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8 questions · timed · auto-graded

Question 15 Marks
Aakash bought vegetables weighing $10 \ kg.$ Out of this $3 \ kg\ 500\ g$ is onions, $2 \ kg\ 75\ g$ is tomatoes and the rest is potatoes. What is weight of the potatoes?
Answer
Weight of vegetables bought $= 10 \ kg$
Weight of onions $= 3 \ kg\ 500\ g = 3 \ kg + 500\ g$
$= 3 \ kg+ \frac{{500}}{{1000}} \ kg = 3 \ kg + 0.500 \ kg ...[\because 1g =\frac{1}{{1000}} \ kg]$
$= (3 + 0.500) \ kg = 3.500\ kg$
Weight of tomatos $= 2 \ kg\ 75\ g = 2 \ kg + 75\ g$
$= 2 \ kg + \frac{{75}}{{1000}} \ kg = 2 \ kg + 0.075 \ kg ...[\because 1g =\frac{1}{{1000}} \ kg]$
$= (2 + 0.075) \ kg = 2 \ kg + 0.075 \ kg$
$\therefore $ Weight of potatoes $= 10 \ kg – 5.575 \ kg = 4.425 \ kg$
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Question 25 Marks
Ravi purchased $5\ kg\ 400\ g$ rice, $2 \ kg\ 20\ g$ sugar and $10\ kg\ 850\ g$ flour. FInd the total weight of his purchases.
Answer
Weight of rice purchased =$ 5 \ kg\ 400\ g = 5 \ kg + 400 g = 5 \ kg + \frac{400}{1000} \ kg ...[\because 1\ g = \frac{1}{1000} \ kg]$
$= 5 \ kg + 0.400 \ kg$
$= (5 + 0.400) \ kg = 5.400 \ kg$
Weight of sugar purchased $= 2 \ kg\ 20\ g = 2 \ kg + 20 g = 2 \ kg + \frac{20}{1000}kg ...[\because 1\ g = \frac{1}{1000} ​kg]$
$= 2 \ kg + 0.020 \ kg$
$= 2.020 \ kg$
Weight of flour purchased $= 10 \ kg\ 850\ g = 10 \ kg + 850 g = 10 \ kg + \frac{850}{1000} \ kg ...[\because 1\ g = \frac{1}{1000} \ kg]$
$= 10 \ kg + 0.850 \ kg$
$= (10 + 0.850) \ kg = 10.850 \ kg$
$\therefore$ Total weight of his purchases is
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Question 35 Marks
Sunita travels $15 \ km\ 268\ m$ by bus, $7 \ km\ 7\ m$ by car and $500\ m$ by foot in order to reach her school. How far is her school from her residence?
Answer
Distance travelled by bus $= 15 \ km\ 268\ m = 15 \ km + 268\ m$
$= 15 \ km + \frac{268}{1000} \ km = 15 \ km + 0.268 \ km ...[\because 1\ m =\;\frac1{1000} \ km]$
$= (15 + 0.268) \ km = 15.268 \ km$
Distance travelled by car $= 7 \ km\ 7\ m = 7 \ km + 7 m$
$= 7 \ km + \frac{7}{1000}km = 7 \ km + 0.007 \ km ...[\because 1\ m =\;\frac1{1000} \ km]$
$= (7 + 0.007) \ km = 7.007 \ km$
Distance travelled by foot $= 500\ m = \frac{500}{1000}km = 0.500 \ km ...[\because 1m =\;\frac1{1000} \ km]$
$\therefore$ Distance of school from residence $= 15.268 \ km + 7.007 \ km + 0.500 \ km = 22.775 \ km$
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Question 45 Marks
Naresh walked $2 \ km\ 35\ m$ in the morning and $1 \ km\ 7\ m$ in the evening. How much distance did he walk in all?
Answer
Distance walked by Naresh in the morning $= 2 \ km + 35\ m = 2 \ km +\;\;\frac{35}{1000} \ km ...[\because 1m =\;\frac1{1000} \ km]$
$= 2 \ km + 0.035 \ km$
$= (2 + 0.035) \ km = 2.035 \ km$
Distance walked by Naresh in the evening $= 1 \ km\ 7\ m = 1 \ km + 7 m$
$= 1 \ km +\;\;\frac7{1000} \ km = 1 \ km + 0.007 \ km ...[\because 1m =\;\;\frac1{1000} \ km]= 1.007 \ km$
$\therefore$ Distance walked in all $= 2.035 \ km + 1.007 \ km = 3.042 \ km.$
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Question 55 Marks
Nasreen bought $3\ m\ 20 \ cm$ cloth for her shirt and $2\ m\ 5 \ cm$ cloth for her another shirt. Find the total length of cloth bought by her.
