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Solution to UTME Mathematics for 2013, Solved by Tawose, Simeon Abiola.
UTME Mathematics 2013
1. Which Question paper type of mathematics is
indicated above is given to you
(A) Type D
(B) Type I
(C) Type B
(D) Type U
Answer: Type I. (Option B)
Please note I am using type D in solving these
questions.
10
3 3 3 3
2. Convert 27 to another number in base three
( ) A 1001 (B) 1010 (C) 1100 (D) 1000
Solution
3 27
3 9 0
3 3 0
3 1 0
0 1
10 3 27 = 1000 (Option D)
3. 3 girls share a number of apple in the ratio 5:3:2.
If the highest share is 40 apples, find the smallest
share.
(A) 36 (B) 24 (C) 16 (D) 38
Solution
Let the total number of apple shared be
The sum of the ratio is 5 3 2 10
The highest got
5 40
10
40 10 80
5
The total number of oranges shared is 80
The smallest share is 2 80 16
10
x
x
x
+ + =
´ =
= ´ =
´ =
4. Evaluate 1.25 0.025, correct to 1 decimal place
0.05
(A) 0.6 (B) 6.2 (C) 6.3 (D) 0.5
´
Solution
2 2
2
2 2 2 2
1.25 0.25 125 10 25 10
0.05 5 10
1.25 0.25 125 25 10 625 10 6.25
0.05 5
1.25 0.25 6.3 to 1 decimal place
5. Calculate the time taken for N3000 to earn N600
if invested at 8% simple interest.
(A) 2 years (B) 3 years (C) 3 years (D) 1 years
Solution
1
2
Time, 100
Given 3000, 600, 8%
100 600 5 2
3000 8 2
T I
P R
P N I N R
T years years
= ´
´
= = =
= ´ = =
´
5
1
1
2 3 5
6. Simplify 3 27
9
( ) 3 (B) 3 (C) 3 (D) 3
n
n
n
A
-
+
- ´
Solution
5 5 5 3 3
1 3( 1)
1 2(1 ) 2 2
5 5 3 3 2 3
1 2 3 (2 2 )
1 2 2 2 2
5
1 2 3 2 2
1
3 27 3 3 3 3
9 3 3
3 27 3 3 3
9 3 3
3 27 3 3
9
n n n n
n n
n n n
n n n n
n n n
n n n
n
n n n
n
- - - +
+ +
- - -
- - + + - +
+ - + - -
- - -
-
+ - + - +
-
´ = ´ = ´
´ = = =
´ = =
1
3
10 10 7. If log 4 0.6021, evaluate log 4
(A) 0.3011 (B) 0.9021 (C) 1.8063 (D) 0.2007
=
Solution
13
1 1
10 3 10 3 log 4 = log 4 = (0.6021) = 0.2007
1 1
5 9
8. Simplify 5( 147 12)
15
(A) 5 (B) (C) (D) 9
-
Solution
( )
( ) ( )
5( 147 12) 5 3 49 4 3
15 3 5
5( 147 12) 5 7 3 2 3 5 5 3
15 3 5 3 5
5( 147 12) 5
on to UTME Mathematics for 2013, Solved by Tawose, Simeon Abiola.
9. , and are subset of the universal set U.
