Free Mathematics Exam Questions and Answers SS3

Free Mathematics Exam Questions and Answers SS3

These Mathematics questions and answers were pulled from our book (Mathematics questions for SS3); Compiled to serve as a reference material to help teachers draw up test and exam questions faster. It could also help students assess their level of exam preparation. Each sample question includes correct answers.
[su_note note_color=”#fcf2e0″]The Questions are based on the current NERDC curriculum (UBE compliant)[/su_note]

Sample Mathematics Exam Questions and Answers

ARITHMETIC AND GEOMETRIC PROGRESSION
1. Johnbull bought an electronic gadget worth ₦208,000. He pays half the worth and agrees to pay the remaining in five instalments with a compound interest of 6% per annum, calculate the annual instalment to the nearest naira.
A. ₦ 26,300
B. ₦ 25.428
C. ₦ 14,380
D. ₦ 24,700

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is D
Explanation:
Worth of item = ₦ 208,000. Amount paid = ₦ half the worth = 208,000  1/2 = 104,000.
Balance = 104,000  [1 + (6/100)]5 = [a  (1.065 – 1)]/[1.06 – 1]; 104,000  [1.06]5
= [a  (1.065 – 1)]/0.06, therefore, a = [104,000  (1.06)5  0.06]/[1.065 – 1] = ₦ 24,700
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2. Calculate the amount arising from annuity of ₦ 750 annually payable in 10 years, if the compund interest payable is at 7% per annum.
A. ₦ 10,363
B. ₦ 9,384
C. ₦ 40,040
D. ₦ 25,313

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A

Explanation:
Annual annuity = ₦ 750, time payable
= 10 years, rate = 7% = 0.07 + 1 = 1.07
Amount = [₦ 750(1.0710 – 1)] [1.07 – 1] = ₦ 10,362.9  ₦ 10,363
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3. Find the amount an annuity of ₦200 payable for 11years at 10% per annum interest.
A. ₦ 7065
B. ₦ 6053
C. ₦ 3706
D. ₦ 1200

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is C

Explanation:
Using the equation S =
[a  (rn – 1)]/[r – 1], where a = ₦ 200, r = 1.1, n = 11.
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CHANGE OF SUBJECT FORMULA
1. Make B the subject of formula: A = [2B + C]/[2BC].
A. B = C/[2A2C – 2]

B. B = C/[2A2C + 2]

C. B = 1/[CA2 + 2]

D. B = 1/[2A2C + 2]

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A

Explanation:
A = [2B + C]/[2BC];
A2 = [2B + C]/[2BC];
2A2BC = 2B + C;
C = 2A2BC – 2B;
C = B[2A2C – 2];
B = C/[2A2C – 2]
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2. Given that A = [12/C – B/3], make C the subject of formula.
A. C = [3A2 + B]/36

B. C = 36/[3A2 + B]

C. C = [36B]/[3A2]

D. C = [3A3 – B]/36

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is B

Explanation:
A = [12/C – B/3];
A2 = 12/C – B/3 = [36 – BC]/3C;
3CA2 = 36 – BC;
36 = 3CA2 + BC;
36 = C[3A2 + B];
C = 36/[3A2 + B] [/su_spoiler]

3. WX + WY + 7/Z = 3, make Y the subject.
A. [3Z – 7 – ZWX]/ZW
B. [7Z – 3 + ZWX]/ZX
C. [ZX – ZWX + 7]/WX
D. [WX + 7Z – ZW]/ZX

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A

Explanation:
WX + WY + 7/Z = 3;
[ZWX + ZWY + 7]/Z = 3;
ZWX + ZWY +7 = 3Z; ZWY = 3Z – 7 – ZWX;
Y = [3Z – 7 – ZWX]/ZW [/su_spoiler]

4. Make P the subject: R = [Q2 – PR]/[Q + P].
A. P = [Q(Q – R)]/2R
B. P = [2R(Q2 – Q)]/R
C. P = [2R(1 – Q2)]/Q
D. P = [2Q(R – Q)]/RQ

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A

Explanation:
R = [Q – PR]/[Q + P]; R[Q + P] = Q2 – PR;
RQ + RP = Q2 – PR; RQ = Q2 – PR – RP;
RQ = Q2 -2PR; 2PR = Q2 – RQ;
P = [Q2 – RQ]/2R = [Q(Q – R)]/2R
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5. Make g the subject of formula from the equation t/L = [(42)/g].
A. g = [43L]/t.
B. g = [4]/[tL]
C. g = [42L2]/t2
D. g = [4L2]/t2

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is C

Explanation:
t/L = [(42)/g], square both sides of the equation,
then, t2/L2 = [42]/g;
gt2 = 42L2;
g =[42L2]/t2.
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Click here to get the complete Mathematics questions for SS3

EQUATIONS AND FORMULAE
1. One stick is 9cm longer than another, 25 of the longer stick is equal to 12 of the shorter stick. Find the length of the longer stick.
A. 45 cm
B. 36 cm
C. 27 cm
D. 54 cm

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A

Explanation:
Let the shorter stick be n;
2/5[n + 9] = n/2  2  2[n + 9] = 5n;
4n + 36 = 5n  5n – 4n = 36,
therefore, n = 36cm.
The longer stick is 36 + 9 = 45cm
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2. The result of adding 3 to x and multiplying the answer by 4 is the same as taking 3 from five times x. Find the value of x.
A. 27
B. 6
C. 15
D. -15

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is C

Explanation:
4[x + 3] = 5x – 3;
4x + 12 = 5x – 3  5x – 4x = 12 + 3;
x = 15.
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3. Somina and Qiana share 191 naira between them so that Qiana get 27 naira less than Somina. Find how much money each gets.
A. ₦ 109; ₦ 82
B. ₦ 109; ₦ 136
C. ₦ 136; ₦ 82
D. ₦ 218; ₦ 191

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A

Explanation:
Let Somina’s share be represented by x;
Qiana receives x – 27; x + x – 27 = 191 2x – 27 = 191; 2x = 191 + 27; 2x = 218.
Therefore, x = 218/2 = ₦ 109.
Qiana’s share = 109 – 27 = ₦ 82.
The answer is ₦ 109 ; ₦ 82
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4. The sum of 8 and one-fourth of n is one more than twice n. Find the value of n.
A. -12
B. 4
C. -312
D. 8

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is B

Explanation:
8 + n/4 = 2n + 1;
multiply through by 4;
32 + n = 8n + 4  8n – n = 32 – 4  7n = 28.
Therefore, n = 28/7 = 4.
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5. A rectangle is one-third as long as it is wide. If its perimeter is 120cm, find the width of the rectangle.
A. 15cm
B. 20cm
C. 30cm
D. 45cm

[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is D

Explanation:
Perimeter of a quadrilateral shape = 2[L + B],
where B = Breadth, L = length = B/3,
Perimeter = 120  2[B/3 + B] = 120;
B/3 + B = 120/2 = 60  B + 3B/3 = 60;
B + 3B = 60  3  4B = 180;
B = 180/4
= 45cm
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[su_column size=”4/5″ center=”no” class=””]Want more questions like this? Get the Complete Mathematics Exam Questions and Answers SS3, with even more questions and answers