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27 L and 500 ml is remain in tank

that's a correct answer and perfect answer

27.5 litre of water is remaining

125 L ÷ 97 L AND ANYTHING ANSWER × BY 500 ML AND THAT IS YOUR ANSWER OK BY LIKE PLZ

the answer is 27 liter 500ml

remains water=125-97.5=27 litre 500ml

the answer is 27 L and 500 ml is remain in tank

A water tank hold =125 litres If 97 l 500ml is used..so,125 l -97 l 500ml = 27 litres 500ml .... is the right answer for this questions......

correct answer is 27L and 500 ML

27l 500ml of water remains in the tank

27 L and 500ml is the correct and best answer for question

Giventotal quantity of water in tank= 125lQuantity of water used= 97l 500mltherefore remaing water = total water - used water =125l - 97l 500ml = 27l 500ml SOLVED

We know that1 litre= 1000 mlso179 l = 179000mlnow179000ml+950ml = 179950ml

the answer is 76%.

please mark me as best for explaination.

Given ratio is 3:2.

The percentage of boys with scholarship is 20% and the percentage of girls with scholarship is 30%.

The percentage of boys without scholarship is 80% and the percentage of girls without scholarship is 70%.

so,

(80/100)(3/5)+(70/100) (2/5)

=48/100+28/100

=76/100

The result is 76%.

Distance covered by plane in 1 hour= 3400km/ 4h = 850km/hso, in 7 hour plane can travel = 850× 7 = 5950km

30 metres of cloth is 810 rupees

=> 1 metre = 810/30 = 27rupees

now let the metres of cloth for 1215 rupees be x

=> cost of the cloth = 27×x but given cost = 1215

=> 27x = 1215=> x= 1215/27 = 45

therefore cloth for 1215 rupees is 45 metres.

It will besh 63.50 c+mythanks

Let's assume the number as x

According to Question,

(5 / 4)⁻³ ÷ x = (15 / 16)⁻²

⇒ (4 / 5)³ = (16 / 15)² × x

⇒ (4)³ / (5)³ = (16)² / (15)² × x

⇒ 64 / 125 = 256 / 225 × x

⇒ 64 / 125 × 225 / 256 = x

⇒ 1 / 5 × 9 / 4 = x

⇒ 9 / 20 = x

⇒ 0.45 ≈ x

Hence, the number is ( 9 / 20 )

One bat cost is 85 rupees

if the bat 1 is 200 then 2nd bat costs 15rs less than half the cost of first onenow half of 200 is 100 and 15, less than 100 is 85 so the bat will cost 85 rupees

total money spent= Rs (4895+1102)=Rs 5997

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X-20x/100= 560x- 0.2x= 5600.8x= 560x= 560/0.8x= 700shplease like the solution 👍 ✔️👍

Let total salary = x

Then,Spent on rent = 3x/8Spent on food = (3/5)*(x - 3x/8) = 3x/8Other expenses= x - 2*3x/8= x - 3x/4= x/4

But, x/4 = 1200x = 4800

Total salary = sh 4800

2)0.273)3.155)0.646)10.5

Volume = lbh = 8*2*4 = 64 cm³. Weight = Density*Volume= 12*64= 768 gmPlease hit the like button if this helped you.

2. 0.2 × 3 = 0.65. 0.8 × 3 = 2.4

1. 8 × 0.2 = 1.62. 0.03 × 9 = 0.273. 0.3 × 50 = 654. 0.16 × 4 = 0.64

area = πr^2 88.2 = 3.142 x r^2r^2 = 28.071292170r = 5.2982348164r is approx 5.30

decimal point is shifted to right

left

Savings % = (100 - 65) % = 35 %Saving = 35 % of Rs. 720= Rs(35/100 ×720) = Rs. 252

why we multiply 3200 and 80 to get total food

when you can find my answer then I will help you

thnxx

I need only with using substitution method

Thanks! Great answer. Regards:Pirzada Najmu Zuha- The Motivational Speaker

let total questions=Xattempted=9X/10

wrong answers=(9X/10)×(3/8)=27X/80

right answers=(9X/10)-(27X/80)=(72X-27X)/80=45X/80=9X/16. he attempted 9/16 right questions.

(7/15)*100 = 140/3 = 46.667%

LCM = 3612x14+7*18-31*6/36168+126-186/36=3

= (28 + 21 - 31) / 6= 3

3 is the best answer

14/3+7/2-31/6

Final result :

3

Step by step solution :

Step1:

31 Simplify —— 6

Equation at the end of step1:

14 7 31 (—— + —) - —— 3 2 6

Step2:

7 Simplify — 2

Equation at the end of step2:

14 7 31 (—— + —) - —— 3 2 6

Step3:

14 Simplify —— 3

Equation at the end of step3:

14 7 31 (—— + —) - —— 3 2 6

Step4:

Calculating the Least Common Multiple :

4.1 Find the Least Common Multiple

The leftdenominatoris : 3

The rightdenominatoris : 2

Number of times each prime factorappears in the factorization of:PrimeFactorLeftDenominatorRightDenominatorL.C.M = Max{Left,Right}31012011Product of allPrime Factors326

Least Common Multiple:6

Calculating Multipliers :

4.2 Calculate multipliers for the two fractions

Denote the Least Common Multiple byL.C.M Denote the Left Multiplier byLeft_M Denote the Right Multiplier byRight_M Denote the Left Deniminator byL_Deno Denote the Right Multiplier byR_Deno

Left_M=L.C.M/L_Deno=2

Right_M=L.C.M/R_Deno=3

Making Equivalent Fractions :

4.3 Rewrite the two fractions intoequivalent fractions

Two fractions are calledequivalentif they have thesame numeric value.

