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Biology
WAEC / WASSCE 2012
Growth, Micro-organisms & Public Health
Solution
(a) Explain the following terms:
(i) disease
(ii) symptoms of diseases.
(b)(i) List two physical and two chemical barriers that prevent pathogens from penetrating the body of an organism.
(ii) Explain how vaccination protects the body from contracting infectious diseases.
(c) Distinguish between an antibody and a antigen.
(d) Name the causative agents of:
(i) Malaria
(ii) Cholera
(iii) AIDS.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a)(i) Disease: A condition in which the normal functions of an organism/part of the body/cell/tissue/organ is impaired/disturbed.
(ii) Symptom: Any observable effect on an organism: Caused by the presence and irritation of a disease-causing agent/pathogen.
(b)(i) Physical barriers:
(i) Skin
(ii) Hairs
(iii) Cuticle
(iv) Epidermis
(v) Bark
(vi) Cell wall lining.
Chemical barriers:
(i) Mucus
(ii) Saliva
(iii) Tears
(iv) Gastric/ HCL and intestinal secretions
(v) Phenolic compound
(vi) Gums
(vii) Tannins
(viii) Wax
(ix) Oily secretions! sebum
(ii) How vaccination protects the body: The vaccine/attenuated pathogen is administered into the body. The body responds by producing copious antibodies. The next time the body is exposed to the same antigen/infection/pathogen antigen, the memory cells allow a more rapid and heightened/vigorous response to antibody production. Protection of the body against future infections.
(c) Differences between antibodies and antigens
(I) Antibodies is a protein molecule. Antigens May not be a protein molecule
(ii) Antibodies is Produced in response to a particular foreign substance/antigen while antigens Stimulates formation of antibodies
(iii) Antibodies Usually found in the plasma fluid of the blood,antigens are Usually produced in/introduced into the body.
(iv) Antibodies are Protective in functions, antigens Leads to an immune response/immunity.
(d) Causative Agent:
(i) Malaria: Plasmodium vivax/Plasmodium/malaria/species
(ii) Cholera: Vibrio cholera.
(ii) AIDS: Human Immuno-deficiency virus/HIV
Source: Myschool Past Questions Archive (Record #52742)
Biology
WAEC / WASSCE 2012
Reproduction in Plants & Animals
Solution
(a) (i) Describe epigeal germination of a seed.
(ii) In a tabular form, state three differences between epige germination and hypogeal germination.
(b)(i) What is seed dormancy?
(ii) State three ways b which dormancy in seeds can be broken.
(c) State six advantages of using contraceptives in puma populations.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a)(i) Description of Epigeal Germination:
(i) Seed absorbs/imbibes water and swells.
(ii) Testa splits and the radicle emerges.
(iii) The hypocotyl of the germinating seed elongates.
(iv) Pushing the seed out of the soil.
(v) The cotyledon(s) become exposed to sunlight.
(vi) Turns green to photosynthesize.
(vii) The plumule develops into a green shoot.
(viii) The cotyledon(s) dry up/shrink and fall off.
(ii) Differences between epigeal and hypogeal germination:
Epigeal germination
Hypogeal germination
Hypocotyl elongates
Epicotyl elongates
cotyledon
cotyledon remains in the soil
cotyledon(s) turns green / photosynthesize
cotyledon does not photosynthesize
plumule does not emerge simultaneously with the radicle
plumule emerges simultaneously with radicle
Energy with the radicle from cotyledon
energy derived from endosperm
(b)(i) Seed dormancy: An inactive period of a seed.. During which growth slows/completely ceases. Due to certain internal or external factors.
(ii) Ways of breaking dormancy: Removal/corrosion of testa. Treatment of seed with acid/digestive enzymes/water. Exposure to alternating high and low temperature. Addition of dormancy-breaking hormones/named hormones. Exposure to fire. Exposure to light.
(c) Advantages of using contraceptives in human population:
(i) It controls birth rate.
(ii) Reduces over-population.
(iii) Promotes maternal health/reduces maternal death.
(iv) Promotes infant survival rate.
(v) Enables parents to determine family size.
(vi) Prevents unwanted pregnancy.
(vii) Enhances good standard of living for the family.
(viii) Some prevent STI or STDs.
Source: Myschool Past Questions Archive (Record #52743)
Biology
WAEC / WASSCE 2012
Variety of Organisms & Classification of Living Things
Solution
Study carefully specimens C, D and E and use them to answer questions 2(a) to 2(c).
