Class 8

In the final stage of revision, Class 8 Maths Question Paper Pdf Kerala Syllabus Set 3 Malayalam Medium help manage time.

Kerala Syllabus Class 8 Maths Model Question Paper Set 3 Malayalam Medium

സമയം : 1½ മണിക്കൂർ
ആകെ സ്കോർ: 40

നിർദ്ദേശങ്ങൾ:

• ആദ്യത്തെ 15 മിനിറ്റ് ചോദ്യങ്ങൾ വായിച്ച് മനസ്സിലാക്കാനും ഉത്തരങ്ങളെക്കുറിച്ച് ചിന്തിക്കാനും ഉപയോഗിക്കുക.
• A, B, C, D എന്നീ നാലു ഭാഗങ്ങളിലായി 16 ചോദ്യങ്ങളാണുള്ളത്.
• എല്ലാ ചോദ്യങ്ങൾക്കും ഉത്തരമെഴുതണം. A അല്ലെങ്കിൽ B എന്ന തരത്തിലുള്ള ചോദ്യങ്ങളിൽ, ഏതെങ്കിലും ഒരു ചോദ്യത്തിന് ഉത്തരമെഴുതിയാൽ മതി.
• ചോദ്യങ്ങൾക്ക് ഏത് ക്രമത്തിലും ഉത്തരങ്ങൾ എഴുതാം. ചോദ്യത്തിന്റെ നമ്പർ കൃത്യമായി എഴുതിയാൽ മതി.
• ആവശ്യമുള്ള സ്ഥലങ്ങളിൽ ഉത്തരങ്ങൾ വിശദീകരിക്കണം.

Section – A
1 സ്കോർ വീതമുള്ള 4 ചോദ്യങ്ങളാണ് ഈ വിഭാഗത്തിലുള്ളത്.

Question 1.
ഒരു വൃത്തത്തിന്റെ മധ്യഭാഗത്തിനു ചുറ്റുമുള്ള കോണിനെ മൂന്ന് തുല്യഭാഗങ്ങളായി തിരിച്ചിരിക്കുന്നു. ആരം വരയ്ക്കുന്നതിലൂടെ ഓരോന്നും 120° ആണ്. ആരത്തിന്റെ അറ്റങ്ങൾ ചേർത്ത് ഒരു ത്രികോണം നിർമ്മിക്കുന്നു. അങ്ങനെ രൂപപ്പെട്ട ത്രികോണം
a) മട്ടത്രികോണം
b) സമപാർശ്വത്രികോണം
c) സമഭുജത്രികോണം
d) ഇവയൊന്നും അല്ല
Answer:
c) സമഭുജത്രികോണം

ഒരു വൃത്തത്തിന്റെ കേന്ദ്രത്തിനു ചുറ്റും ഉള്ള ത്രികോണങ്ങൾ തുല്യമാണ്. ഒരു കോൺ 120, ബാക്കി രണ്ടു കോണുകൾ 30, അതിനാൽ ത്രികോണത്തിന്റെ കോണുകൾ 60°.

Question 2.
5-ൽ അവസാനിക്കുന്ന ഒരു സംഖ്യയുടെ വർഗ്ഗത്തിന്റെ ഒന്നിന്റെയും പത്തിന്റെയും സ്ഥാനത്തെ അക്കങ്ങ ളുടെ തുക ആകാൻ സാധ്യത ചുവടെ തന്നിട്ടുള്ളതിൽ ഏതിനാണ്?
(a) 2
(b) 5
(c) 7
(d) 3
Answer:
(c) 7

Question 3.
ചുവടെയുള്ള പ്രസ്താവനകൾ നോക്കുക.
പ്രസ്താവന I: ഒരു സമഭുജസാമാന്തരികത്തിന്റെ വികർണങ്ങൾ പരസ്പരം 90°ൽ സമഭാഗം ചെയ്യുന്നു.
പ്രസ്താവന II: ഒരു സമഭുജസാമാന്തരികത്തിന്റെ വശങ്ങൾക്ക് തുല്യ നീളമായിരിക്കും.
a) പ്രസ്താവന I. ശരിയാണ്, പ്രസ്താവന II തെറ്റാണ്.
b) പ്രസ്താവന I തെറ്റാണ്, പ്രസ്താവന II ശരിയാണ്.
c) പ്രസ്താവന I ശരിയാണ്, പ്രസ്താവന I-ന്റെ കാരണമാണ് പ്രസ്താവന II.
d) പ്രസ്താവന I ശരിയാണ്, പ്രസ്താവന -ന്റെ കാരണമല്ല പ്രസ്താവന II.
Answer:
d) പ്രസ്താവന . ശരിയാണ്, പ്രസ്താവന – ന്റെ കാരണമല്ല പ്രസ്താവന II.

Question 4.
ചുവടെ കൊടുത്തിരിക്കുന്ന ലംബകത്തിന്റെ പരപ്പളവ് ആയിരിക്കും.

(a) 4.5 cm²
(b) 6 cm²
(c) 5.5 cm²
(d) 6.5 cm²
Answer:
(c) 5.5 cm²
പരപ്പളവ് = \(\frac{1}{2}\) × 2 × (2 + 3.5) = 5.5cm²

Section – B
2 സ്കോർ വീതമുള്ള 4 ചോദ്യങ്ങളാണ് ഈ വിഭാഗത്തിലുള്ളത്.

Question 5.
ഒരു ബഹുഭുജത്തിന് വശങ്ങളുണ്ട്.
(a) അതിന്റെ അകകോണുകളുടെ ആകെത്തുക എത്ര?
Answer:
n വശങ്ങളുള്ള ബഹുഭുജത്തിന്റെ അകക്കോ ണുകളുടെ തുക = 180° × (n – 2)
അകക്കോണുകളുടെ തുക = 180 × 5 = 900

(b) പുറം കോണുകളുടെ ആകെത്തുക എത്ര?
Answer:
360°

Question 6.
രണ്ട് സംഖ്യകളുടെ തുക 60. ഒരു സംഖ്യ മറ്റേ സംഖ്യയുടെ നാല് മടങ്ങാണ്. സംഖ്യ കണക്കാക്കുക.
Answer:
ഒരു സംഖ്യയും അതിന്റെ 4 മടങ്ങും കൂടി ചേരു മ്പോൾ 5 മടങ്ങാകും.
സംഖ്യം 60ന്റെ \(\frac{1}{5}\) ഭാഗമാണ്.
സംഖ്യ = 12

Question 7.
(a) 1 – 2 എന്താണ്
(b) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 കണ്ടെത്തുക.
Answer:
(a) (2 – 1) = -1
(b) -1 -1 -1 -1 -1 = -5

Question 8.
13 സെ.മീ. ആരമുള്ള ഒരു വൃത്തത്തിൽ, ഞാൺ വൃത്തകേന്ദ്രത്തിൽ നിന്ന് 5 സെ.മീ. അകലെയാണ്. ഞാണിന്റെ നീളം കണക്കാക്കുക.
Answer:
ഞാണിന്റെ പകുതി = \(\sqrt{13^2-5^2}\) = 12cm
∴ ഞാണിന്റെ ആകെ നീളം = 24 സെ.മീ.

Section – C
3 സ്കോർ വീതമുള്ള 4 ചോദ്യങ്ങളാണ് ഈ വിഭാഗത്തിലുള്ളത്.

Question 9.
x = 0.333……അന്നെങ്കിൽ
(a) x-ന്റെ ഭിന്നസംഖ്യ രൂപം എഴുതുക.
Answer:
\(\frac{3}{9}\)

(b) x² നെ ഭിന്നസംഖ്യയായി എഴുതുക.
Answer:
\(\frac{1}{9}\)

(c) x² + x -ന്റെ ദശാംശരൂപം എഴുതുക.
Answer:
x² + x = \(\frac{1}{9}+\frac{3}{9}=\frac{4}{9}\) = 0.444…….

Question 10.
x² – y² = (x + y) (x – y) എന്ന സമവാക്യം ഉപയോഗിച്ച്
(a) 2² – 1²
(b) 9² – 10²?
(c) 1² – 2² + 3² – 4² +……….+ 9² – 10² കണക്കാക്കുക.
Answer:
(a) (2+1)(2-1)=3×1=3
(b) -19
(c) – 3 – 7 – 11 – 15 – 19
-(3 + 7 + 11 + 15 + 19) = -55

Question 11.
ഒരു ചതുരത്തിന്റെ ചുറ്റളവ് 50 സെ.മീ. ആണ്. നീളം വീതിയേക്കാൾ 5 സെ.മീ. കൂടുതലാണ്.
(a) നീളം + വീതി എത്ര?
(b) നീളം ‘x’ ആയെടുത്ത് സമവാക്യം രൂപീകരിക്കുക.
(c) വശങ്ങൾ കണക്കാക്കുക.
Answer:
(a) 25 സെ.മീ.
(b) x + x + 5 = 25
2x + 5 = 25, 2x = 20, x=10
(c) വശങ്ങൾ = 10 സെ.മീ, 15 സെ.മീ.

OR

15 വർഷം കഴിയുമ്പോൾ ഒരാളുടെ പ്രായം അയാൾക്ക് 15 വർഷം മുമ്പുണ്ടായിരുന്ന പ്രായത്തിന്റെ രണ്ട് മടങ്ങാണ്.
(a) ഇപ്പോഴത്തെ പ്രായം ‘x’ ആയെടുത്ത് സമവാക്യം എഴുതുക.
(b) ഇപ്പോഴത്തെ പ്രായം കണക്കാക്കുക.
Answer:
(a) 15 വർഷങ്ങൾക്കു ശേഷമുള്ള പ്രായം x + 15 ആണ്.
15 വർഷങ്ങൾക്കു മുൻപുള്ള പ്രായം x – 15 ആണ്.
2(x – 15) = x + 15
2x – 30 = x + 15
x = 45

(b) ഇപ്പോഴത്തെ പ്രായം 45 വയസ്സ്.

Question 12.
ഒരു ത്രികോണത്തിന്റെ വശങ്ങൾ 20 സെ.മീ, 15 സെ.മീ, 25 സെ.മീ. ആണ്. ഇത് ഏതുതരം ത്രികോണം ആണ്? ഇതിന്റെ പരിവൃത്തത്തിന്റെ ആരം കണക്കാക്കുക.
Answer:
15² + 20² = 225 + 400 = 625 = 25²
ഇവിടെ ഒരു മട്ടത്രികോണമാണ് ലഭിക്കുന്നത്. ഒരു മട്ടത്രികോണത്തിന്റെ പരിവൃത്ത കേന്ദ്രം അതിന്റെ കർണത്തിന്റെ മധ്യബിന്ദുവാണ്. അതിനാൽ, പരിവൃത്തത്തിന്റെ ആരം R = കർണ ത്തിന്റെ പകുതി

OR

ലംബകം ABCD-യിൽ സമാന്തരവശങ്ങൾ 8 സെ.മീ, 6 സെ.മീ. ആണ്. വലിയ വശത്തിലെ കോണുകൾ 60, 75° ആണ്.
(a) ഒരു ഏകദേശ ചിത്രം വരയ്ക്കുക.
Answer:

(b) യഥാർത്ഥ അളവുകളിൽ ലംബകം ABCD വരയ്ക്കുക.
Answer:

Section – D
4 സ്കോർ വീതമുള്ള 4 ചോദ്യങ്ങളാണ് ഈ വിഭാഗത്തിലുള്ളത്.

Question 13.
7.5 സെന്റിമീറ്റർ നീളമുള്ള വര വരച്ച് 3.75 സെ.മീ. ലംബവശങ്ങളുടെ നീളമായ മട്ടത്രികോണം വരക്കുക.
Answer:
75 സെ.മീ. നീളമുള്ള ഒരു വര വരയ്ക്കുക. ഈ വരയുടെ ലംബസമഭാജി വരയ്ക്കുക. വരയുടെ പകുതി നീളം ലംബത്തിൽ വരച്ചു മട്ടത്രികോണം പൂർത്തിയാക്കുക.

Question 14.
താഴെ കൊടുത്തിരിക്കുന്ന പാറ്റേൺ നോക്കുക.
1 + 3 = 4 = 2²
1 + 3 + 5 = 9 = 3²
1 + 3 + 5 + 7 = 16 = 4²
(a) ഇതുപോലെ അടുത്ത വരി എഴുതുക.
(b) ആദ്യത്തെ 10 ഒറ്റസംഖ്യകളുടെ തുക എത്ര?
(c) ഇരുപതാമത്തെ ഒറ്റസംഖ്യ ഏത്?
(d) -1 -3 -5 ………. – 39?
Answer:
(a) 1 + 3 + 5 + 7 + 9 = 25 = 5²
(b) 10² = 100
(c) 39
(d) -(1 + 3 + 5 +….+ 39) = -400

Question 15.
ഒരു റബ്ബർ കർഷകന് ഒരു മാസത്തിലെ ഓരോ ദിവസവും ലഭിച്ച റബ്ബർ ഷീറ്റുകളുടെ വിവരം ചുവടെ ചേർക്കുന്നു. ആവൃത്തി പട്ടിക തയ്യാറാക്കി ചുവടെയുള്ള ചോദ്യങ്ങൾക്ക് ഉത്തരമെഴുതുക.

i. ഒരു ദിവസം പരമാവധി എത്ര ഷീറ്റ് ലഭിച്ചു?
ii. 16 ഷീറ്റുകൾ ലഭിച്ച എത്ര ദിവസങ്ങൾ ഉണ്ട്?
iii. ഏറ്റവും കുറവ് ഷീറ്റ് ലഭിച്ച ദിവസങ്ങൾ എത്ര?
iv. 15-ൽ കുറവ് ഷീറ്റ് ലഭിച്ച ദിവസങ്ങളുടെ എണ്ണം?
v. ഈ മാസം ആകെ എത്ര ഷീറ്റുകൾ ലഭിച്ചു?
Answer:

ഷീറ്റുകളുടെ എണ്ണം	ദിവസങ്ങ ളുടെ എണ്ണം
9	3
10	4
11	3
12	3
13	5
14	6
16	6

(i) 16
(ii) 6 ദിവസങ്ങൾ
(iii) 3 ദിവസങ്ങൾ
(iv) 24 ദിവസങ്ങൾ
(v) 85 ഷീറ്റുകൾ

OR

ചുവടെ കൊടുത്തിരിക്കുന്ന സാമാന്തരികത്തിന്റെ പരപ്പളവ് കണക്കാക്കുക.

Answer:

∆ABF ന്റെ പരപ്പളവ് = \(\frac{1}{2}\) × 6 × 3 = 9cm²

ABCD-യുടെ പരപ്പളവ് = \(\sqrt{45} \times \frac{18}{\sqrt{45}}\) = 18cm²
(അല്ലെങ്കിൽ സാമാന്തരികം ABCD-യുടെ പരപ്പ ളവ് = ചതുരത്തിന്റെ പരപ്പളവ് = 6 × 3 = 18 )

Question 16.
വശങ്ങൾ 6 സെ.മീ, 4 സെ.മീ, പരപ്പളവ് 18 ചതുരശ്ര സെന്റീമീറ്ററുമായ ഒരു സാമാന്തരികം വരയ്ക്കുക.
Answer:
6 സെ.മീ. നീളമുള്ള ഒരു വര വരയ്ക്കുക. സാമാ ന്തരികത്തിന്റെ പരപ്പളവ് 18 ചതുരശ്ര സെന്റിമീറ്റർ ആയതിനാൽ അതിന്റെ ഉയരം 3 സെ.മീ. ആയിരി ക്കും. ഈ വരയുടെ ഒരറ്റത്ത് 3 സെ.മീ. ലംബം വരയ്ക്കുക. ഇതിന്റെ മുകളറ്റത്തുകൂടി താഴെ വശ ത്തിനു സമാന്തരമായി ഒരു വര വരയ്ക്കുക. താഴത്തെ വശത്തിന്റെ ഇടത്തേ അറ്റത്തുനിന്ന് മുക ളിലത്തെ വരയിൽ 4 സെ.മീ. അടയാളപ്പെടുത്തി സാമാന്തരികം പൂർത്തിയാക്കുക.

