Tutor MathsGuru: Ask me for your burning Maths questions!

[kwcllf]

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[babyluvv]

Hi MathsGuru,


Pls help.

1) Box A contained 134 marbles and Box B contained 18 marbles at first. An equal number of marbles was added to each box. As a result of this, the number of marbles in A to the number of marbles in B became 5 : 1.
a) How many marbles were added to each box?
b) Next, some marbles were transferred from Box A to Box B so that there was an equal number of marbles in each box. How many marbles were there in each box in the end?

2) 7/12 of the guests in a party are adults. The ratio of the number of boys to the number of girls is 6 : 9. respectively. There are 144 more adults than girls. What is the total number of guests at the party?

3) In an auditorium, the ratio of the number of competitors to the number of non-competitors is 8 : 5. The ratio of the number of male competitors to the number of female competitors is 7 : 4. Given that 3/5 of the non-competitors are males and there are 32 female competitors, how many males and how many females are there in the auditorium?

Thanks,
Cheryl

[blessedami]

babyluvv:

Hi MathsGuru,


Pls help.

1) Box A contained 134 marbles and Box B contained 18 marbles at first. An equal number of marbles was added to each box. As a result of this, the number of marbles in A to the number of marbles in B became 5 : 1.
a) How many marbles were added to each box?
b) Next, some marbles were transferred from Box A to Box B so that there was an equal number of marbles in each box. How many marbles were there in each box in the end?

2) 7/12 of the guests in a party are adults. The ratio of the number of boys to the number of girls is 6 : 9. respectively. There are 144 more adults than girls. What is the total number of guests at the party?

3) In an auditorium, the ratio of the number of competitors to the number of non-competitors is 8 : 5. The ratio of the number of male competitors to the number of female competitors is 7 : 4. Given that 3/5 of the non-competitors are males and there are 32 female competitors, how many males and how many females are there in the auditorium?

Thanks,
Cheryl

Hi,

For the first question, since an equal number of marbles were added to each box, the difference in the beginning and at the end is the same. The difference at the end is;

5 units - 1 unit = 4 units

In the beginning, it was

134-18= 116

So, 4 units is 116, because of the constant difference. Thus 1 unit is
116/4 = 29

1 unit(29) is the number of marbles box B had in the end and at first, it had 18 marbles. So the number of marbles that were added is 11 (part a)

For part b, the total number of marbles in both boxes is still the same since they are not shifted out of those two boxes but still in either one of them. The total number of marbles is 6 units ( 5 units + 1 unit)

6 units --> 6 x 29 = 174

If the number of marbles in each box was the same, that means that the 174 marbles were equally divided into the 2 boxes.

So, there are 87 marbles in each box.

For question 2, 5/12( 1 - 7/12) of the people at the party are children, which is 5 units.

The 5 units, however, is being 15 parts when being further divided into boys and girls. Thus;

5 units = 15 parts
7 units = 21 parts

The number of girls is 9 parts. The difference in adults and girls in parts is 12 parts (21 - 9).

so, 12 parts --> 144
1 part--> 144/12 = 12
total number of guests: 36 parts --> 36 x 12 = 432

[blessedami]

babyluvv:

Hi MathsGuru,


Pls help.

1) Box A contained 134 marbles and Box B contained 18 marbles at first. An equal number of marbles was added to each box. As a result of this, the number of marbles in A to the number of marbles in B became 5 : 1.
a) How many marbles were added to each box?
b) Next, some marbles were transferred from Box A to Box B so that there was an equal number of marbles in each box. How many marbles were there in each box in the end?

2) 7/12 of the guests in a party are adults. The ratio of the number of boys to the number of girls is 6 : 9. respectively. There are 144 more adults than girls. What is the total number of guests at the party?

3) In an auditorium, the ratio of the number of competitors to the number of non-competitors is 8 : 5. The ratio of the number of male competitors to the number of female competitors is 7 : 4. Given that 3/5 of the non-competitors are males and there are 32 female competitors, how many males and how many females are there in the auditorium?

Thanks,
Cheryl

For the third question,

It is stated that the number of female competitors is 4 parts and is 32.

4 parts --> 32
1 part --> 32/4 = 8
Male Competitors: 7 parts --> 7 x 8 = 56

total number of competitors: 4 parts + 7 parts = 11 parts
11 parts --> 11 x 8 = 88

They have also mentioned that the ratio of competitors to non-competitors is 8:5

This shows that 8 units is 88
1 unit --> 88/8 = 11
5 units --> 5 x 11 = 55 ( non- competitors)

3/5 x 55 = 33 ( male non-competitors)
55-33 = 22 ( female non-competitors)

So, the number of males in the stadium is 56 + 33 = 89
No. of females in the stadium is 32 + 22 = 54

Regards,
Blessedami

[Vanilla Cake]

http://postimage.org/image/32rskcgh0/

Pls assist to provide worked solution for this problem sum.
Thks in advance.

[Dharma]

Vanilla Cake:

http://postimage.org/image/32rskcgh0/

Pls assist to provide worked solution for this problem sum.
Thks in advance.

Hi Vanilla Cake,


http://postimage.org/image/ttml6zd0/

[wahwah]

Hi all,

Can someone help me with the following question? Many thanks in advance.

One thousand "2007" are joined together to form a new number 200720072007…20072007(repeated 1000times).
The first m digits and the last n digits of this new number are then removed such that the sum of digits of the resulting number is 7999. Given that both m and n are at least 4, find the greatest value of m-n.

(1) 436
(2) 438
(3) 439
(4) 443
(5) None one the above

[Vanilla Cake]

http://postimage.org/image/vm18njpg/

http://postimage.org/image/18g81qxw/
Thks in advance for your help.

[Dharma]

Vanilla Cake:

http://postimage.org/image/vm18njpg/

http://postimage.org/image/18g81qxw/
Thks in advance for your help.

Dear Vanilla Cake,

Q3.

Area of AOD : Area of COD = OA : OC = 2 : 3
Area of COD : Area of BOC = OD : OB = 3 : 2

Area of AOD/ Area of BOC = 2/3 x 3/2 = 1

Therefore, area of AOD : Area of BOC = 1 : 1

Q28

If you draw the symmetrical lines of the shifted vertical and horizontal diameters to the vertical and horizontal diameters that pass through origin O.

You will find the shaded and unshaded portions cancel off each other leaving a rectangular shaded portion of length 6cm and width 4cm.

Nett area = 6cm x 4cm = 24cm2

[babyluvv]

Hi blessedami,


Thank you for helping!!
:lol: :lol: :lol: :lol: