8.2.2 Basic Measurements, PT3 Practice


Question 6:
The mass of a box containing 6 papayas is 21.32 kg. The mass of the box when it is empty is 1.46 kg.
Calculate the average mass, in g, of a papaya.

Solution:
Mass of 6 papayas
= 21.32 – 1.46
= 19.86 kg

Average mass of a papaya
= 19.86 ÷ 6
= 3.31 kg
= 3.31 × 1000 g
= 3310 g



Question 7:
Louis bought 600 g of cookies. Dennis bought twice the mass of cookies that Jackson bought. They bought 1.35 kg of cookies altogether. Calculate the mass, in g, of cookies bought by Jackson.

Solution:
Let Jackson bought w g of cookies.
600 g + 2 × w + w = 1.35 kg
600 g + 3w = 1.35 kg
600 g + 3w = 1350 g
3w = 1350 g – 600 g
w = 750 g ÷ 3
w = 250 g



Question 8:
If the mass of 4 packets of candies is 2.6 kg, what is the mass of 9 packets of the same candies, in kg?

Solution:
Mass of 4 packets of candies = 2.6 kgMass of 9 packets of candies=2.64×9                                             =5.85 kg

Question 9:
It is given that 14 of fruits is supplied to Juice Stall A and 27 to Juice Stall B. The remaining 133.25 kg is sold to a fruit stall.
Calculate the mass of fruits that has been supplied to Fruit Stall B.

Solution:
14+27=728+828         =1528Remaining fruits sold to fruit stall=11528=1328Total mass of fruits=2813×133.25=287 kgMass of fruits supplied to Juice Stall B=27×287=82 kg

Question 10:
The mixture of metal to produce a piece of 50 sen coin are 23 zinc, 15 nickel and the rest is copper.
If the mass of copper is 1.6 g, find the total mass, in g, of zinc and nickel.

Solution:
Fraction of copper=12315=15151015315=215Portion of copper = 2 gPortion of zinc and nickel = 13 g21.6 gTotal mass of zinc and nickel,1313×1.62=10.4 g