(a) (i) State the mode of nutrition of specimen D.
(ii) State one way by which each of specimens C, D and E is of economic importance.
(b)(i) Classify specimens C, D and E into their phyla and classes.
(ii) List three observable characteristic features of each of the classes to which specimens C, D and E belong.
(iii) In a tabular form, state two observable differences between C and D.
(C) List four observable features of specimen E that are sensory in nature.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a)(i) Mode of nutrition of Specimen D- Parasitic
(ii) One economic importance of each Specimen C, D and E.
Specimen C/Grain Weevil: They destroy grains/reduce their food quality.
Increase production cost for farmers.
Therefore lowers the farmers income.
Specimen D/Tick: They infest cattle/dogs/suck blood thereby making them anemic; and reduce meat quality.
They transmit diseases.
They decrease resistance to diseases in cattle/dog.
Specimen E/Rat: They destroy stored food items/contamination of food. They spread diseases.
(b)(i) Classification of
Specimen
Phylum
Class
C
Arthropoda
Insecta
D
Arthropoda
Arachnida
E
Chordata
Mammalia
(ii) Observable characteristics: Class insecta/ Specimen C: Three body divisions/head, thorax and abdomen. Presence of 6 jointed/'3 pairs of jointed walking legs; presence of compound eyes. Presence of segmented thorax/prothorax, mesothorax and metathorax.
Presence of wings. Presence of body divisions/Cephalothorax and abdomen.
Presence of simple eyes/ocelli
Presence of two appendages for feeding/pedipalps and chelicera.
Class mammalian/specimen E :
Presence of hair/fur on body. Presence of external ears/pinnae. Presence of mammary glands. The presence of four limbs for locomotion.
Heterodont dentition.
(III) Observable differences between C& D:
C/Grain Weevil
D/Tick
Three body division/head, thorax and abdomen
Two body divisions/cephalothorax and abdomen
3 pairs of legs/6 legs
4 pairs of legs/8 legs
Compound eyes
Simple eye present
Presence of wings
Absence of wings
Distinct tnorax / thorax present
Thorax fused with head/cephalothorax
Presence of antennae
Absence of antennae
(c) Observable features of Specimen E/Mammalia that are sensory in nature.
(i) external ears/pinnae (ii) Skin/fur (iii) Whiskers (iv) Eyes (v) Nostrils (vi) Tail
Source: Myschool Past Questions Archive (Record #60742)
Biology
WAEC / WASSCE 2012
Respiration & Gaseous Exchange
Solution
Study carefully specimens J, K and L and use them to answer questions 4(a) to 4(d).
(a)(i) State the class of each of specimens J, K and L
(ii) State three reasons each for placing each of specimens J, K and L in their classes.
(b) State two structural adaptations possessed by each of specimens K and L for escaping predators.
(c) Make a drawing, 8 - 10cm long of the lateral view of specimen J.
(d) State: (i) two similarities; (ii) two observable differences between specimens J and L.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a) Class of Specimen J-Tilapia: Osteichthyes.
Reason: Body is covered with scales. (ii) Presence of paired fins. (iii) Presence of lateral line(s). (iv) Presence of operculum/gill cover.
class of Specimen K/Toad: Amphibian
Reason: (i) Presence or warty skin. (ii) Presence or webbed digits. (iii) Presence of moist skin. (iv) Presence of bulging eyes. (v) Presence of ear drum.
Structural adaptations for escaping predation: Specimen K/Toad:Presence of poison gland on skin which secretes a substance which irritates/repels the predators.
(i) Counter shading of the dorsal and ventral sides/camouflage facilitate escape from predator.
(ii)Long hind limb and short muscular forelimb ror jumping away from predator.
Class of Specimen L/Lizard : Reptilla
Reason: Presence of dry horny scales on skin (ii) presence of fore and hind limbs with clawed digits. (iii) Homodont dentition. (iv) Well developed retractable tongue.
Specimen L/Lizard:Detachable tail to escape from predators. (ii) Skin color blends with surrounding/camouflage to enhance escape from predators. (iii) Eyes for seeing predators. (iv) Limbs for running away from predators.
(c) Drawing Specimen J/Tilapia:
(d)(i) Similarities between Specimen J& L: (i) Presence of eyes. (i) Presence of nostrils. (ii) Presence Or Scales. (v) Presence of head. (v) Presence of tail.
