × \(\frac{60}{360}\)

= \(\frac{22}{7}\) × 6 × 6 1 – \(\frac{60}{360}\)

= \(\frac{22}{7}\) × 36 1 – \(\frac{1}{6}\)

= \(\frac{22}{7}\) × 36 × \(\frac{5}{6}\)
= 94.28 cm²
∴ Area of major sector of circle = 94.28 cm²
Area of equilateral triangle OAB = \(\frac{\sqrt{3}}{4}\) (side)²
= \(\frac{1.73}{4}\) × 12 × 12
= 1.73 × 36 = 62.28 cm²
Shaded Area = Area of equilateral triangle OAB + Area of major sector of circle
= 62.28 + 94.28 = 62.28 cm²
Shaded Area = Area of equilateral triangle OAR + Area of major sectoç of circle
= 62.28 + 94.28 = 156.56
Shaded Area = 156.56 cm².

Question 5.
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in fig. Find the area of the remaining portion of the square.
Solution:
Side of square = 4 cm
Radius of each semi circle cut out (r) = 1 cm

Diameter of circle (R) = 2 cm
.. Radius of circle (R) = 1 cm
Area of square = (Side)²
= (4)² = 16 cm²
Area of 4 quadrants = 4\(\left[\frac{\pi^{2} \theta}{360^{\circ}}\right]\)

= \(\frac{4 \times 90}{360} \times \frac{22}{7} 1 \times 1\)

= 1 × \(\frac{22}{7}\) × 1 × 1 = 3.14 cm²
Area of circle = πR²
= \(\frac{22}{7}\) × 1 × 1
Area of circle = 3.14 cm²
Required area = Area of square – Area of 4 quadrants – Area of circle
= (16 – 3.14 – 3.14) cm² = 9.72 cm²
Required Area = 9.72 cm².

Question 6.
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as
shown in fig. Find the area of the design (shaded region).

Solution:
Radius of table cover (R) = 32 cm
OA = OB = OC = 32 cm

∆ABC is equilateral triangle with AB = AC = BC
∠AOB = ∠BOC = ∠COA = 120°
Now, in ∆BOC,
From O draw, angle bisector of ∠BOC as well as perpendicular bisector 0M of BC.
∴ BM = MC = \(\frac{1}{2}\) BC
Also, OB = OC [radii of the circle]
∴ ∠B = ∠C
∴ ∠O + ∠B + ∠C = 180°
120° + 2∠B = 180°
∠B = 30°
and ∠B = ∠C = 30°
Also, ∠BOM = ∠COM = 60°
∆OMB ≅ ∆OMC [RHS Cong.]
∴ In ∠OMB,
∠OBM = 30° [∠O = 60° and ∠M = 90°]
∴ \(\frac{\mathrm{BM}}{\mathrm{OB}}\) = cos 30°

\(\frac{\mathrm{BM}}{32}=\frac{\sqrt{3}}{2}\)

BM = 16√3 cm.

∴ BC = 2 MB = 32√3 cm
Area of circle = πR² = \(\frac{22}{7}\) × (32)²
= \(\frac{22}{7}\) × 32 × 32 = 3218.28 cm²

Area of ∆ABC = 4 (side)²
= \(\frac{1.73}{4}\) × 32√3 × 32√3 = 1328.64 cm²

∴ Required Area = Area of circle – Area of ∆ABC
= 3218.28 – 1328.64 = 1889.64 cm²

Question 7.
In fig., ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Solution:
Side of square ABCD = 14 cm
Radius of circle (R) = 7 cm
Sector angle (θ) = 90° [Each angle of square 90°]
Area of square = (side)²
= 14 × 14 = 196 cm²
Area of four quadrants = 4 \(\left[\frac{\pi R^{2} \theta}{360}\right]\)
= 4 × \(\frac{22}{7} \times \frac{7 \times 7 \times 90}{360}\)
= 22 × 7 = 154 cm²
∴ Required shaded area = Area of square – Area of 4 quadrants
= 196 – 154 = 42 cm².

 

Question 8.
Fig. depicts a racing track whose left and right ends are semicircular. The distahce between the two inner parallel line segments is 60 m and they are each 106 m long. 1f the track is 10 m wide, find
(i) the distance around the track along its inner edge.
(ii) the area of the track.

Solution:

(i) Here AB = DC = 106 m
AF = BE = CG = HD = 10m
Diameter of inner semicircle (APD and BRC) =60 m
∴ Radius of inner semicircle (APD (r) = 30 m
Radius of outer semicircle (R) = r + 10 = 30 + 10 = 40 m
Distance around the track along inner edge = AB + circumference of semi circle BRC + CD + circumference of semi circle DPA
= 2 AB + 2 [circumference of semi circle BRC]
= 2 (106) + 2(\(\left(\frac{2 \pi r}{2}\right)\))
= 212 + 2πr
= 212 + 2 × \(\frac{22}{7}\) × 30
= 212 + \(\frac{60 \times 22}{7}\)
= 212 + 188.57 = 400.57 m.
∴ Distance around the track along its inner edge = 400.57 m

(ii) Area of track = Area of rectangle ABEF + Area of region BEMGCRB + Area of rectangle CGHD + area of region.
= 2 Area of rectangle ABCD + 2 Area of region (II)
= 2 (AB × AF) + 2
[Area of semi circle with Radius 60 cm – Area of semi circle with radius 30 cm]
= 2 [106 × 10] + 2 [latex]\frac{\pi \mathrm{R}^{2}}{2}-\frac{\pi r^{2}}{2}[/latex]
= 2 × 1060 + \(\frac{2 \pi}{2}\) [R² – r²]
= 2120 + \(\frac{22}{7}\) (40² – 30²)
= 2120 + \(\frac{22}{7}\) [1600 – 900]
= 2120 + \(\frac{22}{7}\) [700]
= 2120 + 2200 = 4320 m²
Area of track = 4320 m²

Question 9.
In Fig., AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution:
Diameter of circle = 14 cm
Radius of circle = 7 cm
Diameter of smaller circle = 7 cm
∴ Radius of smaller circle = \(\frac{7}{2}\) cm
Since AB and CD are to perpendicular the diameters of a circle,
∴ AO ⊥ CD
Area of bigger circle = πR² × 7 × 7 = 154 cm²
Area of bigger semicircle = \(\frac{154}{2}\) = 77 cm²
Area of smaller circle = πr²
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)
= 38.50 cm²

