Class Six Math Chapter 1 : Exercise 1.3 Solution

Class six math chapter 1 : exercise 1.3 solution

Class six math exercise 1.3 solution,Class 6 Math Chapter 1 Exercise 1.3 Solution,Math Exercise 1.3 Solution for Class 6,Class Six Math Chapter 1 Solutions,Exercise 1.3 Solution Class 6 Math,Chapter 1 Exercise 1.3 Class 6 Math Solution,Class 6 Mathematics Chapter 1 Exercise 1.3 Answer,Solved Exercise 1.3 Class 6 Math,Class 6 Chapter 1 Math Solutions Exercise 1.3,Detailed Solution Exercise 1.3 Class 6 Math

Exercise 1.3 Solution

1. Determine the coefficient using basic factors:

(a) 144,240,612

Solution:

By analyzing the prime factors of 144,244 and 612 we get,

2) 144

2) 72

2) 36

2) 18

3)9

2) 240

2) 120

2) 60

2) 30

3) 15

2) 612

2) 306

3) 153

3) 51

here,

144 = 2x2x2x2x3x3

240 = 2x2x2x2x3x5

612 = 2x2x3x3x17

The common prime factors of 144,240,612 are: 2,2,3

So the sum of 144,240 and 612=2x2x3=12

(b) 525,495,570

Solution:

By analyzing the common prime factors of 525, 495 and 612, we get,

3) 525

5) 175

5) 35

3) 495

3) 165

5) 55

2) 570

2) 285

3) 95

here,

525 = 3x5x5x7

495 = 3xx3x5x11

570 = 2x3x5x19

Common prime factors of 525, 495 and 570 are 3.5

So the sum of 525, 495 and 570 = 3×5 = 15.

Solution:

By analyzing the prime product of 2666 and 9699 we get,

2) 2666

31) 1333

3) 9699

53) 3233

here,

2666 = 2x31x43

9699=3x53x61

2666 and 9699 have no common prime factors.

So, HCF of 2666 and 9699 = 1.

2. Determine the mass in the division process:

(a) 105, 165

Solution:

105)165(1

105

60)105(1

45)60(1

15)45(3

Here the last divisor is 15

∴ HCF = 15.

(b) 385,286,818

Solution:

286)418(1).

286

132)286(2).

264

22)132(6).

132

22)385(17

374

11)22(2).

Here the last divisor is 11

∴ HCF = 11.

3. Determine the lasso using prime factorization:

(a) 15, 25, 30

Solution:

15,25,30 by analyzing their prime factors, we get,

3) 15

5) 25

2) 30

3) 15

here,

15 = 3×5

25 = 5×5

30 = 2x3x5

∴ LCM of 15, 25 and 30 = 2x3x5x5=150.

(b) 22, 88, 132, 198

Solution:

22, 88, 132 and 198 by analyzing their prime factors, we get,

2) 22

2) 88

2) 44

2) 22

2) 132

2) 66

3) 33

2) 198

3) 99

3) 33

here,

22 = 2×11

88 = 2x2x2x11

132 = 2x2x3x11

198 = 2x3x3x11

The LCM of 22, 88, 132 and 198 = 2x2x2x3x3x11=792.

Solution:

24, 36, 54, 72 and 96 by analyzing their prime factors, we get,

2) 24

2) 12

2)6

2) 36

2) 18

3)9

2) 54

3) 27

3)9

2) 72

2) 36

2) 18

3)9

2) 96

2) 48

2) 24

2) 12

2)6

here,

24 = 2x2x2x3

36 = 2x2x3x3

54 = 2x3x3x3

72 = 2x2x2x3x3

96 = 2x2x2x2x2x3

∴ The sum of 24, 36, 54, 72 and 96 = 2x2x2x2x2x3x3x3=864.

4. Determine Lasso by Euclidean method:

(a) 96, 120

Sanadhan:

2) 96, 120

2) 48, 60

2) 24, 30

3) 12, 15

4, 5

∴ Required LCM = 2x2x2x3x4x5 = 480.

(b) 35, 49, 91

Solution:

7) 35, 49, 91

5, 7, 13

∴ Required LCM = 7x5x7x13 = 3185.

Solution:

2) 33,55,60,80,90

2) 33,55,30,40,45

3) 33,55,15,20,45

5) 11,55,5,20,15

11) 11,11,1,4,3

1,1,1,4,3

∴ Required LCM =2x2x3x5x11x4x3=7920.

5. Dividing 100 and 184 by which largest number will leave 4 as the divisor each time?

Solution:

Since dividing 100 and 184 by the largest number has 4 divisors each time. So the number to be determined will be the product of (100-4)=96 and (184-4)=180.

now,

96)180(1

84)96(1

12)84(7).

∴HCF of 96 and 180 = 12

∴ Required greatest number = 12.

6. Dividing 27, 40 and 65 by which largest number will have 3, 4, 5 divisors respectively?

Solution:

Since dividing 27, 40, and 65 by the required number leaves remainders of 3, 4, and 5, respectively, the required number will be the HCF of (27 – 3) = 24, (40 – 4) = 36, and (65 – 5) = 60.