OR

‘n’ വശങ്ങളുള്ള ഒരു ബഹുഭുജത്തിന്റെ അകക്കോണുകളുടെ തുക (n – 2) 180° ആണ്. പുറം കോണുകളുടെ തുക എപ്പോഴും 360° ആയിരിക്കും.
(i) 20 വശങ്ങളുള്ള ഒരു ബഹുഭുജത്തിന്റെ അകക്കോണുകളുടെ തുക എത്ര
(ii) 100 വശങ്ങളുള്ള ഒരു ബഹുഭുജത്തിന്റെ പുറംകോണുകളുടെ തുക എത്ര
(ii) അകക്കോണുകളുടെയും പുറംകോണുകളുടെയും തുക തുല്യമായ ബഹുഭുജം ഏതാണ്?
(iv) ഒരു ബഹുഭുജത്തിന്റെ അകക്കോണുകളുടെ തുക 1080 ആകുമോ?
Answer:
(i) (20 – 2) 180 = 18 × 180° = 3240°
(ii) 360°
(ii) ചതുരം
(iv) ആകും, കാരണം 180ന്റെ ഒരു ഗുണിതമാണ് 1080°.

To reduce exam fear, Class 8 Basic Science Question Paper Kerala Syllabus Set 4 provide good practice.

Kerala Syllabus 8th Standard Basic Science Model Question Paper Set 4

Time: 2 hour
Total score:60

Instructions:

1. The first 15 minutes is cool-off time. This time is meant for reading the questions and planning your answers.
2. This question paper includes 10 questions in sections A, B and C
3. You need to answer only one of the options provided for each of the choice-based questions.

PHYSICS

Time: 40 minutes
Total Score: 20

Section – A

Select the correct answer for questions 1 to 2. Answer all questions. (1 score for each question) (2 × 1 = 2)

Question 1.
The figure shows the lines of force between the poles of two bar magnets. Identify the poles represented by A and B.

a) A-North, B-South
b) A-South, B-South
c) A -North, B -North
d) A -South, B -North
Answer:
c) A -North, B -North

Question 2.
Two statement are given below. Examine them and choose the correct answer.
Statement 1: Charging by conduction is the method of charging an object by direct physical contact with a charged object.
Statement 2: Charging by induction is the method of charging an object with direct physical contact of a charged object.
Options:
a) Both statements are correct
b) Only the second statement is correct
c) Only the first statement is correct
d) Both statements are incorrect
Answer:
c) Only the first statement is correct

Section- B

Answer questions 3 to 8 in more than one sentence. Questions 7 and 8 have choices. (2 score for each question) (6 × 2 = 12)

Question 3.
Write down two precautions to be taken for protecting ourselves from lightning.
Answer:
• Do not operate electrical equipment’s.
• Avoid touching metal objects.
• Do not seek shelter under tall or isolated trees. (any two)

Question 4.
Classify the following situations into contact and non-contact forces.
a) Applying break in a bicycle.
b) The earth revolving around the sun.
Answer:
Contact force
a) Applying brake in a bicycle Non-Contact force
b) The earth revolving around the sun

Question 5.
Two identical balloons one filled with water and other with kerosene of equal quantity.
iMG- 1
a) Identify the balloon filled with water.
b) Why is the expansion of balloons appearing to be different in figures (a) and (b)?
Answer:
a) Figure (b)
b) Water has higher density than kerosene. As density increases, liquid pressure increases. So the balloon which is expanded has water filled in it and the other balloon which is small in size has kerosene filled in it.

Question 6.
Lightning conductors are fixed to protect tall buildings from the danger of lightning. Write down how lightning conductors work?
Answer:
A lightning conductor has sharp pointed ends where 9. the electric charge density is highest. It is connected to the earth through a conducting wire, allowing charges to safely flow into the ground. When clouds accumulate a large negative charge, positive charges are induced at the pointed ends of the conductor, which neutralizes the cloud’s charge. If the clouds carry a positive charge, negative charges gather at the pointed ends which neutralizes the cloud’s charge. When lightning occurs charge accumulate at the pointed ends and reaches to earth through the conducting wire. Thus the object or building is protected.

Question 7.
A. Classify the physical quantities given in the box as fundamental quantities and derived quantities.
Length, density, area, mass

Fundamental quantities	Derived quantities

Answer:

Fundamental quantities	Derived quantities
Length, Mass	Density, Area

B. 1 mm is the least count of a metre scale.
a) What do you mean by least count?
b) Arrange the following units in ascending order of their values.
(mm, km, nm. cm)
Answer:
a) The smallest value that can be measured using
an instrument is called its least count.
b) nm < μm < cm < km

Question 8.
A. What would be the radius of curvature of a concave mirror with a focal length of 40 cm?
Answer:
f = 40 cm
f = \(\frac{R}{2}\)
R = f × 2
R = 40 × 2 = 80 cm

OR
B. a) Copy the ray diagram and complete it showing the image formation.

Answer:

b) What are the characteristics of the image formed?
Answer:
Real, inverted and enlarged

Section – C

Questions 9 to 10 carry 3 score each. Question 10 has a choice. (2 × 3 = 6)

Question 9.
Write your inferences.
a) All magnets have same shape.
b) All magnets show N, S direction when suspended freely.
c) All magnets are man made.
Answer:
a) There are different types of magnets.

b) Yes. Because the north pole of magnet turns to geographical south of earth and south pole turns to geographical north of earth.(or Earth acts like a large magnet. Hence the poles of the suspended magnet is opposite to the poles of the earth’s magnet.)

c) No, There are natural and man made magnets.

Question 10.
A. Convert the following units into SI units without changing their values.
a) 3500g
b) 2.5 km
c) 2h
Answer:
a) 3500g = \(\frac{3500}{1000}\) kg = 3.5 kg

b) 2.5km = 2.5 × 1000 m = 2500 m

c) 2h = 2 × 60 × 60s = 7200s.

OR
B. State whether the following symbol are right or wrong. Correct them if wrong.
a) m/s/s
b) 50 kg/m3
c) 100 n
d) 20 kgs
e) 3M
Answer:
a) Wrong. m/s² is the correct form.
b) Right
c) Wrong. 100 N is the correct form.
d) Wrong. 20 kg is the correct form.
e) Wrong. 3 m is the correct form.
f) Wron 1.5 kg or 1500 g is the correct form.

CHEMISTRY

Time: 40 minutes
Total Score: 20

Section – A

Answer all the questions from 1 to 2.1 score for each.

Question 1.
Match the Following

Reaction/Property	Product Basis
i) Acid reacts with a highly reactive metal	a) CO2 gas
ii) Acids react with carbonates	b) H+ ions
iii) Basis of general properties of acids	c) H2 gas
iv) Non-metal oxide reacts with water	d) Acidic properties

A) i – a, ii – c, iii – b, iv – d
B) i – c, ii – a, iii – d, iv – b
C) i – c, ii – a, iii – b, iv – d
D) i – d, ii – b, iii – a, iv – c
Answer:

Reaction/Property	Product Basis
i) Acid reacts with a highly reactive metal	c) H2 gas
ii) Acids react with carbonates	a) CO2 gas
iii) Basis of general properties of acids	b) H+ ions
iv) Non-metal oxide reacts with water	d) Acidic properties

Question 2.
Two statements are given below.
(i) Gold can be coated on a copper bangle by the process of electroplating.
(ii) In this process, the copper bangle is connected to the positive terminal of the battery. Choose the correct one from the options given below.
A. Both the statements (i) and (ii) are incorrect.
B. Statement (i) is correct. However, the statement (ii) is incorrect.
C. Both statements (i) and (ii) are correct.
D. Statement (ii) is correct. However, statement (i) is incorrect.
Answer:
B. Statement (i) is correct. However, the statement (ii) is incorrect.

Section – B

Two questions from 3 to 8 have choices. Each question carries 2 scores. (6 × 2 = 12)

Question 3.
Sodium reacts with Water to form sodium hydroxide and hydrogen gas. Is this a physical change or a chemical change? Why
Answer:
it is a chemical change because new substances are formed.

Question 4.
A) Give a reason for the following:
a) More than 65% of the body of living organisms is made up of water.
b) Ice formed on freezing water floats on it.
Answer:
a) Water has a high specific heat capacity, so it prevents sudden changes in body tempera-ture. This helps plants and animals maintain a stable internal environment and survive in dif-ferent climatic conditions.

b) When water freezes into ice, its volume in¬creases and density decreases. So, ice floats on water.

B. Water expands when it freezes.
a) What change happens to its volume and density?
b) How does this property help aquatic life in cold regions?
Answer:
a) When water freezes, its volume increases and its density decreases. This is why ice is lighter than water and floats on its surface.

b) Ice floats on water because its volume increases (density decreases) when it freezes. Floating ice acts as a protective layer (insulator). This insulation prevents the water below from freezing. This allows fish and aquatic life to survive in cold regions.

Question 5.
a) The symbol for the element hydrogen is H. What is the difference between 2H and H₂?
b) Classify the following into elements and compounds and list them.
[Carbon, Carbon dioxide, Oxygen, Sodium chloride]
Answer:
a) In a 2H refers the two atoms of Hydrogen. H₂ refers to a Hydrogen molecule formed by two atoms.

b)

Elements	Compounds
Carbon	Carbon dioxide
Oxygen	Sodium Chloride

Question 6.
a) How is soda water made?
b) Write the chemical equation for this reaction
Answer:
a) Soda water is made by dissolving carbon dioxide in water under high pressure.
b) CO₂ + H₂O → H₂CO₃

Question 7.
The degree of acidic/basic nature of soil and other substances is determined by calculating their pH value.
a) Who invented the pH scale?
b) What characteristic pH value defines a neutral solution?
Answer:
a) The pH scale was invented by the Danish scientist Sorensen.
b) The characteristic pH value that defines a neutral solution is 7.

Question 8.
A. a) What is the natural source from which natural rubber is obtained?
b) Explain why synthetic rubbers are predominantly used in the manufacture of tyres?
Answer:
a) Natural rubber is obtained from the latex collected from the rubber tree.

b) Synthetic rubber shows less wear and tear compared to natural rubber. That is why synthetic rubbers are used in tyre manufacture.

OR
B. a) What is the difference between thermoplastics and thermosets?
b) Which of the following is a natural polymer?
[Polythene, Nylon, Bakelite, Cellulose]
Answer:
a) Thermoplastics: Soften on heating and can be reshaped many times.
Thermosets: Harden permanently on heating and cannot be remoulded once set.
b) Cellulose.

Section – C

One question from 9 to 10 has a choice. Each question carries 3 scores. (2 × 3 = 6)

Question 9.
One molecule of nitric acid contains one hydrogen atom, one nitrogen atom and three oxygen atoms.
a) Write the chemical formula of nitric acid
b) Find the total number of atoms present in four molecules of nitric acid.
c) How many ammonia (NH3) molecules are required to obtain the same number of atoms present in four molecules of nitric acid?
Answer:
a) HNO₃

b) H = 1
N = 1
O = 3
Total = 1 + 1 + 3 = 5 atoms
For four molecules:
4 × 5 = 20 atoms

c) Atoms in one NH, molecule:
N = 1
H = 3
Total = 4 atoms
Number of ammonia molecules
= 20 ÷ 4 = 5 molecules

Question 10.
A. The stomach produces hydrochloric acid (HCl) for digestion. Excess acid can cause problems, which
are treated with medicines called antacids.
a) Define the term Neutralisation reaction?
b) What is the chemical nature of antacids (acidic or basic)?
c) Write the name of a commonly used antacid?
Answer:
a) The chemical reaction in which an acid and a base react to form water and a salt is known as a neutralisation reaction. Magnesium hydroxide, also known as milk of magnesia

b) Basic

c) Magnesium hydroxide, also known as milk of magnesia

OR
B. Take a pinch of ammonium sulphate in a test tube and mix it well with some water. Pour barium chloride solution into it. A white curdy precipitate is formed.
a) Identify the substance that forms the curdy white precipitate.
b) Write the balanced chemical equation for the reaction between ammonium sulphate ((NH₄)2SO₄) and barium chloride (BaCl).
c) What property of the precipitate is demonstrated when it is observed that adding dilute hydrochloric acid (HCl) causes no change?
Answer:
a) Barium sulphate (BaSO₄)

b) (NH₄)₂ + SO₄ + BaCl₂ → BaSO₄ + 2NH₄Cl

c) Barium sulphate is insoluble in dilute hydrochloric acid.

BIOLOGY

Time: 40 minutes
Total Score: 20

Section – A

Choices are given for questions 7, 8,10. Answer questions 1 and 2. Each question carries 1 score. (2 × 1 = 2)

Question 1.
Which of the following statements is correct regarding electron microscopes?
(A) It helps in seeing up to a thousand times larger.
(B) It helps in detailed observation of living cells, viruses and molecular structures.
(C) It uses an electron beam.
(D) It uses light.
Answer:
(B) and (C)

Question 2.
Choose and write the correct answer from the options given below where the pairs are correctly matched:

Meiosis	Period of intense mental and emotional changes
Adolescence	Fully matured ovum is released from the ovary
Cell plate	Chromosome number is reduced to half
Ovulation	Cytokinesis in plant cell

(a) A-I, B-II, C-III. D-IV
(b) A-II, B-III, C-IV, D-I
(c) A-III, B-I, C-IV, D-II
(d) A-III, B-IV, C-II, D-I
Answer:
(c) A-III, B-I, C-IV, D-II

Section – B

Write answers to questions 3 to 8. Each question carries 2 score. (6 × 2 = 12)

Question 3.
Which pair is correct?
a) Online system – Disease and pest information collection, solution – Up-to-date knowledge about climate change – Market opportunities
b) Family members – financial assistance to farmers, procurement, marketing
c) Local Government Institutions – Financial Assistance, Employment Assistance, Psychological Support.
d) Cooperative societies – Use of barren lands – Approval – Government assistance
e) Krishi Bhavan – Distribution of best planting materials – Scientific guidance and training programs for agriculture
Answer:
a) Online system – Disease and pest information collection, solution – Up-to-date knowledge about climate change – Market opportunities

e) Krishi Bhavan – Distribution of best planting materials – Scientific guidance and training programs for agriculture

Question 4.
Analyse the statements given below and give reasons for each.
a) One of the most important ecological hotspots in India is the Western Ghats.
b) Sacred groves are to be protected and maintained.
Answer:
(a) The Western Ghats is designated as one of the world’s most important ecological hotspots primarily for twp key reasons: high endemicity and significant habitat loss.

Sacred groves are patches of forest or natural vegetation protected and revered by local 8 communities on religious and cultural grounds. Their protection is vital for both biodiversity conservation and cultural heritage.

Question 5.
Complete the table.

Division	Characteristics
Carnivora	____(A)____
____(B)____	Have retractable claws and excellent night vision.
Panthera	____(C)____
Panthera tigris	____(D)____

Answer:
(A) Sharp teeth and claws adapted for hunting and eating meat.
(B) Felidae
(C) Ability to roar, large and powerful body structure.
(D) Orange fur with black stripes and strong large body.

Question 6.
Observe the illustration and answer the questions below.

(a) Which process does the illustration indicate?
(b) Explain the process that the illustration indicates.
Answer:
a) Formation of organelles found in cells.

b) Eukaryotic cell engulfs small aerobic bacteria.
Instead of digesting the small cell, it protects it. Gradually, these become mitochondria. Eukaryotic cell engulfs small photosynthetic bacteria Instead of digesting the photosynthetic bacteria, it protects it. Gradually these become chloroplasts.

Question 7.
A. Complete the flowchart:

Answer:
A – Anther
B – Filament
C – Style
D – Ovary

B. Observe the given figure and answer the following questions:

a) Identify A and B in the given figure.
b) Give an example each for A and B which shows this type of reproduction.
Answer:
a) A- Binary fission, B – Multiple fission
b) Binary fission is seen in Bacteria and amoeba. Multiple fission is seen in Plasmodium.

Question 8.
A. Obsere the picture and answer the questions below.

a) Which organelle is represented by the picture?
b) What is the function of this organelle?
Answer:
a) Endoplasmic Reticulum
b) These organelles appear as a network of tubules within the cell, serve as pathways for conducting materials. It helps in the synthesis and removal of materials required by the cell.

OR
B. Choose the answer to the statements given below from the box and write it down.

Muscle tissue, Connective tissue, Epithelial tissue, Neural tissue

a) Connects different parts of the body.
b) Controls and coordinates bodily functions.
Answer:
a) Connective tissue
b) Neural tissue

Section – C

Answer questions 9 and 10. Each question carries 3 score. (2 × 3 = 6)

Question 9.
The names of Papaya in different regions are given below:
Papaya, Karumusa, Kappalanga
a) How can a person from a different region identify this plant?
b) What is the solution to this problem?
Answer:
a) A person can identify this plant in different regions using its scientific name. Since scientific names are the same worldwide, it helps to avoid the confusion caused by different local names.

b) The solution to this problem is to use the Binomial Nomenclature system. This system gives each organism a unique and universally accepted name. This makes identification clear and unifonn everywhere.

