Convert 112 6 112_6 11 2 6
Worked solution (try it first) Change to base ten using place values:
112 6 = 1 × 36 + 1 × 6 + 2 = 44 112_6 = 1 \times 36 + 1 \times 6 + 2 = 44 11 2 6 = 1 × 36 + 1 × 6 + 2 = 44 .
Divide by 3 repeatedly:
44 = 3 × 14 + 2 44 = 3 \times 14 + 2 44 = 3 × 14 + 2 ,
14 = 3 × 4 + 2 14 = 3 \times 4 + 2 14 = 3 × 4 + 2 ,
4 = 3 × 1 + 1 4 = 3 \times 1 + 1 4 = 3 × 1 + 1 , and
1 = 3 × 0 + 1 1 = 3 \times 0 + 1 1 = 3 × 0 + 1 .
Read the remainders from the bottom up:
1122 3 1122_3 112 2 3 , option D.
Watch out
Read the remainders from the last division to the first. Reading them top-down gives 2211 (option A). Report a problem with this question
Simplify 2 2 3 × 1 1 2 4 4 5 \dfrac{2\frac23 \times 1\frac12}{4\frac45} 4 5 4 2 3 2 × 1 2 1
A 1 1 4 1\frac14 1 4 1 B 1 1 6 1\frac16 1 6 1 C 5 6 \frac56 6 5 D 4 5 \frac45 5 4
Worked solution (try it first) Change the mixed numbers to improper fractions:
2 2 3 = 8 3 2\frac23 = \frac83 2 3 2 = 3 8 ,
1 1 2 = 3 2 1\frac12 = \frac32 1 2 1 = 2 3 and
4 4 5 = 24 5 4\frac45 = \frac{24}{5} 4 5 4 = 5 24 .
Top:
8 3 × 3 2 = 4 \frac83 \times \frac32 = 4 3 8 × 2 3 = 4 .
Divide by flipping:
4 × 5 24 = 20 24 4 \times \frac{5}{24} = \frac{20}{24} 4 × 24 5 = 24 20 = 5 6 = \frac56 = 6 5 , option C.
Watch out
4 4 5 4\frac45 4 5 4 4 + 4 5 = 24 5 4 + \frac45 = \frac{24}{5} 4 + 5 4 = 5 24 4 × 4 5 = 16 5 4 \times \frac45 = \frac{16}{5} 4 × 5 4 = 5 16 4 ÷ 16 5 = 1 1 4 4 \div \frac{16}{5} = 1\frac14 4 ÷ 5 16 = 1 4 1 Report a problem with this question
Evaluate 21 9 \frac{21}{9} 9 21
Worked solution (try it first) Divide:
21 9 = 7 3 = 2.333 … \frac{21}{9} = \frac73 = 2.333\ldots 9 21 = 3 7 = 2.333 … Three significant figures are 2, 3 and 3.
The next figure is 3, so round down: 2.33, option D.
Watch out
Work out the division first: 21 9 = 2.333 … \frac{21}{9} = 2.333\ldots 9 21 = 2.333 … Report a problem with this question
A man earns ₦3,500 per month, out of which he spends 15 % 15\% 15%
A ₦525 B ₦1,025 C ₦1,950 D ₦2,975
Worked solution (try it first) Education is
15 % 15\% 15% of ₦3,500:
0.15 × 3500 = 0.15 \times 3500 = 0.15 × 3500 = ₦525.
He spends
525 + 1950 = 525 + 1950 = 525 + 1950 = ₦2,475 in all.
So he has
3500 − 2475 = 3500 - 2475 = 3500 − 2475 = ₦1,025 left, option B.
Watch out
Take off both amounts. 3500 − 525 = 3500 - 525 = 3500 − 525 = Report a problem with this question
If 27 x + 2 ÷ 9 x + 1 = 3 2 x 27^{x + 2} \div 9^{x + 1} = 3^{2x} 2 7 x + 2 ÷ 9 x + 1 = 3 2 x x x x
Worked solution (try it first) Write both as powers of 3:
27 x + 2 = 3 3 x + 6 27^{x + 2} = 3^{3x + 6} 2 7 x + 2 = 3 3 x + 6 and
9 x + 1 = 3 2 x + 2 9^{x + 1} = 3^{2x + 2} 9 x + 1 = 3 2 x + 2 .
Dividing, subtract the indices:
3 ( 3 x + 6 ) − ( 2 x + 2 ) = 3 x + 4 3^{(3x + 6) - (2x + 2)} = 3^{x + 4} 3 ( 3 x + 6 ) − ( 2 x + 2 ) = 3 x + 4 .
So
x + 4 = 2 x x + 4 = 2x x + 4 = 2 x , which gives
x = 4 x = 4 x = 4 , option B.
Watch out
Subtract the whole second index, 2 x + 2 2x + 2 2 x + 2 3 x + 6 − 2 x − 2 = x + 4 3x + 6 - 2x - 2 = x + 4 3 x + 6 − 2 x − 2 = x + 4 + 2 +2 + 2 − 2 -2 − 2 x + 8 = 2 x x + 8 = 2x x + 8 = 2 x x = 8 x = 8 x = 8 Report a problem with this question
If log 3 x 2 = − 8 \log_3 x^2 = -8 log 3 x 2 = − 8 x x x
A 1 3 \frac13 3 1 B 1 9 \frac19 9 1 C 1 27 \frac{1}{27} 27 1 D 1 81 \frac{1}{81} 81 1
Worked solution (try it first) Change to index form:
x 2 = 3 − 8 x^2 = 3^{-8} x 2 = 3 − 8 .
Take the square root by halving the power:
x = 3 − 4 x = 3^{-4} x = 3 − 4 (taking the positive root).
3 − 4 = 1 3 4 = 1 81 3^{-4} = \frac{1}{3^4} = \frac{1}{81} 3 − 4 = 3 4 1 = 81 1 , option D.
Watch out
Take the square root once, which halves the power from − 8 -8 − 8 − 4 -4 − 4 3 − 2 = 1 9 3^{-2} = \frac19 3 − 2 = 9 1 Report a problem with this question
Simplify ( 6 + 2 ) 2 − ( 6 − 2 ) 2 (\sqrt6 + 2)^2 - (\sqrt6 - 2)^2 ( 6 + 2 ) 2 − ( 6 − 2 ) 2
A 2 6 2\sqrt6 2 6 B 4 6 4\sqrt6 4 6 C 8 6 8\sqrt6 8 6 D 16 6 16\sqrt6 16 6
Worked solution (try it first) Expand the first square:
( 6 + 2 ) 2 = 6 + 4 6 + 4 (\sqrt6 + 2)^2 = 6 + 4\sqrt6 + 4 ( 6 + 2 ) 2 = 6 + 4 6 + 4 , which is
10 + 4 6 10 + 4\sqrt6 10 + 4 6 .
