Average marks scored by Vineet in five subjects are 70. The average marks scored by him (excluding his highest and lowest marks) are 50 and lowest marks scored by him are 45. Find the highest marks scored by him.
Sum of marks scored by him in five subjects = 5 x 70 = 350 Sum of marks scored by him (excluding his highest and lowest marks) = 3 x 50 = 150 Therefore, sum of highest and lowest marks = 350 – 150 = 200 So, highest marks scored by him = 200 – 45 = 155
The income of B is 85% of A and income of C is Rs 3600 more than that of B. The average income of all three together is Rs. 8400. A and B spend 50% of their separate income, while C spends 780 more than what he saves. The savings of C is how much % of average savings of A and B together?
Let the income of A be Rs 100a Than, income of B is = 100 x 0.85a = Rs 85a income of C = Rs (85a + 3600) income of A+B+C = 100a + 85a + 3600 = 270a + 3600 Now , 270a + 3600 = 8400 x 3 270a = 21600 a=80 earning of A= 8000 ; B = 6800 ; C = 10,400 Savings of A= 8000 x ½ = 4000 Savings of B = 6800 x ½ = 3400 Therefore avg savings of A and B = (4000+3400)/2 = Rs 3700 Let savings of C be k Therefore expenditure of C be k+780 Now, k + k + 780 = 10400 2k = 9620 K = 4810 Therefore the % will be = 4810/3700 x 100 = 130%
The average of four numbers (z+16), (y+15), (y-25) and 1.2z is (y-6). If one more number which is 144 is added then the new average will be (y-1). The value of ‘z’ is what percentage of the value of ‘y’?
The average of four numbers (z+16), (y+15), (y-25) and 1.2z is (y-6).
(z+16)+(y+15)+(y-25)+1.2z = 4(y-6) Eq.(i)
If one more number which is 144 is added then the new average will be (y-1).
(z+16)+(y+15)+(y-25)+1.2z+144 = 5(y-1)
Put Eq.(i) in the above equation.
4(y-6)+144 = 5(y-1)
4y-24+144 = 5y-5
5y-4y = 144-24+5
y = 125
Put the value of ‘y’ in Eq.(i).
(z+16)+(125+15)+(125-25)+1.2z = 4(125-6)
(z+16)+140+100+1.2z = 4x119
(z+16)+140+100+1.2z = 476
256+2.2z = 476
2.2z = 476-256 = 220
z = 100
Required percentage = (100/125)x100
= (4/5)x100
= 80%
Average weight of each student, in a class of 70 students, is 86kg. If the ratio of number of boys and girls in the class is 4:3, respectively and average weight of each boy in the class is 92kg, then find the average weight of each girl in the class.
Number of boys in the class = 70 × (4/7) = 40 Number of girls in the class = 70 × (3/7) = 30 Let the weight of each girl in the class be ‘x’ kg. {(40 × 92) + (30 × x)} ÷ {30 + 40} = 86 Or, 3680 + 30x = 6020 Or, 30x = 2340 So, x = 78 Therefore, average weight of each girl in the class = 78kg
The average of three numbers P, Q and R is 1600. R is 75% more than P. The ratio of P and Q is 4:5 respectively. If R is 25% less than S, then find out the number ‘S’.
The average of three numbers P, Q and R is 1600. P+Q+R = 1600x3 P+Q+R = 4800 Eq.(i) The ratio of P and Q is 4:5 respectively. Let’s assume P and Q are ‘4y‘ and ‘5y‘ respectively. R is 75% more than P. R = 175% of P R = 175% of 4y R = 1.75 x 4y R = 7y Put the values of ‘P’, ‘Q’ and ‘R’ in the Eq.(i). 4y+5y+7y = 4800 16y = 4800 y = 300 If R is 25% less than S. R = (100-25)% of S 7y = 75% of S 7y = 0.75xS Put the value of ‘y’ in the above equation. 7x300 = 0.75xS 2100 = 0.75xS 2100/0.75 = S S = 2800
5 friends went to a hotel where all of them had food, out of which four friends contributed Rs.50 each and the 5th friend contributed Rs.80 from the average expenditure per member. What is the average expenditure per member if he contributed more.
Let N = No of the person A = Average of the expenditure T = Total Amount 
The collective number of gentlemen in corporation X equals the mean value of the total number of gentlemen in corporation Y and corporation Z. The combined number of gentlemen in corporations Y and Z is 3700. The count of gentlemen in corporation Z is 650 greater than that in corporation X. Determine the average number of gentlemen in corporations X and Z.
ATQ, Let the number of gentlemen in corporation X = 'x' Number of gentlemen in corporation Y = 'y' Number of gentlemen in corporation Z = 'z' Therefore, x + y = 3700 z = 3700/2 = 1850 y – z = 650 Or, y = 650 + 1850 = 2500. Therefore, x = 3700 – 2500 = 1200 Average number of gentlemen in corporation X and corporation Z = (x + z)/2 = (1200 + 1850)/2 = 1525
Certainly, on a birthday celebration, 500 candies are distributed among 40 guests and 10 hosts. The average number of candies received by each guest is 'x,' while the average number of candies received by each host is (x + 5). Find the value of (x2 - 10x + 50).
ATQ; 40 × (x + 10) × (x + 5) = 500 Or, 40x + 10x = 500 - 50 Or, 50x = 450 Or, x = 9 By putting the value of 'x' in the given equation, we get; x2- 10x + 50 = (9)2 - (10 × 9) + 50 = 81 - 90 + 50 = -9 + 50 = 41
In a classroom, the average score of 15 students on a test is 44 points. If 5 new students, each with an average score of 'x' points, join the class, and 8 students, each with an average score of '(x – 6)' points, leave the class, the new average score of the 12 remaining students becomes '(x – 1)' points. What is the total score of these 12 students?
ATQ, The sum of the score of 15 students on a test is = 44 × 15 = 660 The sum of the the score of 5 students is = 5 × a = 5a The sum of the score of 8 students is = 8(a – 6) = (8a – 48) The sum of the score of 12 students is = 12(a – 1) = (12a – 12) According to the question, 660 + 5a – (8a – 48) = 12a – 12 660 + 5a – 8a + 48 = 12a – 12 660 + 48 + 12 = 12a + 3a 720 = 15a a = 48 a = 48 Average score of 12 students is = 48 – 1 = 47 The sum of the score of 12 students is = 47 × 12 = 564
Determine the average marks achieved by six additional aspirants if the average score of 24 aspirants in a mock test is 67. Adding the marks of these six more aspirants increases the overall average by 2.75. Find the average marks obtained by these additional six aspirants.
ATQ, Total marks scored by 24 aspirants = 24 × 67 = 1608 Total marks scored by 30 aspirants = 30 × 69.75 = 2092.5 Total marks scored by six aspirants who are added later = 2092.5 – 1608 = 484.5 Desired ratio = 484.5/6 = 80.75