Question 10.
A. Gene banks are an Ex-situ conservation method.
a) What is the role of gene banks in the conservation of biodiversity?
b) What are the various levels of Biodiversity?
c) Write the names of Ecological hotspots in India?
Answer:
a) These are research centres with facilities to collect seeds and gametes to preserve them for a long time. Organisms can be recreated out of them whenever required.
b) Ecosystem biodiversity, Species diversity, Genetic diversity.
c) Western Ghats, North-Eastern Himalayas, Indo- Bumia region

B. Logo of an international organisation is given.

a) Write the name of this organisation.
b) What is the obj ective of this organisation?
c) Write two instructions that can be included in the poster to create awareness in people of your locality to protect the environment.
Answer:
a) WWF

b) To understand the importance of biodiversity, i.e,, Conserving Biodiversity, Sustainable Resource Use, and Reducing Human Impact are the major objectives.

c) • Native plants support local pollinators (bees, butterflies) and require less water and maintenance than non-native species, helping conserve water and local biodiversity.

• Carry a reusable water bottle and shopping bag. Local water bodies and parks are often clogged with single-use plastic, harming wildlife and degrading our nature’s beauty.

To reduce exam fear, Class 8 Basic Science Question Paper Kerala Syllabus Set 3 provide good practice.

Kerala Syllabus 8th Standard Basic Science Model Question Paper Set 3

Time: 2 hour
Total score:60

Instructions:

1. The first 15 minutes is cool-off time. This time is meant for reading the questions and planning your answers.
2. This question paper includes 10 questions in sections A, B and C
3. You need to answer only one of the options provided for each of the choice-based questions.

PHYSICS

Time: 40 minutes
Total Score: 20

Section – A

Select the correct answer for questions 1 to 2. Answer all questions. (1 score for each question) (2 × 1 = 2)

Question 1.
The pressure exerted on an area of 10 m₂ by a force of 5 0 N is _____
a) 5 N
b) 5 N/m₂
c) 0.2 N/m₂
d) 0.2 N
Answer:
b) 5 N/m₂
P = \(\frac{\mathrm{F}}{\mathrm{~A}}\), F = 50 N, A = 10₂
P = \(\frac{\mathrm{50}}{\mathrm{10}}\) = 5 N/m₂

Question 2.
Two statements related to image formation in a spherical mirror is given. Examine them and choose the correct answer.
Statement l:When an object is placed at C of concave mirror, size of image formed is as that of object.
Statement 2 :When the object is placed at F of a concave mirror ,we can imagine that the image is formed at infinity.
a) Both statements are correct
b) Only the second statement is correct
c) Only the first statement is correct
d) Both statements are incorrect
Answer:
a) Both statements are correct

Section – B

Answer questions 3 to 8 in more than one sentence. Questions 4 and 6 have choices. (2 score for each question) (6 × 2 = 12)

Question 3.
“Overspeed of vehicles result in accidents”
Write two messages to be written on an awareness board in your school premises to minimise road accidents.
Answer:

1. Speed thrills but kills
2. Don’t mix drink with drive.
3. Wear helmets while driving motor cycles. (Any two)

Question 4.
A. Write down the following units in descending order of their values.
a) 2 cm, 2 m, 2 km, 2 mm
b) 5g, 5 mg, 5 quintal, 5 kg
Answer:
a) 2 km > 2 m> 2 cm > 2 mm
b) 5 quintal > 5 kg > 5 g > 5 mg

OR
B. Complete the table.

Unit	Symbol
Newton	(a)….
Kelvin	(b)….
Ampere	(c)….
Metre	(d)….
Kilogram	(e)….
Second	(f)….

Answer:

Unit	Symbol
Newton	(a) N
Kelvin	(b) K
Ampere	(c) A
Metre	(d) M
Kilogram	(e) K
Second	(f) S

Question 5.
Liquids also exert pressure. Write down any two factors that influence liquid pressure.
Answer:
i) Height of the liquid column
ii) Density of the liquid

Question 6.
A. Find out the magnetic polarity indicated by A and B in the figures given below.

Answer:
Fig(a) A-North pole
Fig(b) B-South pole

B. Some of the devices are mentioned below.
Electric bell, Maglev train, Compass needle, Generators, Magnetic toys
a) From the given devices classify them as electromagnets and permanent magnets?
b) What are the differences between Electromagnets and permanent magnets?
Answer:
a) Electromagnets: Electric bell, Maglev trains, Generators
Permanent magnets: Compass Needle, Magnetic toys

b)

Electromagnets	Permanent magnets
• Poles can be changed.	• Poles are fixed
• Strength can be increased by changing current or number of turns of coil.	• Magnetic strength cannot be increased.
• Magnetism exists only when current flows	• The gained magnetism can be retained for a long time.

Question 7.
A negatively charged rod is brought near an electroscope as shown in the figure.

Copy down the diagram and mark the distribution of electric charge formed at A and B
Answer:

Question 8.
Fill in the blanks.
a) In a spherical mirror, the distance from the pole to the pnncipal focus is called the ______
b) The focal length of a mirror having radius of curvature 18 cm is ______
Answer:
a) Focal length
b) 9 cm

Section – C

Questions 9 to 10 carry 3 score each . Question 10 has a choice. (2 × 3 = 6)

Question 9.
a) Earth is considered as an electron bank why?
b) Observe the figure.

If the plate B is earthed, what change occurs. Draw the picture.
Answer:
a) When a positively charged body is earthed, electrons flows from earth to body. When a negatively charged body is earthed. electrons flows from body to earth. Thus earth receives and gives electrons, so earth is called an electron bank.
b)

Question 10.
A. a) Complete the diagram given below by drawing normal and the path of the reflected rays.

b) Complete the diagram given below by drawing the path of the reflected rays.

c) The principal focus of a concave mirror is real? Why?
Answer:

c) In a concave mirror, the rays that fall parallel to the principal axis converge to the principal focus and it can be caught on a screen. Hence, its principal fochs is real.

OR
Question 10.
B. Classify the following statements as to those related to uses of concave mirrors and convex mirrors and tabulate them accordingly.
a) as make up mirror
b) as rear view mirrors in vehicles
c) in solar concentrators
d) as shaving mirror
e) Dentist mirror
f) Reflectors in street lamps
Answer:
Concave mirrors Convex mirrors
a) as make up mirror
b) as rear view mirrors in vehicles
c) in solar concentrators
f) Reflectors in street lamps
d) as shaving mirror
e) Dentist mirror

CHEMISTRY

Time: 40 minutes
Total Score: 20

Section – A

Answer all the questions from 1 to 2. 1 score each.

Question 1.
Match the following.

Common Name	Chemical formula
i) Caustic soda	a) KOH
ii) Milk of lime	b) NaOH
iii) Caustic potash	c) Ca(OH)2

Choose the correct order.
A) i-b. ii-a, iii-c
B) i-c, ii-b. iii-a
C) i-c. ii-a, iii-b
D) i-b, ii-c, iii-a
Answer:
D) i-b, ii-c, iii-a

Question 2.
Two statements are given.
Statement 1: Soda-lime glass, Borosilicate glass. Flint Glass contain silica.. Statement 2: Silica is the component present in all types of glass.
Select the correct option from the following.
A) Both statements 1 and 2 are correct.
B) Statement 1 is correct. But statement 2 is not correct.
C) Statement 2 is correct. But statement 1 is not correct.
D) Both statements 1 and 2 are not correct.
Answer:
Select the correct option from the following.

Section – B

Two questions from 3 to 8 have choices. Each question carries 2 scores. (6 × 2 = 12)

Question 3.
The observation result of an experiment is given below. Identify the gases formed

Acids	Substance	Gas
Vinegar	Zinc piece	___(a)___
Vinegar	Baking soda	___(b)___

Answer:
a) Hydrogen gas (H₂)
b) Carbon Dioxide (CO₂)

Question 4.
A) Write any two merits and demerits of synthetic fibres.
Answer:
Merits: Less expensive, High durability. Wrinkle-free.
Demerits: Do not absorb water, Less air circulation.

OR
B) Synthetic fibres are generally used by blending them with natural fibres. Why?
Answer:
To give extra strength.
• To give a new texture.
• To bring extra qualities

Question 5.
Water expands when it freezes.
a) What change happens to its volume and density?
b) How does this property help aquatic life in cold regions?
Answer:
When water freezes, its volume increases and
its density decreases. This is why ice is lighter than water and floats on its surface.

b) Ice floats on water because its volume increases (density decreases) when it freezes. Floating ice acts as a protective layer (insulator). This insulation prevents the water below from freezing. This allows fish and aquatic life to survive in cold regions.

Question 6.
Take a small piece of zinc in a test tube. Add a little dilute hydrochloric acid to it. Hold a burning matchstick at the mouth of the test tube.
a) What do you observe?
b) Name the gas formed here?
Answer:
a) When you hold a burning matchstick or splinter near the mouth of the test tube, the gas will ignite and bum with a “pop” sound.

b) Hydrogen gas. (When acids react with highly reactive metals, hydrogen gas is produced.)

Question 7.
a) What are the components of water?
b) How can we separate the components of water?
Answer:
a) Hydrogen, Oxygen
b) Water can be split into its components by passing electricity through it.

Question 8.
A) Which properties of metals are used in the following situations?
a) Copper is used to make bells.
b) Tungsten is used to make the filament of a light bulb.
Answer:
a) Sonority – the ability to produce a specific sound when struck with a hard object.
b) Ductility – the ability to be formed into a very thin sheet.

OR
B) Which properties of metals are utilised in the following cases?
a) Iron is used to make weapons.
b) Aluminium is used to make food packaging.
Answer:
a) Hardness
b) Malleability

Section – C

One question from 9 to 10 have choice. Each question carries 3 scores. (2 × 3 = 6)

Question 9.
Write what type of chemical reaction each of the following is:
a) Fireflies emit light.
b) When zinc and copper nails are inserted into a lemon and connected to an LED through a copper wire, the LED glows.
c) Ammonium chloride decomposes when heated strongly.
Answer:
a) Bioluminescence.
b) Electrochemical Reaction.
c) Thermochemical reactions.

Question 10.
A) A few polymers and their characteristics are given in the table. Complete the table based on their
characteristics.
[Rubber, Fibre, Plastics, PVC (Polyvinyl chloride)]

Characteristics	Polymer
Suitable for making strong threads.	____(a)____
A polymer that can be moulded into various shapes.	Plastic
Polymer with elastic nature.	____(b)____
Polymer with electrical resistance.	____(c)____

Answer:
a) Fibres
b) Rubber
c) PVC (Polyvinyl chloride)

OR
B) Consider the apparatus as shown in the figure. Heat the water in the flask.

a) When does the temperature remain constant?
b) What happens to the water when the temperature remains constant?
c) Why does the temperature not rise even when heat is supplied?
Answer:
a) At 100°C
b) Water starts boiling.
c) The temperature does not rise above 100°C because the heat supplied is being used to overcome the intermolecular forces of attraction between the water molecules and change the water’s state from liquid to gas.

BIOLOGY

Time: 40 minutes
Total Score: 20

Section – A

Choices are given for questions 7, 8, 10. Answer questions 1 and 2. Each question carries 1 score. (2× 1 = 2)

Question 1.
Scientists who advanced the theory of chemical evolution
a) Harold Urey, Stanley Miller
b) Oparin, Haldane
c) Sidney Fox, Joan Oro
d) Watson. Crick
Answer:
b) Oparin, Haldane

Question 2.
Examine the statement and the reason, and choose the correct answer.
Statement (S): Carl Linnaeus is known as the Father of Taxonomy.
Reason (R): Linnaeus introduced the different levels of classification and provided a scientific basis for classification. He also introduced the scientific method of naming organisms called Binomial Nomenclature.
a) Both S and R are true, and R is the correct explanation of S.
b) Both S and R are true, but R is not the correct explanation of S.
c) A is true, but S is false
d) A is false, but S is time
Answer:
a) Both S and R are true, and R is the correct explanation of S.

Section – B

Write answers to questions 3 to 8. Each question carries 2 score.

Question 3.
Over time, complex organisms evolved into the plants and animals we see today.
a) When is life thought to have originated on Earth?
b) What conditions led to the formation of complex organisms?
Answer:
a) Scientists believe that life originated on Earth about 3.5 billion years ago.
b) Some of the simpler organisms released oxygen through photosynthesis, which led to the formation of more complex organisms.

Question 4.
Observe the given figure and answer the following questions:

a) Identify A and B in the given figure.
b) Write about the shape and motility of A and B.
Answer:
a) A – Male gamete (Sperm)
B – Female gamete (Ovum/Egg)

b) Shape of A- Tadpole like shape, Shape of B – Spherical in shape.
Motility of A- Tail helps in the motility, B- It is not motile.

Question 5.
Choose and write the correct pairs.
(A) Matthias Jacob Schleiden discovered that all plants are made of cells
(B) Theodor Schwann – introduced the idea that new cells arise from existing cells
(C) Rudolf Virchow discovered that all animals are made of cells
(D) Ernst Ruska and Max Knoll invented the electron microscope.
Answer:
A and D

Question 6.
A) Complete the table given below:

Ecological Interaction	Examples
Predation	a _____
b _____	Mango tree and vanda
c _____	Mango tree and loranthus

Answer:
A) Deer and Tiger.
b) Commensalism,
c) Parasitism

B) What is trophic level? Which organism belongs to the first trophic level in the food chain
Answer:
Trophic level is the position of an organism in the food chain. l?lants belong to the first trophic level.

Question 7.
A. Differentiate between drip irrigation and wick irrigation.
Answer:
Drip irrigation – This is an irrigation method that uses pipes and valves to deliver water in droplets to the root zone. This minimises water loss and ensures that all plants receive the same amount of water. Wick irrigation This method delivers water directly from the water source to the root zone of plants using a cotton wick. It requires less water than drip irrigation.

OR
B. ‘Farming will be profitable only if all the pests are killed!’
What is your response to this comment from a farmer? How can effective pest control be implemented?
Answer:
I do not agree with this opinion of the farmer. Pest control methods should be selected taking into account the density and nature of the crops. The farmers’ need is not to kill the entire pests, but to control their growth in a way that does not damage the crops. Integrated pest control is a method that minimises the use of pesticides through automated pest control methods using various types of nets and traps, friendly insects, and the cultivation of seeds that are resistant to pests.

Question 8.
A. If there is a mistake in the statements given below, correct the mistake by changing only the underlined word.
a) The cell wall is the hard outer layer of the cell membrane.
b) The cell membrane provides protection and shape to the cell.
c) The cell wall in plants is mainly made of cellulose.
d) Substances enter and leave the cell through the Golgi apparatus.
Answer:
b) The cell wall provides protection and shape to the cell.
d) Substances enter and leave the cell through the cell membrane.

OR
B. Observe the picture and answer the questions below.

a) Which is the complex tissue represented by the picture?
b) What is the function of this complex tissue?
Answer:
a) Xylem
b) The conduction of water and salt to the leaves takes place through the xylem.

Section – C

Answer questions 9 and 10. Each question carries 3 score. (2 × 3 = 6)

Question 9.
Answer the following questions related to twins:
a) Give one characteristic that is always true for Identical twins regarding their sex.
b) Why is the external appearance of Fraternal twins usually different?
c) State the two major consequences (results) of the incomplete division or fusion that defines Conjoined twins.
Answer:
a) They will always belong to the same sex.

b) Fraternal twins develop from two different eggs fertilized by two different sperms. Therefore, they have different genetic material, which results in different external appearances and hereditary traits.

c) Body parts will be joined together, the external appearance and hereditary traits of the offspring will be the same.

Question 10.
A. Observe the illustration and answer the questions below.

a) Who performed the experiment shown in the illustration?
b) What is the source of energy in this experiment?
c) How does this experiment validate the theory of chemical evolution?
Answer:
a) Harold Urey and Stanley Miller
b) Electric spark
c) This experiment was conducted by artificially recreating the conditions of the primitive Earth in the laboratory. This experiment was able to produce amino acids, which are organic molecules. This experiment proved that organic molecules can be formed from inorganic components under suitable conditions.

Question 10.
B. Arrange the timeline properly.

Answer:

In the final stage of revision, Class 8 Maths Question Paper Pdf Kerala Syllabus Set 2 Malayalam Medium help manage time.

