What will be the difference in mass (in g) between two packets each containing 500 sheets of 80 GSM and 70 GSM papers of size 20 cm × 30 cm? (GSM = g/m2)
Paper Mass Difference Calculation
This question asks for the difference in mass between two packets of paper, each containing the same number of sheets but with different paper densities (GSM).
We are given the following information:
- Number of sheets per packet = 500
- Size of each sheet = 20 cm $\times$ 30 cm
- Packet 1 paper density = 80 GSM (g/m²)
- Packet 2 paper density = 70 GSM (g/m²)
We need to find the difference in total mass (in grams) between the two packets.
Paper Area Calculation
First, let's calculate the area of one sheet of paper. The dimensions are given in centimeters (cm), and the GSM is given in grams per square meter (g/m²), so we need to convert the dimensions to meters (m).
- Length = 30 cm = $\frac{30}{100}$ m = 0.3 m
- Width = 20 cm = $\frac{20}{100}$ m = 0.2 m
The area of one sheet is:
\( \text{Area} = \text{Length} \times \text{Width} \)
\( \text{Area} = 0.3 \, \text{m} \times 0.2 \, \text{m} = 0.06 \, \text{m}^2 \)
Mass per Sheet Calculation
GSM (Grams per Square Meter) tells us the mass of one square meter of paper. To find the mass of one sheet, we multiply the GSM by the area of the sheet in square meters.
\( \text{Mass per sheet} = \text{GSM} \times \text{Area per sheet} \)
- Mass of one 80 GSM sheet: \( 80 \, \text{g/m}^2 \times 0.06 \, \text{m}^2 = 4.8 \, \text{g} \)
- Mass of one 70 GSM sheet: \( 70 \, \text{g/m}^2 \times 0.06 \, \text{m}^2 = 4.2 \, \text{g} \)
Total Mass Calculation for each Packet
Each packet contains 500 sheets. To find the total mass of each packet, we multiply the mass per sheet by the number of sheets.
- Total mass of the 80 GSM packet: \( 500 \, \text{sheets} \times 4.8 \, \text{g/sheet} = 2400 \, \text{g} \)
- Total mass of the 70 GSM packet: \( 500 \, \text{sheets} \times 4.2 \, \text{g/sheet} = 2100 \, \text{g} \)
Mass Difference Calculation
Finally, to find the difference in mass between the two packets, we subtract the mass of the lighter packet from the mass of the heavier packet.
\( \text{Mass Difference} = \text{Total mass of 80 GSM packet} - \text{Total mass of 70 GSM packet} \)
\( \text{Mass Difference} = 2400 \, \text{g} - 2100 \, \text{g} = 300 \, \text{g} \)
The difference in mass between the two packets is 300 g.
Important Questions from Numerical Ability
- Arrange in the correct sequence the Percentage method of recipe adjustment.
(i) Determine the total weight of the product needed.
(ii) Calculate the percentage of each ingredient in relation to total weight.
(iii) Convert all ingredients in recipe to gram unit and determine edible portion.
(iv) Check the ratio of ingredients which should be in proper balance.
(v) Multiply each percentage number by the total weight to give exact amount of the ingredient needed.
(vi) Total the weight of ingredients using edible portion.
Choose the correct answer from the code given below : In an examination 10,000 students appeared. The result revealed the number of students who have:
passed in all five subjects = 5583
passed in three subjects only = 1400
passed in two subjects only = 1200
passed in one subject only = 735
failed in English only = 75
failed in Physics only = 145
failed in Chemistry only = 140
failed in Mathematics only = 200
failed in Bio-science only = 157
The number of students passed in at least four subjects is:
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How many hollow spheres having inner radius of 1 cm can be completely filled by transferring water from a completely filled hollow sphere having inner diameter of 20 cm ?
The period of a pendulum is given as T = 2 π (l/g)1/2 where g = 9.81 m/s2 and π = 3.1416. The period of a pendulum of length 1 m correct to the first place of decimal in seconds is
