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 By: sundarshoba (offline)  Friday, October 03 2014 @ 09:12 PM CDT (Read 13386 times)
sundarshoba

Please assist the below question, Sandy has some dark and white chocolates in a container. If she eats 3 dark chocolates, the ratio of number of dark to white chocolates will become 5:4. If she eats 3 white chocolates, the ratio of dark to white chocolates will be 7:5. What is the original number of dark chocolates?, Thanks

Chatty

Registered: 12/31/06
Posts: 53

 By: echeewh (offline)  Saturday, October 04 2014 @ 09:00 PM CDT
echeewh

Hey <sundarshoba>

Following is my worked solution :

Q1.

<If eats 3 dark>
D : W
5 : 4

Apply <work backwards> , we have ...

<before>
D: 5p + 3
W: 4p

<if eats 3 white>
D : W
7 : 5

Apply <work backwards> , we have ...

<before>
D: 7u
W: 5u + 3

Hence, we have ...

D:
5p + 3 = 7u
7u - 5p = 3 -- (1)

W:
4p = 5u + 3
4p - 5u = 3 -- (2)

7u - 5p = 4p - 5u
12u --> 9p
1u --> (9p)/12 = 3p/4 -- (3)

Replace / Sub (3) in (1), we have ...

[7 × (3p/4)] - 5p = 3
(21p/4) - 5p = 3
(1p/4) = 3
p --> 12

D (at first) = 5p + 3
5p + 3 --> (5 × 12) + 3 = 60 + 3
= 63

========

Trust this helps.

Do let me know again if this is different from your <answerkey> or if there's further clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: listener (offline)  Thursday, October 09 2014 @ 10:12 PM CDT
listener

Case 1: Eats 3 dark chocolate
Dark : White : Total
5 : 4 : 9
20 : 16 : 36

Case 2: Eats 3 white chocolate
Dark : White : Total
7 : 5 : 12
21 : 15 : 36

After eating, total number of both chocolates is the same, therefore, use lowest common multiple 36. Compare case 1 and case 2

1 unit --> 3 chocolates
21 units --> 3 x 21 = 63

There are 63 dark chocolate at first.

Regards,
Chit

Newbie

Registered: 12/31/06
Posts: 1

 By: JONAANDR (offline)  Sunday, November 09 2014 @ 10:25 PM CST
JONAANDR

Hi, I need to help. Thanks!

1. When 2 new boys each of height 1.62m joined the group, the average height became 1.67m.
How many pupils were there in the group in the end?

2. Albert, Ben and Cass each had some marbles. Albert gave 1/3 of his marbles to Cass. Ben also gave 1/3 of his marbles to Cass. Cass put all the marbles together. She then gave 1/5 of her marbles to Ben.
As a result, Albert and Cass had 20 marbles each and Ben had 15 marbles at the end. How many marbles did each of them have at first?

3. Mrs Tay baked 200 more walnut cookies than chocolate cookies. She sold 75% of the walnut cookies and 40% of the chocolate cookies. At the end of the day, there were 43 more walnut cookies than chocolate cookies. How many cookies were there at first?

4. Box A, B and C contained some beads. The ratio of the number of beads in Box A to the total number of beads was 2:9. She used 1/3 of the beads in Box B and transferred 1/4 of the beads from Box A to Box C. The ratio of the number of beads left in Box B to the number of beads left in Box C was 4:9. If there were 846 beads in Box B at first, what was the total number of beads left in the end?

Newbie

Registered: 02/27/14
Posts: 6

 By: jo sarah (offline)  Monday, November 10 2014 @ 07:59 AM CST
jo sarah

Hi, "listerner" made a very good observation: since in each scenario, the same no. (3) were eaten, the total remains the same!

Now, let me quickly give some pointers for the new set:

1. When 2 new boys each of height 1.62m joined the group, the average height became 1.67m.
How many pupils were there in the group in the end?

Wonder if there is some information missing in this...

2. Albert, Ben and Cass each had some marbles. Albert gave 1/3 of his marbles to Cass. Ben also gave 1/3 of his marbles to Cass. Cass put all the marbles together. She then gave 1/5 of her marbles to Ben.
As a result, Albert and Cass had 20 marbles each and Ben had 15 marbles at the end. How many marbles did each of them have at first?