Kerala Syllabus Class 8 Maths Model Question Paper Set 2 Malayalam Medium

സമയം : 1½ മണിക്കൂർ
ആകെ സ്കോർ: 40

നിർദ്ദേശങ്ങൾ:

• ആദ്യത്തെ 15 മിനിറ്റ് ചോദ്യങ്ങൾ വായിച്ച് മനസ്സിലാക്കാനും ഉത്തരങ്ങളെക്കുറിച്ച് ചിന്തിക്കാനും ഉപയോഗിക്കുക.
• A, B, C, D എന്നീ നാലു ഭാഗങ്ങളിലായി 16 ചോദ്യങ്ങളാണുള്ളത്.
• എല്ലാ ചോദ്യങ്ങൾക്കും ഉത്തരമെഴുതണം. A അല്ലെങ്കിൽ B എന്ന തരത്തിലുള്ള ചോദ്യങ്ങളിൽ, ഏതെങ്കിലും ഒരു ചോദ്യത്തിന് ഉത്തരമെഴുതിയാൽ മതി.
• ചോദ്യങ്ങൾക്ക് ഏത് ക്രമത്തിലും ഉത്തരങ്ങൾ എഴുതാം. ചോദ്യത്തിന്റെ നമ്പർ കൃത്യമായി എഴുതിയാൽ മതി.
• ആവശ്യമുള്ള സ്ഥലങ്ങളിൽ ഉത്തരങ്ങൾ വിശദീകരിക്കണം.

Section – A
1 സ്കോർ വീതമുള്ള 4 ചോദ്യങ്ങളാണ് ഈ വിഭാഗത്തിലുള്ളത്.

Question 1.
താഴെ കൊടുത്തിരിക്കുന്നതിൽ എല്ലാ കോണുകളും എല്ലാ വശങ്ങളും തുല്യമായ ചതുർഭുജം ഏത്?
a) സമഭുജസാമാന്തരികം
b) ചതുരം
c) സാമാന്തരികം
d) സമചതുരം
Answer:
d) സമചതുരം
എല്ലാ കോണുകളും 90°

Question 2.
ഒരു സിറ്റിയിലെ തുടർച്ചയായ ഏഴ് ദിവസങ്ങളിലെ ഏറ്റവും കൂടിയ താപനില 26°C. ഏറ്റവും കുറഞ്ഞ താപനില 21°C. താപനിലയിലെ വ്യത്യാസം
(a) 6°
(b) 5°
(c) 10°
(d) 9°
Answer:
(b) 5°
26 – 21 = 5

Question 3.
രണ്ട് പ്രസ്താവനകൾ ചുവടെ നൽകിയിരിക്കുന്നു.
p₁ : ഒരു ബഹുഭുജത്തിന്റെ വശങ്ങളുടെ എണ്ണം എന്തുതന്നെയായാലും പുറം കോണുകളുടെ ആകെത്തുക 360° ആണ്.
p₂: വശങ്ങളുടെ എണ്ണം 1 വർദ്ധിക്കുകയാണെങ്കിൽ അകകോണുകളുടെ തുക 180 വർദ്ധിക്കുന്നു.
a) P₁ ഉം P₂, ഉം ശരിയാണ്.
b) P₁ ശരിയാണ് p₂ തെറ്റാണ്.
c) P₁ തെറ്റാണ് p₂, ശരിയാണ്.
d) p₁ ഉം P₂, ഉം തെറ്റാണ്.
Answer:
a) P₁ ഉം P₂, ഉം ശരിയാണ്.

Question 4.
∆ABC 8, AB : BC = 3:4, BC : CA = 8 : 9, എങ്കിൽ AB : BC : CA =
(a) 3 : 5 : 8
(b) 8 : 12 : 15
(c) 6 : 8 : 9
(d) 1 : 4 : 9
Answer:
c) 6:8:9
AB, BC എന്നിവയുടെ യഥാർഥ നീളം 3 സെന്റിമീറ്ററും, 4 സെന്റിമീറ്ററും ഒരേ സംഖ്യ കൊണ്ട് ഗുണിച്ചാൽ കിട്ടുന്ന സംഖ്യയാണ്. ഈ സംഖ്യയെ X എന്ന് സൂചിപ്പിക്കുമ്പോൾ, നമുക്ക് AB = 3x cm, BC = 4x cm ലഭിക്കും. വീണ്ടും BC, CA എന്നിവയുടെ യഥാർഥനീളം 8 സെന്റിമീറ്ററും, 8 സെന്റിമീറ്ററും മറ്റൊരു സംഖ്യ കൊണ്ട് ഗുണിച്ചാൽ കിട്ടുന്ന സംഖ്യ യാണ്. ഈ സംഖ്യയെ y എന്ന് സൂചിപ്പിക്കു മ്പോൾ, നമുക്ക് BC = 8y cm, CA = 9y cm ലഭിക്കും.
ഇപ്പോൾ, 4x = 8y ⇒ y = \(\frac{1}{2}\)x
അപ്പോൾ, CA = 9 × \(\frac{1}{2}\)x = \(\frac{9}{2}\)x
AB: BC: CA = 3x : 4x : \(\frac{9}{2}\)x = 6 : 8 : 9

Section – B
2 സ്കോർ വീതമുള്ള 4 ചോദ്യങ്ങളാണ് ഈ വിഭാഗത്തിലുള്ളത്.

Question 5.
x = \(\frac{1}{2}\) എങ്കിൽ x + x² കണ്ടുപിടിക്കുക. ഇതിനെ ദശാംശരൂപത്തിൽ എഴുതുക.
Answer:

Question 6.
8 സെ.മീ. നീളമുള്ള ഒരു ഞാൺ വൃത്തകേന്ദ്രത്തിൽ നിന്ന് 3 സെന്റിമീറ്റർ അകലെയാണ്. വൃത്തത്തിന്റെ വ്യാസം കണക്കാക്കുക.
Answer:
ഒരു വൃത്തത്തിന്റെ കേന്ദ്രത്തിൽ നിന്ന് വരയ്ക്കുന്ന ലംബം ഞാണിനെ സമഭാഗം ചെയ്യും.

OB² = 3² + 4² = 25
OB = 5 cm
അതിനാൽ, വ്യാസം = 10 സെ.മീ.

Question 7.
ചിത്രത്തിൽ AB = AD, BC = DC

(a) ∠BAC = 20° ആണെങ്കിൽ ∠A എന്താണ്?
Answer:
ABC, ADAC തുല്യതികോണങ്ങളാണ്.
∠BAC = ∠DAC = 20°
∠A = ∠BAC + ∠DAC = 40°

(b) ചിത്രത്തിൽ തുല്യതികോണങ്ങളുടെ പേരുകൾ നൽകുക.
Answer:
∆AOB, ∆AOD തുല്യതികോണങ്ങളാണ്.
∆BOC, ∆DOC തുല്യതികോണങ്ങളാണ്.
∆ABC, ∆ADC തുല്യതികോണങ്ങളാണ്.
∆ABD, ∆CBD തുല്യതികോണങ്ങളാണ്.

Question 8.
23² = 529
(i) 24² എത്രയാണ്?
(ii) 26² എത്രയാണ്?
Answer:
23² = 529
(i) 24² = (23 + 1)²
= 23² + 2 × 23 × 1 + 1²
= 529 + 46 + 1
= 576

(ii) 26² = (23 + 3)²
= 23² + 2 × 23 × 3 + 3²
= 529 + 138 + 9
= 676

Section – C
3 സ്കോർ വീതമുള്ള 4 ചോദ്യങ്ങളാണ് ഈ വിഭാഗത്തിലുള്ളത്.

Question 9.
ഒരു സമബഹുഭുജത്തിന്റെ പുറംകോണത്തിന്റെ അളവ് 2x-ഉം അകകോണത്തിന്റെ അളവ് 4x-ഉം ആണ്.
(a) x കണ്ടെത്താൻ അകവും പുറവുമായ കോണുകൾ തമ്മിലുളള ബന്ധം ഉപയോഗിക്കുക.
Answer:
ഒരു മൂലയിലെ അകക്കോണുകളുടെയും പുറം കോണുകളുടെയും തുക 180.
6x = 180, x = 30

(b) അകവും പുറവുമായ കോണുകളുടെ അളവ് കണ്ടെത്തുക.
Answer:
കോണുകൾ = 60, 120

(c) ബഹുഭൂജത്തിലെ വശങ്ങളുടെ എണ്ണവും ബഹുഭുജത്തിന്റെ തരവും കണ്ടെത്തുക.
Answer:
സമഷഡ്ഭുജം

Question 10.
ഒരു രണ്ടക്ക സംഖ്യയുടെ അക്കങ്ങളുടെ തുക 10 ആണ്. അക്കങ്ങളുടെ സ്ഥാനം മാറ്റിയാൽ രണ്ടക്ക സംഖ്യ ആദ്യത്തെ സംഖ്യയേക്കാൾ 36 കുറവാണ്. ആദ്യത്തെ സംഖ്യ കണക്കാക്കുക.
Answer:
പത്തിന്റെ സ്ഥാനത്ത് അക്കം x ആണ്. ഒറ്റയുടെ
സ്ഥാനത്ത് അക്കം 10 – x ആണ്.
സംഖ്യ = 10x + 10 – x
അക്കങ്ങൾ തിരിച്ചിട്ടാൽ സംഖ്യ 10(10 – x) + x ആയി മാറുന്നു.
10(10 – x) + x = 10x + 10 – x – 36
100 – 10x + x = 9x – 26, 100 – 9x = 9x – 26,
126 = 18x, x = \(\frac{126}{18}\) = 7
സംഖ്യ = 73

Question 11.
താഴെ കൊടുത്തിരിക്കുന്ന ചിത്രത്തിൽ നാല് തുല്യ ചതുരങ്ങൾ ചേർത്തുവെച്ച് സമചതുരമുണ്ടാക്കുന്നു. നീളം + വീതി എത്?

അകത്തെ സമചതുരത്തിന്റെ പരപ്പളവ് പുറത്തെ സമചതുരത്തിന്റെ പരപ്പളവിന്റെ \(\frac{1}{4}\) ഭാഗമാണ്. ചതുര ത്തിന്റെ വലിയ വശത്തിന്റെയും ചെറിയ വശത്തിന്റെയും നീളങ്ങൾ തമ്മിലുള്ള അംശബന്ധം കണക്കാക്കുക.
Answer:
ചതുരത്തിന്റെ വലിയ വശം (നീളം) (ഉം ചെറിയ വശം (വീതി) b ഉം എന്നെടുക്കുക.
അകത്തെ സമചതുരത്തിന്റെ വശം = l – b
അകത്തെ സമചതുരത്തിന്റെ പരപ്പളവ് = (l – b)
പുറത്തെ സമചതുരത്തിന്റെ വശം = l + b
പുറത്തെ സമചതുരത്തിന്റെ പരപ്പളവ് = (l + b)
അകത്തെ സമചതുരത്തിന്റെ പരപ്പളവ് പുറത്ത
സമചതുരത്തിന്റെ പരപ്പളവിന്റെ \(\frac{1}{4}\) ഭാഗമാണ്.
\(\frac{(l-b)^2}{(l+b)^2}=\frac{1}{4}\)
\(\frac{l-b}{l+b}=\frac{1}{2}\)
⇒ 2(l – b) = l + b
⇒ 2l – l = b + 2b
⇒ l = 3b
\(\frac{l}{b}=\frac{3}{1}\)
അതിനാൽ, ചതുരത്തിന്റെ വലിയ വശവും ചെറിയ വശവും തമ്മിലുള്ള അംശബന്ധം 3:1 ആണ്.

OR

രണ്ടു തുല്യ സമഭുജസാമാന്തരികം ഉപയോഗിച്ച് വരച്ച ചിത്രമാണ് ചുവടെ കൊടുത്തിരിക്കുന്നത്. തന്നി രിക്കുന്ന അളവുകൾ ഉപയോഗിച്ച് ചിത്രം വരയ്ക്കുക.

Answer:

ചിത്രത്തിൽ കാണിച്ചിരിക്കുന്നതുപോലെ ആദ്യം AB = 4 സെ.മീ. വരയ്ക്കുക. തുടർന്ന് B യിൽ 60° യും, A യിൽ 120° യും എടുക്കുക. ചിത്രം പൂർത്തി യാക്കുക.

Question 12.
ചുവടെ കൊടുത്തിരിക്കുന്ന പാറ്റേൺ നോക്കൂ. ഇവ ഭിന്നസംഖ്യകളുടെ ദശാംശരൂപമാണ്.
\(\frac{1}{11}\) = 0.0909…….
\(\frac{2}{11}\) = 0.1818…..
\(\frac{3}{11}\) = 0.2727…..
……………………..
(a) അടുത്തവരി എഴുതുക.
Answer:
\(\frac{4}{11}\) = 0.3636………

(b) 0.6363……യുടെ ഭിന്നസംഖ്യ രൂപം എഴുതുക.
Answer:
\(\frac{7}{11}\)

(c) 0.8181… + 0.7272…ന്റെ ദശാംശരൂപം എഴുതുക.
Answer:
\(\frac{9}{11}+\frac{8}{11}=\frac{17}{11}=1+\frac{6}{11}\) = 1.5454…….

OR

ഒരു ത്രികോണത്തിലെ കോണുകളുടെ അംശബന്ധം 2:3:5 ആണ്. ഓരോ കോണും എത്ര വീതമാണ്?
Answer:
180 × \(\frac{2}{10}\) = 36°
180 × \(\frac{3}{10}\) = 54°
180 × \(\frac{5}{10}\) = 90°

Section – D
4 സ്കോർ വീതമുള്ള 4 ചോദ്യങ്ങളാണ് ഈ വിഭാഗത്തിലുള്ളത്.

Question 13.
ഒരു വശത്തിന്റെ നീളം 9.5 സെ.മീ. ആ വശത്തേക്കുള്ള ഉന്നതി 4.75 സെ.മീ. ആയ സമപാർശ്വത്രികോണം വരക്കുക.
Answer:
9.5 സെ.മീ. നീളമുള്ള ഒരു വര വരയ്ക്കുക. ഈ വരയുടെ ലംബസമഭാജി വരയ്ക്കുക. അതിന്റെ ലംബത്തിന്റെ മുകളിൽ വരയുടെ പകുതി നീളം അടയാളപ്പെടുത്തുക. ഈ ബിന്ദുവിൽ നിന്ന് വര യുടെ രണ്ട് അറ്റങ്ങളും ബന്ധിപ്പിച്ച് ഒരു ത്രികോണം രൂപപ്പെടുത്തുക.

Question 14.
വികർണങ്ങളുടെ നീളം 6 സെ.മീ, 8 സെ.മീ. ആയ സമഭുജസാമാന്തരികം വരയ്ക്കുക.
Answer:
6 സെ.മീ. നീളമുള്ള വരയും അതിന്റെ ലംബസമ ഭാജിയും വരയ്ക്കുക. മധ്യബിന്ദുവിൽ നിന്ന് 4 സെ. മീ. അകലെ മുകളിലും താഴെയും രണ്ട് ബിന്ദു ക്കൾ അടയാളപ്പെടുത്തുക. ഈ ബിന്ദുക്കൾ ആദ്യത്തെ വികർണത്തിന്റെ അറ്റങ്ങളുമായി ബന്ധി പ്പിക്കുക. നമുക്ക് സമഭുജസാമാന്തരികം ലഭിക്കും.

Question 15.
ഒരു പ്രദേശത്തെ 30 കുടുംബങ്ങളുടെ വൈദ്യുതി ഉപഭോഗത്തിന്റെ വിവരങ്ങൾ പട്ടികയിൽ നൽകിയിരി ക്കുന്നു. ആവൃത്തി ചതുരം വരയ്ക്കുക.

വൈദ്യുതി ഉപഭോഗം	കുടുംബങ്ങളുടെ എണ്ണം
100 – 200	5
200 – 300	7
300 – 400	8
400 – 500	6
500 – 600	4

Answer:

OR

തന്നിരിക്കുന്ന ചിത്രത്തിൽ ചതുരത്തിന്റെ വശങ്ങളുടെ മധ്യബിന്ദുക്കൾ യോജിപ്പിച്ച് ഒരു സമഭുജസാമാ ന്തരികം ഉണ്ടാക്കിയിരിക്കുന്നു. AB = 6 സെ.മീ, BC = 4 സെ.മീ.

(a) ചതുരത്തിന്റെ പരപ്പളവ് എന്താണ്?
(b) സമഭുജസാമാന്തരികത്തിന്റെ പരപ്പളവ് കണ്ടെത്തുക.
(c) വികർണങ്ങളുടെ നീളം 12 സെന്റിമീറ്ററും 8 സെന്റീമീറ്ററും ആയ സമഭുജസാമാന്തരികത്തിന്റെ പരപ്പ ളവ് കണ്ടെത്തുക.
Answer:
(a) 6 × 4 = 24 cm²
(b) \(\frac{1}{2}\) × 24 = 12 cm²
(c) \(\frac{1}{2}\) × 12 × 8 = 48 cm²

Question 16.
താഴെ കൊടുത്തിരിക്കുന്ന ഗണിതശാസ്ത്ര ആശയം വായിച്ച് തുടർന്നുള്ള നാല് ചോദ്യങ്ങൾക്ക് ഉത്തരം നൽകുക. ഓരോ ചോദ്യത്തിനും 1 സ്കോർ ഉണ്ട്. (4 x 1 = 4)
1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
…………………
…………………
1 മുതൽ ആരംഭിക്കുന്ന തുടർച്ചയായ എണ്ണൽ സംഖ്യകളുടെ ആകെത്തുകയായി എഴുതാൻ കഴിയുന്ന സംഖ്യകളെ ത്രികോണസംഖ്യകൾ എന്ന് വിളിക്കുന്നു.
1² = 1
2² = 4
3² = 9.
…………………
…………………
എണ്ണൽ സംഖ്യകളുടെ വർഗ്ഗങ്ങൾ വർഗ്ഗസംഖ്യകൾ എന്നറിയപ്പെടുന്നു. അതായത്
1, 3, 6, 10 എന്നിവ ത്രികോണസംഖ്യകളാണ്.
1, 4, 9 എന്നിവ വർഗ്ഗ സംഖ്യകളാണ്.
(a) നാലാമത്തെ ത്രികോണസംഖ്യ ഏതാണ്?
(b) അഞ്ചാമത്തെ ത്രികോണസംഖ്യ ലഭിക്കാൻ എത്ര ആദ്യ എണ്ണൽ സംഖ്യകൾ ചേർക്കണം?
(c) ആറാമത്തെ വർഗ്ഗസംഖ്യ ഏതാണ്?
(d) ത്രികോണസംഖ്യയും വർഗ്ഗസംഖ്യയും ആയ രണ്ട് അക്കസംഖ്യ കണ്ടെത്തുക.
(e) ആറാമത്തെ വർഗ്ഗസംഖ്യയും ആറാമത്തെ ത്രികോണസംഖ്യയും തമ്മിലുള്ള വ്യത്യാസം എന്താണ്?
(f) 70-നും 90-നും ഇടയിലുള്ള ഒരു വർഗ്ഗസംഖ്യ എഴുതുക.
Answer:
(a) 10
(b) 5
(c) 36
(d) 36
(e) 15
(f) 81

OR

3² – 1² = 4 × 2
4² – 2² = 4 × 3
5² – 3² = 4 × 4
6² – 4² = 4 × 5
…………………
…………………
(i) ഈ പാറ്റേണിന്റെ അടുത്ത രണ്ട് വരികൾ എഴുതുക.
(ii) ഈ പാറ്റേണിന് പിന്നിലെ തത്വം എന്താണ്?
(iii) ബീജഗണിതം ഉപയോഗിച്ച് ഇത് വിശദീകരിക്കുക.
Answer:
(i) 7² – 5² = 4 × 6
8² – 6² = 4 × 7
(ii) ഒന്നിടവിട്ട് എണ്ണൽ സംഖ്യകളുടെ വർഗ്ഗങ്ങ ളുടെ വ്യത്യാസം എല്ലായ്പ്പോഴും 4-ന്റെ ഗുണി തമായിരിക്കും.
(iii) (n + 2)² − n² = 4(n + 1)

In the final stage of revision, Class 8 Maths Question Paper Pdf Kerala Syllabus Set 1 Malayalam Medium help manage time.

Kerala Syllabus Class 8 Maths Model Question Paper Set 1 Malayalam Medium

സമയം : 1½ മണിക്കൂർ
ആകെ സ്കോർ: 40

നിർദ്ദേശങ്ങൾ:

• ആദ്യത്തെ 15 മിനിറ്റ് ചോദ്യങ്ങൾ വായിച്ച് മനസ്സിലാക്കാനും ഉത്തരങ്ങളെക്കുറിച്ച് ചിന്തിക്കാനും ഉപയോഗിക്കുക.
• A, B, C, D എന്നീ നാലു ഭാഗങ്ങളിലായി 16 ചോദ്യങ്ങളാണുള്ളത്.
• എല്ലാ ചോദ്യങ്ങൾക്കും ഉത്തരമെഴുതണം. A അല്ലെങ്കിൽ B എന്ന തരത്തിലുള്ള ചോദ്യങ്ങളിൽ, ഏതെങ്കിലും ഒരു ചോദ്യത്തിന് ഉത്തരമെഴുതിയാൽ മതി.
• ചോദ്യങ്ങൾക്ക് ഏത് ക്രമത്തിലും ഉത്തരങ്ങൾ എഴുതാം. ചോദ്യത്തിന്റെ നമ്പർ കൃത്യമായി എഴുതിയാൽ മതി.
• ആവശ്യമുള്ള സ്ഥലങ്ങളിൽ ഉത്തരങ്ങൾ വിശദീകരിക്കണം.

Section-A
1 സ്കോർ വീതമുള്ള 4 ചോദ്യങ്ങളാണ് ഈ വിഭാഗത്തിലുള്ളത്.

Question 1.
വിഭാഗങ്ങളും ആവൃത്തിയും കാണിക്കുന്ന പട്ടികയാണ്
a) വൃത്തചിത്രങ്ങൾ
b) ചതുര ചിത്രങ്ങൾ
c) ആവൃത്തി ചതുരം
d) ആവൃത്തി പട്ടിക
Answer:
d) ആവൃത്തി പട്ടിക

Question 2.
ഏത് പ്രസ്താവനയാണ് ശരി?
(a) വൃത്തകേന്ദ്രത്തിൽ നിന്ന് വ്യത്യസ്ത അകലങ്ങളിൽ ഉള്ള ഞാണുകൾക്ക് ഒരേ നീളമാണ്.
(b) വൃത്തകേന്ദ്രത്തിൽ നിന്ന് ഒരേ അകലത്തിൽ ഉള്ള ഞാണുകൾക്ക് ഒരേ നീളമാണ്.
(c) ഒരു ഞാണിന്റെ ലംബസമഭാജി ഒരിക്കലും വൃത്തകേന്ദ്രത്തിലൂടെ കടന്നുപോകില്ല.
(d) ആരം എല്ലായ്പ്പോഴും ഞാണിന് ലംബമായിരിക്കും.
Answer:
(b) വൃത്തകേന്ദ്രത്തിൽ നിന്ന് ഒരേ അകലത്തിൽ ഉള്ള ഞാണുകൾക്ക് ഒരേ നീളമാണ്.

Question 3.
രണ്ട് പ്രസ്താവനകൾ താഴെ കൊടുത്തിരിക്കുന്നു.
(i) എല്ലാ സാമാന്തരികങ്ങളുടേയും വികർണ്ണങ്ങൾ പരസ്പരം ലംബമായി സമഭാഗം ചെയ്യുന്നു.
(ii) സമഭുജസാമാന്തരികത്തിന്റെ വികർണ്ണങ്ങൾ പരസ്പരം ലംബമായി സമഭാഗം ചെയ്യുന്നു.
a) (i) ഉം (ii) ഉം ശരിയാണ്.
b) (i) മാത്രം ശരിയാണ്.
c) (ii) മാത്രം ശരിയാണ്.
d) (i) ഉം (ii) ഉം ശരിയല്ല
Answer:
c) (ii) മാത്രം ശരിയാണ്.

Question 4.
രണ്ട് വശങ്ങൾ തുല്യമായ ത്രികോണത്തിന്റെ ചെറിയ കോൺ 30 ആയാൽ തുല്യമായ കോണുകളിൽ ഒരെണ്ണം.
(a) 100°
(b) 75°
(c) 70°
(d) 50°
Answer:
(b) 75°

കോണുകളുടെ തുക = 180°
ചെറിയ കോൺ = 30
തുല്യമായ കോണുകളുടെ തുക = 150°
തുല്യമായ കോണുകൾ = 75°

Section – B
2 സ്കോർ വീതമുള്ള 4 ചോദ്യങ്ങളാണ് ഈ വിഭാഗത്തിലുള്ളത്.

Question 5.
\(\frac{1}{7}-\frac{3}{7}+\frac{5}{7}-\frac{7}{7}+\ldots \ldots+\frac{21}{7}\) കണക്കാക്കുക.
Answer:
\(\frac{1-3+5-7+9-11+13-15+17-19+21}{7}\)
= \(\frac{-10+21}{7}=\frac{11}{7}\)

Question 6.
ഒരു സംഖ്യയിൽ നിന്നും സംഖ്യയുടെ \(\frac{2}{3}\) ഭാഗം കുറച്ചാൽ 13 കിട്ടും. സംഖ്യ ഏത്?
Answer:
ഒരു സംഖ്യയുടെ \(\frac{1}{3}\) = 13
സംഖ്യ = 3 × 13 = 39

Question 7.
ഒരു ത്രികോണത്തിന്റെ കോണുകൾ തമ്മിലുള്ള അംശബന്ധം 4:5:9 എന്ന അംശബന്ധത്തിലാണ്.
(a) കോണുകൾ x എന്ന സംഖ്യയുടെ ഗുണിതമായി എഴുതുക.
Answer:
കോണുകൾ 4x, 5x, 9x

(b) കോണുകൾ കണക്കാക്കുക.
Answer:
കോണുകളുടെ തുക = 180
⇒ 4x + 5x + 9x = 180
⇒ 18x = 180
⇒ x = 10
അതിനാൽ കോണുകൾ 40°, 50°, 90°

Question 8.
ചുവടെ കൊടുത്തിരിക്കുന്ന ലംബകം വരയ്ക്കുക.

Answer:

Section – C
3 സ്കോർ വീതമുള്ള 4 ചോദ്യങ്ങളാണ് ഈ വിഭാഗത്തിലുള്ളത്.

Question 9.
തുടർച്ചയായ രണ്ട് എണ്ണൽ സംഖ്യകളുടെ തുക 63 ആണ്.
(a) ചെറിയ സംഖ്യ x ആയാൽ മറ്റേ സംഖ്യ ഏത്?
Answer:
സംഖ്യകൾ x, x + 1

(b) സമവാക്യം എഴുതി സംഖ്യകൾ കണക്കാക്കുക.
Answer:
2x + 1 = 63, 2x = 62, x = 31
സംഖ്യകൾ 31, 32

Question 10.
52 വിദ്യാർത്ഥികളുള്ള ഒരു ക്ലാസ്സിൽ ആൺകുട്ടികളുടെ എണ്ണം പെൺകുട്ടികളുടെ എണ്ണത്തിന്റെ 6/7 ആണ്. എത്ര ആൺകുട്ടികളുണ്ട്?
Answer:
ആൺകുട്ടികളുടെ എണ്ണം = x
പെൺകുട്ടികളുടെ എണ്ണം = 52 – x
x = \(\frac{6}{7}\) × (52 – x), 7x = 6(52 − x)
7x = 312 – 6x, 13x = 312, x = 24
ആൺകുട്ടികളുടെ എണ്ണം = 24
പെൺകുട്ടികളുടെ എണ്ണം = 52 – 24 = 28

Question 11.
ഇത് ഒരു സമബഹുഭുജത്തിന്റെ ഭാഗമാണ്.

(a) ത്രികോണത്തിന്റെ കോണുകൾ ഏതൊക്കെയാണ്?
Answer:
18°, 18°, 144°

(b) ഒരു പുറംകോൺ എന്താണ്?
Answer:
180 – 144 = 36°

(c) ഈ ബഹുഭുജത്തിന് എത്ര വശങ്ങളുണ്ട്?
Answer:
\(\frac{360}{36}\) = 10

OR

ഈ ചിത്രം ഒരു സമഭുജത്രികോണവും ഒരു ചതുരവും ചതുരത്തിന്റെ ഒരു വികർണ്ണവും ചേർന്നതാണ്. x-ന്റെ മൂല്യം എന്താണ്?

Answer:
സമഭുജത്രികോണത്തിലെ ഒരു കോൺ 60 സമചതുരത്തിലെ ഒരു കോൺ 90 വികർണം അതിനെ 45 ആയി ഭാഗിക്കുന്നു.
x = 60 + 45 = 105°

Question 12.
ചിത്രത്തിൽ ABCD ഒരു സാമാന്തരികമാണ്. AB = 6 സെ.മീ. AB യും CD യും തമ്മിലുള്ള അകലം 3 സെ. മീ. AD യും BC യും തമ്മിലുള്ള അകലം 2 സെ.മീ. സാമാന്തരികത്തിന്റെ ചുറ്റളവ് കണ്ടെത്തുക.

Answer:
സാമാന്തരികം ABCD യുടെ പരപ്പളവ്
= 6 × 3 = 18 ച. സെ.മീ.
അതുപോലെ തന്നെ, പരപ്പളവ് = BC × 2 = 18
⇒ BC = 18 + 2 = 9cm
AB = CD = 6cm
AD = BC = 9cm
∴ ചുറ്റളവ് = 2 (649) = 30 സെ.മീ.

OR

ഒരു സമഭുജസാമാന്തരികത്തിന്റെ പരപ്പളവ് 120 ചതുരശ്ര സെന്റിമീറ്റർ ആണ്. അതിന്റെ ഒരു വികർണ ത്തിന്റെ നീളം 24 സെ.മീ. ആണ്.
(a) അതിന്റെ രണ്ടാമത്തെ വികർണത്തിന്റെ നീളം കണ്ടെത്തുക.
Answer:
പരപ്പളവ് = 120 ച.സെ.മീ.
⇒ \(\frac{1}{2}\) × d₁ × d₂ = 120
⇒ \(\frac{1}{2}\) × 24 × d₂ = 120
⇒ d₂ = 120 ÷ 12 = 10cm

(b) ചുറ്റളവ് കണ്ടെത്തുക.
Answer:
സമഭുജസാമാന്തരികത്തിന്റെ ഒരു വശം
= \(\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}\) = 13cm
∴ ചുറ്റളവ് = 4 × 13 = 52cm

Section – D
4 സ്കോർ വീതമുള്ള 4 ചോദ്യങ്ങളാണ് ഈ വിഭാഗത്തിലുള്ളത്.

Question 13.
ചുവടെ കൊടുത്തിരിക്കുന്ന ഘട്ടങ്ങൾ പരിശോധിക്കുക.
\(\frac{1}{3}=\frac{1}{10} \times \frac{10}{3}=\frac{1}{10}\left(3+\frac{1}{3}\right)=\frac{3}{10}+\frac{1}{30}\)
\(\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)
(a) രണ്ടു ഘട്ടങ്ങൾ കൂടി എഴുതി \(\frac{1}{3}\) നെ ദശാംശരൂപത്തിൽ എഴുതുക.
Answer:

\(\frac{1}{3}\) ന്റെ ദശാംശരൂപ = 0.3333……….

(b) \(\frac{1}{9}, \frac{2}{9}\), എന്നിവയുടെ ദശാംശരൂപം എഴുതുക.
Answer:
\(\frac{1}{9}\) = 0.111…………
\(\frac{2}{9}\) = 0.222…………

(c) \(\sqrt{0.444 \ldots}\) കണക്കാക്കുക
Answer:
\(\sqrt{0.444 \ldots .}=\sqrt{\frac{4}{9}}=\frac{2}{3}=\frac{6}{9}\)
= 0.666………….

