1984 Mathematics JAMB past Questions And Answers from 1 to 10

1984 mathematics 1 to 10

Click here to see 1983 Mathematics 1 to 10

1. Simplify [(2/3–1/5)–1/3 of 2/5] /(3–1/½)

A. 1/7 B. 7 C. 1/3 D. 3 E. 1/5

Solution:

Clear the Brackets first

[(10 – 3)/15 – 2/15] / [ 3 – (1 x 2/1)]

(7/15 – 2/15)/(3 – 2)

5/15 ÷ 1

= 1/3

2. If 263 + 441 = 714, what number base has been used?

A. 12 B. 11 C. 10 D. 9 E. 8

Solution:

Let the number base be n

Now, let us convert to base 10

(2 x n² + 6 x n¹ + 3 x n⁰) + (4 x n² + 4 x n¹ + 1 x n⁰) = (7 x n² + 1 x n¹ + 4 x n⁰)

2n² + 6n + 3 + 4n² + 4n + 1 = 7n² + n + 4

2n² + 4n² – 7n² + 6n + 4n – n + 3 + 1 – 4 = 0

-n² + 9n = 0

n² = 9n

n = 9

3. 0.00014323/1940000 = k x 10ⁿ where 1 £ k < 10 and n is a whole number. The values of K and n are:

A. 7.381 and –11 B. 2.34 and 10 C. 3.87 and 2 D. 7.831 and –11 E. 5.41 and –2

Solution:

(1.4323 x 10–⁴)/1.94 x 10⁶

0.7383 x 10–¹⁰ = k x 10ⁿ

7.383 x 10–¹¹ = k x 10ⁿ

Therefore, k = 7.383 and n = -11

4. P sold his bicycle to Q at a profit of 10%. Q sold it to R for #209 at a loss of 5%. How much did the bicycle cost P?

A. #200 B. #196 C. #180 D. #205 E. #150

Solution:

Selling price of Q = #209

Cost price = ?

% = 5

% loss = (Cost price – selling price)/ Cost price x 100

5 = (CP – 209)/CP x 100

100CP – 20900 = 5CP

100CP – 5 CP = 20900

95CP = 20900

CP = 20900/95 = #220

Cost price of Q = Selling price of P = #220

P cost price = ?

% = 10

% gain = (SP – CP)/CP x 100

10 = (220 – CP)/CP x 100

22000 – 100CP = 10CP

-100CP – 10CP = – 22000

-110CP = – 22000

CP = 22000/110 = #200

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5. If the price of oranges was raised by 1/2k per orange, the number of oranges customer can buy for #2.40 will be less by 16. What is the present price of an orange?

A. 2½k B. 3½k C. 5½k D. 20k E. 21½k

Solution:

Let P represents price per orange

Q represents the number of oranges that can be bought

Q = 240/P……….i

Now, if the price of oranges is raised by ½k per one, the number of oranges that can be bought for 240k will be less by 16

Q – 16 = 240/(P + ½)

(Q – 16)(P + ½) = 240

(Q – 16)(2P + 1)/2 = 240

(Q – 16)(2P + 1) = 480………ii

Substitute Q = 240/P in equation ii

(240/P – 16)(2P + 1) = 480

(240 – 16P)/P x (2P+1) = 480

(240 – 16P)(2P + 1) = 480P

480P + 240 – 32P² – 16P = 480P

-32P² – 16P + 480P – 480P + 240 = 0

-32P² – 16P + 240 = 0

-2P² – P + 15 = 0

2P² + P – 15 = 0

You can choose to use any quadratic formula to solve for P.

P = 2½k

6. A man invested a total of #50,000 in two companies. If these companies pay dividend of 6% and 8% respectively, how much did he invest at 8% if the total yield is #3700?

A. #15,000 B. #29,600 C. #21,400 D. #27,800 E. #35,000

Solution:

Let the amount invested in the first company be A and the second be (50000 – A)

(6% of A) + [8% of (50000 – A)] = 3700

6A/100 + 8(50000 – A)/100 = 3700

6A + 8(50000 – A) = 370000

6A + 400000 – 8A = 370000

-2A = 370000 – 400000 = – 30000

A = 30000/2 = #15,000

He invested 50000 – 15000 = #35,000 at 8%

7. Thirty boys and x girls sat for a test. The mean of the boys’ scores and that of the girls were respectively 6 and 8. Find x if the total score was 468.

A. 38 B. 24 C .36 D. 22 E. 41

Solution:

Let the total score of the boys be A

A/30 = 6

A = 30 x 6 = 180

Let the total score of the girls be B

B/x = 8

B = 8x

The total score of both = A + B = 180 + 8x = 468

8x = 468 – 180

x = 288/8 = 36

8. The cost of production of an article is made up as follows: Labour #70, Power #15, Materials #30, Miscellaneous #5. Find the angle of the sector representing labour in a pie chart.

A. 210⁰ B. 105⁰ C. 175⁰ D. 150⁰ E. 90⁰

Solution:

Total cost of the Article = 70 + 15 + 30 + 5 = #120

The angle of the sector representing labour in a pie chart = 70/120 x 360

= 70 x 3 = 210⁰

9. Bola chooses at random a number between 1 and 300. What is the probability that the number is divisible by 4?

A. 1/3 B. ¼. C. 1/5 D. 4/300 E. 1/300

Solution:

300/4 = 75

The probability = 75/300

= 1/4

10. Find without using logarithm tables, the value of (Log₃ 27 – Log_¼ 64) ÷ Log₃ 1/81

A. 7/4 B. –7/4 C. –3/2 D. 7/3 E.–1/4

Solution:

Let Log₃ 27 = x

3^x = 27

3^x = 3³

x = 3

Let Log_¼ 64 = y

(1/4)^y = 64

(4^-1 )^y = 4³

4^-y = 4³

-y = 3

y = -3

Let Log₃ 1/81 = z

3^z = 1/81

3^z = 81^-1 = (3⁴)^-1 = 3^-4

z = -4

Therefore, (Log₃ 27 – Log_¼ 64) ÷ Log₃ 1/81 = (x – y) ÷ z

= [3 – (-3)] /-4

6/-4

= -3/2