Area of ∆ABC = \(\frac{1}{2}\) Base × Altitude
= \(\frac{1}{2}\) × 14 × 7 = 49 cm²
∴ Shaded Area = Area of bigger semi circle + Area of smaller circle – Area of triangle
= (77 – 49 + 38.5) cm² = 66.5 cm²

Question 10.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half of the length of the side of the triangle (see Fig.). Find the area of the shaded region.
(Use n = 3.14 and ,√3 = 1.73205)

Solution:
Area of equilateral triangle ABC = 17320.5 cm²

\(\frac{\sqrt{3}}{4}\) (side)² = 17320.5

(side)² = \(\frac{17320.5 \times 4}{1.73205}\)

(side)² = \(\frac{173205}{10} \times \frac{100000 \times 4}{173205}\)

side = \(\sqrt{4 \times 100 \times 100}\)
side = 2 × 100 = 200 cm
AB = BC = AC
Radius of circle (R) = \(\frac{A B}{2}=\frac{200}{2}\) = 100 cm
Sector angle, θ = 60°
Area of sector APN = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{3.14 \times 100 \times 100 \times 60}{360}\)

= \(\frac{15700}{3}\)

Area of three sector = 3 × \(\frac{15700}{3}\) cm²
∴ Required shaded Area = Area of triangle – Area of three sectors
= 17320.5 – 15700 = 1620.5 cm²
∴ Hence, Required shaded Area = 1620.5 cm²

Question 11.
On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig). Find the area of the remaining portion of the handkerchief.

Solution:
Radius of circle (R) = 7 cm
Diameter of circle = 2 × R = 2 × 7 = 14 cm
Since there are three circles along a side of square
∴ side of squrae = 3 [14] = 42 cm
Total area of handkerchief = Area of square = (side)²
= (42)² = 1764 cm².
Area of 9 circular designs = 9πR²
= 9 × \(\frac{22}{7}\) × (7)2
= 9 × \(\frac{22}{7}\) × 7 × 7
= 9 × 154 = 1386 cm²
∴ Required area of remaining portion = Area of square – Area of 9 circular designs
= 1764 – 1386 = 378 cm²
∴ Required area of remaining portion = 378 cm².

Question 12.
In Fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB,
(ii) shaded region.

Solution:

Radius of quadrant (R) = 3.5 cm
Angle of sector (θ) = 90°
OD = 2 cm.

(i) Area of quadrant OACB = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{22}{7} \times \frac{3.5 \times 3.5 \times 90}{360}\) = 9.625 cm².

(ii) Area of ODB = \(\frac{1}{2}\) Base × Altitude
= \(\frac{1}{2}\) × 3.5 × 2 = 3.5 cm²

∴ Shaded Area = Area of quadrant OACB – Area of ∆ODB
= 9.625 – 3.5 = 6.125 cm²
∴ Hence, Shaded Area = 6.125 cm².

Question 13.
In fig., a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.

Solution:
Side of square ABCO = 20 cm
∠AOC = 90°
AB = OA

OB² = OA² + AB²

OB = \(\sqrt{(20)^{2}+(20)^{2}}\)

= \(\sqrt{400+400}\)

= \(\sqrt{800}=\sqrt{400 \times 2}\)
OB = 20√2 cm
Area of square OABC = (side)² = (20)²
∴ Area of square = 400 cm²
Radius of quadrant (R) = 20√2 cm
Sector angle (θ) = 90°
∴ Area of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)
= \(\frac{3.14 \times 20 \sqrt{2} \times 20 \sqrt{2} \times 90}{360}\)
= 2 × 314 cm² = 628 cm²
Required shaded Area = Area of sector – Area of square
= (628 – 400) cm² = 228 cm²

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If ZAOB = 30°, find the area of the shaded region.

Solution:
Radius of sector OBA (R) =21 cm
Radius of sector ODC (r) 7 cm
Sector angle (θ) = 30°
Area of bigger sector (OAB) = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{21 \times 21 \times 30}{360}\) = 115.5 cm²

Area of smaller sector (ODC) = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{7 \times 7 \times 30}{360}\) = 12.83 cm²

Area of smaller sector (ODC) = 12.83 cm²
Now, Shaded Area = Area of bigger sector OAB – Area of smaller sector OCD
= 115.5 – 12.83 = 102.66
Hence, Shaded Area = 102.66 cm².

Question 15.
In fig., ABC is a quadrant of a circle of radius 14 cm and a semi circle is drawn with BC as diameter. Find the area of the shaded region.

Solution:
Radius of quadrant ACPB (r) = 14 cm
Sector angle (θ) = 90°
AB = AC = 14 cm

Area of triangle = \(\frac{1}{2}\) AB × AC
= \(\frac{1}{2}\) × 14 × 14
= 98 cm²

Area of sector ACPB = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{14 \times 14 \times 30}{360}\) = 154 cm²

∴ Area of BOCPB = Area of sector ABPC – Area of \ABC
= 154 cm² – 98 cm² = 56 cm²
In ∆BAC, AB² + AC² = BC²
(14)² + (14)² = BC²
BC = \(\sqrt{196+196}=\sqrt{2(196)}\) = 14√2

∴ Radius of semi circle BOCR = \(\frac{14 \sqrt{2}}{2}\) = 7√2

Area of semi circle = \(\frac{\pi \mathrm{R}^{2}}{2}\)

= \(\frac{22}{7} \times \frac{7 \sqrt{2} \times 7 \sqrt{2}}{2}\)

= \(\frac{22}{7} \times \frac{7 \times 7 \times 2}{2}\)
= 154 cm²

Required Area = Area of semi circle – [Area of sector – Area of ∆BAC]
= 154 – [154 – 98]
= (154 – 56) cm² = 98 cm²
Hence, Shaded Area = 98 cm².

Question 16.
Calculate the area of the designed region in fig. common between the two quadrants of circles of radius 8 cm each.

