Challenging questions – Page 4 – JC A Level H2 Math Tuition Singapore

Challenging Maclaurin’s series problem 2

Challenging Probability problem 1

Problem

A bag contains 10 orange-flavoured, 14 strawberry-flavoured and 16 cherry-flavoured sweets which are of identical shapes and sizes. Benny selects a sweet at random from the bag. If it is not cherry-flavoured, he replaces it and selects another sweet at random. He repeats the process until he obtain a cherry-flavoured sweet. Calculate the probability that

i) the first sweet selected is strawberry-flavoured and the fourth sweet is orange-flavoured;

ii) he selects an even number of sweets

Solution

Challenging Normal Distribution problem 1

Challenging Functions problem 2

Challenging differentiation problem

Use implicit differentiation to differentiate the following with respect to x:

y=x(x+1)(x+2)(x+3)(x+4)(x+5)

Solution

ln y= lnx+ln(x+1)+ln(x+2)+ln(x+3)+ln(x+4)+ln(x+5)

Challenging Permutation question 4

In how many ways can 9 balls of which 4 are red, 4 are white and 1 black be arranged in a line so that no red ball is next to the black?

Answer: 175

Solution

Case 1: 1st ball is black in the whole row of 9 balls

Then the 2nd ball must be white. The rest of the 4 red and 3 white balls can be arranged in = 35 ways

Case 2: last ball is black

Then the 2nd last ball must be white. The rest of the 4 red and 3 white balls can be arranged in = 35 ways

Case 3: Black ball is not the 1st or last ball

Then the black ball must be between 2 white balls. Consider WBW as one unit. Together with 4 red balls and 2 white balls, these can arranged in = 105 ways

Using the addition principle, total number of ways = 35+35+105= 175

Challenging System of Linear Equations problem 1

At Nuts supermarket, a discount is offered on Almonds, Cashew and Walnuts if more than a certain weight of it is bought. On a particular grocery shopping trip, King Kong bought a total of 12 kg of nuts as follows:

Types of Nut	Price per kg ($)	Discount offered
Almond	14	20% for more than 6 kg bought
Cashew	9	15% for more than 3 kg bought
Walnuts	11	10% for more than 2 kg bought

King Kong bought at least 3.5 kg of each type of nut and spent $124.03 after an overall discount of $10.17. Find the weight of each type of nut King Kong bought from the supermarket.

Solution

We know that discount is obtained from the purchase of cashew and walnuts, since the amount bought is at least 3.5 kg of each type. However, we do not know whether any discount is obtained for the purchase of almonds.

Therefore, we make the assumption that the amount of almonds bought is less than or equal to 6kg, and hence no discount. If the solution turns out to be between 3.5 to 6kg inclusive, we accept the solution. Else we reject and change the assumption to almonds bought is greater than 6kg, reformulate and solve the equations again.

Let a, b and c be the number of kg of almonds, cashew and walnuts bought respectively.

Therefore a+b+c= 12 (Eqn 1)

Assume

14a+b(0.85×9)+c(0.9×11)=124.03

14a+7.65b+9.9c=124.03 (Eqn 2)

0.15x9b+0.1x11c=10.17

1.35b+1.1c= 10.17 (Eqn 3)

Solving the 3 equations with GC,

a= 3.8 , b=4.6, c=3.6

Since a is between 3.5 to 6, our assumption and hence solution is valid.

King Kong bought 3.8 kg of almonds, 4.6 kg of cashew and 3.6 kg of walnuts.

Challenging Inequality problem 2

Given that k>0, solve the inequality

Express your answer in terms of k.

Solution

First, we determine the roots of

Then we need to determine whether the roots are bigger or smaller than k, so we can sketch the number line.

Since k> 0

Challenging Permutation question 3

There are a total of 20 amusement rides in a theme park. A child insists on trying at least 5 of the amusement rides. Calculate the number of ways in which this can be done.

Answer: 1042380

Solution

Method 1

20 choose 5 + 20 choose 6 +… 20 choose 20

Using TI 84 OS (2.55)

sum (seq(20 nCr X, X, 5, 20, 1)

= 1042380

Method 2

2^20- (20 choose 0+ 20 choose 1+20 choose 2+20 choose 3+ 20 choose 4)

=1042380