31.
The average age of 35 students in a class is 16
years. The average age of 21 students is 14.what is the average age of
remaining 14 students?
a)
25
b)
27
c)
30
d)
35
32.
16 children are to be divided into two groups A
and B of 10 and 6 children. The average percent marks obtained by the children
of group A is 75 and the average percent marks of all the 16 children is 76. What
is the average percent marks of children of group B?
a)
77 1/3
b)
77 2/3
c)
78 1/3
d)
78 2/3
33.
The average score of a cricketer for ten matches
is 38.9 runs. If the average for the first six matches is 42. Then find the
average for the last four matches.
a)
33.25
b)
33.5
c)
34.25
d)
35.
34.
The average of six numbers is 3.95. The average
of two of them is 3.4, while the average of the other woo is 3.85. What is the
average of the remaining two numbers?
a)
4.5
b)
4.6
c)
4.7
d)
4.8
35.
The batting average for 40 innings of a cricket
player is 50 runs. His highest score exceeds his lowest score by 172 runs. If
these two innings are excluded, the average of the remaining 38 innings is 48
runs. The highest score of the player is:
a)
165 runs
b)
170 runs
c)
172 runs
d)
174 runs
ANSWERS
31.D
32.B
33.C
34.B
35.D
SOLUTION
31.
Sum of the ages of 14 students = (16 x 35) – (14
x 21 ) = 560 -294 = 266.
Required average = (266/14) = 19 years.
32.
Require average = (76 x 16) – (75 x 10 ) / 6 =
(1216 – 750 /6) = 466 / 6 = 233 / 3 = 77
2/3.
33.
Required average = (38.9 x 10) - (42 x 6) / 4 = 137 / 4 = 34.25.
34.
Sum of the remaining two numbers = (3.95 x 6) –
[(3.4 x 2) + (3.85 x 2)]
= 23.70 – (6.8 + 7.7) = 23.70 – 14.5 = 9.20.
Required average = (9.2 / 2) =4.6
35.
Let the highest score be X. then, lowest score =
(X = 172).
Then, (50 x 40) –[X + (X – 172) ] = 38 x 48
2 x = 200 + 172 – 1824 2 x = 348
X = 174.
No comments:
Post a Comment