Paper JAMB 1991 General Maths Objective
Objective paper · 43 questions · partial
JAMB 1991 · UME Topics include Number foundations & fractions, Number bases, Approximation & error, Commercial arithmetic, Logarithms, Surds.
Our copy of this paper is missing questions 8, 12, 17, 27, 29, 41, 47.
Sit this paper Answer every question in order, timed if you like (suggested 30 min). You're marked when you hand in, then you see where to focus and the working for each question.
Or read it here: every question below has a worked solution.
1 2 3 4 5 6 7 9 10 11 13 14 15 16 18 19 20 21 22 23 24 25 26 28 30 31 32 33 34 35 36 37 38 39 40 42 43 44 45 46 48 49 50 Simplify 3 1 3 − 1 1 4 × 2 3 + 1 2 5 3\frac13 - 1\frac14 \times \frac23 + 1\frac25 3 3 1 − 1 4 1 × 3 2 + 1 5 2
A 2 17 30 2\frac{17}{30} 2 30 17 B 3 9 10 3\frac9{10} 3 10 9 C 4 1 10 4\frac1{10} 4 10 1 D 4 11 36 4\frac{11}{36} 4 36 11
Worked solution (try it first) Multiply before you add or subtract (BODMAS):
1 1 4 = 5 4 1\frac14 = \frac54 1 4 1 = 4 5 , and
5 4 × 2 3 = 10 12 \frac54 \times \frac23 = \frac{10}{12} 4 5 × 3 2 = 12 10 Now
10 3 − 5 6 + 7 5 \frac{10}{3} - \frac56 + \frac75 3 10 − 6 5 + 5 7 .
The LCD is 30:
100 30 − 25 30 + 42 30 = 117 30 \frac{100}{30} - \frac{25}{30} + \frac{42}{30} = \frac{117}{30} 30 100 − 30 25 + 30 42 = 30 117 .
117 30 = 3 27 30 \frac{117}{30} = 3\frac{27}{30} 30 117 = 3 30 27 = 3 9 10 = 3\frac{9}{10} = 3 10 9 , option B.
Watch out
Do the multiplication first. Working out 3 1 3 − 1 1 4 3\frac13 - 1\frac14 3 3 1 − 1 4 1 2 3 + 1 2 5 \frac23 + 1\frac25 3 2 + 1 5 2 25 12 × 31 15 = 4 11 36 \frac{25}{12} \times \frac{31}{15} = 4\frac{11}{36} 12 25 × 15 31 = 4 36 11 Also set as JAMB 2016 · UTME · Q5
Report a problem with this question
If 2257 is the result of subtracting 4577 from 7056 in base n n n n n n
Worked solution (try it first) Check by adding:
2257 + 4577 2257 + 4577 2257 + 4577 must give
7056 7056 7056 in base
n n n .
Units column:
7 + 7 = 14 7 + 7 = 14 7 + 7 = 14 is written as 6 carry 1, so
14 = n + 6 14 = n + 6 14 = n + 6 .
So
n = 8 n = 8 n = 8 , option A.
The other columns agree:
5 + 7 + 1 = 13 = 8 + 5 5 + 7 + 1 = 13 = 8 + 5 5 + 7 + 1 = 13 = 8 + 5 , and so on.
Watch out
In base ten, 7056 − 4577 = 2479 7056 - 4577 = 2479 7056 − 4577 = 2479 Report a problem with this question
Find, correct to 3 decimal places, ( 1 0.05 ÷ 1 5.005 ) − ( 0.05 × 2.05 ) \left(\dfrac{1}{0.05} \div \dfrac{1}{5.005}\right) - (0.05 \times 2.05) ( 0.05 1 ÷ 5.005 1 ) − ( 0.05 × 2.05 )
A 99.998 B 98.999 C 89.899 D 9.998
Worked solution (try it first) 1 0.05 = 20 \frac{1}{0.05} = 20 0.05 1 = 20 .
Dividing by
1 5.005 \frac{1}{5.005} 5.005 1 is the same as multiplying by 5.005, so the bracket is
20 × 5.005 = 100.1 20 \times 5.005 = 100.1 20 × 5.005 = 100.1 .
The second bracket is
0.05 × 2.05 = 0.1025 0.05 \times 2.05 = 0.1025 0.05 × 2.05 = 0.1025 .
Subtract:
100.1 − 0.1025 = 99.9975 100.1 - 0.1025 = 99.9975 100.1 − 0.1025 = 99.9975 .
The fourth decimal is 5, so round up: 99.998, option A.
Watch out
1 0.05 \frac{1}{0.05} 0.05 1 10.01 − 0.1025 = 9.9075 10.01 - 0.1025 = 9.9075 10.01 − 0.1025 = 9.9075 Report a problem with this question
Express 62 3 \frac{62}{3} 3 62
Worked solution (try it first) Divide:
62 3 = 20.666 … \frac{62}{3} = 20.666\ldots 3 62 = 20.666 … Three significant figures are 2, 0 and 6.
The next figure is 6, so round up: 20.7, option D.
Watch out
The zero in 20 is a significant figure. Counting only the decimals gives 20.667 or 20.67 (options B and C), which have 5 and 4 significant figures. Report a problem with this question
Factory P produces 20,000 bags of cement per day while factory Q produces 15,000 bags per day. If P reduces production by 5 % 5\% 5% 5 % 5\% 5%
Worked solution (try it first) P loses
5 % 5\% 5% of 20,000:
0.05 × 20 000 = 1000 0.05 \times 20\,000 = 1000 0.05 × 20 000 = 1000 bags.
Q gains
5 % 5\% 5% of 15,000:
0.05 × 15 000 = 750 0.05 \times 15\,000 = 750 0.05 × 15 000 = 750 bags.
The net loss is
1000 − 750 = 250 1000 - 750 = 250 1000 − 750 = 250 bags a day, option A.
Watch out
Q's increase makes up part of P's drop. 1000 bags (option C) is P's loss alone; take away Q's gain of 750. Report a problem with this question
Musa borrows ₦10.00 at 2 % 2\% 2%
A ₦10.80 B ₦10.67 C ₦2.80 D ₦2.67
Worked solution (try it first) Simple interest for 4 months at
2 % 2\% 2% a month:
10 × 0.02 × 4 = 10 \times 0.02 \times 4 = 10 × 0.02 × 4 = ₦0.80.
So after 4 months he owes
10 + 0.80 = 10 + 0.80 = 10 + 0.80 = ₦10.80.
He repays ₦8.00, so he still owes
10.80 − 8 = 10.80 - 8 = 10.80 − 8 = ₦2.80, option C.
Watch out
₦10.80 (option A) is what he owes before paying back ₦8.00. Take the repayment off. Report a problem with this question
If 3 gallons of spirit containing 20 % 20\% 20% 15 % 15\% 15%
A 2 4 5 % 2\frac45\% 2 5 4 % B 16 7 8 % 16\frac78\% 16 8 7 % C 18 1 8 % 18\frac18\% 18 8 1 % D 18 7 8 % 18\frac78\% 18 8 7 %
Worked solution (try it first) Water in the first spirit:
20 % 20\% 20% of 3 gallons is
0.6 0.6 0.6 gallon.
In the second:
15 % 15\% 15% of 5 gallons is
0.75 0.75 0.75 gallon.