For example : 1/2 and2/4are equivalent,y/(y+1)2and(y2+y)/(y+1)3are equivalent as well.

To calculateequivalent fraction, multiply theNumeratorof each fraction, by its respectiveMultiplier.

L. Mult. • L. Num. 14 • 2 —————————————————— = —————— L.C.M 6 R. Mult. • R. Num. 7 • 3 —————————————————— = ————— L.C.M 6

Adding fractions that have a common denominator :

4.4 Adding up the two equivalent fractionsAdd the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

14 • 2 + 7 • 3 49 —————————————— = —— 6 6

Equation at the end of step4:

49 31 —— - —— 6 6

Step5:

Adding fractions which have a common denominator :

5.1 Adding fractions which have a common denominatorCombine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

49 - (31) 3 ————————— = — 6 1

Final result :

3

Processing ends successfully

14/3+7/2-31/6

Final result :

3

Step by step solution :

Step1:

31 Simplify —— 6

Equation at the end of step1:

14 7 31 (—— + —) - —— 3 2 6

Step2:

7 Simplify — 2

Equation at the end of step2:

14 7 31 (—— + —) - —— 3 2 6

Step3:

14 Simplify —— 3

Equation at the end of step3:

14 7 31 (—— + —) - —— 3 2 6

Step4:

Calculating the Least Common Multiple :

4.1 Find the Least Common Multiple

The leftdenominatoris : 3

The rightdenominatoris : 2

Number of times each prime factorappears in the factorization of:PrimeFactorLeftDenominatorRightDenominatorL.C.M = Max{Left,Right}31012011Product of allPrime Factors326

Least Common Multiple:6

Calculating Multipliers :

4.2 Calculate multipliers for the two fractions

Denote the Least Common Multiple byL.C.M Denote the Left Multiplier byLeft_M Denote the Right Multiplier byRight_M Denote the Left Deniminator byL_Deno Denote the Right Multiplier byR_Deno

Left_M=L.C.M/L_Deno=2

Right_M=L.C.M/R_Deno=3

Making Equivalent Fractions :

4.3 Rewrite the two fractions intoequivalent fractions

Two fractions are calledequivalentif they have thesame numeric value.

For example : 1/2 and2/4are equivalent,y/(y+1)2and(y2+y)/(y+1)3are equivalent as well.

To calculateequivalent fraction, multiply theNumeratorof each fraction, by its respectiveMultiplier.

L. Mult. • L. Num. 14 • 2 —————————————————— = —————— L.C.M 6 R. Mult. • R. Num. 7 • 3 —————————————————— = ————— L.C.M 6

Adding fractions that have a common denominator :

4.4 Adding up the two equivalent fractionsAdd the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

14 • 2 + 7 • 3 49 —————————————— = —— 6 6

Equation at the end of step4:

49 31 —— - —— 6 6

Step5:

Adding fractions which have a common denominator :

5.1 Adding fractions which have a common denominatorCombine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

49 - (31) 3 ————————— = — 6 1

Final result : 3

LCM of 3, 2 & 6 is 36=14×12/3×12 + 7×18/2×18 - 31×6/6×6= 168/36 + 126/36 - 186/36=168+126-186/36=108/36=3 is your answer of the given question.

=(28+21 ‐31)/6 = 3 answer

3 is correct answer.

3,2,6 L.C.M. = 6= (28+21-31)/6= (49-31)/6= 18/6=3

LCM=62×14+3×7-1×3128+21-3118

3 is correct answer

lcm = 3612×14+7*18-31*6/36168+126-186/36=3 ANS

3 is best answer for this question

Given:

Edge of a cube= 14 cm

Volume of the largest right circular cylinder = Volume of the cylinder with diameter of base 14 cm and height 14 cm

Radius of cylinder= diameter/2= 14/2=7 cm

Height of cylinder = 14 cm

Volume of cylinder = πr²h

Vol. of largest right circular cylinder = (22/7)×(7)²×14

= 22×(7)²×2

= 22×49×2

= 2156 cm²

Hence, the volume of the largest right circular cylinder which can be cut of a cube, each edge of which is 14 cm long is 2156 cm²

As 2 50 paise in 1 rupeesHence in 9 rupees =2*9=18 coinsnow in 75 paise=50paise +25 paise hence 1 coin more19coins

Sides of right angle triangle arex - 1, x, x + 1

Using pythagoras theoram (x + 1)^2 = x^2 + (x - 1)^2x^2 + 1 + 2x = x^2 + x^2 + 1 - 2xx^2 - 4x = 0x(x - 4) = 0x = 0, 4

Zero value not possible.

So value of x = 4

Sides of triangle are:(x - 1) = 4 - 1 = 3x = 4(x + 1) = 4 + 1 = 5

As we knowa3 +b3 + c3 – 3abc = (a+b+c) (a2 + b2 + c2 – ab – bc – ca)

Given,a + b + c = 0, ab + bc + ca = 5

Therefore, a3 +b3 + c3 – 3abc = 0* (a2 + b2 + c2 – 5)

a3 + b3 + c3 - 3abc = 0

a3 + b3 + c3 = 3abc

no of shirts to each = total no of shirts/no of persons

= 875/5

= 175 shirts

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175 shirts of a person

8 × 8 = 64

8 x8=64law of multiplication

Total pens=25Number of people=5

Number of pens each friend gets=25/5=5

SO first select 4 out of 12 then again 4 out of remaining 8 then again 4 out of remaining 4 It will be 12C4* 8C4 * 4C4 = 12!/4!*4!*4! but here when you notice closely you will find that you counted same case 3! times instead of one This is a problem in this method. To avoid this you divide it by 3! in above expression. = 12!/4!*4!*4!*3! =5775 ways.

bhai thank you very much