MMC 2018 Elimination Round Grade 10 Solutions

MMC 2018 Elimination Round Grade 10
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1. The average of a number and √7 +
√5 is √7–√5. Find the number.
Solution:
Let x be the missing number.
x + (√7 + √5)
2
= √7–√5
x + √7 + √5=2√7– 2√5
x = 2√7–√7– 2√5– √5
x = √7– 3√5
2. Find n if 27.63% of 349 is 2.763% of n.
Solution:
(27.63%)(349) = (2.763%)(n)
(27.63%)(349)
2.763%
= n
n = 10(349) = 3,490
3. Simplify:
101!
99!
Solution:
Using the definition n! = n(n – 1)!,
101!
99!
=
101 × 100 × 99!
99!
= 101 × 100 = 10,100
4. If 3n = 90, find the integer closet to n.
Solution:
The number 90 is between 81 = 34
and 243 = 35. But 90 seems to be
closer to 81 than to 243, so n should
be closest to 4.
5. Find the largest positive integer n
such that (n – 18)4036 ≤ 992018.
Solution:
Taking the 2018th root both sides, we
have
(n – 18)2 ≤ 99
Since we are looking for the largest
positive integer n, we observed that
the largest value of (n – 18)2 less than
99 is 81 since 92 = 81.
Hence, n – 18 = 9  n = 27.
6. How many integers between 60 and
600 (inclusive) are divisible by 2 or
by 3?
Solution:
To get the number of multiples of a
certain number, use the formula of
the arithmetic sequence which is an
= a1 + (n – 1)d.
The number of multiples of 2 from 60
to 600 can be found as follows:
a1 = 60, an = 600, d = 2
an = a1 + (n – 1)d
600 = 60 + (n – 1)(2)
540 = 2(n – 1)
270 = n – 1
n = 271
The number of multiples of 3 from 60
to 600 can be found as follows:
a1 = 60, an = 600, d = 3
an = a1 + (n – 1)d
600 = 60 + (n – 1)(3)
540 = 3(n – 1)
180 = n – 1

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n = 181
But there are some numbers that are
divisible by 2 are also divisible by 3.
Those are the numbers divisible by 6.
The number of such numbers can be
found as follows:
a1 = 60, an = 600, d = 6
an = a1 + (n – 1)d
600 = 60 + (n – 1)(6)
540 = 6(n – 1)
90 = n – 1
n = 91
To get the number of elements in
either sets (A ∪ B), use the Addition
Rule formula n(A ∪ B) = n(A) + n(B) –
n(A ∩ B) where n(A) is the number of
elements in a set. Let n(A) and n(B)
be the number of multiples of 2 and
3, respectively.
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 271 + 181 – 91 = 361
7. How many factors does the product
of 3 different prime factors have?
Solution:
Let a, b and c be the different prime
factors, and abc be their product.
To get the number of factors of a
number, take first the prime
factorization in exponential form.
Take all exponents of all prime
factors, add all of these by 1 and
multiply all of the results.
The exponents of the three different
prime factors are all 1. Adding it
allby 1 gives 2 and thus, we have 2 ×
2 × 2 = 8 factors.
8. A triangle has area 30 cm2, and two
sides are 5 cm and 12 cm long. How
long is the third side?
Solution:
Recall the area of a triangle A =
bh
2
where b and h be the base and
height, respectively. Assume that 12
cm which is the given be the base of
a triangle.
A =
bh
2
30 =
12h
2
60 = 12h
h = 5 cm
The height of a triangle is found to
be 5 cm, which is already given in
the problem. Since the height and
base lines are perpendicular, the
triangle in the problem is a right
triangle whose lengths of both of its
legs are given.
To find the third side, use the
Pythagorean Theorem.
a2 + b2 = c2
52 + 122 = c2
25 + 144 = 169 = c2
c2 = 13 cm
9. If A = {all prime numbers from 1 and
100}, B = {2, 4, 6, …, 98, 100}, and C =
{3, 6, 9, …, 96, 99}, how many
elements does (A ∪ B) ∩ C have?

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Solution:
Notice that B contains all even
numbers up to 100 and C contains
all multiples of 3 less than 100. A ∪ B
must contain all prime OR even
numbers less than or equal to 100.
Let us look the elements of C. We
identify each of them whether it is
prime or even to include in (A ∪ B) ∩
C. Inspecting them one by one in C,
we have the following results:
- 3 is prime
- 6 is even
- 9 is neither prime nor even
- 12 is even
- 15 is neither prime nor even
- 18 is even
- …and so on
According to the results, the set (A ∪
B) ∩ C must contain 3 and all
multiples of 6 less than 100. In other
words, (A ∪ B) ∩ C = {3, 6, 12, 18, 24,
…, 90, 96}
From 6 to 96, we have 96 ÷ 6 = 16
elements, adding it by 1 for the
number 3 gives a total of 17
elements in (A ∪ B) ∩ C.
10. Find the perimeter of a regular
hexagon inscribed inside a circle
whose area is 49π cm2.
Solution:
Recall also the area of a circle A =
πr2 to find the radius.
A = πr2
49π = πr2
r2 = 49
r = 7 cm
Note that the regular hexagon
divides into 6 equilateral triangles
that have three equal sides each.
Therefore, the side of an equilateral
triangle has length 7 cm.
Since hexagon has 6 sides, the
perimeter is 6(7) = 42 cm.
11. Sets A and B are subsets of a set
having 20 elements. If set A has 13
elements, at least how many
elements should set B have to
guarantee that A and B are not
disjoint?
Solution:
Two sets are disjoint if their
intersection is empty set (Ø). There
should be at least 1 element in
intersection to make them non-
disjoint sets.
To guarantee that there’s always an
intersection between two sets, the
combined sets A and B, which isA∪
B, should have 20 elements.
By letting n(A ∩ B) = 1 and using the
Addition Rule, we can solve for n(B),
the number of elements of set B.
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
20 = 13 + n(B) – 1
20 = 12 + n(B)
n(B) = 8 elements