So the 8 gallons of mixture hold
0.6 + 0.75 = 1.35 0.6 + 0.75 = 1.35 0.6 + 0.75 = 1.35 gallons of water.
As a percentage:
1.35 8 × 100 % = 16.875 % \dfrac{1.35}{8} \times 100\% = 16.875\% 8 1.35 × 100% = 16.875% , which is
16 7 8 % 16\frac78\% 16 8 7 % , option B.
Watch out
Don't average the two percentages: 20 + 15 2 = 17.5 % \frac{20 + 15}{2} = 17.5\% 2 20 + 15 = 17.5% 15 % 15\% 15% Report a problem with this question
Simplify 2 log 2 5 − log 72 125 + log 9 2\log\frac25 - \log\frac{72}{125} + \log 9 2 log 5 2 − log 125 72 + log 9
A 1 − 4 log 3 1 - 4\log3 1 − 4 log 3 B − 1 + 2 log 3 -1 + 2\log3 − 1 + 2 log 3 C − 1 + 5 log 2 -1 + 5\log2 − 1 + 5 log 2 D 1 − 2 log 2 1 - 2\log2 1 − 2 log 2
Worked solution (try it first) Move the 2 up as a power:
2 log 2 5 = log 4 25 2\log\frac25 = \log\frac{4}{25} 2 log 5 2 = log 25 4 .
Combine the logs:
log ( 4 25 × 125 72 × 9 ) \log\left(\frac{4}{25} \times \frac{125}{72} \times 9\right) log ( 25 4 × 72 125 × 9 ) .
The number inside is
4500 1800 = 5 2 \frac{4500}{1800} = \frac52 1800 4500 = 2 5 .
5 2 = 10 4 \frac52 = \frac{10}{4} 2 5 = 4 10 , so
log 5 2 = log 10 − log 4 \log\frac52 = \log 10 - \log 4 log 2 5 = log 10 − log 4 .
With
log 10 = 1 \log 10 = 1 log 10 = 1 and
log 4 = 2 log 2 \log 4 = 2\log 2 log 4 = 2 log 2 , the value is
1 − 2 log 2 1 - 2\log 2 1 − 2 log 2 , option D.
Watch out
Subtracting log 72 125 \log\frac{72}{125} log 125 72 72 125 \frac{72}{125} 125 72 125 72 \frac{125}{72} 72 125 72 125 \frac{72}{125} 125 72 Report a problem with this question
Rationalize 2 3 + 3 2 3 2 − 2 3 \dfrac{2\sqrt3 + 3\sqrt2}{3\sqrt2 - 2\sqrt3} 3 2 − 2 3 2 3 + 3 2
A 5 − 2 6 5 - 2\sqrt6 5 − 2 6 B 5 + 2 6 5 + 2\sqrt6 5 + 2 6 C 5 3 5\sqrt3 5 3 D 5
Worked solution (try it first) Multiply the top and bottom by the conjugate of the bottom,
3 2 + 2 3 3\sqrt2 + 2\sqrt3 3 2 + 2 3 .
Bottom:
( 3 2 ) 2 − ( 2 3 ) 2 = 18 − 12 (3\sqrt2)^2 - (2\sqrt3)^2 = 18 - 12 ( 3 2 ) 2 − ( 2 3 ) 2 = 18 − 12 , which is 6.
Top:
( 2 3 + 3 2 ) 2 = 12 + 12 6 + 18 (2\sqrt3 + 3\sqrt2)^2 = 12 + 12\sqrt6 + 18 ( 2 3 + 3 2 ) 2 = 12 + 12 6 + 18 , which is
30 + 12 6 30 + 12\sqrt6 30 + 12 6 .
Divide by 6:
5 + 2 6 5 + 2\sqrt6 5 + 2 6 , option B.
Watch out
( a + b ) 2 = a 2 + 2 a b + b 2 (a + b)^2 = a^2 + 2ab + b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 + 2 × 2 3 × 3 2 = + 12 6 +2 \times 2\sqrt3 \times 3\sqrt2 = +12\sqrt6 + 2 × 2 3 × 3 2 = + 12 6 5 − 2 6 5 - 2\sqrt6 5 − 2 6 Report a problem with this question
Simplify 1 3 + 5 − 1 3 − 5 \dfrac{1}{3 + \sqrt5} - \dfrac{1}{3 - \sqrt5} 3 + 5 1 − 3 − 5 1
A − 1 2 5 -\frac12\sqrt5 − 2 1 5 B 1 2 5 \frac12\sqrt5 2 1 5 C − 1 4 5 -\frac14\sqrt5 − 4 1 5 D 0
Worked solution (try it first) Use the common denominator
( 3 + 5 ) ( 3 − 5 ) = 9 − 5 (3 + \sqrt5)(3 - \sqrt5) = 9 - 5 ( 3 + 5 ) ( 3 − 5 ) = 9 − 5 , which is 4.
The top is
( 3 − 5 ) − ( 3 + 5 ) = − 2 5 (3 - \sqrt5) - (3 + \sqrt5) = -2\sqrt5 ( 3 − 5 ) − ( 3 + 5 ) = − 2 5 .
So the value is
− 2 5 4 = − 1 2 5 \dfrac{-2\sqrt5}{4} = -\frac12\sqrt5 4 − 2 5 = − 2 1 5 , option A.
Watch out
The minus sign goes in front of the whole second bracket: − ( 3 + 5 ) = − 3 − 5 -(3 + \sqrt5) = -3 - \sqrt5 − ( 3 + 5 ) = − 3 − 5 + 1 2 5 +\frac12\sqrt5 + 2 1 5 Report a problem with this question
Evaluate x y 2 − x 2 y x 2 − x y \dfrac{xy^2 - x^2y}{x^2 - xy} x 2 − x y x y 2 − x 2 y x = − 2 x = -2 x = − 2 y = 3 y = 3 y = 3
A − 3 -3 − 3 B − 3 5 -\frac35 − 5 3 C 3 5 \frac35 5 3 D 3
Worked solution (try it first) Factorise the top and bottom:
x y ( y − x ) x ( x − y ) \dfrac{xy(y - x)}{x(x - y)} x ( x − y ) x y ( y − x ) .
y − x = − ( x − y ) y - x = -(x - y) y − x = − ( x − y ) , so after cancelling
x x x and
x − y x - y x − y the fraction is
− y -y − y .
With
y = 3 y = 3 y = 3 the value is
− 3 -3 − 3 , option A.
Watch out
y − x y - x y − x x − y x - y x − y − 1 -1 − 1 y = 3 y = 3 y = 3 Report a problem with this question
A car travels from Calabar to Enugu, a distance of p p p u u u q q q w w w
A p + q u p + w q \dfrac{p + q}{up + wq} u p + w q p + q B u + w u + w u + w C u w ( p + q ) w p + u q \dfrac{uw(p + q)}{wp + uq} w p + u q u w ( p + q ) D w p + u q u + w q \dfrac{wp + uq}{u + wq} u + w q w p + u q
Worked solution (try it first) Time is distance over speed: the two legs take
p u \dfrac pu u p and
q w \dfrac qw w q hours.
Add them over the common denominator
u w uw u w : total time
= w p + u q u w = \dfrac{wp + uq}{uw} = u w w p + u q .
