Video: AQA GCSE Mathematics Higher Tier Pack 1 β€’ Paper 3 β€’ Question 21

AQA GCSE Mathematics Higher Tier Pack 1 β€’ Paper 3 β€’ Question 21

02:23

Video Transcript

π‘Ž to 𝑏 equals three to eight and six 𝑏 equals 11𝑐. Work out π‘Ž to 𝑐 in its simplest form.

We’ll just bring down π‘Ž to 𝑏, but we need to recognize what the ratio of 𝑏 to 𝑐 would be. If six times 𝑏 equals 11 times 𝑐, what’s the ratio between 𝑏 and 𝑐? It’s 11 to six. We know that six times 𝑏 equals 11 times 𝑐. If we divide both sides of this equation by 𝑐, we’ll get six 𝑏 over 𝑐 equals 11.

From there, we divide both sides of the equation by six. On the left, we have 𝑏 over 𝑐 and on the right 11 divided by six or, written as fraction, eleven sixths. The ratio of 𝑏 to 𝑐 equals 11 to six. Let’s look closer at these two ratios.

In one of them, our 𝑏-value is eight, and in the other, our 𝑏-value is 11. We want these two values to be the same. In our π‘Ž-to-𝑏 relationship, we can multiply both sides by 11, which will result in the relationship being 33 to 88. In our second relationship, we’ll multiply both values by eight. 11 times eight equals 88, and six times eight equals 48. The relationship of 𝑏 to 𝑐 could be said to be 88 to 48.

Now that we have equal 𝑏-values, we can say π‘Ž to 𝑏 to 𝑐 equals 33 to 88 to 48. And that means π‘Ž to 𝑐 equals 33 to 48. We’re looking for the relationship of π‘Ž to 𝑐 in its simplest form. Both 33 and 48 are divisible by three. 33 divided by three equals 11, and 48 divided by three equals 16. The relationship of π‘Ž to 𝑐 in its simplest form equals 11 to 16.

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