JAMB Mathematics Past Questions & Answers

751.

Arrange $\frac{3}{5}$ , $\frac{9}{16}$ , $\frac{34}{59}$ and $\frac{71}{97}$ in ascending order of magnitude.

$\frac{3}{5}$ , $\frac{9}{16}$ , $\frac{34}{59}$ , $\frac{71}{97}$

$\frac{9}{16}$ , $\frac{34}{59}$ , $\frac{3}{5}$ , $\frac{71}{97}$

$\frac{3}{5}$ , $\frac{9}{16}$ , $\frac{71}{97}$ , $\frac{34}{59}$

$\frac{9}{16}$ , $\frac{3}{5}$ , $\frac{71}{97}$ , $\frac{34}{59}$

View answer & explanation

Correct answer is B

$\frac{3}{5}$ = 0.60, $\frac{9}{16}$ = 0.56, $\frac{34}{59}$ = 0.58, $\frac{71}{97}$ = 0.73

Hence, $\frac{9}{16} < \frac{34}{59} < \frac{3}{5} < \frac{71}{97}$

752.

Multiply (x + 3y + 5) by (2x2 + 5y + 2)

2x² + 3yx² + 10xy + 15y² + 13y + 10x² + 2x + 10

2x³ + 6yx² + 5xy + 15y² + 31y + 5x² + 2x + 10

2x³ + 6xy² + 5xy + 15y² + 12y + 10x² + 2x = 10

2x² + 6xy² + 5xy + 15y² + 13y + 10x² + 2x + 10

2x³ + 2yx² + 10xy + 10y² + 31y + 5x² + 10

View answer & explanation

Correct answer is B

$(x + 3y + 5)(2x^{2} + 5y + 2)$

= $2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 6y + 10x^{2} + 25y + 10$

= $2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 31y + 10x^{2} + 10$

753.

The sum of $3\frac{7}{8}$ and $1\frac{1}{3}$ is less than the difference between $\frac{1}{8}$ and $1\frac{2}{3}$ by:

3 $\frac{2}{3}$

5 $\frac{1}{4}$

6 $\frac{1}{2}$

8 $\frac{1}{8}$

View answer & explanation

Correct answer is A

$3\frac{7}{8} + 1\frac{1}{3} = 4\frac{21 + 8}{24}$

= $4\frac{29}{24}$

$\equiv 5\frac{5}{24}$

$1\frac{2}{3} - \frac{1}{8} = \frac{5}{3} - \frac{1}{8}$

= $\frac{40 - 3}{24}$

= $\frac{37}{24}$

$5\frac{5}{24} - \frac{37}{24} = \frac{125}{24} - \frac{37}{24}$

= $\frac{88}{24}$

= 3 $\frac{2}{3}$

754.

A rectangular picture 6cm by 8cm is enclosed by a frame $\frac{1}{2}$ cm wide. Calculate the area of the frame

15sq.cm

20sq.cm

13sq.cm

16sq.cm

17sq.cm

View answer & explanation

Correct answer is A

Area of the rectangular picture = L x B = 8 x 6

= 48 sq.cm.

Area of the whole surface (which is gotten by adding $\frac{1}{2}$ on every side to the original picture size) is 9 x 7 = 63 sq. cm

area of the frame is 63 - 48

= 15 sq. cm

755.

A man is standing in the corridor of a 10-storey building and looking down at a tall tree in front of the building. He sees the top of the tree at angle of depression of 30o. If the tree is 200m tall and the man's eyes are 300m above the ground, calculate the angle of depression of the foot tree as seen by the man

30^o

60^o

45^o

25^o

View answer & explanation

Correct answer is B

Let x rep. the angle of depression of the foot of the tree.

tan 30^o = $\frac{y}{100}$

y = 100 tan 30^o

= 57.8

By Pythagoras, AC² = 300² + 58²

= 900 + 3354

tan x = $\frac{opp}{adj}$

= $\frac{58}{300}$

= 0.19

tan x = 0.19

x = tan 0.19

= 60^o