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2017 JAMB Mathematics Past Questions & Answers

1.

Evaluate 1 − ( $\frac{1}{5}$ x 1 $\frac{2}{3}$ ) + (5 + 1 $\frac{2}{3}$ )

A.4
B.3
C. $\frac{22}{3}$
D.3 $\frac{2}{3}$

Show answer

Answer: C. $\frac{22}{3}$

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2.

Given T = { even numbers from 1 to 12 } N = {common factors of 6, 8 and 12} Find T ∩ N

A.{2, 3}
B.{2, 3, 4}
C.{3, 4, 6}
D.{2}

Show answer

Answer: D. {2}

AI explanation

To solve this problem, we need to break it down into two parts: identifying the elements of set T and set N, and then finding their intersection.

1. Analyze the sets

Set T (Even numbers from 1 to 12): Even numbers are integers divisible by 2. Between 1 and 12 (inclusive), these are: T = {2, 4, 6, 8, 10, 12}

Set N (Common factors of 6, 8, and 12): A common factor is a number that divides into all three numbers without leaving a remainder. • Factors of 6: {1, 2, 3, 6} • Factors of 8: {1, 2, 4, 8} • Factors of 12: {1, 2, 3, 4, 6, 12}

Looking at these lists, the numbers that appear in all three are 1 and 2. N = {1, 2}

2. Find the Intersection (T ∩ N)

The intersection (∩) of two sets consists of all elements that are present in both sets.

T = {2, 4, 6, 8, 10, 12}
N = {1, 2}

Comparing the two, the only element they share is 2. Therefore, T ∩ N = {2}

3. Conclusion

Comparing our result with the given options: A. {2, 3} B. {2, 3, 4} C. {3, 4, 6} D. {2}

The calculated result matches option D.

Exam Tip: When dealing with "common factors," always list the factors of the smallest number first. This often narrows down your search quickly, as a common factor cannot be larger than the smallest number in the set.

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3.

Evaluate ( $\sin$ 45º + $\sin$ 30º ) in surd form

A. $\frac{\sqrt{3}}{2\sqrt{2}}$
B.√3 − $\frac{1}{2}$
C. $\frac{1}{2}$ √2
D. $\frac{1 +\sqrt{2}}{2}$

Show answer

Answer: D. $\frac{1 +\sqrt{2}}{2}$

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4.

The value of x + x ( xx) when x = 2 is

A.16
B.10
C.18
D.24

Show answer

Answer: B. 10

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5.

A car dealer bought a second-hand car for 250,000 and spent N 70,000 refurbishing it. He then sold the car for N400,000. What is the percentage gain?

A.60%
B.32%
C.25%
D.20%

Show answer

Answer: C. 25%

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6.

A man covered a distance of 50 miles on his first trip, on a later trip he traveled 300 miles while going 3 times as fast. His new time compared with the old distance was?

A.three times as much
B.the same
C.twice as much
D.half as much

Show answer

Answer: C. twice as much

AI explanation

To solve this problem, we need to compare the time taken for the first trip with the time taken for the second trip.

1. Analyze the Question

Trip 1: Distance ( $d_1$ ) = 50 miles. Let the speed be $v$ . Time ( $t_1$ ) = $\frac{d_1}{v} = \frac{50}{v}$ .
Trip 2: Distance ( $d_2$ ) = 300 miles. Speed is 3 times as fast, so new speed = $3v$ .
Goal: Compare the new time ( $t_2$ ) with the old time ( $t_1$ ).

2. Explain the Concept

The fundamental formula for motion is:

\text{Time} = \frac{\text{Distance}}{\text{Speed}}

We will calculate the time for both trips and find the ratio between them.

3. Show the Solution

Step 1: Calculate the time for the first trip ( $t_1$ ) $ $t_1 = \frac{50}{v}$ $

Step 2: Calculate the time for the second trip ( $t_2$ ) $ $t_2 = \frac{300}{3v}$ $Simplify the fraction:$ $t_2 = \frac{100}{v}$ $

Step 3: Compare $t_2$ with $t_1$ We want to see how $t_2$ relates to $t_1$ : $ $t_2 = \frac{100}{v}$ $Since$ t_1 = \frac{50}{v} $, we can rewrite$ t_2 $as:$ $t_2 = 2 \times \left(\frac{50}{v}\right)$ $ $ $t_2 = 2 \times t_1$ $

The new time is twice the old time.

4. Identify the Answer

Comparing the result to the given options, the new time is twice as much as the old time. This matches option C.

5. Exam Tip

When dealing with "times as fast" or "times as much" problems, always write out the variables clearly. Don't let the large numbers (50 and 300) distract you; focus on the relationship between the variables. If the distance increases by a factor of 6 (from 50 to 300) and the speed increases by a factor of 3, the time must increase by a factor of $6 \div 3 = 2$ .

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7.

In the diagram above MN is a chord of a circle KMN centre O and radius 10cm. If < MON = 140°, find, to the nearest cm, the length of the chord MN.

A.10cm
B.19cm
C.17cm
D.12cm

Show answer

Answer: B. 19cm

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8.

Given m = N $\sqrt{\frac{SL}{T}}$ make T the subject of the formula

A. $\frac{\text{NSL}}{M}$
B. $\frac{N^2SL}{M^2}$
C. $\frac{N^2SL}{M}$
D. $\frac{NSL}{M^2}$

Show answer

Answer: B. $\frac{N^2SL}{M^2}$

AI explanation

To make $T$ the subject of the formula $m = N\sqrt{\frac{SL}{T}}$ , follow these steps:

1. Analyze the equation

We have $m = N\sqrt{\frac{SL}{T}}$ . Our goal is to isolate $T$ on one side of the equation.

2. Explain the concept

To isolate a variable inside a square root, we must first eliminate the root by squaring both sides. Then, we use algebraic manipulation (cross-multiplication and division) to move the variables until $T$ stands alone.

3. Step-by-step solution

Step 1: Divide both sides by $N$

\frac{m}{N} = \sqrt{\frac{SL}{T}}

Step 2: Square both sides to remove the square root

\left(\frac{m}{N}\right)^2 = \left(\sqrt{\frac{SL}{T}}\right)^2

\frac{m^2}{N^2} = \frac{SL}{T}

Step 3: Cross-multiply to bring $T$ to the numerator

T \cdot m^2 = N^2 \cdot SL

T \cdot m^2 = N^2SL

Step 4: Divide both sides by $m^2$ to isolate $T$

T = \frac{N^2SL}{m^2}

(Note: In your question, the variable $m$ is written as lowercase in the equation but uppercase in the options. Based on standard algebraic conventions, we treat them as the same variable.)

