MATHEMATICS
Direction: Observe how the exercises in the following
problems are solved. Then try to see if you can get the formula called for.
Problem 1)
Find how many pieces of cartolina can be bought
for P17.00 at the rate of 2 for P4.25?
a. 8
b. 15
c. 16
d. 75
Solution: P17.00 ÷ P4.25 = 4 x 2 = 8
Problem 2)
In a company with 270 employees, out of every 9
are required to work each Sunday. How many employees are required to work
Sunday?
a. 135
b. 60
c. 3
d. 18
Solution: 270 ÷ 9 = 30 No of 9’s
30 x
2 = 60
Problem 3)
Mrs. Salve made a down payment on a lot and had
a balance of P7, 200. She pays for this over a period of 2 years by making
equal payments every other month. How much is the monthly installment?
a. P600
b. P300
c. P36
d. P3
e. None
of these
Solution: 24 months ÷ 2 = 12
P7,
200 ÷ 12 = P600
Problem 4)
Mr. Ejercita borrowed P200 at 6% per annum for
one (1) month in order to pay a bill of P200 on which he was allowed a discount
of 2 ½ %. The net result of these two transactions was:
a. a
gain of P4
b. a
gain of P5
c. a
gain of P8
d. a
loss of P7
e. none
of these
Solution: P200 x .06 = P12
P12 ÷ 12 = P1 interest for 1 month
P 200 x 2 ½ % = P5
P200 – P1 = P199 borrowed
P200 – P1 = 195 amount par
P 4
gained
Problem 5)
A staff of Basic and Co. of 20 clerks can finish
a certain job in 6 days. How many more clerks should be put on the job in order
to complete it in 4 working days?
a. 7
b. 13
c. 27
d. 30
e. None
of these
Solution: Let x = No. of men to finish in 4 days
20 : x =
4 : 6
4x
= 120
X = 30 men to finish in 4 days
30 – 20 = 10 men to be added
Problem 6)
A Nacida-registered factory employs 40 men at
P18.00 a day, and half as many women, each at half as much pay per day. If half
of the women were given the factory have to pay in a 5 – day week?
a. P300
b. P400
c. P500
d. P600
e. None
of these
Solution: 40 men
at P18.00 per day = P750.00
20 women at P9.00 per day = 180.00
P930.00
x 5
days
P4,650.00
40 men at P18.00 per day = P750.00
10 women at P9.00 / day = 90.00
10 women at P16.00 / day = 160.00
P1,000.00
X 5
P5,000.00
P5,000.00
-4, 650.00
P
350.00 additional amount to be paid in 5 days
Problem 7)
A wallet contains an equal of ten, twenty, and
fifty-centavos coins, and the total value of the coins in the bag is P14.40.
All the coins of the greatest value are worth how much more than the value of
the other coins?
a. P3.60
b. P4.20
c. P5.40
d. P9.00
e. None
of these
Solution: Let x = No. of coins of each denomination
10x
+ 20x + 50x = 1,440
80x = 1,440
X
= 18 no. of coins of each demonstration
P.10
x 18 = P1.80
.20 x 18 =
3.60
.50 x 18 =
9.00
P14.40
1.80 P9.00 value of all coins
of greatest value
+3.60 -5.40 value of all
other coins
P3.60 amount more than other coins
Problem 8)
A poster twice as long as it is high is fastened
to a wall by 36 thumb tacks placed at 3-inches equal intervals on centers along
its perimeter. How long is the poster?
a. 17.5
b. 18
c. 36
d. 35
e. None
of these
Solution:
Let x = the width
2x = the length
P = 2(1 + w)
36 x 3
= 108 inches
2 (x
+ 2x) = 108
2x
+ 4x = 108
6x
= 108
X = 18 inches (width)
2x
= 36 inches (length)
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