Question 14.
ഒരു ത്രികോണത്തിന്റെ ഒരു വശം 6 സെ.മീ. നീളവും, ആ വശത്തിന്റെ അറ്റത്തുള്ള കോണുകൾ 40ഉം 70 ഉം ആണ്. ത്രികോണം വരച്ച് അതിന്റെ പരിവൃത്തം നിർമ്മിക്കുക.
Answer:

Question 15.
(1) 11² + 10² = 121 + 100 = 221 ഇതുപയോഗിച്ച് 442നെ രണ്ട് പൂർണ്ണവർഗ്ഗങ്ങളുടെ തുകയായി എഴുതുക.
(ii) 5ൽ അവസാനിക്കുന്ന ഒരു സംഖ്യയുടെ വർഗം 110pq എന്ന രൂപത്തിലെഴുതിയാൽ pq=
(iii) ചുവടെപ്പറയുന്ന വർഗങ്ങൾ കണക്കാക്കുക.
(a) (7.6)²
(b) (75.4)²
Answer:
(i) 442 = 2 × 221 = 2(11² + 10²)
= (11 + 10)² + (11 – 10)²
442 = 21² + 1²

(ii) p × q = 2 × 5 = 10

(iii) (a) 7.6²
= (7 + 0.6)²
7² + 2 × 7 × 0.6 + (0.6)²
49 + 8.4 + 0.36
= 57.76

(b) (75.4)² = (75 + 0.4)²
= 75² + 2 × 75 × 0.4+ (0.4)²
= 5625 + 60 + 0.16.
= 5685.16

OR

ഒരു ചതുരത്തിന്റെ ചുറ്റളവ് 48 സെന്റിമീറ്ററാണ്. നീളവും വീതിയും തമ്മിലുള്ള അംശബന്ധം 5:3 ആണ്.
(a) നീളം + വീതി
(b) നീളവും വീതിയും കണക്കാക്കുക.
(c) വീതി 1 സെ.മീ. വർദ്ധിപ്പിച്ചാൽ, നീളവും വീതിയും തമ്മിലുള്ള അംശബന്ധം കണ്ടെത്തുക.
Answer:
(a) നീളം + വീതി = \(\frac{48}{2}\) = 24
(b) നീളം : വീതി = 5 : 3
നീളം = 24 × \(\frac{5}{8}\) = 15 cm
വീതി = 24 × \(\frac{3}{8}\) = 9 cm

(c) വീതി 1 സെ.മീ. വർദ്ധിപ്പിച്ചാൽ,
വീതി = 10 സെ.മീ.
നീളം : വീതി = 15: 10 = 3:2

Question 16.
ഒരു പരീക്ഷയിൽ 25 കുട്ടികൾക്ക് ലഭിച്ച മാർക്ക് ചുവടെ ചേർക്കുന്നു. ആവൃത്തി വിഭാഗ പട്ടിക തയ്യാറാക്കി ചുവടെയുള്ള ചോദ്യങ്ങൾക്ക് ഉത്തരമെഴുതുക. (ആകെ മാർക്ക് 40)
25, 30, 33, 18, 36
12, 25, 30, 15, 25
19, 25, 30, 33, 20
35, 37, 29, 31, 30
31, 28, 33, 31, 30
(i) എത്ര കുട്ടികൾക്ക് 15 മാർക്കിൽ താഴെ ലഭിച്ചു?
(ii) എത്ര കുട്ടികൾക്ക് 30 മാർക്കോ അതിൽ കൂടുതലോ ലഭിച്ചു?
(ii) എത്ര കുട്ടികൾക്ക് 25 മാർക്കിൽ കൂടുതൽ ലഭിച്ചു
(iv) എത്ര കുട്ടികൾക്ക് 30നും 40 നു ഇടയിൽ മാർക്ക് ലഭിച്ചു?
Answer:

മാർക്ക്	കുട്ടികളുടെ എണ്ണം
10 – 15	1
15 – 20	3
20 – 25	1
25 – 30	6
30 – 35	11
35 – 40	3
ആകെ	25

(i) 1 കുട്ടി
(ii) 14 കുട്ടികൾ
(iii) 20 കുട്ടികൾ
(iv) 14 കുട്ടികൾ

OR

ചില എണ്ണൽ സംഖ്യകളെ രണ്ട് പൂർണ്ണവർഗ്ഗങ്ങളുടെ വ്യത്യാസമായി എഴുതാം. ഉദാഹരണത്തിന്,
8 = 4 × 2 × 1 = 3² – 1²
12 = 4 × 3 × 1 = 4² – 2²
16 = 4 × 4 × 1 = 5² – 3²
(i) 20നെ രണ്ട് പൂർണ്ണവർഗ്ഗങ്ങളുടെ വ്യത്യാസമായി എഴുതുക.
(i) 8ൽ തുടങ്ങി 4ന്റെ എല്ലാ ഗുണിതങ്ങളും രണ്ട് പൂർണ്ണവർഗങ്ങളുടെ വ്യത്യാസമായി എഴുതുന്ന രീതി ബീജഗണിതം ഉപയോഗിച്ച് വിശദീകരിക്കുക.
Answer:
(a) 20 = 4 × 5 × 1 = 6² −4²
(b) 4 × x × 1 = (x + 1)² -(x – 1)²
(x + 1)² – (x – 1)²
= (x + 1 + x – 1)(x + 1 – [x – 1])
=2x – 2
= 4 × x × 1
(iii) (n + 2)² – n² = 4(n + 1)

In the final stage of revision, Class 8 Maths Question Paper Pdf Kerala Syllabus Set 5 help manage time.

Class 8 Maths Kerala Syllabus Model Question Paper Set 5

Times: 1½ hr.
Score: 40

Instructions

• Use the first 15 minutes to read the questions and think about answers.
• There are 16 questions, split in to 4 parts A, B, C, D
• Answer all questions; but in questions of the type A or B you need answer only one of those.
• You can answer the questions in any order, writing the correct question number.
• Answers must be explained, whenever necessary.

Section – A
This section has 4 questions of score J each. Select the correct answer.

Question 1.
305² = 93025
(30.5)² = ………….
(a) 93.025
(b) 9.3025
(c) 930.25
(d) 9302.5
Answer:
(c) 930.25

Question 2.
One third of a number is 10 less than the number. Which of the following is the number?
(a) 30
(b) 15
(c) 10
(d) 27
Answer:
\(\frac{2}{3}\) of the number is 10
Number is \(\frac{3}{2}\) times 10.
Number = 10 × \(\frac{3}{2}\) = 15

Question 3.
Read the given statements.
Statement I: A mixture of milk and water has a ratio of 3:1. If 5 liters of water are added, the new ratio of milk to water is 3:2. The quantity of milk in the mixture is 15 liters.
Statement II: The addition of water does not change the quantity of milk in the mixture.
A) Statement I is true, Statement II is false
B) Statement I is false, Statement II is true
C) Statement I is true, Statement II is the reason for Statement I
D) Statement I is true, Statement II is not the reason for Statement I
Answer:
C) Statement I is true, Statement II is the reason for Statement I

Milk = 3x
Water = x
If 5 litres of water are added,
Milk = 3x
Water = x + 5
The new ratio of milk to water is 3:2.
That is milk is \(\frac{3}{2}\) of water.
⇒ 3x = \(\frac{3}{2}\) × (x + 5)
⇒ 6x = 3x + 15
⇒ 3x = 15
⇒ x = 5
Milk = 3 × 5 = 15 litres
So, Statement I is true.
Statement II says that adding water to a mixture of milk and water does not change the quantity of milk. This is a fundamental concept of mixture problems where only one component is added or removed. The amount of the other component remains constant. Therefore, Statement II is true.
Hence, Statement I is true, Statement II is the reason for Statement I

Question 4.
If x = 0.333 …, then x² + x =
(a) \(\frac{1}{3}\)
(b) \(\frac{4}{9}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{9}\)
Answer:
(b) \(\frac{4}{9}\)

Section – B
This section has 4 questions of score 2 each.

Question 5.
In the figure AB = AC, BD = CD
(a) Name two equal triangles in the figure
(b) If ∠BAC = 40°, ∠ABD = 30° then what are the angles of triangle ADB ?

Answer:
(a) AB = AC, BD = CD, AD is common
Triangle ABD and triangle ADC are equal.

(b) In triangle ADB, ∠A = 20°, ∠B = 30°, ∠D = 130°

Question 6.
A polygon has inner angle sum is two times the outer angle sum.
(a) What is its inner angle sum ?
Answer:
2 × 360 = 720°

(b) Find the number of sides of the polygon
Answer:
180 × (n – 2) = 720
n – 2 = 4, n = 6

Question 7.
A chord that is 8 cm long lies 3 cm away from the centre of a circle. Calculate the diameter of the circle.
Answer:
The perpendicular drawn from the centre to the chord bisects the chord.

OB² = 3² + 4² = 25
OB = 5 cm
Therefore the diameter = 2 × 5 = 10 cm

Question 8.
In the figure ABCD is a parallelogram. The area of the parallelogram is 36 cm². The distance between parallel sides is 3 cm. Find the length of the side AB.

Answer:
Area = 36 cm²
3 × base = 36
So, base = 36 ÷ 3 = 12 cm
∴ AB = 12 cm

Section – C
This section has 4 questions of socre 3 each.

Question 9.
Given below is an unfinished magic square made up of positive numbers and negative numbers. The sum of the numbers in rows, columns and diagonals are equal.

Find x + y?
Answer:
Consider the number that can be written in the top right side of the column be A.
—7 + 6 + A = x + y + A
⇒ -7 + 6 = x + y
x + y = 6 – 7 = – 1

Question 10.
Draw a square with perimeter is 23 centimeter.
Answer:
Draw a line of length 11.5 centimeter.
Draw the perpendicular bisector of the line. Mark the half length of the line on the perpendicular bisector. Complete the square.
One side = \(\frac{11.5}{2}\) cm, Perimeter = 4 x \(\frac{11.5}{2}\) = 23cm

Question 11.
[A] The perimeter of a rectangle is 28 cm. The ratio between length and breadth is 5:2.
(a) What is the sum of length and breadth?
(b) Find length and breadth.
(c) Find the area of the rectangle.
Answer:
Length = 5x
Breadth = 2x

(a) Perimeter = 28
⇒ 2(length + breadth) = 28
⇒ length + breadth = 14

(b) 5x + 2x = 14
⇒ 7x = 14
⇒ x = 2
Length = 10 cm Breadth = 4 cm

(c) Area = 10 × 4 = 40 sq.cm

OR

[B] Sides of a triangular plot are in the ratio 2:4:6. Perimeter of the plot is 360 metres. Find the length of the sides?
Answer:
Let the sides be 2x, 4x, 6x.
Perimeter = 360 metres
2x + 4x + 6x = 360
⇒ 12x = 360
⇒ x = 30
So, the length of the sides are:
2 × 30 = 60 metre
4 × 30 = 120 metre
6 × 30 = 180 metre

Question 12.
[A] Find the value of (1\(\frac{1}{2}\))² + (2\(\frac{1}{2}\))² + (3\(\frac{1}{2}\))² + (4\(\frac{1}{2}\))²
Answer:

OR

[B] (a) 3² + 2² = 9 + 4 = 13. By using this , write 26 as the sum of the two perfect squares
Answer:
26 = 2 × 13
= 2 (3² + 2²)
= (3 + 2)² + (3 – 2)²
26 = 5² + 1²

(b) (99\(\frac{1}{2}\))² + (100\(\frac{1}{2}\))² = _______________
Answer:
= 99 × 100 + \(\frac{1}{4}\) + 100 × 101 + \(\frac{1}{4}\)
= 9900 + \(\frac{1}{4}\) + 10100 + \(\frac{1}{4}\)
= 20000\(\frac{1}{2}\)
= 20000.5

Section – D
This section has 4 questions of score 4 each.

Question 13.
In the figure D is the midpoint of BC, DL is perpendicular to AB, DM is perpendicular to AC, DL = DM

(a) Prove that triangle BLD and triangle CMD have same side and shape.
Answer:
BD = CD = x, DL = DM = y
Triangle BLD and triangle CMD are right triangles
BD² = BL² + DL²,
CD² = DM² + CM²
Since D is the mid point, BD = CD
BL² + DL² = DM² + CM²,
BL² = CM² ⇒ BL = CM
Three sides of triangle BLD are equal to three sides of triangle CMD. So these triangles are equal

(b) Is ∠B = ∠C?
Answer:
Sides opposite to equal angles are equal, zB = zC

(c) If AB = 12 cm then what is AC?
Answer:
Since ∠B = ∠C we can write AB = AC
⇒ AB = AC = 12 cm

Question 14.
The measure of an outer angle of a regular polygon is 2x, and the measure of an inner angle is 4x.
(a) Use the relationship between inner and outer angles to find x.
Answer:
Sum of the inner angle and outer angle at; corner is 180°
6x = 180, x = 30

(b) Find the measure of one inner and outer angle.
Answer:
Angles are 60°, 120°

(c) Find the number of sides in the polygon and the type of polygon.
Answer:
Regular hexagon

Question 15.
[A] Check the steps given below:

(a) Write two more steps and express \(\frac{1}{9}\) in decimal form.
Answer:

(b) Write the decimal form of \(\frac{1}{3}\) and \(\frac{2}{9}\).
Answer:
\(\frac{1}{3}=\frac{3}{9}\) = 0.333 ………..
\(\frac{2}{9}\) = 0.222 ………..

(c) Write the decimal 0.999… in whole number form.
Answer:
0.999.. = 1

OR

[B] Look at the pattern given below:
1 + 2 + 3 = 2 × 3 = 6
1 + 2 + 3 + 4 + 5 = 3 × 5 = 15
1 + 2 + 3 + 4 + 5 + 6 + 7 = 4 × 7 = 28
(a) Write the sum of the first 9 natural numbers?
Answer:
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
= 5 × 9
= 45

(b) Write the sum of the first 25 natural numbers?
Answer:
13 × 25 = 325

(c) What is – 1 – 2 – 3 …. – 49 ?
Answer:
-(1 + 2 + 3 + .. .+49) = -(25 × 49)
= -1225

Question 16.
[A] The height of students in a class is given below. Prepare a frequency table and draw a histogram. (Heights in cm)

Answer:

Height (cm)	Number of students
148-152	6
152-156	7
156-160	6
160-164	4
164-168	3
168-172	4
Total	30

OR

[B] Detail of rainfall in June and July are given in the table below. Draw a histogram.

Rain fall (mm)	Days
15-25	3
25-35	6
35-45	9
45 – 55	12
55-65	14
65-75	3
75 – 85	2
85 – 95	1
95-105	9

Answer:

In the final stage of revision, Class 8 Maths Question Paper Pdf Kerala Syllabus Set 4 help manage time.

Class 8 Maths Kerala Syllabus Model Question Paper Set 4

Times: 1½ hr.
Score: 40

Instructions

• Use the first 15 minutes to read the questions and think about answers.
• There are 16 questions, split in to 4 parts A, B, C, D
• Answer all questions; but in questions of the type A or B you need answer only one of those.
• You can answer the questions in any order, writing the correct question number.
• Answers must be explained, whenever necessary.

Section – A
This section has 4 questions of score 1 each. Select the correct answer.

Question 1.
Which of the following is a quadrilateral in which all the angles and sides are equal?
(a) Rhombus
(b) Rectangle
(c) Parallelogram
(d) Square
Answer:
(d) A square
All angles are 90 °.

Question 2.
The maximum temperature for seven consecutive days in a city is 26 ° C and the minimum temperature is 21 ° C. Temperature difference is,
(a) 6°
(b) 5°
(c) 10°
(d) 9°
Answer:
(b) 5°
26 – 21 = 5

Question 3.
Two statements are given below
p₁ : Whatever be the number of sides of a polygon the outer angle sum is 360°
p₂ inner angle sum increases by 180° if the number of sides increases by 1.
(a) p₁ and p₂ are true
(b) P₁ is true p₂ is false
(c) p₁ is false p₂ is true
(d) p₁ and p₂ are false.
Answer:
(a) p₁ and p₂ are true

Question 4.
In triangle ABC, AB: BC = 3:4, BC: CA =8:9 then AB: BC: CA =
(a) 3 : 5 : 8
(b) 8 : 12 : 15
(c) 6 : 8 : 9
(d) 1 : 4 : 9
Answer:
(b) 8 : 12 : 15

The actual lengths of AB and BC are 3 centimetres and 4 centimetres multiplied by the same number. Denoting this number as x, we have AB = 3x cm BC = 4x cm
Again, the actual lengths of BC and CA are 8 centimetres and 9 centimetres multiplied by another number. Denoting this number as y, we have BC = 8y cm CA = 9y cm
Now, 4x = 8y ⇒ y = \(\frac{1}{2}\)x
So, CA = 9 × \(\frac{1}{2}\)x = \(\frac{9}{2}\)x
AB : BC : CA = 3x: 4x: \(\frac{9}{2}\)x = 6: 8: 9

Section – B
This section has 4 questions of score 2 each.

Question 5.
If x = \(\frac{1}{2}\) find x + x². Write this in the decimal form.
Answer:

Question 6.
A chord that is 8 cm long lies 3 cm away from the centre of a circle. Calculate the diameter of the circle.
Answer:
The perpendicular drawn from the centre to the chord bisects the chord.

