Test: Pie Charts - 2 - SSC CGL MCQ

Let the total population of all the seven village in 1997 be N.
Then, population of village R in 1997
= 16% of N = 16N/100
and population of village Z in 1997
= 11% of N = 11N/100
Population of village R in 1999 
= { 16N/100 + (10% of 16N/100) }
= 1760N/10000
and population of village Z in 1999
= { 11N/100 - (5% of 11N/100)}
= 1045N/10000
Now, population of village R below poverty line for 1999
= 51% of 1760N/10000 
and population of village Z below poverty line for 1999
= 42% of  1045N/10000 
= 51% of (1760N/10000) / 42% of (1045N/10000)

= 2/1

Population of village Y in 1997 = 30000 (given).

Let the population of village V in 1997 be v.

Then, 15 : 10 = 30000 : v   =>  

15
= 20000.

Now, population of village V in 1998 

= 20000 + (10% of 20000) = 22000.

Therefore Population of village V below poverty line in 1998 

= 58% of 22000 = 12760.

Population of village R = 32000 (given).

Let the population of village Y be y.

Then, 16 : 15 = 32000 : y 
= 30000.

∴ Population of village Y below poverty line 

= 52% of 30000 = 15600

Let N be the total population of all the seven villages.

Then, population of village T below poverty line 

= 46% of (21% of N)

and Population of villages Z below the poverty line

 = 42% of (11% of N)

= 23/11.

Let the population of village X be x.

Then, 38% of x = 12160 
= 32000.
Now, if s be the population of village S, then
16 : 11 = 32000 : s ⇒ s =

Required difference:

= [(16% + 18%) of 5700] - [(8% + 10%) of 8550

= [(34% of 5700) - (18% of 8550)]

= (1938 - 1539)

= 399

Candidates passed from institutes Q and R together
= [(13% + 17%) of 5700]
= 30% of 5700.
Candidates enrolled from institute Q and R together
= [(15% + 10%) of 8550]
= 25% of 8550.
Required Percentage    


= 80%

The percentage of candidates passed to candidates enrolled can be determined for each institute as under:


Highest of these is 86.67% corresponding to institute R.

Required ratio = 

Required percentage

Percentage of proteins and other dry elements in the body

=(16% + 14%)

= 30%

Central angle corresponding to proteins and other dry elements together

= 30% of 360º

= 108º.

Let the body weight be x kg.

Then, weight of skin protein in the body

Quantity of water in the body of a person weighing 50 kg

= (70% of 50)

= 35 kg

Part of the body made of neither bones nor skin
= 1 - (1 + 1) = 11.61015