Express 384.126 to the nearest hundred.
Worked solution (try it first) 384.126 lies between the hundreds 300 and 400.
Look at the tens digit, 8: it is 5 or more, so round up to 400, option B.
Watch out
Nearest hundred, not nearest ten or whole number: 380 (option D) is to the nearest ten and 384 (option C) to the nearest whole number. Report a problem with this question
If A = { natural numbers between 3 and 15 } A = \{\text{natural numbers between 3 and 15}\} A = { natural numbers between 3 and 15 } n ( A ) n(A) n ( A )
Worked solution (try it first) "Between 3 and 15" leaves out 3 and 15, so
A = { 4 , 5 , 6 , … , 14 } A = \{4, 5, 6, \dots, 14\} A = { 4 , 5 , 6 , … , 14 } .
Count them:
14 − 4 + 1 = 11 14 - 4 + 1 = 11 14 − 4 + 1 = 11 , so
n ( A ) = 11 n(A) = 11 n ( A ) = 11 , option D.
Watch out
"Between" leaves out the end numbers. Counting 3 and 15 as well gives 13 (option C). Report a problem with this question
Evaluate 3241 five − 1342 five 3241_{\text{five}} - 1342_{\text{five}} 324 1 five − 134 2 five
A 2341 five 2341_{\text{five}} 234 1 five B 1344 five 1344_{\text{five}} 134 4 five C 1342 five 1342_{\text{five}} 134 2 five D 1324 five 1324_{\text{five}} 132 4 five E 1234 five 1234_{\text{five}} 123 4 five
Worked solution (try it first) Change to base ten:
3241 five = 375 + 50 + 20 + 1 3241_{\text{five}} = 375 + 50 + 20 + 1 324 1 five = 375 + 50 + 20 + 1 = 446 = 446 = 446 and
1342 five = 125 + 75 + 20 + 2 1342_{\text{five}} = 125 + 75 + 20 + 2 134 2 five = 125 + 75 + 20 + 2 Subtract:
446 − 222 = 224 446 - 222 = 224 446 − 222 = 224 .
Change back:
224 = 1 × 125 + 3 × 25 + 4 × 5 + 4 224 = 1 \times 125 + 3 \times 25 + 4 \times 5 + 4 224 = 1 × 125 + 3 × 25 + 4 × 5 + 4 , so the answer is
1344 five 1344_{\text{five}} 134 4 five , option B.
Watch out
If you subtract in columns, each borrow brings 5, not 10, and takes 1 from the next column. Check by adding: 1344 five + 1342 five = 3241 five 1344_{\text{five}} + 1342_{\text{five}} = 3241_{\text{five}} 134 4 five + 134 2 five = 324 1 five Report a problem with this question
Simplify ( 4 3 − 6 ) ( 4 3 + 6 ) (4\sqrt3 - 6)(4\sqrt3 + 6) ( 4 3 − 6 ) ( 4 3 + 6 )
A 12 B 4 C 2 3 2\sqrt3 2 3 D 2 E 3 \sqrt3 3
Worked solution (try it first) The brackets are
( a − b ) ( a + b ) (a - b)(a + b) ( a − b ) ( a + b ) , a difference of two squares:
a 2 − b 2 a^2 - b^2 a 2 − b 2 .
Square each part:
( 4 3 ) 2 = 16 × 3 = 48 (4\sqrt3)^2 = 16 \times 3 = 48 ( 4 3 ) 2 = 16 × 3 = 48 and
6 2 = 36 6^2 = 36 6 2 = 36 .
So the value is
48 − 36 = 12 48 - 36 = 12 48 − 36 = 12 , option A.
Watch out
Square both the 4 and the 3 \sqrt3 3 ( 4 3 ) 2 = 48 (4\sqrt3)^2 = 48 ( 4 3 ) 2 = 48 4 × 3 = 12 4 \times 3 = 12 4 × 3 = 12 12 − 36 = − 24 12 - 36 = -24 12 − 36 = − 24 Report a problem with this question
The third term of a geometric progression is − 5 -5 − 5 − 80 -80 − 80
Worked solution (try it first) The 3rd term is
a r 2 = − 5 ar^2 = -5 a r 2 = − 5 and the 7th is
a r 6 = − 80 ar^6 = -80 a r 6 = − 80 .
Divide:
r 4 = − 80 − 5 = 16 r^4 = \frac{-80}{-5} = 16 r 4 = − 5 − 80 = 16 .
Take the fourth root:
r = ± 2 r = \pm 2 r = ± 2 , and the option is
r = 2 r = 2 r = 2 , option E.
Watch out
16 (option A) is r 4 r^4 r 4 r r r Report a problem with this question
A man with an annual salary of ₦1,300,000.00 is to pay an income tax of 22 % 22\% 22%
A ₦254,760.00 B ₦255,860.00 C ₦258,860.00 D ₦350,140.00 E ₦907,140.00
Worked solution (try it first) Allowances are not taxed, so the taxable income is
1 300 000 − 137 000 = 1\,300\,000 - 137\,000 = 1 300 000 − 137 000 = ₦1,163,000.
The tax is
22 % 22\% 22% of that:
0.22 × 1 163 000 = 255 860 0.22 \times 1\,163\,000 = 255\,860 0.22 × 1 163 000 = 255 860 .
So his tax is ₦255,860.00, option B.
Watch out
The tax is 22 % 22\% 22% 78 % 78\% 78% Report a problem with this question
A wire of length 25 cm 25\text{ cm} 25 cm 24.4 cm 24.4\text{ cm} 24.4 cm
Worked solution (try it first) The error is
25 − 24.4 = 0.6 25 - 24.4 = 0.6 25 − 24.4 = 0.6 cm.
Divide by the true length and multiply by 100:
0.6 25 × 100 % = 0.024 × 100 % \frac{0.6}{25} \times 100\% = 0.024 \times 100\% 25 0.6 × 100% = 0.024 × 100% .
So the percentage error is 2.4%, option C.
Watch out
Divide by the true length, 25 cm, not the measured 24.4 cm. Using 24.4 gives 2.459 … % 2.459\ldots\% 2.459 … % Report a problem with this question
How many years will ₦12,000.00 saved in a bank take to amount to ₦12,960.00 at 2 % 2\% 2%
Worked solution (try it first) The interest is the amount minus the principal:
12 960 − 12 000 = 12\,960 - 12\,000 = 12 960 − 12 000 = ₦960.
One year's interest at
2 % 2\% 2% is
0.02 × 12 000 = 0.02 \times 12\,000 = 0.02 × 12 000 = ₦240.
So it takes
960 ÷ 240 = 4 960 \div 240 = 4 960 ÷ 240 = 4 years, option C.
Watch out
Use the interest, ₦960, not the amount ₦12,960. Dividing the amount by ₦240 gives 54 years, which is nowhere near an option. Report a problem with this question
Rationalize 5 2 + 3 \dfrac{5}{\sqrt2 + \sqrt3} 2 + 3 5
A 5 ( 3 − 2 ) 5(\sqrt3 - \sqrt2) 5 ( 3 − 2 ) B 5 ( 2 + 3 ) 5(\sqrt2 + \sqrt3) 5 ( 2 + 3 ) C 5 ( 2 − 3 ) 5(\sqrt2 - \sqrt3) 5 ( 2 − 3 ) D 3 + 5 2 \sqrt3 + 5\sqrt2 3 + 5 2 E 3 − 2 5 \sqrt3 - 2\sqrt5 3 − 2 5
Worked solution (try it first) Multiply the top and bottom by
3 − 2 \sqrt3 - \sqrt2 3 − 2 .
Bottom:
( 3 + 2 ) ( 3 − 2 ) = 3 − 2 (\sqrt3 + \sqrt2)(\sqrt3 - \sqrt2) = 3 - 2 ( 3 + 2 ) ( 3 − 2 ) = 3 − 2 , which is 1.
So the value is
5 ( 3 − 2 ) 5(\sqrt3 - \sqrt2) 5 ( 3 − 2 ) , option A.
Watch out
If you multiply by 2 − 3 \sqrt2 - \sqrt3 2 − 3 2 − 3 = − 1 2 - 3 = -1 2 − 3 = − 1 5 ( 2 − 3 ) 5(\sqrt2 - \sqrt3) 5 ( 2 − 3 ) Report a problem with this question
In a class of 80 students, every student studies Mathematics or Geography or both. If 65 students study Mathematics and 50 study Geography, how many study both subjects?