Solution:
Side of square = 8 cm
Area of square = (8)² = 64 cm²
Line BD divides square ABCD into the equal parts
Area of ∆ABD = ar of ∆BDC
Sector angle θ = 90°
Area of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{22}{7} \times \frac{8 \times 8 \times 90}{360}\)

Area of sector = 50.28 cm²
Area of ∆ABD = \(\frac{1}{2}\) × AB × AD
= \(\frac{1}{2}\) × 8 × 8
= 32 cm²

∴ Area of segment DMBPD = Area of sector ∆BPD – Area of ∆ABD
= 50.28 – 32 = 18.28 cm²
Hence, Shaded area = 2 area of segment DMBPD = 2 (18.28) = 36.56 cm²

Punjab State Board PSEB 7th Class English Book Solutions English Vocabulary Arrange in Alphabetical Order Exercise Questions and Answers, Notes.

PSEB 7th Class English Vocabulary Arrange in Alphabetical Order

Question 1.
Monkey, picture, bury, dengue, donkey, village, heart, tomb, women, develop.
Answer:
Bury, dengue, develop, donkey, heart, monkey, picture, tomb, village, women.

Question 2.
Able, Wednesday, frozen, return, extremely, disappointed, shivering, wisest, realise, looking.
Answer:
Able, disappointed, extremely, frozen, looking, realise, return, shivering, Wednesday, wisest.

Question 3.
Quick, fine, nice, ancient, bold, clever, deep, mountain, heavy, happy.
Answer:
Ancient, bold, clever, deep, fine, happy, heavy, mountain, nice, quick.

Question 4.
Astronaut, truth, resident, clown, scratch, journey, assigned, disbelief, humble, mission.
Answer:
Assigned, astronaut, clown, disbelief, humble, journey, mission, resident, scratch, truth.

Question 5.
Sheep, bull, horse, bowl, goal, crow, parrot, frog, donkey, duck.
Answer:
Bowl, bull, crow, donkey, duck, frog, goal, horse, parrot, sheep.

Question 6.
Belt, girl, chalk, wife, picture, sister, error, host, duke, actor.
Answer:
Actor, belt, chalk, duke, error, girl, host, picture, sister, wife.

Question 7.
Active, helpful, careful, brother, popular, faithful, famous, difficult, intelligent, polite.
Answer:
Active, brother, careful, difficult, faithful, famous, helpful, intelligent, polite, popular.

Question 8.
Parent, child, sky, plate, madam, doctor, witch, televisiori, bulb, niece.
Answer:
Bulb, child, doctor, madam, niece, parent, plate, sky, television, witch.

Question 9.
Wash, dance, cook, pray, dust, jump, table, father, laugh, help.
Answer:
Cook, dance, dust, father, help, jump, laugh, pray, table, wash.

Punjab State Board PSEB 7th Class English Book Solutions English Grammar Parts of Speech Exercise Questions and Answers, Notes.

PSEB 7th Class English Grammar Parts of Speech

शब्दों को प्रयोग के अनुसार आठ भागों अथवा श्रेणियों में विभाजित किया जाता है, जिन्हें Parts of Speech (शब्द भेद) कहते हैं। ये आठ भाग हैं-

1. Noun
2. Pronoun
3. Adjective
4. Verb
5. Adverb
6. Preposition
7. Conjunction
8. Interjection.

1. Noun-किसी व्यक्ति, स्थान, वस्तु अथवा गुण-दोष के नाम को Noun (संज्ञा) कहते हैं; जैसे Ram, Shyam, Mumbai, Pen, Pencil, Rose, Dog, Beauty, Honesty आदि।

2. Pronoun-जो शब्द किसी संज्ञा के स्थान पर प्रयोग होता है, उसे Pronoun (सर्वनाम) कहते हैं; जैसे I, we, you, they, he, she, it आदि।

3. Adjective-जो शब्द किसी Noun अथवा Pronoun की विशेषता प्रकट करता है, उसे Adjective (विशेषण) कहते हैं; जैसे good, bad, thin, fat, tall, many, useful आदि।

4. Verb-वह शब्द जिससे कार्य का होना अथवा करना प्रकट हो या जो शब्द किसी व्यक्ति, स्थान अथवा वस्तु के बारे में कुछ बतलाए, वह,शब्द Verb (क्रिया) कहलाता है; जैसे-

• The girl reads a book.
• Ram is a good boy.
• Mohan goes to school daily.
• Seema feels sad.
• I helped the poor.

5. Adverb जो शब्द किसी Verb, Adjective अथवा किसी अन्य Adverb की विशेषता प्रकट करे, तो उसे Adverb (क्रिया विशेषण) कहते हैं; जैसे-

• He ran fast. (Verb की विशेषता)
• Rose is very beautiful. (Adjective की विशेषता)
• He works too slowly. (Adverb की विशेषता)

6. Preposition-जो शब्द किसी Noun अथवा Pronoun से पहले प्रयोग किया जाता है और जो उनका सम्बन्ध वाक्य के अन्य शब्दों से जोड़ता है, उसे Preposition (सम्बन्धवाचक शब्द) कहते हैं; जैसे- .

• Mohan is at the gate.
• Ram is proud of his teacher.
• He is fond of music.
• I am writing with a pen.

7. Conjunction–जो शब्द, शब्दों अथवा वाक्यों को जोड़ता है, उसे Conjunction (संयोजक) कहते हैं; जैसे-

• Ram and Shyam are friends.
• You will get through if you work hard.
• He was late because he missed the bus.
• Either he or his brother is at fault.

8. Interjection-अकस्मात् भावना को व्यक्त करने वाला शब्द Interjection (विस्मयादिबोधक अव्यय) कहलाता है जैसे, Alas ! Hurrah ! Oh ! Lo! आदि।

Exercises

I. Write in the space provided the name of the part of speech to which the underlined words belong in the following sentences:

1. Seema is a beautiful girl.
2. Alas! His dog is dead.
3. The sun sets in the west.
4. The lion is a ferocious animal.
5. Delhi is a very big city.
6. Rahim is poor but honest.
7. Honesty is the best policy.
8. The.cat is under the table.
Hints:
1. Adjective
2. Interjection
3. Verb
4. Adjective
5. Adverb
6. Conjunction
7. Noun
8. Preposition.

