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Chemistry
UTME / JAMB 1978
Stoichiometry, Mole Concept & Volumetric Analysis
Which of the following changes is physical?
A
Adding iron filing to aerated water
B
Adding sodium metal to water
C
Cooling a solution of Iron (II) sulphate to obtain the hydrated salt
D
Cooling water to obtain ice
E
Burning the domestic gas (Utilgas) for cooking
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #762)
Chemistry
UTME / JAMB 1978
Organic Chemistry II: Functional Groups, Polymers & Biomolecules
Solution
By means of filtration, one component can be obtained pure from an aqueous mixture of sodium chloride and?
A
potassium nitrate
B
sand
C
lead nitrate
D
sugar (glucose)
E
starch
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
Use enough water to completely submerge the sand-salt mixture. Stir or shake the mixture for a few minutes so that the salt dissolves in the water.
The sand is insoluble, so it will remain visible.
Source: Myschool Past Questions Archive (Record #766)
Chemistry
UTME / JAMB 1978
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
In titration involving sodium hydroxide and dilute hydrochloric acid, where would you place the base?
A
in the beaker
B
in the conical flask
C
in a burette
D
in the standard flask
E
in the measuring cylinder
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
Add a few drops of indicator and put the conical flask on a white tile. Fill the burette with acid and note the starting volume. Slowly add the acid from the burette to the alkali in the conical flask, swirling to mix.
Source: Myschool Past Questions Archive (Record #767)
Mathematics
UTME / JAMB 1978
Trigonometry, Elevation & Bearings
Solution
Evaluate without using tables sin(-1290º)
A
\(\frac{3}{2}\)
B
-\(\frac{3}{2}\)
C
\(\frac{2}{2}\)
D
1
E
\(\frac{1}{2}\)
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
sin(-1290º) = -sin(1290º)
sin([3*360] + 210)
where sin 360 = 0 and sin 210 = 180 + 30 ⇔ -30º
-sin ([3* 0] + [-30])
-sin(-30)
sin30 = \(\frac{1}{2}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #32724)
Mathematics
UTME / JAMB 1978
Indices, Logarithms & Surds
Diagram
Solution
In the Figure FD\\AC, the area of AEF = 6sq.cm. AE = 3cm, BC = 3cm, CD = 5cm, < BCD is an obtuse angle. Find the length of BD.
Attached Figure / Diagram
A
\(\sqrt{33}\)
B
6
C
4
D
2\(\sqrt{13}\)
E
4\(\sqrt{5}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Area of triangle AEF = 6sq. cm
area of triangle = \(\frac{1}{2}\)bh (Line DX makes right angles with the parallel lines)
6 = \(\frac{1}{2}\) x 3 x h
6 = 3h
3h = \(\frac{12}{3}\)
h = 4 = DX
From D, C x D, CX\(^2\) = 5\(^2\)
- 4\(^2\) = 25 - 16 = 9
Cx = 3. From angle B x D, Bx = 6(i.e. 3 + 3)
BD\(^2\) = 4\(^2\) + 6\(^2\)
= 16 + 36 = 52
BD = \(\sqrt{4 \times 13}\)
= 2\(\sqrt{13}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #32731)
Mathematics
UTME / JAMB 1978
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, 0 is the centre of the circle ABC, < CED = 30o, < EDA = 40o. What is the size of < ABC?
Attached Figure / Diagram
A
70o
B
110o
C
130o
D
125o
E
145o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
If < CED = 30º, and < EDA = 40º then
Source: WAEC / JAMB / NECO Classroom Archive (Record #32732)
Mathematics
UTME / JAMB 1978
Indices, Logarithms & Surds
Solution
If x\(^2\) + 4 = 0, then x ?
A
4
B
-4
C
2
D
-2
E
None of the above
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
x\(^2\) + 4 = 0
x\(^2\) = -4
You can't apply a square root to a negative number. So the answer is None of the above.
Source: WAEC / JAMB / NECO Classroom Archive (Record #32692)
Mathematics
UTME / JAMB 1978
Mensuration (Areas, Perimeters & Volumes)
Solution
If a circular paper disc is trimmed in such a way that its circumference is reduced in the ratio 2:5, In what ratio is the surface area reduced?
A
8 : 125
B
2 : 5
C
8 : 25
D
4 : 25
E
4 : 10
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
surface area of formula = \pir\(^2\)
If the radius is reduced then let its radius be x.
Its area is \pix\(^2\) .
x : r = 2 : 5,
so \(\frac{x}{r}\) = \(\frac{2}{5}\)
→ 5x = 2r
Hence, x = 0.4r.
Hence the area of the new circle is \pi (0.4r)\(^2\) = 0.16\pi r\(^2\) .
The ratio of the two areas is
0.16\pir\(^2\) : \pir\(^2\)
= 0.16 : 1 = 16 : 100
= 4 : 25.
Source: WAEC / JAMB / NECO Classroom Archive (Record #32698)
Physics
UTME / JAMB 1978
Newton's Laws, Momentum & Collisions
Solution
The frictional force between two bodies?
A
exists only when there is relative motion or tendency for motion
B
acts so as to oppose the motion
C
depends on the normal reaction between the two surfaces
D
is relatively independent of the area in contact
E
has all of these characteristics
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
Frictional force is the opposing force that is created between two surfaces that try to move in the same direction or that try to move in opposite directions.