Worked solution (try it first) Every student studies at least one subject, so
n ( M ∪ G ) = 80 n(M \cup G) = 80 n ( M ∪ G ) = 80 .
Both subjects:
65 + 50 − 80 = 35 65 + 50 - 80 = 35 65 + 50 − 80 = 35 , option B.
Watch out
80 − 65 = 15 80 - 65 = 15 80 − 65 = 15 Report a problem with this question
The 11th term of an arithmetic progression is 63. Find the first term if its common difference is 3.
Worked solution (try it first) The 11th term of an A.P. is
a + 10 d a + 10d a + 10 d , so
a + 10 × 3 = 63 a + 10 \times 3 = 63 a + 10 × 3 = 63 .
Subtract 30 from both sides:
a = 33 a = 33 a = 33 , option D.
Watch out
The 11th term has 10 d 10d 10 d 11 d 11d 11 d 11 d 11d 11 d a = 63 − 33 = 30 a = 63 - 33 = 30 a = 63 − 33 = 30 Report a problem with this question
Evaluate 15 ⊗ 26 15 \otimes 26 15 ⊗ 26
A 0 ( m o d 5 ) 0 \pmod 5 0 ( mod 5 ) B 1 ( m o d 5 ) 1 \pmod 5 1 ( mod 5 ) C 2 ( m o d 5 ) 2 \pmod 5 2 ( mod 5 ) D 4 ( m o d 5 ) 4 \pmod 5 4 ( mod 5 ) E 5 ( m o d 5 ) 5 \pmod 5 5 ( mod 5 )
Worked solution (try it first) Reduce each number first: 15 is a multiple of 5, so
15 ≡ 0 ( m o d 5 ) 15 \equiv 0 \pmod 5 15 ≡ 0 ( mod 5 ) , and
26 ≡ 1 ( m o d 5 ) 26 \equiv 1 \pmod 5 26 ≡ 1 ( mod 5 ) .
Multiply the remainders:
0 × 1 = 0 0 \times 1 = 0 0 × 1 = 0 .
So
15 ⊗ 26 ≡ 0 ( m o d 5 ) 15 \otimes 26 \equiv 0 \pmod 5 15 ⊗ 26 ≡ 0 ( mod 5 ) , option A.
Watch out
In modulo 5 the only values are 0, 1, 2, 3 and 4, so "5 (mod 5)" (option E) is not a final answer: 5 reduces to 0. Report a problem with this question
Solve 2 3 x = 16 3 4 2^{3x} = 16^{\frac34} 2 3 x = 1 6 4 3
Worked solution (try it first) Write 16 as
2 4 2^4 2 4 :
16 3 4 = ( 2 4 ) 3 4 = 2 3 16^{\frac34} = (2^4)^{\frac34} = 2^3 1 6 4 3 = ( 2 4 ) 4 3 = 2 3 , multiplying the indices.
So
2 3 x = 2 3 2^{3x} = 2^3 2 3 x = 2 3 , and the powers are equal:
3 x = 3 3x = 3 3 x = 3 .
Divide by 3:
x = 1 x = 1 x = 1 , option E.
Watch out
Finish by dividing by 3. Stopping at 3 x = 3 3x = 3 3 x = 3 Report a problem with this question
If Olu, Tony and Tunde share ₦240,000.00 in the ratio 2 : 3 : 5 2 : 3 : 5 2 : 3 : 5
A ₦120,000.00 B ₦80,000.00 C ₦72,000.00 D ₦48,000.00 E ₦40,000.00
Worked solution (try it first) The ratio
2 : 3 : 5 2 : 3 : 5 2 : 3 : 5 has 10 parts, and Tunde has 5 of them.
So Tunde's share is
5 10 × 240 000 = \frac{5}{10} \times 240\,000 = 10 5 × 240 000 = ₦120,000.
Two-thirds of it is
2 3 × 120 000 = 80 000 \frac23 \times 120\,000 = 80\,000 3 2 × 120 000 = 80 000 .
The answer is ₦80,000.00, option B.
Watch out
₦120,000.00 (option A) is Tunde's whole share. The question asks for two-thirds of it. Report a problem with this question
Arrange the fractions 2 3 , 3 5 , 5 12 , 4 15 , 3 10 \frac23, \frac35, \frac5{12}, \frac4{15}, \frac3{10} 3 2 , 5 3 , 12 5 , 15 4 , 10 3
A 4 15 , 3 10 , 5 12 , 3 5 , 2 3 \frac4{15}, \frac3{10}, \frac5{12}, \frac35, \frac23 15 4 , 10 3 , 12 5 , 5 3 , 3 2 B 3 10 , 4 15 , 5 12 , 3 5 , 2 3 \frac3{10}, \frac4{15}, \frac5{12}, \frac35, \frac23 10 3 , 15 4 , 12 5 , 5 3 , 3 2 C 4 15 , 5 12 , 3 10 , 3 5 , 2 3 \frac4{15}, \frac5{12}, \frac3{10}, \frac35, \frac23 15 4 , 12 5 , 10 3 , 5 3 , 3 2 D 4 15 , 5 12 , 3 5 , 3 10 , 2 3 \frac4{15}, \frac5{12}, \frac35, \frac3{10}, \frac23 15 4 , 12 5 , 5 3 , 10 3 , 3 2 E 2 3 , 3 5 , 5 12 , 4 15 , 3 10 \frac23, \frac35, \frac5{12}, \frac4{15}, \frac3{10} 3 2 , 5 3 , 12 5 , 15 4 , 10 3
Worked solution (try it first) The LCM of 3, 5, 12, 15 and 10 is 60, so write every fraction in sixtieths.
2 3 = 40 60 \frac23 = \frac{40}{60} 3 2 = 60 40 ,
3 5 = 36 60 \frac35 = \frac{36}{60} 5 3 = 60 36 ,
5 12 = 25 60 \frac{5}{12} = \frac{25}{60} 12 5 = 60 25 ,
4 15 = 16 60 \frac{4}{15} = \frac{16}{60} 15 4 = 60 16 and
3 10 = 18 60 \frac{3}{10} = \frac{18}{60} 10 3 = 60 18 .
Order the tops, smallest first: 16, 18, 25, 36, 40.
So the order is
4 15 , 3 10 , 5 12 , 3 5 , 2 3 \frac{4}{15}, \frac{3}{10}, \frac{5}{12}, \frac35, \frac23 15 4 , 10 3 , 12 5 , 5 3 , 3 2 , option A.
Watch out
4 15 \frac{4}{15} 15 4 3 10 \frac{3}{10} 10 3 16 60 < 18 60 \frac{16}{60} < \frac{18}{60} 60 16 < 60 18 4 15 \frac{4}{15} 15 4 3 10 \frac{3}{10} 10 3 Report a problem with this question
Find the values of x x x y y y z z z ( x 2 y z 9 ) = ( 4 12 3 9 ) \begin{pmatrix} x & 2y \\ z & 9 \end{pmatrix} = \begin{pmatrix} 4 & 12 \\ 3 & 9 \end{pmatrix} ( x z 2 y 9 ) = ( 4 3 12 9 )
A ( 6 , 4 , 3 ) (6, 4, 3) ( 6 , 4 , 3 ) B ( 4 , 6 , 3 ) (4, 6, 3) ( 4 , 6 , 3 ) C ( 6 , 3 , 4 ) (6, 3, 4) ( 6 , 3 , 4 ) D ( 3 , 4 , 6 ) (3, 4, 6) ( 3 , 4 , 6 ) E ( 4 , 4 , 6 ) (4, 4, 6) ( 4 , 4 , 6 )
Worked solution (try it first) Equal matrices have equal entries in matching positions.
Top-right:
2 y = 12 2y = 12 2 y = 12 , so
y = 6 y = 6 y = 6 .
Bottom-left:
z = 3 z = 3 z = 3 .
So
( x , y , z ) = ( 4 , 6 , 3 ) (x, y, z) = (4, 6, 3) ( x , y , z ) = ( 4 , 6 , 3 ) , option B.
