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Biology
WAEC / WASSCE 2017
Nutrition in Plants & Animals (Photosynthesis & Food Substances)
Solution
(a) Explain briefly the process of perceiving smell in humans.
(b) State the adaptive features of the egg of a domestic fowl
(c) (i) What are autotrophs?
(ii) Name two unicellular organisms that are autotrophs
(d) State five functions of the World Health Organisation
(e) Name two example each of plants and animals that live in a freshwater pond.
(f) In the table below, state five differences between the skin of a mammal and the epidermis of a leaf.
(g) A farmer has two plots of land; plots A and B. on Plot A the farmer planted several seeds of maize. On plot B, the farmer planted several seeds of maize and pepper in a small area. What type of competition would be observed on
(i) Plot A
(ii) Plot B
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a) The process of perceiving smell
Chemical substances/odour carried in the air pass through the air/nasal cavity
The molecules dissolve in the moist membrane of the nasal cavity
Sensory cells are stimulated
Impulses are formed
which pass through the nerves ending to the olfactory nerves
which pass into the cerebrum of the brain/brain
The brain interprets the impulses
And the sensation of smell is produced/smell is perceived
(b)Adaptive features of the egg of a domestic fowl;
Porous: shell; for gaseous exchange/provides shelter/ protects the embryo against mechanical damage/ desiccation
Shell membranes; allow gaseous exchange/for protection
Yolk; contains high level of proteins and fats/provides food for embryonic-growth
Egg white/albumen; consists of water to absorb shock/ proteins/food
Air space; provides surface for gaseous exchange Chalaza; holds the yolk/embryo in position.
(c)i. Autotrophs: Organisms that do not rely on other organisms for food/they manufacture their own food; from using simple inorganic compounds; can be chernosyn-thetic/photosynthetic
ii. Examples of unicellular autotrophs Clamydomonas; Euglena; chemosynthetic; bacteria/ sulphur, bacteria; blue-green; algae/Anabaena; Nostoc/ Oscillatoria; Chlorella diatoms; Nitrosomonas; Nitrobacteria
(d) Functions of WHO
provides grants for medical research
Improves public health/promotes environmental health
prevents the spread of diseases vaccination
provides vaccines/drugs in cases of emergencies
Informs/creates awareness/educates the public of any research findings/outbreak of diseases
Assists by sending medical personnel/logistical sup-port to countries upon request
Helps in the training of medical personnel
keeps health statistics
promotes nutritional/sanitary products.
(e) Examples of plants and animals that live in fresh water pond
Plants: Water arum; diatoms; Ceratophyllum/hornwort; utricularia/bladderwort. Arrowhead/sedge; Pistia; Salvinia/water fern; water lily; wolffish/duckweed any other correctly.
Animals: Cyclops; Daphnia/water flea; Hippopotamus; crocodile; water snake; tadpoles; pond skaters; zooplanktons; mosquito larvae; water bugs; water scorpions; leeches; water snails; catfish/Tilapia fishes etc.
(f) Differences between skin of a mammal and epidermis of leaf
skin of a mammal
Epidermis of leaf
skin is made up of many layers/cells
made up of a single layer/cells
presence of sweat gland
absence of sweat gland
presence of sebaceous gland
absence of sebaceous gland
Absence of cuticle
presence of cuticle
Absence of guard cell stomata
Guard cells are present / stomata
Has blood cells capillaries
absence of blood vessels / capillaries
synthesizes vitamin D
does not make vitamin D
Fatty cells are present
absence of fatty cells
It is a sensory organ
its not a sensory organ
Hair follicle present
hair follicle absent
erector muscle present
erector muscle absent
presence of sweat pore
absence of sweat pore
(g) Type of competition on
(i) Plot A: intra specific/intra species competition
(ii) Plot B: interspecific/interspecies competition
Source: Myschool Past Questions Archive (Record #52662)
Biology
UTME / JAMB 2017
Reproduction in Plants & Animals
Solution
Insects that goes through adult stage without becoming pupa are except ________
A
crickets
B
cockroach
C
grasshopper
D
bees
medium Difficulty
Verified Correct Answer: Option D
Biological Concept & Syllabus Guide
The first stage of incomplete metamorphosis is the egg. During this time, the insect will hatch into a form called a nymph.
The nymph is basically a small version of the adult insect.
Source: Official Exam / Past Question Archive
Biology
WAEC / WASSCE 2017
Variety of Organisms & Classification of Living Things
Solution
Study specimen R and answer questions 4(a) to 4(c).
(a) (i) Name the phylum of specimen R. (ii) State two reasons for the answer in 4(a)(i).