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12. If the 2nd and 5th elements are -2 and
7, respectively, find the 101st term.
Solution:
Using the formula for arithmetic
sequence, we can solve as follows.
If n = 2 and an = -2,
an = a1 + (n – 1)d
-2 = a1 + (2 – 1)d
-2 = a1 + d eq. 1
If n = 5 and an = 7,
an = a1 + (n – 1)d
7 = a1 + (5 – 1)d
7 = a1 + 4d eq. 2
We form a system of equations in 2
variables that can be solved for
them.
-2 = a1 + d  eq. 1
-2 – d = a1
7 = a1 + 4d  eq. 2
7 = -2 – d + 4d
9 = 3d
d = 3
a1 = -2 – d  eq. 1
a1 = -2 – 3 = -5
Therefore, if n = 101,
an = a1 + (n – 1)d
a101 = -5 + (101 – 1)3
a101 = -5 + (100)3
a101 = -5 + 300 = 295
13. Find the sum of the first 71 terms of
the arithmetic sequence above.
Solution:
The sum of terms of an arithmetic
sequence is defined as Sn =
n
2
(2a1 +
(n – 1)d). With results obtained in
problem #12 which is a1 = -5, d = 3
and n = 71, we have
Sn =
n
2
(2a1 + (n – 1)d)
S71 =
71
2
(2(-5) + (71 – 1)3)
S71 =
71
2
(2(-5) + (70)3)
S71 =
71
2
(-10 + 210)
S71 =
71
2
(200)
S71 = 71(100) = 7,100
14. Find k such that k – 2, 2k + 2, and 10k
+ 2 are consecutive terms of a
geometric sequence.
Solution:
If a1, a2, a3 are consecutive terms of
a geometric sentence, then
a2
a1
=
a3
a2
. By
applying this, we have
2k + 2
k – 2
=
10k + 2
2k + 2
2k + 2
k – 2
=
2(5k + 1)
2(k + 1)
2k + 2
k – 2
=
5k + 1
k + 1
(2k + 2)(k + 1) = (5k + 1)(k – 2)
2k2 + 4k + 2 = 5k2 – 9k – 2
2k2 – 5k2 + 4k + 9k + 2 + 2 = 0
-3k2 + 13k + 4 = 0
3k2 – 13k – 4 = 0
By using quadratic formula
x=
-b±√b
2
–4ac
2a
with a = 3, b = -13 and c
= -4, we have

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k=
-b±√b
2
– 4ac
2a
k =
-(-13) ± √(-13)
2
– 4(3)(-4)
2(3)
k =
13 ± √169 + 48
6
=
13 ± √217
6
15. Three numbers form an arithmetic
sequence with common difference
15. If the first is doubled, the second
is unchanged, and the third is
increased by 21, a geometric
sequence is formed. Find the first
number of the geometric sequence.
Solution:
In arithmetic sequence, if a1 = x, then
a2 = x + 15 and a3 = x + 30. The a1 of
a geometric sequence is2x, a2 is still x
+ 15, and a3 is x + 30 + 21 or x + 51.
Applying the same method as
problem #14, we have
x + 15
2x
=
x + 51
x + 15
(x + 15)(x + 15) = (2x)(x + 51)
x2 + 30x + 225 = 2x2 + 102x
x2 – 2x2 + 30x – 102x + 225 = 0
-x2 – 72x + 225 = 0
x2 + 72x – 225 = 0
(x – 3)(x + 75) = 0
x – 3 = 0
x = 3
x + 75 = 0
The first number in a geometric
sequence is 2x. If x = 3, 2x = 2(3) = 6
and if x = -75, 2x = 2(-75) = -150.
16. The sum of an infinite geometric
sequence have sum 14. If their
squares have sum
196
3
, find the sum of
their cubes.
Solution:
The sum of infinite geometric
sequence is defined as S =
a
1 – r
where
|r| < 1.
S =
a
1 – r
14 =
a
1 – r
14(1 – r) = a
14 – 14r = a  eq. 1
The squares of terms have a2 as its
first term and r2 its ratio.
196
3
=
a2
1 –r2
196
3
= (
a
1 – r
) (
a
1 + r
)
But
a
1 – r
=14, so use substitution.
196
3
= 14 (
a
1 + r
)
14
3
=
a
1 + r
14(1 + r) = 3a
14 + 14r = 3a  eq. 2
We have a system of equations
formed. Solving for a and r, we have
14 – 14r = a  eq. 1
14 + 14r = 3a  eq. 2
14 + 14r = 3(14 – 14r)
14 + 14r = 42 – 42r
14r + 42r = 42 – 14
56r = 28

