# LCM & HCF - 1 | Algebra - Number Theory

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**LCM & HCF - 1 | Algebra - Number Theory**

Out of the total number of chocolates he had, Ram gave 1/3^{rd} of the chocolates to his sister, 1/7^{th} to his brother and 1/5^{th} to his friend. How many chocolates did he have in the beginning?

- (a)
15

- (b)
36

- (c)
45

- (d)
70

- (e)
105

Answer: Option E

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**Explanation** :

Let the number of chocolates be N.

Now, since N/3, N/7 and N/5 are integers, N should be divisible by all of 3, 7 and 5.

∴ N should be a multiple of LCM (3, 7, 5) = 105.

Hence, option (d).

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**LCM & HCF - 1 | Algebra - Number Theory**

Two people running in a circle can complete one revolution in 4 hours and 5 hours respectively. After how many hours will they meet at the starting point?

- (a)
20

- (b)
9

- (c)
15

- (d)
18

- (e)
21

Answer: Option A

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**Explanation** :

First person reaches start point after 4, 8, 12, …. (hours)

Second person reaches start point after 5, 10, 15, …. (hours)

Both will meet each other at the starting point after every LCM (4, 5) = 20 hours.

Hence, option (a).

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**LCM & HCF - 1 | Algebra - Number Theory**

Suppose, you have 108 green marbles and 144 red marbles. You decide to separate them into packages of equal number of marbles of same colour. Find the maximum possible number of marbles in each package.

- (a)
4

- (b)
36

- (c)
9

- (d)
24

- (e)
12

Answer: Option B

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**Explanation** :

Maximum possible number of marbles that divides both 108 and 144 is equal to the greatest common divisor of 108 and 144.

∴ 108 = 4 × 27 = 2^{2} × 3^{3}

& 144 = 122 = 2^{4} × 3^{2}

⇒ GCD/HCF of 108 and 144 = 2^{2} × 3^{2} = 36.

Hence, option (b).

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**LCM & HCF - 1 | Algebra - Number Theory**

The LCM of two numbers is 2800 and their HCF is 40. If one of the number is 280, then the other number is?

Answer: 400

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**Explanation** :

We know, product of two numbers = product of their LCM and HCF.

∴ LCM × HCF = Product of Numbers

N_{2} = (2800 × 40)/280 = 400

Hence, 400.

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**LCM & HCF - 1 | Algebra - Number Theory**

The product of two numbers is 8100. If their LCM is 108, then find their HCF.

- (a)
75

- (b)
70

- (c)
80

- (d)
60

- (e)
Data inconsistent

Answer: Option E

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**Explanation** :

We know, product of two numbers = product of their LCM and HCF.

∴ LCM × HCF = Product of Numbers

N_{2} = 8100/108 = 75

But, the data is inconsistent as the LCM is always a multiple of HCF.

Here HCF = 75 and LCM = 108 do not show consistency, i.e. no such numbers are possible.

Hence, option (e).

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**LCM & HCF - 1 | Algebra - Number Theory**

Product of two positive integers is 12615 and their HCF is 29. How many such pairs are possible?

- (a)
1

- (b)
2

- (c)
3

- (d)
4

- (e)
None of these

Answer: Option B

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**Explanation** :

Since HCF is 29

∴ Let numbers be 29a and 29b (where a and b are co-primes)

We have, 29a × 29b = 12615, or ab = 15.

Now, we have to write 15 as a product two co-prime numbers, hence

15 = 1 × 15

15 = 3 × 5

There are only two pairs of co-prime numbers whose product will be 15.

Hence, option (b).

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**LCM & HCF - 1 | Algebra - Number Theory**

The HCF and LCM of two numbers are 26 and 910 respectively. If one of the numbers lies between 150 and 250, then that number is

- (a)
182

- (b)
156

- (c)
234

- (d)
130

Answer: Option A

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**Explanation** :

Since HCF is 26, let us assume that the numbers are 26x and 26y

∴ Product of the two numbers = HCF × LCM = 26x × 26y

⇒ 26x × 26y = 26 × 910

∴ xy = 35

Now, we have to write 35 as a product two co-prime numbers, hence

35 = 1 × 35 (one of x and y is 1 and the other is 35)

35 = 7 × 5 (one of x and y is 7 and the other is 5)

The numbers can be 26 × 1, 26 × 35 or 26 × 5, 26 × 7

Only one number lies within the given limits, 26 × 7 = 182.

Hence, option (a).

Workspace:

**LCM & HCF - 1 | Algebra - Number Theory**

If the HCF of two numbers is 48 and the HCF of two other numbers is 36, what will be the HCF of all the four numbers?

- (a)
4

- (b)
6

- (c)
12

- (d)
8

- (e)
24

Answer: Option C

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**Explanation** :

Since HCF of first two numbers is 48, let the numbers be 48a and 48b.

Also, HCF of next two numbers is 36, let the numbers be 36x and 36y.

Now, HCF of all 4 numbers = HCF(48a, 48b, 36x, 36y).

Since, a and b are co-primes and also x and y are also co primes, a, b, x and y will have not common factor.

⇒ HCF(48a, 48b, 36x, 36y) = HCF(48, 36) = 12.

Note: HCF of all the four numbers = HCF of (HCF of the first 2 numbers and HCF of the other 2 numbers.

Hence, option (c).

Workspace:

**LCM & HCF - 1 | Algebra - Number Theory**

There are three pieces of cake weighing 5/3 lbs,23/4 lbs and 21/15 lbs. The pieces are equally divided and distributed in such a manner that every guest in the party gets one single piece of the cake. Further the weight of each piece of the cake is as heavy as possible. How many guests attended the party?

Answer: 529

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**Explanation** :

Since the weight of the cake was as heavy as possible, it can be obtained by finding the H.C.F. of the three fractions.

To calculate HCF of fractions, first we need to simplify the fractions and then, HCF of fractions = $\frac{HCFofNumerators}{LCMofDenominators}$

∴H.C.F$\left[\left(\frac{5}{3}\right),\left(\frac{23}{4}\right),\mathrm{}\left(\frac{7}{5}\right)\right]$ = $\frac{HCF\left(\mathrm{5,23,7}\right)}{\mathrm{LCM}(3,\mathrm{}4,\mathrm{}5)}$ = $\frac{1}{60}$

Now, since each guest received 1/60 lbs of cake,

The number of guests who attended the party = $\frac{\left[\left(\frac{5}{3}\right)+\left(\frac{23}{4}\right)+\left(\frac{21}{15}\right)\right]}{\left(\frac{1}{60}\right)}$ = 529.

Hence, 529.

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