Please wait...
/
-
Average of a, b and c is 9.
Sum of a, b and c = 9×3 = 27
Average of b and c is 10.
Sum of b and c = 10×2 = 20
Hence, a = 27 – 20 = 7
Let the four numbers are a, b, c, and d.
It is given, (a+b+c+d)/4 =20; → (1)
∴ (a+b+c+d) = 4×20 = 80;
Also, (a+b)/2 = 15 ⇒ (a+b)= 30; → (2)
∴ subtract (2) from (1);
⇒ (a+b+c+d) - (a+b) = 80-30;
⇒ (c+d) = 50;
∴ The required average of the last two number= 50/2= 25
Formula: Average = (Sum of observations)/(Total no. of observations)
Sum of weight of all 18 boys = 18 × 35 = 630 Kg.
∴ Average of all 22 boys = (630+20+22+26+28)/22 = 33 years
Average age of 12 boys = 15 years
Sum of age = 12 × 15 = 180 years
Average age of 18 girls = 12 years.
Sum of ages of 18 girls = 12 × 18 = 216 years
Average of all Girls and Boys = (180 + 216)/30 = 13.2 years
Total number of students = 96
Let the number of boys be denoted by B and girls by G
⇒ G + B = 96
Given that number of girls is 40% more than that of number of boys.
⇒ G = B + 40%B =
Substituting the value in above equation :
B +
⇒ B = 40 and G = 56
Let average marks of Boys = x
Total marks scored by 40 boys = 40x
And average marks of girls = y
Total marks scored by 56 girls = 56y
⇒ x = y +
According to question :
⇒
⇒ 112y = 63 × 96
⇒ y = 6048/112 = 54
Correct (-)
Wrong (-)
Skipped (-)