**Notes**
**NCERT Solutions**
**Assignments**
**Revision Notes**

Given below are the Class 10 Maths Real Numbers Worksheets

a) HCF and LCM problems

b) Prime Factorisation Problemsbr>
c) Division Problems

d) Long answer questions

e) Word Problems

**Question 1)** Use Euclid’s algorithm to find the HCF of 4052 and 12576.

**Question 2)** show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

Solution
Let a be any odd positive integer and b=4.

By division lemma there exist integer q and r such that

a = 4 q + r, where 0 ≤ r < 4

or a = 4q or, a = 4q +1 or, a = 4q + 2 or, a = 4q +3

Now 4q and 4q+2 are even numbers

so a= 4q +1 or, a = 4q +3

Hence , any odd integer is of the form 4q + 1 or, 4q +3.

**Question 3)** Find HCF and LCM of following using Fundamental Theorem of Arithmetic method.

448, 1008 and 168

**Question 4)** Find the HCF and LCM of following using Fundamental Theorem of Arithmetic method 377, 435 and 667.

**Question 5)** Find HCF of numbers 134791, 6341 and 6339 by Euclid’s division algorithm.

**Question 6)** Find the least positive integer which when diminished by 5 is exactly divisible by 36 and 54.

Solution
For this we have to find L.C.M of 36 and 54.

L.C.M. of 36 and 54 = 2*2*3*3*3 = 108

Now,

The least number = (L.C.M. of 36 and 54) + 5

=108+5 = 113

Hence 113 is the least positive integer which on diminished by 5 is exactly divisible by 36 and 54.

**Question 7)** Find HCF and LCM of 12, 63 and 99 using prime factorisation method.

**Question 8)** If the HCF of 144 and 180 is expressed in the form 13m – 3, find the value of m.

**Question 9)** Three alarm clocks ring at intervals of 4, 12 and 20 minutes respectively. If they start ringing together, after how much time will they next ring together?

Solution
Here we need to find the LCM for the numbers 4,12,20

By prime factorisation

4= 2^{2}

12=2^{2}*3

20= 2^{2}*5

LCM = 2^{2}*3*5=60 min

**Question 10)** In sports Day activities of a school, three cyclists start together and can cycle 48 km, 60 km and 72 km a day round the field. The field is circular, whose circumference is 360 km. After how many rounds they will meet again?

Solution
Here we need to find the LCM for the numbers 48,60,72

By prime factorisation

48= 2^{4}*3

60=2^{2}*3*5

72= 2^{3}*3^{2}

LCM = 2^{4}*3^{2}*5=720km

So 2 rounds

**Question 11)** LCM of two numbers is 2295 and HCF is 9. If one of the numbers is 153, find the other number.

**Question 12)** Express 111972 as a product of its prime factors.

**Question 13)** Two tankers contain 850 litres and 680 litres of petrol respectively. Fnd the maximum capacity of tanker which can measure the petrol of either tanker in exact number of times.

Solution
The capacity of required container will be obtained by finding the HCF of the capacity of tankers

By prime factorisation

850 = 2×5^{2}*17

680 = 2^{3}*5*17

HCF = 2*5*17

= 170 Litres

**Question 14)** The traffic lights at three different road – crossing change after every 36 seconds, 60 seconds and 72 seconds. If they change simultaneously at 8 a. m. after, what time will they change again simultaneously?

Solution
Given that traffic light at three different road crossing change after every 36 seconds,60 seconds and 72 seconds respectively.

36 = 2 * 2 * 3 * 3

60 = 2 * 2 * 3 * 5

72 = 2 * 2 * 2 * 3 * 3

Hence LCM of 36, 60 and 72 = (2 * 2 * 2 * 3 * 3 * 5)

= 360

That is after 360 seconds(5 min) they will change simultaneously

Thus the traffic lights change simultaneously at 8:05 am

**Question 15)**Using prime factorization method, find HCF and LCM of 80, 124 and 144. Also, show that HCFX LCM = Product of three numbers.

**Question 17)** Determine the prime factorization of 45470971 positive integers.

**Question 18)** A rectangular courtyard is 18m 72 cm long and 13m 20cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.

Solution
length of rectangular courtyard = 18 m 72 cm =18X100 +72= 1872 cm

Width of rectangular courtyard = 13 m 20 cm=13X100+20= 1320 cm

To find the square tile of maximum side we take the HCF of 1872 and 1320

By Euclid’s division lemma we have

1872 = 1320 * 1 + 552

1320 = 552 * 2 + 216

552 = 216 * 2 + 120

216 = 120 * 1 + 96

120 = 96 * 1 + 24

96 = 24 * 4 + 0

Hence the HCF is 24

Therefore maximum side of the square = 24 cm

Number of tiles required = (Area of rectangular courtyard)/(Area of square tile)

= (1872 * 1320)/(24 * 24)

= 4290

**Question 19)** Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive).

Solution
The least number divisible by all the numbers from 1 to 10 will be the LCM of these numbers.
By prime factorisation

1 = 1

2 = 2 * 1

3 = 3 * 1

4 = 2^{2}

5 = 5 * 1

6 = 2 * 3

7 = 7 * 1

8 = 2^{3}

9 = 3^{2}

10 = 2 * 5

LCM= 1 * 2 * 2 * 2 * 3 * 3 * 5 * 7 = 2520

**Question 20)** What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case?

Other Questions Answer
1) 4

3) 56, 4032

4) 29, 130065

5)1

7) 3, 2772

8)m = 3

15) 4, 22320

17) 7^{2 } 13^{2} 17^{2} 19

20) 3647

Class 10 Maths
Class 10 Science