The following pie-chart shows the sources of funds to be collected by the National
Highways Authority of India (NHAI) for its Phase II projects. Study the pie-chart and answer
the questions that follow.
Total funds to be arranged for Projects (Phase II) =Rs.57,600 crores.
Ans .
External Assistance
20% of the total funds to be arranged = Rs.(20% of 57600) crores
= Rs.11520 crores Rs.11486 crores.
Ans .
(\(187.2^{\circ}\)
Central angle corresponding to Market Borrowing = \([\frac{29952}{57600}*360]^{\circ}\)= (\(187.2^{\circ}\)
Ans .
1:6
Required ratio = \(\frac{4910}{29952}\)= \(\frac{1}{6.1}\) =\(\frac{1}{6}\)
Ans .
6%
Shortage of funds arranged through External Assistance
=Rs.(11486-9695) crores =Rs. 1791 crores.
therefore, Increase required in Market Borrowings =Rs. 1791 crores.
Percentage increase required = \(\frac{1791}{29952}\)* 100 % = 5.98 % = 6%
Ans .
Rs. 5401 crores
Amount permitted = (Funds required from Toll for projects of Phase II ) +(10 % of these funds)
=Rs. 4910 crores + Rs. (10% of 4910) crores
=Rs. (4910 + 491) crores = Rs. 5401 crores.
The pie-chart provided below gives the distribution of land (in a village) under
various food crops. Study the pie-chart carefully and answer the questions that follow.
DISTRIBUTION OF AREAS (IN ACRES) UNDER VARIOUS FOOD CROPS
Ans .
Rice, Wheat and Barley
The total of the central angles corresponding to the three crops which cover 50% of the
total area ,should be 180?.Now, the total of the central angles for the given combinations are:
(i) Wheat,Barley and jowar =\(72^{\circ}\)+ \(36^{\circ}\)+ \(18^{\circ}\) =\(126^{\circ}\)
(ii) Rice,Wheat and Jowar = \(72^{\circ}\)+\(72^{\circ}\)+\(18^{\circ}\)=\(162^{\circ}\)
(iii) Rice,Wheat and Barley = \(72^{\circ}\)+\(72^{\circ}\)+\(36^{\circ}\)=\(180^{\circ}\)
(iv)Bajra,Maize and Rice = \(18^{\circ}\)+ \(45^{\circ}\)+\(72^{\circ}\) = \(135^{\circ}\)
Clearly:(iii) is the required combination.
Ans .
6
The area under any of the food crops is proportional to the angle corresponding to that crop.
Let the area under the rice production be x million acres.
Then, 18:72 = 1.5:x =>x=(72*15/18)=6
Thus, the area under rice production be = 6 million acres.
Ans .
3:1
Let the total production of barley be T tones and let Z acres of land be put under barley production.
Then, the total production of wheat =(6T) tones.
Also,area under wheat production = (2Z) acres.
\(\frac{Area Under Wheat Production}{Area Under Barley Production}\)=\(\frac{72^{\circ}}{36^{\circ}}\)=2
And therefore,Area under wheat = 2*Area under Barley = (2Z)acres
Now, yield per acre for wheat = (6T/2Z) tones/acre = (3T/Z) tones/acre
And yield per acre for barley = (T/Z) tones/acre.
Required ratio = \(\frac{3T/Z}{T/Z}\)= 3:1.
Ans .
33%
Let Z acres of land be put under barley production.
\(\frac{Area Under Wheat Production}{Area Under Barley Production}\)=\(\frac{75^{\circ}}{36^{\circ}}\)=2
Area under rice production = 2 * area under barley production = (2Z) acres.
Now,if p tones be the yield per acre of barley then ,yield per acre of rice=(p+50% of p) tones =(3/2 p) tones.
Total production of rice = (yield per acre) * (area under production)= (3/2 p)*2Z=(3pZ) tones.
And,Total production of barley = (pz) tones.
Percentage production of barley to that rice = (pZ/3pZ *100)%= 33 1/3%
Ans .