II. Complete the following sentences with appropriate ‘Nouns’:

1. ……….. is a good boy.
2. She goes to the ……… everyday.
3. They go for a ……….. daily.
4. ……….. is the Capital of India.
5. ……… is the best policy.
6. The ………. rises in the …….
7. Chandigarh is the ……….. of Punjab and Haryana.
8. Shimla is a beautiful …………
Hints:
1. Mohan
2. temple/Gurudwara
3. walk
4. Delhi
5. Honesty
6. sun, east
7. capital
8. place.

III. Fill in the blanks with suitable “Pronouns’:

1. My son is playing with ……… toys.
2. ……….. father is working in Mumbai.
3. ……….. has gone abroad for higher studies.
4. This is ……… house.
5. The girls are doing ……….. homework.
6. ……….. is the bread-winner of the family.
7. We should respect ……. parents.
8. He loves ……… native place very much.
Hints:
1. his
2. My
3. He
4. our
5. their
6. She
7. our
8. his.

IV. Fill in the blanks with suitable ‘Adjectives’:

1. She is my ……….. friend.
2. Pudding is my ……….. dish.
3. The scenery of Mussoorie is
4. She likes to wear ……… dresses.
5. The Bible is a ……….. book.
6. Cricket is a ………. game.
7. John is an ……….. teacher.
8. Mango is a ………. fruit.
Hints:
1. best
2. favourite
3. beautiful
4. white
5. holy
6. good
7. English
8 juicy.

V. Fill in the blanks with suitable ‘Verbs’:

1. She ……….. ice-cream.
2. They ……….. a lot.
3. My mother ……….. food.
4. John …….. in a factory.
5. They ………. football.
6. We should ……….. a bath everyday.
7. The school peon ……….. the bell.
8. Seema ……….. for a walk daily.
Hints:
1. is eating
2. worked
3. cooks
4. works
5. play
6. take
7. rings
8. goes.

VI. Fill up the blanks with suitable ‘Adverbs’:

1. Sohan walks …………. .
2. Everyone should work ………..
3. She sings ………….
4. His dad is a …………. respectable man.
5. Our teacher speaks ……….. politely.
6. My sister sleeps …………
7. John is a ………. hardworking boy.
8. Girls sang …………. .
Hints:
1. slowly
2. hard
3. sweetly:
4. very
5. very
6. early
7. very
8. nicely.

VII. Complete the following sentences with suitable ‘Prepositions:

1. My grandfather is hard ……….. hearing.
2. My mom is fond ……… music.
3. A burglar broke ……….. our house last night.
4. The students should listen ………. their teachers attentively.
5. Yesterday we went ………. the Rose Garden.
6. Ayushi is playing ……….. the piano.
7. Children have been playing ……….. morning.
8. My uncle has been living ……….. Canada ……….. fifteen years.
9. I will play ……….. finishing my homework.
10. I am standing ……….. Anshu. Anshu is in front of me.
Hints:
1. of
2. of
3. into
4. to
5. to
6. on
7. since
8. in, for
9. after
10. behind.

VIII. Fill in the blanks with suitable ‘Conjunctions’:

1. My younger brother is both intelligent ……….. hardworking.
2. He says ………. he is a doctor.
3. ……….. Seema ………. Rita is at fault.
4. Our servant is poor ……….. honest.
5. He ………. his nephew manage the shop.
6. Ram went on leave ……….. he was injured.
7. You will get the tickets ………..you reach there before 6 o’clock.
8. ……….. he was late, yet he was able to catch the bus.
Hints:
1. and
2. that
3. Either, or
4. but
5. and
6. because
7. if
8. Although

IX. Fill in the blanks with suitable ‘Interjections’:

1. ……….. We have won the game.
2. ……. The man is dead.
3. ……… They have come.

V. Fill in the blanks with suitable ‘Verbs’:

1. She ……….. ice-cream.
2. They ……….. a lot.
3. My mother ……….. food.
4. John …….. in a factory.
5. They ………. football.
6. We should ……….. a bath everyday.
7. The school peon ……….. the bell.
8. Seema ……….. for a walk daily.
Hints:
1. is eating
2. worked
3. cooks
4. works
5. play
6. take
7. rings
8. goes.

VI. Fill up the blanks with suitable ‘Adverbs’:

1. Sohan walks …………. .
2. Everyone should work ………….
3. She sings …………..
4. His dad is a ………….. respectable man.
5. Our teacher speaks ……… politely.
6. My sister sleeps …………..
7. John is a …………. hardworking boy.
8. Girls sang ……….. .
Hints:
1. slowly
2. hard
3. sweetly
4. very
5. very
6. early
7. very
8. nicely.

VII. Complete the following sentences with suitable ‘Prepositions:

1. My grandfather is hard ……….. hearing.
2. My mom is fond ……… music.
3. A burglar broke ……….. our house last night.
4. The students should listen ………. their teachers attentively.
5. Yesterday we went ………. the Rose Garden.
6. Ayushi is playing ……….. the piano.
7. Children have been playing ……….. morning.
8. My uncle has been living ……….. Canada ……….. fifteen years.
9. I will play ……….. finishing my homework.
10. I am standing ……….. Anshu. Anshu is in front of me.
Hints:
1. of
2. of
3. into
4. to
5. to
6. on
7. since
8. in, for
9. after
10. behind.

VIII. Fill in the blanks with suitable ‘Conjunctions’:

1. My younger brother is both intelligent ……….. hardworking.
2. He says ………. he is a doctor.
3. ……….. Seema ………. Rita is at fault.
4. Our servant is poor ……….. honest.
5. He ………. his nephew manage the shop.
6. Ram went on leave ……….. he was injured.
7. You will get the tickets ………..you reach there before 6 o’clock.
8. ……….. he was late, yet he was able to catch the bus.
Hints:
1. and
2. that
3. Either, or
4. but
5. and
6. because
7. if
8. Although

IX. Fill in the blanks with suitable ‘Interjections’:

1. ……….. We have won the game.
2. ………… The man is dead.
3. …………. They have come.
4. …………. Is this the place?
5. …………. Well done.
6. ………….. He has lost all his money in gambling.
7. …………. You are hurt.
8. …………… He has failed again.
Hints:
1. Hurrah !
2. Alas!
3. Lo !
4. Oh !
5. Bravo !
6. Alas !
7. Oh !
8. Alas !

