SSC CGL 20201)The average age of A, B, and C is 20 years, and that of B and C is 25 years. What is the age of A ?
10 years
Average age of A, B and C = 20 years Total age of A, B and C = 60 years
Average age of B and C = 25 years total age of B and C = 50 years
A's age = 60 - 50 = 10 years
SSC CGL 20192)The average of thirteen numbers is 47. The average of the first three numbers is 39 and that of next seven numbers is 49. The 11th number is two times the 12th number and 12th number is 3 less than the 13th number. What is the average of 11th and 13th number?
57
Sum of thirteen numbers = 611
Sum of the first three numbers = 117
Sum of the next seven numbers = 343
let the 12 th number be 'x'
then
11th number = 2x
13th number = x + 3
Sum of the first three numbers + Sum of the next seven numbers + 11th number + 12th number + 13th number = 611
117+343+2x+x+x+3=611
x =37
average of 2x and x+3 = 57
SSC CGL 20193)The number of students in a class is 75, out of which 33\({1 \over 3}\)% are boys and the rest are girls. The average score in mathematics of the boys is 66\({2 \over 3}\)% more than that of the girls. If the average score of the all students is 66, then the average score of the girls is :
54
The number of students in a class = 75;
Number of boys =\( 33\frac{1}{3}\% \)of the total boys = 75/3 = 25;
Number of girls = 75 - 25 = 50;
Let the average score of girls be x.
Total score of girls = 50x;
Average score of boys = x + 2x/3 = 5x/3;
Total score of boys = \(25 \times \frac{5x}{3};\)
Average score of all the students = 66;
\(\frac{50x + 125x/3}{75}\) = 66;
\(\frac{150x + 125x}{75 \times 3}\) = 66;
x = 14850/275 = 54
SSC CGL 20194)The average age of 120 students in a group is 13.56 years. 35% of the number of students are girls and the rest are boys. If the ratio of the average age of boys and girls is 6 : 5, then what is the average age (in years) of the girls?
12
Sum of age of 120 students in a group whos average age is 13.56 = 120 × 13.56 = 1627.2
Number of girls in the group = 120 × (35/100) = 42
Number of boys in the group = 120 – 42 = 78
Let the average of boys and girls are 6x and 5x respectively.
According to the question
78 × 6x + 42 × 5x = 1627.2
⇒ x = 2.4
SSC CGL 20195)The average of 18 numbers is 37.5. If six numbers of average X are added to them, then the average of all the numbers increases by one, The value of x is:
41.5
SSC CGL 20196)The average weight of a certain number of students in a group is 72 kg. If 10 students having an average weight of 78 kg leave and 4 students having an average weight of 80 kg join the group, the average weight of the students in the group decreases by 0.7 kg, The number of students initially in the group is:
46
SSC CGL 20197)The average of 33 numbers is 74. The average of the first 17 numbers is 72.8 and that of the last 17 numbers is 77.2. If the 17th number is excluded, then what will be the average of the remaining numbers (correct to one decimal place)?
72.9
SSC CGL 20208)The average of four terms is 30 and the Ist term is \(\frac{1}{3}rd\) of the sum of the remaining terms. What is the first term?
30
Average of 4 numbers = 30; Sum of these numbers = 30 x 4 = 120; let Ist number is x.
so x = \(\frac{1}{3}(120-x)\); x = 30
SSC CGL 20209)In a class, the average score of thirty students on a test is 69. Later on it was found that the score of one student was wrongly read as 88 instead of 58. The actual average score is:
68
When data read wrongly then total score of thirty students on a test = 69 X 30 = 2070; (Average = sum of all terms/number of terms)
Difference in data = 88 - 58 = 30;
Actual score = 2070 - 30 = 2040;
Actual average score = 2040/30 = 68
SSC CGL 202010)The average score in Mathematics of 90 students of sections A and B together is 49. The number of students in A was 25% more than that of B, and the average score of the students in B was 20% higher than that of the students in A. What is the average score of the students in section A?
45
Let no. of students in section B = x ; A = 1.25x; then x + 1.25x = 90; therefore' x = 40;
Let average of score of section A be y and B be 1.2y;
then 50y + (1.2y x 40) = 90 x 49; y = 45