Safiya, Sarah, and Ramli obtained marks of $1001100_2$, $121_x$, and $1103_4$ respectively in a mathematics exam. (a) Find the sum of Safiya's and Sarah's marks in base 8. (b) Find the difference between Ramli's and Safiya's marks in base 3. (c) Who obtained the highest mark in base 10?

1. Problem Description

Safiya, Sarah, and Ramli obtained marks of $1001100_2$, $121_x$, and $1103_4$ respectively in a mathematics exam.

(a) Find the sum of Safiya's and Sarah's marks in base

8. (b) Find the difference between Ramli's and Safiya's marks in base

3. (c) Who obtained the highest mark in base 10?

2. Solution Steps

(a) Sum of Safiya's and Sarah's marks in base

8. First, convert Safiya's mark ($1001100_2$) to base 10:

$1001100_2 = 1 \cdot 2^6 + 0 \cdot 2^5 + 0 \cdot 2^4 + 1 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 0 \cdot 2^0 = 64 + 8 + 4 = 76_{10}$

Convert $76_{10}$ to base 8:

$76 \div 8 = 9$ remainder $4$

$9 \div 8 = 1$ remainder $1$

$1 \div 8 = 0$ remainder $1$

So, $76_{10} = 114_8$

Next, convert Sarah's mark ($121_x$) to base

1

0. We need to figure out the base $x$ first. Since the digit 2 appears, the base $x$ must be greater than

2. However, the question doesn't give the value of $x$. Let us assume that $x=5$.

Then, $121_5 = 1 \cdot 5^2 + 2 \cdot 5^1 + 1 \cdot 5^0 = 25 + 10 + 1 = 36_{10}$

Convert $36_{10}$ to base 8:

$36 \div 8 = 4$ remainder $4$

$4 \div 8 = 0$ remainder $4$

So, $36_{10} = 44_8$

Sum the marks in base 8: $114_8 + 44_8 = 158_8$.

Since the digit 8 does not exist in base 8, we must carry over.

$114_8 + 44_8 = (1 \cdot 8^2 + 1 \cdot 8^1 + 4 \cdot 8^0) + (4 \cdot 8^1 + 4 \cdot 8^0) = (64 + 8 + 4) + (32 + 4) = 76 + 36 = 112_{10}$

Convert $112_{10}$ to base 8:

$112 \div 8 = 14$ remainder $0$

$14 \div 8 = 1$ remainder $6$

$1 \div 8 = 0$ remainder $1$

So, $112_{10} = 160_8$

(b) Difference between Ramli's and Safiya's marks in base

3. First, convert Ramli's mark ($1103_4$) to base 10:

$1103_4 = 1 \cdot 4^3 + 1 \cdot 4^2 + 0 \cdot 4^1 + 3 \cdot 4^0 = 64 + 16 + 0 + 3 = 83_{10}$

Convert $83_{10}$ to base 3:

$83 \div 3 = 27$ remainder $2$

$27 \div 3 = 9$ remainder $0$

$9 \div 3 = 3$ remainder $0$

$3 \div 3 = 1$ remainder $0$

$1 \div 3 = 0$ remainder $1$

So, $83_{10} = 10002_3$

Convert Safiya's mark ($1001100_2$) to base

3. We already know it is $76_{10}$.

Convert $76_{10}$ to base 3:

$76 \div 3 = 25$ remainder $1$

$25 \div 3 = 8$ remainder $1$

$8 \div 3 = 2$ remainder $2$

$2 \div 3 = 0$ remainder $2$

So, $76_{10} = 2211_3$

Difference: $10002_3 - 2211_3$.

$10002_3 = 1 \cdot 3^4 + 0 \cdot 3^3 + 0 \cdot 3^2 + 0 \cdot 3^1 + 2 \cdot 3^0 = 81 + 2 = 83$

$2211_3 = 2 \cdot 3^3 + 2 \cdot 3^2 + 1 \cdot 3^1 + 1 \cdot 3^0 = 54 + 18 + 3 + 1 = 76$

$83 - 76 = 7$

Convert $7_{10}$ to base 3:

$7 \div 3 = 2$ remainder $1$

$2 \div 3 = 0$ remainder $2$

So, $7_{10} = 21_3$

Alternatively, $10002_3 - 2211_3 = 21_3$

(c) Who obtained the highest mark in base 10?

Safiya: $1001100_2 = 76_{10}$

Sarah: $121_5 = 36_{10}$

Ramli: $1103_4 = 83_{10}$

Ramli has the highest mark in base

1

0.

3. Final Answer

(a) $160_8$

(b) $21_3$

(c) Ramli