Average speed is total distance over total time:
( p + q ) ÷ w p + u q u w = u w ( p + q ) w p + u q (p + q) \div \dfrac{wp + uq}{uw} = \dfrac{uw(p + q)}{wp + uq} ( p + q ) ÷ u w w p + u q = w p + u q u w ( p + q ) , option C.
Watch out
Speeds don't add up: u + w u + w u + w Report a problem with this question
If w w w u v u + v \frac{uv}{u + v} u + v uv w = 8 w = 8 w = 8 u = 2 u = 2 u = 2 v = 6 v = 6 v = 6 u u u v v v w w w
A u v w = 16 ( u + v ) uvw = 16(u + v) uv w = 16 ( u + v ) B 16 u v = 3 w ( u + v ) 16uv = 3w(u + v) 16 uv = 3 w ( u + v ) C u v w = 12 ( u + v ) uvw = 12(u + v) uv w = 12 ( u + v ) D 12 u v w = u + v 12uvw = u + v 12 uv w = u + v
Worked solution (try it first) Inverse variation:
w = k u v / ( u + v ) w = \dfrac{k}{uv/(u + v)} w = uv / ( u + v ) k , which turns over to
w = k ( u + v ) u v w = \dfrac{k(u + v)}{uv} w = uv k ( u + v ) .
Put in
u = 2 u = 2 u = 2 ,
v = 6 v = 6 v = 6 ,
w = 8 w = 8 w = 8 :
8 = 8 k 12 8 = \dfrac{8k}{12} 8 = 12 8 k , so
k = 12 k = 12 k = 12 .
So
w = 12 ( u + v ) u v w = \dfrac{12(u + v)}{uv} w = uv 12 ( u + v ) .
Multiply both sides by
u v uv uv :
u v w = 12 ( u + v ) uvw = 12(u + v) uv w = 12 ( u + v ) , option C.
Watch out
Inversely means u v u + v \frac{uv}{u + v} u + v uv w = 16 u v 3 ( u + v ) w = \frac{16uv}{3(u + v)} w = 3 ( u + v ) 16 uv Report a problem with this question
If g ( x ) = x 2 + 3 x g(x) = x^2 + 3x g ( x ) = x 2 + 3 x g ( x + 1 ) − g ( x ) g(x + 1) - g(x) g ( x + 1 ) − g ( x )
A x + 2 x + 2 x + 2 B 2 ( x + 2 ) 2(x + 2) 2 ( x + 2 ) C 2 x + 1 2x + 1 2 x + 1 D x + 4 x + 4 x + 4
Worked solution (try it first) g ( x + 1 ) = ( x + 1 ) 2 + 3 ( x + 1 ) g(x + 1) = (x + 1)^2 + 3(x + 1) g ( x + 1 ) = ( x + 1 ) 2 + 3 ( x + 1 ) = x 2 + 2 x + 1 + 3 x + 3 = x^2 + 2x + 1 + 3x + 3 = x 2 + 2 x + 1 + 3 x + 3 .
Subtract
g ( x ) = x 2 + 3 x g(x) = x^2 + 3x g ( x ) = x 2 + 3 x : the
x 2 x^2 x 2 and
3 x 3x 3 x cancel, leaving
2 x + 4 2x + 4 2 x + 4 .
So
g ( x + 1 ) − g ( x ) = 2 ( x + 2 ) g(x + 1) - g(x) = 2(x + 2) g ( x + 1 ) − g ( x ) = 2 ( x + 2 ) , option B.
Watch out
The 3 x 3x 3 x 3 ( x + 1 ) − 3 x = 3 3(x + 1) - 3x = 3 3 ( x + 1 ) − 3 x = 3 ( x + 1 ) 2 − x 2 = 2 x + 1 (x + 1)^2 - x^2 = 2x + 1 ( x + 1 ) 2 − x 2 = 2 x + 1 Report a problem with this question
Factorize 1 − ( a − b ) 2 1 - (a - b)^2 1 − ( a − b ) 2
A ( 1 − a − b ) ( 1 − a − b ) (1 - a - b)(1 - a - b) ( 1 − a − b ) ( 1 − a − b ) B ( 1 − a + b ) ( 1 + a − b ) (1 - a + b)(1 + a - b) ( 1 − a + b ) ( 1 + a − b ) C ( 1 − a + b ) ( 1 − a + b ) (1 - a + b)(1 - a + b) ( 1 − a + b ) ( 1 − a + b ) D ( 1 − a − b ) ( 1 + a − b ) (1 - a - b)(1 + a - b) ( 1 − a − b ) ( 1 + a − b )
Worked solution (try it first) This is a difference of two squares, with
1 = 1 2 1 = 1^2 1 = 1 2 :
1 − ( a − b ) 2 = [ 1 − ( a − b ) ] [ 1 + ( a − b ) ] 1 - (a - b)^2 = [1 - (a - b)][1 + (a - b)] 1 − ( a − b ) 2 = [ 1 − ( a − b )] [ 1 + ( a − b )] .
Remove the inner brackets:
1 − ( a − b ) = 1 − a + b 1 - (a - b) = 1 - a + b 1 − ( a − b ) = 1 − a + b and
1 + ( a − b ) = 1 + a − b 1 + (a - b) = 1 + a - b 1 + ( a − b ) = 1 + a − b .
So the factors are
( 1 − a + b ) ( 1 + a − b ) (1 - a + b)(1 + a - b) ( 1 − a + b ) ( 1 + a − b ) , option B.
Watch out
− ( a − b ) = − a + b -(a - b) = -a + b − ( a − b ) = − a + b 1 − a − b 1 - a - b 1 − a − b Also set as JAMB 2016 · UTME · Q6
Report a problem with this question
Which of the following is a factor of r s + t r − p t − p s rs + tr - pt - ps r s + t r − pt − p s
A p − s p - s p − s B s − p s - p s − p C r − p r - p r − p D r + p r + p r + p
Worked solution (try it first) Group the terms with
r r r and those with
p p p :
( r s + r t ) − ( p t + p s ) (rs + rt) - (pt + ps) ( r s + r t ) − ( pt + p s ) .
Take out the common factors:
r ( s + t ) − p ( s + t ) r(s + t) - p(s + t) r ( s + t ) − p ( s + t ) .
Take out the common bracket:
( r − p ) ( s + t ) (r - p)(s + t) ( r − p ) ( s + t ) .
So
r − p r - p r − p is a factor, option C.
Watch out
Taking out − p -p − p − p t − p s = − p ( t + s ) -pt - ps = -p(t + s) − pt − p s = − p ( t + s ) − p ( t − s ) -p(t - s) − p ( t − s ) Report a problem with this question
Find the two values of y y y 3 x + y = 8 3x + y = 8 3 x + y = 8 x 2 + x y = 6 x^2 + xy = 6 x 2 + x y = 6
A − 1 -1 − 1 B − 5 -5 − 5 C 1 and 5 D 1 and 1
Worked solution (try it first) Make
y y y the subject of the linear equation:
y = 8 − 3 x y = 8 - 3x y = 8 − 3 x .
Substitute:
x 2 + x ( 8 − 3 x ) = 6 x^2 + x(8 - 3x) = 6 x 2 + x ( 8 − 3 x ) = 6 , so
− 2 x 2 + 8 x − 6 = 0 -2x^2 + 8x - 6 = 0 − 2 x 2 + 8 x − 6 = 0 .