\(76.8^{\circ}\)
Initially,let t be the total area under considerations.
The area under wheat production initially was =(72/360 * t)acres = (t/5)acres.
Now,if the total area under consideration be increased by 5%,
then the new value of the total area= (105/100 t) acres.
Also,if the area under wheat production be increased by 12%,
then the new value of area under wheat =|\(\frac{t}{5}\) +(12% of \(\frac{t}{5}\))| acres = (112t/500)acres.
Central angle corresponding to wheat in the pie-chart
= \([\frac{Area Under Wheat (new)}{Total area (new)}*360]^{\circ}\) = \([\frac{(112t/500)}{(105t/100)}\)*360]^{\circ}\) =\(76.8^{\circ}\)
The following pie-charts show the distribution of students of graduate and post graduate levels in seven different institute-M,N,P,Q,R,S and T in a town.
DISTRIBUTION OF STUDENTS AT GRADUATE AND POST-GRADUATE LEVELS IN SEVEN INSTITUTES-M,N,P,Q,R,S AND T.
Ans .
(b) 8463
Students of institute M at graduate level = 17% of 27300 = 4641.
Students of institute S at graduate level = 14% of 27300 = 3822
Total number students at graduate level in institutes M and S = 4641+3822=8463.
Ans .
(c) 6669
Required number = (15% of 24700) + (12% of 24700) = 3705 + 2964 = 6669
Ans .
(d) 8372
Required number = (18% of 27300) + (14% of 24700) = 4914 + 3458 = 8372.
Ans .
(d) 19:14
Required ratio =\(\frac{21% of 24700}{14% of 27300}\) = \(\frac{21 * 24700}{14 * 27300}\) = \(\frac{19}{14}\)
Ans .
(d)19:13
Required ratio =\(\frac{21% of 24700}{14% of 27300}\) = \(\frac{(21 * 24700)}{(13* 27300)}\) = \(\frac{19}{13}\)
Study the following pie-chart and the table and the answer the questions based on them.
PROPORTION OF POPULATION OF SEVEN VILLAGES IN 1997
Ans .
22000.
Let the population of village X be x
Then,38% of x=12160 => x = \(\frac{12160 * 100}{38}\) =3200
Now ,if s be the population village S,then
16:11 = 32000 : s => s= \(\frac{11 * 32000}{16}\)= 22000.
Ans .
23:11
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 village Z
below poverty line = 42% of (11% of N)
Required ratio =\(\frac{46percent of 21percent of N}{42percent of 11percent of N}\) = \(\frac{46*21}{42*11}\) = \(\frac{23}{11}\)
Ans .
15600
Population of village R = 32000(given)
Let the population of village Y be y.
Then, 16:15 = 32000 : y => y = \(\frac{15 * 32000}{16}\)= 30000.
Ans .
12760
Population of village Y in 1997 = 30000(given) .
Let the population village V in 1997 be v.
Then, 15:10 = 30000:v => v = \(\frac{10 * 30000}{15}\)= 20000.
Now population of village V in 1998 = 20000 + (10% 0f 20000) = 20000.
Population of village V below poverty line in 1998 = 58% of 22000 = 12760.
Ans .
2:1
Let the total population of all the seven villages in 1997 be N.
Then,population of village R in 1997 = 16% of N = 16/ 100 N
And population of village Z in 1997 = 11% of N =11/100 N
Population of village R in 1999 = {16/100N+(10% of 16/100 N)}=1760/10000 N
and population of village Z in 1999 = {11/100 N-(5% of 11/100 N)} = 1045/10000 N.
Now,population of village R below poverty line for 1999 = 51% of(1760/10000 N)
And population of village Z below poverty line 1999 = 42% of (1045/10000 N)
Required ratio =\(\frac{51 percent of 1760/10000 N}{42 percent of 1045/10000 N}\) = \(\frac{51 * 1760 }{42 * 1045}\) = \(\frac{2}{1}\)