Expand the second:
( 6 − 2 ) 2 = 6 − 4 6 + 4 (\sqrt6 - 2)^2 = 6 - 4\sqrt6 + 4 ( 6 − 2 ) 2 = 6 − 4 6 + 4 , which is
10 − 4 6 10 - 4\sqrt6 10 − 4 6 .
Subtract:
( 10 + 4 6 ) − ( 10 − 4 6 ) = 8 6 (10 + 4\sqrt6) - (10 - 4\sqrt6) = 8\sqrt6 ( 10 + 4 6 ) − ( 10 − 4 6 ) = 8 6 , option C.
Watch out
Subtracting − 4 6 -4\sqrt6 − 4 6 4 6 + 4 6 = 8 6 4\sqrt6 + 4\sqrt6 = 8\sqrt6 4 6 + 4 6 = 8 6 4 6 4\sqrt6 4 6 Report a problem with this question
If P P P Q Q Q P ∩ Q P \cap Q P ∩ Q
A { 3 } \{3\} { 3 } B { 2 , 3 } \{2, 3\} { 2 , 3 } C { 2 , 3 , 5 } \{2, 3, 5\} { 2 , 3 , 5 } D { 1 , 2 } \{1, 2\} { 1 , 2 }
Worked solution (try it first) 30 = 2 × 3 × 5 30 = 2 \times 3 \times 5 30 = 2 × 3 × 5 , so
P = { 2 , 3 , 5 } P = \{2, 3, 5\} P = { 2 , 3 , 5 } .
The factors of 18 less than 10 are
Q = { 1 , 2 , 3 , 6 , 9 } Q = \{1, 2, 3, 6, 9\} Q = { 1 , 2 , 3 , 6 , 9 } .
Both sets contain 2 and 3, so
P ∩ Q = { 2 , 3 } P \cap Q = \{2, 3\} P ∩ Q = { 2 , 3 } , option B.
Watch out
{ 2 , 3 , 5 } \{2, 3, 5\} { 2 , 3 , 5 } P P P Report a problem with this question
In a class of 46 students, 22 play football and 26 play volleyball. If 3 students play both games, how many play neither?
Worked solution (try it first) At least one game:
22 + 26 − 3 = 45 22 + 26 - 3 = 45 22 + 26 − 3 = 45 .
Neither game:
46 − 45 = 1 46 - 45 = 1 46 − 45 = 1 , option A.
Watch out
Subtract the 3 who play both once, from 22 + 26 = 48 22 + 26 = 48 22 + 26 = 48 Report a problem with this question
Make n n n w = v ( 2 + c n ) 1 − c n w = \dfrac{v(2 + cn)}{1 - cn} w = 1 − c n v ( 2 + c n )
A 1 c ( w − 2 v v + w ) \frac1c\left(\dfrac{w - 2v}{v + w}\right) c 1 ( v + w w − 2 v ) B 1 c ( w − 2 v v − w ) \frac1c\left(\dfrac{w - 2v}{v - w}\right) c 1 ( v − w w − 2 v ) C 1 c ( w + 2 v v − w ) \frac1c\left(\dfrac{w + 2v}{v - w}\right) c 1 ( v − w w + 2 v ) D 1 c ( w + 2 v v + w ) \frac1c\left(\dfrac{w + 2v}{v + w}\right) c 1 ( v + w w + 2 v )
Worked solution (try it first) Multiply both sides by
1 − c n 1 - cn 1 − c n and expand:
w − w c n = 2 v + v c n w - wcn = 2v + vcn w − w c n = 2 v + v c n .
Collect the
n n n terms on the right and the rest on the left:
w − 2 v = v c n + w c n w - 2v = vcn + wcn w − 2 v = v c n + w c n , so
w − 2 v = c n ( v + w ) w - 2v = cn(v + w) w − 2 v = c n ( v + w ) .
Divide by
c ( v + w ) c(v + w) c ( v + w ) :
n = 1 c ( w − 2 v v + w ) n = \frac1c\left(\dfrac{w - 2v}{v + w}\right) n = c 1 ( v + w w − 2 v ) , option A.
Watch out
− w c n -wcn − w c n + w c n +wcn + w c n v + w v + w v + w v − w v - w v − w Report a problem with this question
Find the remainder when 2 x 3 − 11 x 2 + 8 x − 1 2x^3 - 11x^2 + 8x - 1 2 x 3 − 11 x 2 + 8 x − 1 x + 3 x + 3 x + 3
A − 871 -871 − 871 B − 781 -781 − 781 C − 187 -187 − 187 D − 178 -178 − 178
Worked solution (try it first) By the remainder theorem, the remainder on dividing by
x + 3 x + 3 x + 3 is
f ( − 3 ) f(-3) f ( − 3 ) .
f ( − 3 ) = 2 ( − 27 ) − 11 ( 9 ) + 8 ( − 3 ) − 1 f(-3) = 2(-27) - 11(9) + 8(-3) - 1 f ( − 3 ) = 2 ( − 27 ) − 11 ( 9 ) + 8 ( − 3 ) − 1 , which is
− 54 − 99 − 24 − 1 -54 - 99 - 24 - 1 − 54 − 99 − 24 − 1 .
So the remainder is
− 178 -178 − 178 , option D.
Watch out
Put in x = − 3 x = -3 x = − 3 ( − 3 ) 3 = − 27 (-3)^3 = -27 ( − 3 ) 3 = − 27 ( − 3 ) 2 = + 9 (-3)^2 = +9 ( − 3 ) 2 = + 9 x = 3 x = 3 x = 3 − 22 -22 − 22 Report a problem with this question
Solve for x x x y y y x 2 − y 2 = 4 x^2 - y^2 = 4 x 2 − y 2 = 4 x + y = 2 x + y = 2 x + y = 2
A x = 0 , y = − 2 x = 0, y = -2 x = 0 , y = − 2 B x = 0 , y = 2 x = 0, y = 2 x = 0 , y = 2 C x = 2 , y = 0 x = 2, y = 0 x = 2 , y = 0 D x = − 2 , y = 0 x = -2, y = 0 x = − 2 , y = 0
Worked solution (try it first) Factorise the difference of two squares:
x 2 − y 2 = ( x + y ) ( x − y ) = 4 x^2 - y^2 = (x + y)(x - y) = 4 x 2 − y 2 = ( x + y ) ( x − y ) = 4 .
Since
x + y = 2 x + y = 2 x + y = 2 , this gives
2 ( x − y ) = 4 2(x - y) = 4 2 ( x − y ) = 4 , so
x − y = 2 x - y = 2 x − y = 2 .