Divide by
− 2 -2 − 2 :
x 2 − 4 x + 3 = 0 x^2 - 4x + 3 = 0 x 2 − 4 x + 3 = 0 .
Factorise:
( x − 1 ) ( x − 3 ) = 0 (x - 1)(x - 3) = 0 ( x − 1 ) ( x − 3 ) = 0 , so
x = 1 x = 1 x = 1 or
x = 3 x = 3 x = 3 .
Then
y = 8 − 3 x y = 8 - 3x y = 8 − 3 x gives
y = 5 y = 5 y = 5 or
y = − 1 y = -1 y = − 1 , option A.
Watch out
The question asks for y y y x x x x x x y y y y y y x x x Report a problem with this question
Find the range of values of x x x x 2 + x 3 + x 4 < 1 \frac x2 + \frac x3 + \frac x4 < 1 2 x + 3 x + 4 x < 1
A x < 12 13 x < \frac{12}{13} x < 13 12 B x < 13 x < 13 x < 13 C x < 9 x < 9 x < 9 D x < 13 12 x < \frac{13}{12} x < 12 13
Worked solution (try it first) Multiply every term by 12, the LCM of 2, 3 and 4:
6 x + 4 x + 3 x < 12 6x + 4x + 3x < 12 6 x + 4 x + 3 x < 12 .
Collect the terms:
13 x < 12 13x < 12 13 x < 12 .
Divide by 13:
x < 12 13 x < \frac{12}{13} x < 13 12 , option A.
Watch out
From 13 x < 12 13x < 12 13 x < 12 12 13 \frac{12}{13} 13 12 13 12 \frac{13}{12} 12 13 Also set as JAMB 2016 · UTME · Q7
Report a problem with this question
Find the positive number n n n
Worked solution (try it first) Write the sentence as an equation:
3 n 2 = 12 n 3n^2 = 12n 3 n 2 = 12 n .
Bring everything to one side and factorise:
3 n ( n − 4 ) = 0 3n(n - 4) = 0 3 n ( n − 4 ) = 0 , so
n = 0 n = 0 n = 0 or
n = 4 n = 4 n = 4 .
The number is positive, so
n = 4 n = 4 n = 4 , option D.
Watch out
"Twelve times the number" is 12 n 12n 12 n 3 n 2 = 12 3n^2 = 12 3 n 2 = 12 n = 2 n = 2 n = 2 Report a problem with this question
Solve the equation ( x − 2 ) ( x − 3 ) = 12 (x - 2)(x - 3) = 12 ( x − 2 ) ( x − 3 ) = 12
A 2, 3 B 3, 6 C − 1 , 6 -1, 6 − 1 , 6 D 1, 6
Worked solution (try it first) Expand the left side:
x 2 − 5 x + 6 = 12 x^2 - 5x + 6 = 12 x 2 − 5 x + 6 = 12 .
Bring everything to one side:
x 2 − 5 x − 6 = 0 x^2 - 5x - 6 = 0 x 2 − 5 x − 6 = 0 .
Factorise:
( x − 6 ) ( x + 1 ) = 0 (x - 6)(x + 1) = 0 ( x − 6 ) ( x + 1 ) = 0 , so
x = − 1 x = -1 x = − 1 or
x = 6 x = 6 x = 6 , option C.
Watch out
You can only set each bracket to zero when the right side is 0. Doing it with 12 on the right gives 2 and 3 (option A). Report a problem with this question
Simplify 1 + x + x 1 + x − x \dfrac{\sqrt{1 + x} + \sqrt x}{\sqrt{1 + x} - \sqrt x} 1 + x − x 1 + x + x
A 1 − 2 x − 2 x ( 1 + x ) 1 - 2x - 2\sqrt{x(1 + x)} 1 − 2 x − 2 x ( 1 + x ) B 1 + 2 x + 2 x ( 1 + x ) 1 + 2x + 2\sqrt{x(1 + x)} 1 + 2 x + 2 x ( 1 + x ) C x ( 1 + x ) \sqrt{x(1 + x)} x ( 1 + x ) D 1 + 2 x − 2 x ( 1 + x ) 1 + 2x - 2\sqrt{x(1 + x)} 1 + 2 x − 2 x ( 1 + x )
Worked solution (try it first) Multiply the top and bottom by the conjugate of the bottom,
1 + x + x \sqrt{1 + x} + \sqrt x 1 + x + x .
Bottom:
( 1 + x ) 2 − ( x ) 2 = ( 1 + x ) − x (\sqrt{1 + x})^2 - (\sqrt x)^2 = (1 + x) - x ( 1 + x ) 2 − ( x ) 2 = ( 1 + x ) − x , which is 1.
Top:
( 1 + x + x ) 2 = ( 1 + x ) + 2 x ( 1 + x ) + x (\sqrt{1 + x} + \sqrt x)^2 = (1 + x) + 2\sqrt{x(1 + x)} + x ( 1 + x + x ) 2 = ( 1 + x ) + 2 x ( 1 + x ) + x , which is
1 + 2 x + 2 x ( 1 + x ) 1 + 2x + 2\sqrt{x(1 + x)} 1 + 2 x + 2 x ( 1 + x ) .
So option B.
Watch out
Multiply by the conjugate of the bottom, which has a plus sign. Using the bottom itself squares the wrong bracket and gives 1 + 2 x − 2 x ( 1 + x ) 1 + 2x - 2\sqrt{x(1 + x)} 1 + 2 x − 2 x ( 1 + x ) Report a problem with this question
Evaluate x 2 ( x 2 − 1 ) − 1 2 − ( x 2 − 1 ) 1 2 x^2(x^2 - 1)^{-\frac12} - (x^2 - 1)^{\frac12} x 2 ( x 2 − 1 ) − 2 1 − ( x 2 − 1 ) 2 1
A ( x 2 − 1 ) 1 2 (x^2 - 1)^{\frac12} ( x 2 − 1 ) 2 1 B x 2 − 1 x^2 - 1 x 2 − 1 C ( x 2 − 1 ) − 1 (x^2 - 1)^{-1} ( x 2 − 1 ) − 1 D ( x 2 − 1 ) − 1 2 (x^2 - 1)^{-\frac12} ( x 2 − 1 ) − 2 1
Worked solution (try it first) Write the second term with the same factor:
( x 2 − 1 ) 1 2 = ( x 2 − 1 ) × ( x 2 − 1 ) − 1 2 (x^2 - 1)^{\frac12} = (x^2 - 1) \times (x^2 - 1)^{-\frac12} ( x 2 − 1 ) 2 1 = ( x 2 − 1 ) × ( x 2 − 1 ) − 2 1 .
Take out
( x 2 − 1 ) − 1 2 (x^2 - 1)^{-\frac12} ( x 2 − 1 ) − 2 1 : the expression is
( x 2 − 1 ) − 1 2 [ x 2 − ( x 2 − 1 ) ] (x^2 - 1)^{-\frac12}\left[x^2 - (x^2 - 1)\right] ( x 2 − 1 ) − 2 1 [ x 2 − ( x 2 − 1 ) ] .