4. Identify the answer

Comparing our result $T = \frac{N^2SL}{m^2}$ with the given options, it matches option B.

5. Exam Tip

When dealing with formulas involving square roots, always remember to square the entire side of the equation, not just the individual terms inside. A common mistake is writing $\frac{m^2}{N}$ instead of $\frac{m^2}{N^2}$ . Always distribute the exponent to every factor in the fraction!

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9.

Make S the subject of the relation p = s + $\frac{sm^2}{nr}$

A.s = $\frac{nrp}{nr + m^2}$
B.s = nr + $\frac{m^2}{mrp}$
C.s = $\frac{nrp}{mr}$ + m2
D.s = $\frac{nrp}{nr}$ + m2

Show answer

Answer: A. s = $\frac{nrp}{nr + m^2}$

AI explanation

Analysis

The goal is to isolate the variable s in the given equation: $p = s + \frac{sm^2}{nr}$

Concept

To make a variable the subject of a formula, we need to group all terms containing that variable on one side of the equation and factor it out. If the variable appears in a fraction, we must find a common denominator or multiply through to clear the fraction.

Step-by-Step Solution

Step 1: Identify the terms with 's' The equation is $p = s + \frac{sm^2}{nr}$ . Both terms on the right side contain $s$ .

Step 2: Factor out 's' Since $s$ is common to both terms, we can factor it out: $p = s(1 + \frac{m^2}{nr})$

Step 3: Simplify the expression inside the bracket To add $1$ and $\frac{m^2}{nr}$ , find a common denominator, which is $nr$ : $1 + \frac{m^2}{nr} = \frac{nr}{nr} + \frac{m^2}{nr} = \frac{nr + m^2}{nr}$

Now substitute this back into the equation: $p = s \left( \frac{nr + m^2}{nr} \right)$

Step 4: Solve for 's' To isolate $s$ , multiply both sides by the reciprocal of the fraction $\left( \frac{nr + m^2}{nr} \right)$ , which is $\left( \frac{nr}{nr + m^2} \right)$ : $s = p \times \left( \frac{nr}{nr + m^2} \right)$

$s = \frac{nrp}{nr + m^2}$

Conclusion

Comparing our result to the given options, we find that it matches option A.

Exam Tip: When you see a variable appearing in multiple terms, always look to factor it out immediately. If you are ever unsure about your algebraic manipulation, try picking simple numbers for $p, n, r,$ and $m$ , calculate $s$ , and plug those values into the options to see which one gives you the same $s$ value!

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10.

In a regular polygon, each interior angle doubles its corresponding exterior angle. Find the number of sides of the polygon

A.8
B.6
C.4
D.3

Show answer

Answer: B. 6

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11.

If $\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}$ = m + n √ 6, find the values of m and n respectively

A.1, − 2
B.− 2, n = 1
C. $\frac{-2}{5}$ , 1
D. $\frac{2}{3}$

Show answer

Answer: B. − 2, n = 1

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12.

Given the quadrilateral RSTO inscribed in the circle with O as centre. Find the size angle x and given RST = 60o

A.100o
B.140o
C.120o
D.10o

Show answer

Answer: B. 140o

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13.

The operation * on the set R of real number is defined by x * y = 3x + 2y − 1, find 3* − 1

A.9
B.- 9
C.6
D.- 6

Show answer

Answer: C. 6

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14.

Find the range of the following set of numbers 0.4, −0.4, 0.3, 0.47, −0.53, 0.2 and −0.2

A.1.03
B.0.07
C.0.03
D.1.0

Show answer

Answer: D. 1.0

AI explanation

Analysis

To find the range of a set of numbers, you must identify the difference between the highest value (maximum) and the lowest value (minimum) in the data set.

Formula:

\text{Range} = \text{Highest Value} - \text{Lowest Value}

Step-by-Step Solution

Step 1: List the numbers and identify the maximum and minimum values. The given set is: $0.4, -0.4, 0.3, 0.47, -0.53, 0.2, -0.2$

Positive numbers: $0.4, 0.3, 0.47, 0.2$
Negative numbers: $-0.4, -0.53, -0.2$

Comparing the positive numbers, the largest is $0.47$ . Comparing the negative numbers, the smallest (the one furthest to the left on the number line) is $-0.53$ .

Step 2: Apply the range formula.

\text{Range} = 0.47 - (-0.53)

Step 3: Perform the calculation. When you subtract a negative number, it becomes addition:

\text{Range} = 0.47 + 0.53

\text{Range} = 1.00

Conclusion

The range of the set is $1.0$ . Comparing this to your options: A. 1.03 B. 0.07 C. 0.03 D. 1.0

The correct option is D.

Exam Tip

Always be careful with signs when dealing with negative numbers! A common mistake is to simply subtract the absolute values (e.g., $0.47 - 0.53$ ). Remember that subtracting a negative is the same as adding the positive equivalent. Always visualize the number line if you are unsure which number is "smaller."

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15.

The curved surface area of a cylinder 5cm high is 110cm2. Find the radius of its base π = $\frac{22}{7}$

A.2.6cm
B.3.5cm
C.3.6cm
D.7.0cm

Show answer

Answer: B. 3.5cm

AI explanation

1. Analyze the question

We are given:

Height of the cylinder ( $h$ ) = 5 cm
Curved surface area ( $A$ ) = 110 cm²
Value of $\pi$ = $\frac{22}{7}$
We need to find the radius ( $r$ ) of the base.

2. Explain the concept

The curved surface area of a cylinder is the area of the side (the "label" part of a tin). The formula for the curved surface area of a cylinder is:

A = 2\pi rh

Where:

$A$ = Curved surface area
$\pi$ = Pi (constant)
$r$ = Radius of the base
$h$ = Height of the cylinder

3. Show the solution

Substitute the given values into the formula:

110 = 2 \times \frac{22}{7} \times r \times 5

Simplify the right side:

110 = \left(2 \times \frac{22}{7} \times 5\right) \times r

110 = \left(\frac{220}{7}\right) \times r

To solve for $r$ , multiply both sides by the reciprocal of $\frac{220}{7}$ (which is $\frac{7}{220}$ ):

r = 110 \times \frac{7}{220}

Divide 110 and 220:

r = \frac{7 \times 110}{220}

r = \frac{7}{2}

r = 3.5 \text{ cm}

4. Identify the answer

Comparing our result to the options provided: A. 2.6cm B. 3.5cm C. 3.6cm D. 7.0cm

The correct option is B.