Answer
Cloth bought for first shirt $= 3\ m\ 20 \ cm = 3\ m + 20 \ cm$
$= 3\ m + \frac{20}{100} m = 3\ m + 0.20\ m ...[\because 1 \ cm =\frac{1}{100} m]$
$= (3 + 0.20)\ m = 3.20\ m$
Cloth bought for another shirt $= 2\ m\ 5 \ cm = 2\ m + 5 \ cm$
$= 2\ m = \frac{5}{100} m = 2\ m + 0.05\ m ...[\because 1 \ cm =\frac{1}{100} m]$
$= (2 + 0.05)\ m = 2.05\ m$
$\therefore$ Total length of cloth bought is
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Question 65 Marks
Make five examples and find the greater number in each of them.
Answer
$1.0.5$ or $0.8$
The whole parts of both the numbers are the same and the tenth part of $0.8$ is greater than the tenth part of $0.5$
Hence,$0.5 < 0.8$
$2.2.0$ or $0.9$
It is clearly seen that the whole part of $2.0$ is greater than that of $0.9$
Therefore,$2.0 > 0.9$
$3.\ 0.042$ or $0.22$
Both the numbers have the same whole parts but the tenth part of $0.22$ is greater than $0.042$
Therefore,$0.042 < 0.22$
$4.3.012$ or $2.99$
It is clearly seen that the whole part of $3.012$ is greater than that of $2.99$
Therefore,$3.012 > 2.99$
$5.\ 0.055$ or $0.15$
Both the numbers have the same whole parts but the tenth part of $0.15$ is greater than $0.055$
Therefore,$0.055 < 0.15$
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Question 75 Marks
Abhishek had $₹7.45.$ He bought toffees for $₹5.30.$ Find the balance amount left with Abhishek.
Answer
Weight of vegetables bought $= 10 \ kg$
Weight of onions $= 3 \ kg\ 500\ g = 3 \ kg + 500 g$
$= 3 \ kg+ \frac{{500}}{{1000}} \ kg = 3 \ kg + 0.500 \ kg ...[\because 1g =\frac{1}{{1000}} \ kg]$
$= (3 + 0.500) \ kg = 3.500\ kg$
Weight of tomatos $= 2 \ kg\ 75\ g = 2 \ kg + 75 g$
$= 2 \ kg + \frac{{75}}{{1000}} \ kg = 2 \ kg + 0.075 \ kg ...[\because 1g =\frac{1}{{1000}} \ kg]$
$= (2 + 0.075) \ kg = 2 \ kg + 0.075 \ kg$
$\therefore $ Weight of potatoes $= 10 \ kg – 5.575 \ kg = 4.425 \ kg$
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Question 85 Marks
Rahul bought $4 \ kg\ 90\ g$ of apples, $2 \ kg\ 60\ g$ of grapes and $5 \ kg\ 300\ g$ of mangoes. Find the total weight of all the fruits he bought.
Answer
Distance travelled by bus $= 15 \ km\ 268\ m = 15 \ km + 268 m$
$= 15 \ km + \frac{268}{1000} \ km = 15 \ km + 0.268 \ km ...[\because 1\ m =\;\frac1{1000} \ km]$
$= (15 + 0.268) \ km = 15.268 \ km$
Distance travelled by car $= 7 \ km\ 7\ m = 7 \ km + 7\ m$
$= 7 \ km + \frac{7}{1000}km = 7 \ km + 0.007 \ km ...[\because 1\ m =\;\frac1{1000} \ km]$
$= (7 + 0.007) \ km = 7.007 \ km$
Distance travelled by foot $= 500\ m = \frac{500}{1000}km = 0.500 \ km ...[\because 1\ m =\;\frac1{1000} \ km]$
$\therefore$ Distance of school from residence $= 15.268 \ km + 7.007 \ km + 0.500 \ km = 22.775 \ km$
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