Exercise

Choose the correct option to determine which part of speech the underlined word belongs to:

Question 1.
Honesty is the best policy.
(a) noun
(b) verb
(c) adjective
(d) preposition
Answer:
(a) noun

Question 2.
The sun rises in the east.
(a) adjective
(b) verb
(c) noun
(d) preposition
Answer:
(c) noun

Question 3.
She is eating ice-cream.
(a) noun
(b) verb
(c) adjective
(d) preposition
Answer:
(b) verb

Question 4.
The school peon rings the bell.
(a) noun
(b) adjective
(c) verb
(d) preposition
Answer:
(c) verb

Question 5.
She likes to wear white dresses.
(a) noun
(b) verb
(c) adjective
(d) preposition
Answer:
(c) adjective

Question 6.
Mango is a juicy fruit.
(a) noun
(b) adjective
(c) verb
(d) preposition
Answer:
(b) adjective

Question 7.
My grandfather is hard of hearing.
(a) noun
(b) verb
(c) adjective
(d) preposition
Answer:
(d) preposition

Question 8.
Ayushi is playing on the piano.
(a) preposition
(b) verb
(c) adjective
(d) noun
Answer:
(a) preposition

Question 9.
My son is playing with his toys.
(a) pronoun
(b) adverb
(c) conjunction
(d) interjection
Answer:
(a) pronoun

Question 10.
We should respect our parents.
(a) adverb
(b) pronoun
(c) interjection
(d) conjunction
Answer:
(b) pronoun

Question 11.
Sohan walks slowly.
(a) pronoun
(b) adverb
(c) conjunction
(d) interjection
Answer:
(b) adverb

Question 12.
Our teacher speaks very politely.
(a) pronoun
(b) interjection
(c) conjunction
(d) adverb
Answer:
(d) adverb

Question 13.
He says that he is a doctor.
(a) pronoun
(b) adverb
(c) conjunction
(d) interjection
Answer:
(c) conjunction

Question 14.
Our servant is poor but honest.
(a) conjunction
(b) adverb
(c) pronoun
(d) interjection
Answer:
(a) conjunction

Question 15.
Hurrah ! We have won the match.
(a) pronoun
(b) adverb
(c) conjunction
(d) interjection
Answer:
(d) interjection

Question 16.
Alas ! He has failed again.
(a) pronoun
(b) interjection
(c) conjunction
(d) adverb
Answer:
(b) interjection

Name the part of speech of the bold words:

A. (i). He lives in a cottage.
(ii) Delhi is very big city.

B. (i) They go for a walk daily.
(ii) She is my best friend.

C. (i) Beauty does not last long.
(ii) The baby plays.

Fill in the blanks with an appropriate noun:

A. Shimla is a beautiful …………….
B. …………….. is the capital of India.
C. …………….. is a good boy.

Name the part of speech of the underlined words:

A. (i) Honesty is the best policy.
(ii) Pudding is my favourite dish.

B. (i) Dey is a good boy.
(ii) The girls are doing their homework.

C. (i) They play with the ball.
(ii) She sings sweetly.

Name the part of speech of the bold words:

A. (i) This is my house.
(ii) Our teacher speaks very politely.

B. (i) He lives in a cottage.
(ii) Josh is very handsome.

C. (i) The sun sets in the west.
(ii) Seema is a beautiful girl.

Choose the correct option to determine which part of speech the underlined word belongs to:

(i) Rohan is a handsome boy
(a) noun
(b) verb
(c) adjective
(d) preposition
Answer:
(c) adjective

(ü) Delhi is a big city.
(a) noun
(b) verb
(c) adjective
(d) preposition
Answer:
(a) noun

Punjab State Board PSEB 7th Class English Book Solutions English Grammar Translation Exercise Questions and Answers, Notes.

PSEB 7th Class English Grammar Translation

A sentence is a well arranged group of words which makes complete sense.
शब्दों का वह सुव्यवस्थित समूह जिसका पूरा अर्थ निकले, Sentence (वाक्य) कहलाता है। जैसे:

1. Mohan is a good player.
2. Where is.your house?
3. Help me, please.
4. Hurrah ! We have won the match.

Kinds of Sentences
वाक्य चार प्रकार के होते हैं-
1. Assertive Sentences. Assertive Sentences वे वाक्य होते हैं जिनमें किसी बात (Positive या Negative) का वर्णन होता है। इन वाक्यों को Statements भी कहते हैं। इनके अन्त में Full stop (.) लगता है।

2. Interrogative Sentences. ऐसे वाक्यों से किसी प्रश्न का बोध होता है। इनके अन्त में प्रश्न चिन्ह (?) लगाया जाता है।

3. Imperative Sentences. ऐसे वाक्यों से आज्ञा, आदेश, उपदेश अथवा विनती का बोध होता है। इनके अन्त में Full stop (.) लगता है।

4. Exclamatory Sentences. ये ऐसे वाक्य होते हैं जो मन के अकस्मात् भावों को अथवा किसी इच्छा को प्रकट करते हैं। इनके अन्त में Sign of exclamation (!) लगता है। जैसे-

• May you live long !
• Hurrah ! We have passed.
• How beautiful the flower is !

Exercise 1

I. Rewrite the following jumbled groups of words as sentences. (Remember to begin with a capital letter and end with a full stop, a question mark or an exclamation mark)

1. faithful is animal the a dog.
2. so am I tired
3. grandparents fond Ali is his of
4. are when leaving for you Patiala
5. best Ravi Harpreet friends are and
6. cooks my food tasty mother
7. got rain we in wet the
8. has lakes Nainital lovely
9. graceful is Radha a dancer
10. circus the enjoyed children the
Answer:
1. The dog is a faithful animal.
2. I am so tired.
3. Ali is fond of his grandparents.
4. When are you leaving for Patiala ?
5. Ravi and Harpreet are best friends.
6. My mother cooks tasty food.
7. We got wet in the rain.
8. Nainital has lovely lakes.
9. Radha is a graceful dancer.
10. The children enjoyed the circus.

