Arithmatic mean = sum/members
=> (1x1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + 7x7) / (1 + 2 + 3 + 4 + 5 + 6 + 7)
=> 140/28 = 5
7 boxes & 24 Balls.
Let, n = No. of boxes ; m = No. of fruits
3n+3 = m---------->(1)
4(n-1)= m---------->(2)
=>4n-4 = 3n+3;
=>4n-3n = 4+3
n = 7
Put n=7 in eqn(1)
=> 3(7)+3 = m
21+3 = m
m = 24;
Number of boxes = 7 and fruits = 24
To join 5 tables we require 4 joins.
Here in the question it is asked that 62 tens, which implies ten times of 62.
=> 62 x 10 = 620.
Hence, the value of 62 tens = 620.
x = ?8 - ?7 = 1/(?8 + ?7) and y = ?6 - ?5 = 1/(?6 + ?5) and z = ?10 - 3 = 1/(?10 + 3)
As (?8 + ?7) > (?6 + ?5), (?8 - ?7) < (?6 - ?5)
As (?10 + 3) > (?8 + ?7) > (?6 + ?5)
z < x < y.
Then, A:B = 2:3 and B:C = 5:8
= (5 x 3/5) : (8 x 3/5) = 3:24/5
A:B:C = 2:3:24/5
= 10:15:24 => B = 98 x 15/49 = 30.
A point has no dimension and a line has one dimension is the statement that best compares a line and a point.
Yes, Every integer is a rational number. A rational number is a number which can be expressed as a ratio of two integers numerator and denominator where (the denominator not being 0 ).
Hence, Every integer can be expressed in ratio of two integers.
If 'k' is number of students in P and 'l' is number of students in Q, then
From the given conditions, we have
=> k - 10 = l + 10 .......(1)
=> k + 20 = 2(l - 20) ....(2)
Solving these eqns, we get
k = 100 and l = 80.
Therefore, number of students in class Q is 80.
No. of students will be 98 x 12 = 1176
let us consider n circles then 1 + 2 + 3 + ....+ n = 1176
n(n+1)/2 = 1176
n = 48 circles.
First ten odd numbers form an arithmetic progression of the form
1,3,5,7,9......
here a = first term = 1
d = common difference = 2
Average of first n numbers = (2a + (n - 1)d)/2
n th term of the AP = a + (n - 1)d
Substituting a = 1, d = 2 and n = 10 in the above formulas
average of first 10 numbers = 10
10 th term of the AP = 19
Therefore average of first 10 terms is 19 - 10 = 9 greater than
the last term. Hence the average is greater than the sum by
9/19 X 100 = 900/19 %
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