pri 6 maths qns
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hi all,
I need help in the following qns. Hope someone can guide me along. thks.
1) Jane has a box containing some black and white counters. When she adds in 15 white counters, 65% of the counters in the box are black. If she adds in another 40 black counters, 75% of the counters in the box are black. How many white counters are there in the box at first?
2) A fruit seller had 2 crates of fruits. Each crate contained mangoes and pears. The ratio of the number of mangoes to the number of pears in crate A was 5:4. In crate B, the number of pears was 2/5 of the number of mangoes. The fruit seller transferred half of the pears from crate A to crate B. In the end, the total number of fruits in crate A became 105 and the ratio of the number of mangoes to number of pears in crate B became 7:4.
a) How many pears were transferred from crate A to crate B
b) How many fruits were there in crate B in the end?
3) Joyce had 42 more beads than Selina. Each of them gave away some of their beads to their friends. The number of beads Selina gave away was 4/7 of the number of beads Joyce had at first. The number of beads Joyce gave away was 2/3 of the number of beads Selina had at first. Both had an equal number of beads left. How many beads did Joyce have at first?
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Originally posted by ultimatenolifer:
hi all,
I need help in the following qns. Hope someone can guide me along. thks.
1) Jane has a box containing some black and white counters. When she adds in 15 white counters, 65% of the counters in the box are black. If she adds in another 40 black counters, 75% of the counters in the box are black. How many white counters are there in the box at first?
2) A fruit seller had 2 crates of fruits. Each crate contained mangoes and pears. The ratio of the number of mangoes to the number of pears in crate A was 5:4. In crate B, the number of pears was 2/5 of the number of mangoes. The fruit seller transferred half of the pears from crate A to crate B. In the end, the total number of fruits in crate A became 105 and the ratio of the number of mangoes to number of pears in crate B became 7:4.
a) How many pears were transferred from crate A to crate B
b) How many fruits were there in crate B in the end?
3) Joyce had 42 more beads than Selina. Each of them gave away some of their beads to their friends. The number of beads Selina gave away was 4/7 of the number of beads Joyce had at first. The number of beads Joyce gave away was 2/3 of the number of beads Selina had at first. Both had an equal number of beads left. How many beads did Joyce have at first?
/* disclaimer: It has been five years since I took PSLE. In this five years, I have already been through four years of secondary school education and a year of tertiary education. Hence, I am not sure if the solutions below are correct and in the correct format for PSLE. */
1)
Let x be number of white counters and y be number of black counters.
y/(x+15+y) = 0.65y = 0.65 (x+15+y)
y = 0.65x + 9.75 + 0.65y
0.35y = 0.65x + 9.75 - (1)
(y+40)/(x+15+y+40) = 0.75
y+40 = 0.75 (x+15+y+40)
y+40 = 0.75x + 11.25 + 0.75y + 30
y - 0.75y = 0.75x + 11.25 + 30 - 40
0.25y = 0.75x + 1.25
y = 3x + 5 - (2)
/* substitute equation (2) into equation (1) */
sub (2) into (1):
0.35 (3x+5) = 0.65x + 9.75
1.05x + 1.75 = 0.65x + 9.75
1.05x - 0.65x = 9.75 - 1.75
0.4x = 8
x = 8/0.4
= 20There are 20 white counters at first.
2a)
mangoes in crate A : pears in crate A
5 : 4
total no. of fruits in crate A = 9 units
4.5 units = 105no. of pears transferred to crate B = 2 units
= (105/4.5) * 2
= 46.667
~ 46/* the ~ sign here is actually the sign with ~ above and _ below when you write it out. */
46 pears were transferred from crate A to crate B.2b)
original mangoes in crate B : original pears in crate B
5 : 2
35 : 14
new mangoes in crate B : new pears in crate B
7 : 4
35 : 20/* since the number of mangoes remains the same, the number of units of new mangoes is the same as the number of units of original mangoes. */
/* new pears - original pears = 20 - 14 = 6 units. */
6 units = 46
new total number of fruits in crate B = 55 units
= (46/6) * 55
= 421.67
~ 421/* the ~ sign here is actually the sign with ~ above and _ below when you write it out. */
There were 421 fruits in crate B in the end.
3)
Let x and y be the original number of beads Joyce and Selina had respectively.
x = 42 + y
y = x - 42 - (1)
/* since the question say both Joyce and Selina had an equal number of beads in the end, we can form two equations {[y - (4/7)(x)] and [x - (2/3)(y)]} from the question and equate both together. */
y - (4/7)(x) = x - (2/3)(y)
y + (2/3)(y) = x + (4/7)(x)
(5/3)y = (11/7)(x) - (2)
/* substitute equation (1) into equation (2). */
sub (1) into (2):
(5/3)(x - 42) = (11/7)(x)
(5/3)(x) - 70 = (11/7)(x)
(5/3)(x) - (11/7)(x) = 70
(35/21)(x) - (33/21)(x) = 70
(2/21)(x) = 70
x = 70 / (2/21)
= 70 * (21/2)
= 735
Joyce had 735 beads at first.
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He is correct :)
I got the same answer
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thks guys,but if i were to not use simultaneous eqns to do the above qns, is there a method? The ans sheet provided the answers as below, nid some advice on why this is so...
1) 21-13 = 8
8u = 40
1u = 5
7u=35
35-15 = 20 white counters
2) 20-14=6
6 x 2=12
15 + 6 =21
21u = 105
6u=30
b) 35 + 20=55
55u= 55 x 5=275
3)7u+42=9u-24
9u-7u=42+24
2u=66
1u=33
21u=693
693+42=735
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Hi,
It is not suitable to solve simultaneous equations at P6 level.
