Average number of pencils
= ( 8 + 10 + 15 ) / 3
=11
Average = (8 + 9 + 12 + 13 + 15 + 9)/6
= 66 / 6
= 11
Given that
x = 1305, y = 4665, z = 6905
Then ,
|x - y| = |1305 - 4665| = 3360
|y - z] = |4665 - 6905| = 2240
|z - x| =|6905 -1305| = 5600
? Required number = HCF of 3360, 2240 and 5600 = 1120
On dividing 1305 by 1120, remainder is 185.
On dividing 4665 by 1120, remainder is 185.
On dividing 6905 by 1120, remainder is 185.
? Common remainder = 185
a2 b4 + 2 a2 b2 = a2 b2 ( b2 +2).......................(i)
and (ab)7 - 4 a2 b9 = a7 b7 - 4 a2 b9 = a2 b2(a5 b5 - 4b7)................(ii)
from Eqs. (i) and (ii)
HCF = a2 b2
Let f(x) = x3 - y2 - 2x = x( x2 - x- 2)
= x( x2 - 2x + x - 2)
= x {x (x - 2) + 1 (x -2)}
= x (x +1) (x - 2)
and g(x) = x3 + x2
= x2 (x +1) = x. x (x +1)
? LCM of [ f(x), g(x)] = x (x +1) . x . (x - 2)
= x2(x + 1)(x - 2)
= x2(x2 - x - 2)
= x4 - x3 - 2x2
Let f (x) = (x4 - y4)
= (x2 - y2) (x2 + y2)
= (x - y) (x + y) (x2 + y2)
and g(x) = (x6 - y6)
= (x3)2 - (y3)2
= (x3 + y3) (x3 - y3)
= (x + y) (x2 - xy + y2) (x - y)(x2 + xy + y2)
= (x - y) (x + y) (x2 - xy + y2)
(x2 + xy + y2)
? HCF of [f(x), g(x)] = (x - y) (x + y)
= x2 - y2
Present average age of family
= 25 year
3 year ago average of family
= 25 - 3 = 22 years
Average = sum of 6 numbers / 6
= ( 13 + 17 + 25 + 11 + 26 + 10 ) / 6
= 102 / 6
= 17
Average salary
= 8000 + 5000 + 11000 + 7000 + 9000 / 5
= Rs. 8000
Actual average
=( 100 + 220 + 300 + ... + 1000 ) / 10
=550
New Average = 550 / 5
=110
Average number of T. V. sets
= ( 20 + 30 + 60 + 80 + 50 ) / 5
= 48
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