(iii) State the habitat of specimen R. (b) (i) State two structural features that adapt specimen R to its habitat. (ii) State three ways in which specimen R is of economic importance.
(c) Make a drawing,8cm- 10cm long of the dorsal view of specimen R and label fully.
Study specimens S and T and answer questions 4(d) to 4(e).
(d) What type of fruits are specimens S and T?
(e) (i) State four observable differences between specimens S and T.
(ii) State three observable similarities between specimens S and T.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a) (i)Phylum of specimen R/Earthworm Annelida.
(ii) Reasons for the classification of specimen R/Earthworm:
-presence of metameric segmentation/segments are separated from each other (by septa) -they are bilaterally symmetrical -segments are separated from each other
-presence of chaetae - body is covered by thin collagen cuticle body is long cylindrical.
Habitat of specimen R/Earthworm: (i) -under decaying leaves -wet/moist soil -in open savanna/savannah beneath the tall grasses - in the forests in decaying tree-stumps.
(i) Features of adaptation of specimen R/Earthworm (b) -bristle-like chaetae/seatae for locomotion moist skin for gaseous exchange
-pointed anterior for burrowing into the soil slimy body reduces friction during movement
-clitellum for attachment during exchange of sperms/reproduction/ secretes cocoon (in which its legs are deposited.
(ii) Economic importance of specimen R/Earthworm -it aerates the soil -It enriches/improves soil fertility.
-its secretions neutralize the acid soil -used bait for fishing as food for some birds.
(d) Type of fruit specimen S/unripe mango fruit - Drupe
Specimen T/tomato fruit-Berry.
(c)(i) Observable differences between specimens Sand T
Specimen S/unripe mango fruit
Specimen T/tomato fruit
-green in colour
red/yellow in colour
one seeded
many seeded
- stony/hard endocarp
fleshy/succulent/soft endocarp
-fibrous mesocarp
succulent mesocarp
- seed large
seed(s) small
basal placentation
axile placentation
mesocarp and endocarp are fused
mesocarp and endocarp are not fused
Similarities between specimens S and T:
-both have seeds -both are fruits -both have three layers/epicard/mesocarp/endocarp/pericarp/ fruit wall
- coloured epicard/pericarp - both have placenta - both have thin epicard both have fleshy mesOcarp
Source: Official Exam / Past Question Archive
Biology
WAEC / WASSCE 2017
Reproduction in Plants & Animals
Solution
Study specimens F and G carefully and answer questions 2(a) and 2(b).
(a) State three observable features of biological importance in: (i) specimen F; (ii) Specimen G.
(b) (i) Classify specimens F and G as either stem tuber or root tuber. (ii) Give two reasons each for the answer in 2(b)(i).
Study specimens H and J and answer questions 2(c) to 2(e).
(C) Classify specimens H and J into the class to which both belong.
(d) (i) state four observable differences between specimens H and J. (ii) State four observable similarities between specimens H and J.
(e) (i) State the feeding habits of each of specimens H and J. (ii) Name two observable features used for feeding in specimen J.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(bi. Classification of Specimen F/Carrot Root tuber
Specimen G / lrish potato Stem tuber
ii. Reasons for classification Specimen F/Carrot -Swollen tap root - lateral roots present has a reduced stem (above the swollen tap root).
Specimen G / irish potato swollen stem/ tuber -presence of bud(s)/eye -presence of scale leaves/leaf -presence of lenticel(s).
(c) Class of specimens H and J Insecta
(d(i). Observable differences between specimens Hand J
H/adult mosquito
J/adult cockroach
- smaller in size
big/large in size
-a pair of wings
two pairs of wings
-absence of hard hind wing
presence of hard hind wing/elytra
-shorter antennae
longer antennae
-proboscis
mandible
absence of maxillary palp/maxilla
presence of maxillary palp/maxilla
- body is cylindrical
body is dorso-ventraly flattened
-thin legs
thick large legs
-absence of spines on legs
presence of spines on legs
ii. Similarities between specimens H adult mosquito and J/ adult cockroach:
-presence of a pair of compound eyes -body is divided into three divisions/head, thorax and abdomen - three pairs of /six (walking) legs
- presence of jointed appendages -presence of a pair of membraneous wings -presence of a pair of antennae
-both have exoskeleton made of chitin -segmented body
(e) Feeding habit of Specimen H / adult mosquito -piercing and sucking Specimen J / adult cockroach Biting and chewing.
