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Dimensions = 50 m and 40 mArea of the rectangular lawn= 50× 40 = 2000 m²Area of the grass surrounding the pond = 1184 m²So, the area of the pond = area of the lawn - area of the grass= 2000 - 1184Area of the pond = 816 m²Let the width of around the pond be 'x' mThen, the length of the pond = (50 - 2x) mand the breadth of the pond = (40 - 2x) mArea of the pond = 816 m²⇒ (50 - 2x) ×(40-2x) = 8162000 - 80x - 100x + 4x² = 8164x² - 180x + 2000 - 816 = 04x² - 180x + 1184 = 0dividing it by 4 we getx² -45x + 296 = 0x² - 37x - 8x + 296 =0x(x-37) - 8(x-37) = 0(x-37) (x-8) = 0x= 37 or x = 8supposeif x = 37, thenthe length of the pond will be 50 - 2*3750 - 74 = -24m, which is not possible because length cannot be negative.Therefore the length of the pond = 50 - 2*850 - 16 = 34 mandBreadth of the pond = 40 - 2*840 - 16 = 24 mLength = 34 m and breadth = 24 m

1 is the correct answer of the given question

(b-c)^2 + (c-a)^2 + (d-b)^2 = (a-d)^2

Let a = p, b = pr , c = pr² and d = pr³

Now, LHS = (b - c)² + (c - a)² + (d - b)² = (pr - pr²)² + (pr² - p)² + (pr³ - pr)² =( p²r² + p²r⁴ - 2p²r³) + (p²r⁴ + p² - 2p²r²) + (p²r^6 + p²r² -2p²r⁴) = p²r^6 - 2p²r³ + p²= (pr³) -2(pr³)(p) + p²= (pr³ - p)²= (p - pr³)² =(a - d)² = RHS

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Mean = 50.ie∑x / 100 = 50 , where x are the observationstherefore∑x = 50 * 100 = 5000now, 100 is actually 110, so

new∑x = 5000 - 100 + 110new∑x = 5010new mean = new∑x/ 100 = 5010/100 = 50.1

new median = 52, as median remains unchanged as it is the middlemost observation, hence 110 has no effect on the median as it is the highest observation

Solution no. 1:-Arithmetic Mean = ∑X/NGiven : Mean = 50 and N = 100Therefore,50 =∑X/100∑X(Wrong) = 5000Correct value = 110Incorrect value = 100Correct Arithmetic Mean = {∑X (Wrong) + Correct valueof observation- Incorrect value of observation}/N⇒ Correct Mean = (5000 + 110 - 100)/100⇒ Correct mean = 5010/100⇒ Correct Mean = 50.1Answer

Solution no. 2 :-In this question, 'the value of thelargest item' should be written instead of 'latest item'.The value of the Median will not change because whatever values are added or subtracted from median, the total observation will remain 100 and median is the centrally located value of a series such that the half of the value or items of the series are above it and the otherhalf arebelow it.Formula of median = Size of(N+1)/2th ItemTotal observations = 100Size of (N+1)/2th item⇒(100+1)/2th Item⇒ 101/2⇒ 50.5th Item110 is corrected observation instead of 100 and 110 is the largest of all the observation. So it will not make any difference in value of median. The value of Median will be 52.Answer

W1 = avg weight initially 25 * W1 = weight of 25 persons in itiallyLet weight of the man who leaves = Wremaining 24 persons weight is = 25 W1 - WWhen new man comes in total weight = 25 W1 - W + 70 Average weight of the new set of 25 persons = (25 W1 - W +70) / 25 = W1 + 1 one more than previous average=> 25W1 - W +70 = 25 W1 + 25=> W = 45 Kganswer

Perimeter=2(l+b)160=2(l+10)80=l+10l=70mArea is l*b=70*10=700m^2

ty

Dimensions = 50 m and 40 mArea of the rectangular lawn= 50× 40 = 2000 m²Area of the grass surrounding the pond = 1184 m²So, the area of the pond = area of the lawn - area of the grass= 2000 - 1184Area of the pond = 816 m²Let the width of around the pond be 'x' mThen, the length of the pond = (50 - 2x) mand the breadth of the pond = (40 - 2x) mArea of the pond = 816 m²⇒ (50 - 2x) ×(40-2x) = 8162000 - 80x - 100x + 4x² = 8164x² - 180x + 2000 - 816 = 04x² - 180x + 1184 = 0dividing it by 4 we getx² -45x + 296 = 0x² - 37x - 8x + 296 =0x(x-37) - 8(x-37) = 0(x-37) (x-8) = 0x= 37 or x = 8supposeif x = 37, thenthe length of the pond will be 50 - 2*3750 - 74 = -24m, which is not possible because length cannot be negative.Therefore the length of the pond = 50 - 2*850 - 16 = 34 mandBreadth of the pond = 40 - 2*840 - 16 = 24 mLength = 34 m and breadth = 24 m

Sinx +sin²x=1sinx=1-sin²xsinx=cos²xsin²x=cos⁴xso,cos^8x+2cos^6x+cos⁴x=cos⁴x(cos⁴x+2cos²x+1)=sin²x (sin²x+2sinx+1)=sin⁴x+2sin³x+sin²x=sin⁴x+sin³x+sin³x+sin²x=sin²x(sin²x+sinx)+ sinx(sin²x+sinx)=sin²x×(1)+ sinx(1)=sin²x+sinx= 1

Since cosx+cos^2x=1, then cosx=1-cos^2x=sin^2x.Thensin^12x+3sin^10x+3 sin^8x+sin^6x+2 sin^4x+2sin^2x-2= cos^6x+3cos^5x+3 cos^4x+cos^3x+2 cos^2x+2cosx-2=cos^6x+3cos^5x+3 cos^4x+cos^3x=cos^4x (cos^2x+3) + cos^3x(3cos^2+1)= cos^4x (1-cosx+3) +cos^3x(3-3cosx+1)= (1-cosx)^2 (4-cosx)+cosx((1-cosx)(4-3cosx)= (1-2cosx+1-cosx)(4-cosx)+cosx(4-7cosx+3-...= (2-3cosx)(4-cosx)+ cosx(7-10cosx)= 8-14cosx+3cos^2x+7cosx-10cos^2x= 8-7cosx-7cos^2x=8-7(1)= 1

cost of each calculator is 14 dollartotal cost for 684 students = 14×684= 9576 dollar

New Sum of Observations = 25*10 - 25= 225New Mean = 225/9 = 25.

As there are 9 observations arranged in ascending order then median is 5th observation

So, 5th observation = 20.5

When largest four observations increased by 2 there will be no change in 5th observation. Therefore, median for new set remain same as of old set which is 20.5

c is correct answer for your questions

a is the right answer

A is the right answer

a is the right answer

756m2 is the right answer

option a is correct answer.

Length of the rectangular garden = 16 m

Breadth of the rectangular garden = 10 m

Area of the rectangular garden = 16*10

= 160 m²

Area of the concrete path of width x meters (given) = 120 m²

Total area = area of rectangular garden + area of the path

⇒ 160 + 120

= 280 m²

The width of the concrete path is uniform (Given)

So,

Length of the whole rectangle (including concrete path) = (16 + 2x) m

Breadth of the whole rectangle (including concrete path) = (10 + 2x) m

⇒ (16 + 2x)*(10 + 2x) = 280

⇒ 4x² + 52x + 160 = 280

⇒ 4x² + 52x + 160 - 280 = 0

⇒ 4x² + 52x - 120 = 0

⇒ Dividing the whole equation by 4, we get.

⇒ x² + 13x - 30 = 0

⇒ x² + 15x - 2x - 30 = 0

⇒ x(x + 15) - 2(x + 15) = 0

⇒ (x - 2) (x + 15) = 0

⇒ x = 2 or x = - 15

As the width cannot be negative. So, the value of x is 2

x = 2 m

Length of the outer rectangle = 16 + (2*2) = 20 mBreadth of the outer rectangle = 10 + (2*2) = 14 m

total height= 15area along height = lb = 15×2= 30 m^2total length of road = 50area along length= 50×2=100n^2total area of path = 100+30=130 m^2cost = 130×100= 13000₹

height÷15, area height=lb=15×2=30 m^2, area along length=50(2)=100m^2, total area of path=100+30=130m^2, cost =130x100=13000 Rs.

(1+ cotx^2)( 1- cosx)( 1+ cosx) = ( cosecx^2)(1- cosx^2) = ( cosecx^2)( sinx^2)= ( sinx^2)( sinx^2)= sinx^4

1. 4 cos4x. 2.-5 sin5x.

(1) 4 cos4x

(2) - 5 sin5x

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(a+b+c)3 - a3-b3-c3 = [(a+b)+c]3 – a3-b3-c3

(a+b)3+c3+3c(a+b)(a+b+c)-a3-b3-c3

A3+b3+3ab(a+b)+c3+3c(a+b)(a+b+c)-a3-b3-c3

A3+b3+c3+3ab(a+b)+3c(a+b)(a+b+c)-a3-b3-c3

3ab(a+b)+3c(a+b)(a+b+c) 3(a+b)[ab+c(a+b+c)]

3(a+b)[ab+ac+bc+c2] 3(a+b)[a(b+c)+c(b+c)]

3(a+b)(b+c)(c+a)

I THINK ITS ANSWER IS 115.5cm2

Given that,19,25,59,48,35,31,30,32,51.

( put all of the values into order, median is the middle value).

19,25,30,31,32,35,48,51,59.

median = 32.

if 25 is replaced by 52, ie,

19,30,31,32,35,48,51,52,59.

median = 35.

c and d are incorrect options

2572 + 2326

= 4898

(b)(11,-6) is correct answer

option b is the right answer

option b is the right answer

option b is the right answer

option b is the right answer

option b is the right answer

the correct answer is B

2 .(11,-6) is the right answer

2r=84m (given)r =84/2=42m

outer circle's radius=42+3.5=45.5minner circle's radius=42m

area of path = (area of outer circle) - (area of inner circle)

Radius of the park = 84/2 = 42marea of the park = (22/7)×42×42 = 5544m^2width of road = 3.5m.•.total radius = 3.5+42 = 45.5mtotal area = (22/7)×45.5×45.5 = 6506.5m^2area cover by the road=6506.5-5544= 1062.5m^2total cost of constructing the road=1062.5×240=Rs255000

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The focal length (f) of a lens is the distance between the center of the lens and the point at which the reflected light, of a beam of light travelling parallel to the center line, meets the center line (principal axis).

The radius of curvature (r) is the radius of the lens that forms a complete sphere.

Looking at the attached diagram we can say the following:

The red line represents the light hitting the lens (AB) and reflecting off of the lens (BF). We know that the green line, RB, which represents the radius line of the circle, bisects the angle ABF because it is always at right angles to the lens. Therefore angles ABR and RBF are equal. We also know because of the rule of alternate angles that angle ABR is equal to angle BRF. Therefore, triangle BRF is an isosceles triangle. Consequently the lines BF and RF are equal.

We also know that BF and FC are approximately equal for lens that are small.

Therefore:

RF=BF=FC and

RC=RF+FC=FC+FC=2FC or

r=2f

Therefore, the radius of curvature is twice the focal length.

Then, m+n=?