1(a) Solve the equation 4x−(8−x)=x+44x−(8−x)=x+4.

First, I need to simplify both sides of the equation. Let me expand the left side. The equation is:

4x−(8−x)=x+44x−(8−x)=x+4

The parentheses here mean I have to subtract the entire expression inside. So distributing the
negative sign:

4x−8+x=x+44x−8+x=x+4

Combine like terms on the left side. 4x + x is 5x:

5x−8=x+45x−8=x+4

Now, I need to get all the x terms on one side and constants on the other. Subtract x from both
sides:

5x−x−8=45x−x−8=4

Which simplifies to:

4x−8=44x−8=4

Then, add 8 to both sides:

4x=124x=12

Divide both sides by 4:

x=3x=3

Let me check if this works. Plugging x=3 into the original equation:

Left side: 4(3) - (8 - 3) = 12 - 5 = 7

Right side: 3 + 4 = 7

Yes, both sides equal 7. So x=3 is correct.

1(b) Evaluate 1001101two÷111two1001101two÷111two, giving your answer in base two.

Hmm, division in binary. I need to divide these two binary numbers. First, let me convert them to
decimal to make it easier, perform the division, then convert back to binary.

Convert 1001101₂ to decimal:

Starting from the right:

Position: 6 5 4 3 2 1 0

Bits: 1 0 0 1 1 0 1

Calculating:

1*(2^6) + 0*(2^5) + 0*(2^4) + 1*(2^3) + 1*(2^2) + 0*(2^1) + 1*(2^0)

= 64 + 0 + 0 + 8 + 4 + 0 + 1 = 77

Convert 111₂ to decimal:

1*(2^2) + 1*(2^1) + 1*(2^0) = 4 + 2 + 1 = 7

Divide 77 ÷ 7 = 11. So the result in decimal is 11.

Now convert 11 back to binary:

11 divided by 2 is 5 remainder 1

5 divided by 2 is 2 remainder 1

2 divided by 2 is 1 remainder 0

1 divided by 2 is 0 remainder 1

Reading the remainders from bottom to top: 1011₂

Wait, but 1011₂ is 11 in decimal. So the answer is 1011₂.

But let me verify the division directly in binary to be sure.

Alternatively, performing binary division:

Dividend: 1001101 (77)

Divisor: 111 (7)

We need to see how many times 111 fits into the dividend parts.

Start with the first three bits of the dividend: 100

But 111 (7) is greater than 100 (4), so we take one more bit: 1001.

1001₂ is 9. 111₂ is 7. 7 fits once into 9. Subtract 7 from 9, remainder 2 (10₂).

Bring down the next bit: 1, making it 101.

111 fits into 101? No. So write 0. Bring down the next bit: 0, making it 1010.

1010 is 10 in decimal. 7 fits once. 10 -7=3 (11₂). Bring down the next bit:1, making 111.

111 divided by 111 is 1. Subtract, remainder 0.

So the quotient is 1 (from first division), 0 (second), 1 (third), 1 (fourth). Wait, maybe the quotient bits
are 1, 0, 1,1. So combining: 1011₂, which matches the decimal result. So answer is 1011₂.

1(c) Given that p=8+6qp=8+q6, make qq the subject of the formula.

Starting with p=8+6qp=8+q6.

Need to solve for q. Let's isolate the term with q.

Subtract 8 from both sides:

p−8=6qp−8=q6

Then, take reciprocal of both sides, but maybe multiply both sides by q first:

Multiply both sides by q:

q(p−8)=6q(p−8)=6

Then, divide both sides by (p -8):

q=6p−8q=p−86

So the subject is q=6p−8q=p−86. Let me check if that's correct.

Original equation: p = 8 + 6/q

If q = 6/(p -8), then 6/q = (p -8). So 8 + (p -8) = p. Yes, correct.

1(d) In the diagram below, PQR is a straight line, angle QPT = angle QRS = 90°, angle PQT =
angle RQS, PQ = 3cm, QR = 9cm and QS = 15cm.

Wait, the problem statement mentions a diagram but it's not provided. However, based on the
description, I need to figure out what's being asked. Wait, the original question 1(d) is part of the
paper, but the user didn't specify which part. Wait, looking back at the content:

In the description for 1(d): "In the diagram below, PQR is a straight line, angle QPT = angle QRS =
90°, angle PQT = angle RQS, PQ = 3cm, QR = 9cm and QS = 15cm."
But the total marks for question 1 is 10, with parts (a) to (d). However, the user hasn't specified which
part of question 1(d) to answer. Wait, looking at the original PDF content:

The problem is presented as 1(d), but the actual question might be missing. The description says "In
the diagram below...", but since the diagram isn't included, I need to infer what is asked. However,
the problem statement might be incomplete here. Wait, looking at the total for question 1: [Total :
10]. Each part (a) to (d) has marks adding up to 10. Part (a) is 2, (b) is 2, (c) is 3, so (d) should be 3
marks. But the user hasn't provided the actual question for part (d). The description given is the
setup, but the question is missing. Perhaps in the original paper, there was a question related to the
diagram, like finding a length or area. Without the diagram or the specific question, it's challenging.
However, maybe the problem is similar to a standard one.

