Psle Math Questions And Answers: 2017

Time needed = ( \frac{\text{remaining}}{\text{rate}} = \frac{\frac{5}{12}}{\frac{5}{12}} = 1 ) hour.

Total ribbon: [ 0.8S + 1.2(S + 3) = 5.4 ] [ 0.8S + 1.2S + 3.6 = 5.4 ] [ 2.0S + 3.6 = 5.4 ] [ 2.0S = 1.8 ] [ S = 0.9 ]

Mrs. Lim had a roll of ribbon that was 5.4 m long. She used 0.8 m of ribbon to wrap each small gift box and 1.2 m to wrap each large gift box. She wrapped 3 more large boxes than small boxes. After wrapping all the boxes, she had no ribbon left. How many small boxes did she wrap? Step-by-Step Solution Let the number of small boxes = ( S ). Number of large boxes = ( S + 3 ). 2017 psle math questions and answers

If total ribbon = 8.4 m, 0.8 m per small, 1.2 m per large, large = small + 3: ( 0.8S + 1.2(S+3) = 8.4 ) → ( 2S + 3.6 = 8.4 ) → ( 2S = 4.8 ) → ( S = 2.4 ) (still not integer).

Remaining empty: ( 1 - \frac{7}{12} = \frac{5}{12} ) tank. She used 0

Wait — this gives ( S = 0.9 ), which is impossible (boxes must be whole numbers). This suggests that my reconstructed numbers are off; the actual 2017 question had different values (e.g., 8.4 m total ribbon, etc.), but the remains the same: form a linear equation from the total length.

In 2 hours: ( 2 \times \frac{7}{24} = \frac{14}{24} = \frac{7}{12} ) tank filled. How many small boxes did she wrap

Ribbon used for small boxes: ( 0.8 \times S ) metres. Ribbon used for large boxes: ( 1.2 \times (S + 3) ) metres.

After turning off C, rate of A + B = ( \frac{1}{6} + \frac{1}{4} = \frac{4}{24} + \frac{6}{24} = \frac{10}{24} = \frac{5}{12} ) tank/hour.