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# HCF and LCM Questions for CAT

HCF and LCM for CAT are one of the key topics in the CAT Quant section. You can check out these HCF and LCM CAT Previous year questions.  In this article, we will look into some important HCF and LCM Questions for CAT. These are a good source for practice; If you want to practice these questions, you can download this CAT HCF and LCM Questions PDF, which is completely Free.

HCF and LCM CAT Questions form an important part of CAT Quant Number Systems. Applications of LCM and HCF questions in CAT were asked in the past years. These CAT questions will be very useful for quantitative aptitude for CAT.

Take a free CAT mock test and also try to solve CAT previous questions to get good understanding of the LCM and HCF CAT questions. This topic comes under the number system for CAT. Though the Least Common Multiple and Highest Common Factor is a very basic topic, their applications are very important.

You can download the HCF and LCM CAT Problems or you can go through the questions below.

Question 1: Abhi is a very superstitious person. He went to an astrologer who suggested him that the next lottery winning ticket will be a 2-digit number which is less than 100, not a multiple of 2,3 or 5 and is not a perfect square or a perfect cube. Abhi wants to buy all the tickets which have a possibility of becoming the winning ticket. How many tickets should he buy ?

a) 22

b) 21

c) 24

d) 23

e) 26

Solution:

To find all the numbers that suits Abhi, we just subtract the ones that are not possible from the total number of possibilities.

Since it is a 2-digit number, so the number of possibilities = 99-9 = 90

Thus, within all the 2-digit numbers :

Number of multiples of 2 = 49 – 4 = 45

Number of multiples of 3 = 33 – 3=30

Number of multiples of 5 = 19 – 1 = 18

Number of perfect squares = 1 (only 49 matches the requirements)

Number of perfect cubes = 0 (all the cubes are already multiples of the prohibited numbers)

Here, we need to account for using a few numbers from the set more than once.

Thus, multiples of 6,10,15,30 (LCM of (2,3),(2,5),(3,5),(2,3,5) respectively) have been counted more than once.

Multiples of 6 = 16 – 1 = 15

Multiples of 10 = 9

Multiples of 15 = 6

Multiples of 30 = 3

Thus, we need to add the multiples of 6,10,15 to the sum and deduct the multiples of 30 to get the actual number of numbers.

Thus, number of numbers = 90 – (45+30+18+1) + (15+9+6) – 3 = 23 numbers

Alternatively,
There are 25 primes less than 100, 21 of which are 2 digit primes
These primes are relatively co-prime to 2, 3 and 5
We know the prime numbers are 2,3,5,7,11,13,17….
The lowest possible number which is not a prime,
not divisible by 2,3, and 5 and not a perfect square is 7*11=77
The only other possible number is 7*13=91
Hence, total such numbers = 21+2 = 23

Question 2: Three friends P, Q and R started running on a circular track of length 128 m at the same time and from the same spot. P and R were running in the clockwise direction with speeds 4 m/sec and 8 m/sec respectively. Q was running in counter clockwise direction with speed 6 m/sec. What is the time taken by all of them to meet together for the first time?

a) 32 seconds

b) 96 seconds

c) 64 seconds

d) 192 seconds

Solution:

Ratio of speeds of P and R = 4 : 8 = 1 : 2

Both P and R are running in same direction. Hence, number of meeting points = |2-1| = 1

They meet at one point only, i.e they meet at starting point

Therefore, three of them will meet at starting point for the first time

Time taken by P to complete one round $\frac{128}{4}$ = 32 seconds

Time taken by R to complete one round = $\frac{128}{8}$ = 16 seconds

Time taken by Q to complete one round = $\frac{128}{6}$ = $\frac{64}{3}$ seconds

Time taken by all of them to meet together for the first time = LCM(32, 16, $\frac{64}{3}$) = $\ \frac{\ LCM\left(32,\ 16,\ 64\right)}{HCF\left(1,1,3\right)}$ = $\ \frac{64}{1}$ = 64 seconds

Alternate Explanation:

Length of the circular track = 128m

Speed of P, Q and R is 4 m/sec, 6 m/sec and 8 m/sec

Relative speed of P and R = 8 – 4 = 4 m/sec

Time taken by P and R to meet, t1 = $\frac{128}{4} =32$seconds

Relative speed of R and Q = 8 + 6 = 14 m/sec

Time taken by Q and R to meet, t2 = $\frac{128}{14}$ seconds

Time taken by all the three friends to meet together for the first time = $LCM(t1, t2)$= $LCM(\frac{128}{4}, \frac{128}{14})$= $\frac{LCM(128, 128)}{HCF(4, 14)}$=$\frac{128}{2}$ = 64 seconds

Hence, option C is the right choice.

Question 3: If LCM of two numbers 28!*32! and 30!*31! is $\frac{a!*b!}{c!}$, where a, b (a > b) are two-digit numbers and c is one digit number. Find out the value of a -(b+c)?

Solution:

Let us first calculate HCF of 28!*32! and 30!*31!

$\Rightarrow A = 28! *(32 *31!)$

$\Rightarrow A = 32*28! *31!$              … (1)

$\Rightarrow B = 30!*31!$

$\Rightarrow B = 30*29*28!*31!$           … (2)

From equation 1 and 2 it is clear that HCF = 2*28!*31!

We know that: Product of 2 number = LCM * HCF

$\Rightarrow LCM = \frac{Product of A and B}{HCF of A and B}$

$\Rightarrow LCM = \frac{(28!*32!)*(30!31!)}{2*28!*31!}$

$\Rightarrow LCM = \frac{32!*30!}{2} = \frac{32!*30!}{2!}$

Hence we can say that a = 30, b = 30 and c = 2

Hence a-(b+c) = 32 -(30 +2) = 0 (Answer :0)

Question 4: Ram has a certain number of chocolates with him. If he divided these chocolates among 12 children, he would be left with 9 chocolates. If he divides the chocolates among 17 children then he would be left with 14 chocolates. Similarly, if he decides to divide the chocolates among 7 people, he will be left with 4 chocolates. What is the minimum possible number of chocolates with Ram?

a) 116

b) 1425

c) 1723

d) 201

Solution:

Let the total number of chocolates with Ram be ‘n’.
Then ‘n’ can be expressed as
n = 12k + 9 = 17m + 14 = 7p + 4
We can see that in each of the above cases, the remainder is -3.
Hence, the minimum number of chocolates with Ram will be LCM (12, 7, 17) – 3
Thus, the required number will be 12*7*17 – 3 = 1428 – 3 = 1425.

Question 5: Three wheels can complete 60, 36 and 24 revolutions per minute. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?

a) $\frac{5}{2}$ s

b) $\frac{5}{3}$ s

c) $5$ s

d) $7.5$ s

Solution:

The first wheel completes a revolution in $\frac{60}{60}=1$ second
The second wheel completes a revolution in $\frac{60}{36}=1\frac{2}{3}$ second
The third wheel completes a revolution in $\frac{60}{24}=2\frac{1}{2}$ second

The three wheels touch the ground simultaneously at time which are multiples of the above three times.
Hence, the required number is $LCM(1,\frac{5}{3},\frac{5}{2}) = 5$ seconds.

So, the correct option is option (c)

Question 6: What is the least number that must be subtracted from 2015 so that the resultant number when divided by 9, 12 and 16 leaves the same remainder 7?

a) 143

b) 136

c) 164

d) 150

Solution:

The LCM of 9,12 and 16 is 144.
The closest multiple of 144 which is less than 2015 is 144*13 = 1872.
To get a remainder of 7, we need to add 7 to the multiple of the LCM, in this case 1872+7=1879.
Hence, the required number to be subtracted equals 2015-1879=136

Question 7: Car A starts from city P towards city Q. At the same time, another car B starts from the city Q towards city P. They will meet at point R if the ratio of the speed of car A and car B is 4:5. They meet at point S if the ratio of the speed of car A and car B is 3:4. The ratio of the distance between R and S and the distance between P and Q is m:n where m and n are co-primes. Find the value of m+n.

