IB myp 4-5 MATHEMATICS – Practice Questions- All Topics
Topic :Number–Number systems notation
Topic :Number- Weightage : 21 %
All Questions for Topic : Absolute Values,Representing and Solving Inequalities,Including compound and double inequalities,Irrational numbers,Surds, roots and radicals, including simplifying,Standard form (scientific notation),Laws of exponents, including integer and negative exponents,Number systems notation,Direct and Inverse Proportion,Number Sequence(Prediction ,Description)
Question (a) : 2 marks
A three-digit number can be written in terms of sum of multiples of its digits. For example, 437 can be written in the format shown below.
$\rm{X}$ is a three-digit number $a b c$. Write down $\rm{X}$ and $\rm{X}^{\prime}$ as a sum of multiples of their digits.
▶️Answer/Explanation
Ans:
To represent a three-digit number $X = abc$ as a sum of multiples of its digits, we can use the place value notation. Each digit is multiplied by the corresponding power of 10 to represent its place value.
For the number $X = abc$, we can write it as:
$X = (100 \times a) + (10 \times b) + (1 \times c)$
Similarly, the inverse of $X$, denoted as $X’ = cba$, can be written as:
$X’ = (100 \times c) + (10 \times b) + (1 \times a)$
Let’s take an example to illustrate this. Consider the number $X = 437$:
$X = (100 \times 4) + (10 \times 3) + (1 \times 7) = 400 + 30 + 7 = 437$
The inverse of $X$, denoted as $X’ = 734$, can be written as:
$X’ = (100 \times 7) + (10 \times 3) + (1 \times 4) = 700 + 30 + 4 = 734$
Therefore, for $X = 437$, we have:
$X = (100 \times 4) + (10 \times 3) + (1 \times 7)$
$X’ = (100 \times 7) + (10 \times 3) + (1 \times 4)$
Note that this representation holds for any three-digit number $X = abc$, where $a$, $b$, and $c$ represent the digits of the number.
Question (b) : 2 marks
Using your answer from part (b), determine the difference $D$ in terms of $a$ and $c$.
▶️Answer/Explanation
Ans:
To find the difference $D$ between $X’$ and $X$ in terms of $a$ and $c$, we subtract $X$ from $X’$ and take the absolute value of the result.
From the previous answer, we have:
$X = (100 \times a) + (10 \times b) + (1 \times c)$
$X’ = (100 \times c) + (10 \times b) + (1 \times a)$
Substituting the values, we get:
$X’ – X = [(100 \times c) + (10 \times b) + (1 \times a)] – [(100 \times a) + (10 \times b) + (1 \times c)]$
Simplifying this expression, we have:
$X’ – X = 100c + 10b + a – 100a – 10b – c$
Combining like terms, we get:
$X’ – X = -99a + 99c$
To find the absolute value of $X’ – X$, we take the absolute value of the expression $-99a + 99c$. Therefore, the difference $D$ is given by:
$D = |X’ – X| = |(-99a + 99c)| = |-99(a – c)| = 99|a – c|$
Hence, the difference $D$ in terms of $a$ and $c$ is equal to $99|a – c|$.