(ii) Observable features used for feeding in specimen J adult /cockroach Mandible; maxillae; labium; labrum
Source: Myschool Past Questions Archive (Record #60289)
Biology
WAEC / WASSCE 2017
Basic Ecological Concepts, Ecosystems & Biomes
Solution
Study specimens C, D and E carefully and answer questions 1(a) to 1(C).
(a) (i) Name the phylum to which specimens C and E belong. (ii) State two reasons for the answer in 1(a)(i).
(b) State three observable structural differences between the following specimens. (i)C and D (ii) C and E.
(C) What is the relationship between specimens C and D?
(d)(i)) Name the habitat of specimen D. (ii) State two ways in which specimen D is adapted to its habitat.
(C) Make a drawing, 8cm - 10cm long of the dorsal view of specimen C and label fully.
medium Difficulty
Verified Correct Answer: Option
Biological Concept & Syllabus Guide
(a)(i) Phylum of specimens C / Butterfly and E / Grasshopper - Arthropoda
(ii) Reasons : Metameric segmentation / segmented bodies, jointed appendages, presence of chitinoud exoskeleton made of chittin, bilateral symmetry
(b) Differences:
C / Butterfly
D / Caterpillar
wings are present
wings are absent
proboscis present
mandibles present
- proleg absent
proleg present
- legs are longer
legs are shorter
- claspers are absent
claspers are present
- has one pair of compound eyes
has simple eyes
- has antennae
absence of antenna
- Osmeterium absent
- Osmeterium absent
(ii) C and E
C(Butterfly) -
E (Grasshopper)
-presence of proboscis
presence of mandibles
clubbed/rounded/knobbed antennae
not clubbed/rounded/knobbed antennae/tapering antennae
-wing is membraneous/soft
harder/less membraneous wings
surface of wing is powdery/have Scales/eye spots
surface of wing is not powdery absence of scales/eye spots
-wings are pigmented/colourful
wings have uniform colour green/brown
hind limbs/legs smaller/shorter/less muscular
hind limbs/legs/larger and elongated/ more muscular
- abdomen is hairy
abdomen is not hairy/smooth
(c) Relationship between specimens C/Butterfly and D/caterpillar:
(i) C/Butterfly is the adult/imago of D/caterpillar:
(ii) D/Caterpillar is the larva/larva stage of C/Butterfly.
(d) (i) Habitat of specimen D/Caterpillar: Garden/citrus leaves/vegetables/leaves/fruits/green plant.
(ii) Ways in which Specimen D/Caterpillar is adapted to its habitat: (i) Mandibles; for chewing: (ii) It has three pairs of true legs with claws; for locomotion:
(iii) It has claspers; for attachment to vegetation/twig (IV) Presence of four pairs of proleg/false legs; for climbing:
(V) Its coloured pattern; enables it to blend with its environment/camouflage/escape predators: (VI) It has spiracles: for gaseous exchange;
(VI) It has Simple eyes; for vision; (Vii) Has osmeterium; as defense mechanism/which emits foul smell to scare away predators.
(e) Diagram/drawing of dorsal view of specimen C/Butterfly
Source: Myschool Past Questions Archive (Record #60290)
Mathematics
UTME / JAMB 2016
Sets, Logic & Venn Diagrams
Diagram
Solution
The venn diagram shows a class of 50 students with the games they play. How many students play only two games?
Attached Figure / Diagram
A
15
B
16
C
20
D
18
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Number of students that plays only two games = 4 + 8 + 3 = 15
Source: WAEC / JAMB / NECO Classroom Archive (Record #44756)
Mathematics
UTME / JAMB 2016
Statistics (Data Representation, Central Tendency & Dispersion)
Diagram
Solution
mark
2
3
4
Frequency
4
4
y
The table above shows the frequency distribution of marks obtained by a group of students. If the total mark is 48, find the value of y.
Attached Figure / Diagram
A
6
B
8
C
7
D
5
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
(2 x 4) + (3 x 4) + (4 x y) = 48
8 + 12 + 4y = 48
4y = 48 - 20
\(\frac{4y}{4}\) = \(\frac{28}{4}\)
y = 7
Source: WAEC / JAMB / NECO Classroom Archive (Record #44776)
Physics
UTME / JAMB 2016
Circular Motion, Gravitation & Simple Harmonic Motion
Solution
A satellite revolving around the earth is kept on its orbit by?
A
Centripetal forces only
B
Centripetal and frictional forces
C
Centrifugal forces only
D
Centripetal and centrifugal forces
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The concept of centrifugal force can be applied in rotating devices, such as centrifuges, centrifugal pumps, centrifugal governors, and centrifugal clutches, and in centrifugal railways, planetary orbits and banked curves, when they are analyzed in a rotating coordinate system.