The Venn diagram show showing the relationship
( ) is
P Q R
P Ç Q È R
Answer: Option C
10. If { : is odd, 1 20} and
{ : is prime, 2 25}, find
(A) {3,5,7,11,17,19}
(B) {3,5,11,13,17,19}
(C) {3,5,7,11,13,17,19}
(D) {2,3,5,7,11,13,17,19}
P x x x
Q y y y P Q
= - < £
= - < £ Ç
Solution
{ : is odd, 1 20}
{1,3,5,7,9,11,13,15,17,19}
{ : is prime, 2 25}
{2,3,5,7,11,13,17,19}
{3,5,7,13,17,19} (option B)
P x x x
P
Q y y y
Q
P Q
= - < £
=
= - < £
=
Ç =
2
2 2
11. If 4 4, find in terms of
(A) 2 (B) 2 (C) 2 (D) 2
S t t t S
S S S S
= - +
- + - +
Solution
1
2
2
2 2
4 4
( 2) ( 2)
2
2
S t t
S t t
S t
t S
= - +
= - = éë - ùû
= -
= +
12. If 4 is a factor of 2 , then is
(A) 4 (B) 12 (C) 20 (D) 2
x - x - x - k k
Solution
2
2
Let ( ) ,
if 4 is a factor of ( ) then (4) 0
4 4 0
16 4 0
12
f x x x k
x f x f
k
k
k
= - -
- =
- - =
- - =
=
13. The remainder when 6 3 2 47 30 is
divided by 3 is
(A) 21 (B) 42 (C) 63 (D) 18
p p p
p
- - +
-
Solution
3 2
3 2
Let ( ) 6 47 30
The remainder when ( ) is divided by 3 is (3)
(3) 6(3) 3 47(3) 30
(3) 162 9 141 30 42
The remainder is 3
f p p p p
f p p f
f
f
= - - +
-
= - - +
= - - + =
8
5
128 288
5 5
14. varies jointly as and , varies inversely as
Given that 4, 3 and 2 when 1,
find the value of when 6, 4 and
( ) (B) 15 (C) 10 (D)
P m u q
p m u q
p m u q
A
= = = =
= = =
Solution
8
5
Given 4, 3, 2 and 1
{ is the proportionality constant}
4 1 2
3 2 3
so when 6, 4 and will be
2 6 4 5
3 8
10
p mu
q
p m u q
p kmu k
q
k pq
mu
k
m u q p
p kmu
q
p
m
= = = =
=
=
= ´ =
´
= = =
= = ´ ´ ´
=
Solution to UTME Mathematics for 2013, Solved by Tawose, Simeon Abiola.
2
2
15. If varies inversely as the square root of and
how does vary with and .
( ) varies directly as and
( ) varies inversely as and
( ) varies inversely as and
( ) varie
r s t
s r t
A s r t
B s r t
C s r t
D s s directly as r2 and t2
Solution
2
2
2
2
2
2
2
2
2
2
2
1
( proportionality constant)
Square both sides
multiply both sides by
Divide both sides by
Let (where is a constant)
1
varies inversely as the
r
st
r k k
st
r k
st
s
sr k
t
r
s k
r t
k A A
s A
r t
s
r t
s r
m
= =
=
=
=
=
=
m
and t
16. Evaluate 3( 2) 6( 3)
( ) 4 ( ) 4 ( ) 4 ( ) 4
x x
A x B x C x D x
+ > +
> < > - < -
Solution
3( 2) 6( 3)
3 6 6 18
3 6 18 6
3 12
4
x x
x x
x x
x
x
+ > +
+ > +
- > -
- >
< -
17
Solution
2
The roots of the equation are 1 and 2
The quadratic equation will be
( 1)( 2)
2
x x
x x y
y x x
= - =
+ - =
= - -
18. Solve for : 2 3
( ) 1 ( ) 5 ( ) 2 3 (D) 1 5
x x
A x B x C x x
- <
< < - < < - < <
Solution
2 3
3 2 3
Add 2 to both side of the inequality
3 2 3 2
1 5
x
x
x
x
- <
- < - <
- + < < +
- < <
19. If the sum of the first term of G.P is 3, and the
sum of the second and the third term is 6,
find the sum of the first term and the common ratio
(A) 5 (B) 2 (C) 3 (D) 5
-
- - -
Solution
1
1
2
1 2 3
1 2
1 2
2
2 3
2 3
The term of a G.P is given as
, ,
3
(1 ) 3 ( )
6
(1 ) 6 ( )
Divide equation ( ) by ( )
(1 ) 6
(1 ) 3
2
Substitute 2 i
th n
n
n
n ar
T ar
T a T ar T ar
T T a ar
T T a r i
T T ar ar
T T ar r ii
ii i
ar r
a r
r
r
-
-
-
=
= = =
+ = + =
+ = + = - - -
+ = + = -
+ = + = - - - -
+ = -
+
= -
= - nto equation (i)
(1 2) 3
3
3
Sum of first term and common ratio ( 2 3) 5
2 3 5
a
a
a
a r
- =
- =
= -
= - - = -
+ = - - = -
7 10 13 42
3 4 4
3 1 1 3 3 1 3 1
1 1 1 1
20. The term of the progression , , , , is
( ) n ( ) n ( ) n (D) n
n n n n
nth
A - B - C + +
+ + + -
´´´
Solution
Solution to UTME Mathematics for 2013, Solved by Tawose, Simeon Abiola.
