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New answer posted

10 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

P1 + P2 + P3 = 59

P1 and P2 minimize

Then P3 minimum i.e. 47

P1

P2

P3

5

7

47

5

11

43

3

13

43

5

13

41

5

17

37

3

19

37

5

23

31

7

23

29

So, only possible value of P3 can take 29, 31, 37, 41, 43, and 47

Total 6 values

New answer posted

10 months ago

0 Follower 18 Views

A
alok kumar singh

Contributor-Level 10

Time period of Oscillation, T = 2 π I M B

1 4 = 2 π 9 . 8 * 1 0 6 M * 0 . 0 4 9

1 1 6 = 4 π 2 * 9 . 8 * 1 0 6 M * 4 9 * 1 0 3

M = 4 π 2 * 9 . 8 * 1 0 6 4 9 * 1 0 3 * 1 6

= 4 π 2 * 9 . 8 * 1 6 * 1 0 3 4 9

= 1 2 . 8 π 2 * 1 0 3 * 1 0 2 * 1 0 2

=1280π2*105Am2

New answer posted

10 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

1x=1156y

1x=y9015y

x=15yy90 ______equation 1

i.e. y = 90+y,

1≤ y1 9

y1 is odd

Equation 1 becomes

x=15 (90+y1)y1

=15+90*15y1

y1 can take 1,3,5,9

y=91, 93, 95, 99

New answer posted

10 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Sol1

20percent (0.6)

3L

Sol2                    

40percent (0.4)

1L

Sol3                    

1L

4L

Sol4                    

XL

3L

Sol5                    

3.5L

7L

1 + x = 3.5

x= 2.5

Required percent y = 2 . 5 3 * 1 0 0  = 83.33 percent

New answer posted

10 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

x, a, b, y are in GP

a = xr

b = xr2

y = xr3

r = ( y x ) 1 / 3

rewrite,

a = x2/3.y1/3

b = x1/3.y2/3

a * b = x * y

a3 + b3 = x2y + y2 x

= xy (x+y)

= xy (2A) …………… [ x + y 2 = A ]

Hence, n = 2

New answer posted

10 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Let the side of = a

CD = a/2

Let EC = x

In GEC

GE/EC = tan60 = 3

GE = x 3

DE = a/2 – x = FG = HF

GE = HG

x3=a2x+a2x

3 = a – 2x

x = a3+2

GE = x 3 = a33+2

r = side of square ÷ 2

r= a34+23

ratio = a/r

a (4+23)a3

4+23)3=43+2

New answer posted

10 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

The potential V at any point, at distance r from centre of dipole = KPcosθ r 2

At axial point where  θ=0? , V=KPr2=9*109*4*10622=9*103 V

At axial point where  θ=180? , V=KPr2=9*103 V

New answer posted

10 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

(d):3515.2 – 3380 = 135.2 = SI on Rs. 3380 for one year.

By SI formula, 135.2 = 3 3 8 0 * R * 1 1 0 0  or R = 4 %

Now we have 3380 = P (1.04)2 or P = Rs. 3125

New answer posted

10 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

(c):Let speed of passenger train be x m/s.

[ ( 6 6 + 8 8 8 . 3 3 + x ) ] = 0 . 1 6 8 * 6 0

[ 3 0 * 5 1 8 = 8 . 3 3 ]

x = 6.94 m/sec = 25 kmph.

New answer posted

10 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

(c):We have 4 men = 5 women 1 man = 5 4 women, 2 women = 4 boys, 1 women = 2 boys 5 4  women = 2 * 5 4  boys = 5 2  or 1 man = 5 4 women = 3 2  boys

Now 2M + 3W+ 4B = 2 * B + 3 * 2B + 4B = 15 B Boys

Or 15 Boys do the work in 10 days (10 hectares) 6M + 4W + 7B = (6 * + 4 * 2 + 7) B = 30

Boys 30 boys will do 16 hectares of work in 8 days.

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