II. Make sentences of your own with the following words. The first one has been done for you:

1. fox – The fox is a cunning animal.
2. clever – He is very …………
3. beauty – There is ……….. all around us.
4. run – Don’t …………. to the water.
5. sweetly – She sings ………… .
6. behind – There is a tree …………. the hut.
7. they – ………… are good boys.
8. and – Ram ……….. Sham are good friends.
9. bouquet – This ………….. is very beautiful.
10. the – ………….. Taj is in Agra.
11. children – Nehru loved ………. .
Answer:
1. fox – The fox is a cunning animal.
2. clever – He is very clever.
3. beauty – There is beauty all around us.
4. run – Don’t run to the water.
5. sweetly – She sings sweetly.
6. behind – There is a tree behind the hut.
7. they – They are good boys.
8. and – Ram and Sham are good friends.
9. bouquet – This bouquet is very beautiful.
10. the – The Taj is in Agra.
11. children – Nehru loved children.

Exercise 2

I. Read the following sentences and say whether they are Assertive (A), Interrogative (In), Imperative (Im) or Exclamatory (E) sentences:

1. What a sweet fhango this is !
2. Where are you going for a holiday ?
3. I need your help to do this sum.
4. Please close the door.
5. Mr. Gupta is going for a walk.
6. Do you play tennis everyday ?
7. How terribly hot it is !
8. Be attentive when I teach.
9. The rainbow has seven colours.
10. Have you had your bath ?
Answer:
1. What a sweet fhango this is ! – (E)
2. Where are you going for a holiday ? – (In)
3. I need your help to do this sum. – (A)
4. Please close the door. – (Im)
5. Mr. Gupta is going for a walk. – (A)
6. Do you play tennis everyday ? – (In)
7. How terribly hot it is ! – (E)
8. Be attentive when I teach. – (Im)
9. The rainbow has seven colours. – (A)
10. Have you had your bath ? – (In)

II. Rewrite the following sentences inserting suitable punctuation marks. The first one has been done for you.

1. doesn’t the nightingale sing sweetly	Doesn’t the nightingale sing sweetly ?
2. nepal is a beautiful country	Nepal is a beautiful country.
3. how green the grass is	How green the grass is !
4. leave the room immediately	Leave the room immediately.
5. will you get me a glass of water	Will you get me a glass of water ?
6. sukhwinder is a great singer	Sukhwinder is a great singer.
7. what a naughty boy he is	What a naughty boy he is !
8. what is the answer to this question	What is the answer to this question ?
9. have your breakfast immediately	Have your breakfast immediately.
10. gandhiji is called the father of the nation	Gandhiji is called the Father of the Nation.
11. why do you want to go out now	Why do you want to go out now ?
12. oh god what a beautiful sight	Oh God ! what a beautiful sight!
1. doesn’t the nightingale sing sweetly	Doesn’t the nightingale sing sweetly ?

III. Turn the following statements into negative and questions. The first two have been done for you.

Statements	Negative	Interrogative
1. She is a clever girl.	She is not a clever eirl.	Is she a clever eirl ?
2. Ranjit likes mangoes.	Raniit does not like mangoes.	Does Raniit like maneoes ?
3. Radha is happy.	Radha is not happy.	Is Radha happy ?
4. They have played badly.	They have not played badly.	Have they played badly ?
5. Rina bought a car.	Rina did not buy a car.	Did Rina buy a car ?
6. We are going for a picnic ?	We are not going for a picnic.	Are we going for a picnic ?
7. They went to Shimla.	They did not go to Shimla.	Did they go to Shimla ?
8. You must go home now.	You must not go home now.	Must you go home now ?
9. We returned home in the evening.	We did not return home in the evening.	Did we return home in the evening ?
10. They can speak Punjabi.	They cannot speak Punjabi.	Can they speak Punjabi ?

Punjab State Board PSEB 7th Class English Book Solutions English Vocabulary Punctuation Exercise Questions and Answers, Notes.

PSEB 7th Class English Vocabulary Punctuation

Punctuation का अर्थ है-लिखने में प्रयुक्त विराम चिन्ह। जब कभी हम लिखते हैं तो हमें अपने भावों को स्पष्ट करने के लिए कई स्थानों पर रुकना पड़ता है। इसके लिए कुछ विशेष नियम हैं और उन्हीं के अनुसार हम विराम चिन्हों का वाक्यों में प्रयोग करते हैं। ग़लत चिन्ह लगाने से वाक्यों का अर्थ भी बदल जाता है। यह बात नीचे लिखे वाक्यों से स्पष्ट हो जाएगी।

Examples:

1. “The Sun”, said the Moon, “moves round the Earth.”
2. The Sun said, “The Moon moves round the Earth.”

ऊपर दिए गए दोनों वाक्य एक जैसे शब्द समूहों से बने हैं, परन्तु विराम चिन्ह अलग-अलग ढंग से लगाए गए हैं जिससे कि इनके अर्थों में अन्तर आ गया है।

पहले वाक्य का यह अर्थ निकलता है-
चन्द्रमा ने कहा, “सूर्य पृथ्वी के चारों ओर घूमता है।”
इसके विपरीत दूसरे वाक्य का अर्थ निकलता है
सूर्य ने कहा, “चन्द्रमा पृथ्वी के चारों ओर घूमता है।”

Capital Letters and Important Punctutation Marks

1. Capital Letters. [ABC]
2. Full Stop [.]
3. Comma [,]
4. Mark of Interrogation [?]
5. Inverted Commas [” “]
6. Apostrophe [‘]
7. Mark of Exclamation [!]

1. CAPITAL LETTERS

अंग्रेजी भाषा ही एकमात्र ऐसी भाषा है, जिसमें दो प्रकार के Letters (अक्षर) होते हैं

1. Capital letters (बड़े अक्षर) जैसे- A, B, C…….
2. Small letters (छोटे अक्षर) जैसे-a, b, c, ……

इन बड़े तथा छोटे अक्षरों का प्रयोग करने के विशेष नियम हैं1. प्रत्येक वाक्य के पहले शब्द का पहला letter (अक्षर) capital (बड़ा) होता है; जैसे,
Once a fox was hungry.
The teachers are teaching.