1 2
We calculate for the term for the numerator
and denominator separately and we later merge two
together
For the numerator we have
4, 7, 10, 13,
4, 7
common difference 7 4 3
( 1)
4 (
th
n
n
n
T T
d
T a n d
T n
-
´´´
= =
= = - =
= + -
= + -
1 2
1)3 4 3 3 3 1
The denominator, the sequence is 2, 3, 4 5,
2, 3, 3 2
2 ( 1)1 2 1 1
The T of the sequence will be 3 1
1
n
n
n n
T T d
T n n n
n
n
= + - = +
´´´
= = = -
= + - = + - = +
+
+
21. If a binary operation is defined by
2 , find 2 (3 4)
( ) 26 ( ) 24 ( ) 16 ( ) 14
x y x y
A B C D
*
* = + * *
Solution
Given the binary operation that ( ) 2
3 4 3 2(4) 11
2 (3 4) 2 11 2 2(11) 24
2 (3 4) 24
x * y = x + y
* = + =
* * = * = + =
* * =
5 3 4 2
22. If and , find 2
2 1 3 5
8 14 7 7
( ) ( )
7 7 14 8
14 8 7 7
( ) ( )
7 7 8 14
P Q P Q
A B
C D
é ù é ù
= ê ú = ê ú +
ë û ë û
é ù é ù
ê ú ê ú
ë û ë û
é ù é ù
ê ú ê ú
ë û ë û
Solution
5 3 4 2
and ,
2 1 3 5
5 3 4 2 10 6 4 2
2 2
2 1 3 5 4 2 3 5
10 4 6 2 14 8
2 (option C)
4 3 2 5 7 7
P Q
P Q
P Q
é ù é ù
= ê ú = ê ú
ë û ë û
é ù é ù é ù é ù
+ = ê ú + ê ú = ê ú + ê ú
ë û ë û ë û ë û
é + + ù é ù
+ = ê ú = ê ú ë + + û ë û
3 3
2 2
5 5
2 2
3 3
2 2
5 5
2 2
5 3
23. Find the inverse of
6 4
2 2
( ) ( )
3 3
2 2
( ) ( )
3 3
A B
C D
é ù
ê ú
ë û
é ù é - ù
ê - ú ê - - ú ë û ë û
é - ù é ù
ê - ú ê - - ú ë û ë û
Solution
1
1
3 3 42
1 2 2
6 5 5
2 2 2
Given that
1
5 3
6 4
1 4 3 1 4 3
5 4 3 6 6 5 20 18 6 5
1 4 3 2
2 6 5 3
a b
A
c d
d b
A
ad bc c a
A
A
A
-
-
-
é ù
= ê ú
ë û
é - ù
= ê ú - ë - û
é ù
= ê ú
ë û
é - ù é - ù
= ê ú = ê ú ´ - ´ ë - û - ë - û
é - ù é - ù é - ù
= ê ú = ê ú = ê ú ë - û ë - û ë - û
24.
In the diagram above, find the value of x
(A) 15o (B) 30o (C) 40o (D) 45o
Solution
Consider
15 {Base angle of Isosceles }
180 {sum of angles in }
180 15 15 150
180 {Sum of angles on a straight line}
150 180
30
Consider
CDE
CED CDE
ECD CED CDE
ECD
ECD BCA
BCA
BCA
ABC
B
D
Ð = Ð = D
Ð = - Ð - Ð D
Ð = - - =
Ð + Ð =
+ Ð =
Ð =
D
Ð
o
o
o o o o
o
o o
o
180
110 30 180
40
AC ABC BCA
x
x
+ Ð + Ð =
+ + =
=
o
o o o
o
25
The value of x in the figure above is
110o
x
15o
110o
x
15o
A
B C D
E
P
P
T
Q
S
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