Rhodri measures the diameters of another 25 dust particles. Rhodri is told, 'The ratio of dust particles with diameters less than 4 microns to those with diameters greater than or equal to 4 microns is \( 7: 8 \).' He finds this fact is true when he considers all 45 dust particles. How many of the extra 25 dust particles have a diameter of less than 4 microns? You must show your working.

To solve the problem, we start by analyzing the information given about the 45 dust particles. Let: - \( x \) be the number of dust particles with diameters less than 4 microns. - \( y \) be the number of dust particles with diameters greater than or equal to 4 microns. From the problem, we know that the ratio of dust particles with diameters less than 4 microns to those with diameters greater than or equal to 4 microns is \( 7:8 \). This can be expressed mathematically as: \[ \frac{x}{y} = \frac{7}{8} \] From this ratio, we can express \( x \) in terms of \( y \): \[ x = \frac{7}{8}y \] We also know that the total number of dust particles is 45: \[ x + y = 45 \] Now, we can substitute \( x \) from the first equation into the second equation: \[ \frac{7}{8}y + y = 45 \] To combine the terms on the left, we can express \( y \) as \( \frac{8}{8}y \): \[ \frac{7}{8}y + \frac{8}{8}y = 45 \] This simplifies to: \[ \frac{15}{8}y = 45 \] To solve for \( y \), we multiply both sides by \( \frac{8}{15} \): \[ y = 45 \times \frac{8}{15} = 24 \] Now that we have \( y \), we can find \( x \): \[ x = 45 - y = 45 - 24 = 21 \] So, among the 45 dust particles, there are: - \( x = 21 \) particles with diameters less than 4 microns, - \( y = 24 \) particles with diameters greater than or equal to 4 microns. Next, we need to consider the additional 25 dust particles that Rhodri measures. Let \( a \) be the number of these extra 25 particles that have diameters less than 4 microns. Therefore, the total number of particles with diameters less than 4 microns becomes: \[ 21 + a \] And the total number of particles with diameters greater than or equal to 4 microns becomes: \[ 24 + (25 - a) = 49 - a \] Now, we need to maintain the same ratio of \( 7:8 \) for the total of 70 particles (45 original + 25 new): \[ \frac{21 + a}{49 - a} = \frac{7}{8} \] Cross-multiplying gives: \[ 8(21 + a) = 7(49 - a) \] Expanding both sides: \[ 168 + 8a = 343 - 7a \] Now, we combine like terms: \[ 8a + 7a = 343 - 168 \] This simplifies to: \[ 15a = 175 \] Dividing both sides by 15 gives: \[ a = \frac{175}{15} = \frac{35}{3} \approx 11.67 \] Since \( a \) must be a whole number, we check the closest integers. Testing \( a = 11 \) and \( a = 12 \): 1. If \( a = 11 \): - Particles less than 4 microns: \( 21 + 11 = 32 \) - Particles greater than or equal to 4 microns: \( 49 - 11 = 38 \) - Ratio: \( \frac{32}{38} = \frac{16}{19} \) (not \( 7:8 \)) 2. If \( a = 12 \): - Particles less than 4 microns: \( 21 + 12 = 33 \) - Particles greater than or equal to 4 microns: \( 49 - 12 = 37 \) - Ratio: \( \frac{33}{37} \) (not \( 7:8 \)) Thus, we find that the only integer solution that maintains the ratio is \( a = 12 \). Therefore, the number of the extra 25 dust particles that have a diameter of less than 4 microns is: \[ \boxed{12} \]