PSLE
Kimberly had two ribbons of the same length. She cut the first ribbon into equal parts of length 40 cm and to each part she tied 2 gold balls. After that, she cut the second ribbon into equal parts of length 60 cm and to each part she tied 4 black balls. When she finished tying, she counted that there were 10 more black balls than gold balls. How many balls were there altogether?

	First ribbon	Second ribbon	Difference	Total
Length of each part	40 cm	60 cm
Number of balloons tied to each part	2 gold balloons	4 black balloons
LCM	120 cm	120 cm
Number of equal parts	120 ÷ 40 = 3	120 ÷ 60 = 2
Number of balloons for each ribbon per set	3 x 2 = 6	4 x 2 = 8	8 - 6 = 2	6 + 8 = 14

LCM of 40 and 60 = 120

Number of sets of 2 gold balls in a length of 120 cm
= 120 ÷ 40
= 3

Number of gold balls in a length of 120 cm
= 3 x 2
= 6

Number of sets of 4 black balls in a length of 120 cm
= 120 ÷ 60
= 2

Number of black balls in a length of 120 cm
= 2 x 4
= 8

Extra number of black balls in a length of 120 cm
= 8 - 6
= 2

Number of sets of balls
= 10 ÷ 2
= 5

Total number of balls in one set of 120 cm of gold balloons and one set of 120 cm of black balloons
= 6 + 8
= 14

Total number of balls of the two ribbons
= 5 x 14
= 70

Answer(s): 70