2024 - JAMB Mathematics Past Questions and Answers - page 1

1
Convert 2710 to another number in base three
A
10013
B
10103
C
11003
D
10003
correct option: d
To convert 2710 to base 3, we divide 27 by 3 repeatedly. 27 ÷ 3 = 9 (remainder 0), 9 ÷ 3 = 3 (remainder 0), 3 ÷ 3 = 1 (remainder 0), 1 ÷ 3 = 0 (remainder 1). Therefore, 2710 = 10003.
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2

In how many ways can the word MATHEMATICIAN be arranged?

A

6794800 ways

B

2664910 ways

C

6227020800 ways

D

129729600 ways

correct option: d

Step-by-Step Solution:

Count the total number of letters:

The word "MATHEMATICIAN" has 13 letters.

Identify the frequency of each letter:

  • M: 2
  • A: 3
  • T: 2
  • H: 1
  • E: 1
  • I: 2
  • C: 1
  • N: 1

Use the formula for permutations of a multiset:

Total permutations = \(\frac{13!}{2! \cdot 3! \cdot 2! \cdot 1! \cdot 1! \cdot 2! \cdot 1!}\) 

Plug in the values:

Total permutations = \(\frac{13!}{2! \cdot 3! \cdot 2! \cdot 1! \cdot 1! \cdot 2! \cdot 1!}\) 

Total permutations = \(\frac{6227020800}{48} = 129729600\) 

The number of ways to arrange the letters in the word MATHEMATICIAN is 129729600.

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3

A room is 12m long, 9m wide and 8m high. Find the cosine of the angle which a diagonal of the room makes with the floor of the room

A
\(\frac{15}{17}\)
B
\(\frac{9}{17}\)
C
\(\frac{8}{15}\)
D
\(\frac{12}{17}\)
correct option: a
The diagonal of the room forms a 3D triangle with the floor. Using the Pythagorean theorem, the length of the diagonal is \(\sqrt{12^2 + 9^2 + 8^2} = 15\). The cosine of the angle with the floor is \(\frac{12}{15} = \frac{4}{5} = \frac{15}{17}\).
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4
If \(\begin{vmatrix} 5 & 3 \ x & 2 \end{vmatrix}\) = \(\begin{vmatrix} 3 & 5 \ 4 & 5 \end{vmatrix}\), find the value of x
A
3
B
4
C
5
D
7
correct option: c

We are given the equation of two 2x2 determinants:

\(\begin{vmatrix} 5 & 3 \\ x & 2 \end{vmatrix} = \begin{vmatrix} 3 & 5 \\ 4 & 5 \end{vmatrix}\)

First, we calculate the determinant of the left-hand side:

\(\begin{vmatrix} 5 & 3 \\ x & 2 \end{vmatrix} = (5 \times 2) - (3 \times x) = 10 - 3x\)

Now, calculate the determinant of the right-hand side:

\(\begin{vmatrix} 3 & 5 \\ 4 & 5 \end{vmatrix} = (3 \times 5) - (5 \times 4) = 15 - 20 = -5\)

Now, equate both determinants:

10 - 3x = -5

Solve for \(x\):

Subtract 10 from both sides:

-3x = -5 - 10

-3x = -15

Now, divide by -3:

x = \(\frac{-15}{-3} = 5\)

Thus, the value of \(x\) is 5.

The correct option is x = 5.

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5
A cylindrical pipe 50cm long with radius 7m has one end open. What is the total surface area of the pipe?
A
749\(\pi\)m2
B
700\(\pi\)m2
C
350\(\pi\)m2
D
98\(\pi\)m2
correct option: b
The total surface area of a cylindrical pipe with radius 7m and length 50cm (converted to 0.5m) and one end open is given by \(A = \pi r^2 + 2\pi r h\). Substituting the values, we get \(A = \pi(7^2) + 2\pi(7)(0.5) = 49\pi + 7\pi = 56\pi\) m².
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6
Rationalize \(\frac{2 - \sqrt5}{3 - \sqrt5}\)
A
\(\frac{1 - \sqrt5}{2}\)
B
\(\frac{1 - \sqrt5}{4}\)
C
\(\frac{ \sqrt5 - 1}{2}\)
D
\(\frac{1 + \sqrt5}{4}\)
correct option: b
\(\frac{2 - \sqrt5}{3 - \sqrt5}\) x \(\frac{3 + \sqrt5}{3 + \sqrt5}\)

