JAMB Mathematics 1988 past questions and answers

49 questions. Try each one, then open the answer to check yourself and read the explanation.

Question 1

Fractions, Decimals and Approximation

Simplify
1½/(2 ÷ 1/4 of 32)

  1. A.3/256
  2. B.3/32
  3. C.6
  4. D.85
Show the answer and explanation

Answer: C (6)

Explanation

"Of" comes first: 1/4 of 32 = 8. Then 2 ÷ 8 = 1/4, and 1½ ÷ 1/4 = 3/2 × 4 = 6.

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Question 2

Fractions, Decimals and Approximation

If x is the addition of prime numbers between 1 and 6; and y the H.C.F. of 6, 9, 15, find the product of x and y.

  1. A.27
  2. B.30
  3. C.33
  4. D.90
Show the answer and explanation

Answer: B (30)

Explanation

The primes between 1 and 6 are 2, 3 and 5, so x = 10. The HCF of 6, 9 and 15 is 3. So xy = 30.

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Question 3

Ratio, Percentages and Rates

A 5.0g of salt was weighed by Tunde as 5.1g. What is the percentage error?

  1. A.20
  2. B.2
  3. C.1
  4. D.0.2
Show the answer and explanation

Answer: B (2)

Explanation

The error is 5.1 − 5.0 = 0.1g. Percentage error = 0.1/5.0 × 100 = 2%.

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Question 4

Fractions, Decimals and Approximation

Find correct to one dec. place, 0.24633 ÷ 0.0306

  1. A.0.8
  2. B.1.8
  3. C.8.0
  4. D.8.1
Show the answer and explanation

Answer: D (8.1)

Explanation

0.24633 ÷ 0.0306 = 24633 ÷ 3060 = 8.05, which is 8.1 to one decimal place.

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Question 5

Commercial Arithmetic

Two sisters, Taiwo and Kehinde, own a store. The ratio of Taiwo's share to Kehinde's is 11:9. Later Kehinde sells 2/3 of her share to Taiwo for N720.00 find the value of the store.

  1. A.N1,080.00
  2. B.N2,400.00
  3. C.N3,000.00
  4. D.N3,600.00
Show the answer and explanation

Answer: B (N2,400.00)

Explanation

Kehinde owns 9/20 of the store, and 2/3 of that is 6/20 = 3/10 of the store. So 3/10 of the value is N720, and the store is worth 720 × 10/3 = N2,400.

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Question 6

Ratio, Percentages and Rates

A basket contains green, black and blue balls in the ratio 5:2:1. If there are 10 blue balls, find the corresponding new ratio when 10 green and 10 black balls are removed from the basket.

  1. A.1:1:1
  2. B.4:2:1
  3. C.5:1:1
  4. D.4:1:1
Show the answer and explanation

Answer: D (4:1:1)

Explanation

With 10 blue balls there are 50 green and 20 black. Removing 10 green and 10 black leaves 40 : 10 : 10 = 4 : 1 : 1.

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Question 7

Commercial Arithmetic

A tax payer is allowed 1/8th of his income tax-free and pays 20% on the remainder. If he pays N490.00 tax, what is his income?

  1. A.N560.00
  2. B.N2,450.00
  3. C.N2,800.00
  4. D.N3,920.00
Show the answer and explanation

Answer: C (N2,800.00)

Explanation

20% of the taxable part is N490, so the taxable part is N2,450. That is 7/8 of his income, so the income is 2,450 × 8/7 = N2,800.

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Question 8

Indices

Evaluate (81/3 × 52/3)/102/3

  1. A.2/5
  2. B.5/3
  3. C.∛5
  4. D.∛2
Show the answer and explanation

Answer: D (∛2)

Explanation

81/3 = 2 and 102/3 = 22/3 × 52/3. The 52/3 cancels, leaving 2 ÷ 22/3 = 21/3 = ∛2.

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Question 9

Logarithms

If log102 = 0.3010 and log103 = 0.4771, evaluate without using logarithm tables, log104.5.

  1. A.0.3010
  2. B.0.4771
  3. C.0.6532
  4. D.0.9542
Show the answer and explanation

Answer: C (0.6532)

Explanation

4.5 = 9/2 = 3 × 3 ÷ 2. So log 4.5 = 2 × 0.4771 − 0.3010 = 0.6532.