Add to
x + y = 2 x + y = 2 x + y = 2 :
2 x = 4 2x = 4 2 x = 4 , so
x = 2 x = 2 x = 2 and
y = 0 y = 0 y = 0 , option C.
Watch out
Option D, x = − 2 x = -2 x = − 2 y = 0 y = 0 y = 0 x 2 − y 2 = 4 x^2 - y^2 = 4 x 2 − y 2 = 4 x + y = − 2 x + y = -2 x + y = − 2 Report a problem with this question
If y y y n \sqrt n n y = 4 y = 4 y = 4 n = 4 n = 4 n = 4 y y y n = 1 7 9 n = 1\frac79 n = 1 9 7
A 17 \sqrt{17} 17 B 4 3 \frac43 3 4 C 8 3 \frac83 3 8 D 2 3 \frac23 3 2
Worked solution (try it first) Put in
y = 4 y = 4 y = 4 ,
n = 4 n = 4 n = 4 :
4 = 2 k 4 = 2k 4 = 2 k , so
k = 2 k = 2 k = 2 .
1 7 9 = 16 9 1\frac79 = \frac{16}{9} 1 9 7 = 9 16 , so
n = 4 3 \sqrt n = \frac43 n = 3 4 .
So
y = 2 × 4 3 = 8 3 y = 2 \times \frac43 = \frac83 y = 2 × 3 4 = 3 8 , option C.
Watch out
4 3 \frac43 3 4 n \sqrt n n k = 2 k = 2 k = 2 y y y Report a problem with this question
U U U V V V U = 81 U = 81 U = 81 V = 2 V = 2 V = 2 U U U V = 3 V = 3 V = 3
Worked solution (try it first) U = k V 3 U = \dfrac{k}{V^3} U = V 3 k , so
k = U V 3 k = UV^3 k = U V 3 , which is
81 × 8 = 648 81 \times 8 = 648 81 × 8 = 648 .
When
V = 3 V = 3 V = 3 :
U = 648 27 = 24 U = \dfrac{648}{27} = 24 U = 27 648 = 24 , option A.
Watch out
Use the cube, not the square: with V 2 V^2 V 2 k = 324 k = 324 k = 324 U = 324 9 = 36 U = \frac{324}{9} = 36 U = 9 324 = 36 Report a problem with this question
The values of y y y 1 5 y + 1 5 < 1 2 y + 2 5 \frac15y + \frac15 < \frac12y + \frac25 5 1 y + 5 1 < 2 1 y + 5 2
A y > 2 3 y > \frac23 y > 3 2 B y < 2 3 y < \frac23 y < 3 2 C y > − 2 3 y > -\frac23 y > − 3 2 D y < − 2 3 y < -\frac23 y < − 3 2
Worked solution (try it first) Multiply every term by 10, the LCM of 5 and 2:
2 y + 2 < 5 y + 4 2y + 2 < 5y + 4 2 y + 2 < 5 y + 4 .
Subtract
2 y 2y 2 y and 4 from both sides:
− 2 < 3 y -2 < 3y − 2 < 3 y .
Divide by 3:
y > − 2 3 y > -\frac23 y > − 3 2 , option C.
Watch out
Watch the sign when you move the 4: 2 − 4 = − 2 2 - 4 = -2 2 − 4 = − 2 − 2 < 3 y -2 < 3y − 2 < 3 y + 2 +2 + 2 y > 2 3 y > \frac23 y > 3 2 Report a problem with this question
Find the range of values of m m m ( m − 3 ) ( m − 4 ) < 0 (m - 3)(m - 4) < 0 ( m − 3 ) ( m − 4 ) < 0
A 2 < m < 5 2 < m < 5 2 < m < 5 B − 3 < m < 4 -3 < m < 4 − 3 < m < 4 C 3 < m < 4 3 < m < 4 3 < m < 4 D − 4 < m < 3 -4 < m < 3 − 4 < m < 3
Worked solution (try it first) The roots are
m = 3 m = 3 m = 3 and
m = 4 m = 4 m = 4 , where each bracket is zero.
The product is negative only between the roots, where one bracket is positive and the other negative.
So
3 < m < 4 3 < m < 4 3 < m < 4 , option C.
Watch out
m − 3 = 0 m - 3 = 0 m − 3 = 0 m = + 3 m = +3 m = + 3 − 3 -3 − 3 m = 0 m = 0 m = 0 ( − 3 ) ( − 4 ) = 12 (-3)(-4) = 12 ( − 3 ) ( − 4 ) = 12 Report a problem with this question
The shaded region is represented by the inequality
A y ≤ 4 x + 2 y \le 4x + 2 y ≤ 4 x + 2 B y ≥ 4 x + 2 y \ge 4x + 2 y ≥ 4 x + 2 C y ≤ − 4 x + 4 y \le -4x + 4 y ≤ − 4 x + 4 D y ≤ 4 x + 4 y \le 4x + 4 y ≤ 4 x + 4
Worked solution (try it first) The line through
( 0 , 4 ) (0, 4) ( 0 , 4 ) and
( 1 , 0 ) (1, 0) ( 1 , 0 ) has gradient
0 − 4 1 − 0 = − 4 \frac{0 - 4}{1 - 0} = -4 1 − 0 0 − 4 = − 4 and
y y y -intercept 4, so
y = − 4 x + 4 y = -4x + 4 y = − 4 x + 4 .
The origin is in the shaded region, and
0 ≤ − 4 ( 0 ) + 4 = 4 0 \le -4(0) + 4 = 4 0 ≤ − 4 ( 0 ) + 4 = 4 is true.
So the region is
y ≤ − 4 x + 4 y \le -4x + 4 y ≤ − 4 x + 4 , option C.
Watch out
The line falls from ( 0 , 4 ) (0, 4) ( 0 , 4 ) ( 1 , 0 ) (1, 0) ( 1 , 0 ) y = 4 x + 4 y = 4x + 4 y = 4 x + 4 ( − 1 , 0 ) (-1, 0) ( − 1 , 0 ) Report a problem with this question
The n n n n 2 − 6 n − 4 n^2 - 6n - 4 n 2 − 6 n − 4
A 24 B 23 C − 24 -24 − 24 D − 25 -25 − 25
Worked solution (try it first) Put
n = 3 n = 3 n = 3 :
T 3 = 9 − 18 − 4 = − 13 T_3 = 9 - 18 - 4 = -13 T 3 = 9 − 18 − 4 = − 13 .
Put
n = 4 n = 4 n = 4 :
T 4 = 16 − 24 − 4 = − 12 T_4 = 16 - 24 - 4 = -12 T 4 = 16 − 24 − 4 = − 12 .
So the sum is
− 13 + ( − 12 ) = − 25 -13 + (-12) = -25 − 13 + ( − 12 ) = − 25 , option D.