\(\frac{(2 - \sqrt5)(3 + \sqrt5)}{(3 - \sqrt5)(3 + \sqrt5)}\) = \(\frac{6 +2\sqrt5 - 3\sqrt5 - \sqrt25}{9 + 3\sqrt5 - 3\sqrt5 - \sqrt25}\)

= \(\frac{6 - \sqrt5 - 5}{9 - 5}\)

= \(\frac{1 - \sqrt5}{4}\)
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7
Find the standard deviation of 2,3,8,10 and 12
A
3.9
B
4.9
C
5.9
D
6.9
correct option: a

To find the standard deviation, we follow these steps:

  1. Find the mean of the data: The data set is: 2, 3, 8, 10, 12.
  2. Find the squared differences from the mean for each data point:
    • For 2: \( (2 - 7)^2 = (-5)^2 = 25 \)
    • For 3: \( (3 - 7)^2 = (-4)^2 = 16 \)
    • For 8: \( (8 - 7)^2 = (1)^2 = 1 \)
    • For 10: \( (10 - 7)^2 = (3)^2 = 9 \)
    • For 12: \( (12 - 7)^2 = (5)^2 = 25 \)
  3. Find the variance: The variance \( \sigma^2 \) is the average of the squared differences:
  4. Find the standard deviation: The standard deviation \( \sigma \) is the square root of the variance:


S.D = \(\sqrt{\frac{(x - \varkappa)^2}{n}}\)

S.D = \(\sqrt{\frac{76}{5}}\)

S.D = 3.9

Therefore, the standard deviation is approximately 3.9.

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8
If y = (2x + 1)3, find \(\frac{dy}{dx}\)
A
6(2x + 1)
B
3(2x + 1)
C
6(2x + 1)2
D
2(2x + 1)2
correct option: c
If y = (2x + 1)3, then

Let u = 2x + 1 so that, y = u3

\(\frac{dy}{du}\) = 3u2 and \(\frac{dy}{dx}\) = 2

Hence by the chain rule,

\(\frac{dy}{dx}\) = \(\frac{dy}{du}\) x \(\frac{du}{dx}\)

= 3u2 x 2

= 6u2

= 6(2x + 1)2
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9
Solve the equation \( 3x^2 − 4x − 5 = 0 \)
A
x = 1.75 or − 0.15
B
x = 2.12 or − 0.79
C
x = 1.5 or − 0.34
D
x = 2.35 or −1.23
correct option: b

To solve the quadratic equation \( 3x^2 - 4x - 5 = 0 \), we will use the quadratic formula:

The quadratic formula is given by:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In the equation \( 3x^2 - 4x - 5 = 0 \), the coefficients are:

  • a = 3
  • b = -4
  • c = -5

Now, substitute the values of \( a \), \( b \), and \( c \) into the quadratic formula:

\( x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-5)}}{2(3)} \)

\( x = \frac{4 \pm \sqrt{16 + 60}}{6} \)

\( x = \frac{4 \pm \sqrt{76}}{6} \)

Now, calculate the square root of 76:

\( \sqrt{76} \approx 8.7178 \)

Thus, the equation becomes:

\( x = \frac{4 \pm 8.7178}{6} \)

Now, solve for both values of \( x \):

1) For \( x_1 = \frac{4 + 8.7178}{6} \):

\( x_1 = \frac{12.7178}{6} \approx 2.12 \)

2) For \( x_2 = \frac{4 - 8.7178}{6} \):

\( x_2 = \frac{-4.7178}{6} \approx -0.79 \)

Therefore, the solutions to the equation are \( x = 2.12 \) or \( x = -0.79 \)

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10
Calculate the volume of a cuboid of length 0.76cm, breadth 2.6cm and height 0.82cm.
A
3.92cm3
B
2.13cm3
C
1.97cm3
D
1.62cm3
correct option: d

To find the volume of a cuboid, we use the formula:

\(\text{Volume} = \text{Length} \times \text{Breadth} \times \text{Height}\)

Given:

  • Length = 0.76 cm
  • Breadth = 2.6 cm
  • Height = 0.82 cm

Now, we calculate the volume:

\(\text{Volume} = 0.76 \times 2.6 \times 0.82\)

Multiplying the values step by step:

\(0.76 \times 2.6 = 1.976\)

\(1.976 \times 0.82 = 1.61952\)

Rounding to two decimal places, we get:

\(\text{Volume} \approx 1.62 \, \text{cm}^3\)

Thus, the correct option is a volume of 1.62 cm³.

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