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Question 10

Surds

Find m such that (m + √3)(1 − √3)2 = 6 − 2√3.

  1. A.1
  2. B.2
  3. C.3
  4. D.4
Show the answer and explanation

Answer: C (3)

Explanation

(1 − √3)² = 4 − 2√3. Expanding (m + √3)(4 − 2√3) = 4m − 6 + (4 − 2m)√3. Matching with 6 − 2√3 gives 4m − 6 = 6, so m = 3 (and 4 − 2m = −2 checks).

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Question 11

Fractions, Decimals and Approximation

The thickness of an 800-page book is 18mm. Calculate the thickness of one leaf of the book giving your answers in metres and in standard form.

  1. A.2.25 × 10−4m
  2. B.4.50 × 10−4m
  3. C.2.25 × 10−5m
  4. D.4.50 × 10−5m
Show the answer and explanation

Answer: D (4.50 × 10−5m)

Explanation

18mm = 0.018m. Each leaf carries two pages, so the book has 400 leaves. One leaf = 0.018 ÷ 400 = 0.000045 = 4.50 × 10−5m.

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Question 12

Algebraic Expressions

Simplify
(x + 2)/(x + 1) − (x − 2)/(x + 2)

  1. A.3/(x + 1)
  2. B.(3x + 2)/((x + 1)(x + 2))
  3. C.(5x + 6)/((x + 1)(x + 2))
  4. D.(2x2 + 5x + 2)/((x + 1)(x + 2))
Show the answer and explanation

Answer: C ((5x + 6)/((x + 1)(x + 2)))

Explanation

Over the common denominator (x + 1)(x + 2), the top is (x + 2)² − (x − 2)(x + 1) = x² + 4x + 4 − (x² − x − 2) = 5x + 6.

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Question 13

Algebraic Expressions

If 1/p = (a2 + 2ab + b2)/(a − b) and 1/q = (a + b)/(a2 − 2ab + b2) find p/q

  1. A.(a + b)/(a − b)
  2. B.1/(a2 − b2)
  3. C.(a − b)/(a + b)
  4. D.a2 − b2
Show the answer and explanation

Answer: B (1/(a2 − b2))

Explanation

p = (a − b)/(a + b)² and q = (a − b)²/(a + b). Dividing, p/q = (a − b)/(a + b)² × (a + b)/(a − b)² = 1/((a + b)(a − b)) = 1/(a² − b²).

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Question 14

Variation

If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3.

  1. A.1/3
  2. B.2/3
  3. C.8/27
  4. D.4/9
Show the answer and explanation

Answer: C (8/27)

Explanation

x = k/∛y. With x = 1, y = 8: k = 2. When x = 3, ∛y = 2/3, so y = (2/3)³ = 8/27.

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Question 15

Algebraic Expressions

If a = −3, b = 2, c = 4, evaluate
(a3 − b3 − c1/2)/(b − a − c)

  1. A.37
  2. B.−37/5
  3. C.37/5
  4. D.−37
Show the answer and explanation

Answer: D (−37)

Explanation

Top: −27 − 8 − 2 = −37. Bottom: 2 + 3 − 4 = 1. So the value is −37.

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Question 16

Algebraic Expressions

If g(y) = (y − 3)/11 + 11/(y2 − 9) what is g(y + 3)?

  1. A.y/11 + 11/(y(y + 6))
  2. B.y/11 + 11/(y(y + 3))
  3. C.(y + 30)/11 + 11/(y(y + 3))
  4. D.(y + 3)/11 + 11/(y(y − 6))
Show the answer and explanation

Answer: A (y/11 + 11/(y(y + 6)))

Explanation

Replace y by y + 3: (y + 3 − 3)/11 + 11/((y + 3)² − 9) = y/11 + 11/(y² + 6y) = y/11 + 11/(y(y + 6)).

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Question 17

Algebraic Expressions

Factorize completely (x2 + x)2 − (2x + 2)2

  1. A.(x + 1)(x + 2)(x − 2)
  2. B.(x + 1)2(x − 2)2
  3. C.(x + 1)2(x + 2)2
  4. D.(x + 1)2(x + 2)(x − 2)
Show the answer and explanation

Answer: D ((x + 1)2(x + 2)(x − 2))

Explanation

Use difference of two squares: (x² + 3x + 2)(x² − x − 2) = (x + 1)(x + 2)(x − 2)(x + 1) = (x + 1)²(x + 2)(x − 2).