The main purpose of a frictional force is to create resistance to the motion of one surface over the other surface.
Source: WAEC / JAMB / NECO Classroom Archive (Record #29232)
Physics
UTME / JAMB 1978
Calorimetry, Latent Heat & Change of State
Solution
The following are some units i. Ns ii. Mm iii. Nm\(^-2\) iv. JKg\(^-1\). What are the units of latent heat?
A
i
B
ii
C
iii
D
iv
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
=
specific latent heat of a substance
L = \(\frac{Q}{M}\)
i.e unit is JKg\(^-1\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #29278)
Mathematics
UTME / JAMB 1978
Plane Geometry, Polygons & Circle Theorems
Solution
A rectangular picture 6cm by 8cm is enclosed by a frame \(\frac{1}{2}\)cm wide. Calculate the area of the frame
A
15sq.cm
B
20sq.cm
C
13sq.cm
D
16sq.cm
E
17sq.cm
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Area of the rectangular picture = L x B = 8 x 6
= 48 sq.cm.
Area of the whole surface (which is gotten by adding \(\frac{1}{2}\) on every side to the original picture size) is 9 x 7 = 63 sq. cm
area of the frame is 63 - 48
= 15 sq. cm
Source: Myschool Past Questions Archive (Record #32685)
Mathematics
UTME / JAMB 1978
Plane Geometry, Polygons & Circle Theorems
Solution
The sum of \(3\frac{7}{8}\) and \(1\frac{1}{3}\) is less than the difference between \(\frac{1}{8}\) and \(1\frac{2}{3}\) by:
A
3\(\frac{2}{3}\)
B
5\(\frac{1}{4}\)
C
6\(\frac{1}{2}\)
D
8
E
8\(\frac{1}{8}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(3\frac{7}{8} + 1\frac{1}{3} = 4\frac{21 + 8}{24}\)
= \(4\frac{29}{24}\)
\(\equiv 5\frac{5}{24}\)
\(1\frac{2}{3} - \frac{1}{8} = \frac{5}{3} - \frac{1}{8}\)
= \(\frac{40 - 3}{24}\)
= \(\frac{37}{24}\)
\(5\frac{5}{24} - \frac{37}{24} = \frac{125}{24} - \frac{37}{24}\)
= \(\frac{11}{3}\)
= 3\(\frac{2}{3}\)
Source: Myschool Past Questions Archive (Record #32686)
Mathematics
UTME / JAMB 1978
Linear, Simultaneous & Quadratic Equations
Solution
Multiply (x + 3y + 5) by (2x2 + 5y + 2)
A
2x2 + 3yx2 + 10xy + 15y2 + 13y + 10x2 + 2x + 10
B
2x3 + 6yx2 + 5xy + 15y2 + 31y + 10x2 + 2x + 10
C
2x3 + 6xy2 + 5xy + 15y2 + 12y + 10x2 + 2x + 10
D
2x2 + 6xy2 + 5xy + 15y2 + 13y + 10x2 + 2x + 10
E
2x3 + 2yx2 + 10xy + 10y2 + 31y + 5x2 + 10
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\((x + 3y + 5)(2x^{2} + 5y + 2)\)
= \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 6y + 10x^{2} + 25y + 10\)
= \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 31y + 10x^{2} + 10\)
Source: Myschool Past Questions Archive (Record #32687)
Mathematics
UTME / JAMB 1978
Fractions, Decimals, Percentages & Approximations
Solution
Arrange \(\frac{3}{5}\),\(\frac{9}{16}\), \(\frac{34}{59}\) and \(\frac{71}{97}\) in ascending order of magnitude.
A
\(\frac{3}{5}\), \(\frac{9}{16}\), \(\frac{34}{59}\), \(\frac{71}{97}\)
B
\(\frac{9}{16}\), \(\frac{34}{59}\), \(\frac{3}{5}\) , \(\frac{71}{97}\)
C
\(\frac{3}{5}\), \(\frac{9}{16}\), \(\frac{71}{97}\), \(\frac{34}{59}\)
D
\(\frac{9}{16}\), \(\frac{3}{5}\), \(\frac{71}{97}\), \(\frac{34}{59}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{3}{5}\) = 0.60, \(\frac{9}{16}\) = 0.56, \(\frac{34}{59}\) = 0.58, \(\frac{71}{97}\) = 0.73
Hence, \(\frac{9}{16} < \frac{34}{59} < \frac{3}{5} < \frac{71}{97}\)
Source: Myschool Past Questions Archive (Record #32688)
Mathematics
UTME / JAMB 1978
Sequences & Series (AP & GP)
Solution
The sum of the progression is 1 + x + x2 + x3 + ......
A
\(\frac{1}{1 - x}\)
B
\(\frac{1}{1 + x}\)
C
\(\frac{1}{x - 1}\)
D
\(\frac{1}{x}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Sum of n terms of Geometric progression is \(S_{n} = \frac{a(1 - r^n)}{1 - r}\)
In the given series, a (the first term) = 1 and r (the common ratio) = x.
\(S_{n} = \frac{1(1 - x^{n})}{1 - x}\)
a = 1, and as n tends to infinity
\(S_{n} = \frac{1}{1 - x}\)
Source: Myschool Past Questions Archive (Record #32689)