collection.
2. A merchant paid P48,000 for some dresses and shoes. He paid P400.00
for each dress and P1,000.00 for each pair of shoes. He sold the dresses at a
profit of 20% and the shoes at a profit of 50%. If his total profit was P18,000.
How many dresses and shoes did he buy?
3. The sum of two numbers is 115. The difference is -21. Find the numbers.
4. Two angles are complementary, one angle is 42 degrees more than half
the other. Find the angles.
5. A two digit number is 6 times the sum of its digits. The tens digit is 1
more than the units digit. Find the original number.
6. The sum of two numbers is 29. One number is 7 less than twice the other
number. Find the two numbers.
7. Eugene and Erick went to Jollibee to buy french-fries and hamburgers.
Eugene bought 6 hamburgers and 4 french-fries French fries for P330.00.
Erick bought 4 hamburgers and 9 french fries for P505.00. What is the cost
of each french-fries and each hamburgers?
8. Jeny has P1,490.00 in P20 bills and P50 bills. If she has 40 pieces in all,
how many of each denomination does she have?
9. A vending machine takes only nickels and dimes. There are 5 times as
many dimes as nickels in the machine. The face value of the coins is $4.40.
How many of each coin are in the machine.
10. A chemistry experiment calls for a 30% sulfuric acid solution. If the lab
supply room has only 50% and 20% sulfuric acid solution on hand, how much
of each should be mixed to obtain twelve liters of a 30% solution.
SOLUTION
1. A collection of 105 coins consists of 1.00 peso and 5.00 pesos. If the total value is 205.00 pesos, find the number of coins of each denomination in the collection.
Solution by elimination: Let x = number of 1 peso coins Y= number of 5 peso coins
(
)
(
)
= + = + 2 205 5 1 105 equation y x equation y xSubtract the first equation from the second equation( equation 2 minus equation 1) , the result is
100 4y =
25
=
⇒y , substitute in the first equation X + 25 = 105
80
= ⇒x
Answer # 1: There are 80 →1 peso coins and 25 →5 peso coins.
2. A merchant paid P48,000 for some dresses and shoes. He paid P400.00 for each dress and P1,000.00 for each pair of shoes. He sold the dresses at a profit of 20% and the shoes at a profit of 50%. If his total profit was P18,000. How many dresses and shoes did he buy?
Solution by elimination method: Let x = number of dresses
Y = number of shoes Equation:
( )(
) ( )(
)
= + = + 18000 1000 5 . 0 400 2 . 0 : 2 equation 48000 1000 400 : 1 equation y x y xDivide equation 1 by 100 the result is 4x+10y=480, call this equation 3
Simplify equation 2, the result is
18000 500
80x+ y= , divide this by 20, the result is 900
25
⇒y=28, substitute in equation 3 and solve for x. 480 10 4x+ y = becomes
( )
28 480 10 4x+ = 200 4 = ⇒ x 50 = ⇒xAnswer #2: He buy 50 dresses and 28 shoes
3. The sum of two numbers is 115. The difference is -21. Find the numbers.
Solution by elimination method: Let x and y be the numbers
− = − = + 21 : 2 equation 115 : equation1 y x y x
Subtract equation 2 from equation 1, the result is
(
)
136 2 21 115 = − = − − = + y y x y x 68 =⇒y , substitute this in equation 1 and solve for x.
115 = +y x becomes 115 68= + x 47 = ⇒x
Answer #3: The numbers are 68 and 47
4. Two angles are complementary, one angle is 42 degrees more than half the other. Find the angles.
Solution by substitution method: Let x and y be the measure of the angles
+
=
=
+
y
x
y
x
2
1
42
:
equation2
90
:
equation1
Substitute equation 2 in equation 1, the result is 90 2 1 42 + =
180 3 84+ = ⇒ y 96 3 = ⇒ y 32 =
⇒y substitute this in equation 2
( )
32 58 2 1 42+ = = xAnswer # 4: The measure of the angles are 32 and 58 degrees.
