Topic : Proportion Method
Question : Nicole enlarged the size of rectangle to a height of 28 in. What is the new width if it was originally 6 inch wide and 7 inch tall?
a). 24in. b). 112 in. c) 2in. d) 7in.
Solution :
Forming the proportion :
6 : 7 = X :28
Using law of proportion, that product of extremes is equal to the product of means
6 x 28 = 7 x X
X = 28 x 6 /7
X = 168 / 7
X = 24
So, the width is 24 inch.
Friday, April 10
Monday, April 6
Question on Percentage of Sugar Solution
Topic : Word Problem on Percentage
Question : How many gallons of 15% sugar solution must be mixed with 5 gallons of 40% sugar solution to make 30% sugar solution?
Solution :
%sugar ---gallons
15% -------- x
40% -------- 5
30 % ------- x+5
so, 15% of x+ 40% of 5 = (x+5) 30%
0.15x+ 2 = 0.3x+15
x= 0.5/0.15
x= 10/3
10/3 gallons of 15% sugar solution is mixed with 5 gallons of 40% sugar solution to
make 30% sugar solution.
Question : How many gallons of 15% sugar solution must be mixed with 5 gallons of 40% sugar solution to make 30% sugar solution?
Solution :
%sugar ---gallons
15% -------- x
40% -------- 5
30 % ------- x+5
so, 15% of x+ 40% of 5 = (x+5) 30%
0.15x+ 2 = 0.3x+15
x= 0.5/0.15
x= 10/3
10/3 gallons of 15% sugar solution is mixed with 5 gallons of 40% sugar solution to
make 30% sugar solution.
Thursday, April 2
Thursday, March 26
Monday, March 23
Wednesday, March 18
Upper Bound and Lower Bound of Polynomials
Friday, March 13
Problem to Find the Unknown values of Geometrical Figure
Topic : Square Based Triangle
Question : The Square based pyramid depicted below has averticle height of 13cm. K is the mid point of CD and angle AKL =50 degrees. To the nearest tenth of degree, what is the measure of angle ACK?
Solution :

Length of Square be x
Tan 50º = 1.19175
= 13 / (x/2)
x = 21.81659041
x = 21.82
Sin 50º = (x/2) / AK
AK = (x/2) / 0.766
AK = 14.24
Tan ∟AKC = AK /(x/2) = 14.24/ 10.91 = 1.31
∟AKC = tanˉ¹(1.31)
∟AKC = 53º
Question : The Square based pyramid depicted below has averticle height of 13cm. K is the mid point of CD and angle AKL =50 degrees. To the nearest tenth of degree, what is the measure of angle ACK?
Solution :
Length of Square be x
Tan 50º = 1.19175
= 13 / (x/2)
x = 21.81659041
x = 21.82
Sin 50º = (x/2) / AK
AK = (x/2) / 0.766
AK = 14.24
Tan ∟AKC = AK /(x/2) = 14.24/ 10.91 = 1.31
∟AKC = tanˉ¹(1.31)
∟AKC = 53º
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