Question

please anyone solve these questions and if you do not know please don't answer useless answers otherwise I will report them​

Answer 1

Step-by-step explanation:

8)

amt. of calcium=10*1000/100

=100gm

amt. of carbon=3*1000/100

=30gm

amt. of oxygen=12*1000/100

=120gm

9)

Answer 2

Answer:

Q8

mass of calcium = 100g

mass of carbon = 30g

mass of oxygen = 120g

Q9

maximum marks = 200

minimum passing marks = 80

Q10

percentage of boys = 33.33%

Q11

candidate won 358800 votes

Step-by-step explanation:

Q8: 10% Ca ,  3% C , 12% O

weight of chalk = 1 Kg = 1000 g

mass of calcium = 10% of 1000 g = $\frac{10*1000}{100} = 100$ g

mass of Carbon = 3% of 1000 g = $\frac{3 *1000}{100} = 30$ g

mass of Oxygen = 12% of 1000g =$\frac{12*1000}{100}=120$ g

Q9: Let maximum marks in examination be x and minimum pass marks be y

Case 1

Student gets 25% marks and fails by 30 marks

Therefore marks scored by student = 25% of x = $\frac{25x}{100} = \frac{x}{4}$ marks

also we know he fails by 30 marks

So , marks scored + 30 = passing marks

$\frac{x}{4} +30 =y\\\frac{x +120}{4}=y\\x+120=4y\\x-4y=-120...(1)$

Case 2

Student scores 50% marks and gets 20 marks more than passing marks

Therefore marks scored by student = 50% of x = $\frac{50x}{100} = \frac{x}{2}$ marks

also we know that he got 20 marks more than passing marks

So, marks scored - 20 = passing marks

$\frac{x}{2}-20=y\\\frac{x-40}{2}=y\\x-40=2y\\x-2y=40...(2)$

Now we will work on equations (1) and (2) to get the values of x and y

(1) - (2)

$x-4y-x+2y=-120-40\\-2y = -160\\y=\frac{-160}{-2} = 80$

In (2) putting value of y

$x - 2y = 40\\x - (2*80) = 40\\x - 160=40\\x =40+160=200$

Therefore maximum marks in the examination were = x = 200

minimum passing marks in the examination were = y = 80

Q10: Number of Girls = 120

Number of Boys = 57

5% girls leave

Number of girls left = 5% of 120 = $\frac{5*120}{100} = \frac{12}{2} = 6$ girls

Number of girls left = 120 - 6 = 114

%of Boys in school = $\frac{Number \ of \ boys}{Total \ number \ of \ students \ in \ school}*100$

= $\frac{57}{57+114}*100 = \frac{57}{171}*100 = 33.33%$%

Therefore 33.33%

Q11: Total number of votes = 6,00,000

Total number of invalid votes = 8% of 6,00,000

=$\frac{600000*8}{100} = 6000*8 = 48000$ votes

Total number of valid votes = 600000-48000 = 552000

Since candidate got 65% of total valid votes

number of votes received by the candidate = 65% of 552000

$\frac{65*552000}{100} = 65*5520 = 358800$ votes

Therefore candidate received 3,58,800 votes