HCF and LCM Practice Quiz: WBPSC Clerkship 2024
A little practice. A step forward.
12 questions on HCF and LCM at the level of the Clerkship Part-I arithmetic paper, with worked explanations and common traps. 📖 Read the detailed topic
What is the HCF of 48 and 72?
48 $= 2^{4} \times 3$ and 72 $= 2^{3} \times 3^{2}$. Taking the lower powers gives $2^{3} \times 3 = 24$. 144 is the LCM; 12 and 36 are common factors but not the greatest.
- HCF = product of common prime factors with the lowest powers
- LCM = highest powers (144 here)
- HCF × LCM = 24 × 144 = 3456 = 48 × 72
- Stopping at any common factor: 12 and 36 divide both but are not the greatest
- HCF = lower power, LCM = higher power
| Number | Prime factors |
|---|---|
| 48 | $2^{4} \times 3$ |
| 72 | $2^{3} \times 3^{2}$ |
| HCF | $2^{3} \times 3 = 24$ |
| LCM | $2^{4} \times 3^{2} = 144$ |
What is the LCM of 12, 18 and 30?
$12 = 2^{2} \times 3$, $18 = 2 \times 3^{2}$, $30 = 2 \times 3 \times 5$. Taking the highest powers gives $2^{2} \times 3^{2} \times 5 = 180$. 90 leaves remainder 6 with 12, 60 leaves remainder 6 with 18, and 360 is a common multiple but not the least.
- For three numbers take the highest power of every prime
- 180 divided by 12, 18, 30 gives 15, 10, 6
- Stopping at any common multiple: 360 is one, but not the least
- Check the answer by dividing it by each number
The product of two numbers is 2160 and their HCF is 12. What is their LCM?
Rule: HCF × LCM = product. So LCM $= 2160 \div 12 = 180$. 2148 comes from subtracting (2160 − 12) by mistake; 144 and 360 fail the rule too: 12 × 144 = 1728 and 12 × 360 = 4320, neither equal to 2160.
- The rule holds for two numbers only
- LCM = product ÷ HCF
- The rule does not work for three numbers
- The product is shared between HCF and LCM
For example the numbers 36 and 60: product 2160, HCF 12, so LCM $= 2160 \div 12 = 180$.
The HCF of two numbers is 16 and their LCM is 96. If one number is 32, what is the other?
Product $= 16 \times 96 = 1536$. The other number $= 1536 \div 32 = 48$. Check: HCF of 32 and 48 is 16, LCM is 96. With 96 the HCF would be 32, with 64 the LCM would be 64, with 24 the HCF would be 8.
- Other number = (HCF × LCM) ÷ the known number
- Always check that the HCF and LCM match
- Taking the LCM itself as the other number
- HCF divides both numbers; LCM is a multiple of both
What is the smallest number that leaves remainder 5 when divided by 6, 8 and 12?
The LCM of 6, 8 and 12 is 24. With remainder 5 the number is $24 + 5 = 29$; dividing 29 by 6, 8 and 12 leaves 5 each time. 24 leaves remainder 0, 19 leaves 1 when divided by 6, and 53 is larger than the smallest possible (it is $24 \times 2 + 5$).
- Remainder r in every case: number = LCM + r
- For the smallest number use the first multiple of the LCM
- Giving the LCM (24) itself as the answer
- "Leaves a remainder" means add it to the LCM
What is the greatest 4-digit number exactly divisible by 12, 15 and 20?
The LCM of 12, 15 and 20 is 60. Dividing 9999 by 60 gives quotient 166 and remainder 39, so the number is $9999 - 39 = 9960$. 9990 leaves remainder 6 with 12; 9900 is divisible but not the greatest; 9980 leaves remainder 8 with 12.
- Subtract the remainder from the greatest n-digit number
- The greatest 4-digit number is 9999
- Subtracting the LCM instead of the remainder
- 9999 divided by the LCM, drop the remainder
- $9999 \div 60 = 166$, remainder $39$
- $166 \times 60 = 9960$ (divided by 12, 15, 20 gives 830, 664, 498)
A floor of 540 cm × 420 cm is to be covered with equal square tiles without cutting any. What is the largest possible side of a tile?