Watch out
Work each term in full: 9 − 18 − 4 = − 13 9 - 18 - 4 = -13 9 − 18 − 4 = − 13 − 12 -12 − 12 T 3 T_3 T 3 T 4 T_4 T 4 − 24 -24 − 24 Report a problem with this question
The sum to infinity of a geometric progression is − 1 10 -\frac{1}{10} − 10 1 − 1 8 -\frac18 − 8 1
A − 1 5 -\frac15 − 5 1 B − 1 4 -\frac14 − 4 1 C − 1 3 -\frac13 − 3 1 D − 1 2 -\frac12 − 2 1
Worked solution (try it first) Use
S ∞ = a 1 − r S_\infty = \dfrac{a}{1 - r} S ∞ = 1 − r a , so
1 − r = a S ∞ 1 - r = \dfrac{a}{S_\infty} 1 − r = S ∞ a .
Divide:
( − 1 8 ) ÷ ( − 1 10 ) = 10 8 \left(-\frac18\right) \div \left(-\frac{1}{10}\right) = \frac{10}{8} ( − 8 1 ) ÷ ( − 10 1 ) = 8 10 = 5 4 = \frac54 = 4 5 , so
1 − r = 5 4 1 - r = \frac54 1 − r = 4 5 .
So
r = 1 − 5 4 = − 1 4 r = 1 - \frac54 = -\frac14 r = 1 − 4 5 = − 4 1 , option B.
Watch out
It is a ÷ S ∞ a \div S_\infty a ÷ S ∞ 1 − r 1 - r 1 − r S ∞ ÷ a S_\infty \div a S ∞ ÷ a 1 − r = 4 5 1 - r = \frac45 1 − r = 5 4 r = 1 5 r = \frac15 r = 5 1 Report a problem with this question
The binary operation ∗ * ∗ p ∗ q = p q + p − q p * q = pq + p - q p ∗ q = pq + p − q 2 ∗ ( 3 ∗ 4 ) 2 * (3 * 4) 2 ∗ ( 3 ∗ 4 )
Worked solution (try it first) Work out the bracket first:
3 ∗ 4 = 12 + 3 − 4 = 11 3 * 4 = 12 + 3 - 4 = 11 3 ∗ 4 = 12 + 3 − 4 = 11 .
Then
2 ∗ 11 = 2 × 11 + 2 − 11 = 22 − 9 2 * 11 = 2 \times 11 + 2 - 11 = 22 - 9 2 ∗ 11 = 2 × 11 + 2 − 11 = 22 − 9 .
So
2 ∗ ( 3 ∗ 4 ) = 13 2 * (3 * 4) = 13 2 ∗ ( 3 ∗ 4 ) = 13 , option B.
Watch out
11 (option A) is only the value of the bracket. You still have to combine it with 2: 2 ∗ 11 = 13 2 * 11 = 13 2 ∗ 11 = 13 Report a problem with this question
A binary operation on the set of real numbers is defined by m ∗ n = m n 2 m * n = \frac{mn}{2} m ∗ n = 2 mn m , n ∈ R m, n \in \mathbb R m , n ∈ R − 5 -5 − 5
A − 4 5 -\frac45 − 5 4 B − 2 5 -\frac25 − 5 2 C 4 4 4 D 5 5 5
Worked solution (try it first) The inverse
x x x of
− 5 -5 − 5 combines with it to give the identity 2:
− 5 ∗ x = 2 -5 * x = 2 − 5 ∗ x = 2 .
Use the rule:
− 5 x 2 = 2 \frac{-5x}{2} = 2 2 − 5 x = 2 .
Multiply both sides by 2:
− 5 x = 4 -5x = 4 − 5 x = 4 .
Divide by
− 5 -5 − 5 :
x = − 4 5 x = -\frac45 x = − 5 4 , option A.
Watch out
Keep the division by 2 in the rule. Writing − 5 x = 2 -5x = 2 − 5 x = 2 x = − 2 5 x = -\frac25 x = − 5 2 Report a problem with this question
If ∣ 5 3 x 2 ∣ = ∣ 3 5 4 5 ∣ \begin{vmatrix} 5 & 3 \\ x & 2 \end{vmatrix} = \begin{vmatrix} 3 & 5 \\ 4 & 5 \end{vmatrix} 5 x 3 2 = 3 4 5 5 x x x
Worked solution (try it first) A
2 × 2 2 \times 2 2 × 2 determinant is
a d − b c ad - bc a d − b c .
The left side is
5 × 2 − 3 x = 10 − 3 x 5 \times 2 - 3x = 10 - 3x 5 × 2 − 3 x = 10 − 3 x .
The right side is
3 × 5 − 5 × 4 = 15 − 20 = − 5 3 \times 5 - 5 \times 4 = 15 - 20 = -5 3 × 5 − 5 × 4 = 15 − 20 = − 5 .
So
10 − 3 x = − 5 10 - 3x = -5 10 − 3 x = − 5 .
Take 10 from both sides:
− 3 x = − 15 -3x = -15 − 3 x = − 15 .
Divide by
− 3 -3 − 3 :
x = 5 x = 5 x = 5 , option C.
Watch out
Take the products in the right order, leading diagonal first: 15 − 20 = − 5 15 - 20 = -5 15 − 20 = − 5 20 − 15 = 5 20 - 15 = 5 20 − 15 = 5 x = 5 3 x = \frac53 x = 3 5 Report a problem with this question
Given that I 3 I_3 I 3 ∣ I 3 ∣ |I_3| ∣ I 3 ∣
Worked solution (try it first) I 3 I_3 I 3 has 1s on the leading diagonal and 0s everywhere else.
For a matrix with zeros off the diagonal, the determinant is the product of the diagonal entries:
1 × 1 × 1 = 1 1 \times 1 \times 1 = 1 1 × 1 × 1 = 1 , option C.
Watch out
The zeros off the diagonal don't make the determinant 0 (option B). It is 1 × 1 × 1 = 1 1 \times 1 \times 1 = 1 1 × 1 × 1 = 1 Report a problem with this question
In the diagram, Q R ∥ T U QR \parallel TU QR ∥ T U ∠ P Q R = 80 ∘ \angle PQR = 80^\circ ∠ P QR = 8 0 ∘ ∠ P S U = 95 ∘ \angle PSU = 95^\circ ∠ P S U = 9 5 ∘ ∠ S U T \angle SUT ∠ S U T
A 15 ∘ 15^\circ 1 5 ∘ B 25 ∘ 25^\circ 2 5 ∘ C 30 ∘ 30^\circ 3 0 ∘ D 80 ∘ 80^\circ 8 0 ∘
Worked solution (try it first) Draw a line through
S S S parallel to
Q R QR QR and
T U TU T U .