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r =
28
56
=
1
2
a = 14 + 14(
1
2
) eq. 1
a = 14 – 7 = 7
Therefore, the sum of the cubes is
S =
a3
1 – r3
S =
7
3
1 – (
1
2
)
3
S =
343
1 –
1
8
=
343
7
8
S =343 (
8
7
) = 392
17. A sequence {an} has two terms a1 = 3
and a2 = 2. For every n ≥ 3, an is the
sum of all the preceding terms of the
sequence. Find a12.
Solution:
Using the rule stated in the problem,
we have the terms obtained below:
a1 = 3
a2 = 2
a3 = 3 + 2 = 5
a4 = 3 + 2 + 5 = 10
a5 = 3 + 2 + 5 + 10 = 20
a6 = 3 + 2 + 5 + 10 + 20 = 40
…and so on
Notice that the terms starting a3form
a geometric sequence. The
common ratio is 2, the first term a3 =
5 so a12 should be the 10th term (n =
10). Therefore,
a12 = 5(2)10 – 1 = 5(2)9
a12= 5(512) = 2,560
18. Let r and s be the roots of x2 – 9x + 7
= 0. Find (r + 1)(s + 1).
Solution:
Sum of roots: r + s =-(-9) = 9
Product of roots: rs = 7
(r + 1)(s + 1)
= rs + r + s + 1
= 7 + (-9) + 1 = -1
19. If P(x + 2) = 3x2 + 5x + 4, find P(x).
Solution:
To get P(x), take the inverse of x + 2
to “undo” the expression and
change to x. The inverse of x + 2 is x –
2.
Replacing all x to x – 2, we have
P(x + 2) = 3x2 + 5x + 4
P((x – 2) + 2) = 3(x – 2)2 + 5(x – 2) + 4
P(x – 2 + 2) = 3(x2 – 4x + 4) + 5(x – 2) +
4
P(x) = 3x2 – 12x + 12 + 5x – 10 + 4
P(x) = 3x2 – 7x + 6
20. Find the remainder if P(x) = 27x3 +
81x2 + 5x + 20 is divided by x + 3.
Solution:
We use the Remainder Theorem to
find the remainder by finding P(c) if
P(x) is divided by x – c. Thus we find
P(-3) for this problem because of the
divisor x – (-3).
P(-3) = 27(-3)3 + 81(-3)2 + 5(-3) + 20
= 27(-27) + 81(9) + 5(-3) + 20

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= -729 + 729 – 15 + 20 = 5
21. Find the quotient of the quotient
when p(x) = 6x4 – 8x3 + x2 + 9x + 7 is
divided by x +
2
3
.
Solution:
Using synthetic division, we have
-
2
3
6 -8 1 9 7
-4 8 6 10
6 -12 9 15 17
The quotient is 6x3 – 12x2 + 9x + 15
and hence, its constant term is 15.
22. Find the constant k if x + 3 is a factor
of f(x) = 2x4 + 8x3 + (k2+ 1)x2 + kx +
15.
Solution:
Since x + 3 is a factor of f(x), f(-3) = 0
by Remainder Theorem. With x = -3
and f(-3) = 0, we have
f(x) = 2x4 + 8x3 + (k2 – 1)x2 + kx + 15
0 = 2(-3)4 + 8(-3)3 + (k2 + 1)(-3)2 + k(-3)
+ 15
0 = 2(81) + 8(-27) + 9(k2 + 1) + k(-3) +
15
0 = 162 – 216 + 9k2 + 9 – 3k + 15
0 = 9k2 – 3k – 30
0 = 3k2 – k – 10
0 = (k + 3)(3k – 10)
0 = k + 3
k = -3
0 = 3k – 10
10 = 3k
k =
10
3
23. Find the largest real root of 2x3 – 5x2 –
2x + 2 = 0.
Solution:
We use Rational Roots Theorem to
find all possible rational roots of a
polynomial equation. Let p be the
factor of a constant term, q be the
factor of a leading coefficient (the
term with the highest power) and
p
q
be the possible rational root.
Applying this, we have
p = factors of 2 = ±1, ±2
q = factors of 2 = ±1, ±2
p
q
= ±
1
2
, ±1, ±2
Testing somepossible values of
p
q
, we
have
1
2
as a root because
1
2
2 -5 -2 2
1 -2 -2
2 -4 -4 0
We have depressed equation 2x2 –
4x – 4 = 0 which is equivalent to x2 –
2x – 2 = 0. Solving it by quadratic
formula, we have
x=
-b±√b
2
– 4ac
2a
x =
-(-2) ± √(-2)
2
– 4(1)(-2)
2(1)
x =
2 ± √4 + 8
2(1)
=
2 ± √12
2
x =
2 + 2√3
2
,
2 – 2√3
2
x = 1 + 1√3, 1 – 1√3

