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Latihan · Cabar Minda
Latihan

Latihan tambahan

5 soalan untuk mengukuhkan unit Cabar Minda. Jika salah, baca petunjuk dan cuba lagi.

Soalan 1 / 5Latihan tambahan
Kertas A ada 10 petak, 8 berlorek. Jalur B ada 5 bahagian. Berapakah bahagian jalur B perlu dilorek supaya sama besar?
A
B
分母变成 5 = penyebut menjadi 5 先写 A 的分数,再把分子分母同时除以同一个数,让分母变成 B 的格数 5,新的分子就是要涂的格数。 例子:9 格涂 3 格,长条 3 格 → 3/9 = 1/3 → 涂 1 格。 注意:B 一共只有 5 格,涂 5 格就是全部涂满了。
turn the denominator into 5 = penyebut menjadi 5 Write paper A's fraction, then divide the top and bottom by the same number so that the denominator becomes 5, the number of parts in strip B. The new numerator is how many to shade. Example: 3 of 9 squares shaded, strip of 3 parts → 3/9 = 1/3 → shade 1. Watch out: B has only 5 parts, so shading 5 would fill the whole strip.
Soalan 2 / 5Latihan tambahan
Kertas P ada 6 petak, 2 berlorek. Jalur Q ada 3 bahagian. Berapakah bahagian jalur Q perlu dilorek?
P
Q
三格的长条 = jalur tiga bahagian 把 P 的分数约分,分母变成 3 时,看分子是几。也可以把 P 的每一列小格合成长条的一大格来比较。 例子:长条 3 格涂 2 格,纸有 6 小格 → 2/3 = 4/6 → 纸上涂 4 格。 注意:Q 的每一格比 P 的小格大一倍,不要照抄 P 涂的小格数。
a strip of three parts = jalur tiga bahagian Simplify paper P's fraction; when the denominator becomes 3, read the numerator. You can also join each column of small squares in P into one big part of the strip and compare. Example: 2 of 3 parts shaded, paper of 6 squares → 2/3 = 4/6 → shade 4 squares. Watch out: each part of Q is twice as big as a square of P, so do not copy P's count.
Soalan 3 / 5Latihan tambahan
Kertas R ada 8 petak, 4 berlorek. Jalur S ada 2 bahagian. Berapakah bahagian jalur S perlu dilorek?
R
S
一半 = separuh 先把 R 的分数约到最简,看看是不是一半。S 只分成 2 格,再想一想它的一半是几格。 例子:10 格涂 5 格,长条 4 格 → 5/10 = 1/2 = 2/4 → 涂 2 格。 注意:把 S 全部涂满,涂色部分就比 R 大了。
half = separuh Simplify R's fraction to its simplest form and see whether it is a half. S is cut into only 2 parts, so think about how many of its parts make a half. Example: 5 of 10 squares shaded, strip of 4 parts → 5/10 = 1/2 = 2/4 → shade 2. Watch out: if all of S is shaded, its shaded part becomes bigger than R's.
Soalan 4 / 5Latihan tambahan
Kertas T ada 10 petak, 6 berlorek. Jalur U ada 5 bahagian. Berapakah bahagian jalur U perlu dilorek?
T
U
分母从 10 变 5 = penyebut 10 menjadi 5 6 和 10 都是双数,可以同时除以 2,分母就变成 U 的格数。看图时,T 的每一列 2 小格就等于 U 的 1 格。 例子:长条 5 格涂 2 格,纸有 10 小格 → 2/5 = 4/10 → 纸上涂 4 格。 注意:6 是 T 的小格数,U 一共只有 5 格。
denominator 10 becomes 5 = penyebut 10 menjadi 5 6 and 10 are both even, so divide both by 2 and the denominator becomes the number of parts in U. In the picture, each column of 2 small squares in T matches 1 part of U. Example: 2 of 5 parts shaded, paper of 10 squares → 2/5 = 4/10 → shade 4 squares. Watch out: 6 counts T's small squares, but U has only 5 parts altogether.
Soalan 5 / 5Latihan tambahan
Jalur V ada 4 bahagian, 1 berlorek. Kertas W ada 8 petak. Berapakah petak kertas W perlu dilorek supaya sama besar?
V
W
反过来想 = fikir secara terbalik 这次是从大格变成小格。看 V 的 1 格对着 W 的几小格,再把分子和分母同时乘这个数;分母变成 8 时,分子就是 W 要涂的格数。 例子:长条 5 格涂 3 格,纸有 10 格 → 3/5 = 6/10 → 涂 6 格。 注意:W 的格子比较小,只涂 1 格会比 V 少。
think in reverse = fikir secara terbalik This time we go from big parts to small squares. See how many squares of W sit under 1 part of V, then multiply the top and bottom by that number. When the denominator is 8, the numerator tells how many squares of W to shade. Example: 3 of 5 parts shaded, paper of 10 squares → 3/5 = 6/10 → shade 6. Watch out: W's squares are smaller, so shading only 1 would be less than V.

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