38
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:

54, 60 - LCM is ____

 A) 600 B) 540 C) 60 D) 54

Explanation:

Here the LCM of 54, 60 is

=> 2 x 3 x 3 x 3 x 10 = 540

5 102
Q:

If the HCF of 210 and 55 is expressible in the form of 210 x 5 + 55P, then value of P = ?

 A) -23 B) 27 C) 16 D) -19

Explanation:

HCF of 210 and 55 is 5

Now, 210x5 + 55P = 5

=> 1050 + 55P = 5

=> 55P = -1045

=> P = -1045/55

=> P = -19.

8 349
Q:

A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is  ?

 A) 63 B) 31 C) 16 D) 27

Explanation:

To get the least number of coconuts :
LCM = 30 => 30 + 1 = 31

12 1315
Q:

A drink vendor has 368 liters of Maaza, 80 liters of Pepsi and 144 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required ?

 A) 47 B) 46 C) 37 D) 35

Explanation:

The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = 368/16 = 23
Number of cans of Pepsi = 80/16 = 5
Number of cans of Sprite = 144/16 = 9
The total number of cans required = 23 + 5 + 9 = 37 cans.

13 1076
Q:

H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 and 16 x 81 x 5 x 11 x 49 is :

 A) 360 B) 180 C) 90 D) 120

Explanation:

4 x 27 x 3125 = $\inline \fn_jvn \small 2^{2}x3^{3}x5^{5}$ ;

8 x 9 x 25 x 7 = $\inline \fn_jvn \small 2^{3}x3^{2}x5^{2}x7$

16 x 81 x 5 x 11 x 49 = $\inline \fn_jvn \small 2^{4}x3^{4}x5x7^{2}x11$

H.C.F = $\inline \fn_jvn \small 2^{2}x3^{2}x5$ = 180.