5. Tip

When dealing with geometry problems involving $\pi = \frac{22}{7}$ , always look for numbers in the equation that are multiples of 11 or 7. This often allows you to cancel out terms early, making your calculation much faster and reducing the risk of arithmetic errors during your JAMB exam!

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16.

Simplify 3 $^{n − 1}$ × $\frac{27^{n + 1}}{81^n}$

A.3 $^{2n}$
B.9
C.3n
D.3 $^{n + 1}$

Show answer

Answer: B. 9

AI explanation

1. Analyze the question

We are asked to simplify the expression: $3^{n-1} \times \frac{27^{n+1}}{81^n}$

To solve this, we need to express all the bases as powers of a common base. Since 3, 27, and 81 are all powers of 3, we will convert everything to base 3.

2. Explain the concept

We use the Laws of Indices:

$a^m \times a^n = a^{m+n}$
$a^m \div a^n = a^{m-n}$
$(a^m)^n = a^{m \times n}$

Conversion to base 3:

$3 = 3^1$
$27 = 3^3$
$81 = 3^4$

3. Show the solution

Step 1: Substitute the base 3 equivalents into the expression. $3^{n-1} \times \frac{(3^3)^{n+1}}{(3^4)^n}$

Step 2: Simplify the powers using the power rule $(a^m)^n = a^{m \times n}$ . $3^{n-1} \times \frac{3^{3(n+1)}}{3^{4n}}$ $3^{n-1} \times \frac{3^{3n+3}}{3^{4n}}$

Step 3: Combine the terms using the multiplication and division rules. First, multiply the terms in the numerator: $3^{(n-1) + (3n+3)} \div 3^{4n}$ $3^{n + 3n - 1 + 3} \div 3^{4n}$ $3^{4n + 2} \div 3^{4n}$

Next, perform the division: $3^{(4n + 2) - 4n}$ $3^{4n - 4n + 2}$ $3^2$

Step 4: Calculate the final value. $3^2 = 9$

4. Identify the answer

Comparing our result to the given options: A. $3^{2n}$ B. 9 C. $3n$ D. $3^{n+1}$

The correct option is B.

5. Tip

Whenever you see indices problems involving numbers like 9, 27, 81, 125, or 625, immediately check if they can be written as powers of a smaller prime number (like 2, 3, or 5). This makes the indices much easier to add or subtract!

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17.

Evaluate 1 - ( $\frac{1}{5}$ x $\frac{2}{3}$ ) + ( 5 + $\frac{2}{3}$ )

A.4
B.3
C.2 $\frac{2}{3}$
D. $\frac{98}{15}$

Show answer

Answer: D. $\frac{98}{15}$

AI explanation

To solve this expression, we must follow the BODMAS rule (Brackets, Of, Division, Multiplication, Addition, and Subtraction).

Expression: $1 - (\frac{1}{5} \times \frac{2}{3}) + (5 + \frac{2}{3})$

Step 1: Solve the operations inside the brackets

First bracket: $(\frac{1}{5} \times \frac{2}{3}) = \frac{1 \times 2}{5 \times 3} = \frac{2}{15}$
Second bracket: $(5 + \frac{2}{3})$ • To add these, find a common denominator: $5 = \frac{15}{3}$ • $\frac{15}{3} + \frac{2}{3} = \frac{17}{3}$

Step 2: Substitute the results back into the expression

Now the expression looks like this: $1 - \frac{2}{15} + \frac{17}{3}$

Step 3: Simplify the expression

To perform the addition and subtraction, find the Least Common Multiple (LCM) of the denominators (1, 15, and 3), which is 15.

$1 = \frac{15}{15}$
$\frac{2}{15} = \frac{2}{15}$
$\frac{17}{3} = \frac{17 \times 5}{3 \times 5} = \frac{85}{15}$

Now combine them: $\frac{15}{15} - \frac{2}{15} + \frac{85}{15}$

$= \frac{15 - 2 + 85}{15}$ $= \frac{13 + 85}{15}$ $= \frac{98}{15}$

Conclusion

Comparing our result with the provided options, we see that it matches option D.

Exam Tip: When dealing with fractions in JAMB, always convert whole numbers to fractions with the same denominator as the other terms. It makes addition and subtraction much faster and reduces the risk of calculation errors.

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18.

Evaluate $\frac{0.00000231}{0.007}$ and leave the answer in standard form

A.3.3 x 10-4
B.3.3 x 10-3
C.3.3 x 10-5
D.3.3 x 10-8

Show answer

Answer: A. 3.3 x 10-4

AI explanation

To solve this problem, we need to express the numbers in standard form (scientific notation) to make the division easier to handle.

1. Analyze the question

We are asked to evaluate the expression:

\frac{0.00000231}{0.007}

And ensure the final result is in standard form ( $a \times 10^n$ , where $1 \leq a < 10$ ).

2. Explain the concept

Standard form allows us to write very small or very large numbers as a product of a number between 1 and 10 and a power of 10.

For $0.00000231$ : Move the decimal point 6 places to the right to get $2.31$ . Since we moved to the right, the exponent is negative: $2.31 \times 10^{-6}$ .
For $0.007$ : Move the decimal point 3 places to the right to get $7$ . The exponent is negative: $7 \times 10^{-3}$ .

3. Show the solution

Now, substitute these into the expression:

\frac{2.31 \times 10^{-6}}{7 \times 10^{-3}}

Step 1: Divide the coefficients

\frac{2.31}{7} = 0.33

Step 2: Divide the powers of 10 Using the law of indices ( $\frac{a^m}{a^n} = a^{m-n}$ ):

10^{-6} \div 10^{-3} = 10^{-6 - (-3)} = 10^{-6 + 3} = 10^{-3}

Step 3: Combine the results

0.33 \times 10^{-3}

Step 4: Convert to proper standard form The coefficient $0.33$ is not between 1 and 10. We must move the decimal one place to the right to get $3.3$ . Moving the decimal to the right decreases the exponent by 1:

0.33 \times 10^{-3} = 3.3 \times 10^{-1} \times 10^{-3} = 3.3 \times 10^{-4}

4. Identify the answer

Comparing our result to the given options, we find it matches option A.