OB² = 3² + 4² = 25
OB =5 cm
Therefore the diameter = 2 × 5 = 10 cm

Question 7.
In the diagram AB = AD, BC = DC

(a) If ∠BAC = 20° then what is ∠A ?
(b) Name pairs of equal triangles in the figure
Answer:
(a) Triangle BAC and triangle DAC are equal ∠BAC = ∠DAC = 20°
∠A = ∠BAC + ∠DAC = 20° + 20° = 40°

(b) Triangle AOB and triangle AOD are equal Triangle BOC and triangle DOC are equal Triangle ABC and triangle ADC are equal Triangle ABD and triangle CBD are equal

Question 8.
23² = 529
(i) What is 24²?
(ii) What is 26²?
Answer:
23² = 529
(i) 24² = (23 + 1)²
= 23² + 2 × 23 × 1 + 1²

= 529 + 46 + 1 = 576
(ii) 26² = (23 + 3)²
= 23² + 2 × 23 × 3 + 3²
= 529 + 138 + 9 = 676

Section – C
This section has 4 questions of score 3 each.

Question 9.
The measure of an outer angle of a regular polygon is 2x, and the measure of an inner angle is 4x.
(a) Use the relationship between inner and outer angles to find x.
(b) Find the measure of one inner and outer angle.
(c) Find the number of sides in the polygon and the type of polygon.
Answer:
(a) Sum of the inner angle sum at a comer is 180° 6x = 180, x = 30
(b) Angles are 60°, 120°
(c) Regular hexagon

Question 10.
Sum of the digits of a two digit number is 10. On reversing the digits it is found that the number decreases by 36. Find the number.
Answer:
Digit in ten’s place is x. Digit in one’s place is 10 – x
Number is 10x + 10 – x
When the digits are reversed number becomes 10(10 – x) + x
It is given that 10(10 – x) + x = 10x + 10 – x – 36
100 – 10x + x = 9x – 26, 100 – 9x = 9x – 26, 126 = 18x,
x = \(\frac{126}{18}\) = 7
Number is 73

Question 11.
[A] In the following figure, four equal rectangles joined to form a square.

Area of the inner square is \(\frac{1}{4}\) part of the area of the outer square. Find the ratio between larger and smaller sides of the rectangle.
Answer:
Let l be the larger side (length) and b be the smaller side(breadth) of the rectangle.
Side of the inner square = l – b
Area of the inner square = (l – b)²
Side of the outer square = l + b
Area of the outer square = (l + b)²
Area of the inner square is \(\frac{1}{4}\) part of the area of the outer square.
\(\frac{(l-b)^2}{(l+b)^2}=\frac{1}{4}\)
\(\frac{l-b}{l+b}=\frac{1}{2}\)
2 (l – b) = l + b
⇒ 2l – l = b + 2b
⇒ l = 3b
⇒ \(\frac{l}{b}=\frac{3}{1}\)
So, the ratio between larger and smaller sides of the rectangle is 3:1

OR

[B] The figure shown below is draw using 2 equal rhombuses. Draw the figure as per the given measures.

Answer:
Draw AB 4 cm first as in the fig. then take 60° at B and 120° at A, on each side of AB there complete the fig:

Question 12.
[A] Look at the pattern given below. These are the decimal forms of fractions:
\(\frac{1}{11}\) = 0.0909 …….
\(\frac{2}{11}\) = 0.1818……..
\(\frac{3}{11}\) = 0.2727…….
……………………………….
(a) Write the next line.
Answer:
\(\frac{4}{11}\) = 0.3636…….

(b) Write the fraction form of 0.6363…
Answer:
\(\frac{7}{11}\)

(c) Write the decimal form of 0.8181… + 0.7272…
Answer:
\(\frac{9}{11}+\frac{8}{11}=\frac{17}{11}\)
= 1 + \(\frac{6}{11}\)
= 1.5454…….

OR

[B] The angles of a triangle are in the ratio 2:3:5. How much is each angle?
Answer:
180 × \(\frac{2}{10}\) = 36°
180 × \(\frac{3}{10}\) = 54°
180 × \(\frac{5}{10}\) = 90°

Section – D
This section has 4 questions of score 4 each.

Question 13.
Draw an isosceles triangle whose length of its one side is 9.5 centimetres, and the height to this side is 4.75 centimetre.
Answer:
Draw a line of length 9.5 cm. draw the perpendicular bisector of this line.
Mark the half length of the line on the top of its perpendicular bisector.
From this point join the both ends of the line and form a triangle.

Question 14.
Draw a rhombus of diagonals 6 cm and 8 cm.
Answer:
Draw 6 cm long line and its perpendicular bisector. Mark two points upper and lower 4 cm from the midpoint. Join these points with the ends of the first diagonal. We get rhombus.

Question 15.
[A] Electricity consumption of 30 families in a locality is given in the table below. Draw a histogram.

Electricity Consumption	Number of families
100 – 200	5
200 – 300	7
300 – 400	8
400 – 500	6
500 – 600	4

Answer:

OR

[B] In the given figure a rhombus is formed by joining the midpoints of the sides of the rectangle. AB =6 centimetres. BC = 4 centimetres.

(a) What is the area of the rectangles?
Answer:
6 × 4 = 24 cm²

(b) Find the area of the rhombus
Answer:
\(\frac{1}{2}\) × 24 = 12cm²

(c) Find the area of rhombus with diagonals 12 centimetres and 8 centrimetres
Answer:
\(\frac{1}{2}\) × 12 × 8 = 48 cm²

Question 16.
[A] Read the mathematical concept given below and answer any four questions that follow. Each question carries 1 score. (4 × 1 = 4)
1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
……………….
……………….
The numbers which can be expressed as the sum of consecutive natural numbers starting from 1 are known as trianglular numbers.
1² = 1
2² = 4
3² = 9
………………..
………………..
The squares of natural numbers are known as square numbers.
That is,
1, 3, 6, 10, ……… Are trianglular numbers.
1, 4, 9, ………… Are square numbers.
(a) Which is the fouth triangular number?
(b) How many first natural numbers are added to get 5th triangular number?
(c) Which is the 6th square number?
(d) Find the two digit number which is both triangular and square number
(e) What is the difference between the 6th square number and the 6th triangular number?
(f) Write a square number between 70 and 90
Answer:
(a) 10
(b) 5
(c) 36
(d) 36
(e) 15
(f) 81

OR

[B] 3² – 1² = 4 × 2
4² – 2² = 4 × 3
5² – 3² = 4 × 4
6² – 4² = 4 × 5
(i) Write the next twolines of this pattern.
(ii) What is the principle behind this pattern.
(iii) Explain it using algebra.
Answer:
i) 7² – 5² = 4 × 6
8² – 6² = 4 × 7

ii) The difference in the squares of alternate natural numbers always be a multiple of 4

iii) (n + 2)² – n² = 4(n + 1)

In the final stage of revision, Class 8 Maths Question Paper Pdf Kerala Syllabus Set 3 help manage time.

Class 8 Maths Kerala Syllabus Model Question Paper Set 3

Times: 1½ hr.
Score: 40

Instructions

• Use the first 15 minutes to read the questions and think about answers.
• There are 16 questions, split in to 4 parts A, B, C, D
• Answer all questions; but in questions of the type A or B you need answer only one of those.
• You can answer the questions in any order, writing the correct question number.
• Answers must be explained, whenever necessary.

Section-A
This section has 4 questions of score 1 each. Select the correct answer.

Question 1.
The angle around the centre of a circle is divided into three equal parts, each 120° by drawing radii. The ends of radii are joined to make a triangle. The triangle so formed is
(a) Right triangle
(b) Isosceles triangle
(c) Equilateral triangle
(d) None of these
Answer:
(c) Equilateral triangle

Triangles formed around the centre of the circle are equal, one angle 120°Other two angles are 30°. So angles of the triangle will be 60°

Question 2.
From the following, which of the number will be the sum of the digits in the ones and tens place a square of a number which ends in 5
(a) 2
(b) 5
(c) 7
(d) 3
Answer:
(c) 7

Question 3.
Read the given statements.
Statement I: The diagonals of a rhombus bisect each other at 90°.
Statement II: The sides of a rhombus are equal in length
A) Statement I is true, Statement II is true
B) Statement I is false, Statement II is false
C) Statement I is true, Statement II is the reason for Statement I
D) Statement I is true, Statement II is not the reason for Statement I
Answer:
D) Statement I is true, Statement II is not the reason for Statement I
the reason for Statement I The diagonals of a rhombus do bisect each other at right angles. However, the reason given that the sides are equal does not directly cause the diagonals to bisect at 90°.

Question 4.
Area of the trapezium shown in the following figure is ___________.

(a) 4.5 cm²
(b) 6 cm²
(c) 5.5 cm²
(d) 6.5 cm²
Answer:
(c) 5.5 cm²

Area = \(\frac{1}{2}\) × 2 × (2+ 3.5) = 5.5 cm²

Section – B
This section has 4 questions of score 2 each.

Question 5.
A polygon has 7 sides
(a) What is the sum of its inner angles Y.
(b) What is the sum of the outer angles?
Answer:
(a) Angle sum of a polygon of n sides is 180° × (n – 2)
Angle sum is 180 × 5 = 900°

(b) 360°

Question 6.
Sum of two numbers is 60. One number is four times the other. Find the number.
Answer:
When number and four times the number is added we get five times the number
Number is \(\frac{1}{5}\) part of 60.
Number is = 12

Question 7.
(a) What is 1 – 2 ?
Answer:
-(2 – 1) = -1

(b) Find 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
Answer:
-1 – 1 – 1 – 1 – 1 = -5

Question 8.
In a circle of radius 13 cm, a chord is at a distance of 5 cm from the centre. Calculate the length of the chord.
Answer:
Half of the chord = \(\sqrt{13^2-5^2}\) = 12 cm
Therefore, the full length of the chord = 24 cm.

Section – C
This section has 4 questions of socre 3 each.

Question 9.
If x = 0.333 … then,
(a) Write the fractional form of x ?
Answer:
\(\frac{3}{9}\)

(b) Write x² as a fraction
Answer:
\(\frac{1}{9}\)

(c) Write the decimal form of x² + x
Answer:
x² + x = \(\frac{1}{2}\) = 0.444 ……..

Question 10.
Using this equation x² – y² = (x + y)(x – y)
(a) What is 2² – 1²?
Answer:
(2 + 1) (2 – 1) = 3 x 1 = 3

(b) What is 9² – 10²?
Answer:
19

(c) Find 1² – 2² + 3² – 4² + ….. + 9² – 10² ?
Answer:
-3 – 7 – 19 – 11 – 15 – 19 = -(3 + 7 + 11 + 15 + 19)
= -55

Question 11.
[A] Perimeter of a rectangle is 50 cm. Length is 5 cm more than breadth
(a) What is length + breadth ?
(b) If x is the length then write the equation
(c) Find the sides
Answer:
(a) 25 cm

(b) x + x + 5 = 25
2x + 5 = 25, 2x = 20, x = 10

(c) Sides are 10 cm, 15 cm

OR

[B] A man’s age after 15 years will be two times his age before 15 years.
(a) If the present age is x then write the equation.
(b) Find the present age?
Answer:
(a) Age after 15 years is x + 15, age before 15 years is x – 15
2(x – 15) = x + 15
2x – 30 = x + 15
⇒ x = 45

(b) Present age 45years

Question 12.
[A] A triangle has sides 20 cm, 15 cm, and 25 cm. What type of triangle is it? Calculate the radius of its circumcircle.
Answer:
15² + 20² = 225 + 400 = 625 = 25²
Here we get a right angled triangle.
For a right triangle the circumcentre is the midpoint of the hypotenuse.
So the circumradius R = half the hypotenuse 25
R = \(\frac{25}{2}\) = 12.5 cm

OR

[B] In a trapezium ABCD, the parallel sides are 8 cm, 6 cm. The angles in the largest side is 60°, 75°.
(a) Draw a rough figure
Answer:

(b) Draw trapezium ABCD in actual measures.
Answer:

Section – D
This section has 4 questions of score 4 each.

Question 13.
Draw a 7.5 centimeter long line and construct a right triangle whose perpendicular sides are of 3.75 centimetre long.
Answer:
Draw a 7.5 cm long line.
Draw the perpendicular bisector of this line. Complete the right triangle by drawing the half length of the line on the perpendicular line.

Question 14.
Look at the pattern given below:
1 + 3 = 4 = 2²
l + 3 + 5 = 9 = 3²
1 + 3 + 5 + 7 = 16 = 4²
(a) Complete the next line.
(b) Find the sum of the first 10 odd numbers.
(c) What is the 20th odd number.
(d) What is —1 — 3 — 5 …….. 39
Answer:
(a) 1 + 3 + 5 + 7 + 9 = 25 = 5²
(b) 10² = 100
(c) 39
(d) -(1 + 3 + 5 + ………. + 39) = -400

Question 15.
[A] A rubber farmer records the number of rubber sheets obtained on each day of a month. Using thi data, prepare a frequency table and answer the following questions:

i. What is the maximum number of sheets obtained in a single day?
ii. How many days did he get 16 sheets?
iii. How many days did he get the minimum number of sheets?
iv. How many days did he get less than 15 sheets?
v. What is the total number of sheets obtained in the whole month?
Answer:

Number of sheets	Number of days
9	3
10	4
11	3
12	3
13	5
14	6
16	6

i. 16
ii. 6 days
iii. 3 days
iv. 24 days
v. 85 sheets

OR

[B] Find the area of the parallelogram in the figure.

Answer:

Area of ∆ABF = \(\frac{1}{2}\) × 6 × 3 = 9 cm
AB = \(\sqrt{6^2+3^2}\)
= \(\sqrt{36+9}=\sqrt{45}\)
\(\frac{1}{2}\) × AB × h
\(\frac{1}{2}\) × \(\sqrt{45}\) × h = 9
h = \(\frac{9 \times 2}{\sqrt{45}}=\frac{18}{\sqrt{45}}\)
Area of ABCD = \(\sqrt{45} \times \frac{18}{\sqrt{45}}\) = 18 cm
[or Area of Parallelogram ABCD = Area of the rectangle = 6 × 3 = 18]

Question 16.
[A] Draw a parallelogram of sides 6 cm, 4cm and area 18 cm².
Answer:
First draw a horizontal line of length 6 cm as the bottom side of the parallelogram. Since area of the parallelogram is 18 sq.cm, its distance to. the opposite side (height) must be 3 cm.
So, draw a perpendicular of height 3 cm from one end of the line. Draw the line through its top end, parallel to the first line, then the top side of the parallelogram is a part of this line.
Mark 4 cm from left end of the bottom line on the top side and complete the parallelogram.

OR

[B] Sum of innerangle of a polygon with ‘n’ sides is (n – 2) 180°. The sum of outer angles will be always 360°.
(i) What is the sum of innerangles of a polygon with 20 sides?
(ii) What is the sum of outer angles of a polygon with 100 sides. .
(iii) Which is the polygon having sum of inner angles and outer angles equal.
(iv) Can 1080° be the sum of inner angles of a polygon?
Answer:
i) (20 – 2) 180 = 18 × 180° = 3240°
ii) 360°
iii) Rectangle
iv) Yes, because 1080° is a multiple of 180

In the final stage of revision, Class 8 Maths Question Paper Pdf Kerala Syllabus Set 2 help manage time.

Class 8 Maths Kerala Syllabus Model Question Paper Set 2

Times: 1½ hr.
Score: 40

Instructions

• Use the first 15 minutes to read the questions and think about answers.
• There are 16 questions, split in to 4 parts A, B, C, D
• Answer all questions; but in questions of the type A or B you need answer only one of those.
• You can answer the questions in any order, writing the correct question number.
• Answers must be explained, whenever necessary.

Section – A
This section has 4 questions of score 1 each. Select the correct answer.

