Say answer is "8 hour X minute". According as proposition, the angle between the minute hand and "mark 4" of the watch is equal to the angle between the hour hand and "mark 8" of the watch.
We know in 60 minutes the minute hand makes 360 degrees (360/60=6 degrees per minute) and the hour hand makes 360/12=30 degrees (30/60=1/2 degrees per minute).
Therefore, (20-X) minutes corresponds to 6(20-X) degrees (this is the angle between the minute hand and "mark 4").
And in X minutes the hour hand makes X/2 degrees with "mark 8".
Thus, X/2=6(20-X) gives X=18 minutes 27 and 9/13 second.
So, the answer is 8 hour, 18 minutes, 27 9/13 second.
Thursday, November 26, 2009
A Question of Time
The hour and minute hands are at equal distance from the 6 hour, what time will it be exactly?
Wednesday, November 4, 2009
Sunday, October 25, 2009
Sunday, October 4, 2009
Saturday, October 3, 2009
Soalan Modul JPN kelantan paper 2 (3.1)
3.1 Yusoff started working for a company on 1 January 1995 with an initial annual salary of RM 21 000. Every January, the company increased his salary by 10% of the previous year’s salary.
Calculate
(a) his annual salary, to the nearest RM, for the year 2008,
[ 3 marks]
(b) the minimum value of n such that his annual salary in the n th year will exceed RM 70 000,
[ 2 marks]
(c) the total salary, to the nearest RM, paid to him by the company, for the years 1995 to 2008.
[ 2 marks]
Calculate
(a) his annual salary, to the nearest RM, for the year 2008,
[ 3 marks]
(b) the minimum value of n such that his annual salary in the n th year will exceed RM 70 000,
[ 2 marks]
(c) the total salary, to the nearest RM, paid to him by the company, for the years 1995 to 2008.
[ 2 marks]
Monday, September 28, 2009
Friday, September 4, 2009
Solution to Covent Garden Problem
The mixed apples were sold of at the rate of five apples for two pence. So they must would have had a multiple of five i.e. 5, 10, 15, 20, 25, 30,…, 60, 65,… etc apples.
But the minimum number of apples they could have together is 60; so that 30 would have been of Mrs. Smith's that would fetch her 10 (an integer) pence and the other 30 of Mrs. Jones's that would fetch her 15 (also an integer) pence.
When sold separately it would fetch them 10+15=25 pence altogether. But when sold together it would fetch them 60X2/5=24 pence i.e. a loss of one (25-24=1) pence.
Since they lost 7 pence altogether; they had altogether 60X7=420 apples that fetched them only 420X2/5=168 pence and they shared 84 pence each of them. But Mrs. Jones could sell her 420/2=210 apples for 210/2=105 pence so she lost "21 pence".
But the minimum number of apples they could have together is 60; so that 30 would have been of Mrs. Smith's that would fetch her 10 (an integer) pence and the other 30 of Mrs. Jones's that would fetch her 15 (also an integer) pence.
When sold separately it would fetch them 10+15=25 pence altogether. But when sold together it would fetch them 60X2/5=24 pence i.e. a loss of one (25-24=1) pence.
Since they lost 7 pence altogether; they had altogether 60X7=420 apples that fetched them only 420X2/5=168 pence and they shared 84 pence each of them. But Mrs. Jones could sell her 420/2=210 apples for 210/2=105 pence so she lost "21 pence".
Covent Garden Problem
Here is a puzzle known as the Covent Garden Problem, which appeared in London half a century ago, accompanied by the somewhat surprising assertion that it had mystified the best mathematicians of England:
Mrs. Smith and Mrs. Jones had equal number of apples but Mrs. Jones had larger fruits and was selling hers at the rate of two for a penny, while Mrs. Smith sold three of hers for a penny.
Mrs. Smith was for some reason called away and asked Mrs. Jones to dispose of her stock. Upon accepting the responsibility of disposing her friend's stock, Mrs. Jones mixed them together and sold them of at the rate of five apples for two pence.
When Mrs. Smith returned the next day the apples had all been disposed of, but when they came to divide the proceeds they found that they were just seven pence short, and it is this shortage in the apple or financial market which has disturbed the mathematical equilibrium for such a long period.
Supposing that they divided the money equally, each taking one-half, the problem is to tell just how much money Mrs. Jones lost by the unfortunate partnership?
Mrs. Smith and Mrs. Jones had equal number of apples but Mrs. Jones had larger fruits and was selling hers at the rate of two for a penny, while Mrs. Smith sold three of hers for a penny.
Mrs. Smith was for some reason called away and asked Mrs. Jones to dispose of her stock. Upon accepting the responsibility of disposing her friend's stock, Mrs. Jones mixed them together and sold them of at the rate of five apples for two pence.
When Mrs. Smith returned the next day the apples had all been disposed of, but when they came to divide the proceeds they found that they were just seven pence short, and it is this shortage in the apple or financial market which has disturbed the mathematical equilibrium for such a long period.
Supposing that they divided the money equally, each taking one-half, the problem is to tell just how much money Mrs. Jones lost by the unfortunate partnership?
Sunday, August 2, 2009
SBP Trial Paper
Any student interested in SBP ADD MAT trial paper, please send me your email in comments after August
Saturday, July 11, 2009
Wednesday, July 1, 2009
Monday, May 25, 2009
Sunday, May 24, 2009
Wednesday, April 8, 2009
Monday, April 6, 2009
Simultaneous Equation
Chapter 4 – Simultaneous Equations
In this chapter, students will be taught to solve simultaneous equations using the substitution method and solve simultaneous equation involving real-life situations.
1. Solve simultaneous equations in two unknown: one linear equation and one non-linear equation.
1.1 Solve simultaneous equations using the substitution method
The following simultaneous equation consists of a linear equation and a non-linear equation (up to second degree only).
Solving the equations means finding 2 sets of solutions that satisfy both the equation given, a linear equation and a non-linear equation. A general method of solution is to make one of the unknowns the subject of the linear equation, then substitute this value in the non-linear equation.From (i) we obtain y = 2x – 3 (iii)
Substitute y= 2x - 3 into (ii)
In this chapter, students will be taught to solve simultaneous equations using the substitution method and solve simultaneous equation involving real-life situations.
1. Solve simultaneous equations in two unknown: one linear equation and one non-linear equation.
1.1 Solve simultaneous equations using the substitution method
The following simultaneous equation consists of a linear equation and a non-linear equation (up to second degree only).
Solving the equations means finding 2 sets of solutions that satisfy both the equation given, a linear equation and a non-linear equation. A general method of solution is to make one of the unknowns the subject of the linear equation, then substitute this value in the non-linear equation.From (i) we obtain y = 2x – 3 (iii)
Substitute y= 2x - 3 into (ii)
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