The side must divide both 540 and 420, and the largest such number is the HCF. $540 = 2^{2} \times 3^{3} \times 5$, $420 = 2^{2} \times 3 \times 5 \times 7$, HCF $= 60$. 30 and 20 fit but are not the largest; 120 does not divide 540.
- Tiles, pieces, equal parts: HCF
- With 60 cm tiles you need 9 × 7 = 63 tiles
- Calculating the LCM (3780) instead
- "No cutting + largest" means HCF
| Tile side | Along 540 cm | Along 420 cm | Total tiles |
|---|---|---|---|
| 60 cm | 9 | 7 | 63 |
| 30 cm | 18 | 14 | 252 |
| 20 cm | 27 | 21 | 567 |
Three lights flash every 15, 20 and 25 seconds. After flashing together, when will they next flash together?
We need the LCM: $15 = 3 \times 5$, $20 = 2^{2} \times 5$, $25 = 5^{2}$, LCM $= 2^{2} \times 3 \times 5^{2} = 300$ seconds $= 5$ minutes. 3 minutes (180 s) and 2 minutes (120 s) are not divisible by 25; 10 minutes (600 s) is the second meeting, not the first.
- "Together again" means LCM
- Check whether the answer must be in minutes
- Writing 300 seconds and forgetting to convert the unit
- Bells, lights, trains: events that coincide use the LCM
What is the HCF of $\frac{2}{3}$, $\frac{4}{9}$ and $\frac{8}{15}$?
The fractions are in lowest terms. The HCF of the numerators (2, 4, 8) is 2; the LCM of the denominators (3, 9, 15) is 45. So the HCF is $\frac{2}{45}$. Check: $\frac{2}{3} \div \frac{2}{45} = 15$, $\frac{4}{9} \div \frac{2}{45} = 10$, $\frac{8}{15} \div \frac{2}{45} = 12$, all whole numbers.
- HCF of fractions = HCF of numerators ÷ LCM of denominators
- Reduce the fractions to lowest terms first
- Using the HCF of the denominators (3) and writing $\frac{2}{3}$
- HCF uses the LCM of the denominators: the reverse
| Numerators | Denominators | |
|---|---|---|
| Numbers | 2, 4, 8 | 3, 9, 15 |
| HCF | 2 | 3 |
| LCM | 8 | 45 |
Two numbers are in the ratio 5 : 7 and their HCF is 8. What is their sum?
Let the numbers be $5k$ and $7k$; as 5 and 7 are coprime the HCF is $k = 8$. The numbers are 40 and 56, with sum 96. Check: LCM $= 35 \times 8 = 280$ and $8 \times 280 = 2240 = 40 \times 56$. A sum of 84, 72 or 120 would give $k = 7, 6, 10$, not 8.
- For ratio a : b in lowest terms the numbers are ak and bk, where k = HCF
- LCM = a × b × k
- Treating 8 as the LCM
- The multiplier k of the ratio is the HCF
What is the smallest number that leaves remainder 4 when divided by 6, 9 and 12 and is exactly divisible by 7?
The LCM of 6, 9 and 12 is 36, so the number has the form $36k + 4$. For $k = 1, 2, 3, 4$ we get 40, 76, 112, 148, which leave remainders 5, 6, 0, 1 with 7. So the first number divisible by 7 is 112 $= 7 \times 16$.
- Same remainder r: number $= \text{LCM} \times k + r$
- Then fit the extra condition (divisible by 7) by trying k
- Stopping at the first number, 40, without checking the extra condition
- List 36k + 4 and test the condition for 7
| k | $36k + 4$ | Remainder with 7 |
|---|---|---|
| 1 | 40 | 5 |
| 2 | 76 | 6 |
| 3 | 112 | 0 |
| 4 | 148 | 1 |
What is the HCF of 84, 126 and 210?
$84 = 2^{2} \times 3 \times 7$, $126 = 2 \times 3^{2} \times 7$, $210 = 2 \times 3 \times 5 \times 7$. The common factors give $2 \times 3 \times 7 = 42$. 21 and 14 are common factors but not the greatest; 126 is not divisible by 84.
- For three numbers take the factors present in all three
- 5 occurs only in 210, so it is dropped
- Finding the HCF of two numbers and not checking the third
- What is common to all is the HCF
📊 Topic Analysis
12 questions on HCF and LCM at the level of the Clerkship Part-I arithmetic paper, with worked explanations and common traps.
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