Cathy, Mary and Penelope have equal number of pencils. Cathy packs all her pencils equally into 6 packets. Mary packs all her pencils equally into 3 packets. Penelope packs all her pencils equally into 10 packets. 3 packets of Cathy's pencils, 2 packets of Mary's pencils and 4 packets of Penelope's pencils add up to 94 pencils. How many pencils do they have altogether?

	Cathy	Mary	Penelope
Number of packets	6	3	10
Number of pencils	30 u	30 u	30 u
Number of pencils in each packet	5 u	10 u	3 u

All the pencils can be put into the packets without remainder.

All the children have equal numbers of pencils. 
Make the number of pencils that each child has the same. LCM of 6, 3 and 10 = 30

Number of pencils that each child has = 30 u

Number of pencils in 1 packet of Cathy's pencils = 30 u ÷ 6 = 5 u
Number of pencils in 1 packet of Mary's pencils = 30 u ÷ 3 = 10 u
Number of pencils in 1 packet of Penelope's pencils = 30 u ÷ 10 = 3 u

Number of pencils in 3 packets of Cathy's pencils, 2 packets of Mary's pencils and 4 packets of Penelope's pencils
= (3 x 5 u) + (2 x 10 u) + (4 x 3 u)
= 15 u + 20 u + 12 u
= 47 u

47 u = 94

1 u = 94 ÷ 47 = 2

Total number of pencils that they have
= 3 x 30 u
= 90 u
= 90 x 2
= 180

Answer(s): 180