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10% of 450 - 12% of 500 + 8% of 500 = 10×450/100 - 12×500/100 + 8×500/100 = 45 - 60 + 40 = 85 - 60 = 25

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Let Ruchi have money = nThen money Sindhu have = 2n

According to the given conditionn - 60 = 2n - 2102n - n = 210 - 60n = 150

Money Ruchi have at the beginning = Rs 150Money Sindhu have at the beginning = Rs 300

we have identity cos (A-B)=cosA cosB +sinA sinBsosin3A sinA+cos3A cosA=cos (3A-A)=cos2Atherefore, sin3A sinA+cos3A cosA+2sin^2 A=cos2A+2sin^2 A=1-2sin^2 A+2sin^2 A....... (using cos2A=1-2sin^2 A)=1

Number of tiles = Area of floor/ Area of a tile= (1080*100*100)/(24*10)= 45000.

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Gain of 10%=110x/100, if x is the averageSo68.20=110x100.........x=682/1168.20=62

STEP II : Use weighted average method to find the ratio60....62........65So 60:65=65−62/62−60=3/2=3:2

This years enrolment = (last year's enrolment + 15/100 of last years' enrolment) 552 = 115/ 100 * last years' enrolment

last year's enrolment = 55200/115 = 480 students.

rest suger is sold in 13 percent profit

1+2 =3total par 3×12%=36%1part par 1×10%=10%

He must sold the remaining Suger at 13% gain so as to get 12% on taotal

he must be sell there 2/3 to 13% profit...

rest suger is sold 13% profit

1+2 =3 total par 3× 12%=36%1part par 1×10%=10%

rs 45001/3 of total sugar =1/3*4500=150010% of 1500= xso, x% of 3000 = 3390-3000

Therefore,he must sold the remaining sugar at 13% gain so as to get 12% gain on total sugar

rest suger is sold in 13 percent profit

this question is right answer 13 precent

rest Suger is solid in13%

cp of sugar =4500. without profit one third of the gain =10%. cp on one third of the sugar -Rs=4500÷3=Rs 1500Rupisj

CPof sugar = Rs 4500Profit on one-third of the sugar = 10%

CP of one-third of the sugar = Rs45003=Rs.150045003=Rs.1500

SPofone−thirdofthesugar=100+gain%100×CP=Rs110100×1500=Rs1650SPofone-thirdofthesugar=100+gain%100×CP=Rs110100×1500=Rs1650

Now, profit= Rs1650−15001650−1500= Rs 150

At a profit of 12%, we have:SPofsugar=100+gain%100×CP=Rs112100×4500=Rs5040SPofsugar=100+gain%100×CP=Rs112100×4500=Rs5040

∴ Gain= Rs5040−45005040−4500= Rs 5400

Profit on the remaining amount of sugar = Rs540−150540−150= Rs 390CP of the remaining sugar = Rs4500−15004500−1500= Rs 3000

Gainpercentage=(gainCP×100)%=(3903000×100)%=13%Gainpercentage=gainCP×100%=3903000×100%=13%

Therefore, the profit on the remaining amount of sugar is 13%.

1+2=3total par 3×12%=36%1part par 1×10%=10%

the remaining sugar will be sold at 13%profit

13%rest sugar profit in this grocer

a grace bought 300 PC's but don't mind they can you solve these problem

The remaining sugar must be sold to have a gain of 12 per cent on the whole at 13 per cent

he must sold tha remaining sugar at 13%gain so as to get 12% gain total

this question is Right answer 13% sugar

I did not understand

13

13% is the profit...........

rest sugar is sold in 13 percent profit

1+2=3total per 3×12%=36%1part per 1×10%=10%

13% remaining sugar sold

RS 3000gain percentage =13%

Area of a equilateral triangle=√3a^2/4or 81√3=√3a^2/4or a^2=81√3×4/√3or a=√81×4.•.a=18

then, height of equilateral triangle=√3a/2=18√3/2=9√3 answer

the median of a triangle joins to the vertex to the midpoint of the opposite side

| a-b| = √a²+b²-2abcos∅ = 5

=> 9+16 -2*3*4cos∅ = 25=> cos∅ = 0

so, |a+b| = √a²+b² +2abcos(90) = √25 = 5

Overall gain = (6/100)×(15000) - (5/100)×(15000)= 900 - 750.= ₹150.

please give percentage

For right circular cylinder

Diameter = 12 cm

Radius(R1) = 12/2= 6 cm & height (h1) = 15 cm

Volume of Cylindrical ice-cream container= πr1²h1= 22/7 × 6× 6× 15= 11880/7 cm³

Volume of Cylindrical ice-cream container=11880/7 cm³

For cone,

Diameter = 6 cm

Radius(r2) =6/2 = 3 cm & height (h2) = 12 cmRadius of hemisphere = radius of cone= 3 cm

Volume of cone full of ice-cream= volume of cone + volume of hemisphere

= ⅓ πr2²h2 + ⅔ πr2³= ⅓ π ( r2²h2 + 2r2³)

= ⅓ × 22/7 (3²× 12 + 2× 3³)

= ⅓ × 22/7 ( 9 ×12 + 2 × 27)

= 22/21 ( 108 +54)

= 22/21(162)

= (22×54)/7

= 1188/7 cm³

Let n be the number of cones full of ice cream.