5. A two digit number is 6 times the sum of its digits. The tens digit is 1 more than the units digit. Find the original number.
Solution by elimination method: Let x = units digit Y = ten’s digit The value of the number is x+10y .
(
)
+ = + = + x y y x y x 1 : 2 equation 6 10 : 1 equationSimplify equation 1 it becomes,
y x y x+10 =6 +6 0 4 5 − = ⇒ x y y x 4 5 = ⇒
(
x)
x= + ⇒5 41 we substitute equation 2. x x 4 4 5 = + ⇒ 4 = ⇒x , therefore y=5.Answer: The original number is 54.
6. The sum of two numbers is 29. One number is 7 less than twice the other number. Find the two numbers.
Solution by substitution method: Let x and y be the numbers
− = = + 7 2 : 2 equation 29 : 1 equation y x y x
Substitute equation 2 in equation 1, the result is
(
2y−7)
+y=29 36 3 = ⇒ y⇒x=17
Answer # 6 : The numbers are 17 and 12.
7. Eugene and Erick went to Jollibee to buy french-fries and hamburgers. Eugene bought 6
hamburgers and 4 french-fries French fries for P330.00. Erick bought 4 hamburgers and 9 frenchfriyes for P505.00. What is the cost of each french-fries and each hamburgers?
Solution by elimination method: Let x = number of hamburgers Y = number of French fries
= + ⇒ × = + = + ⇒ × = + 4 equation this call , 1515 27 12 3 505 9 4 : 2 equation 3 equation this call , 660 8 12 2 330 4 6 : 1 equation y x y x y x y x
Subtract equation 3 from equation 4, the result is
(
)
855 19 660 8 12 27 1515 12 = = + − + = y y x y x 45 =
⇒y , substitute this in equation 1 and solve for x,
( )
45 330 4 6x+ = 150 6 = ⇒ x 25 = ⇒xAnswer #7: The cost of each hamburger is P25.00 and the cost of each French fries is P45.00
8. Jeny has P1,490.00 in P20 bills and P50 bills. If she has 40 pieces in all, how many of each denomination does she have?
Solution by elimination method: Let x= number of P20 bills Y=number of P50 bills = + = + 1490 50 20 : 2 equation 40 : 1 equation y x y x
Multiply equation 1 by 20, the result is
800 20
(
)
690 30 800 20 20 50 1490 20 = = + − + = y y x y x 23 =⇒y , substitute this in equation 1, and solve for x. 40 23= + x 17 = ⇒x
Answer #8: There are 17 P20 bills and 23 P50 bills
9. A vending machine takes only nickels and dimes. There are 5 times as many dimes as nickels in the machine. The face value of the coins is $4.40. How many of each coin are in the machine. Solution by elimination method: Let x = number of nickels
Y= number of dimes = + = 40 . 4 10 . 05 . : equation2 x 5 y : equation1 y x
Multiply equation 2 by 100, the result is,
440 10
5x+ y = , then use equation 1 the result is
( )
5 440 10 5x+ x = 440 55 = ⇒ x 8 =⇒x , substitute in equation 1 and solve for y
( )
8 5 = y 40 = ⇒yAnswer #9: There are 8 nickels and 40 dimes.
10. A chemistry experiment calls for a 30% sulfuric acid solution. If the lab supply room has only 50% and 20% sulfuric acid solution on hand, how much of each should be mixed to obtain twelve liters of a 30% solution.
Solution by elimination method: Let x = amount of 50% sulfuric acid solution in liters
Y = amount of 20% sulfuric acid solution in liters
( )
( )
( )
= + = + 12 3 . 0 2 . 0 5 . 0 : equation2 12 : equation1 y x y xMultiply equation 2 by 10, the result is,
36 2
Subtract equation 5 from equation 4, the result is
(
)
12 3 24 2 2 36 2 5 = = + − + = x y x y x 4 =⇒x , substitute in equation 1 and solve for y, the result is
12 4+y=
8
= ⇒y
Answer #10: They need 4 liters of 50% sulfuric acid and 8 liters of 20% sulfuric acid.