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Question 18

Algebraic Expressions

Simplify (x − y)/(x1/3 − y1/3)

  1. A.x2 + xy + y2
  2. B.x2/3 + x1/3y1/3 + y2/3
  3. C.x2/3 − x1/3y1/3 − y2/3
  4. D.x2 − xy + y2
Show the answer and explanation

Answer: B (x2/3 + x1/3y1/3 + y2/3)

Explanation

Let a = x1/3, b = y1/3. Then x − y = a³ − b³ = (a − b)(a² + ab + b²), so the fraction is a² + ab + b² = x2/3 + x1/3y1/3 + y2/3.

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Question 19

Quadratic Equations

Solve the following equation for x.
x2 + (2/r2)x + 1/r4 = 0

  1. A.r2
  2. B.1/r2
  3. C.−1/r2
  4. D.1/r
Show the answer and explanation

Answer: C (−1/r2)

Explanation

The left side is a perfect square: (x + 1/r²)² = 0. So x = −1/r².

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Question 20

Inequalities

List the integral values of x which satisfy the inequality −1 < 5 − 2x ≤ 7

  1. A.−1, 0, 1, 2
  2. B.0, 1, 2, 3.
  3. C.−1, 0, 1, 2, 3.
  4. D.−1, 0, 2, 3.
Show the answer and explanation

Answer: A (−1, 0, 1, 2)

Explanation

Subtract 5: −6 < −2x ≤ 2. Divide by −2 and reverse the signs: 3 > x ≥ −1. The integers are −1, 0, 1, 2.

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Question 21

Linear and Simultaneous Equations

Given that 3x − 5y − 3 = 0
2y − 6x + 5 = 0
the value of (x, y) is

  1. A.(−1/8, 19/24)
  2. B.(8, 24/19)
  3. C.(−8, 24/19)
  4. D.(19/24, −1/8)
Show the answer and explanation

Answer: D ((19/24, −1/8))

Explanation

Doubling the first equation gives 6x = 10y + 6. Put this into the second: 2y − 10y − 6 + 5 = 0, so y = −1/8. Then 3x = 5(−1/8) + 3 = 19/8, so x = 19/24.

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Question 22

Quadratic Equations

The solution of the quadratic equation px2 + qx + b = 0 is

  1. A.(−b ± √(b2 − 4ac))/2a
  2. B.(−b ± √(p2 − 4pb))/2a
  3. C.(−q ± √(q2 − 4bp))/2p
  4. D.(−q ± √(p2 − 4bp))/2p
Show the answer and explanation

Answer: C ((−q ± √(q2 − 4bp))/2p)

Explanation

In the formula x = (−B ± √(B² − 4AC))/2A, here A = p, B = q and C = b. So x = (−q ± √(q² − 4bp))/2p.

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Question 23

Algebraic Expressions

Simplify
1/(x2 + 5x + 6) + 1/(x2 + 3x + 2)

  1. A.(x + 3)/((x + 1)(x + 2))
  2. B.1/((x + 1)(x + 2)(x + 3))
  3. C.2/((x + 1)(x + 3))
  4. D.4/((x + 1)(x + 3))
Show the answer and explanation

Answer: C (2/((x + 1)(x + 3)))

Explanation

The denominators are (x + 2)(x + 3) and (x + 1)(x + 2). The sum is (x + 1 + x + 3)/((x + 1)(x + 2)(x + 3)) = 2(x + 2)/(…) = 2/((x + 1)(x + 3)).

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Question 24

Algebraic Expressions

Simplify (4a2 − 49b2)/(2a2 + 5ab − 7b2)

  1. A.(a − b)/(2a + b)
  2. B.(2a + 7b)/(a − b)
  3. C.(2a − 7b)/(a + b)
  4. D.(2a − 7b)/(a − b)
Show the answer and explanation

Answer: D ((2a − 7b)/(a − b))

Explanation

The top is (2a − 7b)(2a + 7b) and the bottom is (2a + 7b)(a − b). Cancelling (2a + 7b) leaves (2a − 7b)/(a − b).