a) 62

b) 63

c) 64

d) 65

Solution:

Assuming the distance between P and Q is 63 units (LCM of 4+5=9 and 3+4=7)
The fraction of the distance travelled by each car will be proportional to their speeds because the time taken is the same in each case.
Assume the speed of car A in the first case to be 4x and 5x
Hence, the distance of R from P = Distance travelled by A from P to R = Speed of A $\times$ Time taken to meet= $\dfrac{63}{4x+5x}\times\ 4x$ = 28 units
Similarly, let the speed of car A in the second case be 3x and 4x

Then, the distance of S from P = Distance travelled by A from P to S = Speed of A $\times$ Time taken to meet = $\dfrac{63}{3x+4x}\times\ 3x$ = 27 units

Hence, the distance between R and S = 1 unit
The ratio $\ \dfrac{\ m}{n}=\ \dfrac{\ 1}{63}$
Hence, the sum = 1+63 = 64

Question 8: Krishna and Shyam can finish work together in 18 days. After 6 days, Krishna left and Shyam finished the work alone. If the total time taken to finish the work in this way is 24 days, what is the ratio of work done by Krishna and Shyam?

a) 3:8

b) 4:9

c) 1:8

d) 2:9

Solution:

It is given that Krishna and Shyam can finish work in 18 days. So after 6 days, one-third of work is finished.

Shyam finishes 2/3 of the work in 24-6=18 days.

Hence, Shyam can finish total work in 18*3/2 = 27 days

Assuming the total work to be 54 units.  (LCM(27,18)=54)

Work done by both Krishna and Shyam in 1 day = 54/18 = 3 units

Work done by Shyam in 1 day = 54/27 = 2 units

Hence, work done by Krishna = 3-2 = 1 units in 1 day

Then work done by Krishna in 6 days = 6 units

Work done by Shyam in 24 days = 24*2=48 units

The ratio = 6:48= 1:8

Question 9: Find the sum of the terms which are common to the following series:
S1: 4, 11, 18 …… 704
S2: 2, 13, 24…… 827

a) 2187

b) 4374

c) 2836

d) 3186

Solution:

The given series are:

S1: 4, 11, 18 …… 704
S2: 2, 13, 24…… 827

The common difference of the first series is 7 and the common difference of the second series is 11.

The terms which are common to them will form an AP with a common difference of LCM(7, 11) = 77

The first common term of these two series is 46.

The last term of the common series cannot be greater than 704.

Let’s find the number of common terms.

704 = 46 + ( n – 1) 77

Taking only the integral value, n = 9

The sum of the series = $\frac{9}{2}*[2*46 + (8) * 77] = 3186$

Question 10: Find the largest number which leaves the same remainder when it divides 722, 1097 and 1347?

a) 5

b) 45

c) 125

d) None of these

Solution:

Let the greatest number which leaves the same remainder when it divides 722, 1097 and 1347 be N and let the remainder be R.

So, N divides 722-R, 1097-R and 1347-R.
It also divides (1097-R) – (722-R) and (1347-R) – (722-R).
So, N divides 1097-722 and 1347-722.
So, N divides 375 and 625.
The HCF of 375 and 625 is 125 and hence, the greatest possible value of N is 125.

The remainder when 125 divides 722, 1097 and 1347 is 97.

Question 11: 3 friends A, B and C run around a circular track of length 120 m continuously. All of them start at the same point. A and B run in the clockwise direction while C runs in the anti-clockwise direction. The speeds of A, B and C are in the ratio 13:5:7. At how many distinct points will all three of them meet?

a) 6

b) 4

c) 3

d) 5

Solution:

Length of the track is 120 m.
Speeds of A and B are in the ratio 13:5. They are running in the same direction.
Therefore, they will meet at 13-5 = 8 distinct points.
Let us use the distance from the starting point to label the meeting points.
A and B will meet at points at a distance of 0 m, 15 m, 30 m, 45 m, 60 m, 75 m, 90 m, and 105 m from the starting point.