The bracket is
x 2 − x 2 + 1 = 1 x^2 - x^2 + 1 = 1 x 2 − x 2 + 1 = 1 , so the result is
( x 2 − 1 ) − 1 2 (x^2 - 1)^{-\frac12} ( x 2 − 1 ) − 2 1 , option D.
Watch out
Take out the lower power, − 1 2 -\frac12 − 2 1 x 2 − 1 x^2 - 1 x 2 − 1 − 1 -1 − 1 + 1 +1 + 1 2 x 2 − 1 2x^2 - 1 2 x 2 − 1 Report a problem with this question
Find the gradient of the line passing through the points ( − 2 , 0 ) (-2, 0) ( − 2 , 0 ) ( 0 , − 4 ) (0, -4) ( 0 , − 4 )
Worked solution (try it first) Gradient is the change in
y y y over the change in
x x x , taken in the same order:
− 4 − 0 0 − ( − 2 ) \dfrac{-4 - 0}{0 - (-2)} 0 − ( − 2 ) − 4 − 0 .
The top is
− 4 -4 − 4 and the bottom is
0 + 2 = 2 0 + 2 = 2 0 + 2 = 2 .
So the gradient is
− 4 2 = − 2 \frac{-4}{2} = -2 2 − 4 = − 2 , option C.
Watch out
Subtract in the same order on the top and the bottom. Taking − 4 − 0 -4 - 0 − 4 − 0 − 2 − 0 -2 - 0 − 2 − 0 − 4 − 2 = 2 \frac{-4}{-2} = 2 − 2 − 4 = 2 Report a problem with this question
What is the n n n 27 , 9 , 3 , … 27, 9, 3, \dots 27 , 9 , 3 , …
A 27 ( 1 3 ) n − 1 27\left(\frac13\right)^{n - 1} 27 ( 3 1 ) n − 1 B 3 n + 2 3^{n + 2} 3 n + 2 C 27 + 18 ( n − 1 ) 27 + 18(n - 1) 27 + 18 ( n − 1 ) D 27 + 6 ( n − 1 ) 27 + 6(n - 1) 27 + 6 ( n − 1 )
Worked solution (try it first) Each term is a third of the one before:
9 ÷ 27 = 1 3 9 \div 27 = \frac13 9 ÷ 27 = 3 1 and
3 ÷ 9 = 1 3 3 \div 9 = \frac13 3 ÷ 9 = 3 1 .
So this is a G.P. with
a = 27 a = 27 a = 27 and
r = 1 3 r = \frac13 r = 3 1 .
The
n n n th term of a G.P. is
a r n − 1 ar^{n - 1} a r n − 1 .
So the
n n n th term is
27 ( 1 3 ) n − 1 27\left(\frac13\right)^{n - 1} 27 ( 3 1 ) n − 1 , option A.
Watch out
The ratio is 9 ÷ 27 = 1 3 9 \div 27 = \frac13 9 ÷ 27 = 3 1 3 n + 2 3^{n + 2} 3 n + 2 27 , 81 , 243 , … 27, 81, 243, \dots 27 , 81 , 243 , … Report a problem with this question
In the figure, P Q ∥ T S PQ \parallel TS P Q ∥ T S ∠ P Q R = 110 ∘ \angle PQR = 110^\circ ∠ P QR = 11 0 ∘ ∠ R S T = 120 ∘ \angle RST = 120^\circ ∠ R S T = 12 0 ∘ x = ∠ Q R S x = \angle QRS x = ∠ QR S
A 130 ∘ 130^\circ 13 0 ∘ B 110 ∘ 110^\circ 11 0 ∘ C 100 ∘ 100^\circ 10 0 ∘ D 90 ∘ 90^\circ 9 0 ∘
Worked solution (try it first) Draw a line through
R R R parallel to
P Q PQ P Q and
T S TS T S .
∠ P Q R \angle PQR ∠ P QR and the angle between
R Q RQ R Q and that line are co-interior, so the upper part of
x x x is
180 ∘ − 110 ∘ = 70 ∘ 180^\circ - 110^\circ = 70^\circ 18 0 ∘ − 11 0 ∘ = 7 0 ∘ .
∠ R S T \angle RST ∠ R S T and the angle between
R S RS R S and that line are co-interior too, so the lower part of
x x x is
180 ∘ − 120 ∘ = 60 ∘ 180^\circ - 120^\circ = 60^\circ 18 0 ∘ − 12 0 ∘ = 6 0 ∘ .
So
x = 70 ∘ + 60 ∘ = 130 ∘ x = 70^\circ + 60^\circ = 130^\circ x = 7 0 ∘ + 6 0 ∘ = 13 0 ∘ , option A.
Watch out
Subtract each given angle from 180 ∘ 180^\circ 18 0 ∘ 110 ∘ + 120 ∘ 110^\circ + 120^\circ 11 0 ∘ + 12 0 ∘ 230 ∘ 230^\circ 23 0 ∘ R R R Report a problem with this question
The angles of a quadrilateral are 5 x − 30 5x - 30 5 x − 30 4 x + 60 4x + 60 4 x + 60 60 − x 60 - x 60 − x 3 x + 61 3x + 61 3 x + 61
A 5 x − 30 5x - 30 5 x − 30 B 4 x + 60 4x + 60 4 x + 60 C 60 − x 60 - x 60 − x D 3 x + 61 3x + 61 3 x + 61
Worked solution (try it first) The angles of a quadrilateral add up to
360 ∘ 360^\circ 36 0 ∘ :
( 5 x − 30 ) + ( 4 x + 60 ) + ( 60 − x ) + ( 3 x + 61 ) = 360 (5x - 30) + (4x + 60) + (60 - x) + (3x + 61) = 360 ( 5 x − 30 ) + ( 4 x + 60 ) + ( 60 − x ) + ( 3 x + 61 ) = 360 .
Collect terms:
11 x + 151 = 360 11x + 151 = 360 11 x + 151 = 360 , so
11 x = 209 11x = 209 11 x = 209 and
x = 19 x = 19 x = 19 .
The angles are
65 ∘ 65^\circ 6 5 ∘ ,
136 ∘ 136^\circ 13 6 ∘ ,
41 ∘ 41^\circ 4 1 ∘ and
118 ∘ 118^\circ 11 8 ∘ .
The smallest is
60 − x 60 - x 60 − x , option C.
Watch out
Find x x x 5 x − 30 5x - 30 5 x − 30 65 ∘ 65^\circ 6 5 ∘ Report a problem with this question
The area of a square is 144 cm 2 144\text{ cm}^2 144 cm 2
A 11 3 11\sqrt3 11 3 B 12 cm C 12 2 12\sqrt2 12 2 D 13 cm
Worked solution (try it first) The side is
144 = 12 \sqrt{144} = 12 144 = 12 cm.
The diagonal is the hypotenuse of a right-angled triangle with two sides of 12:
12 2 + 12 2 = 12 2 \sqrt{12^2 + 12^2} = 12\sqrt2 1 2 2 + 1 2 2 = 12 2 cm, option C.
Watch out
12 cm (option B) is the side. The diagonal is longer: 12 2 12\sqrt2 12 2 Report a problem with this question
One angle of a rhombus is 60 ∘ 60^\circ 6 0 ∘
A 8 3 8\sqrt3 8 3 B 16 3 \frac{16}{\sqrt3} 3 16 C 5 3 5\sqrt3 5 3 D 10 3 \frac{10}{\sqrt3} 3 10
Worked solution (try it first) The short diagonal joins the two
120 ∘ 120^\circ 12 0 ∘ corners and cuts the rhombus into two equilateral triangles, so each side is 8 cm.