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Question 25

Graphs of Functions

Use the graph of y = (x2 + 3)/(x + 4) and y = 1 to answer this question. What is the solution of the equation x2 − x − 1 = 0?

Diagram for this question
  1. A.x = 1.6 and x = −0.6
  2. B.x = −1.6 and x = 0.6
  3. C.x = 1.6 and x = 0.6
  4. D.x = −1.6 and x = −0.6.
Show the answer and explanation

Answer: A (x = 1.6 and x = −0.6)

Explanation

Setting (x² + 3)/(x + 4) = 1 gives x² + 3 = x + 4, i.e. x² − x − 1 = 0. So the roots are where the curve meets the line y = 1, at about x = 1.6 and x = −0.6.

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Question 27

Graphs of Functions

The solution of x2 − 2x − 1 = 0 are the points of intersection of two graphs. If one of the graphs is y = 2 + x − x2, find the second graph.

  1. A.y = 1 − x
  2. B.y = 1 + x
  3. C.y = x − 1
  4. D.y = 3x + 3
Show the answer and explanation

Answer: A (y = 1 − x)

Explanation

We need 2 + x − x² = g(x) to be the same as x² − 2x − 1 = 0. So g(x) = 2 + x − x² + (x² − 2x − 1) = 1 − x.

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Question 28

Progressions

If the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is −6, find the common difference

  1. A.4
  2. B.8
  3. C.6⅔
  4. D.9⅓
Show the answer and explanation

Answer: D (9⅓)

Explanation

(a + 7d) + (a + 8d) = 72 gives 2a + 15d = 72, and a + 3d = −6 gives 2a + 6d = −12. Subtracting, 9d = 84, so d = 9⅓.

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Question 29

Progressions

If 7 and 189 are the first and fourth terms of a geometric progression respectively, find the sum of the first three terms of the progression.

  1. A.182
  2. B.91
  3. C.63
  4. D.28.
Show the answer and explanation

Answer: B (91)

Explanation

7r³ = 189, so r³ = 27 and r = 3. The first three terms are 7, 21 and 63, which add up to 91.

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Question 30

Circle Theorems

In the figure above, PQRS is a circle. If chords QR and RS are equal, calculate the value of x

Diagram for this question
  1. A.80°
  2. B.60°
  3. C.45°
  4. D.40°
Show the answer and explanation

Answer: D (40°)

Explanation

Angle QRS = 180° − 120° = 60°, and since QR = RS, triangle QRS is equilateral, so angle QSR = angle QPR = 60°. Angle PSR = 180° − 100° = 80° and angle QPS = 180° − 60° = 120°, so angle SPR = 60°. In triangle PSR, x = 180° − 80° − 60° = 40°.

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Question 31

Angles and Polygons

In the figure above, PQ is parallel to ST and QR̂S = 40°. Find the value of x

Diagram for this question
  1. A.55
  2. B.60
  3. C.65
  4. D.75
Show the answer and explanation

Answer: A (55)

Explanation

Following the turns of the path P–Q–R–S–T between the parallel lines gives 3x + x − 40° = 180°. So 4x = 220° and x = 55°.

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Question 32

Angles and Polygons

For which of the following exterior angles is a regular polygon possible? i. 35° ii. 18° iii. 15°

  1. A.i and ii
  2. B.ii only
  3. C.ii and iii
  4. D.iii only
Show the answer and explanation

Answer: C (ii and iii)

Explanation

A regular polygon has 360° ÷ (exterior angle) sides, which must be a whole number. 360 ÷ 18 = 20 and 360 ÷ 15 = 24 work, but 360 ÷ 35 is not whole.

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Question 33

Mensuration of Plane Shapes

In the figure above, PS = 7cm and RY = 9cm. If the area of parallelogram PQRS is 56cm2, find the area of trapezium PQTS.

Diagram for this question
  1. A.56cm2
  2. B.112cm2
  3. C.120cm2
  4. D.176cm2
Show the answer and explanation

Answer: C (120cm2)

Explanation

The height is 56 ÷ 7 = 8cm. QT = QR + RY + YT = 7 + 9 + 7 = 23cm. Area of the trapezium = ½(23 + 7) × 8 = 120cm².

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Question 34

Mensuration of Plane Shapes

A quadrant of a circle of radius 6cm is cut away from each corner of rectangle 25cm long and 18cm wide. Find the perimeter of the remaining figure.

