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Maths 3 Marks Question

Decimals · Jharkhand Board (JAC) STD 7 (Jharkhand - English Medium). 15 questions.

Questions

3 Marks Question

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15 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Each bag of cement weighs $48.5kg$. How many such bags will weigh $223\ kg$?
Answer
Total weight of all the bags $= 2231kg$
Weight of each bag $= 48.5kg$
Number of bags $=\frac{\text{Totalweight}}{\text{Weight of each bag}}$
$=\Big(\frac{2231}{48.5}\Big)\text{kg}$
$=\Big(\frac{2231\times10}{48.5\times10}\Big)=\frac{22310}{485}=46$
Hence, $46$ bags of cement will weigh $2231kg$
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Question 23 Marks
A tin holds $16.5$ litres of oil. How many such tins will be required to hold $478.5$ litres of oil?
Answer
Oil in one tin $= 16.5$ litres Total oil $= 478.5$ litres
$\therefore$ Number of tins required $=\frac{478.5}{16.5}=\frac{478.5\times10}{16.5\times10}$
$=\frac{4785}{165}=29\text{ tins}$
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Question 33 Marks
The total of $24$ chairs is $Rs. 9255.60$. Find the cost of each chair.
Answer
Cost of $24$ chairs $= Rs. 9255.60$
$\therefore$ Cost of one chair $=\text{Rs. }\Big(\frac{9255.60}{24}\Big)$
$=\text{Rs. }\Big(\frac{9255.60\times10}{24\times10}\Big)$
$=\text{Rs. }\Big(\frac{92556}{240}\Big)$
$=\text{Rs. }385.65$

Hence, the cost of one chair is $Rs. 385.65$
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Question 43 Marks
How many pieces of plywood, each of $0.45\ cm$ thick, are required to make a pile $2.43\ m$ high?
Answer
Thickness of the pile of plywoods $= 2.43m = 2.43 \times 100\ cm = 243cm$
Thickness of one piece of plywood $= 0.45\ cm$
$\therefore$ Required number of pieces of plywood
$=\frac{243}{0.45}=\frac{243\times100}{0.45\times100}$
$=\frac{24300}{45}=540$

Hence, the required number of pieces of plywood is $540$
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Question 53 Marks
A car covers a distance of $22.8\ km$ in $2.4$ litres of petrol. How much distance will it cover in $1$ litre of petrol?
Answer
Distance covered by the car with $2.4$ litres of petrol $= 22.8\ km$
$\therefore$ Distance covered with $1$ litre of petrol $=\Big(\frac{22.8}{2.4}\Big)\text{km}$
$=\Big(\frac{228}{24}\Big)\text{km}=\Big(\frac{228\div12}{24\div12}\Big)\text{km}$
$=\Big(\frac{19}{2}\Big)=9\frac{1}{2}\text{km}$
Hence, the distance covered by the car with $1$ litre of petrol is $9\frac{1}{2}\text{km}$
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Question 63 Marks
Find the product $0.68 \times 6.5 \times 0.04$
Answer
First, we will find the product $68 \times 65 \times 4$
Now, $68 \times 65 \times 4 = 4420 \times 4 = 17680$
Sum of decimal places in the given decimals $= (2 + 1 + 2) = 5$
So, the product have five decimal places $0.68 \times 6.5 \times 0.04 = 0.17680 = 0.1768$
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Question 73 Marks
The product of two decimals is $261.36$. If one of them is $17.6$, find the other.
Answer
Product of two decimals $= 261.36$ One decimal $= 17.6$
Second decimal $= 261.36 ÷ 17.6$ $=\frac{261.36}{17.6}=\frac{261.36\times10}{17.6\times10}$
$=\frac{2613.6}{176}=14.85$
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Question 83 Marks
If $69$ buckets of equal capacity can be filled with $586.5$ litres of water, what is the capacity of eacch bucket?
Answer
Capacity of $69$ buckets of water $= 586.5$ litres
$\therefore$ Capacity of one such bucket $=\Big(\frac{585.5}{69}\Big)\text{ liters}=8.5\text{ liters}$
Hence, the capacity of each water bucket is $8.5$ litres.
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Question 93 Marks
How many pieces of plywood, each $0.35cm$ thick, are required to make a pile $1.89m$ high?
Answer
Total thickness $= 1.89m = 189\ cm$
Thickness of one piece $= 0.3 5\ cm$
Number of pieces $= 189 ÷ 0.35$
$=\frac{189}{0.35}=\frac{189\times100}{0.35\times100}$
$=\frac{18900}{35}=540\text{ pieces}$
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Question 103 Marks
Each side of a polygon is $3.8\ cm$ in length and its perimeter is $22.8\ cm$. How many sides does the polygon have?
Answer
Let the number of sides of the polygon be $n$ Length of each side of the polygon $= 3.8\ cm$
$\therefore$ Perimeter of the polygon = $(3.8 \times n)\ cm$ But it is given that its perimeter is $22.8\ cm$
$\therefore$ $(3.8 \times n)\ cm = 22.8\ cm$
$\Rightarrow\text{n}=\Big(\frac{22.8}{3.8}\Big)=\frac{228}{38}=6$
Hence, the given polygon has six sides.
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Question 113 Marks
The product of two decimals is $1.824$. If one of them is $0.64$, find the other.
Answer
Product of the given decimals $= 1.824$
One decimal $= 0.64$
The other decimal $= 1.824 ÷ 0.64$
$=\frac{1.824}{0.64}=\Big(\frac{1.824\times100}{0.64\times100}\Big)$
$=\frac{182.4}{64}=2.85$
Hence, the other decimal is $2.85$
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Question 123 Marks
Mr. Soni bought some bags of cement, each weighing $49.8\ kg$. If the total weight of all the bags is $1792.8\ kg$, how many bags did he buy?
Answer
Total weight of cement $= 1792.8\ kg$
Cement in one bag $= 49.8\ kg$
Number of bags $= 1792.8 ÷ 49.8$
$=\frac{1792.8}{49.8}=\frac{1792.8\times10}{49.8\times10}$
$=\frac{17928}{498}=36\text{ bags}$
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Question 133 Marks
Monica cuts 46m of cloth into pieces of $1.15m$ each. How many pieces does she get?
Answer
Number of pieces in $1.15m = 1$
$\therefore$ Number of pieces in $46\text{m}=\frac{46}{1.15}$
$=\frac{46\times100}{1.15\times100}$
$=\frac{4600}{115}=40\text{ pieces}$
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Question 143 Marks
The weight of $37$ bags of sugar is $3644.5\ kg$. If all the bags weigh equally, what is the weight of each bag?
Answer
Weight of 37 bags of sugar = 3644.5kg
$\therefore$ Weight of $1$ bag of suga $=\Big(\frac{3644.5}{37}\Big)=98.5\text{kg}$

Hence, each bag of sugar weighs $98.5\ kg.$
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Question 153 Marks
1.8m of cloth is required for a shirt. How many such shirts can be made from a piece of cloth 45m long?
Answer
Length of cloth for one shirt = 1.8m Total length of piece of cloth = 45m Number of shirts will be = 45 ÷ 1.8 $=\frac{45}{1.8}=\frac{45\times10}{1.8\times10}$ $=\frac{450}{18}=25\text{ shirts}$
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