5. Exam Tip

When dividing decimals, it is often easier to multiply both the numerator and denominator by a power of 10 (like $10^6$ or $10^7$ ) to remove the decimals entirely before dividing. Also, always double-check your sign when subtracting negative exponents—many students mistakenly calculate $-6 - 3$ instead of $-6 - (-3)$ .

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19.

Find the number of ways that the letters of the word EXCELLENCE be arranged

A. $\frac{10!}{2!2!2!}$
B. $\frac{10!}{4!2!}$
C. $\frac{10!}{4!2!2!}$
D. $\frac{10!}{2!2!}$

Show answer

Answer: C. $\frac{10!}{4!2!2!}$

AI explanation

Analysis

To find the number of ways to arrange the letters of a word, we use the formula for permutations of objects with repetitions:

\text{Number of arrangements} = \frac{n!}{n_1! \times n_2! \times \dots \times n_k!}

Where:

$n$ is the total number of letters.
$n_1, n_2, \dots$ are the frequencies of the repeating letters.

Concept

Count the total letters ( $n$ ): The word EXCELLENCE has 10 letters.
Identify repeating letters: • E appears 3 times. • X appears 1 time. • C appears 2 times. • L appears 2 times. • N appears 1 time.

Wait, let's re-count carefully: E - X - C - E - L - L - E - N - C - E

E: 4 times (E, E, E, E)
X: 1 time
C: 2 times (C, C)
L: 2 times (L, L)
N: 1 time

Total letters ( $n$ ) = 10. Frequencies: E=4, C=2, L=2.

Solution

Using the formula:

\text{Arrangements} = \frac{10!}{4! \times 2! \times 2!}

$10!$ represents the total permutations if all letters were unique.
$4!$ accounts for the arrangements of the four 'E's that look identical.
$2!$ accounts for the arrangements of the two 'C's.
$2!$ accounts for the arrangements of the two 'L's.

Comparing this to your options: A. $\frac{10!}{2!2!2!}$ B. $\frac{10!}{4!2!}$ C. $\frac{10!}{4!2!2!}$ D. $\frac{10!}{2!2!}$

The calculation matches Option C.

Tip

When solving permutation problems with repeated letters, always write out the word and tally the count of each letter individually before plugging them into the formula. This prevents missing a letter or miscounting the frequency of a repeating one!

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20.

If a rod 10cm in length was measured as 10.5cm, calculate the percentage error

A.5%
B.10%
C.8%
D.7%

Show answer

Answer: A. 5%

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21.

Find the gradient of the line joining the points (3, 2) and (1, 4)

A.3/2
B.2/1
C.-1
D.3/2

Show answer

Answer: C. -1

AI explanation

Analysis

To find the gradient (also known as the slope) of a straight line passing through two points, we use the coordinates of the points $(x_1, y_1)$ and $(x_2, y_2)$ .

Concept

The gradient ( $m$ ) of a line is defined as the "rise over run," or the change in $y$ divided by the change in $x$ . The formula is:

m = \frac{y_2 - y_1}{x_2 - x_1}

Step-by-Step Solution

Identify the coordinates: • Point 1: $(x_1, y_1) = (3, 2)$ • Point 2: $(x_2, y_2) = (1, 4)$

Substitute the values into the formula: $ $m = \frac{4 - 2}{1 - 3}$ $

Perform the subtraction: • Numerator: $4 - 2 = 2$ • Denominator: $1 - 3 = -2$

Calculate the final value: $ $m = \frac{2}{-2} = -1$ $

Conclusion

Comparing our result to the provided options, the correct value is $-1$ .

Exam Tip: Always be careful with the signs in the denominator. A common mistake is to calculate $3 - 1$ instead of $1 - 3$ . If you start with the second point for $y$ , you must start with the second point for $x$ to keep the ratio correct.

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22.

If temperature t is directly proportional to heat h, and when t = 20oC, h = 50 J, find t when h = 60J

A.24oC
B.20oC
C.34oC
D.30oC

Show answer

Answer: A. 24oC

AI explanation

Analysis

The question states that temperature ( $t$ ) is directly proportional to heat ( $h$ ). This means that as one increases, the other increases at a constant rate. Mathematically, this is expressed as $t \propto h$ , which converts to the equation $t = kh$ , where $k$ is the constant of proportionality.

Concept

To solve this, we follow two steps:

Find the constant ( $k$ ): Use the initial values provided ( $t = 20$ and $h = 50$ ) to determine the value of $k$ .
Solve for the unknown: Use the value of $k$ and the new value of $h$ ( $60$ ) to calculate the new temperature ( $t$ ).

Step-by-Step Solution

Step 1: Find $k$ Given: $t = 20$ , $h = 50$ Using the formula $t = kh$ : $20 = k \times 50$ $k = \frac{20}{50}$ $k = 0.4$

Step 2: Find $t$ when $h = 60$ Using the formula $t = kh$ with our known $k$ : $t = 0.4 \times 60$ $t = 24$

The temperature is $24^\circ\text{C}$ .

Conclusion

Comparing our result to the given options, the correct answer is A.

Exam Tip

When dealing with direct proportion questions like this, you can also use the ratio method:

\frac{t_1}{h_1} = \frac{t_2}{h_2}

Substituting the values:

\frac{20}{50} = \frac{t_2}{60}

t_2 = \frac{20 \times 60}{50} = \frac{1200}{50} = 24

This method is often faster and helps you avoid errors when calculating the constant $k$ separately!

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23.

The base in which the operation was performed was

A.6
B.2
C.4
D.5

Show answer

Answer: C. 4

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24.

In how many ways can the word MATHEMATICS be arranged?

A. $\frac{11!}{9!2!}$
B. $\frac{11!}{9!2!2!}$
C. $\frac{11!}{2!2!2!}$
D. $\frac{11!}{2!2!}$

Show answer

Answer: C. $\frac{11!}{2!2!2!}$

AI explanation

Analysis

To find the number of ways to arrange the letters of a word, we use the formula for permutations of a multiset (arrangements with repeated items). The formula is:

\text{Number of arrangements} = \frac{n!}{n_1! \times n_2! \times \dots \times n_k!}

where:

$n$ is the total number of letters.
$n_1, n_2, \dots$ are the frequencies of the repeated letters.

