26
Q:

# L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :

 A) -2 B) -1 C) 1 D) 2

Explanation:

H. C. F of two prime numbers is 1.  Product of numbers = 1 x 161 = 161.

Let the numbers be a and b . Then , ab= 161.

Now, co-primes with  product 161 are (1, 161) and (7, 23).

Since x and y are prime numbers and x >y , we have x=23 and y=7.

$\inline&space;\fn_jvn&space;\therefore$ 3y-x = (3 x 7)-23 = -2

Q:

Find the greatest number that will divide 964, 1238 and 1400 leaving remainder of 41,31 and 51 respectively ?

 A) 64 B) 69 C) 71 D) 58

Explanation:

To find the greatest number which divides the numbers 964, 1238 and 1400 leaving the remainders 41, 31 and 51 is nothing but the HCF of (964 - 41), (1238 - 31), (1400 - 51).

Therefore, HCF of 923, 1207 and 1349 is 71.

2 41
Q:

A student can divide her books into groups of 5, 9 and 13. what is the smallest possible number of the books ?

 A) 487 B) 585 C) 635 D) 705

Explanation:

The smallest possible number of books = L.C.M of 5, 9 and 13.
Therefore, L.C.M of 5,9 and 13 is = 5 x 9 x 13 = 585.

4 38
Q:

If HCF of two numbers is 8,which of the following can never be their LCM?

 A) 32 B) 48 C) 60 D) 152

Explanation:

60 is not a multiple of 8

6 54
Q:

In a palace, three different types of coins are there namely gold, silver and bronze. The number of gold, silver and bronze coins is 18000, 9600 and 3600 respectively. Find the minimum number of rooms required if in each room should give the same number of coins of the same type ?

 A) 26 B) 24 C) 18 D) 12

Explanation:

Gold coins = 18000 , Silver coins = 9600 , Bronze coins = 3600

Find a number which exactly divide all these numbers

That is HCF of 18000, 9600& 3600

All the value has 00 at end so the factor will also have 00.

HCF for 180, 96 & 36.

Factors of

180 = 3 x 3 x 5 x 2 x 2

96 = 2 x 2 x 2 x 2 x 2 x 3

36 = 2 x 2 x 3 x 3

Common factors are 2x2×3=12

$\inline \fn_jvn \small \therefore$ Actual HCF is 1200

Gold Coins $\inline \fn_jvn \small \left ( \frac{18000}{1200} \right )$ will be in 15 rooms

Silver Coins $\inline \fn_jvn \small \left ( \frac{9600}{1200} \right )$ will be in 8 rooms

Bronze Coins $\inline \fn_jvn \small \left ( \frac{3600}{1200} \right )$ will be in 3 rooms

Total rooms will be (15+8+3)  =  26 rooms.

5 245
Q:

Three numbers are in the ratio of 3:4:5 and their L.C.M is 3600.Their HCF is:

 A) 40 B) 60 C) 100 D) 120

Explanation:

Let the numbers be 3x, 4x, 5x.

Then, their L.C.M = 60x.

So, 60x=3600 or x=60.

$\therefore$ The numbers are (3 x 60), (4 x 60), (5 x 60).

Hence,required H.C.F=60