Question 1.
A table showing classes and frequencies is called ________
(a) Pie chart
(b) Bar graph
(c) Histogram
(d) Frequency table
Answer:
(d) Frequency table

Question 2.
Which statement is TRUE?
(a) Equal chords are at unequal distances from the centre
(b) Chords at equal distances from the centre are equal
(c) Perpendicular bisector of a chord never passes through the centre
(d) Radius is always perpendicular to the chord
Answer:
(b) Chords at equal distances from the centre are equal

Question 3.
Two statements are given below:
(i) The diagonals of all parallelograms bisects each other perpendicularly.
(ii) The diagonal of a rhombus bisects each other perpendicularly.
(a) (i) and (ii) are true
(b) Only (i) is true
(c) Only (ii) is true
(d) (i) and (ii) is not true
Answer:
(c) Only (ii) is true

Question 4.
Two sides of a triangle are equal. Small angle is 30°. What is the one of the equal angles?
(a) 100°
(b) 75°
(c) 70°
(d) 50°
Answer:
(b) 75°

Angle sum 180
Smallest is 30°. Sum of the equal angles is 150°
Equal angle is 75°

Section – B
This section has 4 questions of score 2 each.

Question 5.
Calculate \(\frac{1}{7}-\frac{3}{7}+\frac{5}{7}-\frac{7}{7}+\cdots+\frac{21}{7}\)
Answer:
\(\frac{1-3+5-7+9-11+13-15+17-19+21}{7}\)
= \(\frac{-10+21}{7}\)
= \(\frac{11}{7}\)

Question 6.
When \(\frac{2}{3}\) of a number is subtracted from a number we get 13 . What is the number?
Answer:
\(\frac{1}{3}\) of the number is 13.
Number is 3 × 13 = 39

Question 7.
The angles of a triangle are in the ratio 4:5:9.
(a) Write the angles as multiples of x.
Answer:
Angles are 4x, 5x, 9x

(b) Find the angles.
Answer:
Sum of angles =180
⇒ 4x + 5x + 9x = 180
⇒ 18x = 180
⇒ x = 10
So, angles are 40°, 50°, 90°

Question 8.
Draw the following trapezium.

Answer:

Section – C
This section has 4 questions of socre 3 each.

Question 9.
Sum of two consecutive natural numbers is 63
(a) If the smaller is x then what is the other number.
Answer:
x, x + 1 are the numbers

(b) Write the equations and find the numbers.
Answer:
2x + 1 = 63, 2x = 62, x = 31
Numbers are 31, 32

Question 10.
In a class of 52 students, number of boys is \(\frac{6}{7}\) of number of girls. How many are boys ?
Answer:
Number of boys x, number of girls 52 — x 6
x = \(\frac{6}{7}\) × (52 – x),7x = 6(52 – x)
7x = 312 – 6x, 13x = 312, x – 24
Boys 24 girls 28

Question 11.
[A] This is the section of a regular polygon.

(a) What are the angles of triangle ?
Answer:
18°, 18°, 144°

(b) What is an outer angle ?
Answer:
180 – 144 = 36°

(c) How many sides does this polygon have ?
Answer:
\(\frac{360}{36}\) = 10

OR

[B] The diagram shows an equilateral triangle, a square, and one diagonal of the square. What is the value of ?

Answer:
Equilateral triangle has one inner angle 60°
Square has inner angle 90°
A diagonal divides it into 45° each
x = 60 + 45 = 105°

Question 12.
[A] In the figure ABCD is a parallelogram. AB = 6 cm. The distance between AB and CD is 3 cm and the distance between AD and BC is 2 cm. Find the perimeter of the parallelogram.

Answer:
Area of the parallelogram ABCD = 6 × 3 = 18 cm²
Also, area – BC × 2 = 18
⇒ BC = 18 + 2 = 9 cm
AB = CD = 6 cm
AD = BC = 9 cm
∴ Perimeter = 2(6 + 9) = 30 cm

OR

[B] The area of a rhombus is 120 cm² and the length of one of its diagonals is 24 cm.
(a) Find the length of its second diagonal.
Answer:
Area =120 cm²
⇒ \(\frac{1}{2}\) × d₁ × d₂ = 120
⇒ \(\frac{1}{2}\) × 24 × d₂ = 120
⇒ d₂ = 120 ÷ 12 = 10 cm

(b) Find the perimeter.
Answer:
Length of side of the rhombus
= \(\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}\) = 13 cm
∴ Perimeter = 4 × 13 = 52 cm

Section – D
This section has 4 questions of score 4 each.

Question 13.
Check the steps given below:
\(\frac{1}{3}=\frac{1}{10} \times \frac{10}{3}\)
= \(\frac{1}{10}\left(3+\frac{1}{3}\right)\)
= \(\frac{3}{10}+\frac{1}{30}\)
\(\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)

(a) Write two more steps and convert \(\frac{1}{3}\) into decimal form.
Answer:
Step 1:

Decimal form of \(\frac{1}{3}\) is 0.3333….

(b) Write the decimal form of \(\frac{1}{9}\) and \(\frac{2}{9}\)
Answer:
\(\frac{1}{9}\) = 0.111 …..
\(\frac{2}{9}\) = 0.222……

(c) Calculate \(\sqrt{0.444} \ldots\)
Answer:
\(\sqrt{0.444} \ldots\) = \(\sqrt{\frac{4}{9}}=\frac{2}{3}=\frac{6}{9}\) = 0.666 ………..

Question 14.
One side of a triangle has a length of 6 cm, and the angles at the ends of that side are 40° and 70°. Draw the triangle and construct its circumcircle.
Answer:

Question 15.
[A] (i) 11² + 10² = 121 +100 = 221 By using this , write 442 as the sum of the two perfect squares.
(ii) If the square of a number ending in 5 is of the form 110pq. Then p × q = ________.
(iii) (a) 7.6² = ________
(b) 75.4² = ________
Answer:
(i) 442 = 2 × 221
= 2(11² + 10²)
= (11 + 10)² + (11 – 10)²
442 = 21² + 1²

(ii) p × q = 2×5 = 10

(iii) (a) 7.6² = 72 + 2 × 7 × 0.6 + (0.6)²
= 49 + 8.4 + 0.36
= 57.76

(b) 75.4² = 75² + 2 × 75 × 0.4 + (0.4)²
= 5625 + 60 + 0.16
= 5685.16

OR

[B] The perimeter of a rectangle is 48 centimetres. The ratio between the length and breadth is 5:3,
(a) Length + breadth =
(b) Calcualte the length and breadth.
(c) If the breadth is increased by 1 centimetre, then find the ratio between length and breadth
Answer:
a) Length + breadth = \(\frac{48}{2}\) = 24 cm

b) Length : breadth = 5:3
∴ Length = 24 × \(\frac{5}{8}\) = 15 cm
Breadth = 24 × \(\frac{3}{8}\) = 9 cm 8

c) If the breadth is increased by 1 cm
Means
New breadth = 9 + 1 = 10 cm
∴ Length : breadth = 15:10 = 3:2

Question 16.
[A] The marks obtained by 25 students in an examination are given below. Prepare a grouped frequency table and answer the following questions, (maximum mark: 40)
25, 30, 33, 18, 36
12, 25, 30, 15, 25
19, 25, 30, 33, 20
35, 37, 29, 31, 30
31, 28, 33, 31, 30
(i) How many students scored less than 15 marks?
(ii) How many students scored 30 marks or more?
(iii) How many students scored more than 25 marks?
(iv) How many students scored between 30 and 40 marks?
Answer:

Class interval	Number of students
10-15	1
15-20	3
20-25	1
25-30	6
30-35	11
35-40	3
Total	25

i. 1 student
ii. 14 students
iii. 20 students
iv. 14 students

OR

[B] Some natural numbers can be, written as the difference of two perfect squares. For example,
8 = 4 × 2 × 1 = 3² – 1²
12 = 4 × 3 × 1 = 4² – 2²
16 = 4 × 4 × 1 = 5² – 3² ……..
(a) Write 20 as the difference of two perfect squares.
Answer:
20 = 4 × 5 × 1 = 6² – 4²

(b) Explain using algebra, the method of writng all multiples of 4, starting with 8 as the difference of two perfect squares.
Answer:
4 × x × 1 = (x + 1)² – (x – 1)²
(x + 1)² – (x – 1)²
= (x + 1 + x – 1)(x + 1 – [x – 1])
= 2x × 2 = 4 × x × 1

In the final stage of revision, Class 8 Maths Question Paper Pdf Kerala Syllabus Set 1 help manage time.

Class 8 Maths Kerala Syllabus Model Question Paper Set 1

Times: 1½ hr.
Score: 40

Instructions

• Use the first 15 minutes to read the questions and think about answers.
• There are 16 questions, split in to 4 parts A, B, C, D
• Answer all questions; but in questions of the type A or B you need answer only one of those.
• You can answer the questions in any order, writing the correct question number.
• Answers must be explained, whenever necessary.

Section – A
This section has 4 questions of score 1 each. Select the correct answer.

Question 1.
In a right angled triangle with its perpendicular sides 9 cm and 12 cm, the radius of its circumcircle is:
(a) 6 cm
(b) 7.5 cm
(c) 10 cm
(d) 15cm
Answer:
(b) 7.5 cm

Hypotenuse =15 cm
⇒ Radius = \(\frac{15}{2}\) = 7.5 cm

Question 2.
A histogram is drawn for …
(a) Discrete data
(b) Continuous data
(c) Alphabetical data
(d) Tally marks
Answer:
(b) continuous data

Question 3.
Four statements are given below Identify the true statement
(a) Equal triangles have equal area
(b) Equiangular triangles (Triangles having equal angles) are equal triangles
(c) Equality one angle and any two sides of two triangles make them equal
(d) One side and angle at the ends can fix a triangle

(a) (a) and (b) are true
(b) (a) and (c) are true
(c) (b) and (c) are true
(d) (a) and (d) are true
Answer:
(d) (a) and (d) are true
Equality means identical or having same shape and side

Question 4.
101² – 100²
(a) 1
(b) 201
(c) 100
(d) none of these
Answer:
101² – 100² = 101 + 100 = 201

Section – B
This section has 4 questions of score 2 each.

Question 5.
Find the sum of the angles marked in the star.

Answer:

In triangle BQD, sum of the marked angles is the outer angle at Q
In triangle PCE, the marked angle sum is the outer angle at P
All marked angles make the sum 180°

Question 6.
The age of Ammu’s mother is now four times of Ammu. After 20 years mothers age will be two times Ammu’s age
(a) If Ammu’s age is x then write the equation.
Answer:
Ammu’s age x, Mothers age 4x

(b) Find the age of Ammu and her mother.
Answer:
After 20 years Ammu’s age = x +
20,mothers age = 4x + 20
4x + 20 = 2(x + 20), 4x + 20
= 2x + 40, x = 10
Ammu’s age 10, Mother age 40

Question 7.
Write the following in the simplified form
\(\frac{12-11+10-9+8-7}{6-5+4-3+2-1}\)
Answer:
\(\frac{12-11+10-9+8-7}{6-5+4-3+2-1}=\frac{1+1+1}{1+1+1}=\frac{3}{3}\)
= 1

Question 8.
The area of a rectangle is 2100 sq.cm.
(a) Write length and breadth as multiple of x.
Answer:
Length = 7x
Breadth = 3x

(b) Find the length and breadth of the rectangle.
Answer:
Area = 2100 sq.cm
length × breadth = 2100
7x × 3x = 2100
21x² = 2100
x² = 100
x = 10
Length = 70 cm
Breadth = 30 cm

Section – C
This section has 4 questions of socre 3 each.

Question 9.
Draw a square with side length 4 cm and construct its circumcircle.
Answer:

Question 10.
Draw a rectangle of diagonal 5.5 cm with the angle between the diagonals 50°.
Answer:
Draw a diagonal 5.5 cm long. Draw its perpendicular bisector. Take angle 50° from this midpoint. Distance on either side of the mid point on the first diagonal is same as on the second diagonal. Join the points. We get the required rectangle.

Question 11.
[A] i. Five angles of a hexagon have measures 100°, 110°, 120°, 130°, 140°. What is the measure of the sixth angle?
ii. Squares are drawn on the sides of a regular hexagon.

Find x
Answer:
(i) Inner angle sum of the hexagon is 720°
Sixth angle is
= 720 – (100 + 110 + 120 + 130 + 140)
= 720 – 600
= 120°

(ii) One inner angle of regular hexagon is 120°
Two squares in a comer make the angles 90°.
Angle around the comer is 360°
x = 360 – (120 + 90 + 90) = 60

OR

[B] Draw 22 \(\frac{1}{2}\) cone using scale and compass.
Answer:
Draw a line and mark a point on it. Draw a perpendicular line through the point.
90° angle is formed. Draw the angle bisctor. 45° angle is formed. Draw the angle bisctor.
22 \(\frac{1}{2}\) angle is formed.

Question 12.
[A] (a) Write the following as powers of 10: \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}\)
Answer:
\(\frac{1}{2}\)

(b) Write their decimal forms.
Answer:
\(\frac{1}{2}\) = 0.5, \(\frac{1}{4}\) = 0.25, \(\frac{1}{8}\) = 0.125

(c) Calculate: 1 – \(\frac{1}{2}+\frac{1}{4}-\frac{1}{8}\)
Answer:
1 – \(\frac{1}{2}=\frac{1}{2}, \frac{1}{4}-\frac{1}{8}=\frac{1}{8}\)
\(\frac{1}{2}+\frac{1}{8}=\frac{4}{8}+\frac{1}{8}=\frac{5}{8}\) = 0.625

OR

[B] Four numbers forming a square in a calendar are given below:

(a) Find the difference of its diagonal products.
(b) Is it the same for all squares of four numbers in a calendar? Explain using algebra.
Answer:
(a) (5 × 11) – (4 × 12) = 7
(b) Yes

x	x + 1
x + 7	x + 8

Difference of diagonal product
= ((x + 1)(x + 7)) – (x(x + 8))
= x² + 8x + 7 – (x² + 8x)
= 7

Section – D
This section has 4 questions of score 4 each.

Question 13.
Draw a right triangle with two angles 45° and length of its hypotneous as 4.75 centimetres.
Answer:
Draw a line of length 9.5 cm. Draw its perpendicular bisector.
The part of the line on one side of the perpendicular bisector will be 4.75 cm.
The angle between the line and the perpendicular bisector is 90 0 Draw the angle bisector.
Draw a perpendicular from one end of the line to the angle bisector. It forms a right triangle.

Question 14.
(i) Draw an equilateral triangle with side length 4 cm and construct its circumcircle.
Answer:

(ii) The angles of a triangle are in the ratio 1:2:3. If the longest side is 18 cm, what is the radius of the circumcircle?
Answer:
It is a right-angled triangle.
The radius of the circumcircle is 9 cm, which is half of the hypotenuse.

Question 15.
[A] (i) In the figure AB = CF, EF = BD, ∠DBC = ∠AFE

(a) Are the triangles AFE and triangle BDC equal ? How do you know this?
Answer:
AB = CF
∴ AB + BF = CF + BF
ie., AF = CB,
BD = EF,
∠DBC = ∠AFE
Triangles are equal

(b) What are the equal sides and equal angles of these triangles?
Answer:
CD = AE, EF = BD, ∠DBC = ∠AFE

(ii) In the figure AB = AC, E is the mid point of AB, F is the mid point of AC .
(a) What are the equal triangles in the figure ?
Answer:
AB = AC ⇒ ∠B = ∠C
BE = CF, BC is common
Triangle BEC and triangle CFB are equal

(b) If BF = 12 cm then what is CE ?
Answer:
Sides opposite to equal angles are equal. CE = 12 cm

OR

[B] ABCDEF is a polygon

(a) What is the suitable name for this polygon?
(b) How many diagonals can be drawn from any one of the vertices?
(c) What is the sum of inner angles of this polygon
(d) What is the sum of inner anlgles of a 7 sided polygon?
Answer:
(a) Hexagon
(b) 3 w
(c) 180 × 4 = 720°
(d) 720 + 180 = 900°

Question 16.
[A] The scores of 30 students in a test are given below.
21, 32, 5, 12, 36, 24, 31, 15, 27, 34
44, 39, 37, 25, 7, 20, 23, 12, 9, 40
46, 42, 28, 16, 47, 27, 37, 18, 49, 26

(a) Prepare a frequency table as shown below.
Answer:

(b) How many students scored more than 30?
Answer:
7 + 6 = 13 students

OR

[B] In figure, ∠AOB = 71° OC is the bisector of ∠AOB

(a) What is the measure of ∠AOC ?
Answer:
∠AOC = \(\frac{71}{2}\) = 35\(\frac{1^0}{2}\)

(b) Draw triangle PQR with PQ = 6 centimetres,
∠P = 35\(\frac{1^0}{2}\), ∠Q = 60°
Answer:
Draw OQ = 6 cm, ∠P = 71° and
∠Q = 60°. Draw bisector of ∠P.
Now in ∆PQR, ∠P = \(\frac{71}{2}\) = 35.5°