Word Problem: Factors and Multiples - 1 | Mathematics for Class 5 PDF Download

Q.1. Find the biggest number that can divide 280 and 1245 so that when you divide 280, you get 4 left over, and when you divide 1245, you get 3 left over.

Solution: We need to find the largest number that divides both 280 and 1245 with remainders of 4 and 3 respectively. To solve this, we subtract these remainders from the original numbers:

• For 280, subtract 4: $280-4=276280\; -\; 4\; =\; 276$
• For 1245, subtract 3: $1245-3=12421245\; -\; 3\; =\; 1242$

Now, we find the HCF (Highest Common Factor) of 276 and 1242 using prime factorization or the division method.

• $276=2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 3\times 23276\; =\; 2^2\; \backslash times\; 3\; \backslash times\; 23$
• $1242=2\times 3\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 231242\; =\; 2\; \backslash times\; 3^2\; \backslash times\; 23$

The common factors are $2\times 3\times 23=1382\; \backslash times\; 3\; \backslash times\; 23\; =\; 138$

So, the largest number that divides both is 138.

Q.2. A room is 7 meters and 35 centimeters wide and 11 meters and 55 centimeters long. You want to cover the floor with square tiles. What is the length of the biggest square tile you can use?

Solution: First, convert the dimensions into centimeters:

• 7 meters 35 centimeters = $7\times 100+35=$735 cm
• 11 meters 55 centimeters = $11\times 100+55$ = 1155 cm

Now, we need to find the HCF of 735 and 1155 to determine the largest tile size.

Using the division method:

• $1155÷735=11155\; \backslash div\; 735\; =$ (remainder 420)
• $735÷420=$1 (remainder 315)
• $420÷315=$1 (remainder 105)
• $315÷105=$3 (remainder 0)

So, the HCF is 105 cm. The largest square tile you can use is 105 cm.

Q.3. You have three ropes: one is 2 meters 76 centimeters, another is 12 meters 42 centimeters, and the last one is 1 meter 38 centimeters. What is the longest piece of tape you can use to measure all of them exactly?

Solution: First, convert all the lengths into centimeters:

• 2 meters 76 centimeters = $2\times 100+76=$276 cm
• 12 meters 42 centimeters = $12\times 100+42=$1242 cm
• 1 meter 38 centimeters = $1\times 100+38=$138 cm

Now, find the HCF of 276, 1242, and 138.

Using prime factorization:

• $276=2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 3\times 23$
• $1242=2\times 3\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 231242\; =\; 2\; \backslash times\; 3^2\; \backslash times\; 23$
• $138=2\times 3\times 23138\; =\; 2\; \backslash times\; 3\; \backslash times\; 23$

The common factors are $2\times 3\times 23=1382\; \backslash times\; 3\; \backslash times\; 23\; =\; 138$
So, the longest tape that can measure all the lengths is 138 cm.

Q.4. In a seminar, there are 106 people for Hindi, 159 people for English, and 265 people for Math. What is the largest number of people that can sit in each room if each room has the same number of people and only one subject in it?

Solution: We need to find the HCF of 106, 159, and 265.

Using the division method:

• $265÷159=$1 (remainder 106)
• $159÷106=$1 (remainder 53)
• $106÷53=$2 (remainder 0)

So, the HCF is 53. The largest number of people that can sit in each room is 53.

Q.5. Find the smallest number that, when divided by 21, 28, 36, and 45, leaves a remainder of 9 in each case.

Solution: We need to find the LCM of 21, 28, 36, and 45, and then add 9 to get the smallest number.

• Prime factors of 21: $3\times 73\; \backslash times\; 7$
• Prime factors of 28: $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 72^2\; \backslash times\; 7$
• Prime factors of 36: $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 3\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}2^2\; \backslash times\; 3^2$
• Prime factors of 45: $3\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 53^2\; \backslash times\; 5$

The LCM is $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 3\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 5\times 7=12602^2\; \backslash times\; 3^2\; \backslash times\; 5\; \backslash times\; 7\; =\; 1260$

The smallest number that leaves a remainder of 9 is $1260+9=12691260\; +\; 9\; =\; 1269$

Q.6. Find the smallest number which, when 9 is added to it, becomes divisible by 12, 15, 20, and 27.

Solution: We need to find the LCM of 12, 15, 20, and 27, and then subtract 9 from it to get the smallest number.

• Prime factors of 12: $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 32^2\; \backslash times\; 3$
• Prime factors of 15: $3\times 53\; \backslash times\; 5$
• Prime factors of 20: $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 52^2\; \backslash times\; 5$
• Prime factors of 27: $3\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">3</sup>}$

The LCM is $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 3\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">3</sup>}\times 5=5402^2\; \backslash times\; 3^3\; \backslash times\; 5\; =\; 540$.

The smallest number is $540-9=531.540\; -\; 9\; =\; 531$

Q.7. What is the smallest number you must subtract from 3000 so that the result is divisible by 32, 36, and 48?

Solution: We need to find the LCM of 32, 36, and 48, and then find how much to subtract from 3000 to make it divisible by this LCM.

• Prime factors of 32: $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">5</sup>}2^5$
• Prime factors of 36: $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 3\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}$
• Prime factors of 48: $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">4</sup>}\times 32^4\; \backslash times\; 3$

The LCM is $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">5</sup>}\times 3\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}=2882^5\; \backslash times\; 3^2\; =\; 288$

When we divide 3000 by 288, the remainder is 144. So, we need to subtract $144144$ from 3000 to make it divisible by 288.

The answer is $3000-144=2856.$

Q.8. Find the smallest 4-digit number that, when divided by 24, 16, and 12, leaves a remainder of 5.

Solution: We need to find the LCM of 24, 16, and 12, and then add 5 to the smallest multiple of this LCM that is a 4-digit number.

• Prime factors of 24: $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">3</sup>}\times 3$
• Prime factors of 16: $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">4</sup>}2^4$
• Prime factors of 12: $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}\times 32^2\; \backslash times\; 3$

The LCM is $2\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">4</sup>}\times 3=482^4\; \backslash times\; 3\; =\; 48$

The smallest multiple of 48 that is a 4-digit number is 48 × 209 = 10032.

Adding 5, the number is $10032+5=1003710032\; +\; 5\; =\; 10037$

Q.9. The LCM of 285 and 63 is 252. Find their HCF.

Solution: We use the formula:

So, the HCF is 71.

Q.10. The LCM of two numbers is 1344, and their HCF is 4. One of the numbers is 84. What is the other number?

Solution: We use the formula:

So, the other number is 64.

Q.11. Nairitee visits the library every 4 days, and Nitheeya visits every 6 days. On which days will they both meet at the library in December?

Solution: We need to find the LCM of 4 and 6, which is 12. So, they will meet every 12 days. In December (31 days), they will meet on days.