By corresponding angles,
S P SP S P makes
80 ∘ 80^\circ 8 0 ∘ with it, the same as
∠ P Q R \angle PQR ∠ P QR .
∠ P S U = 95 ∘ \angle PSU = 95^\circ ∠ P S U = 9 5 ∘ , so
S U SU S U lies
95 ∘ − 80 ∘ = 15 ∘ 95^\circ - 80^\circ = 15^\circ 9 5 ∘ − 8 0 ∘ = 1 5 ∘ beyond the parallel line.
By alternate angles with
T U TU T U ,
∠ S U T = 15 ∘ \angle SUT = 15^\circ ∠ S U T = 1 5 ∘ , option A.
Watch out
Don't copy 80 ∘ 80^\circ 8 0 ∘ S S S 80 ∘ 80^\circ 8 0 ∘ 95 ∘ 95^\circ 9 5 ∘ x x x 15 ∘ 15^\circ 1 5 ∘ Report a problem with this question
The angles of a pentagon are x x x 2 x 2x 2 x 3 x 3x 3 x 4 x 4x 4 x 5 x 5x 5 x x x x
A 24 ∘ 24^\circ 2 4 ∘ B 30 ∘ 30^\circ 3 0 ∘ C 33 ∘ 33^\circ 3 3 ∘ D 36 ∘ 36^\circ 3 6 ∘
Worked solution (try it first) The interior angles of a pentagon add up to
( 5 − 2 ) × 180 ∘ = 540 ∘ (5 - 2) \times 180^\circ = 540^\circ ( 5 − 2 ) × 18 0 ∘ = 54 0 ∘ .
So
x + 2 x + 3 x + 4 x + 5 x = 540 x + 2x + 3x + 4x + 5x = 540 x + 2 x + 3 x + 4 x + 5 x = 540 , which is
15 x = 540 15x = 540 15 x = 540 .
Divide by 15:
x = 36 ∘ x = 36^\circ x = 3 6 ∘ , option D.
Watch out
Use 540 ∘ 540^\circ 54 0 ∘ 360 ∘ 360^\circ 36 0 ∘ 360 ∘ 360^\circ 36 0 ∘ x = 24 ∘ x = 24^\circ x = 2 4 ∘ Report a problem with this question
In the diagram, P Q R PQR P QR O O O P R PR P R ∠ Q P R = x ∘ \angle QPR = x^\circ ∠ QP R = x ∘ ∠ Q R P \angle QRP ∠ QR P
A x ∘ x^\circ x ∘ B ( 90 − x ) ∘ (90 - x)^\circ ( 90 − x ) ∘ C ( 90 + x ) ∘ (90 + x)^\circ ( 90 + x ) ∘ D ( 180 − x ) ∘ (180 - x)^\circ ( 180 − x ) ∘
Worked solution (try it first) P R PR P R is a diameter, so
∠ P Q R = 90 ∘ \angle PQR = 90^\circ ∠ P QR = 9 0 ∘ (angle in a semicircle).
The angles of triangle
P Q R PQR P QR add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ Q R P = 180 ∘ − 90 ∘ − x ∘ \angle QRP = 180^\circ - 90^\circ - x^\circ ∠ QR P = 18 0 ∘ − 9 0 ∘ − x ∘ .
So
∠ Q R P = ( 90 − x ) ∘ \angle QRP = (90 - x)^\circ ∠ QR P = ( 90 − x ) ∘ , option B.
Watch out
Include the right angle at Q Q Q ( 180 − x ) ∘ (180 - x)^\circ ( 180 − x ) ∘ Q Q Q Report a problem with this question
Find the area of the trapezium P Q R S PQRS P QR S P Q = 7 cm PQ = 7\text{ cm} P Q = 7 cm S R = 13 cm SR = 13\text{ cm} S R = 13 cm P S = 6 cm PS = 6\text{ cm} P S = 6 cm
A 91 cm 2 91\text{ cm}^2 91 cm 2 B 78 cm 2 78\text{ cm}^2 78 cm 2 C 60 cm 2 60\text{ cm}^2 60 cm 2 D 19 cm 2 19\text{ cm}^2 19 cm 2
Worked solution (try it first) Area of a trapezium
= 1 2 ( a + b ) h = \frac12(a + b)h = 2 1 ( a + b ) h , with parallel sides 7 cm and 13 cm and height 6 cm.
So the area is
1 2 × 20 × 6 = 60 cm 2 \frac12 \times 20 \times 6 = 60\text{ cm}^2 2 1 × 20 × 6 = 60 cm 2 , option C.
Watch out
Use the average of the two parallel sides, not the longer one. 13 × 6 = 78 cm 2 13 \times 6 = 78\text{ cm}^2 13 × 6 = 78 cm 2 Report a problem with this question
A circular arc subtends an angle of 150 ∘ 150^\circ 15 0 ∘ 12 cm 12\text{ cm} 12 cm
A 30 π cm 2 30\pi\text{ cm}^2 30 π cm 2 B 60 π cm 2 60\pi\text{ cm}^2 60 π cm 2 C 120 π cm 2 120\pi\text{ cm}^2 120 π cm 2 D 150 π cm 2 150\pi\text{ cm}^2 150 π cm 2
Worked solution (try it first) Area of a sector
= θ 360 × π r 2 = \frac{\theta}{360} \times \pi r^2 = 360 θ × π r 2 , with
150 360 = 5 12 \frac{150}{360} = \frac{5}{12} 360 150 = 12 5 and
r 2 = 144 r^2 = 144 r 2 = 144 .
So the area is
5 12 × 144 π = 60 π cm 2 \frac{5}{12} \times 144\pi = 60\pi\text{ cm}^2 12 5 × 144 π = 60 π cm 2 , option B.
Watch out
Square the radius: r 2 = 144 r^2 = 144 r 2 = 144 π r \pi r π r π r 2 \pi r^2 π r 2 5 π 5\pi 5 π Report a problem with this question
Calculate the volume of a cuboid of length 0.76 cm 0.76\text{ cm} 0.76 cm 2.6 cm 2.6\text{ cm} 2.6 cm 0.82 cm 0.82\text{ cm} 0.82 cm
A 3.92 cm 3 3.92\text{ cm}^3 3.92 cm 3 B 2.13 cm 3 2.13\text{ cm}^3 2.13 cm 3 C 1.97 cm 3 1.97\text{ cm}^3 1.97 cm 3 D 1.62 cm 3 1.62\text{ cm}^3 1.62 cm 3
Worked solution (try it first) Volume of a cuboid = length × breadth × height.
0.76 × 2.6 = 1.976 0.76 \times 2.6 = 1.976 0.76 × 2.6 = 1.976 , then
1.976 × 0.82 = 1.62032 1.976 \times 0.82 = 1.62032 1.976 × 0.82 = 1.62032 .
