Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
primary 5 | Maths
| Whole Numbers
One Answer Below
Anyone can contribute an answer, even non-tutors.
Hi pls help
Date Posted: 4 years ago
Views: 338
a) Large box required 0.5+0.2 = 0.7m of string than small box.
b) 7 small box = 7u meter of string
5 large box = 5u + (5 x 0.7) = 5u + 3.5
Total string for 5 large and 7 small
= 7u + 5u + 3.5
= 12u + 3.5
12u + 3.5 = 8.9
12u = 5.4
1u = 0.45
Therefore, large box required = 0.45+0.7 = 1.15, small box required = 0.45
Total string Mikka have 8.9 + (1.15-0.5) = 9.55m
Or 8.9 + (0.45 + 0.2)= 9.55m.
Hope this helps
b) 7 small box = 7u meter of string
5 large box = 5u + (5 x 0.7) = 5u + 3.5
Total string for 5 large and 7 small
= 7u + 5u + 3.5
= 12u + 3.5
12u + 3.5 = 8.9
12u = 5.4
1u = 0.45
Therefore, large box required = 0.45+0.7 = 1.15, small box required = 0.45
Total string Mikka have 8.9 + (1.15-0.5) = 9.55m
Or 8.9 + (0.45 + 0.2)= 9.55m.
Hope this helps
See 1 Answer
a) Large box required 0.5+0.2 = 0.7m of string than small box.
b) 7 small box = 7u meter of string
5 large box = 5u + (5 x 0.7) = 5u + 3.5
Total string for 5 large and 7 small
= 7u + 5u + 3.5
= 12u + 3.5
12u + 3.5 = 8.9
12u = 5.4
1u = 0.45
Therefore, large box required = 0.45+0.7 = 1.15, small box required = 0.45
Total string Mikka have 8.9 + (1.15-0.5) = 9.55m
Or 8.9 + (0.45 + 0.2)= 9.55m.
Hope this helps
b) 7 small box = 7u meter of string
5 large box = 5u + (5 x 0.7) = 5u + 3.5
Total string for 5 large and 7 small
= 7u + 5u + 3.5
= 12u + 3.5
12u + 3.5 = 8.9
12u = 5.4
1u = 0.45
Therefore, large box required = 0.45+0.7 = 1.15, small box required = 0.45
Total string Mikka have 8.9 + (1.15-0.5) = 9.55m
Or 8.9 + (0.45 + 0.2)= 9.55m.
Hope this helps
360 Tutoring Program