P6 Maths Worksheets Singapore - Problem Questions
Revision 3
Name : ____________________
Date : _____________
Score : _____________
Write down your correct answer
and show your workings:
1) At a party, Sarah
distributed 18 sweets to each of the boys and 20 sweets to each of the
girls and had 48 sweets left.
If she had distributed 20 sweets to each of the boys and 18 sweets to each
of the girls instead, she would have 32 sweets left.
If Sarah had 40 pupils in her class,
a) How many more boys than girls were there in her class?
b) How many sweets did she have at first?
2) There were 2 identica
flights of steps. For the first flights of steps, Siti walked up some steps
adn ran 4 steps and
took a total of 75 seconds. For the second flight of steps, she walked
up some steps and ran 11 steps and took a total
of 40 seconds. How long will Siti take if she had walked up both flight
of steps? Leave your answer in seconds.
3) May earned twice as much
money as her brother. After May spent $32 and her brother spent $50, the
ratio of money
May had left to the amount of money her brother had left was 7:3.
Find the amount of money May earned.
4) The club has 1150
members. 20% of men and 25% of women are non professional photographers.
A total of 250 members are non professional photographers.
How many more men than women are there in the club?
5) Bobby, Edward and kelvin
made a bottle of paper stars.
The number
of paper stars Bobby made was 21 more than 1/4 the total number of paper
stars in the bottle.
The number
of paper stars Edward made was 30 more than 1/4 of the remaining number
of paper stars in the bottle.
Kelvin
made the remaining 114 paper stars.
How many
paper stars did they make in all?
Answers:
1)
Scenario 1:
Boys : 18 sweets each
Girls : 18 sweets + 2 sweets each
Left : 48 sweets
Scenario 2 :
Boys : 18 sweets + 2 sweets each
Girls : 18 sweets each
Left : 32
So, no. of sets of 18 sweets will be the same in both
scenarios.
The only difference is the sets of "2 sweets".
Difference in sets of 2 sweets = 48 - 32 = 16
16 = 8 sets of 2 sweets
(a) There are 8 more boys than girls.
(b) Total no. of pupils = 40
no. of girls = 16
no. of boys = 24
no. of sweets = (18 * 24) + (20*16) + 48 = 800
2)
Assuming Siti walked and ran at the same constant speeds:
Time difference in running up (11-4) 7 steps = 75-40 =
35s
So additional time taken to walk up 1 step = 35/7 = 5s
Additional time taken to walk up 4 steps = 4x5 = 20s
Total time taken to walk up 2 flights of steps
= 2 x (20+75) = 190
or = 2 x (11x5 + 40) = 190
3)
May ____________________________ ____________________________
May {---------32-------}{----------------------7 units-----------------------------------}
Bro ____________________________
Bro {------50-------------}{----3units----}
So the equation is:
32 + 7 units = 2(50 + 3 units)
32 + 7 units = 100 + 6 units
1 unit = 100-32 = 68
Hence May earned 7(68 )+32 = $508
4)
This question is changing % into fractions as it is easier
to think in 1 unit than as 20% or 25%.
20% = 1 in 5 parts
25% = 1 in 4 parts
it is known total people = 1150
total non-photographers = 250
20% men and 25% women = 250 so in fraction:
1 unit of men + 1 unit of women = 250
since 1 unit of men and 1 unit of women are different
things, it is hard to make comparison, however, we know that if we take
away 4 units from men and women, the remainder is 1 unit of men.
hence 250*4 = 1000
1150-1000 = 150
Only then you will know 1 unit of men(=20% of men) =
150
so total men(100% of men) = 150*5 = 750
1 unit of women(=25% of women) = 250-150 = 100
so total women (100% of women) = 100*4 = 400
hence difference between men and women = 750-400=350
5)
Start backwards from Edward. Since Edward made 30 more
than a quarter of the stars and Kelvin made the rest, it means that Kelvin's
stars plus 30 makes up 3/4 of the stars,
ie. 114 + 30 = 144 is 3/4 of the stars
So the sum of Edward's and Kelvin's stars is 144 x 4/3
= 192
We apply the same principle with Bobby's stars now.
3/4 of the total stars = 192 + 21 = 213
So the total number of stars made by all 3 children is
213 x 4/3 = 284
(A+B+C)/3 = (A+B)/2 + 5
A+B+C = 3(A+B)/2 + 15
A+B+38 = 3(A+B)/2 + 15
A+B+23 = 3(A+B)/2
2(A+B)+46 = 3(A+B)
=> A+B = 46
=> A+B+C = 46 + 38 = 84 |