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Q:

The H.C.F and L.C.M of two numbers are  84 and 21 respectively.  If the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is

 A) 12 B) 48 C) 84 D) 108

Answer:   C) 84

Explanation:

Let the numbers be x and 4x. Then,  $\inline&space;\fn_jvn&space;x\times&space;4x&space;=84\times&space;21\:&space;\:&space;\Leftrightarrow&space;\:&space;\:&space;x^{2}=\frac{84\times&space;21}{4}&space;\:&space;\:&space;\Leftrightarrow&space;\:&space;\:&space;x=21$

Hence Larger Number = 4x = 84

Q:

54, 60 - LCM is ____

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

Answer & Explanation Answer: B) 540

Explanation:

Here the LCM of 54, 60 is

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

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5 131
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

Answer & Explanation Answer: D) -19

Explanation:

HCF of 210 and 55 is 5

Now, 210x5 + 55P = 5

=> 1050 + 55P = 5

=> 55P = -1045

=> P = -1045/55

=> P = -19.

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8 373
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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12 1330
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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13 1096
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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8 870