2. Inverted Commas (” “) के भीतर लिखे जाने वाले प्रत्येक वाक्य के पहले शब्द का पहला letter capital होता है; जैसे,
He said to me, “Who is he?”

3. Proper Noun तथा Proper Adjective का पहला letter capital होता है; जैसे,-
I met Mohan. (Proper Noun) He is an Indian. (Proper Adjective) I visited Japan. (Proper Noun) I like English people. (Proper Adjective).

4. दिनों, महीनों, त्योहारों, पुस्तकों, जातियों आदि के नाम सदा capital letter से आरम्भ होते हैं; जैसे,
He came here on Monday. (Day)
I read the Ramayana. (Book)
He was happy on Diwali. (Festival)

5. शब्दों का संक्षिप्त रूप (Abbreviations) प्रस्तुत करते समय capital letters का प्रयोग होता है; जैसे,
M.P., M.L.A., M.A., B.A.

6. भाषाओं तथा विषयों के नाम का पहला अक्षर capital होता है; जैसे,
Hindi, English, History, Science.

7. God तथा God के लिए प्रयोग किए जाने वाले pronoun (सर्वनाम) He, His, Him सदा capital letter से आरम्भ होते हैं; जैसे,
I pray to God. I pray to Him daily. I love His ways.

II. Use of Full Stop (.)

1. Exclamatory तथा Interrogative sentences को छोड़कर सभी वाक्यों के अन्त में Full Stop का प्रयोग होता है।
Ram ate food.
Come here.
He is a great man.

2. Abbreviations के अन्त में Full Stop का प्रयोग होता है; जैसे,
He is an M.A.
I met Mr. R.R. Sharma.
Sh. Tulsi Ram is an M.L.A.

III. Use Of Comma (,)

Comma का प्रयोग निम्नलिखित ढंग से किया जाता है:

1. एक ही Part of speech के शब्दों को पृथक करने के लिए; जैसे,-
(i) I bought a chair, a table and a fan.
(ii) He is kind, gentle and brave.

2. Apposition (व्याख्या करने वाले शब्दों) से पहले तथा बाद में; जैसे,
Ram, our friend, is going to England.

3. Reported Speech को शेष वाक्य से अलग करने के लिए; जैसे,
Ram said to me, “You are a good player.”

4. No तथा Yes के पश्चात्; जैसे,
No, I cannot go there.
Yes, I shall see him today.

5. तिथि को वर्ष से अलग करने के लिए; जैसे,
March 23, 20…. .

6. Adverb Clause को Principal Clause से अलग करने के लिए; जैसे,
When he came here, I met him.

IV. Sign of Interrogation (?)

प्रश्नवाचक चिन्ह (Sign of Interrogation) का प्रयोग प्रश्नवाचक वाक्य के अन्त में होता है; जैसे,
How are you?
Do you love me?

V. Use of Inverted Commas (” “)

(a) किसी व्यक्ति की बात को उसी के द्वारा कहे हुए शब्दों में व्यक्त करने के लिए Inverted Commas (“”) का प्रयोग किया जाता है; जैसे,
He said to me, “Where are you going ?”
Ram said to him, “How are you now ?”

(b) यदि किसी व्यक्ति द्वारा कहे गए शब्द दो भागों में हों, तो Inverted Commas दो बार आरम्भ होते हैं और दो बार बन्द होते हैं; जैसे,-
“He”, said Mohan, “is a foolish boy.”

VI. Use of Apostrophe (‘)

(a) Possessive case बनाने के लिए Noun के साथ प्रयोग किया जाता है; जैसे,
Sita’s house.
Ram’s house.

(b) किसी लुप्त अक्षर के स्थान पर भी इसका प्रयोग किया जाता है; जैसे,
Don’t = Do not
Can’t = cannot
Didn’t = Did not
I’ve = I have
That’s = That is
Haven’t = Have not

VII. Sign of Exclamation (!)

1. विस्मयादि बोधक (Exclamatory) शब्दों या शब्द-समूहों के बाद; जैसे,-
What a beautiful flower !
How fine the building is !

2. सम्बोधन सूचक शब्दों के बाद; जैसे,-
Thief ! Thief ! Catch the thief.
Mother ! Mother ! I am alone.
Foolish ! Do not do it.

3. सुख-दुःख आदि प्रकट करने वाले शब्दों के साथ; जैसे,-
Alas ! I am undone.
Hurrah ! We have won the match.

4. शुभ-कामना प्रकट करने वाले वाक्यों के अन्त में; जैसे,-
May you live long !
May you succeed !

Exercises

I. Rewrite the sentences using commas and capitals, where necessary. The first one has been done for you:

1. will you get the radio for us rachna ?
Will you get the radio for us, Rachna ?

2. the gita the bible and the quran are all holy books.
The Gita, the Bible and the Quran are all holy books.

3. the stars the moon and the sun are heavenly bodies.
The stars, the moon and the sun are heavenly bodies.

4. akbar the great and birbal were great admirers of each other.
Akbar the great and Birbal were great admirers of each other.

5. on reaching the banks of ganga he offered his prayers.
On reaching the banks of Ganga, he offered his prayers.

6. saying thus he breathed his last.
Saying thus, he breathed his last.

II. Rewrite the following sentences using appropriate marks of punctuation. One has been done for you:

1. what shall I bring you when I come back home said the king to his sisters.
“What shall I bring you when I come back home ?” said the king to his sisters.

2. the traveller said can you tell me the way to the nearest inn yes said the man
The traveller said, “Can you tell me the way to the nearest inn ?” “Yes”, said the man.

3. dont be afraid little boy I wont harm you I want to be your friend said the giant.
“Don’t be afraid, little boy. I won’t harm you. I want to be your friend,” said the giant.

4. will your mother not be sorry to see you without jewellery gaurav asked
“Will your mother not be sorry to see you without jewellery ? Gaurav asked.

5. how have you found so much wealth asked ali babas wife
“How have you found so much wealth ?” asked Ali Baba’s wife.