Watch out
Match each letter to its own position: x x x y y y ( 6 , 4 , 3 ) (6, 4, 3) ( 6 , 4 , 3 ) x x x y y y Report a problem with this question
If P = ( − 1 2 3 1 ) P = \begin{pmatrix} -1 & 2 \\ 3 & 1 \end{pmatrix} P = ( − 1 3 2 1 ) Q = ( 2 3 2 1 ) Q = \begin{pmatrix} 2 & 3 \\ 2 & 1 \end{pmatrix} Q = ( 2 2 3 1 ) P Q PQ P Q
A ( 2 1 − 8 − 10 ) \begin{pmatrix} 2 & 1 \\ -8 & -10 \end{pmatrix} ( 2 − 8 1 − 10 ) B ( 2 1 8 10 ) \begin{pmatrix} 2 & 1 \\ 8 & 10 \end{pmatrix} ( 2 8 1 10 ) C ( − 2 1 8 10 ) \begin{pmatrix} -2 & 1 \\ 8 & 10 \end{pmatrix} ( − 2 8 1 10 ) D ( − 2 − 1 8 10 ) \begin{pmatrix} -2 & -1 \\ 8 & 10 \end{pmatrix} ( − 2 8 − 1 10 ) E ( 2 − 1 8 10 ) \begin{pmatrix} 2 & -1 \\ 8 & 10 \end{pmatrix} ( 2 8 − 1 10 )
Worked solution (try it first) Multiply each row of
P P P by each column of
Q Q Q .
Row 1 of
P P P is
( − 1 , 2 ) (-1, 2) ( − 1 , 2 ) : with column 1 it gives
− 2 + 4 = 2 -2 + 4 = 2 − 2 + 4 = 2 , and with column 2,
− 3 + 2 = − 1 -3 + 2 = -1 − 3 + 2 = − 1 .
Row 2 of
P P P is
( 3 , 1 ) (3, 1) ( 3 , 1 ) : with column 1 it gives
6 + 2 = 8 6 + 2 = 8 6 + 2 = 8 , and with column 2,
9 + 1 = 10 9 + 1 = 10 9 + 1 = 10 .
So
P Q = ( 2 − 1 8 10 ) PQ = \begin{pmatrix} 2 & -1 \\ 8 & 10 \end{pmatrix} P Q = ( 2 8 − 1 10 ) , option E.
Watch out
Watch the sign in the top-right entry: ( − 1 ) ( 3 ) + 2 ( 1 ) = − 3 + 2 = − 1 (-1)(3) + 2(1) = -3 + 2 = -1 ( − 1 ) ( 3 ) + 2 ( 1 ) = − 3 + 2 = − 1 + 1 +1 + 1 Report a problem with this question
Evaluate log 4 16 + log 3 27 − log 8 4096 \log_4 16 + \log_3 27 - \log_8 4096 log 4 16 + log 3 27 − log 8 4096
A 1 9 \frac19 9 1 B 1 3 \frac13 3 1 C 1 D 2 E 3
Worked solution (try it first) 4 2 = 16 4^2 = 16 4 2 = 16 , so
log 4 16 = 2 \log_4 16 = 2 log 4 16 = 2 .
And
3 3 = 27 3^3 = 27 3 3 = 27 , so
log 3 27 = 3 \log_3 27 = 3 log 3 27 = 3 .
8 4 = 4096 8^4 = 4096 8 4 = 4096 , so
log 8 4096 = 4 \log_8 4096 = 4 log 8 4096 = 4 .
So the value is
2 + 3 − 4 = 1 2 + 3 - 4 = 1 2 + 3 − 4 = 1 , option C.
Watch out
8 3 = 512 8^3 = 512 8 3 = 512 8 4 = 4096 8^4 = 4096 8 4 = 4096 log 8 4096 = 4 \log_8 4096 = 4 log 8 4096 = 4 2 + 3 − 3 = 2 2 + 3 - 3 = 2 2 + 3 − 3 = 2 Report a problem with this question
Solve the equation 4 x 5 − 7 3 = 5 x 12 \dfrac{4x}{5} - \dfrac73 = \dfrac{5x}{12} 5 4 x − 3 7 = 12 5 x
A − 6 2 23 -6\frac2{23} − 6 23 2 B − 3 1 2 -3\frac12 − 3 2 1 C 5 2 23 5\frac2{23} 5 23 2 D 6 2 23 6\frac2{23} 6 23 2 E 6 3 23 6\frac3{23} 6 23 3
Worked solution (try it first) Clear the fractions by multiplying every term by 60, the LCM of 5, 3 and 12:
48 x − 140 = 25 x 48x - 140 = 25x 48 x − 140 = 25 x .
Take
25 x 25x 25 x from both sides and add 140:
23 x = 140 23x = 140 23 x = 140 .
Divide by 23:
23 × 6 = 138 23 \times 6 = 138 23 × 6 = 138 leaves remainder 2, so
x = 6 2 23 x = 6\frac{2}{23} x = 6 23 2 , option D.
Watch out
Moving − 140 -140 − 140 + 140 +140 + 140 − 140 -140 − 140 x = − 6 2 23 x = -6\frac{2}{23} x = − 6 23 2 Report a problem with this question
If f ( x ) = 3 x 2 − 9 x − 5 f(x) = 3x^2 - 9x - 5 f ( x ) = 3 x 2 − 9 x − 5 f ( − 3 ) f(-3) f ( − 3 )
Worked solution (try it first) Put
x = − 3 x = -3 x = − 3 into
f ( x ) = 3 x 2 − 9 x − 5 f(x) = 3x^2 - 9x - 5 f ( x ) = 3 x 2 − 9 x − 5 .
( − 3 ) 2 = 9 (-3)^2 = 9 ( − 3 ) 2 = 9 , so
3 x 2 = 27 3x^2 = 27 3 x 2 = 27 .
And
− 9 × ( − 3 ) = + 27 -9 \times (-3) = +27 − 9 × ( − 3 ) = + 27 .
So
f ( − 3 ) = 27 + 27 − 5 = 49 f(-3) = 27 + 27 - 5 = 49 f ( − 3 ) = 27 + 27 − 5 = 49 , option D.
Watch out
− 9 x -9x − 9 x x = − 3 x = -3 x = − 3 − 9 × ( − 3 ) = + 27 -9 \times (-3) = +27 − 9 × ( − 3 ) = + 27 − 27 -27 − 27 27 − 27 − 5 = − 5 27 - 27 - 5 = -5 27 − 27 − 5 = − 5 Report a problem with this question
Find u u u f f f v v v 1 v = 1 f − 1 u \dfrac1v = \dfrac1f - \dfrac1u v 1 = f 1 − u 1
A u = − f v − f u = \dfrac{-f}{v - f} u = v − f − f B u = f v f − v u = \dfrac{fv}{f - v} u = f − v f v C u = f v v + f u = \dfrac{fv}{v + f} u = v + f f v D u = f v v − f u = \dfrac{fv}{v - f} u = v − f f v E u = f v 2 ( f + v ) u = \dfrac{fv}{2(f + v)} u = 2 ( f + v ) f v
Worked solution (try it first) Add
1 u \frac1u u 1 and subtract
1 v \frac1v v 1 to get
1 u \frac1u u 1 alone:
1 u = 1 f − 1 v \frac1u = \frac1f - \frac1v u 1 = f 1 − v 1 .
Combine over
f v fv f v :
1 u = v − f f v \frac1u = \frac{v - f}{fv} u 1 = f v v − f .
Turn both sides upside down:
u = f v v − f u = \dfrac{fv}{v - f} u = v − f f v , option D.
Watch out
1 f − 1 v = v − f f v \frac1f - \frac1v = \frac{v - f}{fv} f 1 − v 1 = f v v − f v v v f − v f - v f − v Report a problem with this question
Find the quadratic equation whose roots are − 1 -1 − 1
A x 2 − 4 x − 5 = 0 x^2 - 4x - 5 = 0 x 2 − 4 x − 5 = 0 B x 2 − 4 x + 1 = 0 x^2 - 4x + 1 = 0 x 2 − 4 x + 1 = 0 C x 2 − 4 x + 5 = 0 x^2 - 4x + 5 = 0 x 2 − 4 x + 5 = 0 D x 2 + 4 x − 5 = 0 x^2 + 4x - 5 = 0 x 2 + 4 x − 5 = 0 E x 2 + 4 x + 5 = 0 x^2 + 4x + 5 = 0 x 2 + 4 x + 5 = 0
Worked solution (try it first) Roots
− 1 -1 − 1 and 5 give the factors
( x + 1 ) (x + 1) ( x + 1 ) and
( x − 5 ) (x - 5) ( x − 5 ) .