To 2 decimal places that is
1.62 cm 3 1.62\text{ cm}^3 1.62 cm 3 , option D.
Watch out
Multiply all three lengths. Stopping after two gives 1.976 ≈ 1.97 1.976 \approx 1.97 1.976 ≈ 1.97 Report a problem with this question
The locus of a point equidistant from the intersection of the lines 3 x − 7 y + 7 = 0 3x - 7y + 7 = 0 3 x − 7 y + 7 = 0 4 x − 6 y + 1 = 0 4x - 6y + 1 = 0 4 x − 6 y + 1 = 0
A line parallel to 7 x + 13 y + 8 = 0 7x + 13y + 8 = 0 7 x + 13 y + 8 = 0 B circle C semicircle D bisector of the line 7 x + 13 y + 8 = 0 7x + 13y + 8 = 0 7 x + 13 y + 8 = 0
Worked solution (try it first) Two lines that are not parallel meet at one point.
Solving
3 x − 7 y = − 7 3x - 7y = -7 3 x − 7 y = − 7 and
4 x − 6 y = − 1 4x - 6y = -1 4 x − 6 y = − 1 gives
( 3.5 , 2.5 ) (3.5, 2.5) ( 3.5 , 2.5 ) .
Points all at the same distance from one fixed point form a circle centred on that point.
So the locus is a circle, option B.
Watch out
The intersection of two lines is a single point, and the locus of points a fixed distance from a point is a circle. The line 7 x + 13 y + 8 = 0 7x + 13y + 8 = 0 7 x + 13 y + 8 = 0 Report a problem with this question
The gradient of the straight line joining the points P ( 5 , − 7 ) P(5, -7) P ( 5 , − 7 ) Q ( − 2 , − 3 ) Q(-2, -3) Q ( − 2 , − 3 )
A 1 2 \frac12 2 1 B 2 5 \frac25 5 2 C − 4 7 -\frac47 − 7 4 D − 2 3 -\frac23 − 3 2
Worked solution (try it first) Gradient is the change in
y y y over the change in
x x x , in the same order:
− 3 − ( − 7 ) − 2 − 5 \dfrac{-3 - (-7)}{-2 - 5} − 2 − 5 − 3 − ( − 7 ) .
The top is
4 4 4 and the bottom is
− 7 -7 − 7 .
So the gradient is
− 4 7 -\frac47 − 7 4 , option C.
Watch out
Subtract in the same order top and bottom. Taking Q − P Q - P Q − P P − Q P - Q P − Q + 4 7 +\frac47 + 7 4 Report a problem with this question
The distance between the point ( 4 , 3 ) (4, 3) ( 4 , 3 ) y = 2 x + 4 y = 2x + 4 y = 2 x + 4 y = 7 − x y = 7 - x y = 7 − x
A 13 \sqrt{13} 13 B 3 2 3\sqrt2 3 2 C 26 \sqrt{26} 26 D 10 5 10\sqrt5 10 5
Worked solution (try it first) At the intersection both
y y y -values are equal:
2 x + 4 = 7 − x 2x + 4 = 7 - x 2 x + 4 = 7 − x , so
3 x = 3 3x = 3 3 x = 3 and
x = 1 x = 1 x = 1 .
Then
y = 7 − 1 = 6 y = 7 - 1 = 6 y = 7 − 1 = 6 , so the lines meet at
( 1 , 6 ) (1, 6) ( 1 , 6 ) .
From
( 1 , 6 ) (1, 6) ( 1 , 6 ) to
( 4 , 3 ) (4, 3) ( 4 , 3 ) the changes are 3 and
− 3 -3 − 3 , so the distance is
9 + 9 = 18 = 3 2 \sqrt{9 + 9} = \sqrt{18} = 3\sqrt2 9 + 9 = 18 = 3 2 , option B.
Watch out
Put x = 1 x = 1 x = 1 y = 6 y = 6 y = 6 y y y 18 \sqrt{18} 18 9 × 2 = 3 2 \sqrt9 \times \sqrt2 = 3\sqrt2 9 × 2 = 3 2 Report a problem with this question
Find the equation of the line through the points ( − 2 , 1 ) (-2, 1) ( − 2 , 1 ) ( − 1 2 , 4 ) \left(-\frac12, 4\right) ( − 2 1 , 4 )
A y = 2 x − 3 y = 2x - 3 y = 2 x − 3 B y = 2 x + 5 y = 2x + 5 y = 2 x + 5 C y = 3 x − 2 y = 3x - 2 y = 3 x − 2 D y = 2 x + 1 y = 2x + 1 y = 2 x + 1
Worked solution (try it first) Gradient: the change in
y y y is
4 − 1 = 3 4 - 1 = 3 4 − 1 = 3 and the change in
x x x is
− 1 2 − ( − 2 ) = 3 2 -\frac12 - (-2) = \frac32 − 2 1 − ( − 2 ) = 2 3 , so
m = 3 ÷ 3 2 = 2 m = 3 \div \frac32 = 2 m = 3 ÷ 2 3 = 2 .
Through
( − 2 , 1 ) (-2, 1) ( − 2 , 1 ) :
y − 1 = 2 ( x − ( − 2 ) ) y - 1 = 2(x - (-2)) y − 1 = 2 ( x − ( − 2 )) , so
y − 1 = 2 x + 4 y - 1 = 2x + 4 y − 1 = 2 x + 4 .
So
y = 2 x + 5 y = 2x + 5 y = 2 x + 5 , option B.
Watch out
x − ( − 2 ) x - (-2) x − ( − 2 ) x + 2 x + 2 x + 2 y − 1 = 2 ( x − 2 ) y - 1 = 2(x - 2) y − 1 = 2 ( x − 2 ) y = 2 x − 3 y = 2x - 3 y = 2 x − 3 Report a problem with this question
If angle θ \theta θ 135 ∘ 135^\circ 13 5 ∘ cos θ \cos\theta cos θ
A 1 2 \frac12 2 1 B 2 2 \frac{\sqrt2}{2} 2 2 C − 2 2 -\frac{\sqrt2}{2} − 2 2 D − 1 2 -\frac12 − 2 1
Worked solution (try it first) 135 ∘ = 180 ∘ − 45 ∘ 135^\circ = 180^\circ - 45^\circ 13 5 ∘ = 18 0 ∘ − 4 5 ∘ , so it is in the second quadrant with reference angle
45 ∘ 45^\circ 4 5 ∘ .
Cosine is negative in the second quadrant, so
cos 135 ∘ = − cos 45 ∘ \cos135^\circ = -\cos45^\circ cos 13 5 ∘ = − cos 4 5 ∘ = − 2 2 = -\frac{\sqrt2}{2} = − 2 2 , option C.
