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Given the sides of the triangular park are 120m, 80m and 50m.

perimeter of the triangular park = 120m + 80m + 50m= 250m

therefore it's semi-perimeter = 250/2= 125mhence, the gardener have to plant grass in 1452.36m²Area of triangular park by heron formula=√s(s-a)(s-b)(s-c)=√125(125-120)(125-80)(125-50)=√2109375=1452.36m^2

now we have to find the cost of fencing the field with a barbed wire at the rate of rs 20 per m leaving a space of 3m wide for a gate.

therefore the gardener have to fence = 250 - 3= 247m

so total cost of fencing at the rate of rs 20 per meter = 247 × 20= rs 4940

thanks bhaiya

ix). 3p-7=0 3p=7 p=7/3. ans.

vi). n/7=(-3) n=(-21) ans.

viii). 3X+1=16 3X=16-1 3X=15 X=5. ans.

let the present age of Albert be 5x then present age of Salim=9x5 years hence age of Albert=5x+5age of Salim=9x+5According to the question(5x+5)/(9x+5)=3/53(9x+5)=5(5x+5)27x+15=25x+2527x-25x=25-152x=10x=10/2x=5

Hence present age of Albert=5×5 = 25 yearspresent age of Salim=9×5=45 years

Let a=120m b=80m c=50m

s=a+b+c/2

=120+80+50/2

=250/2

=125m

From heron's formula

Area of triangle=√s(s-a)(s-b)(s-c)

=√125*5*45*75

=√5*5*5*5*5*3*3*5*5*3

=375√15m2

Perimeter of trianglular park=250m

Space left for fencing=250-3

=247m

Cost of fencing 247m=247*20

=Rs.4940

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Total surface area=2(lb+bh+hl)=2(3*2+2*1.5+1.5*3)=2(6+3+4.5)=2(13.5)=27m²

Cost of painting 1m²=Rs20Cost of painting 27m²=Rs540

When CFCs reach the upper atmosphere, they are exposed to ultraviolet rays, which causes them to break down into substances that include chlorine. The chlorine reacts with the oxygen atoms inozoneand rips apart theozonemolecule. In other regions, theozone layerhas deteriorated by about 20 percent.

4 is the correct answer

4 is the best answer

your answer is right. please accepted my answer

f) 8×5=40g)8×7=56h)9×7=63I)6×4=24j)9×8=72K)8×8=64l)6×6=36m)7×3=21is correct answer

S.I=P×R×T/100 = 400×5×2/100×2 =20 A=P+S.I = 400+20 = 420THIS IS THE BEST ANSWER

your fact is very amazing

8*5=407*8=569*7=634*6=249*8=728*8=646*6=367*3=21

8×5=40 7×8=56 9×7=73 6×4=24 8×9=72 8×8=64 6×6=36 7×3=21

8*5=407*8=569*7=636*4=249*8=728*8=646*6=367*3=21

1) 8*5=402)7*8=563)9*7=634)4*6=245)9*8=726)8*8=647)6*6=368)7*3=21

f 40g 56h 63i 24j 72k 64l 36m 21

1)8*5=40 2)7*8=563)9*7=634)4*6=245)9*8=726)8*8=647)6*6=368)7*3=21

8×5=408×7=569×7=636×4=249×8=728×8=646×6=367×3=21are the correct result.

From where you all get answere

Cost paid = 4242.6 * 4000/10000 = 1697.04

Area of square = S²

total cost to plough the field = area x cost720 = S² x 5S= 12m

As the perimeter is equalso, perimeter of square = perimeter of triangle4S = 3 x side of triangle48 = 3aa = 16

area of equilateral triangle =√3/ 4a²=√3 / 4 x 16²= 108.8 m²

total cost = 10 x 108.8= 1088 rs

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let height =hlength =5hbreadth=5h-10

let height,=h length=5h breadth=5h-10 anwer

let height = hlength = 5hbreadth = 5h - 10

let hight=hlength=5hbreadth=5h-10 answer.