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The roots of 2x3 – 5x2 – 2x + 2 = 0 are
1
2
, 1 + 1√3 and 1 – 1√3. The value of
the root1 + 1√3 is greater than 1,
which is greater than the other root
1
2
.
Therefore, the largest root is 1 + 1√3.
24. If p(x) is a 3rd degree polynomial,
p (
5
3
) = p(4) = p(-5) = 0, and p(0) = -
200, find p(2).
Solution:
If p(c) = 0, then we can say that c is
a zero of p(x) and x – c is a factor of
p(x).
In the problem above, we say that
5
3
,
4 and -5 are all zeros of p(x), and
p(x)= a (x –
5
3
) (x – 4)(x + 5) where a is
the constant. With x = 0 and p(x) = -
200, we can solve for a.
p(x)= a (x –
5
3
) (x – 4)(x + 5)
-200 = a (0 –
5
3
) (0 – 4)(0 + 5)
-200 = a (-
5
3
) (-4)(5)
-200 =
100
3
a
-600 = 100a
a = -6
Thus,
p(x)= -6 (x –
5
3
) (x – 4)(x + 5)
p(2)= -6 (2 –
5
3
) (2 – 4)(2 + 5)
p(2)= -6 (
1
3
) (-3)(7)
p(2) = -6(-1)(7) = 42
25. The midpoint of P(1, 2) and A is Q;
the midpoint of P and C is S; the
midpoint of Q and S is R(5, 1.5). Find
the coordinates of B, the midpoint of
A and C.
Solution:
The midpoint of two points (x1, y1)
and (x2, y2) is defined as (
x1 + x2
2
,
y1 + y2
2
).
This is known as Midpoint Formula.
Let (a, b) be the coordinates of Q. If
the midpoint of P(1, 2) and A(xA, yA)
is Q(a, b), then A can be computed
as follows.
(
xP + xA
2
,
yP + yA
2
) = (xQ, yQ)
(
1 + xA
2
,
2 + yA
2
) = (a, b)
(1 + xA, 2 + yA) = (2a, 2b)
(xA, yA) = (2a – 1, 2b – 2)
The midpoint of Q(a, b) and S(xS, yS)
is R(5, 1.5). S is computed as follows.
(
xQ + xS
2
,
yQ + yS
2
) = (xR, yR)
(
a + xS
2
,
b + yS
2
) = (5, 1.5)
(a + xS, b + yS) = (10, 3)
(xS, yS) = (10 – a, 3 – b)
The midpoint of P(1, 2) and C(xC, yC)
is S(10 – a, 3 – b). C is computed as
follows.
(
xP + xC
2
,
yP + yC
2
) = (xS, yS)
(
1 + xC
2
,
2 + yC
2
) = (10 – a, 3 – b)
(1 + xC, 2 + yC) = (20 – 2a, 6 – 2b)
(xC, yC) = (19 – 2a, 4 – 2b)

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Finally, the midpoint of A(2a – 1, 2b –
2) and C(19 – 2a, 4 – 2b) is B that we
are looking for.
(
xA + xC
2
,
yA + yC
2
) = (xB, yB)
= (
2a – 1 + 19 – 2a
2
,
2b – 2 + 4 – 2b
2
)
= (
18
2
,
2
2
)
= (9, 1)
26. Find the radius of the circle with
equation x2 + y2 + 4x – 3y + 5 = 0.
Solution:
The equation of the circle in center-
radius form is (x – h)2 + (y – k)2 = r2
where (h, k) be the coordinates of
the center and r the length of the
radius.
Rewrite the given equation into
center-radius form.
x2 + y2 + 4x – 3y + 5 = 0
x2 + 4x + y2 – 3y = -5
x2 + 4x + 4 + y2 – 3y +
9
4
= -5 + 4 +
9
4
(x + 2)2
+ (y –
3
2
)
2
=
5
4
= (
√5
2
)
2
The radius is
√5
2
.
27. Give the equation (in center-radius
form) of the circle having as a
diameter the segment with endpoints
(-2, 10) and (4, 2).
Solution:
The endpoints of the diameter have
given coordinates, so their midpoint
is the center of the circle. Using
Midpoint Formula, the center has an
order pair of
(
x1 + x2
2
,
y1 + y2
2
)
= (
-2 + 4
2
,
10 + 2
2
)
= (
2
2
,
12
2
)
= (1, 6)
To compute for the radius, find the
distance between the center and
either of the points (-2, 10) or (4, 2) by
using Distance Formula d =
√(x2– x1)
2
+ (y2
– y1
)
2
. We use the
center and (4, 2) for our purpose.
d = √(x2– x1)
2
+ (y2
– y1
)
2
r = √(4 – (-2))
2
+ (2 – 10)
2
r = √(4 + 2)
2
+ (2 – 10)
2
= √6
2
+ (-8)
2
r = √36+ 64 = √100
r = 10
Using the center and (-2, 10) yields
also the same result which is r = 10.
Therefore, the equation of the circle
is (x – 1)2 + (y – 6)2 = 102 or (x – 1)2 +
(y – 6)2 = 100 in center-radius form.
28. The center of a circle is on the x-axis.
If the circle passes through (0, 5) and
(6, 4), find the coordinates of its
center.
Solution:
Since the center lies on x-axis, the
coordinates must be (x, 0). We know
that the distance between the