Watch out
Attach the quadrant's sign: cos 45 ∘ = 2 2 \cos45^\circ = \frac{\sqrt2}{2} cos 4 5 ∘ = 2 2 90 ∘ 90^\circ 9 0 ∘ 180 ∘ 180^\circ 18 0 ∘ Report a problem with this question
A man stands on a tree 150 cm 150\text{ cm} 150 cm 74 ∘ 74^\circ 7 4 ∘
A 52 cm 52\text{ cm} 52 cm B 43 cm 43\text{ cm} 43 cm C 40 cm 40\text{ cm} 40 cm D 15 cm 15\text{ cm} 15 cm
Worked solution (try it first) The angle of elevation of the man from the boat is also
74 ∘ 74^\circ 7 4 ∘ (alternate angles), so
tan 74 ∘ = 150 d \tan74^\circ = \frac{150}{d} tan 7 4 ∘ = d 150 .
Rearrange:
d = 150 tan 74 ∘ d = \frac{150}{\tan74^\circ} d = t a n 7 4 ∘ 150 = 150 3.487 = \frac{150}{3.487} = 3.487 150 .
So
d ≈ 43 d \approx 43 d ≈ 43 cm, option B.
Watch out
Divide the height by tan 74 ∘ \tan74^\circ tan 7 4 ∘ 150 tan 74 ∘ ≈ 523 150\tan74^\circ \approx 523 150 tan 7 4 ∘ ≈ 523 Report a problem with this question
If y = x 2 − 1 x y = x^2 - \dfrac1x y = x 2 − x 1 d y d x \dfrac{dy}{dx} d x d y
A 2 x − 1 x 2 2x - \frac{1}{x^2} 2 x − x 2 1 B 2 x + x 2 2x + x^2 2 x + x 2 C 2 x − x 2 2x - x^2 2 x − x 2 D 2 x + 1 x 2 2x + \frac{1}{x^2} 2 x + x 2 1
Worked solution (try it first) Write
y = x 2 − x − 1 y = x^2 - x^{-1} y = x 2 − x − 1 .
Differentiate:
x 2 x^2 x 2 gives
2 x 2x 2 x , and
− x − 1 -x^{-1} − x − 1 gives
− ( − 1 ) x − 2 = x − 2 -(-1)x^{-2} = x^{-2} − ( − 1 ) x − 2 = x − 2 .
So
d y d x = 2 x + 1 x 2 \frac{dy}{dx} = 2x + \frac{1}{x^2} d x d y = 2 x + x 2 1 , option D.
Watch out
Two minus signs make a plus: − x − 1 -x^{-1} − x − 1 + x − 2 +x^{-2} + x − 2 Report a problem with this question
Find d y d x \dfrac{dy}{dx} d x d y y = cos x y = \cos x y = cos x
A sin x \sin x sin x B − sin x -\sin x − sin x C tan x \tan x tan x D − tan x -\tan x − tan x
Worked solution (try it first) The derivative of
cos x \cos x cos x is
− sin x -\sin x − sin x , option B.
Check: the graph of
cos x \cos x cos x slopes downwards just after
x = 0 x = 0 x = 0 , where
sin x \sin x sin x is positive, so the derivative must be negative there.
Watch out
cos x \cos x cos x − sin x -\sin x − sin x sin x \sin x sin x + cos x +\cos x + cos x sin x \sin x sin x Report a problem with this question
Evaluate ∫ 1 2 ( x 2 − 4 x ) d x \displaystyle\int_1^2 (x^2 - 4x)\,dx ∫ 1 2 ( x 2 − 4 x ) d x
A 11 3 \frac{11}{3} 3 11 B 3 11 \frac{3}{11} 11 3 C − 3 11 -\frac{3}{11} − 11 3 D − 11 3 -\frac{11}{3} − 3 11
Worked solution (try it first) Integrate:
[ x 3 3 − 2 x 2 ] 1 2 \left[\frac{x^3}{3} - 2x^2\right]_1^2 [ 3 x 3 − 2 x 2 ] 1 2 .
At
x = 2 x = 2 x = 2 :
8 3 − 8 = − 16 3 \frac83 - 8 = -\frac{16}{3} 3 8 − 8 = − 3 16 .
At
x = 1 x = 1 x = 1 :
1 3 − 2 = − 5 3 \frac13 - 2 = -\frac53 3 1 − 2 = − 3 5 .
Subtract:
− 16 3 + 5 3 = − 11 3 -\frac{16}{3} + \frac53 = -\frac{11}{3} − 3 16 + 3 5 = − 3 11 , option D.
Watch out
Keep the sign. The curve is below the x x x 11 3 \frac{11}{3} 3 11 Report a problem with this question
Evaluate ∫ 0 π / 4 sec 2 θ d θ \displaystyle\int_0^{\pi/4} \sec^2\theta\,d\theta ∫ 0 π /4 sec 2 θ d θ
Worked solution (try it first) sec 2 θ \sec^2\theta sec 2 θ is the derivative of
tan θ \tan\theta tan θ , so its integral is
tan θ \tan\theta tan θ .
[ tan θ ] 0 π / 4 = tan π 4 − tan 0 \left[\tan\theta\right]_0^{\pi/4} = \tan\frac\pi4 - \tan0 [ tan θ ] 0 π /4 = tan 4 π − tan 0 Watch out
The integral of sec 2 θ \sec^2\theta sec 2 θ tan θ \tan\theta tan θ sec θ \sec\theta sec θ 2 − 1 \sqrt2 - 1 2 − 1 Report a problem with this question
The grades of 36 students in a class test are shown in the pie chart (Pass 120 ∘ 120^\circ 12 0 ∘ 80 ∘ 80^\circ 8 0 ∘ 90 ∘ 90^\circ 9 0 ∘
Worked solution (try it first) The angles add up to
360 ∘ 360^\circ 36 0 ∘ , so Excellent is
360 ∘ − ( 120 ∘ + 80 ∘ + 90 ∘ ) = 70 ∘ 360^\circ - (120^\circ + 80^\circ + 90^\circ) = 70^\circ 36 0 ∘ − ( 12 0 ∘ + 8 0 ∘ + 9 0 ∘ ) = 7 0 ∘ .
Its share of the 36 students:
70 360 × 36 = 7 \frac{70}{360} \times 36 = 7 360 70 × 36 = 7 , option D.
Watch out
Find the Excellent angle first. 90 360 × 36 = 9 \frac{90}{360} \times 36 = 9 360 90 × 36 = 9 Report a problem with this question
The bar chart shows the distribution of marks in a class test. If the pass mark is 5, what percentage of students failed the test?