The long diagonal bisects the
60 ∘ 60^\circ 6 0 ∘ angles and meets the short one at right angles.
In one right-angled quarter, the hypotenuse is 8 and the angle is
30 ∘ 30^\circ 3 0 ∘ , so half the long diagonal is
8 cos 30 ∘ = 4 3 8\cos30^\circ = 4\sqrt3 8 cos 3 0 ∘ = 4 3 .
So the long diagonal is
2 × 4 3 = 8 3 2 \times 4\sqrt3 = 8\sqrt3 2 × 4 3 = 8 3 cm, option A.
Watch out
Multiply by cos 30 ∘ \cos30^\circ cos 3 0 ∘ 8 cos 30 ∘ = 16 3 \frac{8}{\cos30^\circ} = \frac{16}{\sqrt3} c o s 3 0 ∘ 8 = 3 16 Report a problem with this question
If the exterior angles of a pentagon are x ∘ x^\circ x ∘ ( x + 5 ) ∘ (x + 5)^\circ ( x + 5 ) ∘ ( x + 10 ) ∘ (x + 10)^\circ ( x + 10 ) ∘ ( x + 15 ) ∘ (x + 15)^\circ ( x + 15 ) ∘ ( x + 20 ) ∘ (x + 20)^\circ ( x + 20 ) ∘ x x x
A 118 ∘ 118^\circ 11 8 ∘ B 72 ∘ 72^\circ 7 2 ∘ C 62 ∘ 62^\circ 6 2 ∘ D 36 ∘ 36^\circ 3 6 ∘
Worked solution (try it first) The exterior angles of any polygon add up to
360 ∘ 360^\circ 36 0 ∘ :
x + ( x + 5 ) + ( x + 10 ) + ( x + 15 ) + ( x + 20 ) = 360 x + (x + 5) + (x + 10) + (x + 15) + (x + 20) = 360 x + ( x + 5 ) + ( x + 10 ) + ( x + 15 ) + ( x + 20 ) = 360 .
Collect terms:
5 x + 50 = 360 5x + 50 = 360 5 x + 50 = 360 , so
5 x = 310 5x = 310 5 x = 310 .
Divide by 5:
x = 62 x = 62 x = 62 , option C.
Watch out
The question gives exterior angles, which add up to 360 ∘ 360^\circ 36 0 ∘ 118 ∘ 118^\circ 11 8 ∘ 180 ∘ − 62 ∘ 180^\circ - 62^\circ 18 0 ∘ − 6 2 ∘ x x x Report a problem with this question
In the figure, P M N PMN P M N P Q R PQR P QR M Q T R N MQTRN M QT R N P T PT P T P M = 5 PM = 5 P M = 5 P N = 12 PN = 12 P N = 12 P Q = 4.8 PQ = 4.8 P Q = 4.8 P R PR P R P T PT P T
P Q R N M T Not to scale: the same figure serves both questions.
A 7.3, 5.9 B 7.7, 12.5 C 12.5, 7.7 D 5.9, 7.3
Worked solution (try it first) For secants from
P P P :
P Q × P R = P M × P N PQ \times PR = PM \times PN P Q × P R = P M × P N .
Here
P M × P N = 5 × 12 = 60 PM \times PN = 5 \times 12 = 60 P M × P N = 5 × 12 = 60 .
So
4.8 × P R = 60 4.8 \times PR = 60 4.8 × P R = 60 and
P R = 60 ÷ 4.8 = 12.5 PR = 60 \div 4.8 = 12.5 P R = 60 ÷ 4.8 = 12.5 cm.
For the tangent:
P T 2 = P M × P N = 60 PT^2 = PM \times PN = 60 P T 2 = P M × P N = 60 , so
P T = 60 ≈ 7.7 PT = \sqrt{60} \approx 7.7 P T = 60 ≈ 7.7 cm.
In the order asked,
P R PR P R then
P T PT P T : 12.5, 7.7, option C.
Watch out
The question asks for P R PR P R P T PT P T Report a problem with this question
In the figure, P M N PMN P M N P Q R PQR P QR M Q T R N MQTRN M QT R N P T PT P T ∠ P N R = 110 ∘ \angle PNR = 110^\circ ∠ P N R = 11 0 ∘ ∠ P M Q = 55 ∘ \angle PMQ = 55^\circ ∠ P M Q = 5 5 ∘ ∠ M P Q \angle MPQ ∠ M P Q
P Q R N M T Not to scale: the same figure serves both questions.
A 40 ∘ 40^\circ 4 0 ∘ B 30 ∘ 30^\circ 3 0 ∘ C 25 ∘ 25^\circ 2 5 ∘ D 15 ∘ 15^\circ 1 5 ∘
Worked solution (try it first) M Q R N MQRN M QR N is a cyclic quadrilateral.
An exterior angle of a cyclic quadrilateral equals the interior angle opposite it, so
∠ Q R N = ∠ P M Q = 55 ∘ \angle QRN = \angle PMQ = 55^\circ ∠ QR N = ∠ P M Q = 5 5 ∘ .
P P P ,
Q Q Q and
R R R are on a straight line, so in triangle
P N R PNR P N R the angle at
R R R is
55 ∘ 55^\circ 5 5 ∘ and the angle at
N N N is
110 ∘ 110^\circ 11 0 ∘ .
The angles add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ M P Q = 180 ∘ − 110 ∘ − 55 ∘ \angle MPQ = 180^\circ - 110^\circ - 55^\circ ∠ M P Q = 18 0 ∘ − 11 0 ∘ − 5 5 ∘ = 15 ∘ = 15^\circ = 1 5 ∘ , option D.
Watch out
The exterior angle at M M M R R R 55 ∘ 55^\circ 5 5 ∘ ∠ Q R N \angle QRN ∠ QR N 180 ∘ − 55 ∘ = 125 ∘ 180^\circ - 55^\circ = 125^\circ 18 0 ∘ − 5 5 ∘ = 12 5 ∘ P N R PNR P N R 180 ∘ 180^\circ 18 0 ∘ Report a problem with this question
In the figure, the two horizontal lines are parallel. Find the value of y y y
A 28 ∘ 28^\circ 2 8 ∘ B 122 ∘ 122^\circ 12 2 ∘ C 150 ∘ 150^\circ 15 0 ∘ D 152 ∘ 152^\circ 15 2 ∘
Worked solution (try it first) Angles on a straight line at the bend: the upper slanted segment makes
180 ∘ − 152 ∘ = 28 ∘ 180^\circ - 152^\circ = 28^\circ 18 0 ∘ − 15 2 ∘ = 2 8 ∘ with the horizontal.
Draw a horizontal line through the apex.
By alternate angles the upper segment makes
28 ∘ 28^\circ 2 8 ∘ with it, so the downward segment makes
28 ∘ + 30 ∘ = 58 ∘ 28^\circ + 30^\circ = 58^\circ 2 8 ∘ + 3 0 ∘ = 5 8 ∘ with it.
