Numbers - 03

Explanation:-

LCM = 45 HCF
LCM + HCF = 1150
45*HCF + HCF = 1150
46 × HCF = 1150
HCF = 25
Then,
LCM = 45 × 25 = 1125.
Now, use the formula,
1st number × second Number = LCM × HCF
125 × second number = 1125 × 25
Second Number = 225

Explanation:-

In every 30 minutes, Virus doubles it self. It means in 49.5 hours, it would be half empty and it will be full in 50 hours.

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{two}}\,{\text{numbers}}\,{\text{be}}\,X\,{\text{and}}\,Y. \cr & {\text{According}}\,{\text{to}}\,{\text{question}}, \cr & X \times \left( {\frac{3}{5}} \right) = 70\% \,of\,Y \cr & \frac{{3X}}{5} = \frac{{70Y}}{{100}} \cr & \frac{X}{Y} = \frac{{\left( {70 \times 5} \right)}}{{\left( {100 \times 3} \right)}} = \frac{7}{6} \cr & X:Y = 7:6 \cr}$$

Explanation:-

Let he has solved correctly X no. of sums. Therefore incorrect no. of sums = 2X .
Now,
X + 2X = 48
3X = 48
X = 16 sums he has done correctly.

Explanation:-

Assume x is the required number.
Step 1 Amitabh 2x Sashi 2x + 50
Step 2 Amitabh 4x + 100. Sashi 4x + 150
Step 3 Amitabh 8x + 300. Sashi 8x + 350
Step 4 Amitabh 16x + 700 Sashi 16x + 750
Now, 16x + 750 > 1000
And, x can be 16
So 7 is the sum

Explanation:-

Let number be X.
According to question,
7X - 3X = 36
4X = 36
X = 9
Required Number is 9

Explanation:-

Required Number can be given by:

HCF of (3026 - 11) and (5053 - 13)
HCF of 3015 and 5040 = 15
To find HCF, we break the given numbers in their prime Factors
3015 = 3 × 3 × 5 × 67
5040 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7
We take common multiples in these two given numbers to get required HCF
And Common multiples are: 3 × 5
So, Required HCF = 15

Explanation:-

$$\eqalign{ & {\text{Total}}\,{\text{Marks}} = 90 \cr & {\text{Marks}}\,{\text{in}}\,{\text{QA}} = \frac{7}{9}\,of\,90 \cr & = \frac{{\left( {7 \times 90} \right)}}{9} \cr & = 7 \times 10 \cr & = 70 \cr}$$

Explanation:-

The only possible number which is both perfect square and perfect cube is 729 which is cube of 9 and square of 27.
So, 7 × 2 × 9 = 126.

Explanation:-

$$\eqalign{ & \frac{{{{75}^{{{75}^{75}}}}}}{{37}} \cr & {\text{When 75 is divided by 37, it leaves remainder of }}1. \cr & {\text{So, the expression will become}}, \cr & \frac{{{1^{{{75}^{75}}}}}}{{37}} \cr & {\text{Now,}}\,{\text{for}}\,{\text{any}}\,{\text{power}}\,{\text{of}}\,1, \cr & {\text{we}}\,{\text{always}}\,{\text{get}}\,{\text{1}}\,{\text{and}}\,{\text{expression}}\,{\text{becomes}}, \cr & \frac{1}{{37}}\,{\text{that}}\,{\text{leaves}}\,{\text{remainder}}\,1 \cr}$$