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center and any point on a circle is its
radius. So, the coordinate (0, 5) has
the same distance from the center
(x, 0) as (6, 4).
Recall also the distance formula d =
√(x2– x1)
2
+ (y2
– y1
)
2
which is the
distance between two points.
Distance between (x, 0) and (0, 5) =
Distance between (x, 0) and (6, 4)
√(0 – x)
2
+ (5 – 0)
2
= √(6 – x)
2
+ (4 – 0)
2
(-x)2 + 52 = (6 – x)2 + 42
x2 + 25 = 36 – 12x + x2 + 16
25 = 52 – 12x
-12x = -27
x =
-27
-12
=
9
4
Therefore, the center has
coordinates (
9
4
, 0).
29. If θ is an angle in a triangle, and sec
θ =
7
5
, find tan θ.
Solution:
By using trigonometric identity sec θ
=
1
cos θ
, we have
sec θ =
1
cos θ
=
7
5
5 = 7 cos θ
cos θ =
5
7
Recall the memory aid SOH-CAH-
TOA that is used to remember the
ratios of sine, cosine and tangent of
a certain angle. Since we have cos
θ, we use CAH.
CAH cos θ =
adjacent
hypotenuse
=
5
7
To get the opposite side of a right
triangle, we use Pythagorean
Theorem.
(opposite)2 + (adjacent)2 =
(hypotenuse)2
(opp)2 + 52 = 72
(opp)2 + 25 = 49
(opp)2 = 24 opp = √24=2√6
To find tan θ, use again the memory
aid above.
TOA  tan θ =
opposite
adjacent
Therefore, tan θ =
2√6
5
.
30. In ΔABC, BC = 6, CA = 7, and AB = 8.
Find
sin B+ sin C
sin A
.
Solution:
In solving triangles (sides and
angles), use the formula for Law of
Sines.
BC
sin A
=
AC
sin B
=
AB
sin C
Substituting the given values, we
have
6
sin A
=
7
sin B
=
8
sin C
To make ratios equal, we assign the
values of sin A, sin B and sin C. To
reduce all 3 fractions to
1
n
(where n ≠
0) for our simplification purposes, we

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Page11
assign sin A = 6n, sin B = 7n and sin C
= 8n.
Substituting the values into
sin B+ sin C
sin A
,
we have
7n + 8n
6n
=
15n
6n
=
5
2
31. In ΔABC, AB = 3, AC = 2√3, and ∠A =
30°. Find BC.
Solution:
Notice that ∠A is between AB and
AC, so we use Law of Cosines.
BC2 = AB2 + AC2 – 2(AB)(AC)cos A
have
BC2 = 32 + (2√3)2 – 2(3)(2√3)cos 30°
BC2 = 32 + (2√3)2 – 2(3)(2√3)(
√3
2
)
BC2 = 9 + 12 – 6(3)
BC2 = 9 + 12 – 18 = 3
BC = √3
32. In a circle, chord PQ is bisected by
chord RS at T. If RT = 3 and ST = 6, find
PQ.
Solution:
Use the formula (TP)(TQ) = (TR)(TS) for
two intersecting chords.
Since RS bisects PQ at T, TP = TQ.
have
(TP)(TP) = (3)(6)
TP2 = 18
TP = √18 = 3√2 = TQ
Note that TP + TQ = PQ. Therefore,
PQ = 3√2+ 3√2= 6√2
See the figure above.
33. Rectangle ABCD is inscribed in a
circle, BC = 2√3, and CD = 2. A point
P is chosen on arc AB. Find ∠APB.
Solution:

mbalvero
Page12
Showing the figure above, we have
a right triangle BCD with right angle
C. To get the degree measure of
∠BDC, we use tangent function with
opposite side BC. Thus, we have tan
θ =
opposite
adjacent
=
2√3
2
= √3. Solving for θ, θ
= ∠BDC = 60°.
If ∠BDC = 60°, ∠BDA must be 30°
because ∠BDC and ∠BDA are
complementary angles.
Notice that APBD is an inscribed
quadrilateral whose opposite angles
are supplementary. Therefore, the
sum of ∠BDA and ∠APB is 180°.
If ∠BDA =30°, therefore, ∠APB should
be 150°.
34. Pentagon ABCDE is inscribed in a
circle, BC = CD = DE, and AB = AE. If
∠BAE = 96°, find ∠BDA.
Solution:
If AB = AE, then their intercepted arc
AB and arc AE are equal.
The measure of an inscribed angle is
one-half of its intercepted arc. If
∠BAE = 96°, then the major arc BCE =
2(96) = 192°. Remember that the
degree measure of a whole circle is
360°. Therefore, if arc BCE = 192°,
then arc BAE = 360 – 192 = 168°.
But arc BAE = arc AB + arc AE and
arc AB = arc AE. Therefore, arc AB =
168 ÷ 2 = 84°.
The intercept of ∠BDA is arc AB and
hence, ∠BDA = 84 ÷ 2 = 42°.
35. A line through a point A outside a
circle is tangent to the circle at D.
Another line through A intersects the
circle at points B (closer to A) and C.
If BC/AB = 2 and AD = 6, find AC.
Solution:

mbalvero
Page13
If a tangent and a secant lines are
drawn to a circle with their common
point outside the circle, then we use
the formula AD2 = (AB)(AC) in this
case.
The ratio of BC to AB is 2, so BC = 2x
and AB = x for some number x. Note
that AC = AB + BC.
Substituting into the formula, we
have
AD2 = (AB)(AC)
AD2 = (AB)(AB + BC)
62 = x(x + 2x)
36 = x(3x)
36 = 3x2
12 = x2
x = √12 = 2√3
AB = 2√3
BC = 2(2√3) = 4√3
Therefore, 2√3 + 4√3 = 6√3
36. In a circle, chords AB and CD
intersect at E. If arc AC = 120° and
arc BD = 20°, find ∠AEC.
Solution:
In two intersecting chords, the arc
AC and arc BD are the intercepts of
the vertical angles ∠AEC and ∠BED,
respectively, and ∠AEC = ∠BED =
1
2
(arc AC + arc BD). Therefore, ∠AEC
=
1
2
(120 + 20) =
1
2
(140) = 70°.
37. Find the area (in square units) of the
region between two concentric
circles if a chord of the larger circle
which is tangent to the smaller circle
has length 7.
Solution:
Draw the radius of the smaller circle
that is perpendicular to the chord
and draw the radius of the larger
circle that meets one of the
endpoints of a chord. We formed a

mbalvero
Page14
right triangle with these two radii and
a chord with half of its length (which,
in the given, is
7
2
).
Let x be the radius of the smaller
circle. Using Pythagorean Theorem,
we can find the radius of the larger
circle which is the hypotenuse.
a2 + b2 = c2
(
7
2
)
2
+ x2 = c2
49
4
+ x2 = c2
c = √
49
4
+ x2
Recall the area of the circle A = πr2.
The area of a smaller circle (with r =
x) is πx2 while the area of the larger
circle (with r =√
49
4
+ x2) is
π(√
49
4
+ x2)
2
= π(
49
4
+ x2)
To get the area of the said region,
subtract the two areas.
Area = (area of larger circle) – (area
of the smaller circle)
= π(
49
4
+ x2) – πx2
= π(
49
4
+x2
–x2
)
= π(
49
4
) =
49
4
π square units
38. A triangle having sides 3√7, 9, and
12 units is inscribed in a circle. Find
the circumference of the circle.
Solution:
To determine what kind of triangle
given its side lengths, we use the
relation a2 + b2 ??c2 such that ??
can be replaced with > which is an
obtuse triangle, < an acute triangle
and = a right triangle. The longest
side should be the value of c which
is 12.
a2 + b2 ?? c2
(3√7)2 + 92 ?? 122
63 + 81 ?? 144
144 = 144
Since 144 = 144, the inscribed
triangle is a right triangle. The
hypotenuse is the diameter of the
circle which has the length 12.
The radius of the circle is 6 which
gives the area of a circle of π(6)2 =
36 square units.
39. In the figure, quadrilateral ABCD is
inscribed in a circle. Lines AB and CD
intersect at P, while lines AD and BC
intersect at Q. If ∠APD = 50° and
∠AQB = 34°, find ∠PAQ.
Solution:
Let x be the measure of ∠PAQ

mbalvero
Page15
Notice that ΔBAQ and ΔDAP are
seen in the figure. The sum of the
interior angles of a triangle is 180°. In
ΔBAQ, ∠BQA = 34° and ∠BAQ = x.
∠QBA should be 180 – 34 – x = 146 –
x.
On the other hand, the ΔDAP has
measures ∠DPA = 50° and ∠DAP = x.
PDA should be 180 – 50 – x = 130 – x.
Take note that ABCD is inscribed in a
circle. ∠CBA and ∠CDA are
supplementary angles. ∠CBA =
∠QBA and ∠CDA = ∠PDA.
∠QBA + ∠PDA = 180
(146 – x) + (130 – x) = 180
276 – 2x = 180
-2x = -96
x = ∠PAQ = 48°
40. How many books (all different) are
there if there are 5,040 ways of
arranging them in a shelf?
Solution:
If a set of n objects are arranged,
the number of ways in arranging
them is n!. By trial and error method,
we have 7 different books because
7! = 5,040.
41. Lila and Dolf each counted the
number of three-digit positive
integers that can be formed using
the digits 1, 2, 3, 4, and 5. Lila
assumed a digit to be repeated,
while Dolf assumed no digit can be
repeated. If they counted correctly,
what is the difference of their
answers?
Solution:
Lila counts the 3-digit numbers
whose any digit can be repeated.
The hundreds digit can have 5
possible digits, the tens has also 5
and ones has 5 which give 5 × 5 × 5
= 125 such numbers.
Dolf counts also 3-digit numbers but
no digit id repeated. Here, there are
5 possible digits for hundreds digit.
Since no digit will be repeated, there
must be 4 possible digits in tens digit
because one digit is used. Ones digit
should have 3 possible digits
because 2 digits are used. There
should be 5 × 4 × 3 = 60 such
numbers for Dolf.
Hence, the difference of their
answers is 125 – 60 = 65.
42. In how many ways can a
photographer arrange 7 students for
a club picture if Ana, Carol and Iris
must always be together, but Rey
and Ruel should not beside each
other?
Solution:
Since Ana, Carol and Iris should be
together in a picture, they will count
as a single unit. There are now 5 units
instead of 7. And since Rey and Ruel
refused to be together, they should
have at least 1 unit between them.

