1.Find HCF of 216 and 1176 using common division method.
2.Find the product using suitable properties 1005 ×168
3.Arrange the given fractions in descending order
. 4.rearrange the number and perform the operation.
4.Shekar bought a house for ₹25,84,770and then spent₹2,94,732on renovating the house. If he later sold the house for ₹29,47,952,how much money did he gain in this transaction?
5.The number of common prime factors of 75, 60 and 105 is ______________
6.Write the divisibility rule of 6.
B) Determine whether 639210 is divisible by 6 or not? Prove your answer with proper working.
7.(a)Estimate the product by rounding off each of the numbers to nearest hundreds.758 × 6784
(b) Estimate the sum by rounding off each number to nearest hundreds.17692 + 13843
8. The distance between Akash’s school and home is 2km 385m. Every day, he walks to and from the school. Calculate the total distance covered by him from Monday to Friday.
9.From a cloth of length 
two smaller parts of its length 
and 
are cut out. Find the length of remaining cloth.
10.An auditorium has a capacity of 4000 seats. There are 400 seats allotted for ₹2500, 1000 seats for ₹1800, 1200 seats for ₹1000 and the rest for ₹800. If all tickets are sold, how much money would be collected in all?
11.How often will five bells toll together in one hour, if they start together and toll at intervals of 5, 6, 8, 12, 20 seconds, respectively?
12.Find the length of the longest tape that can be used to measure exactly the lengths 4m 20cm, 2m 40cm, 3m 60 cm.(
13.Find the product of greatest 3-digit number and the greatest 4-digit number using properties of whole numbers.
Answers
1.Find HCF of 216 and 1176 using common division method
Steps to find GCF
- Find the prime factorization of 216
216 = 2 × 2 × 2 × 3 × 3 × 3
- Find the prime factorization of 1176
1176 = 2 × 2 × 2 × 3 × 7 × 7
- To find the GCF, multiply all the prime factors common to both numbers:
- Therefore, GCF = 2 × 2 × 2 × 3
GCF = 24
2.Find the product using suitable properties 1005 ×168
1005 × 168
= (1000 + 5) × 168
= 1000 × 168 + 5 × 168 (using distributive property)
= 168000 + 840
= 168840
5.The number of common prime factors of 75, 60 and 105 is
105 has 3 prime factors- 3, 5 and 7. Now, we can see that out of these 3 factors of 105, 3 and 5 are common with that of 60 and 75. Thus, we can say that 60, 75 and 105 have 2 common prime factors, namely, 3 and 5.
6.Write the divisibility rule of 6.
The Rule for 6: The prime factors of 6 are 2 and 3. So for a number to be divisible by 6, it must also be divisible by 2 and 3.
B) Determine whether 639210 is divisible by 6 or not? Prove your answer with proper working.
For number 639210, sum of digits = 6 +3+9+2+1+0= 21 which is multiple of 3. Therefore, 639210 is divisible by 3. We showed above that it is also divisible by 2. Hence, it is also divisible by 6.
7.(a)Estimate the product by rounding off each of the numbers to nearest hundreds.758 × 6784
758 can be rounded of to 800 and 6784 can be rounded off to 7000
Thus, product =800×7000=5600000
(b) Estimate the sum by rounding off each number to nearest hundreds.17692 + 13843
Nearest hundreds round off 17677 is 17700 and round off 13589 is 13600
So, rounded the total nearest is =17700+13600=31300
8. The distance between Akash’s school and home is 2km 385m. Every day, he walks to and from the school. Calculate the total distance covered by him from Monday to Friday.
Distance between Akashś School and home = 2 km 385 m
Distance covered in one day = 2×2km385m = 4 km 770 m
No of days from Monday to Friday = 5
∴Total Distance covered in 5 days = 5×4km770m = 23 km 850m
11.How often will five bells toll together in one hour, if they start together and toll at intervals of 5, 6, 8, 12, 20 seconds, respectively?
L.C.M. of 6, 5, 7, 10 and 12 is 420.
So, the bells will toll together after every 420 seconds i.e. 7 minutes.
Now, 7 x 8 = 56 and 7 x 9 = 63.
Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.
12.Find the length of the longest tape that can be used to measure exactly the lengths 4m 20cm, 2m 40cm, 3m 60 cm.
First convert the lengths into cm. We know that,
1 m = 100 cm
Now,
4 m 20 cm = 420 cm
2 m 40 cm = 240 cm
3 m 60 cm = 360 cm
To find the length of the longest tape we just have to find the HCF of 420, 240 and 360.
Writing down all the numbers as a product of their prime factors:
The prime factors of 420 are: 2 2 3 5 7
The prime factors of 240 are: 2 2 2 2 3 5
The prime factors of 360 are: 2 2 2 3 3 5
Product of all common prime factors: 2 2 3 5 = 60
HCF (420, 240, 360) = 60
13.Find the product of greatest 3-digit number and the greatest 4-digit number using properties of whole numbers.
13.Find the product of greatest 3-digit number and the greatest 4-digit number using properties of whole numbers.Distributive law:
Let a, b, c be three numbers. If
a (b + c) = ab + ac, then this property is called distributive law.
Solution:
The largest three digit number is 999 and the largest four digit number is 9999.
We have to find their products using distributive law.
Now, 999 x 9999
= 999 x (10000 - 1),
since 9999 = 10000 - 1
= (999 × 10000) - (999 × 1), using distribution law
= {(1000 - 1) × 10000} - 999,
X
since 999 = 1000 - 1
= (1000 x 10000) - (1 × 10000) - 999,
using distributive law
= 10000000 - 10000 - 999
= 9989001
Answer:
Hello Pradhanmadhumita
I am Vaishnavi Singh
7th Class
Age 12+
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