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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

Answer:   A) -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:

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

Answer & Explanation Answer: B) 31

Explanation:

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

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6 400
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

Answer & Explanation Answer: C) 37

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.

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8 475
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

Answer & Explanation Answer: B) 180

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.

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5 375
Q:

The difference of two numbers is 14. Their LCM and HCF are 441 and 7. Find the two numbers ?

 A) 63 and 49 B) 64 and 48 C) 62 and 46 D) 64 and 49

Answer & Explanation Answer: A) 63 and 49

Explanation:

Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y

Difference = 14
=> 7x - 7y = 14
=> x - y = 2

product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63

Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7

The numbers are 7x and 7y
=> 63 and 49

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8 617
Q:

If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to ?

 A) 13/125 B) 14/57 C) 11/120 D) 16/41

Answer & Explanation Answer: C) 11/120

Explanation:

Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120

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9 420