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HCF and LCM questions for IPMAT, with solutions

10 free HCF and LCM practice questions for IPMAT Indore, Rohtak and JIPMAT, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in HCF and LCM on Learn.

1. Question 1Type the answerWritten by IPMATwayHCF and LCM Basics

   The LCM of $36$ and $48$ is

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   Answer: 144

   $36=2^2\cdot3^2$ and $48=2^4\cdot3$. The LCM takes each prime at its highest power: $2^4\cdot3^2=144$. (HCF $=2^2\cdot3=12$, and $12\times144=1728=36\times48$.)

2. Question 2Type the answerWritten by IPMATwayHCF and LCM Basics

   The LCM of $33$, $45$ and $60$ is

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   Answer: 1980

   Factorise: $33=3\cdot11$, $45=3^2\cdot5$, $60=2^2\cdot3\cdot5$. The LCM takes the highest power of each prime: $2^2\cdot3^2\cdot5\cdot11=4\cdot9\cdot5\cdot11=1980$.

3. Question 3Type the answerWritten by IPMATwayHCF and LCM Basics

   The product of two numbers is $84$ and their HCF is $2$. Their LCM is

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   Answer: 42

   For two numbers, HCF $\times$ LCM $=$ product. So LCM $=\frac{84}{2}=42$. (Possible pairs: $2$ and $42$, or $6$ and $14$; both have HCF $2$ and LCM $42$.)

4. Question 4Type the answerWritten by IPMATwayHCF and LCM Basics

   The LCM of two numbers is $1890$ and their HCF is $30$. If one of them is $270$, the other is

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   Answer: 210

   Product rule: other number $=\frac{\text{HCF}\times\text{LCM}}{270}=\frac{30\times1890}{270}=\frac{56700}{270}=210$. Check: $270=30\cdot9$, $210=30\cdot7$, LCM $=30\cdot63=1890$.

5. Question 5Type the answerWritten by IPMATwayHCF and LCM Basics

   The HCF and LCM of two numbers are $12$ and $144$ respectively. If one of the numbers is $36$, the other number is

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   Answer: 48

   Product rule: other number $=\frac{12\times144}{36}=\frac{1728}{36}=48$. Check: $36=12\cdot3$, $48=12\cdot4$, $\gcd(3,4)=1$, LCM $=12\cdot12=144$.

6. Question 6Type the answerWritten by IPMATwayHCF and LCM Applications

   The greatest $5$-digit number which leaves a remainder of $5$ when divided by $8$ and also when divided by $9$ is

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   Answer: 99941

   $N-5$ is a multiple of LCM$(8,9)=72$. The largest five-digit number is $99999=72\times1388+63$, so the largest five-digit multiple of $72$ is $99999-63=99936$. Then $N=99936+5=99941$, still five-digit.

7. Question 7Type the answerWritten by IPMATwayHCF and LCM Applications

   Four bells toll together at $9{:}00$ a.m. They toll after $7$, $8$, $11$ and $12$ seconds respectively. How many times will they toll together again in the next $3$ hours?

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   Answer: 5

   The bells toll together every LCM$(7,8,11,12)$ seconds. $8=2^3$, $12=2^2\cdot3$, so LCM $=2^3\cdot3\cdot7\cdot11=1848$ s ($30$ min $48$ s). Three hours is $10800$ s, and $10800\div1848\approx5.84$. So they toll together $5$ more times.

8. Question 8Type the answerWritten by IPMATwayHCF and LCM Applications

   Three drums contain $1653$ litres, $2261$ litres and $2527$ litres of petrol. The greatest capacity of a measuring vessel, in litres, that can measure the petrol in each drum exactly, every time completely filled, is

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   Answer: 19

   $2261-1653=608=2^5\cdot19$; $2527-2261=266=2\cdot7\cdot19$; HCF of the differences is $38$, and $1653$ is odd, so the HCF is $19$ ($1653=19\cdot87$, $2261=19\cdot119$, $2527=19\cdot133$).

9. Question 9Type the answerWritten by IPMATwayHCF and LCM Applications

   The least number which when divided by $2$, $3$, $4$, $5$ and $6$ leaves the remainder $1$ in each case, but when divided by $7$ leaves no remainder, is

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   Answer: 301

   Remainder $1$ on division by $2,3,4,5,6$ means $N-1$ is a multiple of LCM$(2,3,4,5,6)=60$. So $N=60k+1$: $61,121,181,241,301,\dots$ Test divisibility by $7$: $61,121,181,241$ fail; $301=7\times43$. So $N=301$.

10. Question 10Type the answerWritten by IPMATwayHCF and LCM Applications

    What is the largest four-digit number that can be added to $7249$ so that the sum is divisible by each of $12$, $14$, $21$, $33$ and $54$?

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    Answer: 9383

    LCM of $12=2^2\cdot3$, $14=2\cdot7$, $21=3\cdot7$, $33=3\cdot11$, $54=2\cdot3^3$ is $2^2\cdot3^3\cdot7\cdot11=8316$. $7249+n$ must be a multiple of $8316$. $8316-7249=1067$, $16632-7249=9383$, $24948-7249=17699$ (five digits). The largest four-digit $n$ is $9383$.