Page 60 - Mathematics

Example 3.19 Solution

Estimate the value of 256.5 63.5÷ 58.34 109.8× ≈ 60 110×
2.23 47.56× 2 50×
Solution 6600
Round off both numbers to the nearest ≈ 100
ones. 66

256.5 256≈ to the nearest ones and ≈
63.5 64≈ to the nearest ones. It Example 3.22
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follows that,
256.5 63.5 264 64÷ ≈ ÷ A box can contain 7 pencils. There are
4≈ 251 pencils to be packed in these boxes.
Approximate the required number of
Therefore, 256.5 63.5÷ ≈ 4.
boxes and determine the number of
pencils which will be left unpacked.

Example 3.20 Solution
The number of boxes can be obtained by
3.485× 35.82
Estimate the value of dividing the total number of pencils to be
351.25 packed by the number of pencils which a
correct to 1 decimal place. box can hold.

Solution Thus, number of boxes = 251

3.485× 35.82 ≈ 3.5× 36 7
351.25 350 35.857=
3.5 6× If the number of boxes is
≈ approximated to 36, then
350
36 × 7 = 252 which exceeds the
0.06≈ available number of pencils. If the

0.1≈ correct to 1 decimal place. number of boxes is rounded down
Mathematics Form One Example 3.21 58.34 109.8× . unpacked.
to 35, it gives 245 packed pencils,
which means 6 pencils will remain

Therefore, 35 boxes are required and 6
Estimate the value of
2.23 47.56×
pencils will be left unpacked.

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25/10/2024 09:51:15
Mathematics form 1.indd 54 25/10/2024 09:51:15
Mathematics form 1.indd 54