Concept

When some items in a set are identical, swapping them does not create a new unique arrangement. Therefore, we divide the total factorial ( $n!$ ) by the factorial of the count of each repeated letter to remove these redundant arrangements.

Step-by-Step Solution

Count the total number of letters ( $n$ ): M-A-T-H-E-M-A-T-I-C-S There are 11 letters in total. So, $n = 11$ .

Identify the frequency of repeated letters: • M appears 2 times. • A appears 2 times. • T appears 2 times. • H appears 1 time. • E appears 1 time. • I appears 1 time. • C appears 1 time. • S appears 1 time.

Apply the formula: Substitute the values into the permutation formula: $ $\text{Arrangements} = \frac{11!}{2! \times 2! \times 2!}$ $(We divide by$ 2! $for M,$ 2! $for A, and$ 2!$ for T).

Simplify: The expression is $\frac{11!}{2!2!2!}$ .

Conclusion

Comparing our result to the given options, we can see that it matches option C.

Exam Tip: Always write out the word and tally the frequency of each letter carefully before plugging them into the formula. A common mistake is forgetting to count the letters that appear only once (though they don't affect the denominator since $1! = 1$ , it helps ensure you haven't missed any letters).

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25.

Factorize completely X2+2XY+Y2+3X+3Y-18

A.(x + y + 6)(x + y -3)
B.(x - y - 6)(x - y + 3)
C.(x - y + 6)(x - y - 3)
D.(x + y - 6)(x + y + 3)

Show answer

Answer: A. (x + y + 6)(x + y -3)

AI explanation

Analysis

The expression to factorize is: $x^2 + 2xy + y^2 + 3x + 3y - 18$ . This is a quadratic-like expression involving two variables ( $x$ and $y$ ). We can simplify this by grouping terms and recognizing a perfect square trinomial.

Concept

Perfect Square Trinomial: Recall that $(x + y)^2 = x^2 + 2xy + y^2$ .
Substitution: We can replace the group $(x + y)$ with a single variable (let's say $u$ ) to make the quadratic easier to factorize.
Factorization of Quadratics: Once in the form $u^2 + bu + c$ , we look for two numbers that multiply to give $c$ and add to give $b$ .

Step-by-Step Solution

Step 1: Group the terms. Notice that the first three terms form a perfect square: $(x^2 + 2xy + y^2) + 3x + 3y - 18$

Step 2: Simplify the expression. Substitute $(x + y)^2$ for the first part and factor out $3$ from the next two terms: $(x + y)^2 + 3(x + y) - 18$

Step 3: Let $u = (x + y)$ . Now the expression becomes: $u^2 + 3u - 18$

Step 4: Factorize the quadratic. We need two numbers that multiply to $-18$ and add to $+3$ . These numbers are $+6$ and $-3$ . $(u + 6)(u - 3)$

Step 5: Substitute $(x + y)$ back in for $u$ . Replace $u$ with $(x + y)$ : $((x + y) + 6)((x + y) - 3)$ $= (x + y + 6)(x + y - 3)$

Conclusion

Comparing our result to the given options, we find that it matches option A.

Tip

Whenever you see an expression with $x^2, y^2,$ and $2xy$ , always look for the $(x+y)^2$ or $(x-y)^2$ pattern first. It significantly reduces the complexity of the problem and saves you time during the JAMB exam!

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26.

Simplify ( $\sqrt[3]{64a^3})^{-1}$

A.4a
B. $\frac{1}{8a}$
C.8a
D. $\frac{1}{4a}$

Show answer

Answer: D. $\frac{1}{4a}$

AI explanation

1. Analyze the Question

We are asked to simplify the expression $(\sqrt[3]{64a^3})^{-1}$ . This involves two main operations: finding the cube root of the terms inside the bracket and then applying the negative exponent rule.

2. Explain the Concept

Cube Root ( $\sqrt[3]{x}$ ): This is equivalent to raising a number to the power of $1/3$ . So, $\sqrt[3]{64a^3} = (64a^3)^{1/3}$ .
Power of a Product Rule: $(xy)^n = x^n \cdot y^n$ . We apply this to both the constant (64) and the variable ( $a^3$ ).
Negative Exponent Rule: $x^{-n} = \frac{1}{x^n}$ . This rule tells us that a negative exponent indicates the reciprocal of the base.

3. Show the Solution

Step 1: Simplify the expression inside the bracket. We need to find the cube root of $64a^3$ :

\sqrt[3]{64a^3} = \sqrt[3]{64} \times \sqrt[3]{a^3}

Since $4 \times 4 \times 4 = 64$ , the cube root of 64 is 4.
The cube root of $a^3$ is simply $a$ (because $a \times a \times a = a^3$ ).

So, the expression inside the bracket simplifies to $4a$ .

Step 2: Apply the negative exponent. Now we substitute this back into the original expression:

(4a)^{-1}

Using the negative exponent rule ( $x^{-1} = \frac{1}{x}$ ):

(4a)^{-1} = \frac{1}{4a}

4. Identify the Answer

Comparing our result to the given options, we can see that our solution matches option D.

5. Tip

When dealing with cube roots of variables, remember that $\sqrt[3]{x^n} = x^{n/3}$ . If the power is a multiple of 3, it will always simplify perfectly to a whole number power. For constants like 64, always look for the number that, when multiplied by itself three times, gives you the value inside the root!

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27.

In how many ways can the word MACICITA be arranged?

A. $\frac{8!}{2!}$
B. $\frac{8!}{3! 2!}$
C. $\frac{8!}{2! 2! 2!}$
D.8!

Show answer

Answer: C. $\frac{8!}{2! 2! 2!}$

AI explanation

1. Analyze the question

The problem asks for the number of distinct ways to arrange the letters in the word MACICITA. This is a classic problem of permutations with repeated objects.

2. Explain the concept

When arranging a set of $n$ objects where some objects are identical, we use the formula:

\text{Number of arrangements} = \frac{n!}{n_1! \times n_2! \times \dots \times n_k!}

Where:

$n$ is the total number of letters.
$n_1, n_2, \dots$ are the frequencies of the repeated letters.

3. Show the solution

Step 1: Count the total number of letters ( $n$ ). The word is M-A-C-I-C-I-T-A. Counting them: M(1), A(2), C(3), I(4), C(5), I(6), T(7), A(8). So, $n = 8$ .

Step 2: Identify the frequency of each letter.