A 10 % 10\% 10% B 20 % 20\% 20% C 50 % 50\% 50% D 60 % 60\% 60%
Worked solution (try it first) Read the bars for marks 0 to 10 and add:
2 + 3 + 4 + 6 + 5 + 4 + 6 + 5 + 1 + 3 + 1 = 40 2 + 3 + 4 + 6 + 5 + 4 + 6 + 5 + 1 + 3 + 1 = 40 2 + 3 + 4 + 6 + 5 + 4 + 6 + 5 + 1 + 3 + 1 = 40 students.
Failing means a mark below 5 (marks 0 to 4):
2 + 3 + 4 + 6 + 5 = 20 2 + 3 + 4 + 6 + 5 = 20 2 + 3 + 4 + 6 + 5 = 20 students.
As a percentage:
20 40 × 100 = 50 % \frac{20}{40} \times 100 = 50\% 40 20 × 100 = 50% , option C.
Watch out
A mark of 5 is a pass. Counting the 4 students who scored 5 as failures gives 24 40 = 60 % \frac{24}{40} = 60\% 40 24 = 60% Report a problem with this question
The mean of seven numbers is 96. If an eighth number is added, the mean becomes 112. Find the eighth number.
Worked solution (try it first) Total of the seven numbers:
7 × 96 = 672 7 \times 96 = 672 7 × 96 = 672 .
Total of all eight:
8 × 112 = 896 8 \times 112 = 896 8 × 112 = 896 .
The eighth number is the difference:
896 − 672 = 224 896 - 672 = 224 896 − 672 = 224 , option D.
Watch out
The new mean is over 8 numbers. Using 7 × 112 = 784 7 \times 112 = 784 7 × 112 = 784 784 − 672 = 112 784 - 672 = 112 784 − 672 = 112 Report a problem with this question
Find the median of 2, 3, 7, 3, 4, 5, 8, 9, 9, 4, 5, 3, 4, 2, 4 and 5.
Worked solution (try it first) Put the 16 numbers in order: 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 7, 8, 9, 9.
With an even count, the median is halfway between the 8th and 9th.
Both are 4.
So the median is 4, option D.
Watch out
Order the list first. The 8th and 9th numbers as written are both 9, which gives 9 (option A). Report a problem with this question
Find the range of 4, 9, 6, 3, 2, 8, 10 and 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
Check the whole list for the smallest number: it is 2, not 3. Using 3 gives 8 (option C), and 11 alone (option A) is just the largest number. Similar: NECO 2024 · Paper 1 · Q52
Report a problem with this question
Find the standard deviation of 2, 3, 8, 10 and 12.
Worked solution (try it first) The numbers add up to 35, so the mean is 7.
The deviations are
− 5 -5 − 5 ,
− 4 -4 − 4 , 1, 3, 5.
Their squares are 25, 16, 1, 9, 25, which add up to 76.
The variance is
76 5 = 15.2 \frac{76}{5} = 15.2 5 76 = 15.2 , so the standard deviation is
15.2 ≈ 3.9 \sqrt{15.2} \approx 3.9 15.2 ≈ 3.9 , option A.
Watch out
Divide by the number of values, 5. Dividing 76 by 4 gives 19 ≈ 4.4 \sqrt{19} \approx 4.4 19 ≈ 4.4 Report a problem with this question
Evaluate n + 1 C n − 2 {}^{n + 1}C_{n - 2} n + 1 C n − 2 n = 15 n = 15 n = 15
Worked solution (try it first) With
n = 15 n = 15 n = 15 ,
n + 1 C n − 2 = 16 C 13 ^{n + 1}C_{n - 2} = {}^{16}C_{13} n + 1 C n − 2 = 16 C 13 .
Choosing 13 is the same as leaving out 3:
16 C 13 = 16 C 3 ^{16}C_{13} = {}^{16}C_3 16 C 13 = 16 C 3 = 16 × 15 × 14 3 × 2 × 1 = \dfrac{16 \times 15 \times 14}{3 \times 2 \times 1} = 3 × 2 × 1 16 × 15 × 14 .
That is
3360 6 = 560 \frac{3360}{6} = 560 6 3360 = 560 , option D.
Watch out
Divide by 3 ! = 6 3! = 6 3 ! = 6 16 × 15 × 14 = 3360 16 \times 15 \times 14 = 3360 16 × 15 × 14 = 3360 16 P 3 ^{16}P_3 16 P 3 Report a problem with this question
In how many ways can the letters of the word TOTALITY be arranged?
Worked solution (try it first) TOTALITY has 8 letters, so start with
8 ! = 40 320 8! = 40\,320 8 ! = 40 320 .
T appears 3 times (T, O, T, A, L, I, T, Y) and every other letter once, so divide by
3 ! = 6 3! = 6 3 ! = 6 .
40 320 ÷ 6 = 6720 40\,320 \div 6 = 6720 40 320 ÷ 6 = 6720 , option A.
Watch out
Count the T's: there are three, so divide by 3 ! 3! 3 ! 2 ! 2! 2 ! Report a problem with this question
The probability that a student passes a physics test is 2 3 \frac23 3 2
A 4 9 \frac49 9 4 B 6 9 \frac69 9 6 C 4 27 \frac{4}{27} 27 4 D 2 27 \frac{2}{27} 27 2
Worked solution (try it first) One order, pass, pass, fail, has probability
2 3 × 2 3 × 1 3 = 4 27 \frac23 \times \frac23 \times \frac13 = \frac{4}{27} 3 2 × 3 2 × 3 1 = 27 4 .
The failed test can be the first, second or third, so there are 3 such orders.
So the probability is
3 × 4 27 = 4 9 3 \times \frac{4}{27} = \frac{4}{9} 3 × 27 4 = 9 4 , option A.
Watch out
Count every order of two passes and one fail. Using only one order gives 4 27 \frac{4}{27} 27 4 Report a problem with this question
The probabilities that a man and his wife live to 80 years are 2 3 \frac23 3 2 3 5 \frac35 5 3
A 2 15 \frac{2}{15} 15 2 B 3 15 \frac{3}{15} 15 3 C 7 15 \frac{7}{15} 15 7 D 13 15 \frac{13}{15} 15 13
Worked solution (try it first) "At least one" is the complement of "neither".
The chances of not living to 80 are
1 − 2 3 = 1 3 1 - \frac23 = \frac13 1 − 3 2 = 3 1 and
1 − 3 5 = 2 5 1 - \frac35 = \frac25 1 − 5 3 = 5 2 .
Neither lives to 80:
1 3 × 2 5 = 2 15 \frac13 \times \frac25 = \frac{2}{15} 3 1 × 5 2 = 15 2 .
So at least one does:
1 − 2 15 = 13 15 1 - \frac{2}{15} = \frac{13}{15} 1 − 15 2 = 15 13 , option D.
Watch out
2 15 \frac{2}{15} 15 2 Report a problem with this question