Expand:
( x + 1 ) ( x − 5 ) = x 2 − 5 x + x − 5 (x + 1)(x - 5) = x^2 - 5x + x - 5 ( x + 1 ) ( x − 5 ) = x 2 − 5 x + x − 5 = x 2 − 4 x − 5 = x^2 - 4x - 5 = x 2 − 4 x − 5 .
So the equation is
x 2 − 4 x − 5 = 0 x^2 - 4x - 5 = 0 x 2 − 4 x − 5 = 0 , option A.
Watch out
A root of − 1 -1 − 1 x + 1 x + 1 x + 1 x − 5 x - 5 x − 5 ( x − 1 ) ( x + 5 ) = x 2 + 4 x − 5 (x - 1)(x + 5) = x^2 + 4x - 5 ( x − 1 ) ( x + 5 ) = x 2 + 4 x − 5 Report a problem with this question
The product of two numbers is 40 and their sum is 13. Find the numbers.
A 2 and 20 B 2 and 8 C 4 and 10 D 4 and 8 E 5 and 8
Worked solution (try it first) Call the numbers
x x x and
13 − x 13 - x 13 − x , so their sum is 13.
Their product is
x ( 13 − x ) = 40 x(13 - x) = 40 x ( 13 − x ) = 40 .
Rearrange:
x 2 − 13 x + 40 = 0 x^2 - 13x + 40 = 0 x 2 − 13 x + 40 = 0 , which factorises as
( x − 5 ) ( x − 8 ) = 0 (x - 5)(x - 8) = 0 ( x − 5 ) ( x − 8 ) = 0 .
So the numbers are 5 and 8, option E.
Watch out
Check both conditions. 4 and 10 (option C) multiply to 40 but add to 14, and 4 and 8 (option D) add to 12. Report a problem with this question
If ( x + 6 ) (x + 6) ( x + 6 ) x 2 + 4 x − 12 x^2 + 4x - 12 x 2 + 4 x − 12
A ( x − 2 ) (x - 2) ( x − 2 ) B ( x − 6 ) (x - 6) ( x − 6 ) C ( x + 2 ) (x + 2) ( x + 2 ) D ( x + 4 ) (x + 4) ( x + 4 ) E ( x + 6 ) (x + 6) ( x + 6 )
Worked solution (try it first) The other factor is
( x + b ) (x + b) ( x + b ) , where
6 × b = − 12 6 \times b = -12 6 × b = − 12 (the constant term).
Check the middle term:
6 + ( − 2 ) = 4 6 + (-2) = 4 6 + ( − 2 ) = 4 , which matches
4 x 4x 4 x .
So
x 2 + 4 x − 12 = ( x + 6 ) ( x − 2 ) x^2 + 4x - 12 = (x + 6)(x - 2) x 2 + 4 x − 12 = ( x + 6 ) ( x − 2 ) and the other factor is
( x − 2 ) (x - 2) ( x − 2 ) , option A.
Watch out
The constant is − 12 -12 − 12 ( x + 6 ) ( x + 2 ) (x + 6)(x + 2) ( x + 6 ) ( x + 2 ) + 12 +12 + 12 Report a problem with this question
Calculate the mid-point of the line joining ( 8 , − 3 ) (8, -3) ( 8 , − 3 ) ( − 2 , 3 ) (-2, 3) ( − 2 , 3 )
A ( − 3 , 0 ) (-3, 0) ( − 3 , 0 ) B ( 0 , 5 ) (0, 5) ( 0 , 5 ) C ( 3 , 0 ) (3, 0) ( 3 , 0 ) D ( 0 , 3 ) (0, 3) ( 0 , 3 ) E ( 5 , 0 ) (5, 0) ( 5 , 0 )
Worked solution (try it first) The midpoint is the average of the ends: add the coordinates and halve.
x x x :
8 + ( − 2 ) 2 = 3 \frac{8 + (-2)}{2} = 3 2 8 + ( − 2 ) = 3 .
y y y :
− 3 + 3 2 = 0 \frac{-3 + 3}{2} = 0 2 − 3 + 3 = 0 .
So the midpoint is
( 3 , 0 ) (3, 0) ( 3 , 0 ) , option C.
Watch out
Add the coordinates, don't subtract them: 8 − ( − 2 ) 2 = 5 \frac{8 - (-2)}{2} = 5 2 8 − ( − 2 ) = 5 ( 5 , 0 ) (5, 0) ( 5 , 0 ) Report a problem with this question
Calculate the gradient of the line joining ( − 2 , − 5 ) (-2, -5) ( − 2 , − 5 ) ( 4 , 8 ) (4, 8) ( 4 , 8 )
Worked solution (try it first) Gradient is the change in
y y y over the change in
x x x :
8 − ( − 5 ) 4 − ( − 2 ) = 13 6 \dfrac{8 - (-5)}{4 - (-2)} = \frac{13}{6} 4 − ( − 2 ) 8 − ( − 5 ) = 6 13 .
13 ÷ 6 = 2.1666 … 13 \div 6 = 2.1666\ldots 13 ÷ 6 = 2.1666 … The second decimal is 6, so round up: 2.2, option D.
Watch out
Round, don't cut off: 2.166 … 2.166\ldots 2.166 … Report a problem with this question
Given the statements p p p q q q
A p ∨ q p \vee q p ∨ q B p ∧ q p \wedge q p ∧ q C p ∧ ∼ q p \wedge \sim q p ∧ ∼ q D ∼ p ∨ q \sim p \vee q ∼ p ∨ q E ∼ p ∧ q \sim p \wedge q ∼ p ∧ q
Worked solution (try it first) "Not guilty" is the negation of
p p p , written
∼ p \sim p ∼ p . "Criminals" is
q q q .
"But" joins the two parts like "and", so use
∧ \wedge ∧ .
So the statement is
∼ p ∧ q \sim p \wedge q ∼ p ∧ q , option E.
Watch out
"But" means "and" (∧ \wedge ∧ ∨ \vee ∨ ∼ p ∨ q \sim p \vee q ∼ p ∨ q Report a problem with this question
Simplify x + 1 3 x + 1 2 \dfrac{x + \frac13}{x + \frac12} x + 2 1 x + 3 1
A 2 ( 3 x + 1 ) 3 ( 2 x + 1 ) \dfrac{2(3x + 1)}{3(2x + 1)} 3 ( 2 x + 1 ) 2 ( 3 x + 1 ) B 2 + x 3 + x \dfrac{2 + x}{3 + x} 3 + x 2 + x C x + 1 x − 1 \dfrac{x + 1}{x - 1} x − 1 x + 1 D 3 x + 1 2 x − 1 \dfrac{3x + 1}{2x - 1} 2 x − 1 3 x + 1 E x 2 x − 1 \dfrac{x^2}{x - 1} x − 1 x 2
Worked solution (try it first) Multiply the top and the bottom by 6, the LCM of 3 and 2:
6 x + 2 6 x + 3 \dfrac{6x + 2}{6x + 3} 6 x + 3 6 x + 2 .
Take out common factors:
6 x + 2 = 2 ( 3 x + 1 ) 6x + 2 = 2(3x + 1) 6 x + 2 = 2 ( 3 x + 1 ) and
6 x + 3 = 3 ( 2 x + 1 ) 6x + 3 = 3(2x + 1) 6 x + 3 = 3 ( 2 x + 1 ) .
So the fraction is
2 ( 3 x + 1 ) 3 ( 2 x + 1 ) \dfrac{2(3x + 1)}{3(2x + 1)} 3 ( 2 x + 1 ) 2 ( 3 x + 1 ) , option A.
Watch out
Multiply every term by 6, including the x x x x + 2 x + 3 \frac{x + 2}{x + 3} x + 3 x + 2 Report a problem with this question
Find the equation of a line whose gradient is 6 and y y y − 7 -7 − 7
A y = 8 + 7 x y = 8 + 7x y = 8 + 7 x B y = − 7 − 6 x y = -7 - 6x y = − 7 − 6 x C y = 7 − 6 x y = 7 - 6x y = 7 − 6 x D y = − 7 + 6 x y = -7 + 6x y = − 7 + 6 x E y = 7 + 6 x y = 7 + 6x y = 7 + 6 x
Worked solution (try it first) A line with gradient
m m m and
y y y -intercept
c c c is
y = m x + c y = mx + c y = m x + c .