M: appears 1 time
A: appears 2 times (at positions 2 and 8)
C: appears 2 times (at positions 3 and 5)
I: appears 2 times (at positions 4 and 6)
T: appears 1 time

Step 3: Apply the formula. Total arrangements = $\frac{8!}{1! \times 2! \times 2! \times 2! \times 1!}$

Since $1! = 1$ , we can simplify this to:

\frac{8!}{2! \times 2! \times 2!}

4. Identify the answer

Comparing our result to the given options: A. $\frac{8!}{2!}$ B. $\frac{8!}{3! 2!}$ C. $\frac{8!}{2! 2! 2!}$ D. 8!

The correct option is C.

5. Exam Tip

Whenever you see a question asking for "arrangements" of a word, always write out the word and count the frequency of each letter immediately. A common mistake is to forget to count one of the repeated letters. Always ensure the sum of the frequencies in your denominator equals the total number of letters in the numerator!

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28.

Find the principal which amounts to ₦ 5,500 at a simple interest in 5 years at 2% per annum

A.₦ 4,900
B.₦ 5,000
C.₦ 4,700
D.₦ 4,000

Show answer

Answer: B. ₦ 5,000

AI explanation

Analysis

To solve this problem, we need to understand the relationship between the Principal ( $P$ ), the Amount ( $A$ ), the Rate ( $R$ ), and the Time ( $T$ ) under simple interest.

Concept

In simple interest, the Amount ( $A$ ) is the sum of the Principal ( $P$ ) and the Simple Interest ( $I$ ).

The formula for Simple Interest is: $I = \frac{P \times R \times T}{100}$
The formula for Amount is: $A = P + I$
Substituting the interest formula into the amount formula gives: $A = P + \frac{PRT}{100}$
We can factor out $P$ : $A = P(1 + \frac{RT}{100})$

Solution

Given:

Amount ( $A$ ) = ₦ 5,500
Time ( $T$ ) = 5 years
Rate ( $R$ ) = 2%
Principal ( $P$ ) = ?

Using the formula $A = P(1 + \frac{RT}{100})$ :

$5500 = P(1 + \frac{2 \times 5}{100})$

$5500 = P(1 + \frac{10}{100})$

$5500 = P(1 + 0.1)$

$5500 = P(1.1)$

Now, solve for $P$ : $P = \frac{5500}{1.1}$

$P = \frac{55000}{11}$

$P = 5000$

The principal is ₦ 5,000.

Conclusion

Comparing our result with the given options, the correct choice is B.

Exam Tip

When dealing with simple interest questions, always check if the question asks for the "Interest" or the "Amount." If it asks for the "Amount," remember that you must add the interest back to the principal. A common mistake is to calculate only the interest and stop there!

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29.

If y = x Sin x, find $\frac{dy}{dx}$ when x = $\frac{\pi}{2}$

A. $\frac{- \pi}{2}$
B.-1
C.1
D. $\frac{ \pi}{2}$

Show answer

Answer: C. 1

AI explanation

Analysis

To solve this problem, we need to find the derivative of the function $y = x \sin x$ with respect to $x$ . Since $y$ is the product of two functions of $x$ ( $x$ and $\sin x$ ), we must use the Product Rule of differentiation.

Concept: The Product Rule

The product rule states that if $y = u \cdot v$ , where $u$ and $v$ are functions of $x$ , then:

\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

In this case:

$u = x$
$v = \sin x$

Step-by-Step Solution

Step 1: Differentiate the individual parts.

$\frac{du}{dx} = \frac{d}{dx}(x) = 1$
$\frac{dv}{dx} = \frac{d}{dx}(\sin x) = \cos x$

Step 2: Apply the Product Rule formula.

\frac{dy}{dx} = (x)(\cos x) + (\sin x)(1)

\frac{dy}{dx} = x \cos x + \sin x

Step 3: Substitute $x = \frac{\pi}{2}$ into the derivative.

\frac{dy}{dx} \text{ at } x = \frac{\pi}{2} = \left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right)

Step 4: Evaluate the trigonometric functions.

Recall that $\cos(\frac{\pi}{2}) = 0$
Recall that $\sin(\frac{\pi}{2}) = 1$

Substituting these values:

\frac{dy}{dx} = \left(\frac{\pi}{2} \cdot 0\right) + 1

\frac{dy}{dx} = 0 + 1

\frac{dy}{dx} = 1

Conclusion

The value of the derivative at $x = \frac{\pi}{2}$ is 1. This matches option C.

Exam Tip

When dealing with trigonometric functions in calculus, always ensure your calculator is in Radian mode if you are calculating values, though knowing your unit circle values ( $\sin 90^\circ=1, \cos 90^\circ=0$ ) is much faster and prevents errors during JAMB exams!

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30.

If α and β are the roots of the equation 3x2 + 5x - 2 = 0, find the value of 1/α + 1/β

A. $\frac{-5}{3}$
B. $\frac{-2}{3}$
C. $\frac{1}{2}$
D. $\frac{5}{2}$

Show answer

Answer: D. $\frac{5}{2}$

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31.

If U = {x : x is an integer and 1 ≤ x ≤ 20 } E1 = {x: x is a multiple of 3} E2 = {x: x is a multiple of 4} and an integer is picked at random from U, find the probability that it is not in E2

A. $\frac{3}{4}$
B. $\frac{3}{10}$
C. $\frac{1}{4}$
D. $\frac{1}{20}$

Show answer

Answer: A. $\frac{3}{4}$

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32.

If two graphs Y = px2 + q and y = 2x2 − 1 intersect at x =2, find the value of p in terms of q

A.q − $\frac{8}{7}$
B.7 − $\frac{q}{4}$
C.8 − $\frac{q}{2}$
D.7 + $\frac{q}{8}$

Show answer

Answer: B. 7 − $\frac{q}{4}$

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33.

Find the sum to infinity of the series $\frac{1}{4}$ , $\frac{1}{8}$ , $\frac{1}{16}$ ,..........

A. $\frac{1}{2}$
B. $\frac{3}{5}$
C. $\frac{-1}{5}$
D. $\frac{73}{12}$

Show answer

Answer: A. $\frac{1}{2}$

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34.

A cylindrical tank has a capacity of 3080m3. What is the depth of the tank if the diameter of its base is 14m? Take pi = 22/7.

