Test: LCM & GCD - 2 - GMAT MCQ

Result 1:
The product of two positive integers ‘a’ and ‘b’ is equal to the product of the LCM (a, b) and HCF (a, b).
i.e., a * b = LCM (a, b) * HCF (a, b)

Result 2:
HCF is a factor of both the positive integers.
LCM is a multiple of both the positive integers.
So, it is evident that the LCM of the two positive integers has to be a multiple of the HCF of the two numbers.

Combining these two results, we have to find out which among the pairs has a product of 144 such that the LCM is a multiple of the HCF.

I. LCM : 24 and HCF : 6. Product of the LCM and HCF = 24 * 6 = 144. The LCM 24 is a multiple of the HCF 6. Hence, this is a possible pair.

II. LCM : 18 and HCF : 8. Product of the LCM and HCF = 18 * 8 = 144. However, the LCM 18 is NOT a multiple of the HCF 8. Hence, this one is not a possible pair.

III. LCM : 16 and HCF : 9. Product of the LCM and HCF = 16 * 9 = 144. However, the LCM 16 is NOT a multiple of the HCF 9. Hence, this one is also not a possible pair.

We want 4 digit number, so option A is not the answer
Now we want greatest number. So out of remaining options, 9600 is greatest
9600 is divisible by 25, 75, 40 and 15
So answer is C i.e. 9600

Solution

LCM of 3, 5, 6, 8, 10 and 12=2×2×2×3×5=120
∴Required number=(120k+2)
(120k+2) is divisible by 13 then k=8
Required number =120×8+2=962

In such cases and sums, simply follow these easy steps
Number is divided by 893. Remainder =193.
Also, we observe that 893 is exactly divisible by 47.
So now simply divide the remainder by 47. 

Highest Common Factor (HCF) is 13, that means
the numbers are multiples of 13.
We need sum of numbers to be 156.
In such sums, first write multiples of 13 till 156 

Here we need to exactly measure these lengths.
Also, lm = 100 cm; ... 1m 35cm = 135cm and 1m 20cm = 120cm
Option 1- 5 cm can measure 30cm, 90cm, 1m 20cm and 1m 35 cm
Option 2 - 10 cm cannot measure 1m 35cm i.e. 135cm
Option 3 -15 cm can measure30cm, 90cm, 1m 20cm and 1m 35 cm
Option 4 -30 cm cannot measure 1m 35cm
Since we want greatest length 15 > 5
∴ Answer = 15cm 

90 is not a perfect square.
90=3x3x2x5
Here there are two 3's (this gives square of 3) but only one 2 and one 5
if we multiply by 2 and 5 then we get, 3 x3x 2x2x5x 5= 900
900 is perfect square = number of marbles

∴ LCM = 9x3x5x7 = 945  =  They meet after 945 minutes.

Here least possible 4 digit number is needed that means we need the LCM We must first find LCM of 12, 16, 18 and 20 .

We need the next instances when the alarm rings.
That means the Least Common Multiple (LCM) of 3, 4, 8, 10, 12 3, 4 divide 12 so neglect them.
LCM of 8, 10 and 12 

∴ LCM = 4x2x5x3= 120 = They ring after 2 hours
After starting, they ring together ist time in 2 hours.
Then 2nd time in 2+2 = 4 hours.
Then 3rd time in 6 hours.
And 4th time in 8 hours. 

Here least number is needed that means we need the LCM
We must first find LCM of 36, 24 and 16 

(x-4) is HCF i.e. a factor of both equations
∴ The equations must get satisfied for x=4
Also, when x=4, both equations are equal in value.
Putting x=4 in both equations
4² - 8(4) + 15 = 4²- k(4) +1

∴ 31-32 = 17- 4k
∴ k=4 

Here least number is needed that means we need
the Least Common Multiple i.e. LCM
We must now first find LCM of 20, 36 and 48 

:.LCM=4x3 x5 x 3 x 4 = 720
But this is not the mnverbecause there are remainders as per the given condition.
If we observe dosely, the difference between the given numbers and remainders is same
20-13 = 7; 48-1 = 7; 36-29 = 7
Difference is same = 7
So simply subtract this difference from LCM.
Number = 720 - 7 = 713

Here greatest number that can divide means the HCF
Remainders are different so simply subtract remainders from numbers
17-4 = 13; 42-3 = 39; 93-15 = 78
Now let's find HCF of 13, 39 and 78
By direct observation we can see that all numbers are divisible by 13.
∴ HCF =13 = required greatest number 

What to do when we don't lcnow the remainder and the largest number which divides?
In such cases subtract the 3 numbers from each other 
47-35 = 12
35-27 = 8
27-47 = -20 (Ignore minus sign) 
Largest number is needed that means we need HCF of 12, 8 and 20