Number System-Test 2

Answer: Option D

Explanation

Let one number be 'x'

Then, another number is x+18

Now, product of two numbers= HCF×LCM

⇒ x×(x+18)=6×168

⇒ x²×18x=1008

⇒ x²×18x-1008=0

⇒ x²×42x-24x-1008=0

⇒ x(x-24)+42(x-24)=0

⇒ x=24

∴ Another number is 42.

Required sum =(24)²+(42)²

=576+1764=2340.

Hence, Option D is correct.

Answer: Option D

Explanation

Here, LCM of two numbers=120

HCF of these two numbers=10

Let the two numbers be 10x and 10y.

Then, LCM of 10x and 10y=120

=10xy =120

= xy =12

Here, xy can be (3,4)

So, numbers are 10×3 and 10×4.

Sum of these two numbers =30+40=70

Hence, Option D is correct.

Answer: Option B

Explanation

Here, we have to find maximum number.

Hence, we have to find out the HCF of 2923 and 3239

So, find the HCF of these two numbers by factorisation method

i.e, 2923=1, 37, 79

      3239=1, 41, 79

∴ HCF=79

Hence, maximum number of girls =79.

Hence, Option B is correct.

Answer: Option A

Explanation

Here, the sum of two numbers =45

or    x+y=45

And x×y=500

So, going through the option 5 is the only common  factor which satifies both the equations.

Hence, Option A is correct.

Answer: Option C

Explanation

Let LCM of the two numbers= x

Then, HCF of the two numbers= $\frac{x}{28}$

Now, 

$$\begin{gathered} \Rightarrow x+\frac{x}{28}=1740 \\ \Rightarrow \frac{29x}{28}=1740 \\ \Rightarrow x=\frac{1740\times 28}{29}=1680 \end{gathered}$$

So, LCM of these two numbers =1680

Again, 240×y=1680×60

[First number × Second number]= [LCM×HCF]

or y=7×60y=420.

Hence, the number is 420 or sum of digits =4+2+0=6.

Hence, Option C is correct.