A.23m
B.25m
C.20m
D.22m

Show answer

Answer: C. 20m

AI explanation

1. Analyze the Question

We are given the volume (capacity) of a cylindrical tank and the diameter of its base. We need to find the depth (height) of the tank.

Volume ( $V$ ): $3080 \, \text{m}^3$
Diameter ( $d$ ): $14 \, \text{m}$
$\pi$ : $22/7$
Find: Height ( $h$ )

2. Explain the Concept

A cylinder's volume is calculated using the formula:

V = \pi r^2 h

Where:

$r$ is the radius of the base.
$h$ is the height (or depth) of the cylinder.

Since we are given the diameter, we must first find the radius. Remember that the radius is half of the diameter ( $r = d / 2$ ).

3. Show the Solution

Step 1: Find the radius ( $r$ )

r = \frac{d}{2} = \frac{14}{2} = 7 \, \text{m}

Step 2: Substitute the known values into the volume formula

3080 = \left(\frac{22}{7}\right) \times (7)^2 \times h

Step 3: Simplify the equation

3080 = \left(\frac{22}{7}\right) \times 49 \times h

Divide 49 by 7:

3080 = 22 \times 7 \times h

3080 = 154 \times h

Step 4: Solve for $h$

h = \frac{3080}{154}

h = 20 \, \text{m}

4. Identify the Answer

Comparing our result to the given options, the correct answer is 20m, which corresponds to option C.

5. Exam Tip

When dealing with geometry problems involving $\pi = 22/7$ , always look for ways to cancel out the denominator early. Notice how the $7$ in the denominator canceled out perfectly with the $7^2$ (which is $7 \times 7$ ), making the multiplication much easier and faster for your JAMB exam!

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35.

What is the product of 2x2 − x + 1 and 3 − 2x

A.4x3 − 8x2 + 5x + 3
B.−4x3 + 8x2 − 5x + 3
C.−4x3 − 8x2 + 5x + 3
D.4x3 + 8x2 − 5x + 3

Show answer

Answer: B. −4x3 + 8x2 − 5x + 3

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36.

y is inversely proportional to x and y is 6 when x = 7. Find the constant of the variation

A.47
B.42
C.54
D.46

Show answer

Answer: B. 42

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37.

The pie chart shows the allocation of money to each sector in a farm. The total amount allocated to the farm is ₦ 80 000. Find the amount allocated to fertilizer

A.₦ 35, 000
B.₦ 40,000
C.₦ 25,000
D.₦ 20,000

Show answer

Answer: B. ₦ 40,000

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38.

Divide 4x3 - 3x + 1 by 2x - 1

A.2x2 -x + 1
B.2x2 - x -1
C.2x2 + x + 1
D.2x2 + x -1

Show answer

Answer: D. 2x2 + x -1

AI explanation

To solve the division of the polynomial $4x^3 - 3x + 1$ by $2x - 1$ , we will use the Long Division method for polynomials.

1. Analyze the problem

We are dividing a cubic polynomial by a linear binomial.

Dividend: $4x^3 + 0x^2 - 3x + 1$ (Note: We insert $0x^2$ as a placeholder to keep the columns aligned).
Divisor: $2x - 1$

2. Step-by-Step Long Division

Step 1: Divide the first term of the dividend by the first term of the divisor.

$4x^3 \div 2x = 2x^2$
Multiply $2x^2$ by $(2x - 1)$ : $2x^2(2x - 1) = 4x^3 - 2x^2$
Subtract this from the dividend: $(4x^3 + 0x^2) - (4x^3 - 2x^2) = 2x^2$

Step 2: Bring down the next term (-3x) and repeat.

New expression: $2x^2 - 3x$
Divide the first term: $2x^2 \div 2x = +x$
Multiply $x$ by $(2x - 1)$ : $x(2x - 1) = 2x^2 - x$
Subtract: $(2x^2 - 3x) - (2x^2 - x) = -3x + x = -2x$

Step 3: Bring down the final term (+1) and repeat.

New expression: $-2x + 1$
Divide the first term: $-2x \div 2x = -1$
Multiply $-1$ by $(2x - 1)$ : $-1(2x - 1) = -2x + 1$
Subtract: $(-2x + 1) - (-2x + 1) = 0$

3. Conclusion

The quotient is $2x^2 + x - 1$ .

Comparing this result to your options: A. $2x^2 - x + 1$ B. $2x^2 - x - 1$ C. $2x^2 + x + 1$ D. $2x^2 + x - 1$

The result matches option D.

Exam Tip

When performing polynomial long division, always ensure your dividend is written in descending order of powers (e.g., $x^3, x^2, x^1, \text{constant}$ ). If a power is missing (like the $x^2$ term in this question), always write it as $0x^2$ . This prevents alignment errors that lead to common mistakes!

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39.

What is the next number in the series 2, 1, $\frac{1}{2}$ , $\frac{1}{4}$ ...

A. $\frac{1}{3}$
B. $\frac{2}{8}$
C. $\frac{3}{7}$
D. $\frac{1}{8}$

Show answer

Answer: D. $\frac{1}{8}$

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40.

Find ∫(x2 + 3x − 5)dx

A. $\frac{x_3}{3}$ - $\frac{3x_2}{2}$ - 5x + k
B. $\frac{x_3}{3}$ - $\frac{3x_2}{2}$ + 5x + k
C. $\frac{x_3}{3}$ + $\frac{3x_2}{2}$ - 5x + k
D. $\frac{x_3}{3}$ + $\frac{3x_2}{2}$ + 5x + k

Show answer

Answer: C. $\frac{x_3}{3}$ + $\frac{3x_2}{2}$ - 5x + k

AI explanation

To solve the indefinite integral $\int (x^2 + 3x - 5) \, dx$ , we apply the Power Rule for Integration.

1. The Concept

The Power Rule states that for any term $ax^n$ :

\int ax^n \, dx = \frac{ax^{n+1}}{n+1} + C

(Where $C$ or $k$ is the constant of integration).

When integrating a polynomial, we integrate each term individually.

2. Step-by-Step Solution

We integrate each part of the expression $(x^2 + 3x - 5)$ separately:

For $x^2$ : Increase the power by 1 ( $2+1=3$ ) and divide by the new power: $ $\int x^2 \, dx = \frac{x^3}{3}$ $

For $3x$ : The power of $x$ is 1. Increase it by 1 ( $1+1=2$ ) and divide by the new power: $ $\int 3x^1 \, dx = \frac{3x^2}{2}$ $

For $-5$ : A constant $a$ integrates to $ax$ . So, $-5$ becomes $-5x$ : $ $\int -5 \, dx = -5x$ $

Adding the constant: Since this is an indefinite integral, we add the constant of integration, $k$ .