Differences
(ii)
J/Tilapia
L/Lizard
Presence of fins
Presence of limbs
Absence of tympanum
Presence of tympanum
Lateral line present
Lateral line absent
Short tail
Long tail
Presence of operculum/gill cover
No operculum/gill cover
Source: Myschool Past Questions Archive (Record #60743)
Mathematics
WAEC / WASSCE 2011
Plane Geometry, Polygons & Circle Theorems
Solution
One of the factors of (mn - nq - n\(^2\) + mq) is (m - n). The other factor is?
A
(n - q)
B
(q - n)
C
(n + q)
D
(q - m)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
mn - nq - n\(^2\) + mq
mn - n\(^2\) - nq + mq
(mn - n\(^2\)) (mq - nq)
n(m - n) + q(m - n)
= (n +q) (m - n)
Source: WAEC / JAMB / NECO Classroom Archive (Record #36639)
Mathematics
WAEC / WASSCE 2011
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
(a)
In the diagram, PQRST is a quadrilateral. PT // QS, < PTQ = 42°, < TSQ = 38° and < QSR = 30°. If < QTS = x and < POT = y, find: (i) x ; (ii) y. (SEE THE DIAGRAMS ABOVE)
(b)
In the diagram, PQRS is a circle centre O. If POQ = 150°, < QSR = 40° and < SQP = 45°, calculate < RQS.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)(i) < PTS = 38^\circ (alternate angle)
42^\circ + x + 38^\circ = 180^\circ (angles on a straight line)
80^\circ + x = 180^\circ
x = 180^\circ - 80^\circ = 100^\circ
(ii) < SQT = 42^\circ (alternate angle)
< SQR = 60^\circ (sum of angles in a triangle)
y + 42^\circ + 60^\circ = 180^\circ (angle on a straight line)
y + 102^\circ = 180^\circ
y = 180^\circ - 102^\circ = 78^\circ
(b) \(< QSP = \frac{150}{2}\) (angle at the centre is twicw angle at the circumference)
\(< QSP = 75^\circ\)
\(\hat{R} + \hat{P} = 180^\circ\) (opposite angles in a cyclic quad are supplementary)
\(\hat{P} + 75^\circ + 45^\circ = 180^\circ\) (sum of angle in a triangle)
\(\hat{P} = 180^\circ - 120^\circ\)
\(\hat{P} = 60^\circ\)
\(\hat{R} + \hat{P} = 180^\circ\)
\(\hat{R} = 180^\circ - 60^\circ\)
\(\hat{R} = 120^\circ\)
\(< RQS + < QRS + < QSR = 180^\circ\) (sum of angle in a triangle)
\(< RQS = 180^\circ - (40^\circ + 120^\circ)\)
= \(180^\circ - 160^\circ\)
\(< RQS = 20^\circ\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #52748)
Mathematics
UTME / JAMB 2011
Inequalities & Linear Programming
Solution
Solve the inequality x2 + 2x > 15.
A
x 5
B
-5 < x < 3
C
x 5
D
x > 3 or x < -5
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x2 + 2x > 15
x2 + 2x - 15 > 0
(x2 + 5x) - (3x - 15) > 0
x(x + 5) - 3(x + 5) >0
(x - 3)(x + 5) > 0
therefore, x = 3 or -5
i.e. x< 3 or x > -5
Source: WAEC / JAMB / NECO Classroom Archive (Record #29387)
Mathematics
WAEC / WASSCE 2011
Mensuration (Areas, Perimeters & Volumes)
Solution
John pours 96 litres of red oil into a rectangular container with length 220cm and breadth 40cm. Calculate, correct to the nearest cm, the height of the oil in the container
A
11cm
B
18cm
C
21cm
D
34cm
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
volume of rectangular container = L x B x H
1 litre = 1000cm3
96 litres = \(\frac{1000cm^3}{1 litres}\) x 96 litres
volume = 96000cm3
96000 = 220 x 40 x 11
96--- = 8800H
H = \(\frac{96000}{8800}\)
= 10.97cm
= 11cm
Source: WAEC / JAMB / NECO Classroom Archive (Record #36688)
Physics
WAEC / WASSCE 2011
Waves, Sound & Vibrations
Solution
Which of the following statements about the electromagnetic waves is correct?
