Question:I buy 8 kg of cooking apples at 10p per kg and 2 kg of eating apples at 15p per kg. Find the average cost of 1 kg of apples.
8 kg at 10p per kg cost 80 new pence 2 kg at 15p per kg cost 30 new pence Total cost of 10 kg is 110 new pence The average cost per kg is `110/10` = 11p.
Question:A merchant blends two brands of tea costing 60 new pence per kg an d80 new pence per kg in ratio 3:1. What is the cost of 1 kg of the mixture?
Suppose he mixes 3 kg of 60p tea with 1 kg of 80p tea. Cost of 3 kg at 60p per kg is 180 new pence Cost of 1 kg at 80p per kg is 80 new pence Total cost of 4 kg of tea is 260 new pence Average cost per kg is 65 new pence.
Question:A boy cycles 10 km at 15 km/h and a further 20 km at 20 km/h. Find his average speed over the whole journey.
The time taken to go d k at v km/h. = `d/v` hours. The time taken to cycle 10 km at 15 km/h. = `10/15` = `2/3` hour. The time taken to cycle 20 km at 20 km/h. = `20/20` = 1 hour. The total time taken to cycle 30 km = `1 2/3` hours. The average speed = distance gone/time taken = `30/(1 2/3) = 90/5 = 18 km/h`
Question:A motor-cyclist mixes oil and petrol in the ratio 1:20. If petrol costs 9p per litre and oil 19 `1/2`p per litre, find the cost of a litre of the mixture.
Consider any convenient volumes in the ratio 1:20, e.g 1 litre of oil and 20 litre of petrol. 20 liters of petrol cost 20 x 9p = £1.80 1 liters of oil costs 1 x 19`1/2`p = £`0.19 1/2` Total cost of 21 liters of mixture = £1.99`1/2` Average cost per litre of mixture = `(£1.99 1/2)/21 = (199.1/2p)/21 = 9 1/2p`
Question:In the triangles ABC and XYZ, the angle A equals the angle Y and the angle B equal the angle X. Given that AB=6 cm, AC=4 cm, BC=7 cm and that XY=4 cm, find the lengths of XZ and YZ.
Since `hat(A)=hat(B)` and `hat(B) = hat(X)` the angles C and Z are equal.
Therefore the triangles `{("ABC"),("YXZ"):}` are similar.
`:. (AB)/(YX)=(BC)/(XZ)=(AC)/(YZ)`
`:. 6/4=7/(XZ)=4/(YZ)`
`:. XZ=4 2/3 cm; YZ=2 2/3 cm`Question:In the following fig. XY is parallel to BC. Prove that the triangles ABC and AXY are similar, and hence find two expressions equal to `(XY)/(BC)`. figure Line
Since XY is parallel to BC,
`hat(X)=hat(B)` (corresponding).
Similarly `hat(Y)=hat(C)`.
The triangle `{("AXY"),("ABC"):}` are equiangular and therefore similar.
`:. (XY)/(BC)=(AX)/(AB)=(AY)/(AC)`
(Notice that any ratio of lengths on AB is equal to the corresponding ratio of lengths on AC. For example, `(AX)/(XB)=(AY)/(YC)`. The ratio of the transversals, XY and BC is, however, equal only to the ratio of a side of the small triangle AXY to the corresponding side of the large triangle ABC.)Question:The sides of the triangle ABC are 3 cm, 3 cm and 5 cm. The sides of the triangle XYZ are 9 cm, 12 cm and 15 cm. If the area of the triangle ABC is 6 `cm^2` , find the area of the triangle XYZ is 54 `cm^2` .
Each side of the triangle XYZ is three times the corresponding side of the triangle ABC. The area of XYZ is therefore nine times as great as the area of ABC. Therefore the area of the triangle XYZ is 54 `cm^2` .
Question:A bottle holds 1 litre. How much would a similar bottle, in which each length is doubled, hold?
The ratio of the linear lengths is 2:1. The ratio of the volumes is `2^3:1^3` or 8:1. The second bottle therefore holds 8 litres.
Question:In the triangle ABC, the angle B=`90^0`, the angle A=`42^0` and AC=10 cm. Find the length of BC.
Considering the angle A, the side BC is the opposite side and AC is the hypotenuse. Therefore sin `42^0`=`BC/10`
`:.` CB=10 sin `42^0` cm = `10 xx 0.6691 cm = 6.691 cm`
= 6.69 cm (correct to 3 sig. fig.).Question:In the triangle ABC, the angle B=`90^@` and the angle A=`38^@`, Given that BC=4 cm, calculate the Length of AB.
The tangent of the angle A=`(BC)/(AB)` . The unknown side is the denominator of the fraction. Therefore consider the angle C.
The angle C = `90^@` - `38^@` = `52^@`
`:. tan 52^@ = (AB)/4`
`:.` 4 tan `52^@` cm = `4 xx 1.2799 cm`
= 5.1196 cm
= 5.12 cm (correct to 3 sig. fig.).