[**Q1**. There is a missing dimension which needs to be given, but is not there in the question.

The required area is found by: Area of ASEB + Area of ARS

which is = (Area of ATB - Area of STE) + Area of ARS

Area of ARS = 0.5.x 3 cm x 3 cm

The others require the missing info. Nevertheless, note that VT = EB

**Q2.**

Let Hollywood (H) has 8u guests at first.

Then, after the checkout, H has 2u left.

And Broadway (B] has 2u + 260, which is 2/5 of what B has at first.

So, B has 5u + (260/2 x 5) = 5u + 650 guests at first.

Therefore, 8u + 5u + 650 = 5200

That is, 13u = 4550

And, 1u = 350

Thus, H has 8 x 350 = **2800** **guests** at first.

**Q3.**

Day 1:

Let there be 10u adults.

So there are 10u + 120 children.

Day 2:

There are 9.5u adults. [note: it is perfectly ok to have fractions of units]

And, there are 12.5u + (1.25 x 120) = 12.5u + 150 children

Total = 2570. So that, 9.5u + 12.5u + 150 = 2570

That is, 22u = 2420

And, 1u = 110

(a) there are **1100** adults on day 1.

(b) [now that you have the no. of units for all the people on both days, surely you can compute the sum of money collected, right? ]

**Q4.**

..................$1................50cts

Before:......2p..................5p

Change:...+70..............-140..............[140 x 50cts = $70 = 70 x $1]

After:.........5u..................2u

So, for the $1 coins: 2p + 70 = 5u

For the 50 cts coins: 5p - 140 = 2u

So, 1u = 2.5p - 70

And 5u = 12.5p - 350

Therefore,

And, 12.5p - 350 = 2p + 70

1p = 40

So, there are 5 x 40 = 200 50-cts coins at first, and their value is **$****100**.

Cheerio, hope this helps.