Let the two reqrd nos.be x and 48-x3x=4*(48-x)-103x +4x = 1827x= 182x=26thereforenos arex and 48-xx=26and smaller one is 22...ie.48-26

sinA * (1 + tan) + cosA * (1 + cotA) = secA + cosecA

LHS = sinA * (1 + sinA/cosA) + cosA * (1 + cosA/sinA)

= sinA + sin^2A/cosA + cosA + cos^2A/sinA => rewrite as:

= sinA + cos^2A/sinA + cosA + sin^2A/cosA

= (sin^2A + cos^2A)/sinA + (cos^2A + sin^2A)/cosA

= 1/sinA + 1/cosA

= cosecA + secA

= secA + cosecA = RHS

According to Heron's formulaA=√[s (s-a)(s-b)(s-c)]

Here s=(a+b+c)/2s = half of ∆ 's perimeter

half of the triangle perimeter

Heron's formula: Area = √(s)(s-a)(s-b)(s-c); where s is the semi perimeter of the triangle. s = (40+24+32)/2= 48 m. Area = √[ 48*(48-40)(48-24)(48-32)]= √ (48*8*24*16)= √147456= 384 m².

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(a) To provide low resistance possible to the ground

Interest = PRT

.

Given that interest for 5 years at 10% is Rs 196 less than 4 years at 16%

⇒ P(0.16(4) - P(0.1)(5) = 196

.

Solve for P:

P(0.16(4) - P(0.1)(5) = 196

0.64P - 0.5P = 196

0.14P = 196

P = Rs 1400

.

Answer: The principal amount is Rs 1400

Given Radha lent a certain sum of money at 4% per annum and in 8 years the interest amounted to rs 340 less than the sum lent, to find the sum she lent.

Let the sum be x, then amount is Rs 340 less so it will be x - 340.

We have t = 8 yrs , r = 4%

We know that SI = P x R x T /100

x - 340 = x x 4 x 8 / 100

x - 340 = 32 x / 100

x = 32 x / 100 + 340

100 x = 34,000 + 32 x

68 x = 34,000

x = 34,000 / 68

x = rs 500

So Radha lent a sum of rs 500

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an Arithmetic progression(AP) isa sequence of numbers in which each differs from the preceding one by a constant quantityinwhich a(first term) = 100 cmd(common difference) = -5cmAn(nth term) = 0cmn = ?thereforeAn = a + (n-1)d0 = 100 -5n + 5-105 = - 5n21 = nhence in the 21 st year the tree will stop growing

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Age of Mansi is 7 years old.

As her brother is twice old as Mansi

Then age of her brother = 2*7= 14 years

Mansi=7 yrsBrother=2*mansi=2*7=14yrs

Mansi= 7 yearsBrother= twice old as mansi = 2x7 =14

Answer= 14 years

thank you very much

16)TheDistance Formula. You know that thedistancebetween two points in a plane with Cartesian coordinates A ( x 1 , y 1 ) and B ( x 2 , y 2 ) is given by the followingformula: A B = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2. Thedistance formulais really just the Pythagorean Theorem in disguise.

17)The section formula tells us the coordinates of the point which divides a given line segment into two parts such that their lengths are in the ratio m:n . ... The section formula is helpful in coordinate geometry; for instance, it can be used to find out the centroid, incenter and excenters of a triangle

Consider f(x,y) = x² + y² - 25

So out (6,8)f(6,8) = 6² + 8² - 25 = 36 + 64 - 25 = 75 > 0So, (6,8) is outside

Amount = P(1+R/100)^nwhere n = number of yearsR = rate of interest= P= 150000

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let Charu's age be x and soPankajis 3x years oldNow 3 years ago Pankajage was 3x - 3And Charu was x - 3By the sum it is given as (3x -3) = 4 *(x - 3) - 2==> 3 x - 3 = 4 x - 12 -2==> x = 11Charu's age is 11 and Pankaj is 33 years old