A
radio waves have longer frequencies than x-rays
B
x-rays have the longest wavelength
C
some of the waves travel at 3.0 x 108ms-1
D
white light occupies a region beyond ultraviolet rays
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
All electromagnetic waves: are transverse waves; can travel through a vacuum ; travel at exactly the same speed in a vacuum, the speed of light, 300,000,000 m/s.
The common designations are radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV) light, X-rays and gamma-rays.
ultraviolet rays lie beside violet of visible light and has a lesser frequency than visible light.
Source: WAEC / JAMB / NECO Classroom Archive (Record #35481)
Chemistry
WAEC / WASSCE 2011
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
The colour of phenolphthalein indicator in alkaline solution at the end-point of an acid-base titration is?
A
colourless
B
orange
C
pink
D
yellow
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
Phenolphthalein is one of the most commonly-used indicators in neutralisation titration.
Phenolphthalein is naturally colorless but turns pink in alkaline solutions.
Source: WAEC / JAMB / NECO Classroom Archive (Record #37613)
Chemistry
UTME / JAMB 2011
Acids, Bases, pH & Salts
Solution
Which of the following is a deliquescent compound?
A
NaNO3
B
CaCl2
C
CuO
D
Na2CO3.10H2O
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
Deliquescent substances are referred to known as those substances which absorbs moisture from the surrounding and dissolves themselves with the moisture or water to form a solution.
Examples are sodium hydroxide(NaOH), calcium chloride(CaCl2), potassium hydroxide(KOH), ammonium chloride(NH4Cl) , etc
Source: WAEC / JAMB / NECO Classroom Archive (Record #4779)
Chemistry
WAEC / WASSCE 2011
Periodic Table & Periodicity
Solution
Which of the following ions has the electron configuration 2,8,8?
A
Na+
B
Mg2+
C
F-
D
CI-
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
Chlorine(Cl) has configuration 2,8,7 but Chlorine ion (Cl-) has 18 electrons and configuration 2,8,8.
Here is the reason:
Chlorine has 17 electrons, and these are distributed in shells (or energy levels) as follows: 2 in the first shell, 8 in the second, and 7 in the third.This is represented as 2,8 7. Chlorine can gain one electron to achieve a stable electron configuration, resembling that of the noble gas Argon.
When chlorine gains an electron, its electron configuration becomes 2, 8, 8, making it a chloride ion (Cl⁻). The 2,8,8 configuration means the outermost shell is full with 8 electrons, which is the octet rule, making the ion stable.
Source: WAEC / JAMB / NECO Classroom Archive (Record #37554)
Mathematics
UTME / JAMB 2011
Plane Geometry, Polygons & Circle Theorems
Solution
If 2q35 = 778, find q
A
2
B
1
C
4
D
O
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
2q35 = 778
2 x 52 + q x 51 + 3 x 50 = 7 x 81 + 7 x 80
2 x 25 + q x 5 + 3 x 1 = 7 x 8 + 7 x 1
50 + 5q + 3 = 56 + 7
5q = 63 - 53
q = \(\frac{10}{5}\)
q = 2
Source: Myschool Past Questions Archive (Record #27410)
Mathematics
UTME / JAMB 2011
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify \(\frac{3\frac{2}{3} \times \frac{5}{6} \times \frac{2}{3}}{\frac{11}{15} \times \frac{3}{4} \times \frac{2}{27}}\)
A
\(5\frac{2}{3}\)
B
30
C
\(4\frac{1}{3}\)
D
50
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{3\frac{2}{3} \times \frac{5}{6} \times \frac{2}{3}}{\frac{11}{15} \times \frac{3}{4} \times \frac{2}{27}}\)
\(\frac{\frac{11}{3} \times \frac{5}{6} \times \frac{2}{3}}{\frac{11}{15} \times \frac{3}{4} \times \frac{2}{27}}\)
\(\frac{110}{54} \div \frac{66}{1620}\)
50
Source: Myschool Past Questions Archive (Record #27413)
Mathematics
UTME / JAMB 2011
Fractions, Decimals, Percentages & Approximations
Solution
A man invested N5,000 for 9 months at 4%. What is the simple interest?
A
N150
B
N220
C
N130
D
N250
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
S.I. = \(\frac{P \times R \times T}{100}\)
If T = 9 months, it is equivalent to \(\frac{9}{12}\) years
S.I. = \(\frac{5000 \times 4 \times 9}{100 \times 12}\)
S.I. = N150
Source: Myschool Past Questions Archive (Record #27414)