Put in
m = 6 m = 6 m = 6 and
c = − 7 c = -7 c = − 7 :
y = 6 x − 7 y = 6x - 7 y = 6 x − 7 , which is
y = − 7 + 6 x y = -7 + 6x y = − 7 + 6 x , option D.
Watch out
The intercept is − 7 -7 − 7 + 7 +7 + 7 Report a problem with this question
Find the sum of the roots of the quadratic equation x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0
A 10 B 5 C − 2 -2 − 2 D − 3 -3 − 3 E − 5 -5 − 5
Worked solution (try it first) For
a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 , the sum of the roots is
− b a -\frac ba − a b .
Here
a = 1 a = 1 a = 1 and
b = − 5 b = -5 b = − 5 , so the sum is
− − 5 1 = 5 -\frac{-5}{1} = 5 − 1 − 5 = 5 .
Check:
( x − 2 ) ( x − 3 ) = 0 (x - 2)(x - 3) = 0 ( x − 2 ) ( x − 3 ) = 0 gives roots 2 and 3, which add to 5.
So the answer is option B.
Watch out
The sum is − b a -\frac ba − a b b a \frac ba a b − 5 -5 − 5 Report a problem with this question
Find the roots of the equation 8 x 2 − 6 x − 9 = 0 8x^2 - 6x - 9 = 0 8 x 2 − 6 x − 9 = 0
A x = 3 2 x = \frac32 x = 2 3 − 4 3 -\frac43 − 3 4 B x = 3 2 x = \frac32 x = 2 3 − 3 4 -\frac34 − 4 3 C x = 3 2 x = \frac32 x = 2 3 3 4 \frac34 4 3 D x = − 3 2 x = -\frac32 x = − 2 3 3 4 \frac34 4 3 E x = − 3 2 x = -\frac32 x = − 2 3 − 3 4 -\frac34 − 4 3
Worked solution (try it first) Find two numbers with product
8 × ( − 9 ) = − 72 8 \times (-9) = -72 8 × ( − 9 ) = − 72 and sum
− 6 -6 − 6 : they are
− 12 -12 − 12 and 6.
Split and group:
8 x 2 − 12 x + 6 x − 9 = 4 x ( 2 x − 3 ) + 3 ( 2 x − 3 ) 8x^2 - 12x + 6x - 9 = 4x(2x - 3) + 3(2x - 3) 8 x 2 − 12 x + 6 x − 9 = 4 x ( 2 x − 3 ) + 3 ( 2 x − 3 ) = ( 2 x − 3 ) ( 4 x + 3 ) = (2x - 3)(4x + 3) = ( 2 x − 3 ) ( 4 x + 3 ) .
Set each factor to zero:
2 x − 3 = 0 2x - 3 = 0 2 x − 3 = 0 gives
x = 3 2 x = \frac32 x = 2 3 and
4 x + 3 = 0 4x + 3 = 0 4 x + 3 = 0 gives
x = − 3 4 x = -\frac34 x = − 4 3 , option B.
Watch out
4 x + 3 = 0 4x + 3 = 0 4 x + 3 = 0 x = − 3 4 x = -\frac34 x = − 4 3 x x x − 4 3 -\frac43 − 3 4 Report a problem with this question
Expand ( 2 x − 3 ) ( 3 x + 4 ) (2x - 3)(3x + 4) ( 2 x − 3 ) ( 3 x + 4 )
A 6 x 2 − 17 x − 12 6x^2 - 17x - 12 6 x 2 − 17 x − 12 B 6 x 2 − x − 12 6x^2 - x - 12 6 x 2 − x − 12 C 6 x 2 − x + 12 6x^2 - x + 12 6 x 2 − x + 12 D 6 x 2 + x − 12 6x^2 + x - 12 6 x 2 + x − 12 E 6 x 2 + 17 x − 12 6x^2 + 17x - 12 6 x 2 + 17 x − 12
Worked solution (try it first) Multiply each term in the first bracket by each in the second:
6 x 2 + 8 x − 9 x − 12 6x^2 + 8x - 9x - 12 6 x 2 + 8 x − 9 x − 12 .
Collect the
x x x terms:
8 x − 9 x = − x 8x - 9x = -x 8 x − 9 x = − x .
So
( 2 x − 3 ) ( 3 x + 4 ) = 6 x 2 − x − 12 (2x - 3)(3x + 4) = 6x^2 - x - 12 ( 2 x − 3 ) ( 3 x + 4 ) = 6 x 2 − x − 12 , option B.
Watch out
The x x x 2 x × 4 = 8 x 2x \times 4 = 8x 2 x × 4 = 8 x − 3 × 3 x = − 9 x -3 \times 3x = -9x − 3 × 3 x = − 9 x − x -x − x 6 x 2 + x − 12 6x^2 + x - 12 6 x 2 + x − 12 Report a problem with this question
The difference between the present ages of two brothers is 6 and their product is 135. What is the sum of their ages?
Worked solution (try it first) Let the younger brother be
x x x , so the older is
x + 6 x + 6 x + 6 .
Their product is
x ( x + 6 ) = 135 x(x + 6) = 135 x ( x + 6 ) = 135 .
Rearrange:
x 2 + 6 x − 135 = 0 x^2 + 6x - 135 = 0 x 2 + 6 x − 135 = 0 , which factorises as
( x − 9 ) ( x + 15 ) = 0 (x - 9)(x + 15) = 0 ( x − 9 ) ( x + 15 ) = 0 .
An age is positive, so
x = 9 x = 9 x = 9 and the ages are 9 and 15.
So the sum of their ages is
9 + 15 = 24 9 + 15 = 24 9 + 15 = 24 , option B.
Watch out
15 (option E) is the older brother's age. The question asks for the sum of both ages. Report a problem with this question
In the diagram, P Q S PQS P QS O O O R S T RST R S T S S S ∠ S O P = 120 ∘ \angle SOP = 120^\circ ∠ S O P = 12 0 ∘ ∠ P S T \angle PST ∠ P S T
A 64 ∘ 64^\circ 6 4 ∘ B 60 ∘ 60^\circ 6 0 ∘ C 35 ∘ 35^\circ 3 5 ∘ D 31 ∘ 31^\circ 3 1 ∘ E 29 ∘ 29^\circ 2 9 ∘
Worked solution (try it first) O S = O P OS = OP O S = O P (radii), so triangle
O S P OSP O S P is isosceles:
∠ O S P = 180 ∘ − 120 ∘ 2 \angle OSP = \dfrac{180^\circ - 120^\circ}{2} ∠ O S P = 2 18 0 ∘ − 12 0 ∘ A radius meets a tangent at
90 ∘ 90^\circ 9 0 ∘ , so
∠ O S T = 90 ∘ \angle OST = 90^\circ ∠ O S T = 9 0 ∘ .
So
∠ P S T = 90 ∘ − 30 ∘ \angle PST = 90^\circ - 30^\circ ∠ P S T = 9 0 ∘ − 3 0 ∘ = 60 ∘ = 60^\circ = 6 0 ∘ , option B.
Watch out
30 ∘ 30^\circ 3 0 ∘ ∠ O S P \angle OSP ∠ O S P ∠ P S T \angle PST ∠ P S T O S OS O S Report a problem with this question
In the diagram, A A A B B B C C C D D D O O O B A BA B A M M M ∠ M A D = 82 ∘ \angle MAD = 82^\circ ∠ M A D = 8 2 ∘ ∠ A D O = 74 ∘ \angle ADO = 74^\circ ∠ A D O = 7 4 ∘ ∠ A B O \angle ABO ∠ A B O
A 24 ∘ 24^\circ 2 4 ∘ B 74 ∘ 74^\circ 7 4 ∘ C 82 ∘ 82^\circ 8 2 ∘ D 98 ∘ 98^\circ 9 8 ∘ E 164 ∘ 164^\circ 16 4 ∘
Worked solution (try it first) O A = O D OA = OD O A = O D (radii), so
∠ O A D = ∠ O D A = 74 ∘ \angle OAD = \angle ODA = 74^\circ ∠ O A D = ∠ O D A = 7 4 ∘ .