Here, 24=2x2x2x3

36 = 2x2x3x3

60 = 2x2x3x5

∴ The HCF of 24, 36, and 60 = 2 x 2 x 3 = 12
∴ The largest required number = 12.

7. What is the smallest number that, when divided by 8, 12, 18, and 24, leaves a remainder of 5 each time?

Solution:
The smallest number divisible by the given numbers is their LCM.
Therefore, adding 5 to the LCM of 8, 12, 18, and 24 will give the smallest required number.

2) 8,12,18,24

2) 4,6,9,12

2) 2,3,9,6

3) 1,3,9,3

1,1,3,1

∴ LCM = 2 x 2 x 2 x 3 x 3 = 72
∴ The smallest required number = 72 + 5 = 77.

8. What is the smallest number that, when divided by 20, 25, 30, 36, and 48, leaves remainders of 15, 20, 25, and 43, respectively?

Solution:
The difference between the given divisors and remainders

20-15 = 5

25-20 = 5

30-25 = 5

36-31=5

48-43=5, which is the same specific number in each case.
Therefore, the smallest required number will be the LCM of 20, 25, 30, 36, and 48 minus 5.

2) 20,25,30,36,48

2)10,25,15,18,24

3) 5,25,15,9,12

5) 5,25,5,3,4

1,5,1,3,4

∴ The LCM of 20, 25, 30, 36, and 48 = 2 x 2 x 3 x 5 x 5 x 3 x 4 = 3600
∴ The smallest required number = 3600 – 5 = 3595.

9. A sheet of iron and a sheet of copper have lengths of 672 cm and 960 cm, respectively. What will be the length of the largest identical piece that can be cut from both sheets? Also, determine the number of pieces from each sheet.

Solution:
The length of the largest identical piece that can be cut from both sheets will be the HCF of the given lengths of the iron and copper sheets.

672)960(1

672

288)672(2).

576

96)288(3

288

∴ The HCF of 672 and 960 = 96
∴ The length of each piece to be cut = 96 cm.
∴ Number of pieces from the iron sheet = (672 ÷ 96) = 7 pieces.
And number of pieces from the copper sheet = (960 ÷ 96) = 10 pieces.

10. What is the smallest four-digit number that is exactly divisible by 12, 15, 20, and 35?

Solution:
A number exactly divisible by the LCM of the given numbers will be divisible by each of them without a remainder.

2) 12,15,20,35

2) 6,15,10,35

3) 3,15,5,35

5) 1,5,5,35

1,1,1,7

∴ LCM of 12, 15, 20, 35 = 2 x 2 x 3 x 5 x 7 = 420.
The smallest four-digit number = 1000

420)1000(2

840

160

It turns out that the number 1000 is not divisible by 420. Dividing by 420 leaves a remainder of 160. If we subtract 160 from 1000, the number would be divisible by 420, but then it becomes a three-digit number (1000 – 160 = 840). On the other hand, if we add (420 – 160) or 260 to 1000, the number will be divisible by 420.

Thus, the required number = 1000 + (420 – 160) = 1000 + 260 = 1260.

11. What is the largest five-digit number that, when divided by 16, 24, 30, and 36, leaves a remainder of 10 each time?

Solution:
First, find the LCM of the given divisors 16, 24, 30, and 36:

2) 16,24,30,36

2) 8,12,15,18

2) 4,6,15,9

3) 2,3,15,9

2,1,5,3

∴ Required LCM = 2 x 2 x 2 x 3 x 2 x 5 x 3 = 720
We know, the largest five-digit number = 99999

720)99999(138

720

2799

2160

6399

5760

639

In the above division process, it is seen that the number 99999 is not divisible by 720. Dividing by 720 leaves a remainder of 639. If we subtract 639 from 99999, the resulting number will be divisible by 720.
Again, if we add (720 – 639) = 81, the number will also be divisible by 720. However, (99999 + 81) = 100080, which is a six-digit number.
∴ The largest five-digit number divisible by 720 = (99999 – 639) = 99360.
But according to the question, the remainder must be 10.
∴ The required largest five-digit number = 99360 + 10 = 99370.

12. From a bus stand, four buses travel a certain distance of 10 km, 20 km, 24 km, and 32 km, respectively, at set intervals. What is the minimum distance they must travel to meet at the same point?

Solution:
The required distance will be the LCM of 10, 20, 24, and 32 in kilometers.

2) 10, 20, 24, 32

2) 5, 10, 12, 16

2) 5,5,6,8

5) 5,5,3,4

1,1,3,4

∴ LCM = 2 x 2 x 2 x 5 x 3 x 4 = 480.
∴ Required distance = 480 km.

13. The product of two numbers is 3380, and their HCF is 13. Find the LCM of the two numbers.

Solution:
Given, the product of the two numbers is 3380, and the HCF is 13.
We know,
Product of two numbers = HCF x LCM of the two numbers

Or, 3380=13 x LCM

Or, LCM =3380 ÷ 13

Or, LCM = 260.

Therefore, LCM = 260.

Post Views: 2,306

Class six math chapter 1 : exercise 1.3 solution