There are five prime numbers between 30 and 50. They are 31,37,41,43 and 47.Therefore the required average=\(\frac{(31+37+41+43+47)}{5}\)=\(\frac{199}{5}\) =39.8.
sum of first n natural numbers=n(n+1)/2;
So,sum of 40 natural numbers=\(\frac{(40*41)}{2}\)=820.
Therefore the required average=\(\frac{820}{40}\)=20.5.
Required average =7*\(\frac{(1+2+3+…….+20)}{20}\)=\(\frac{(7*20*21)}{(20*2)}\)=\(\frac{147}{2}\)=73.5.
let the numbers be x,x+2,x+4 and x+6. Then,
\(\frac{(x+(x+2)+(x+4)+(x+6))}{4}\) = 27
\(\frac{(4x+12)}{4}\)= 27
x+3=27
x=24.
Therefore the largest number=(x+6)=24+6=30.
Total weight of(36+44) students=(36*40+44*35)kg =2980kg. Therefore weight of the total class=(2980/80)kg =37.25kg.
Let the average expenditure of all nine be Rs.x Then 12*8+(x+8)=9x or 8x=104 or x=13. Total money spent = 9x = Rs.(9*13)= Rs.117.
Let the third number be x.
Then second number = 3x.
First number=3x/2.
Therefore x+3x+(3x/2)=(44*3) or x=24 So largest number= 2nd number=3x=72.
Clearly 13th result=(sum of 25 results)-(sum of 24 results)
=(18*25)-(14*12)+(17*12) =450-(168+204) =450-372 =78
6th result = (58*6+63*6-60*11)=66
Let A,B,c represent their individual wgts. Then, A+B+C=(45*3)Kg=135Kg A+B=(40*2)Kg=80Kg & B+C=(43*2)Kg=86Kg B=(A+B)+(B+C)-(A+B+C) =(80+86-135)Kg =31Kg.
Total age of 39 persons = (39 * 15) years = 585 years.
Average age of 40 persons= 15 yrs 3 months = 61/4 years.
Total age of 40 persons = ((61/4 )*40) years= 610 years.
Age of the teacher = (610 - 585) years=25 years.
Total weight increased =(1.8 x 10) kg =18 kg.
Weight of the new man =(53 + 18) kg =71 kg.
Let the original average expenditure be Rs. x. Then, 42 (x - 1) - 35x = 42 --> 7x = 84 --> x = 12 --> Original expenditure = Rs. (35 x 12) = Rs. 420.
Let the average after 17th inning = x. Then, average after 16th inning = (x - 3) 16 (x - 3) + 87 = 17x or x = (87 - 48) = 39
Required average speed = ((2xy)/(x+y)) km / hr =(2 x 84 x 56)/(84+56)km/hr = (2*84*56)/140 km/hr =67.2 km/hr.