  1. A.38cm
  2. B.(38 + 12π)cm
  3. C.(86 − 12π)cm
  4. D.(86 − 6π)cm
Show the answer and explanation

Answer: B ((38 + 12π)cm)

Explanation

The rectangle's perimeter is 86cm. Each corner loses 6 + 6 = 12cm of straight edge (48cm in all) and gains a quarter arc of 2π × 6 ÷ 4 = 3π. Perimeter = 86 − 48 + 12π = (38 + 12π)cm.

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Question 35

Triangles and Quadrilaterals

In the figure above, ST̂Q = SR̂P, PT = TQ = 6cm and QS = 5cm. Find SR.

Diagram for this question
  1. A.47/5
  2. B.5
  3. C.32/5
  4. D.22/5
Show the answer and explanation

Answer: A (47/5)

Explanation

Triangles QST and QPR are similar (common angle Q and equal angles at T and R). So QS/QP = QT/QR: 5/12 = 6/QR, giving QR = 72/5. SR = 72/5 − 5 = 47/5.

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Question 36

Angles and Polygons

Four interior angles of a pentagon are 90° − x°, 90° + x°, 110° − 2x°, 110° + 2x°, Find the fifth interior angle.

  1. A.110°
  2. B.120°
  3. C.130°
  4. D.140°.
Show the answer and explanation

Answer: D (140°.)

Explanation

A pentagon's angles add up to 540°. The four given angles add up to 400° (the x terms cancel), so the fifth is 140°.

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Question 37

Triangles and Quadrilaterals

In the figure above, PS = RS = QS and QŜR = 50°. Find QP̂R.

Diagram for this question
  1. A.25°
  2. B.40°
  3. C.50°
  4. D.65°
Show the answer and explanation

Answer: A (25°)

Explanation

S is equally far from P, Q and R, so it is the centre of a circle through them. Angle QPR stands on the same arc QR as the angle at the centre QSR, so it is half of 50°, i.e. 25°.

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Question 38

Circle Theorems

In the figure above, XR and YQ are tangents to the circle YZXP. If ZX̂R = 45° and YẐX = 55°, find ZŶQ

Diagram for this question
  1. A.135°
  2. B.125°
  3. C.100°
  4. D.90°
Show the answer and explanation

Answer: C (100°)

Explanation

The angle between a tangent and a chord equals the angle in the alternate segment. So angle ZYX = angle ZXR = 45°, and in triangle XYZ angle YXZ = 180° − 45° − 55° = 80°. The angle between YZ and the tangent on the other side of Y is 80°, so angle ZYQ = 180° − 80° = 100°.

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Question 39

Trigonometry

From a point 14√3 metres away from a tree, a man discovers that the angle of elevation of the tree is 30°. If the man measures this angle of elevation from a point 2 metres above the ground, how high is the tree?

  1. A.12m
  2. B.14m
  3. C.14√3m
  4. D.16m
Show the answer and explanation

Answer: D (16m)

Explanation

Above eye level the tree rises 14√3 × tan 30° = 14√3 × 1/√3 = 14m. Adding the 2m eye height gives 16m.

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Question 40

Bearings

Alero starts a 3km walk from P on a bearing 023°. She then walks 4km on a bearing 113° to Q. What is the bearing of Q from P?

  1. A.26°52'
  2. B.53°8'
  3. C.76°8'
  4. D.90°
Show the answer and explanation

Answer: C (76°8')

Explanation

The two legs are at right angles (113° − 23° = 90°). At P the angle between the first leg and PQ is tan⁻¹(4/3) = 53°8'. The bearing of Q is 23° + 53°8' = 76°8'.

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Question 41

Trigonometry

If cot θ = x/y, find cosec θ

  1. A.(1/y)(x2 + y2)
  2. B.√(x/y)
  3. C.(1/y)√(x2 + y2)
  4. D.y/x
Show the answer and explanation

Answer: C ((1/y)√(x2 + y2))

Explanation

Take adjacent side x and opposite side y, so the hypotenuse is √(x² + y²). cosec θ = hypotenuse/opposite = √(x² + y²)/y.

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Question 42

Trigonometry

In triangle PQR, PQ = 1cm, QR = 2cm and PQ̂R = 120°. Find the longest side of the triangle.

