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 By: Kuro (offline)  Friday, September 13 2013 @ 02:30 AM CDT (Read 717 times)
Kuro

Q1. Pls see attachment.
Ans: 43.5

Q2. Hotel Hollywoods and Hotel Broadway had a total of 5200 guests. After 3/4 of the guests in Hotel Hollywood and 3/5 of the guests in Hotel Broadway checked out, Hotel Broadway had 260 more guests than Hotel Hollywood. How many guests did Hotel Hollywood have at first? (Ans: 2800)

Q3. NY Primary School organised a 2-day Charity Funfair. Admission tickets were sold to every visitor at a cost of \$40.On the first day,120 more children visited the funfair than adults.The number of children on the second day was 25% more than the number of children on the first day.The number of adults on the second day was 5% less than the number of adults on the first day.There were a total of 2570 people at the funfair on the second day.
(a)Find the number of adults at the funfair on the first day. (Ans:1100)
(b)What was the total sum of money collected from the sale of the admission tickets for both days? (Ans:195600)

Q4. Hanson had some one-dollar coins and some fifty-cent coins.The ratio of the number of one-dollar coins to the number of fifty-cent coins he had was 2 : 5.Hanson took 140 fifty-cent coins to the bank and changed these fifty-cent coins for the same value of one-dollar coins.In the end, the number of one-dollar coins to the number of fifty-cent coins he had became 5 : 2.Find the value of the fifty-cent coins Hanson had at first. (Ans: \$100)

Junior

Registered: 12/31/06
Posts: 27

 By: jo sarah (offline)  Friday, September 13 2013 @ 09:03 AM CDT
jo sarah

[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.

Regular Member

Registered: 03/20/12
Posts: 111

 By: echeewh (offline)  Sunday, September 15 2013 @ 10:54 PM CDT
echeewh

Hi <Kuro>

This is it .

Attached pls find the worked solution for Q1.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 625

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