Tutor MathsGuru: Ask me for your burning Maths questions!

[jieheng]

Chan09:

Have another query:

A spent 3/10 of her salary on food, she spent 1/4 of the remainder on a luggage and $120 more on a toy than luggage. She had $468 left. How much was her salary?
Ans: $1680

Salary [-----][-----][-----][------][----][-----][----][---][---][---]

As 7units is not divisible by 4 , change [----] to [-4u-]

Salary [-4u-][-4u][-4u-][-4u-][-4u][-4u-][-4u-][4u][4u][4u]

Salary [--------------21--------------][----7u----][4u][4u][4u]

Salary [$468][$120][----7u----][----7u----][4u][4u][4u]

From the model ,

14u -----> 468 + 120
14u ----->588
1u -------> 42

Her salary was = 40 u = 40 * 42 = $1680 (Ans)

[Jcong]

Please help me.


Beaker A and Beaker B contained 9 litres of water altogether at first. Jamie poured 1/3 of the water from Beaker A to Beaker B. Next she poured 3/8 of the water from Beaker B to Beaker A. She then had the same volume of water in the 2 beakers. How much water was there in each beaker at first? Express your answer in ml.

Thanks for your help.

[MathIzzzFun]

Vanilla Cake:

Pls help to solve the following Maths sums:


Question
(19/20+1919/2020+191919/202020+.......1919...19/2020...20)x20/2011=?
Note: 1919...19/2020...20 means 2011 \"19\"/2011 \"20\"
(4 marks)

Question
(100xa+10xb+c)x(a+b+c)=1926
What is the a+b+c?
(6 marks)

Question 29
Sn where S is the sum of the digits of n.
Example: S1 -> S=1 and S29 -> S=2+9=11
What is the sum of S1+S2+S3+.......S2010+S2011?
(6 marks)

Note:
Sn-> n is a subscript
S1-> 1 is a subscript
S2-> 2 is a subscript
S3-> 3 is a subscript
S29-> 29 is a subscript
S2010-> 2010 is a subscript
S2011-> 2011 is a subscript

Thanks in advance for your time and effort to provide the worked solutions.

Q1.
(19/20+1919/2020+191919/202020+.......1919...19/2020...20)x20/2011
>> number of terms within the brackets = 2011

19/20 = 19/20 x 1
1919/2020 = 19/20 x 101/101 = 19/20 x 1
191919/202020 = 19/20 x 10101/10101 = 19/20

so, (19/20+1919/2020+191919/202020+.......1919...19/2020...20)x20/2011
= (19/20 x 1 x 2011 ) x 20/2011
= 19

Q2.
1926 < 2000 (= 100 x 20) so a + b + c is less than 20

list the factors ie
1926 = 1 x 1926 = 2 x 963 = 3 x 642 = 6 x 321 = 9 x 214 = 18 x 107

so a+ b + c = 6 (a = 3, b = 2, c =1)

cheers.

[jieheng]

Jcong:

Please help me.


Beaker A and Beaker B contained 9 litres of water altogether at first. Jamie poured 1/3 of the water from Beaker A to Beaker B. Next she poured 3/8 of the water from Beaker B to Beaker A. She then had the same volume of water in the 2 beakers. How much water was there in each beaker at first? Express your answer in ml.

Thanks for your help.

To solve this type of question , we need to count backwards .

A [------------------]

B [------------------]

Jamie poured 3/8 of the water from Beaker B to Beaker A ,

the volume of water remained in breaker B is 5/8

A [--][--][--][--][--]

B [--][--][--][--][--]

Transferred back 3u (3/8 of B) from A to B

A [--][--]

B [--][--][--][--][--][--][--][--]

Jamie poured 1/3 of the water from Beaker A to Beaker B

A left 2u as 1u was poured to B

Transferred back 1u from B to A

A [--][--][--] (3u)

B [--][--][--][--][--][--][--] (7u)

9 litres = 9000 ml

total 10u -----> 9000
1u ------> 900

the amount of water in breaker A at first = 3u = 3*900 = 2700 ml (Ans)

the amount of water in breaker B at first = 7u = 7*900 = 6300 ml (Ans)

[charken]

hi mathsguru,

1st question

I have a very tough question for you.Matthew used 1/5 of a box of sugar
for cooking and 3/4 of the remainder to make bread.The rest were packed
equally into 5 packets.what fraction of the total amount of sugar was in each packet?
2nd question
Serene filled up7/8 of her petrol tank for a trip .She used 6/11of the petrol by the end of the trip.The capacity of her petrol tank was 70literes.
how much did she use for the trip.
express the answer as a decimal correct to 1 decimal place.
(only for this question.)
ths