mbalvero
Page16
For Problems 43 and 44:
A committee is to be formed from 7
City A residents and 9 City B residents.
Here are possible positions for Rey (X)
and Ruel (Y):
X __ Y __ __
X __ __ Y __
X __ __ __ Y
__ X __ Y __
__ X __ __Y
__ __ X __ Y
There are 6 possible positions for Rey
and Ruel. The single unit of three girls
(Ana, Carol and Iris) can be placed
in any of the blanks in each position.
The remaining two students can be
arranged freely.
We have
- 6 positions for Rey and Ruel,
- 2! = 2 ways to arrange Rey and
Ruel their positions,
- 3! = 6 arrangements for the 3 units
(a single unit of three girls and
other two students), and
- 3! = 6 ways to arrange Ana, Carol
and Iris their positions in a single
unit.
Therefore, the number of ways (or
possible club pictures) is (6)(2)(6)(6) =
432.
43. How many 6-member committees
with 3 members from each city can
be formed?
Solution:
This means that 3 members belong
to City A while another 3 belong to
City B.
Solving the number ways for each
city using Combination formula C(n,
r) =
n!
r!(n – r)!
, we have
City A: n = 7, r = 3
C(7, 3) =
7!
3!(7 – 3)!
=
(7)(6)(5)(4!)
6(4!)
= (7)(5) = 35
City B: n = 9, r = 3
C(9, 3) =
9!
3!(9 – 3)!
=
(9)(8)(7)(6!)
6(6!)
= (3)(4)(7) = 84
Since two cities are independent,
multiply the number of ways from
both cities. Hence, (35)(84) = 2,940.
44. How many 5-member committees
with more City A than City B
members can be formed?
Solution:
Since there are more members from
City A from City B, we have multiple
cases to deal with. Here are those
cases:
- Case 1: 3 from City A, 2 from City
B
- Case 2: 4 from City A, 1 from City
B
- Case 3: 5 from City A

mbalvero
Page17
For Problems 45 and 46:
The numbers 1, 2, …, 25 are written
on slips of paper which are placed in
a box.
Solving for the 3 cases in the same
manner as the previous one, we
have the following solutions below.
Case 1: 3 from City A, 2 from City B
C(7, 3) C(9, 2) =
7!
3!(7 – 3)!
×
9!
2!(9 – 2)!
=
(7)(6)(5)(4!)
6(4!)
×
(9)(8)(7!)
2(7!)
= (7)(5) × (9)(4) = 1,260
Case 2: 4 from City A, 1 from City B
C(7, 4) C(9, 1) =
7!
4!(7 – 4)!
×
9!
1!(9 – 1)!
=
(7)(6)(5)(4)(3!)
24(3!)
×
(9)(8!)
1(8!)
= (7)(5) × 9 = 315
Case 1: 5 from City A
C(7, 5) =
7!
5!(7 – 5)!
=
(7)(6)(5)(4)(3)(2!)
120(2!)
= (7)(3) = 21
Adding all 3 cases, we have 1,260 +
315 + 21 = 1,596 such committees.
45. If two slips are picked at the same
time from the box, find the
probability that one number is a
multiple of 7 and the other is a
multiple by 8.
Solution:
First, we’re going to find the number
of ways to pick 2 slips of paper at the
same time. Using Combination
formula with n = 25 and r = 2, we
have
C(25, 2) =
25!
2!(25 – 2)!
=
(25)(24)(23!)
2(23!)
= (25)(12) = 300 ways
From 1 to 25, there are 3 multiples of
7 (7, 14, 21) and also 3 multiples of 8
(8, 16, 24). Using counting principle,
there are 3 × 3 = 9 ways to pick such
pair of slips.
Hence, the probability is
9
300
=
3
100
or
3%.
46. If three slips are picked at the same
time from the box, find the
probability that the sum of the
numbers is odd.
Solution:
Using Combination formula again,
we’re going to find the number of
ways to pick 3 slips at the same time.
With n = 25 and r = 3, we have
C(25, 3) =
25!
3!(25 – 3)!
=
(25)(24)(23)(22!)
6(22!)
= (25)(4)(23) = 2,300 ways
One way to express an odd number
is the sum of 3 odd numbers (ex. 3 +
5 + 7 = 15). An odd number can also
be the sum of one odd and two
even numbers(ex. 2 + 4 + 5 = 11).
We have two cases to express an
odd number as a sum.