By alternate angles again, the downward segment makes
58 ∘ 58^\circ 5 8 ∘ with the lower line on its right-hand side.
y y y is on the left, so
y = 180 ∘ − 58 ∘ = 122 ∘ y = 180^\circ - 58^\circ = 122^\circ y = 18 0 ∘ − 5 8 ∘ = 12 2 ∘ , option B.
Watch out
28 ∘ 28^\circ 2 8 ∘ y y y 180 ∘ − 58 ∘ 180^\circ - 58^\circ 18 0 ∘ − 5 8 ∘ Report a problem with this question
In the figure, P Q = P R = P S PQ = PR = PS P Q = P R = P S Q R QR QR T T T ∠ S R T = 68 ∘ \angle SRT = 68^\circ ∠ S R T = 6 8 ∘ ∠ Q P S \angle QPS ∠ QP S
A 136 ∘ 136^\circ 13 6 ∘ B 124 ∘ 124^\circ 12 4 ∘ C 112 ∘ 112^\circ 11 2 ∘ D 68 ∘ 68^\circ 6 8 ∘
Worked solution (try it first) P Q = P R = P S PQ = PR = PS P Q = P R = P S , so
Q Q Q ,
R R R and
S S S lie on a circle with centre
P P P .
Angles on a straight line:
∠ Q R S = 180 ∘ − 68 ∘ \angle QRS = 180^\circ - 68^\circ ∠ QR S = 18 0 ∘ − 6 8 ∘ It stands on the major arc
Q S QS QS , so the reflex angle at
P P P is
2 × 112 ∘ = 224 ∘ 2 \times 112^\circ = 224^\circ 2 × 11 2 ∘ = 22 4 ∘ .
Angles at a point add up to
360 ∘ 360^\circ 36 0 ∘ :
∠ Q P S = 360 ∘ − 224 ∘ \angle QPS = 360^\circ - 224^\circ ∠ QP S = 36 0 ∘ − 22 4 ∘ = 136 ∘ = 136^\circ = 13 6 ∘ , option A.
Watch out
112 ∘ 112^\circ 11 2 ∘ ∠ Q R S \angle QRS ∠ QR S 360 ∘ − 2 × 112 ∘ 360^\circ - 2 \times 112^\circ 36 0 ∘ − 2 × 11 2 ∘ Report a problem with this question
A flagstaff stands on top of a vertical tower. A man standing 60 m from the tower observes that the angles of elevation of the top and bottom of the flagstaff are 64 ∘ 64^\circ 6 4 ∘ 62 ∘ 62^\circ 6 2 ∘
A 60 ( tan 62 ∘ − tan 64 ∘ ) 60(\tan62^\circ - \tan64^\circ) 60 ( tan 6 2 ∘ − tan 6 4 ∘ ) B 60 ( cot 64 ∘ − cot 62 ∘ ) 60(\cot64^\circ - \cot62^\circ) 60 ( cot 6 4 ∘ − cot 6 2 ∘ ) C 60 ( cot 62 ∘ − cot 64 ∘ ) 60(\cot62^\circ - \cot64^\circ) 60 ( cot 6 2 ∘ − cot 6 4 ∘ ) D 60 ( tan 64 ∘ − tan 62 ∘ ) 60(\tan64^\circ - \tan62^\circ) 60 ( tan 6 4 ∘ − tan 6 2 ∘ )
Worked solution (try it first) Here the distance, 60 m, is adjacent to the angle and the height is opposite, so each height is
60 tan θ 60\tan\theta 60 tan θ .
The top of the flagstaff is
60 tan 64 ∘ 60\tan64^\circ 60 tan 6 4 ∘ high and the bottom is
60 tan 62 ∘ 60\tan62^\circ 60 tan 6 2 ∘ high.
The flagstaff is the top minus the bottom:
60 ( tan 64 ∘ − tan 62 ∘ ) 60(\tan64^\circ - \tan62^\circ) 60 ( tan 6 4 ∘ − tan 6 2 ∘ ) , option D.
Watch out
You are finding heights from a known distance, so multiply by tan \tan tan Report a problem with this question
Simplify cos 2 x ( sec 2 x + sec 2 x tan 2 x ) \cos^2x(\sec^2x + \sec^2x\tan^2x) cos 2 x ( sec 2 x + sec 2 x tan 2 x )
A tan x \tan x tan x B tan x sec x \tan x\sec x tan x sec x C sec 2 x \sec^2x sec 2 x D cosec 2 x \text{cosec}^2x cosec 2 x
Worked solution (try it first) Take out the common factor
sec 2 x \sec^2 x sec 2 x in the bracket:
cos 2 x sec 2 x ( 1 + tan 2 x ) \cos^2 x \sec^2 x(1 + \tan^2 x) cos 2 x sec 2 x ( 1 + tan 2 x ) .
cos 2 x sec 2 x = 1 \cos^2 x \sec^2 x = 1 cos 2 x sec 2 x = 1 , because
sec x = 1 cos x \sec x = \frac{1}{\cos x} sec x = c o s x 1 .
That leaves
1 + tan 2 x 1 + \tan^2 x 1 + tan 2 x .
By the identity
1 + tan 2 x = sec 2 x 1 + \tan^2 x = \sec^2 x 1 + tan 2 x = sec 2 x , the answer is
sec 2 x \sec^2 x sec 2 x , option C.
Watch out
The identity is 1 + tan 2 x = sec 2 x 1 + \tan^2 x = \sec^2 x 1 + tan 2 x = sec 2 x cosec 2 x \text{cosec}^2 x cosec 2 x 1 + cot 2 x 1 + \cot^2 x 1 + cot 2 x Report a problem with this question
From a point Z Z Z X X X 60 3 60\sqrt3 60 3 Y Y Y Y Y Y X X X
A 030 ∘ 030^\circ 03 0 ∘ B 045 ∘ 045^\circ 04 5 ∘ C 060 ∘ 060^\circ 06 0 ∘ D 090 ∘ 090^\circ 09 0 ∘
Worked solution (try it first) Y Y Y is 60 m north and
60 3 60\sqrt3 60 3 m east of
X X X , so there is a right angle at
Z Z Z .
The angle at
X X X , east of north, has
tan θ = 60 3 60 = 3 \tan\theta = \frac{60\sqrt3}{60} = \sqrt3 tan θ = 60 60 3 = 3 , so
θ = 60 ∘ \theta = 60^\circ θ = 6 0 ∘ .
Measured clockwise from north, the bearing of
Y Y Y from
X X X is
060 ∘ 060^\circ 06 0 ∘ , option C.
Watch out
Take the angle at X X X 30 ∘ 30^\circ 3 0 ∘ Y Y Y Report a problem with this question
A surveyor walks 500 m up a hill which slopes at an angle of 30 ∘ 30^\circ 3 0 ∘
A 250 m B 500 3 3 \frac{500\sqrt3}{3} 3 500 3 C 250 2 250\sqrt2 250 2 D 250 3 250\sqrt3 250 3
Worked solution (try it first) The 500 m walked up the slope is the hypotenuse, and the height is the side opposite the
30 ∘ 30^\circ 3 0 ∘ angle.
So the height is
500 sin 30 ∘ = 500 × 1 2 500\sin30^\circ = 500 \times \frac12 500 sin 3 0 ∘ = 500 × 2 1 = 250 = 250 = 250 m, option A.