[jieheng]

charken:

hi mathsguru,

1st question

I have a very tough question for you.Matthew used 1/5 of a box of sugar
for cooking and 3/4 of the remainder to make bread.The rest were packed
equally into 5 packets.what fraction of the total amount of sugar was in each packet?
2nd question
Serene filled up7/8 of her petrol tank for a trip .She used 6/11of the petrol by the end of the trip.The capacity of her petrol tank was 70literes.
how much did she use for the trip.
express the answer as a decimal correct to 1 decimal place.
(only for this question.)
ths

Q1)

[-----][-----][-----][-----][-----]

[-----] for cooking

[-----][-----][-----] for making bread

[-----] -----> 1/5

The fraction of the total amount of sugar was in each packet = (1/5) / 5 = 1/25 (Ans)

Q2)

the amount of petrol in 7/8 of the petrol tank = ( 7/8 ) * 70 = 61.25

the amount of petrol she used for the trip was = ( 6/11 ) * 61.25 = 33.4 litres (Ans)

[jieheng]

Vanilla Cake:

Pls help to solve the following Maths sums:



Question 29
Sn where S is the sum of the digits of n.
Example: S1 -> S=1 and S29 -> S=2+9=11
What is the sum of S1+S2+S3+.......S2010+S2011?
(6 marks)

Note:
Sn-> n is a subscript
S1-> 1 is a subscript
S2-> 2 is a subscript
S3-> 3 is a subscript
S29-> 29 is a subscript
S2010-> 2010 is a subscript
S2011-> 2011 is a subscript

Thanks in advance for your time and effort to provide the worked solutions.

Sum of S1 to S9 = (1+2+3+4+5+6+7+8+9) = 45

Sum of S10 to S19 = Sum of all ten-digits + Sum of all one-digits = 1*10 + 45

Sum of S20 to S29 = Sum of all ten-digits + Sum of all one-digits = 2*10 + 45

Sum of S1 to S99 = (1+2+3+4+5+6+7+8+9)*10 + 45*10 = 450 + 450 =900

S100 = 1

Sum of S101 to S199 = Sum of all hundred-digits + 900 = 1*99 + 900

S200 = 2

Sum of S201 to S299 = Sum of all hundred-digits + 900 = 2*99 + 900

Sum of S1 to S999 = (1+2+3+4+5+6+7+8+9)*99 + 900*10 + (S100+S200+.......................+S900) = 45*99 + 900*10 + 45 = 13500

S1000 = 1

Sum of S1001 to S1999 = 1*999 + 13500

Sum of S1 to S1999 = 13500 + 1 + 1*999 + 13500 = 28000

S2000 = 2

Sum of S2001 to S2011 = 2*11 + (1+2+.....+9) + 1 + 2 = 25 + 45 = 70

Sum of S1 to S2011 = 28000 + 2 + 70 = 28072 (Ans)

[Jcong]

John has 993 tables and chairs at first. After he sold 2/5 of the tables and 5/8 of the chairs, he had 459 tables and chairs left. How many tables did he sell?

[jieheng]

Jcong:

John has 993 tables and chairs at first. After he sold 2/5 of the tables and 5/8 of the chairs, he had 459 tables and chairs left. How many tables did he sell?

Note : There is no direct relationship between 1 unit of table and 1 unit of chair
T [---][---][---][---][---] = (5[---])

C [*][*][*][*][*][*][*][*] = (8[*])

5[---] + 8[*] -----> 993

T [---][---][---] = (3[---])

C [*][*][*] = (3[*])

3[---] + 3[*] -----> 459

1[---] + 1[*] -----> 459/3 = 153

8[---] + 8[*] -----> 153*8 = 1224

3[---] + 5[---] + 8[*] -----> 1224

3[---] + 993 -----> 1224

3[---] -----> (1224 - 993) = 231

1[---] -----> 231/3 = 77

2[---] -----> 77*2 = 154

Number of tables that he sold = 154 (Ans)

[wahwah]

Hi jieheng,

Thanks for the solution, but can explain some of the workings? Thanks!

4. Mr Tan bought three times as many badges as toy cars ad spent $144 in total. He spent $84 more on toy cars than on badges. given that a toy car costs $10.40 more than a badge, what is the cost of a badge?

B [-----][-----][-----]

C [-----]

cost of the badges = (144 - 84) / 2 = 30
Why divide by 2?