B and C run in the opposite directions. Their speeds are in the ratio 5:7.
Therefore, they will meet at 5+7 = 12 distinct points.
They will meet at points at a distance of 0 m, 10 m , 20 m, 30 m, 40 m, 50 m, 60 m, 70 m ,80 m, 90 m, 100 m and 110 m.

A and C run in the opposite directions. Their speeds are in the ratio 13:7.
Therefore, they will meet at 13+7 = 20 distinct points.
They will meet at points at a distance of 6 m, 12 m, 18 m, … 114 m.

All 3 of them will meet at points that are multiples of the LCM of 15, 10 and 6.
LCM (15,6,10) = 30 m.

Alternatively, Number of distinct points = HCF(8, 12, 20)

Therefore, in a sufficiently large number of rounds, all 3 of them will meet at points at a distance of 0 m, 30 m, 60 m and 90 m from the starting point. There are 4 points in total. Therefore, option D is the right answer.

Question 12: Anil, Sunil, and Ravi run along a circular path of length 3 km, starting from the same point at the same time, and going in the clockwise direction. If they run at speeds of 15 km/hr, 10 km/hr, and 8 km/hr, respectively, how much distance in km will Ravi have run when Anil and Sunil meet again for the first time at the starting point?

a) 4.8

b) 4.6

c) 5.2

d) 4.2

Solution:

Anil and Sunil will meet at a first point after LCM ( $\frac{3}{15},\frac{3}{10}$) = 3/5 hr

In the mean time, distance travelled by ravi = 8 * 3/5 = 4.8 km

Question 13: A, B, and C can do a work in 15 days, 20 days, and 12 days respectively. They started the work together, but C left the work two days before the completion and B left the work two days before C left. If the work was completed in m days, what is the value of 12*m (in days)?

Solution:

Let the total work be 120 units (LCM of 15, 20 and 12).
A can complete the work in 15 days.
=> A’s 1 day’s work = 8 units
B can complete the work in 20 days.
=> B’s 1 day’s work = 6 units
C can complete the work in 12 days.
=> C’s 1 day’s work = 10 units
Let the total time required to complete the work be x days.
Then, A worked for x days, C worked for (x – 2) days, and B worked for (x – 4) days.
Therefore, 8*x + 6 * (x – 4) + 10 * (x – 2) = 120
On solving, we get x = $\frac{41}{6}$ days.
So, m = $\frac{41}{6}$ days.
Therefore, 12*m = 82 days.
Hence, 82 is the correct answer.

Question 14: A class which has x students had birthdays of three students – Seema, Raghav and Anand on a particular day. Seema bought 212 chocolates and distributed them equally among the other students and then had some chocolates left. Raghav bought 142 chocolates and distributed equally among the other students. He was also left with the same number of chocolates as Seema. Anand bought 352 chocolates did the same thing and had the same number of chocolates as the other two. What could be the maximum value of x ?

Solution:

Let n be the number of chocolates left with each of them after distributing equally and c1, c2 and c3 be the number of chocolates distributed by Seema, Raghav and Anand respectively. Let y = x – 1 (This is because each person distributing the chocolates distributes them only among the other children. The distributor himself does not take any chocolate)
According to the question, c1y + n = 212 -(1)
c2y + n = 142 -(2)
and c3y + n = 352 – (3)
subtracting (2) from (1) and (3) and (1) from (2) we get,
(c1 – c2)y = 70, (c3 – c2 )y = 210 and (c3 – c1)y = 140
The maximum value of y will occur at the HCF of (70, 140, 210)
Thus, the maximum value of y is 70. Therefore, the total no. of students is y + 1 = 71

Question 15: A number when divided by $7, 8$ and $9$ leaves remainders $1, 2$ and $3$, respectively. What is the difference between the remainders when the number is divided by $24$ and $21$, respectively?

a) $7$

b) $3$

c) $15$

d) $6$

Solution:

The number leaves remainders $1, 2$ and $3$ when divided by $7, 8$ and $9$, respectively. Since the difference between the divisor and the remainder is constant $(7-1 = 8-2 = 9-3)$, the required number is given by

$LCM(7, 8, 9)\times n – LCM(1, 2, 3) = 504n – 6$

We have to find the remainders when the number is divided by $24$ and $21$.