Solution:
Using only letters A, B, C for each person; and in this case, since we know what each has in the end, we work backwards starting from the no. in the end:

...A..........B..........C
...20........15........20 .........................(this came after C gave 1/5 to B; meaning C is left with 4/5, ie, the 20 is 4/5 => 1/5 is 5)
...............-5.........+5 ........................(so B return 5 to C: we are working back towards the beginning)
----------------------------
...20........10........25 ........................ (this is after B gave 1/3 to C; meaning B is left with 2/3, ie, the 10 is 2/3 => 1/3 is 5)
..............+5.........-5 .........................(here, C returns 5 to B )
---------------------------
...20........15........20 ........................(this from A giving 1/3 to C; ie, A is left with 2/3, making the 1/3 to be 10)
.+10...................-10 .........................(C returns 10 to A)
--------------------------
...30........15........10

That is, A has 30 marbles, B has 15 and C has 10 marbles at first.

[no need answer key; just start with these and follow the question to see if you get the no. that they each has at the end. If so, you are right!]

3. Mrs Tay baked 200 more walnut cookies than chocolate cookies. She sold 75% of the walnut cookies and 40% of the chocolate cookies. At the end of the day, there were 43 more walnut cookies than chocolate cookies. How many cookies were there at first?

Solution:
Note: 75% = 3/4; and 40% = 2/5. So in order to have a no. that divides by 4 and 5, we choose 20u in the following:
..................walnut....................choc
at 1st:......200 + 20u.................20u
sold:........150 + 15u................. 8u
left:...........50 + 5u....................12u

and there are 43 more walnut than choc, so:
...........( 50 + 5u ) = 12u + 43
so that, .....7u = 7
that is, ......1u = 1

there were 200 + 20u + 20u = 200 + 40 = 240 cookies at first.

4. Box A, B and C contained some beads. The ratio of the number of beads in Box A to the total number of beads was 2:9. She used 1/3 of the beads in Box B and transferred 1/4 of the beads from Box A to Box C. The ratio of the number of beads left in Box B to the number of beads left in Box C was 4:9. If there were 846 beads in Box B at first, what was the total number of beads left in the end?

Solution:
at start, we have:
....................A.................B.................C
....................4u.....................(14u)..........................from fact that A : (B+C) = 2 : 9 as given in question
................................................................................this means, if transfer 1/4 from A, (B+C) must have a total of 15u
..............................................................and notice the final ratio of B : C = 4 : (8+1), where the "4" comes from after using 1/3 in B, ie, "4" is 2/3 of B
...............................................................so returning 1/3 of B which is "2", the ratio of B : C at first is 6 : 8
so that the ratio of A : B : C is 4 : 6 : 8 at first.
Now, B has 846 at first.
That is, 6u = 846 beads
giving, 1u = 846 / 6 = 141 beads
the total no. of beads in the end is 16u = 16 x 141 = 2256 beads.

[again, use these nos. to start and work through the sentences in question, and you shall know the answers are correct.]

Regular Member

Registered: 03/20/12
Posts: 111

 By: JONAANDR (offline)  Monday, November 10 2014 @ 09:38 PM CST
JONAANDR

Quote by: jo sarah

Hi, "listerner" made a very good observation: since in each scenario, the same no. (3) were eaten, the total remains the same!

Now, let me quickly give some pointers for the new set:

1. When 2 new boys each of height 1.62m joined the group, the average height became 1.67m.
How many pupils were there in the group in the end?

Wonder if there is some information missing in this...

2. Albert, Ben and Cass each had some marbles. Albert gave 1/3 of his marbles to Cass. Ben also gave 1/3 of his marbles to Cass. Cass put all the marbles together. She then gave 1/5 of her marbles to Ben.
As a result, Albert and Cass had 20 marbles each and Ben had 15 marbles at the end. How many marbles did each of them have at first?

Solution:
Using only letters A, B, C for each person; and in this case, since we know what each has in the end, we work backwards starting from the no. in the end:

...A..........B..........C
...20........15........20 .........................(this came after C gave 1/5 to B; meaning C is left with 4/5, ie, the 20 is 4/5 => 1/5 is 5)
...............-5.........+5 ........................(so B return 5 to C: we are working back towards the beginning)
----------------------------
...20........10........25 ........................ (this is after B gave 1/3 to C; meaning B is left with 2/3, ie, the 10 is 2/3 => 1/3 is 5)
..............+5.........-5 .........................(here, C returns 5 to B )
---------------------------
...20........15........20 ........................(this from A giving 1/3 to C; ie, A is left with 2/3, making the 1/3 to be 10)
.+10...................-10 .........................(C returns 10 to A)
--------------------------
...30........15........10

That is, A has 30 marbles, B has 15 and C has 10 marbles at first.

[no need answer key; just start with these and follow the question to see if you get the no. that they each has at the end. If so, you are right!]