B A M BAM B A M is a straight line:
∠ B A D = 180 ∘ − 82 ∘ \angle BAD = 180^\circ - 82^\circ ∠ B A D = 18 0 ∘ − 8 2 ∘ = 98 ∘ = 98^\circ = 9 8 ∘ , so
∠ B A O = 98 ∘ − 74 ∘ \angle BAO = 98^\circ - 74^\circ ∠ B A O = 9 8 ∘ − 7 4 ∘ O A = O B OA = OB O A = O B (radii), so
∠ A B O = ∠ B A O = 24 ∘ \angle ABO = \angle BAO = 24^\circ ∠ A B O = ∠ B A O = 2 4 ∘ , option A.
Watch out
98 ∘ 98^\circ 9 8 ∘ ∠ B A D \angle BAD ∠ B A D 74 ∘ 74^\circ 7 4 ∘ ∠ O A D \angle OAD ∠ O A D ∠ B A O \angle BAO ∠ B A O Report a problem with this question
Calculate the area of trapezium A B C D ABCD A B C D A B ∥ D C AB \parallel DC A B ∥ D C ∣ A B ∣ = 8 cm |AB| = 8\text{ cm} ∣ A B ∣ = 8 cm ∣ D C ∣ = 16 cm |DC| = 16\text{ cm} ∣ D C ∣ = 16 cm ∣ B C ∣ = 12 cm |BC| = 12\text{ cm} ∣ B C ∣ = 12 cm ∠ B C D = 30 ∘ \angle BCD = 30^\circ ∠ B C D = 3 0 ∘
A 27 cm 2 27\text{ cm}^2 27 cm 2 B 36 cm 2 36\text{ cm}^2 36 cm 2 C 72 cm 2 72\text{ cm}^2 72 cm 2 D 92 cm 2 92\text{ cm}^2 92 cm 2 E 144 cm 2 144\text{ cm}^2 144 cm 2
Worked solution (try it first) In the right-angled triangle at
C C C , the height is
h = 12 sin 30 ∘ h = 12\sin30^\circ h = 12 sin 3 0 ∘ .
As
sin 30 ∘ = 1 2 \sin30^\circ = \frac12 sin 3 0 ∘ = 2 1 ,
h = 6 h = 6 h = 6 cm.
Area of a trapezium
= 1 2 ( a + b ) h = \frac12(a + b)h = 2 1 ( a + b ) h = 1 2 ( 8 + 16 ) × 6 = \frac12(8 + 16) \times 6 = 2 1 ( 8 + 16 ) × 6 .
So the area is
12 × 6 = 72 cm 2 12 \times 6 = 72\text{ cm}^2 12 × 6 = 72 cm 2 , option C.
Watch out
Keep the 1 2 \frac12 2 1 ( 8 + 16 ) × 6 = 144 cm 2 (8 + 16) \times 6 = 144\text{ cm}^2 ( 8 + 16 ) × 6 = 144 cm 2 Report a problem with this question
A boy walks 4 km 4\text{ km} 4 km 214 ∘ 214^\circ 21 4 ∘
A 5.1 km B 8.3 km C 11.1 km D 15.7 km E 17.4 km
Worked solution (try it first) Put the start at the origin.
After the first leg he is at
( − 4 , 0 ) (-4, 0) ( − 4 , 0 ) .
On
214 ∘ 214^\circ 21 4 ∘ (that is,
34 ∘ 34^\circ 3 4 ∘ west of south), walking
t t t km adds
( − t sin 34 ∘ , − t cos 34 ∘ ) (-t\sin34^\circ, -t\cos34^\circ) ( − t sin 3 4 ∘ , − t cos 3 4 ∘ ) .
South-west of the start means as far south as west:
4 + 0.5592 t = 0.8290 t 4 + 0.5592t = 0.8290t 4 + 0.5592 t = 0.8290 t , so
t ≈ 14.83 t \approx 14.83 t ≈ 14.83 km.
He is then about 12.29 km west and 12.29 km south, so his distance is
12.29 2 ≈ 17.4 12.29\sqrt2 \approx 17.4 12.29 2 ≈ 17.4 km, option E.
Watch out
South-west means equal distances west and south (a bearing of 225 ∘ 225^\circ 22 5 ∘ Report a problem with this question
Two points P P P Q Q Q P P P 70 ∘ 70^\circ 7 0 ∘ Q Q Q 55 ∘ 55^\circ 5 5 ∘
A 15 ∘ 15^\circ 1 5 ∘ B 25 ∘ 25^\circ 2 5 ∘ C 45 ∘ 45^\circ 4 5 ∘ D 65 ∘ 65^\circ 6 5 ∘ E 125 ∘ 125^\circ 12 5 ∘
Worked solution (try it first) Both latitudes are north of the equator, so subtract them.
The difference is
70 ∘ − 55 ∘ = 15 ∘ 70^\circ - 55^\circ = 15^\circ 7 0 ∘ − 5 5 ∘ = 1 5 ∘ , option A.
Watch out
Add only when one latitude is north and the other south. Adding here gives 125 ∘ 125^\circ 12 5 ∘ Report a problem with this question
A bird on top of a building 35 m 35\text{ m} 35 m 25 m 25\text{ m} 25 m
A 75.00 ∘ 75.00^\circ 75.0 0 ∘ B 60.00 ∘ 60.00^\circ 60.0 0 ∘ C 54.46 ∘ 54.46^\circ 54.4 6 ∘ D 35.25 ∘ 35.25^\circ 35.2 5 ∘ E 10.00 ∘ 10.00^\circ 10.0 0 ∘
Worked solution (try it first) The angle of depression equals the angle of elevation of the bird from the prey (alternate angles).
The height 35 m is opposite the angle and 25 m is adjacent:
tan θ = 35 25 = 1.4 \tan\theta = \frac{35}{25} = 1.4 tan θ = 25 35 = 1.4 .
So
θ = tan − 1 1.4 ≈ 54.46 ∘ \theta = \tan^{-1}1.4 \approx 54.46^\circ θ = tan − 1 1.4 ≈ 54.4 6 ∘ , option C.
Watch out
Put the height on top: 35 25 \frac{35}{25} 25 35 25 35 \frac{25}{35} 35 25 35.54 ∘ 35.54^\circ 35.5 4 ∘ Report a problem with this question
Calculate the length of an arc which subtends an angle of 66 ∘ 66^\circ 6 6 ∘ 8 cm 8\text{ cm} 8 cm
A 3.4 cm B 5.2 cm C 7.4 cm D 9.2 cm E 12.3 cm
Worked solution (try it first) Arc length is
θ 360 ∘ × 2 π r \frac{\theta}{360^\circ} \times 2\pi r 36 0 ∘ θ × 2 π r .
Put in
θ = 66 ∘ \theta = 66^\circ θ = 6 6 ∘ and
r = 8 r = 8 r = 8 : arc
= 66 360 × 2 × 22 7 × 8 = \frac{66}{360} \times 2 \times \frac{22}{7} \times 8 = 360 66 × 2 × 7 22 × 8 To one decimal place the arc is 9.2 cm, option D.
Watch out
Use the circumference 2 π r 2\pi r 2 π r 66 360 \frac{66}{360} 360 66 π r \pi r π r Report a problem with this question
Find the value of y y y
A 48 ∘ 48^\circ 4 8 ∘ B 57 ∘ 57^\circ 5 7 ∘ C 105 ∘ 105^\circ 10 5 ∘ D 132 ∘ 132^\circ 13 2 ∘ E 198 ∘ 198^\circ 19 8 ∘
Worked solution (try it first) Angles on a straight line: the interior angle at the top vertex is
180 ∘ − 123 ∘ = 57 ∘ 180^\circ - 123^\circ = 57^\circ 18 0 ∘ − 12 3 ∘ = 5 7 ∘ .
y y y is an exterior angle of the triangle, so it equals the sum of the two opposite interior angles:
y = 75 ∘ + 57 ∘ = 132 ∘ y = 75^\circ + 57^\circ = 132^\circ y = 7 5 ∘ + 5 7 ∘ = 13 2 ∘ , option D.
Watch out
123 ∘ 123^\circ 12 3 ∘ 57 ∘ 57^\circ 5 7 ∘ 75 ∘ + 123 ∘ 75^\circ + 123^\circ 7 5 ∘ + 12 3 ∘ 198 ∘ 198^\circ 19 8 ∘ Report a problem with this question
The base radius and height of a cone are 4 cm 4\text{ cm} 4 cm 6 cm 6\text{ cm} 6 cm
A 75 cm 3 75\text{ cm}^3 75 cm 3 B 86 cm 3 86\text{ cm}^3 86 cm 3 C 98 cm 3 98\text{ cm}^3 98 cm 3 D 101 cm 3 101\text{ cm}^3 101 cm 3 E 110 cm 3 110\text{ cm}^3 110 cm 3
Worked solution (try it first) Volume of a cone:
1 3 π r 2 h = 1 3 × 22 7 × 16 × 6 \frac13\pi r^2h = \frac13 \times \frac{22}{7} \times 16 \times 6 3 1 π r 2 h = 3 1 × 7 22 × 16 × 6 .
That is
22 7 × 32 ≈ 100.6 cm 3 \frac{22}{7} \times 32 \approx 100.6\text{ cm}^3 7 22 × 32 ≈ 100.6 cm 3 .
To the nearest whole number,
101 cm 3 101\text{ cm}^3 101 cm 3 , option D.
Watch out
Square the radius and keep the 1 3 \frac13 3 1 π r h \pi rh π r h 75 cm 3 75\text{ cm}^3 75 cm 3 Report a problem with this question
Find the distance between the points ( 3 , − 4 ) (3, -4) ( 3 , − 4 ) ( − 5 , 2 ) (-5, 2) ( − 5 , 2 )
Worked solution (try it first) The changes are
− 5 − 3 = − 8 -5 - 3 = -8 − 5 − 3 = − 8 in
x x x and
2 − ( − 4 ) = 6 2 - (-4) = 6 2 − ( − 4 ) = 6 in
y y y .
By Pythagoras the distance is
8 2 + 6 2 = 100 = 10 \sqrt{8^2 + 6^2} = \sqrt{100} = 10 8 2 + 6 2 = 100 = 10 , option E.
Watch out
Square the changes, add, then take the square root. Just adding them gives 8 + 6 = 14 8 + 6 = 14 8 + 6 = 14 Report a problem with this question
Calculate the surface area of a sphere with diameter 10.4 cm 10.4\text{ cm} 10.4 cm π = 3.142 \pi = 3.142 π = 3.142
A 439 cm 2 439\text{ cm}^2 439 cm 2 B 400 cm 2 400\text{ cm}^2 400 cm 2 C 340 cm 2 340\text{ cm}^2 340 cm 2 D 339 cm 2 339\text{ cm}^2 339 cm 2 E 338 cm 2 338\text{ cm}^2 338 cm 2
Worked solution (try it first) Radius
= 10.4 ÷ 2 = 5.2 = 10.4 \div 2 = 5.2 = 10.4 ÷ 2 = 5.2 cm.
The surface area of a sphere is
4 π r 2 4\pi r^2 4 π r 2 .
4 × 3.142 × 5.2 2 = 4 × 3.142 × 27.04 4 \times 3.142 \times 5.2^2 = 4 \times 3.142 \times 27.04 4 × 3.142 × 5. 2 2 = 4 × 3.142 × 27.04 To 3 significant figures that is
340 cm 2 340\text{ cm}^2 340 cm 2 , option C.
Watch out
Round, don't cut off: 339.84 rounds up to 340. Dropping the decimals gives 339 (option D). Report a problem with this question
Obi walks 400 m 400\text{ m} 400 m 30 ∘ 30^\circ 3 0 ∘
A 430 m B 400 3 400\sqrt3 400 3 C 200 3 200\sqrt3 200 3 D 200 m E 100 m
Worked solution (try it first) The 400 m walked up the slope is the hypotenuse, and the height is opposite the
30 ∘ 30^\circ 3 0 ∘ angle.
So the height is
400 sin 30 ∘ = 400 × 1 2 400\sin30^\circ = 400 \times \frac12 400 sin 3 0 ∘ = 400 × 2 1 = 200 = 200 = 200 m, option D.
Watch out
The height is opposite the angle, so use sine. Cosine gives the horizontal distance, 400 cos 30 ∘ = 200 3 400\cos30^\circ = 200\sqrt3 400 cos 3 0 ∘ = 200 3 Report a problem with this question
The volume of a spherical ball is 114 cm 3 114\text{ cm}^3 114 cm 3
A 8 cm B 6 cm C 5 cm D 4 cm E 3 cm
Worked solution (try it first) Volume of a sphere:
4 3 π r 3 = 114 \frac43\pi r^3 = 114 3 4 π r 3 = 114 .
So
r 3 = 3 × 114 4 π r^3 = \dfrac{3 \times 114}{4\pi} r 3 = 4 π 3 × 114 Take the cube root:
r ≈ 3 r \approx 3 r ≈ 3 cm, option E.
Watch out
Take the cube root of 27.2, not the square root. The square root is about 5.2 and gives 5 cm (option C). Report a problem with this question
Find the equation of a line with gradient − 1 4 -\frac14 − 4 1 ( 3 , 2 ) (3, 2) ( 3 , 2 )
A y + x = 3 y + x = 3 y + x = 3 B y − x = 1 y - x = 1 y − x = 1 C 5 y + 2 x = 2 5y + 2x = 2 5 y + 2 x = 2 D 4 y + x = 11 4y + x = 11 4 y + x = 11 E 3 y + 4 x = 7 3y + 4x = 7 3 y + 4 x = 7
Worked solution (try it first) Use
y − y 1 = m ( x − x 1 ) y - y_1 = m(x - x_1) y − y 1 = m ( x − x 1 ) with
( 3 , 2 ) (3, 2) ( 3 , 2 ) :
y − 2 = − 1 4 ( x − 3 ) y - 2 = -\frac14(x - 3) y − 2 = − 4 1 ( x − 3 ) .
Multiply both sides by 4:
4 y − 8 = − x + 3 4y - 8 = -x + 3 4 y − 8 = − x + 3 .
Collect terms:
4 y + x = 11 4y + x = 11 4 y + x = 11 , option D.
Watch out
Multiply every term by 4, including the 2. Leaving it gives 4 y + x = 5 4y + x = 5 4 y + x = 5 4 ( 2 ) + 3 = 11 4(2) + 3 = 11 4 ( 2 ) + 3 = 11 Report a problem with this question
Calculate the value of y y y
A 3 3 3\sqrt3 3 3 B 2 3 2\sqrt3 2 3 C 3 \sqrt3 3 D 3 3 2 \frac{3\sqrt3}{2} 2 3 3 E 2 \sqrt2 2
Worked solution (try it first) The 6 cm side is the hypotenuse, and
y y y is the side next to the
30 ∘ 30^\circ 3 0 ∘ angle, so use cosine:
cos 30 ∘ = y 6 \cos30^\circ = \dfrac{y}{6} cos 3 0 ∘ = 6 y .
So
y = 6 cos 30 ∘ y = 6\cos30^\circ y = 6 cos 3 0 ∘ = 6 × 3 2 = 6 \times \frac{\sqrt3}{2} = 6 × 2 3 = 3 3 = 3\sqrt3 = 3 3 cm, option A.
Watch out
y y y 30 ∘ 30^\circ 3 0 ∘ 6 sin 30 ∘ = 3 6\sin30^\circ = 3 6 sin 3 0 ∘ = 3 y y y Report a problem with this question
Given the numbers 11, 8, 9, 6, 4, 3, 10, 2, 6, 5, calculate the mean.
Worked solution (try it first) Add the ten numbers:
11 + 8 + 9 + 6 + 4 + 3 + 10 + 2 + 6 + 5 = 64 11 + 8 + 9 + 6 + 4 + 3 + 10 + 2 + 6 + 5 = 64 11 + 8 + 9 + 6 + 4 + 3 + 10 + 2 + 6 + 5 = 64 .
Divide by 10: the mean is
64 10 = 6.4 \frac{64}{10} = 6.4 10 64 = 6.4 , option C.
Watch out
Divide by the count, 10. 6.0 (option D) is the mode, which occurs twice, not the mean. Report a problem with this question
Given the numbers 11, 8, 9, 6, 4, 3, 10, 2, 6, 5, find the median.
Worked solution (try it first) Put the 10 numbers in order: 2, 3, 4, 5, 6, 6, 8, 9, 10, 11.
With an even count, the median is halfway between the 5th and 6th.
Both are 6.
So the median is 6, option B.
Watch out
Order the list first. The 5th and 6th numbers as written are 4 and 3, which gives 3.5, not an option. Report a problem with this question
Given the numbers 11, 8, 9, 6, 4, 3, 10, 2, 6, 5, find the mode.
