1)
Daniel and Garett were playing marbles. In the 1st round, Daniel lost 1/3 of his marbles to Garett. In the 2nd
round, Garett lost 1/4 of the marbles he had then to Daniel. In the last round, Daniel lost 1/5 of the marbles
he had then to Garett. In the end, Daniel had 760 marbles and Garett had 1324 marbles. How many marbles
did Daniel have at first?
For this problem you need to work backwards. Add back the number to the one who lost and minus from the
one who won.
Lost
End
Last round
2
nd
round
st
1 round
At first
Daniel
760
Daniel -1/5 (4/5760)
190
-----950
Gareth – ¼ (3/41134) -378
-----572
Daniel – 1/3 (2/3572) +286
-----858
------
Garett
1324
-190
------1134
+378
-------1512
-286
-------1222
---------
Ans: 858 marbles
2)
Irfan and Hannah read a total of 30 books in November. In December, Irfan read 9 books fewer than
Hannah who had read thrice as many as she did in November. If both of them read a total of 45 books, how
many books did Irfan read in the 2 months?
I assumed that both read 45 books in Dec.
Let u be the no. of book read in Nov by Hannah. In Dec Hannah read 3X = 3u. Irfan read 9 books less than
Hannah which is 3u -9. Put these data into the table below to see a better picture.
November
December
Irfan
?
3u-9
Hannah
u
3u
Total
30
45
-------75
--------
Solve the equation for Dec. (I assumed that you have been taught algebra)
3u-9+3u=45
3u+3u=45+9
6u=54
u=9
(Books read by Hannah in Nov)
3u=27 (Books read by Hannah in Dec)
3u-9 = 18 (Books read by Irfan in Dec)
Compute books read by Irfan in November
30-9= 21
21 + 18=39
Ans: 39 books
3) At a motocycle exhibition, 55 motocycles were parked from one end of the road to the other at the equal
distance apart. The length of motocycle is 279cm. The distance between 2 motocycles is 4m. Find the
length of the road?
Convert 4m to cm = 400cm
400+279=679 (This is the length of each motorcycle plus the parking distance)
679 x 55 = 37345
The last motorocycle does not need the parking distance, so have to minus 279cm from the total
37345-279=37066
37066 cm = 370.66 m
Ans: 370.66 m
4) Mr Bashir went for dinner with his family members with 23 more five-dollar notes than two-dollar notes.
After paying $90 for the meal with some five-dollar notes, he has 1.5 times as many five-dollar notes as twodollar notes. How many five-dollar notes did he have at first?
Let n be the no. of two-dollar notes
n+23 = five-dollar notes
He uses only $5 notes to pay for the $90 meal
90 divide by 5 = 18
n + (23-18) = n +5 (no. of five-dollar notes left)
The ratio between five-dollars notes and two-dollars notes at the end is 1.5:1, or up it to whole number which
is 3:2 for easy computation.
n=2u
n+5=3u
u=5
n+23=5+23=28
Ans: 28 five-dollar notes
5) 3/8 of Amy's money is equal to 1/4 of Margaret's money. After each of them spent $40, Margaret had $64
more than Amy. How much money does Margaret have left after spending the money?
First thing to do is to find out what is 1 whole(8/8) of Amy’s money = what fraction of Margaret’s money.
3/81/4
8/88/3 x ¼ = 2/3
1/364 (Margaret has $64 more than Amy)
2/3128 (Amy has 2/3 of Margaret’s money, which is $128 at first)
128-40=88
Ans: $88
6) Bernice spent 2/3 of her money on party hats and 3/5 of remainder on 2 banners for a Christmas party.
Each banner cost 6 times as much as a party hat. How many party hats did she buy?
Remainder = 3/3-2/3 = 1/3
3/5 of remainder to buy 2 banners (spilt each unit into 5 smaller units)
Hat
Remainder
3/5 of remainder = 3/15 of the total sum of money
3/152 banners
Each banner costs 6 times a hat
3/152 x 6=12 (3/15 can buy 12 hats)
10/15  10/3 x 12 = 40
Ans: 40 hats