mbalvero
Page18
Case 1: Three numbers are odd.
Notice that from 1 to 25, 13 numbers
are odd (1, 3, 5, …, 25). The number
of ways to pick 3 numbers from 13
odd numbers is
C(13, 3) =
13!
3!(13 – 3)!
=
(13)(12)(11)(10!)
6(10!)
= (13)(2)(11) = 286
Case 2: An odd number and two
even numbers
Again, there are 13 odd numbers. 12
numbers are even (2, 4, 6, …, 24). We
can find the number of ways to pick
such numbers. Since they are
independent, we multiply the two
numbers.
C(13, 1) C(12, 2) =
13!
1!(13 – 1)!
×
12!
2!(12 – 2)!
=
(13)(12!)
1(12!)
×
(12)(11)(10!)
2(10!)
= 13 × (6)(11) = 858
Adding both cases we have 286 +
858 = 1,144 and therefore, the
probability is
1,144
2,300
=
286
575
.
47. Two fair dice are rolled. Find the
probability that one die is 2 while the
other is a prime number.
Solution:
There are only 5 possible outcomes
for the event namely (2, 2), (2, 3), (3,
2), (2, 5) and (5, 2).
Note there are 6 × 6 = 36 total
outcomes for rolling a pair of dice.
Hence, the probability is
5
36
.
48. In a class of Grade 10 students, the
probability that a random chosen
student likes dogs is 0.72, that a
student likes cats is 0.54, and a
student likes neither is 0.11. Find the
probability that a student likes both
cats and dogs.
Solution:
Let D be the students who likes dogs,
P(D) = 0.54, while C are those who
like cats, P(C) = 0.72.
If D ∪ C are those who like dogs or
cats, (D ∪ C)C don’t (maybe they like
other animals). If P((D ∪ C)C) = 0.11,
then P(D ∪ C) = 1 – 0.11 = 0.89.
Using the Addition Rule with two non-
mutually exclusive events (events
that have intersection), we can solve
for P(D ∩ C) where D ∩ C likes both
cats and dogs.
P(D ∪ C) = P(D) + P(C) – P(D ∩ C)
0.89 = 0.72 + 0.54 – P(D ∩ C)
0.89 = 1.26 – P(D ∩ C)
-P(D ∩ C) = -0.37
P(D ∩ C) = 0.37
49. In a company, the men who do not
wear glasses are thrice as many as
the men who do, while the women
who do not wear glasses are ten
more than the women who do. The
probability that an employee wears
glasses is
3
10
. If a selected employee
wears glasses, the probability that
the employee is a woman is
2
3
. How

mbalvero
Page19
many are the employees of the
company?
Solution:
Let M and F be the number of male
and female employees, both
wearing glasses, respectively. The
table below shows the number of
employees in a company.
With
glasses
Without
glasses
Total
Male M 3M 4M
Female F F + 10 2F + 10
Total M + F 3M + F +
10
4M + 2F
+ 10
The total number of employees in a
company seems to be 4M + 2F + 10.
Out of them, M + F wears glasses.
Thus, the probability that an
employee wears glasses is
M + F
4M + 2F + 10
which is equivalent to
3
10
.
Therefore, we have
M + F
4M + 2F + 10
=
3
10
as
the first equation. Simplifying this, we
have
M + F
4M + 2F + 10
=
3
10
10(M + F) = 3(4M + 2F + 10)
10M + 10F = 12M + 6F + 30
-2M + 4F = 30
2M – 4F = -30 eq. 1
There are M + F employees who
wear glasses, and F of them are
women. Thus, the probability is
F
M + F
which is also
2
3
. Therefore, we have
F
M + F
=
2
3
as the second equation and
this can be simplified as
F
M + F
=
2
3
3F = 2(M + F)
3F = 2M + 2F
F = 2M eq. 2
We have formed a system of
equations in 2 variables. To solve for
M and F, substitute the second
equation into first one.
2M – 4F = -30  eq. 1
F = 2M  eq. 2
2M – 4(2M) = -30
2M – 8M = -30
-6M = -30
M = 5
F = 2(5) = 10
As said earlier, the total number of
employees is 4M + 2F + 10. Therefore,
the number of employees is 4(5) +
2(10) + 10 = 20 + 20 + 10 = 50.
50. The number of ways of selecting 6
objects from n distinct objects is the
same as the number of ways of
selecting and arranging 3 objects
from n distinct objects. Find n.
Solution:
The keyword “selecting”means we
use combination while “selecting
and arranging” means we use
permutation. We have the following
working equation and its solution
below.
C(n, 6) = P(n, 3)
n!
6!(n – 6)!
=
n!
(n – 3)!

mbalvero
Page20
n!(n – 3)! = 720n!(n – 6)!
n!(n – 3)(n – 4)(n – 5)(n – 6)! = 720n!(n
– 6)!
(n – 3)(n – 4)(n – 5) = 720
n3 – 12n2 + 47n – 60 = 720
n3 – 12n2 + 47n – 780 = 0
(n – 13)(n2 + n + 60) = 0
n – 13 = 0
n = 13

MMC 2018 Elimination Round Grade 10 Solutions

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