We can observe that $504$ is a multiple of both $24$ and $21$. Thus, both will completely divide the $504n$ part of the number.

Thus, the remainder when the given number is divided by $24=(-6)=24-6=18$.

Similarly, on division by $21$, we get the remainder as $(-6)=21-6=15$

Thus, the required difference between the remainders = $18-15=3$

Hence, the correct answer is option B.

Question 16: A can complete a task in 12 days, B can complete a task in 16 days, and C is 50% less efficient than A. A, B and C start working on a task together. A and C work on day 2, B and C work on day 1 and all of them work on day 3; this continues till they finish the task. Find the amount earned by A if it is known that C earned Rs 13,500 on completing the task.

a) Rs 18,900

b) Rs 16,200

c) Rs 43,200

d) Rs 27,000

Solution:

A can complete the task in 12 days, and B can complete the task in 16 days.

It is mentioned that C is 50% less efficient than A; this implies that C takes twice the time taken by A

C can complete the task in 24 days

Let the total amount of work be 48 units (LCM 12,16,24)

Work done by A in 1 day = $\frac{48}{12}\$ = 4 units

Work done by B in 1 day = $\frac{48}{16}\$ = 3 units

Work done by C in 1 day = $\frac{48}{24}\$ = 2 units

Work done by B and C on day 1 = 3 + 2 = 5 units

Work done by A and C on day 2 = 4 + 2 = 6 units

Work done by all of them on day 3 = 4 + 3 + 2 = 9 units

Work done in 3 days = 5 + 6 + 9 = 20 units

Work done in 6 days = $2\times20$ = 40 units

On day 7, B and C will work and finish 5 units

work left = 48 – 40 – 5 = 3 units

A and C will work for half day on day 8 to finish the work

Total time taken = $7\frac{1}{2}$ days

C works on all the day, work done by C = $\frac{15}{2}\times\ 2$ = 15 units

B works for 5 days, work done by B = $5\times\ 3$ = 15 units

A works for 4 days and a half day, work done by A = 4(4) + 2 = 18 units

Amount earned by A = $\frac{18}{15}\times\ 13,500$ = Rs 16,200

Question 17: The LCM of two number is 1617. If their HCF is a prime number (except 7), which of the following cannot be the difference between the two numbers?

a) 876

b) 114

c) 1606

d) 506

Solution:

Let the numbers be Ha and Hb, where H is the HCF of the two numbers and a and b are co-prime.
Now, $1617 = 3 \times 7^2 \times 11$
So, the LCM must be Hab.
This means that H can be 3, 7, or 11 as it is a prime number.
If H=3, a X b is $7^2 \times 11$. This means a = 49, b = 11 or a=539 and b=1
So, the difference in this case can be 3(49-11) = 3 X 38 = 114. Similarly, in the second case, the difference is 3(539-1) = 538 X 3 = 1614
If H = 11, a X b is $7^2 \times 3$. This means a = 49, b = 3 or a=147 and b=1
So, the difference in this case can be 11*(49-3) = 11 X 46 = 506. Similarly, in the second case, the difference is 11*(147-1) = 11*146 = 1606
So, all except A are possible differences.

Question 18: Three arithmetic sequences $S_{1}$, $S_{2}$ and $S_{3}$ are given below.
$S_{1}$ = 1, 4, 7, 10, …., 991
$S_{2}$ = 2, 6, 10, 14, …., 1014
$S_{3}$ = 3, 8, 13, 18, …., 1008
If a sequence $S_{x}$ contains all terms which are common in all three sequences, then find the sum of terms of the sequence $S_{x}$.

