Percent and Percentage Test - 6

Consider x as the number

16 2/3% of x = 30

By further calculation

50/ (3 × 100) of x = 30

So we get

1/6 of x = 30

Here

x = 30 × 6 = 180

So the number is 180.

Salary of a man = ₹ 1200

Increased salary of a man = ₹ 1700

So the amount of increase = 1700 – 1200 = ₹ 500

Here the percentage increase = (500 × 100)/ 1200

We get

= 125/3

=41.67%

20 by 15%

Increase on 20 by 15% = 20 × 15/100 = 3

So the increased number = 20 + 3 = 23

Consider x as the number

0.2% of x = 5

By further calculation

2/ (10 × 100) of x = 5

So we get

1/500 of x = 5

Here

x = (5 × 500)/ 1 = 2500

So the number is 2500.

In an alloy

Solid = 13 parts

Liquid = 7 parts

Gas = 5 parts

So the total object = 13 + 7 + 5 = 25 parts

Percentage of Solid = 13/25 × 100 = 52%

Percentage of liquid = 7/25 × 100 = 28%

Percentage of gas = 5/25 × 100 = 20%

300 by10%

Decrease on 300 by 10% = 300 × 10/100 = 30

So the decreased number = 300 – 30 = 270

60 by 5%

It is given that

Rate of increase = 5%

So the total increase = 5% of 60

We can write it as

= 5/100 × 60

= 3

Here the increased number = 60 + 3 = 63

Petrol in the cask = 900 litres

Petrol lost by leakage and evaporation = 16%

So the petrol lost = 16% of 900 litres

We can write it as

= (16 × 900)/ 100

= 154 litres

Petrol left in the cask = 900 – 154 = 746 litres

Consider 100 as the number

So the increase = 20% = 20

Increased number = 100 + 20 = 120

If the increased number is 120 then the original number = 100

If the increased number is 96 then the original number = 100/120× 96 = 80

Marks for first division = 60%

A student gets 428 marks and misses the first division by 112marks

Marks for first division = 428 + 112 = 540

60% of total marks = 540

We can write it as

60/100 × total marks = 540

So we get

Total marks = (540 × 100)/ 60 = 900

Petrol in the cask = 225 litres

Petrol lost by leakage and evaporation = 10%

So the petrol lost = 10% of 225 litres

We can write it as

= (10 × 225)/ 100

= 22.5 litres

Petrol left in the cask = 225 – 22.5 = 202.5 litres

48 by 12 ½ %

Increase on 48 by 12 ½ % = 48 × 25/2%

We can write it as

= 48 × 25/ (2 × 100)

By further calculation

= 48 × 1/8

= 6

So the increased number = 48 + 6 = 54

50 by 12.5%

Decrease on 50 by 12.5% = 50 × 12.5/100

We can write it as

= (50 × 125)/ (10 × 100)

= 25/4

= 6.25

Sothe decreased number = 50 – 6.25 = 43.75

Weight of the machine = 20 kg

Error weight of the machine = 20.8 kg

Error in weight = 20.4 – 20 = 0.4 kg

So the error percent = (0.4 × 100)/ 20

We can write it as

= (4 × 100)/ (10 × 20)

= 2%

80 by 20%

Decrease on 80 by 20% = 80 × 20/100 = 16

So the decreased number = 80 – 16 = 64