MATH 1 - REVISION WORKSHEET SOLVED - FACTORS AND MULTIPLES - CH 2
Factors and Multiples Worksheet Solutions
1. Find the HCF of 112, 168 and 420 by prime factorization.
Prime Factorization:
112 = 2 X 2 X 2 X 2X7 =2^4X7
168 =2 X 2 X 2 X 3X7 = 2^3 x 3 x 7
420 = 2 X 2 X 3 X 5 X 7 = 2^2 x 3 x 5 x 7
Common Factors: 2^2 x 7 = 4 x 7
Answer: 28
2. Find the HCF of 252, 324 and 594 by long division method.
Step 1: Divide 324 by 252. Remainder is 72. Divide 252 by 72. Remainder is 36. Divide 72 by 36. Remainder is 0. HCF(252, 324) = 36.
Step 2: Divide 594 by 36. Remainder is 18. Divide 36 by 18. Remainder is 0.
Answer: 18
3. Traffic lights change after every 48, 72, and 108 seconds. If they change simultaneously at 8 a.m., after how much time will they change again?
Method: Find the LCM of 48, 72, and 108.
Calculation: LCM = 432 seconds.
Conversion: 432 / 60 = 7 minutes and 12 seconds.
Answer: They will change again at 8:07:12 a.m.
4. Reduce 289/391 to the lowest terms.
Step 1: Find HCF of 289 and 391.
Answer:= 17/23
5. Find the largest number which divides 630 and 940 leaving remainders 6 and 4 respectively.
Step 1: Subtract remainders:
11. The HCF and LCM are 131 and 8253. If one number is 917, find the other.
Calculation: $\frac{131 \times 8253}{917}$
Step: $917 \div 131 = 7$. Then $8253 \div 7 = 1179$.
Answer: 1179
12. Reduce 517/799 to the lowest terms.
Step 1: Find HCF of 517 and 799 using division.
Answer: 11/17
1. Find the HCF of 112, 168 and 420 by prime factorization.
Prime Factorization:
112 = 2 X 2 X 2 X 2X7 =2^4X7
168 =2 X 2 X 2 X 3X7 = 2^3 x 3 x 7
420 = 2 X 2 X 3 X 5 X 7 = 2^2 x 3 x 5 x 7
Common Factors: 2^2 x 7 = 4 x 7
Answer: 28
-------------------------------------------
2. Find the HCF of 252, 324 and 594 by long division method.
Step 1: Divide 324 by 252. Remainder is 72. Divide 252 by 72. Remainder is 36. Divide 72 by 36. Remainder is 0. HCF(252, 324) = 36.
Step 2: Divide 594 by 36. Remainder is 18. Divide 36 by 18. Remainder is 0.
Answer: 18
-------------------------------------------
3. Traffic lights change after every 48, 72, and 108 seconds. If they change simultaneously at 8 a.m., after how much time will they change again?
Method: Find the LCM of 48, 72, and 108.
Calculation: LCM = 432 seconds.
Conversion: 432 / 60 = 7 minutes and 12 seconds.
Answer: They will change again at 8:07:12 a.m.
---------------------------------------
4. Reduce 289/391 to the lowest terms.
Step 1: Find HCF of 289 and 391.
Since 289 = 17 x17 and
391 = 17 x 23,
the HCF is 17.
Step 2: Divide both by 17.
Step 2: Divide both by 17.
289 ÷ 17
391 ÷17
= 17
23
Answer:= 17/23
---------------------------------------
5. Find the largest number which divides 630 and 940 leaving remainders 6 and 4 respectively.
Step 1: Subtract remainders:
630 - 6 = 624 and 940 - 4 = 936$.
Step 2: Find HCF of 624 and 936.
Answer: 312
6. The product of two numbers is 2160 and their HCF is 12. Find the LCM.
Formula: $\text{HCF} \times \text{LCM} = \text{Product of numbers}$
Calculation: $12 \times \text{LCM} = 2160 \implies \text{LCM} = 2160 \div 12$
Answer: 180
7. The HCF of two numbers is 144 and their LCM is 2880. If one number is 576, find the other.
Formula: $\text{Other number} = \frac{\text{HCF} \times \text{LCM}}{\text{Given number}}$
Calculation: $\frac{144 \times 2880}{576} = \frac{2880}{4}$
Answer: 720
8. Find the smallest number which when diminished by 3 is divisible by 21, 28, 36 and 45.
Step 1: Find LCM of 21, 28, 36, and 45. LCM = 1260.
Step 2: Since the number is "diminished by 3," add 3 to the LCM. $1260 + 3 = 1263$.
Answer: 1263
9. Find the smallest number which when divided by 15, 20, 25 and 30 leaves remainder 5 in each case.
Step 1: Find LCM of 15, 20, 25, and 30. LCM = 300.
Step 2: Add the common remainder: $300 + 5 = 305$.
Answer: 305
10. Three boys step off together. Their steps measure 36 cm, 48 cm and 54 cm. At what distance will they again step together?
Method: Find the LCM of 36, 48, and 54.
Calculation: LCM = 432.
Answer: 432 cm (or 4.32 meters).
Step 2: Find HCF of 624 and 936.
Answer: 312
6. The product of two numbers is 2160 and their HCF is 12. Find the LCM.
Formula: $\text{HCF} \times \text{LCM} = \text{Product of numbers}$
Calculation: $12 \times \text{LCM} = 2160 \implies \text{LCM} = 2160 \div 12$
Answer: 180
7. The HCF of two numbers is 144 and their LCM is 2880. If one number is 576, find the other.
Formula: $\text{Other number} = \frac{\text{HCF} \times \text{LCM}}{\text{Given number}}$
Calculation: $\frac{144 \times 2880}{576} = \frac{2880}{4}$
Answer: 720
8. Find the smallest number which when diminished by 3 is divisible by 21, 28, 36 and 45.
Step 1: Find LCM of 21, 28, 36, and 45. LCM = 1260.
Step 2: Since the number is "diminished by 3," add 3 to the LCM. $1260 + 3 = 1263$.
Answer: 1263
9. Find the smallest number which when divided by 15, 20, 25 and 30 leaves remainder 5 in each case.
Step 1: Find LCM of 15, 20, 25, and 30. LCM = 300.
Step 2: Add the common remainder: $300 + 5 = 305$.
Answer: 305
10. Three boys step off together. Their steps measure 36 cm, 48 cm and 54 cm. At what distance will they again step together?
Method: Find the LCM of 36, 48, and 54.
Calculation: LCM = 432.
Answer: 432 cm (or 4.32 meters).
----------------------------------------
11. The HCF and LCM are 131 and 8253. If one number is 917, find the other.
Calculation: $\frac{131 \times 8253}{917}$
Step: $917 \div 131 = 7$. Then $8253 \div 7 = 1179$.
Answer: 1179
-------------------------------------------
12. Reduce 517/799 to the lowest terms.
Step 1: Find HCF of 517 and 799 using division.
799 ÷ 517 (rem 282)
517 ÷ 282 (rem 235)
282 ÷ 235 (rem 47)
235 ÷ 47 = 5.
HCF is 47.
Step 2:
Step 2:
517 ÷ 47
799 ÷ 47
11/17
Answer: 11/17
-------------------------------------
Comments
Post a Comment