x - ?121 = 0
? x = ?121 ? x = ± 11
and y2 - 121 = 0 ? y2 = 121
? y = ?121 = ± 11
? x = y
Given, lateral surface area = 94.2 sq cm
? 2?rh = 94.2
? r = 94.2/(2?h) = 94.2/(2 x 3.14 x 5) = 3 cm
According to the formula
Percentage increase in volume
= (2k + k2/100)% = ( 2 x 25 + 252/100)%
= (50 + 625/100)% = (50 + 6.25)%
= 56.25%
Here, A = -8%, B = 4%
According to the formula,
Net effect on volume = [2A + B + A2 + 2AB/100 + A2B/1002]%
= [2 x (-8) + 4 + (-8)2 + 2 x (-8) x 4 /100 + (-8)2 x 4 /1002]
= [-16 + 4 + 64 - 64/100 + 256/104]%
=[-12 + 0 + 0.0256]%
= -11.9744% (decrease)
Area of 4 wails = Lateral surface area perimeter = 2(l + b) = 250
? l + b = 250/2 = 125 m
Area to be painted = Cost/Rate = 15000/10 = 1500 sq m
Area of 4 wails = 2(l + b)h = 250 h
? 250h = 1500
? h = 1500/250 = 6 m
Volume of cone = (1/3)?r2h
= (1/3) x (22/7) x 10 x 10 x 21 = 2200 cm3
Given, l = 9 m
and diameter = 14 m ? r = 14/2 = 7 m
Now, volume = (1/3)?r2h
= (1/3)? x 49 x ?l2 - r2
= (1/3)? x 49 x ?81 - 49
= (1/3) x 49? x ?32 = 49??32 /3 m3
Let level of water will be increased by h cm
? x (15)2 x h = (4/3)?(10)3
? h = [(4/3) x 10 x 10 x 10] / [15 x 15]
= 525/27 cm
Let the side of the cube be 3k, 4k and 5k, respectively.
So, volumes of these cubes are 27 k3, 64k3, 125 k3 respectively.
? Volume of the new bigger cube = 27 k3 + 64k3 + 125 k3
= 216k3
So, side of the new cube = 6k
Since, diagonal of the cube = ?(6k)2 + (6k)2 + (6k)2
= 12?3
? 108k2 = 432
?k = 2
So, sides of the three cubes were 6 cm, 8 cm and 10 cm respectively.
Series pattern
12, 32, 52, 72, 92, 112
? Missing term = 12 = 1
The first 100 terms of this series is
(1 - 2 - 3) + (2 - 3 - 4) + ... + (33 - 34 - 35) + 34
The first 33 terms of the above series will given an AP as
-4, -5, -6, ... -36
? Answer = 33 x (-20 ) + 34 = - 626
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