3. Combining the results

Putting it all together:

\frac{x^3}{3} + \frac{3x^2}{2} - 5x + k

Comparing this result to your options, it matches option C.

Exam Tip

Always remember that integration is the reverse of differentiation. If you are ever unsure of your answer during an exam, quickly differentiate your result. If you differentiate $\frac{x^3}{3} + \frac{3x^2}{2} - 5x + k$ , you will get back to the original expression $(x^2 + 3x - 5)$ , confirming your answer is correct!

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41.

The locus of a point which is equidistant from the line PQ forms a

A.circle centre P
B.pair of parallel lines each opposite to PQ
C.circle centre Q
D.perpendicular line to PQ

Show answer

Answer: B. pair of parallel lines each opposite to PQ

AI explanation

Analysis

The question asks for the locus (the set of all points that satisfy a specific condition) of a point that maintains a constant distance from a fixed line $PQ$ .

Concept Explanation

In geometry, the locus of points equidistant from a fixed line is a pair of lines running parallel to the original line.

Think of it this way: if you have a straight road (line $PQ$ ), and you want to walk along a path where you are always exactly 5 meters away from that road, you would have to walk in a straight line parallel to the road. Because you could be 5 meters to the "left" or 5 meters to the "right" of the road, there are two such parallel lines—one on each side.

Step-by-Step Solution

Define the fixed object: We have a line segment $PQ$ .
Define the condition: The distance from any point $(x, y)$ on the locus to the line $PQ$ must be a constant, let's call it $d$ .
Visualize the path: • If you move along a path that is always distance $d$ from the line $PQ$ , you are tracing a line parallel to $PQ$ . • Since the distance can be measured on either side of the line $PQ$ , there are two such lines, one on each side, both parallel to $PQ$ .
Evaluate the options: • A & C: A circle is the locus of points equidistant from a fixed point (the center), not a line. • D: A perpendicular line would move closer to or further from the line $PQ$ as you travel along it, so it cannot be the locus. • B: This correctly describes the two parallel lines (one on each side) that maintain a constant distance from the original line $PQ$ .

Conclusion

The locus of a point equidistant from a fixed line consists of two lines parallel to the given line, one on each side.

Exam Tip: Always distinguish between a fixed point and a fixed line.

Equidistant from a point = Circle.
Equidistant from a line = Pair of parallel lines.
Equidistant from two points = Perpendicular bisector.

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42.

Find the sum of the range and the mode of the set of numbers 10, 9, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 7, 10, 6, 5

A.16
B.14
C.12
D.10

Show answer

Answer: A. 16

AI explanation

To solve this problem, we need to determine the range and the mode of the given data set and then find their sum.

1. Analyze the Data

First, let's list the numbers in ascending order to make them easier to work with: 4, 5, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10

2. Find the Range

The range is the difference between the highest value and the lowest value in a data set.

Highest value = 10
Lowest value = 4
Range = $10 - 4 = 6$

3. Find the Mode

The mode is the number that appears most frequently in the data set. Let's count the frequency of each number:

4: appears 1 time
5: appears 1 time
6: appears 2 times
7: appears 3 times
8: appears 3 times
9: appears 4 times
10: appears 6 times

Since 10 appears the most (6 times), the mode is 10.

4. Calculate the Sum

The question asks for the sum of the range and the mode:

Sum = Range + Mode
Sum = $6 + 10 = 16$

Conclusion

Comparing our result to the options provided: A. 16 B. 14 C. 12 D. 10

The calculated sum is 16, which matches option A.

Exam Tip: When dealing with a long list of numbers, always arrange them in ascending order first. It prevents you from missing a number or miscounting the frequency, which is the most common cause of errors in statistics questions.

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43.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.

44.

Simplify 4 $\sqrt{27}$ + 5 $\sqrt{12}$ − 3 $\sqrt{75}$

A.7
B.− 7
C.− 7 $\sqrt{3}$
D.7 $\sqrt{3}$

Show answer

Answer: D. 7 $\sqrt{3}$

AI explanation

Analysis

To simplify this expression, we need to express each square root in its simplest surd form. The goal is to break down the numbers inside the square roots (the radicands) into factors where at least one factor is a perfect square.

Concept: Simplifying Surds

A surd like $\sqrt{n}$ can be simplified if $n$ has a factor that is a perfect square (e.g., 4, 9, 16, 25, 36...). We use the rule:

\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}

By identifying a perfect square factor, we can extract its square root and leave the remaining prime factor under the radical.

Step-by-Step Solution

Step 1: Simplify each term individually.

For $4\sqrt{27}$ : $27 = 9 \times 3$ $4\sqrt{27} = 4\sqrt{9 \times 3} = 4 \times 3\sqrt{3} = 12\sqrt{3}$

For $5\sqrt{12}$ : $12 = 4 \times 3$ $5\sqrt{12} = 5\sqrt{4 \times 3} = 5 \times 2\sqrt{3} = 10\sqrt{3}$

For $3\sqrt{75}$ : $75 = 25 \times 3$ $3\sqrt{75} = 3\sqrt{25 \times 3} = 3 \times 5\sqrt{3} = 15\sqrt{3}$

Step 2: Substitute the simplified terms back into the original expression. The expression becomes:

12\sqrt{3} + 10\sqrt{3} - 15\sqrt{3}

Step 3: Combine the like terms. Since all terms now have $\sqrt{3}$ , we simply add and subtract the coefficients:

(12 + 10 - 15)\sqrt{3}

(22 - 15)\sqrt{3} = 7\sqrt{3}

Conclusion

Comparing our result with the given options, we find that $7\sqrt{3}$ matches option D.

Exam Tip

When you see multiple surds in an addition or subtraction problem, they almost always share a common "root" (in this case, $\sqrt{3}$ ). If you get stuck, try dividing the numbers inside the roots by small prime numbers like 2, 3, or 5 to see which one leaves you with a perfect square!

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45.

Find the equation of the locus of a point p (x, y) such that pv = pw, where v= (1, 1) and w = (3, 5)

A.2x + 2y = 9
B.2x + 3y = 8
C.2x + y = 9
D.x + 2y = 8

Show answer

Answer: D. x + 2y = 8

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