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) = 0.65
y = 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
= 20
There 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 = 105
no. 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.
