# Mathematics 2017 Past Questions | JAMB

Study the following Mathematics past questions and answers for JAMB, WAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.

**31. **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.**\(\frac{11!}{2!2!}\)

**Correct Option: Answer is C**

MATHEMATICS is an eleven letter word = 11!

There are 2Ms and 2As and 2Es

Divide the number of repeating letters

= \(\frac{11!}{2!2!2!}\)

**32. **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!

**Correct Option: Answer is C**

MACICITA is an eight letter word = 8!

Since we have repeating letters, we have to divide to remove duplicates accordingly. There are 2A, 2C, 2I

∴ \(\frac{8!}{2! 2! 2!}\)

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

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

**Correct Option: Answer is B**

Y ∝ \(\frac{1}{2}\)

Y = 6, X = 7

Y = \(\frac{k}{x}\) where k is constant

6 = \(\frac{k}{7}\)

k = 42

**34**.

**A.**123^{o}**B.**170^{o}**C.**117^{o}**D.**137^{o}

**Correct Option: Answer is C**

MN || PQ || RS

MN = PQ = RS (parallel lines)

Label the angle in the lines

a = i (corresponding angles are equal)

b = x (corresponding angles are equal)

If |MN| = |RS|

If a = i

and a = 63 = i

a + b = 180 (Adjacent interior angles are supplementary i.e add to 180)

∴ i + x = 180

63 + x = 180

x = 180 – 63

x = 117°

**35**. 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

**Correct Option: Answer is D**

The locus of a point p(x, y) such that pv = pw where v = (1, 1)

and w = (3, 5). This means that the point p moves so that its distance from v and w are equidistance

\(\sqrt{(x − x_1)^2 + (y − y_1)^2}\) = \(\sqrt{(x − x_2)^2 + (y − y_2)^2}\)

\(\sqrt{(x -1)^2 + (y – 1)^2}\) = \(\sqrt{(x – 3)^2 + (y – 5)^2}\)

square both sides

(x – 1)^{2} + (y – 1)^{2} = (x – 3)^{2} + (y – 5)^{2}

x^{2} – 2x + 1 + y2 – 2y + 1 = x^{2} – 6x + 9 + y2 – 10y + 25

x^{2} + y^{2} -2x -2y + 2 = x^{2} + y^{2} – 6x – 10y + 34

Collecting like terms

x^{2} – x^{2} + y^{2} – y^{2} – 2x + 6x -2y + 10y = 34 – 2

4x + 8y = 32

Divide through by 4

**36. **Find ∫(x^{2} + 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

**Correct Option: Answer is A**

∫x^{n}dx = \(\frac{x_{n + 1}}{n + 1}\)

∫dx = x + k

where k is constant

∫(x^{2} + 3x − 5)dx

∫x^{2} dx + ∫3xdx − ∫5dx

\(\frac{2_{2 + 1}}{2 + 1}\) + \(\frac{3x^{1 + 1}}{1 + 1}\) − 5x + k

\(\frac{x_3}{3}\) + \(\frac{3x_2}{2}\) − 5x + k

**37.**

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.**18cm**C.**17cm**D.**12cm

**Correct Option: Answer is **

From the diagram

Sin 70°

x = 10 Sin 70°

= 9.3969

Hence, length of chord MN = 2x

= 2 × 9.3969

= 18.79

= 19cm (nearest cm)

**38. **If m * n = [m_{n} − n_{m}] for m, n belong to R, evaluate − 3 * 4

**A.**3**B.**4**C.**5**D.**6

**Correct Option: Answer is C**

**39. **Factorize completely x2 + 12xy + 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)

**Correct Option: Answer is A**

\(x^{2} + 2xy + y^{2} + 3x + 3y – 18\)

\(x^{2} + 2xy + 3x + y^{2} + 3y -18\)

\(x^{2} + 2xy – 3x + 6x + y^{2} -3y + 6y -18\)

\(x^{2} + 2xy -3x + y^{2} -3y + 6x + 6y -18\)

\(x^{2} + xy -3x + xy + y^{2} – 3y + 6x + 6y -18\)

x(x + y – 3) + y(x + y – 3) + 6(x + y – 3)

= (x + y – 3)(x + y + 6)

= (x + y + 6)(x + y -3)

**40. **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 = \(\frac{nrp}{mr}\) + m^{2}**D.**s = \(\frac{nrp}{nr}\) + m^{2}

**Correct Option: Answer is A**

##### Explanation

p = s + \(\frac{sm^2}{nr}\)

p = s + ( 1 + \(\frac{m^2}{nr}\))

p = s (1 + \(\frac{nr + m^2}{nr}\))

nr × p = s (nr + m2)

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