  1. A.√3cm
  2. B.√7cm
  3. C.3cm
  4. D.7cm
Show the answer and explanation

Answer: B (√7cm)

Explanation

The longest side is PR, opposite the 120° angle. By the cosine rule PR² = 1² + 2² − 2(1)(2)cos 120° = 5 + 2 = 7, so PR = √7cm.

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Question 43

Trigonometry

If cos2θ + 1/8 = sin2θ, find tan θ.

  1. A.3
  2. B.3√7/7
  3. C.3√7
  4. D.√7
Show the answer and explanation

Answer: B (3√7/7)

Explanation

Replace sin²θ by 1 − cos²θ: 2cos²θ = 7/8, so cos²θ = 7/16 and sin²θ = 9/16. tan θ = 3/√7 = 3√7/7.

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Question 44

Mensuration of Solids

If a metal pipe 10cm long has an external diameter of 12cm and a thickness of 1cm, find the volume of the metal used in making the pipe.

  1. A.120πcm3
  2. B.110πcm3
  3. C.60πcm3
  4. D.50πcm3
Show the answer and explanation

Answer: B (110πcm3)

Explanation

Outer radius 6cm, inner radius 5cm. Volume = π(6² − 5²) × 10 = 110π cm³.

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Question 45

Mensuration of Solids

In the figure above, a solid consists of a hemisphere surmounted by a right circular cone, with radius 3.0cm and height 6.0cm. Find the volume of the solid.

Diagram for this question
  1. A.18πcm3
  2. B.36πcm3
  3. C.54πcm3
  4. D.108πcm3
Show the answer and explanation

Answer: B (36πcm3)

Explanation

Cone: ⅓π × 3² × 6 = 18π. Hemisphere: ⅔π × 3³ = 18π. Total = 36π cm³.

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Question 46

Loci

PQR is a triangle in which PQ = 10cm and QP̂R = 60°. S is a point equidistant from P and Q. Also, S is a point equidistant from PQ and PR. If U is the foot of the perpendicular from S on PR, find the length SU in cm to one decimal place.

  1. A.2.7
  2. B.2.9
  3. C.3.1
  4. D.3.3
Show the answer and explanation

Answer: B (2.9)

Explanation

S is on the perpendicular bisector of PQ, so it is above the midpoint, 5cm from P. S is also on the bisector of angle P, which makes 30° with PQ. Its distance from PQ (equal to SU) is 5 tan 30° = 2.89 ≈ 2.9cm.

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Question 47

Data Presentation

In a class of 150 students, the sector in a pie chart representing the students offering Physics has angle 12°. How many students are offering Physics?

  1. A.18
  2. B.15
  3. C.10
  4. D.5
Show the answer and explanation

Answer: D (5)

Explanation

Number = 12/360 × 150 = 5.

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Question 48

Measures of Central Tendency

If x and y represent the mean and the median respectively of the following set of numbers; 11, 12, 13, 14, 15, 16, 17, 18, 19, 21. Find x/y correct to one decimal place.

  1. A.1.6
  2. B.1.2
  3. C.1.1
  4. D.1.0.
Show the answer and explanation

Answer: D (1.0.)

Explanation

The sum is 156, so the mean is 15.6. The median is (15 + 16)/2 = 15.5. x/y = 15.6/15.5 = 1.006 ≈ 1.0.

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Question 49

Measures of Central Tendency

Score (x): 0, 1, 2, 3, 4, 5, 6
Frequency (f): 7, 11, 6, 7, 7, 5, 3
In the distribution above, the mode and median respectively are

  1. A.1, 3
  2. B.1, 2
  3. C.3, 3
  4. D.0, 2.
Show the answer and explanation

Answer: B (1, 2)

Explanation

The highest frequency (11) is at score 1, so the mode is 1. There are 46 scores; the 23rd and 24th fall in score 2 (cumulative 7, 18, 24), so the median is 2.

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Question 50

Probability

If two dice are thrown together, what is the probability of obtaining at least a score of 10?

  1. A.1/6
  2. B.1/12
  3. C.5/6
  4. D.11/12
Show the answer and explanation

Answer: A (1/6)

Explanation

Totals of 10 or more come from (4,6), (6,4), (5,5), (5,6), (6,5) and (6,6): 6 of the 36 outcomes. Probability = 6/36 = 1/6.

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