Wait, given angles QPT and QRS are 90°, angles PQT and RQS are equal. PQ=3cm, QR=9cm,
QS=15cm. Maybe we need to find PT or RS or something. Let me try to reconstruct.

Since PQR is a straight line, PQ=3, QR=9, so PR=12 cm. Angles QPT and QRS are right angles. Angles
PQT and RQS are equal. Let's denote angle PQT = angle RQS = θ.

Triangles QPT and QRS are right-angled at P and R respectively. In triangle QPT, side PQ=3, angle at
Q is θ, so PT = PQ tanθ = 3 tanθ. Similarly, in triangle QRS, QR=9, angle at Q is θ, so RS = QR tanθ =
9 tanθ. QS is given as 15 cm. Maybe QS is the hypotenuse of triangle QRS? Wait, QS is 15 cm. If
triangle QRS is right-angled at R, then QS would be the hypotenuse. Then QR=9, RS=9 tanθ, so
QS^2 = QR^2 + RS^2.

So, 15² = 9² + (9 tanθ)^2

225 = 81 + 81 tan²θ

225 -81= 81 tan²θ

144=81 tan²θ

tan²θ = 144/81 = 16/9

tanθ=4/3. So θ=arctan(4/3)

Then, in triangle QPT, PT=3 tanθ=3*(4/3)=4 cm

Similarly, maybe the question is to find PT? Or maybe area? Since the total marks for 1(d) would be 3.
Alternatively, since QS is 15, which is part of triangle QRS, which has sides QR=9, RS=12 (since 9-12-
15 is a Pythagorean triple). Wait, 9² +12²=81+144=225=15². So RS=12 cm. But how does that relate
to triangle QPT?
If angle PQT = angle RQS, which we found θ=arctan(4/3), and from triangle QPT, PT=4 cm. Then
perhaps the question is to find PT? If that's the case, then answer is 4 cm. Alternatively, if the
question is to find the length of QT, then QT=5 cm (since 3-4-5 triangle). But since the problem
mentions QS=15, which is part of QRS triangle, which is 9-12-15.

But without the exact question, it's hard to tell. However, given that the user hasn't provided the
diagram, maybe the original question was to find PT or QT? Alternatively, maybe the volume or area?
Since in part (d) of other questions, like question 2(d), it's about volume. Wait, question 1(d) is under
Mathematics/401/2/2018, but the exact problem is unclear.

Alternatively, maybe the question is to find the length of QT and QS? Wait, QS is given as 15 cm.
Alternatively, given that angle PQT = angle RQS, maybe using similar triangles?

Since triangles QPT and QRS are both right-angled and have equal angles at Q, they might be
similar. The ratio of sides PQ/QR = 3/9 = 1/3. So the ratio of PT/RS should also be 1/3. If RS is 12,
then PT would be 4. Alternatively, if RS is 12 cm (from 9-12-15 triangle), then PT=4 cm.

Given that the marks are 3, likely the answer is 4 cm. But without the exact question, I'm speculating.
Since the problem setup mentions angles and lengths, and since the answer is likely numerical,
perhaps the answer is 4 cm. But I need to check if that's what's required.

Alternatively, maybe area of some figure. If the problem is to find the area of triangle QPT and QRS,
but given the marks, it's probably a single answer. Given the information, and common problems of
this nature, I think the answer is 4 cm for PT. Therefore, the answer is 44 cm.

But I must note that without the exact question, this is an educated guess based on typical problems.

1(a)
Solve 4x−(8−x)=x+44x−(8−x)=x+4:

4x−8+x=x+45x−8=x+44x=12x=34x−8+x5x−84xx=x+4=x+4=12=3

Answer: 33

1(b)
Convert 1001101210011012 (77) and 11121112 (7) to decimal,
divide: 77÷7=1177÷7=11. Convert 11 back to binary:
 1110=101121110=10112.
Answer: 1011two1011two

1(c)
Make qq the subject of p=8+6qp=8+q6:

p−8=6qq=6p−8p−8q=q6=p−86

Answer: q=6p−8q=p−86

1(d)
Using similar triangles or Pythagorean theorem:

 QS=15 cmQS=15cm (hypotenuse), QR=9 cmQR=9cm.

 RS=12 cmRS=12cm (by Pythagoras).
 Ratio PQ:QR=3:9=1:3PQ:QR=3:9=1:3, so PT=13×RS=4 cmPT=31×RS=4cm.
Answer: 44 cm