Worked solution (try it first) Count each number: 6 appears twice and every other number once.
The mode is the number that occurs most often: 6, option C.
Watch out
The mode is the most frequent number, not the largest (11, option A) or the first in the list. Report a problem with this question
Find the range of 4, 9, 6, 3, 2, 8, 10, 7, 11.
Worked solution (try it first) The largest number is 11 and the smallest is 2.
The range is
11 − 2 = 9 11 - 2 = 9 11 − 2 = 9 , option B.
Watch out
The smallest number is 2, not the first number 4. Using 4 gives 11 − 4 = 7 11 - 4 = 7 11 − 4 = 7 Similar: JAMB 2012 · UTME · Q45
Report a problem with this question
Calculate the variance of 30, 28, 35, 25, 37.
A 31.0 B 25.1 C 19.6 D 19.5 E 15.3
Worked solution (try it first) The numbers add up to 155, so the mean is 31.
The deviations are
− 1 -1 − 1 ,
− 3 -3 − 3 , 4,
− 6 -6 − 6 , 6.
Their squares are 1, 9, 16, 36, 36, which add up to 98.
The variance is
98 5 = 19.6 \frac{98}{5} = 19.6 5 98 = 19.6 , option C.
Watch out
Don't stop at the mean: 31.0 (option A) is the mean, not the variance. Report a problem with this question
Tickets numbered 1 to 16 inclusive are mixed up and a ticket is drawn at random. What is the probability that the number is a multiple of 2 or 3?
A 5 16 \frac5{16} 16 5 B 1 2 \frac12 2 1 C 5 8 \frac58 8 5 D 11 16 \frac{11}{16} 16 11 E 13 16 \frac{13}{16} 16 13
Worked solution (try it first) From 1 to 16 there are 8 multiples of 2 and 5 multiples of 3 (3, 6, 9, 12, 15).
6 and 12 are in both lists, so take them off once:
8 + 5 − 2 = 11 8 + 5 - 2 = 11 8 + 5 − 2 = 11 .
So the probability is
11 16 \frac{11}{16} 16 11 , option D.
Watch out
6 and 12 are multiples of both 2 and 3, so count them once. Adding 8 + 5 8 + 5 8 + 5 13 16 \frac{13}{16} 16 13 Report a problem with this question
The lucky numbers in a raffle draw are 4, 5, 12, 20, 2, 8, 3, 6, 10, 9, 7, 25, 12, 10, 14, 27. If a number is picked at random, what is the probability that it is a perfect cube?
A 11 16 \frac{11}{16} 16 11 B 5 8 \frac58 8 5 C 7 16 \frac7{16} 16 7 D 3 8 \frac38 8 3 E 1 8 \frac18 8 1
Worked solution (try it first) There are 16 numbers in the list.
A perfect cube is a whole number cubed: 8 is
2 3 2^3 2 3 and 27 is
3 3 3^3 3 3 .
No other number in the list is a cube.
So the probability is
2 16 = 1 8 \frac{2}{16} = \frac18 16 2 = 8 1 , option E.
Watch out
Don't mix up cubes and squares: 4, 9 and 25 are perfect squares, not cubes. Only 8 and 27 count. Report a problem with this question
The probabilities that it will rain in Lagos and Oyo on the same day are 3 4 \frac34 4 3 1 2 \frac12 2 1
A 1 12 \frac1{12} 12 1 B 1 8 \frac18 8 1 C 1 6 \frac16 6 1 D 1 4 \frac14 4 1 E 1 2 \frac12 2 1
Worked solution (try it first) It does not rain in Lagos with probability
1 − 3 4 = 1 4 1 - \frac34 = \frac14 1 − 4 3 = 4 1 , and not in Oyo with probability
1 − 1 2 = 1 2 1 - \frac12 = \frac12 1 − 2 1 = 2 1 .
No rain in either town means both of these happen, so multiply:
1 4 × 1 2 = 1 8 \frac14 \times \frac12 = \frac18 4 1 × 2 1 = 8 1 , option B.
Watch out
Multiply the chances of no rain. Adding them, 1 4 + 1 2 = 3 4 \frac14 + \frac12 = \frac34 4 1 + 2 1 = 4 3 Report a problem with this question
The table shows the ages of students in a class. What is the probability that a student chosen at random is less than 18 years old?
Age (years)
16
17
18
No. of students
4
8
6
A 9 10 \frac9{10} 10 9 B 4 5 \frac45 5 4 C 2 3 \frac23 3 2 D 1 2 \frac12 2 1 E 1 3 \frac13 3 1
Worked solution (try it first) There are
4 + 8 + 6 = 18 4 + 8 + 6 = 18 4 + 8 + 6 = 18 students.
Less than 18 years means 16 or 17:
4 + 8 = 12 4 + 8 = 12 4 + 8 = 12 students.
So the probability is
12 18 = 2 3 \frac{12}{18} = \frac23 18 12 = 3 2 , option C.
Watch out
6 18 = 1 3 \frac{6}{18} = \frac13 18 6 = 3 1 Report a problem with this question
A particle moves a distance of S S S t t t S = 5 t 3 − 12 t 2 + 7 S = 5t^3 - 12t^2 + 7 S = 5 t 3 − 12 t 2 + 7
A 0.8 sec B 1.2 sec C 1.6 sec D 2.0 sec E 2.5 sec
Worked solution (try it first) Velocity is
v = d S d t = 15 t 2 − 24 t v = \frac{dS}{dt} = 15t^2 - 24t v = d t d S = 15 t 2 − 24 t .
Acceleration is
a = d v d t = 30 t − 24 a = \frac{dv}{dt} = 30t - 24 a = d t d v = 30 t − 24 .
Set
a = 0 a = 0 a = 0 :
30 t = 24 30t = 24 30 t = 24 , so
t = 0.8 t = 0.8 t = 0.8 s, option A.
Watch out
Acceleration is the second derivative. Setting the velocity to zero instead gives 15 t 2 = 24 t 15t^2 = 24t 15 t 2 = 24 t t = 1.6 t = 1.6 t = 1.6 Report a problem with this question
A particle moves a distance of S S S t t t S = 5 t 3 − 12 t 2 + 7 S = 5t^3 - 12t^2 + 7 S = 5 t 3 − 12 t 2 + 7
A 70 m/s B 63 m/s C 50 m/s D 40 m/s E 30 m/s
Worked solution (try it first) Velocity is
v = d S d t = 15 t 2 − 24 t v = \frac{dS}{dt} = 15t^2 - 24t v = d t d S = 15 t 2 − 24 t .
At
t = 3 t = 3 t = 3 :
15 × 9 − 24 × 3 = 135 − 72 = 63 15 \times 9 - 24 \times 3 = 135 - 72 = 63 15 × 9 − 24 × 3 = 135 − 72 = 63 m/s, option B.
Watch out
Differentiate before substituting. S S S t = 3 t = 3 t = 3 135 − 108 + 7 = 34 135 - 108 + 7 = 34 135 − 108 + 7 = 34 Report a problem with this question
A particle moves a distance of S S S t t t S = 5 t 3 − 12 t 2 + 7 S = 5t^3 - 12t^2 + 7 S = 5 t 3 − 12 t 2 + 7
A 400 m/s 2 400\text{ m/s}^2 400 m/s 2 B 360 m/s 2 360\text{ m/s}^2 360 m/s 2 C 336 m/s 2 336\text{ m/s}^2 336 m/s 2 D 300 m/s 2 300\text{ m/s}^2 300 m/s 2 E 180 m/s 2 180\text{ m/s}^2 180 m/s 2
Worked solution (try it first) Velocity is
v = 15 t 2 − 24 t v = 15t^2 - 24t v = 15 t 2 − 24 t , so acceleration is
a = d v d t = 30 t − 24 a = \frac{dv}{dt} = 30t - 24 a = d t d v = 30 t − 24 .
At
t = 12 t = 12 t = 12 :
30 × 12 − 24 = 360 − 24 30 \times 12 - 24 = 360 - 24 30 × 12 − 24 = 360 − 24 = 336 m/s 2 = 336\text{ m/s}^2 = 336 m/s 2 , option C.
Watch out
Don't drop the − 24 -24 − 24 − 24 t -24t − 24 t − 24 -24 − 24 360 m/s 2 360\text{ m/s}^2 360 m/s 2 Report a problem with this question