Volume of Cylindrical ice-cream container =n × Volume of one cone full with ice cream

11880/7 = n × 1188/7

11880 = n × 1188

n = 11880/1188= 10

n = 10

Hence, the required Number of cones = 10

CP=15000010% profitso SP should be 165000one third sold at loss of 6%so SP=50000*0.94=47000so remianing S0=118000so remaining have to sell at 18%

CP=15000010% profitso SP should be 165000one third sold at loss of 6%so SP=50000*0.94=47000so remianing S0=118000so remaining have to sell at 18%

Thankew

4 no =a-3d,a-d,a+d,a+3dsum , 4a=56a=14(a-3d)(a+3d)/(a-d)(a+d)=5:66(a²-9d²)=5(a²-d²)a²=54d² - 5d²14² =49d²d²=14² /7²d=±14/7=±2for d=2, a-3d=14-6=8a-d=14-2=12a+d=14+2=16a+3d=14+6=20Numbers are 8,12,16,20

population of the city = 2100000

ratio of male to females = 4:3

let the ratio factor be X.

number of males = 4X

number of females = 3X

A/Q

4X + 3X = 2100000

7X = 2100000

X = 2100000 / 7

X = 300000

number of females = 3X = 3 x 300000 = 900000

number of females in fertility age group = 55% of 900000 = (55 x 900000)/100 = 495000

ER. RAVI KUMAR ROY

since GRF is 40, that means 40%

crude Birth rate = 40% of 495000 = (40 x 495000)/100 = 198000 children (ANS)

Answer is 30%...Marked price is ₹ 19, 500.Cost price is ₹ 15, 000.

See calculations in the picture enclosed.

Total Selling Price for 2 bullocks = 24000 + 24000 = 48000.

Given that, on the 1st bullock, he gains 25%.

Cost Price of the 1st bullock = 24000 * 100/125

= 19200.

Given that, on the 2nd bullock, he lost 20%.

Cost price of the 2nd bullock = 24000 * 100/80

= 30000.Total cost price of 2 bullocks = 19200 + 30000

= 49200.

We know that Loss = CP - SP

= 49200 - 48000

= 1200.

AND,

Loss% = (Loss/CP) * 100

= (1200 * 100/49200)2.43%

Given -

Kms driven : 100km

Fuel consumed : 8 litres

To find : how much it will travel in 26 litres of fuel

Solution : First of all find out fuel economy of car how much it traveled in 1litre and multiply your answer with 26litres of fuel

Find per litre fuel consumption

Distance Traveled/fuel consumption = Fuel economy

100/8 = 12.5

The car travels 12.5 km in 1litre of fuel.

2. To find distance covered in 26litre

Sincr car can travel 12.5 in 1 litre of fuel

In 26litre it will cover

Fuel consumption X fuel economy per litre = distance traveled

26 X 12.5 = 325

Therefore car can travel 325 kms in 26litres of fuel.

40 litre of milk contains 10% water so water is 40×10/100 = 4 litre if we add 5 litre of water then total water 4+5 = 9 litre and total solution 40+5 = 45 so water percentage 9*100/45 = 20% so 5 litre water need to add .

pq=h; qr=10m; V3=h/10; h=10V3 x V3/ V3=10×3/V3=30/V3

Hope this will help you25 liters = 90 km1 km = 25/90288km = 288 x 25/90 liters = 80 liters

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Let the cost price of the article be Rs 100.

Given that Renu sold it at loss of 8%.

Hence the selling price = Rs 100 – Rs 8 = Rs 92

It is given that if she had purchased it at 10 % less

The new CP = Rs 100 – Rs 10 = Rs 90.

Given that she will get a gain of 20% on it.

Hence new SP = Rs 90 + Rs 18 = Rs 108

Therefore the difference in SP = Rs 108 – Rs 92 = Rs 16

Given the difference in SP = Rs 36

Initial CP =100 * (36/16) = 225

Thus the initial cost price is Rs 225.

Let the increased rate be x. Then, (100+20)/100 * cost = ((100+x)/100)*(90/100)*cost=> 6/5= (100+x/100)*(9/10)=> 100+x = 600/5 * 10/9=> 100+x = 400/3=> x = 400/3 - 100=> x =100/3 =33(1/3)%

answer is wrong

give me correct answer

The answer has now been corrected.