6. the staff room needs sweeping take the broom and sweep it said the principal.
“The staff room needs sweeping. Take the broom and sweep it,” said the Principal.

7. O king replied buddha this is the custom we always observe.
“O king”, replied Buddha, “this is the custom we always observe.”

8. mrs prasad said how much did you pay for this picture.
Mrs. Prasad said, “How much did you pay for this picture ?”

9. gopal sita radha and neelu went to ludhiana.
Gopal, Sita, Radha and Neelu went to Ludhiana.

10. the farmer said I cant give you any money
The farmer said, “I can’t give you any money.”

III. Use an apostrophe where necessary. Some may not need. One has been done for you:

1. The clowns hat had a feather.	The clown’s hat had a feather.
2. Bulbul’s eyes are brown.	Bulbul’s eyes are brown.
3. The boys ties were flying about.	The boys’ ties were flying about.
4. My sisters essay got the first prize.	My sister’s essay got the first prize.
5. The childrens books are torn.	The children’s books are torn.
6. Lets go to the zoo this evening.	Let’s go to the zoo this evening.
7. Crows nests are made of twigs.	Crows’ nests are made of twigs.
8. On Sunday therell be a fair.	On Sunday there’ll be a fair.
9. The babies dresses were on sale.	The babies dresses were on sale.

IV. Punctuate the following with commas and inverted commas, where needed.

1. Rajan said Goodbye.
2. Shiela says in our house there are chairs tables fans and radios.
3. Whose books are these ? said Rani.
4. I looked for my pen everywhere said Ali.
5. Give me the books, shouted Ranjit.
Answer:
1. Rajan said, “Goodbye.”
2. Shiela says, “In our house, there are chairs, tables, fans and radios.”
3. “Whose books are these ?” said Rani.
4. “I looked for my pen everywhere,” said Ali.
5. “Give me the books,” shouted Ranjit.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 4 Integers Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 4 Integers Ex 4.1

1. Write two examples from day-to-day life in which we can use positive and negative integers.
Solution:
1. If positive represents above sea level, then negative represents below sea level.
2. If positive represents a deposit, negative represents a withdrawal.

2. Write the opposite of the following:

Question (a)
A profit of ₹ 500
Solution:
A loss of ₹ 500

Question (b)
A withdrawal of ₹ 70 from bank account
Solution:
Deposit of ₹ 70 in bank account

Question (c)
A deposit of ₹ 1000
Solution:
Withdrawal of ₹ 1000

Question (d)
326 B.C
Solution:
326 AD

Question (e)
500 m below sea level
Solution:
500 m above sea level

Question (f)
25° above 0°C.
Solution:
25° below 0°C.

3. Represent the situations mentioned in integers.
Solution:
(a) + 500
(b) – 70
(c) + 1000
(d) – 326
(e) – 500 m
(f) + 25.

4. Represent the following situations in integers.

Question (a)
A deposit of ₹ 500.
Solution:
+ 500

Question (b)
An Aeroplane is flying at a height two thousand metre above the sea level.
Solution:
+ 2000

Question (c)
A withdrawal of ₹ 700 from Bank Account.
Solution:
– 700

Question (d)
A diver dives to a depth of 6 feet below ground level.
Solution:
– 6.

5. Represent the following numbers on number line.

Question (i)
(a) – 5
(b) + 6
(c) o
(d) + 1
(e) – 9
(f) – 4
(g) + 8
(h) + 3.
Solution:

6. Integers are represented on a horizontal number line as shown where A represents – 2. With reference to the number line, answer the following questions:

(a) Which point represent – 3?
(b) Locate the point which represents the opposite of B and name it P.
(c) Write integers for the points C and E.
(d) Which point marked on the number line has the least value?
Solution:

(a) Point B represents – 3.
(b) Point P represents + 3.
(c) Point C represents -7 and Point E represents + 4.
(d) Point C has the least value – 7.

7. In each of the following pairs, which number is to the right of other on the number line?

Question (i)
(a) 2 9
(b) -3, -8
(c) 0, -5
(d) -11, 10
(e) -9, 9
(f) 2, – 200.
Solution:
(a) 9
(b) – 3
(c) 0
(d) 10
(e) 9
(f) 2

8. Write all the integers between the given pairs (write them in increasing order)

Question (a)
0 and -6
Solution:
-5, -4, -3, -2, -1

Question (b)
-6 and +6
Solution:
-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

Question (c)
-9 and -17
Solution:
-16, -15, -14, -13, -12, -11, -10

Question (d)
-19 and -5.
Solution:
-18, -17, -16, -15, -14, -13, -12, -11, -10, -9, -8, -7, -6.

9.

Question (a)
Write five negative integers greater than ‘-15’.
Solution:
Five negative integers greater than ‘-15’ are:
-14, -13, -12, -11, -10

Question (b)
Write five integers smaller than ‘-20’.
Solution:
Five integers smaller than ‘-20’ are:
-21, -22, -23, -24, -25

Question (c)
Write five integers greater than 0.
Solution:
Five integers greater than 0 are:
1,2, 3, 4, 5

Question (d)
Write five integers smaller than 0.
Solution:
Five integers smaller than 0 are:
-1, -2, -3, -4, -5.

10. Encircle the greater integer in each given pair.

(a) -5, -7
(b) 0,-3
(e) 5, 7
(d) -9, 0
(e) -9, -11
(f) -4, 4
(g) -10, -100
(h) 10, 100.
Solution:
(a) -5
(b) 0
(c) 7
(d) 0
(e) -9
(f) 4
(g) -10
(h) 100.

11. Arrange the following integers in ascending order:

Question (a)
0, -7, -9, 5, -3, 2, -4
Solution:
Ascending order of given integers is:
-9, -7, -4, -3, 0, 2, 5

Question (b)
8, -3, 7, 0, -9, -6.
Solution:
Ascending order of given integers is:
-9, -6, -3, 0, 7, 8.

12. Arrange the following integers in descending order:

Question (a)
-9, 3, 4, -6, 8, -3
Solution:
8, 4, 3, -3 -6, -9

Question (b)
4, 8,-3,-2, 5, 0.
Solution:
8, 5, 4, 0, -2, -3.