Solution:

Let us assume $S_{y}$ is the series that contains all the common terms in $S_{1}$ and $S_{2}$.

Common difference of the series $S_{1}$ = 3

Common difference of the series $S_{2}$ = 4

We can see that the first term of $S_{y}$ = 10. Also common difference of the arithmetic sequence $S_{y}$ = LCM of (3, 4) = 12.

Hence, we can say that $S_{y}$ = 10, 22, 34, 46, 58, …

Now we can say that $S_{x}$ is the series that contains all the common terms in $S_{x}$ and $S_{3}$.

Common difference of the series $S_{y}$ = 12

Common difference of the series $S_{3}$ = 5

We can see that the unit digit of each number in $S_{3}$ is either 3 or 8. We also know that all terms in $S_{x}$ are even. Hence, the first common term of both the series will be the one which if the first term in $S_{x}$ that ends with 8 which is 58.

Hence, we can say that the first term of $S_{x}$ = 58. Also common difference of the arithmetic sequence $S_{y}$ = LCM of (12, 5) = 60.

Therefore, $S_{x}$ = 58, 118, 178, …

We know that the last term of $S_{1}$ is 991. Hence we can say that the last term of $S_{x}$ will be $\leq$ 991.

Let ‘p’ be the number of terms in the series $S_{x}$. Then we can say that,

$\Rightarrow$ 58 + (p – 1)*60 $\leq$ 991

$\Rightarrow$ p $\leq$ 16.55

Hence, the number of terms in the series $S_{x}$ = 16.

Therefore, the sum of all the terms of series $S_{x}$ = $\dfrac{16}{2}(2*58+(16-1)60)$ = 8128

Question 19: There are 2 mixtures of brass and zinc, first with a ratio of brass and zinc 3:7 and second one with a ratio of 3:5. In what ratio should these mixtures be mixed to get a mixture with a ratio of brass and zinc of 7:13?

a) 3 : 7

b) 2 : 5

c) 1 : 4

d) 1 : 2

Solution:

Let us examine 40 gms (LCM of 10 and 8) of sample of both the mixtures.

In the first mixture, amount of brass, $b_1 = \dfrac{3}{3+7} \times 40 = 12$ gms

In the second mixture, amount of brass, $b_2 = \dfrac{3}{3+5} \times 40 = 15$ gms

In the first mixture, amount of zinc, $z_1 = \dfrac{7}{3+7} \times 40 = 28$ gms

In the second mixture, amount of zinc, $z_2 = \dfrac{7}{3+5} \times 40 = 25$ gms

Let the ratio of mixture of the first and the second mixture be $k$

Total amount of brass in the mixture = $(12 \times k) + 15 = 12k+15$

Total amount of zinc in the mixture = $(28 \times k) + 25 = 28k+25$

Ratio of brass and zinc in the mixture = $\dfrac{12k+15}{28k+25}$

$\Rightarrow \dfrac{7}{13} = \dfrac{12k+15}{28k+25}$

$\Rightarrow 196k+175 = 156k+195$

$\Rightarrow 40k = 20$

$\Rightarrow k = \dfrac{20}{40} = \dfrac{1}{2}$

Thus the required ratio is 1:2

Question 20: What will be the 23rd common term between the following two sequences?
3, 9, 15, 21. . . . . . . . 903
7, 12, 17, 22 . . . . . . . 952

Solution:

The two given sequences are
3, 9, 15, 21. . . . . . . . 903
7, 12, 17, 22 . . . . . . . 952

If we add one more term to each of the sequences then we can observe that the first term which will be common between the two sequences will be 27.
We can see that both the sequences are in AP with common difference of 6 and 5 respectively.
Hence the next common term in the sequence would be
27 + LCM (5, 6) = 57
Similarly the next common term would be
57 + 30 = 87
So we can see that the sequence of common terms form an AP with a common difference of 30 and first term as 27.
Hence the 23rd term of this AP would be
27 + 22*30 = 687