For Q1, we could draw a table like this:
Scenario Black White Ratio
Add 15 white 65% 35% 13 : 7
Add 40 black 75% 25% 3 : 1 = 21 : 7
Now 21u - 13u = 8u
8u -> 40
1u -> 5
7u -> 35
Number of white counters at first = 35 - 15 = 20.
There were 20 white counters at first.
Thanks!
P.S. It is not encouraged to present the steps like in the answer sheet.
Cheers,
Wen Shih -
hi guys, need to clarify on 2 more qns
1)Mr lim had a total of 1448 fiction, and non fiction books in his bookstore. He sold 80% of the fiction books and 75% of the non fiction books and found that he had 52 more fiction books than non fiction books left. How many non fiction books did he sell?
2)Mrs Tan bought 1/3 as many chocolates as sweets. She gave each of her students 4 chocolates and 3 sweets, after which she had 6 chocolates and 180 sweets left.
a) How many students received the chocolates and sweets?
b) How many sweets did she buy?
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Originally posted by ultimatenolifer:
1)Mr lim had a total of 1448 fiction, and non fiction books in his bookstore. He sold 80% of the fiction books and 75% of the non fiction books and found that he had 52 more fiction books than non fiction books left. How many non fiction books did he sell?
Total no. of books = Non Fiction + Fiction
Non Fiction + Fiction = 1448
After selling 80% of fiction and 75% of non fiction, Mr Lim have 52 more fiction than non fiction.
Thus:
(Fiction - 80%) - (Non Fiction - 75%) = 52
Let Fiction be F, and non fiction be NF,
(F - 0.8F) - (NF - 0.75NF) = 52; F means 1F, NF means 1NF
(1F - 0.8F) - (1NF - 0.75NF) = 52
0.2F - 0.25NF = 52 -- (Equation 1)
Amount of Non Fiction Book = Total no. of books - Fiction,
Thus: Non Fiction Book = 1448 - Fiction, or:
NF = 1448 - F -- (Equation 2)
Substitute Equation 2 into Equation 1, we get:
0.2F - 0.25(1448 - F) = 52
0.2F - (362 - 0.25F) = 52
0.2F - 362 + 0.25F = 52 -- (I have Opened the bracket)
0.45F - 362 = 52
0.45F = 52 + 362
0.45F = 414
F=414/45
F=920
There are 920 Fiction books.
Thus, Nonfiction = 1448 - 920 = 528.
There are 528 Non fiction books
Thus, he sold 75% of 528 Nonfiction books, = 396
DOuble check:
528 + 920 must equal 1448
And the equation (F - 0.8F) - (NF - 0.75NF) must equal to 52
Sub 920 fiction and 528 non fiction into the equation, we get:
[920 - 0.8(920)] - [528 - 0.75(528)]
184 - 132 = 52 -
Originally posted by ultimatenolifer:
hi guys, need to clarify on 2 more qns
2)Mrs Tan bought 1/3 as many chocolates as sweets. She gave each of her students 4 chocolates and 3 sweets, after which she had 6 chocolates and 180 sweets left.
a) How many students received the chocolates and sweets?
b) How many sweets did she buy?
Let total no. of chocolate be C, Total no. of sweet be S, and Total no. of students be y.
Total no. of Chocolate is 1/3 Total no. of Sweets.
--> C = 1/3(S) -- eqn 1After giving out 4C and 3S each to students, she left 6 chocolates and 180 sweets.
Total no. of Chocolate minus (4 each x no. of students) = 6
--> C - 4y = 6 -- eqn 2
Total no. of sweet minus (3 each x no. of students) = 180
--> S - 3y = 180 -- eqn 3
From eqn 3, total no of sweet = 180 + (3 each x no. of students)
--> 180 + 3y -- eqn 4
Total no. of Chocolate = 1/3(Total no. of sweets)
--> C = 1/3(180 + 3y)
C = 60 + y -- eqn 5
Substitude eqn 5 into eqn 2, we get:
(60 + y) - ( 4y) = 6
60 - 3y = 6
- 3y = 6-60
-3y = -54
y= 54/3
y= 18
(a) There are 18 students
Substitute y = 18 into eqn 4 to find no. of sweets, we get:
Total no. of sweets = 180 + 3y
Total no. of sweets = 180 + 3(18)
Total no. of sweets = 234.
(b) She bought 234 sweets
Double check:
Total no. of chocolate = 6 + 4y, <-- sub y =18 inside
6 + 4(18) = 78 chocolates
If no. of sweet is 234, and mrs tan bought 1/3 as many chocolate as sweets, she should have bought 1/3 x 234 = 78 chocolates. -
Is there any way not to use simultaneous eqns because p6 cant use simultaneous eqns
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Hi,
You may refer to my P6 resources on solving strategies:
http://wenshih.wordpress.com/primary-mathematics
Thanks!
Cheers,
Wen Shih -
Originally posted by ultimatenolifer:
Is there any way not to use simultaneous eqns because p6 cant use simultaneous eqns
After reading through and learning from Mr Wen Shih's link, I have used the model method to solve the "fiction and nonfiction books" question. I have listed down my thought process as well. Really helps me to think from a non algebra method
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Most parents are not familiar with the model drawing method and they teach their kids to use algebra instead. Is algebraic method acceptable to primary school examiners?
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Algebra is ok as long as working is shown.