3. Mrs Tay baked 200 more walnut cookies than chocolate cookies. She sold 75% of the walnut cookies and 40% of the chocolate cookies. At the end of the day, there were 43 more walnut cookies than chocolate cookies. How many cookies were there at first?

Solution:
Note: 75% = 3/4; and 40% = 2/5. So in order to have a no. that divides by 4 and 5, we choose 20u in the following:
..................walnut....................choc
at 1st:......200 + 20u.................20u
sold:........150 + 15u................. 8u
left:...........50 + 5u....................12u

and there are 43 more walnut than choc, so:
...........( 50 + 5u ) = 12u + 43
so that, .....7u = 7
that is, ......1u = 1

there were 200 + 20u + 20u = 200 + 40 = 240 cookies at first.

4. Box A, B and C contained some beads. The ratio of the number of beads in Box A to the total number of beads was 2:9. She used 1/3 of the beads in Box B and transferred 1/4 of the beads from Box A to Box C. The ratio of the number of beads left in Box B to the number of beads left in Box C was 4:9. If there were 846 beads in Box B at first, what was the total number of beads left in the end?

Solution:
at start, we have:
....................A.................B.................C
....................4u.....................(14u)..........................from fact that A : (B+C) = 2 : 9 as given in question
................................................................................this means, if transfer 1/4 from A, (B+C) must have a total of 15u
..............................................................and notice the final ratio of B : C = 4 : (8+1), where the "4" comes from after using 1/3 in B, ie, "4" is 2/3 of B
...............................................................so returning 1/3 of B which is "2", the ratio of B : C at first is 6 : 8
so that the ratio of A : B : C is 4 : 6 : 8 at first.
Now, B has 846 at first.
That is, 6u = 846 beads
giving, 1u = 846 / 6 = 141 beads
the total no. of beads in the end is 16u = 16 x 141 = 2256 beads.

[again, use these nos. to start and work through the sentences in question, and you shall know the answers are correct.]

1.The average height of a group of pupils was 1.68m. When 2 new boys each of height 1.62m joined the group, the average height became 1.67m.
How many pupils were there in the group in the end?

Thks!

Newbie

Registered: 02/27/14
Posts: 6

 By: echeewh (offline)  Monday, November 10 2014 @ 10:57 PM CST
echeewh

Hey <JONAANDR>

Appreciate that you start a new fresh thread/post [New Topic] since you are having a new set of questions. Thanks!!

Following is my worked solution for Q1:

<Before - Original>
Let number of pupils in the original group be 1u.

Total height of (1u) pupils = 1.68 × (1u)
= 1.68u -- (1)

<After 2 new boys joined>
Number of Pupils = 1u + 2
Height of 2B = 2 × 1.62 = 3.24
Total height of (1u + 2) pupils = 1.67 × (1u + 2)
= 1.67u + 3.34
Total height of (1u) pupils = 1.67u + 3.34 - 3.24
= 1.67u + 0.1 -- (2)

Since (1)=(2), we have ...
1.68u = 1.67u + 0.1
1.68u - 1.67u = 0.1
0.01u = 0.1
1u = 0.1 ÷ 0.01 = 10

Number of pupils (end) = 10 + 2 = 12

=======

Trust this helps.

Do let me know again if this is different from your <Answerkey> or if there's further clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: JONAANDR (offline)  Monday, November 10 2014 @ 11:55 PM CST
JONAANDR

Quote by: echeewh

Hey <JONAANDR>

Appreciate that you start a new fresh thread/post [New Topic] since you are having a new set of questions. Thanks!!

Following is my worked solution for Q1:

<Before - Original>
Let number of pupils in the original group be 1u.

Total height of (1u) pupils = 1.68 × (1u)
= 1.68u -- (1)

<After 2 new boys joined>
Number of Pupils = 1u + 2
Height of 2B = 2 × 1.62 = 3.24
Total height of (1u + 2) pupils = 1.67 × (1u + 2)
= 1.67u + 3.34
Total height of (1u) pupils = 1.67u + 3.34 - 3.24
= 1.67u + 0.1 -- (2)

Since (1)=(2), we have ...
1.68u = 1.67u + 0.1
1.68u - 1.67u = 0.1
0.01u = 0.1
1u = 0.1 ÷ 0.01 = 10

Number of pupils (end) = 10 + 2 = 12

=======

Trust this helps.

Do let me know again if this is different from your <Answerkey> or if there's further clarification.

Cheers,
Edward

Thanks a lot!

Newbie

Registered: 02/27/14
Posts: 6

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