Explore topic-wise InterviewSolutions in Current Affairs.

Explain

Explain

As equation 2 is just double of equation 1. If we multiple eq. 1 with 2 we get the same result.

multiply 1st equation by 2 then we will get 2nd equationmeans both line coincide or both are same equation so for this we will say ( infinite soultion is possible)

Comparing the given equations with the standard form of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

a₁/a₂ = 9/18 = 1/2b₁/b₂ = 3/6 = 1/2c₁/c₂ = 12/24 = 1/2

=> a₁/a₂ = b₁/b₂ = c₁/c₂=> Lines are coincident=> The system of equations have infinite solutions.

𝙞𝙣𝙛𝙞𝙣𝙞𝙩𝙚 𝙨𝙤𝙡𝙪𝙩𝙞𝙤𝙣 𝙞𝙨 𝙥𝙤𝙨𝙨𝙞𝙗𝙡𝙚

Multiply the given equation throughout with (2x - 3)²

px + q = (2x - 3) + 3px + q = 2x(p - 2)x + q = 0Comparing coeficients of x on both sides,p - 2 = 0 p = 2Comparing constant terms on both sides, q = 0

From the given options find out the quadratic equation whose zeroes are 0 and 2.

x² - 2x = 0=> x(x - 2) = 0=> x = 0, x = 2

Thus, p and q are the roots of x² - 2x = 0

3x + 2y = 12 9x - 2y = 12 3x + 2y + 9x - 2y = 12 + 12 12x = 24 x = 2 6 + 2y = 12 2y = 6 y = 3

answer is I think it is 104

no

not seen clearly

Petrol used inOne km of distance =115/20=5.75.Therefore petrol used in covering 345km =5.75×345=1983.75

1983.75 is the best best answer

The answer is given by step by step..kilometers. litres of petrol.115. 20345. ?by cross multiplication it we get answer.345×20÷115=60hope this answer will help you.....

Plz like my answer and accepted as best Plz

Sex Jeff for fudged Fri ddhc fjcdhbvf fhjxcb

petrol in one km of distance =115/20=5.75, petrol in covering 345 km = 5.75 x 345= 1983. 75

2x-5y=4.........equation(1)3x-8y=5..............equation(2)

Equation(1)

2x-5y=42x=4+5y

x=(4+5y)/2............equation(A)

Equation(2)3x-8y=5.........equa.(2)

Put value of (A) in (2)

3[(4+5y)/2]-8y=5(12+15y)/2-8y=5(12+15y-16y)/2=5..........(BY TAKING LCM)12-y=5×212-y=10y=12-10y=2Equation(1)2x-5y=4

Put value of y in any equation.

2x-5y=42x-5×2=42x-10=42x=4+102x=14x=14/2x=7

3x -5y=4)×2(9x-2y= 7)×5

6x-10y=845-10y= 35

now , subtract.so , we get39x= 27x = 27/39 = 9/13so , y = -5/13 .

Using unitary method;20 litres of petrol is used to cover 500 km∴24 litres of petrol is used to cover distance x.

∴x=(24×500)/20=600 km.

1. 1200 scooter in 16 daysSo number of scooters in 1 day = 1200/16 = 75

2. 2500 cycles in 20 daysSo in one day it produce 2500/20 = 125 cycles So in 30 days it produce 125*30 = 3750 cycles.

In 5 Litre 57.5 km is covered so in 1 Litre it will cover 57.5 km /5 L= 11.5 Km/Lso in 11 litres it will cover 11.5 × 11 = 126.5 km

distance travelled per K.M = 67.5/4.5 = 15 K.M

so, with 26.4 L Petrol , distance covered will be 15*(26.4) = 396 KM

thnks bro 😎for answering

suppose other fraction is x 9 3/5 = 48/5 3 4/10 = 34/10 so 34/10 * x = 48/5 17/2 * x = 24 so x = 48/17 = 2 14/17

sugar = 2×60= 120tea leaves = 0.5×320= 160

sugar cost 2×60=120tea leaves cost 1/2×320=160

1 kilogram sugar = 60then, 2 kilogram sugar = 2×60 = 120

1000 gram or 1 kg of tea leaves = 320therefore, 500 gram of tea leaves = 320÷2 = 170