Watch out
The height is opposite the angle, so use sine. Cosine gives the horizontal distance, 500 cos 30 ∘ = 250 3 500\cos30^\circ = 250\sqrt3 500 cos 3 0 ∘ = 250 3 Report a problem with this question
In the figure, P Q R S PQRS P QR S U Q = 4 UQ = 4 U Q = 4 Q V = 6 QV = 6 Q V = 6 W R = 2 WR = 2 W R = 2 △ U V W \triangle UVW △ U V W
A 64 cm 2 64\text{ cm}^2 64 cm 2 B 54 cm 2 54\text{ cm}^2 54 cm 2 C 50 cm 2 50\text{ cm}^2 50 cm 2 D 10 cm 2 10\text{ cm}^2 10 cm 2
Worked solution (try it first) Find the pieces of the square outside
△ U V W \triangle UVW △ U V W .
△ U Q V \triangle UQV △ U Q V :
1 2 × 4 × 6 = 12 cm 2 \frac12 \times 4 \times 6 = 12\text{ cm}^2 2 1 × 4 × 6 = 12 cm 2 .
△ V R W \triangle VRW △ V R W has
V R = 8 − 6 = 2 VR = 8 - 6 = 2 V R = 8 − 6 = 2 :
1 2 × 2 × 2 = 2 cm 2 \frac12 \times 2 \times 2 = 2\text{ cm}^2 2 1 × 2 × 2 = 2 cm 2 .
P U W S PUWS P U W S is a trapezium with parallel sides
P U = 4 PU = 4 P U = 4 and
S W = 8 − 2 = 6 SW = 8 - 2 = 6 S W = 8 − 2 = 6 , and height 8:
1 2 ( 4 + 6 ) × 8 = 40 cm 2 \frac12(4 + 6) \times 8 = 40\text{ cm}^2 2 1 ( 4 + 6 ) × 8 = 40 cm 2 .
Take them from the square:
64 − ( 12 + 2 + 40 ) = 64 − 54 64 - (12 + 2 + 40) = 64 - 54 64 − ( 12 + 2 + 40 ) = 64 − 54 = 10 cm 2 = 10\text{ cm}^2 = 10 cm 2 , option D.
Watch out
54 cm² (option B) is the total of the three outside pieces. Take it away from the square's 64 cm² to get the triangle. Report a problem with this question
Find the total surface area of a solid cylinder whose base radius is 4 cm and height is 5 cm.
A 56 π cm 2 56\pi\text{ cm}^2 56 π cm 2 B 72 π cm 2 72\pi\text{ cm}^2 72 π cm 2 C 96 π cm 2 96\pi\text{ cm}^2 96 π cm 2 D 192 π cm 2 192\pi\text{ cm}^2 192 π cm 2
Worked solution (try it first) A solid cylinder has a curved surface
2 π r h 2\pi rh 2 π r h and two circular ends
2 π r 2 2\pi r^2 2 π r 2 : a total of
2 π r ( r + h ) 2\pi r(r + h) 2 π r ( r + h ) .
2 π × 4 × ( 4 + 5 ) = 72 π cm 2 2\pi \times 4 \times (4 + 5) = 72\pi\text{ cm}^2 2 π × 4 × ( 4 + 5 ) = 72 π cm 2 , option B.
Watch out
A solid cylinder has two ends. With only one, 40 π + 16 π = 56 π cm 2 40\pi + 16\pi = 56\pi\text{ cm}^2 40 π + 16 π = 56 π cm 2 Report a problem with this question
Find the volume of the figure: a cone of height x x x y y y a a a
A π a 2 3 \frac{\pi a^2}{3} 3 π a 2 B π a 2 y \pi a^2y π a 2 y C π a 2 3 ( y + x ) \frac{\pi a^2}{3}(y + x) 3 π a 2 ( y + x ) D π a 2 ( x 3 + y ) \pi a^2\left(\frac x3 + y\right) π a 2 ( 3 x + y )
Worked solution (try it first) Cone:
1 3 π a 2 x \frac13\pi a^2x 3 1 π a 2 x .
Cylinder:
π a 2 y \pi a^2y π a 2 y .
Add them and take out the common factor
π a 2 \pi a^2 π a 2 :
π a 2 ( x 3 + y ) \pi a^2\left(\frac x3 + y\right) π a 2 ( 3 x + y ) , option D.
Watch out
The 1 3 \frac13 3 1 y y y Report a problem with this question
Fifty boxes, each of 50 balls, were inspected for the number which were defective, with the results below. The mean and the median of the distribution are respectively
No. of defective per box
4
5
6
7
8
9
No. of boxes
2
7
17
10
8
6
A 6.7, 6 B 6.7, 6.5 C 6, 6.7 D 6.5, 6.7
Worked solution (try it first) Total defective:
4 ( 2 ) + 5 ( 7 ) + 6 ( 17 ) + 7 ( 10 ) + 8 ( 8 ) + 9 ( 6 ) = 333 4(2) + 5(7) + 6(17) + 7(10) + 8(8) + 9(6) = 333 4 ( 2 ) + 5 ( 7 ) + 6 ( 17 ) + 7 ( 10 ) + 8 ( 8 ) + 9 ( 6 ) = 333 , so the mean is
333 50 = 6.66 ≈ 6.7 \frac{333}{50} = 6.66 \approx 6.7 50 333 = 6.66 ≈ 6.7 .
With 50 boxes the median is halfway between the 25th and 26th values.
Running totals: 2, 9, 26.
The 10th to 26th values are all 6, so the median is 6.
So the mean and median are 6.7 and 6, option A.
Watch out
The median is the middle of the 50 boxes, not the middle of the column 4 to 9, which is 6.5 (option B). Running totals show the 25th and 26th boxes both have 6. Report a problem with this question
Using the same table, find the percentage of boxes containing at least 5 defective balls each.
Worked solution (try it first) "At least 5" means 5 or more.
Only the 2 boxes with 4 defective fall short, so
50 − 2 = 48 50 - 2 = 48 50 − 2 = 48 boxes qualify.
As a percentage:
48 50 × 100 = 96 \frac{48}{50} \times 100 = 96 50 48 × 100 = 96 , option A.
Watch out
"At least 5" includes the boxes with exactly 5. Leaving out those 7 boxes gives 41 50 = 82 % \frac{41}{50} = 82\% 50 41 = 82% Report a problem with this question
A crate of soft drinks contains 10 bottles of Coca-Cola, 8 of Fanta and 6 of Sprite. If one bottle is selected at random, what is the probability that it is NOT a Coca-Cola bottle?
A 5 12 \frac5{12} 12 5 B 1 3 \frac13 3 1 C 3 4 \frac34 4 3 D 7 12 \frac7{12} 12 7
Worked solution (try it first) There are
10 + 8 + 6 = 24 10 + 8 + 6 = 24 10 + 8 + 6 = 24 bottles.
Not Coca-Cola means Fanta or Sprite:
8 + 6 = 14 8 + 6 = 14 8 + 6 = 14 bottles.
So the probability is
14 24 = 7 12 \frac{14}{24} = \frac{7}{12} 24 14 = 12 7 , option D.
Watch out
10 24 = 5 12 \frac{10}{24} = \frac{5}{12} 24 10 = 12 5 Also set as JAMB 2016 · UTME · Q8
Report a problem with this question