Source: WAEC / JAMB / NECO Classroom Archive (Record #44729)
Physics
WAEC / WASSCE 2016
Atomic, Quantum, Nuclear Physics & Electronics
Solution
In a p-type semiconductor, the?
A
number of holes are equal to the number of electrons
B
electrical resistivity increases
C
electrons are the majority charge carriers
D
holes are the majority charge carriers
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Properties of p type semiconductor
This type of semiconductors form due to the doping trivalent atoms in a pure semiconductor crystal.
The majority carriers in p-type semiconductors are holes which have positive charges.
Minority carriers of this type semiconductors are free electrons.
Source: WAEC / JAMB / NECO Classroom Archive (Record #45692)
Chemistry
WAEC / WASSCE 2016
Non-Metals & Their Compounds
Solution
Which of the following arrangements shows increasing order of reactivity of the halogens?
A
F2 > Cl2 > Br2 >I2
B
I2 2 22
C
F2 2 22
D
I2 > Br2 > Cl2 > F2
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
The reactivity of halogens decreases as you move down Group 7.
Halogens become less reactive as you move down the group as the atomic radius of the element increases which means that the distance between the outer shell of electrons and the nucleus increases.
Source: WAEC / JAMB / NECO Classroom Archive (Record #45091)
Mathematics
UTME / JAMB 2016
Integration & Applications
Solution
Integrate \(\frac{2x^3 + 2x}{x}\) with respect to x
A
\(\frac{2x^3}{3}\) - 2x + k
B
x3 + 2x + k
C
\(\frac{2x^3}{3}\) + 2x + k
D
x3 - 2x + k
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\int\) \(\frac{2x^3 + 2x}{x}\) = \(\int\) \(\frac{2x^3}{x}\) + \(\frac{2x}{x}\) = \(\int\) (2x2 + 2) dx
\(\frac{2x^2+1}{2 + 1}\) + 2x + k
= \(\frac{2x^3}{3}\)+ 2x + k
Source: Myschool Past Questions Archive (Record #44744)
Mathematics
UTME / JAMB 2016
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
If the mean of 4, y, 8 and 10 is 7. Find Y?
A
6
B
10
C
7
D
9
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{4 + y + 8 + 10}{4}\) = 7
22 + y = 4 x 7
22 + y = 28
y = 28 - 22 = 6
Source: Myschool Past Questions Archive (Record #44745)
Mathematics
UTME / JAMB 2016
Differentiation & Applications
Solution
Find the derivative of y = ( \(\frac{1}{3}\)X + 6)\(^2\)
A
2( \(\frac{1}{3}\)X + 6
B
\(\frac{2}{3}\) ( \(\frac{1}{3}\)X + 6)
C
\(\frac{2}{3}\) ( \(\frac{1}{3}\)X + 6)2
D
\(\frac{1}{3}\) ( \(\frac{1}{3}\)X + 6)2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Y = ( \(\frac{1}{3}\)X + 6)\(^2\)
\(\frac{dy}{dx}\) = 2( \(\frac{1}{3}\)X + 6) \(\frac{1}{3}\)
= \(\frac{2}{3}\) ( \(\frac{1}{3}\)X + 6)
Source: Myschool Past Questions Archive (Record #44746)
Mathematics
UTME / JAMB 2016
Sets, Logic & Venn Diagrams
Solution
In a school of 150 students, 80 offer French while 60 offer Arabic and 20 offer neither. How many students offer both subjects?
A
45
B
10
C
35
D
30
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
80 - x + x + 60 - x + 20 = 150
160 - x = 150
x = 160 - 150 = 10
Source: Myschool Past Questions Archive (Record #44747)
Mathematics
UTME / JAMB 2016
Sequences & Series (AP & GP)
Solution
If the 2nd term of a G.P is \(\frac{8}{9}\) and the 6th term is 4\(\frac{1}{2}\). Find the common ratio.
A
2
B
\(\frac{3}{2}\)
C
\(\frac{2}{3}\)
D
3
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
T2 = \(\frac{8}{9}\), T6 = 4\(\frac{1}{2}\)
ar = \(\frac{8}{9}\)
ar5 = \(\frac{9}{2}\)
\(\frac{ar}{ar^5}\) = \(\frac{8}{9}\) x \(\frac{2}{9}\)
\(\frac{1}{r^4}\) = \(\frac{16}{81}\)
r = \(\sqrt[4]{\frac{18}{16}}\) = \(\frac{3}{2}\)
Source: Myschool Past Questions Archive (Record #44748)