1 kilogram sugar =60than, 2 kilogram sugar=2×60=12

1000 gram or 1 kg of tea leaves =320there fore 500 gram of tea leaves =320÷2

sugar= 2×60=120 tea leaves=0.5×320=160

cosec^6 A = cot^6 A + 3cot^2 A cosec^2 A + 1

Then, we will prove it as :

To Prove:

cosec^6 A = cot^6 A + 3 cot^2 A cosec^2 A + 1

i.e. cosec^6A – cot^6A – 3 cot^2 A cosec^2A = 1

Taking L.H.S. ,

cosec^6A – cot^6A – 3 cot^2 A cosec^2 A

=(cosec^2 A)^3– (cot^2 A)^3– 3 cot^2 A cosec^2 A

={(cosec^2 A – cot^2 A) ((cosec^2 A)^2+ (cot^2 A)^2+ cosec^2 A cot^2 A)}

={1 ((cosec^2 A)^2+ (cot^2 A)^2– 2 cosec^2 A cot^2 A + 2 cosec^2 A cot^2 A + cosec^2A cot^2 A)} – 3 cot^2 A cosec^2 A

={(cosec^2 A – cot^2 A)^2+ 3 cosec^2Acot^2A} – 3 cosec^2A cot^2 A

=(cosec^2 A – cot^2 A)^2

=(1)2

= 1

Hence,

cosec^2 A – cot^6A – 3 cot^2A cosec^2 A = 1

⇒ cosec^2 A -cot^6 A =3 cot^2A cosec^2A + 1

thats not the question

Cosec A + cot A - 1 / cot A - cosec A + 1we know that,cosec ² A - cot ² A = 1substituting this in the numeratorcosec A + cot A -(cosec ² A - cot ² A) / (cot A - cosec A + 1)x²-y²= (x+y)(x-y)cosec A + cot A - (cosec A + cot A) (cosec A - cot A) / (cot A - cosec A + 1)taking common(cosec A + cot A)(1-cosec A + cot A) / (cot A - cosec A + 1)cancelling like terms in numerator and denominatorwe are left with cosec A + cot A= 1/sin A + cos A/sin A= (1+cos A) / sin A

Total investment = Rs. 7500rate of dividend = 10%annual income = Rs. 500nominal value = Rs. 100

therefore,no. of shares = 500 × (100/10) × (1/100)= 50 shares

market value = 7500/50= Rs. 150

Total investment = Rs. 11440rate of dividend = 10%annual income = Rs. 520nominal value = Rs. 50

therefore, no. of shares = 520 × (100/10) × (1/50)= 104 shares

market value = 11440/104= Rs. 110

Total dividend = r/100*N.V*no. of shares

= 15/100*110*4000

Rs.66,000

Annual Income = r/100 * N.V * No. of shares

= 15/100 * 110 * 88

Rs.1,452

Return Percent = 10%

Rate of dividend/100* N.V = Rate of Return/100 * M.V

15/100 * 110 = 10/100 * x

x = Rs.165

Thankzz a lot.

Total capital of stock = 450000 Rs. dividend = 4%(1) Number of share = 135Value of each share = 10 Rs.value of 135 shares = 135 x 10 = 1350 Rs. dividend he earn = 4% of 1350 = (4 x 1350) / 100 = 54 Rs..

total capital stock = 450000 number is share = 135 value each share = Rs. 10; 135 shares =135 ×10=1350 rs. divided 4°/• of 1350= 4 x1350/100=Rs. 54

a. 0.25b. 0.3333333c.0.5d.0.4e.2/4=1/2=0.5f.0. 6666666g.0. 5555555