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A dealer 25kg of rice at the rate of rupees 6perVkg and 35kg of rice at the rate of rupees 7per kg he mixed the two and sold the mixture at the rate of rupees 6.75per kg what was his gain/loss in this transaction

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Price of Rice_{1} = 6/ per Kg & the Quantity = 25 Kg => the total cost = Price * Units = 6 * 25 = 150/
Price of Rice2 = 7/ per Kg & the Quantity = 35 Kg => the total cost = Price * Units = 7 * 35 = 245/
Total CP = 150 +245 = 395/
SP for 1 Kg of Mixed Rice = 6.75 & Total Quantity = 25 + 35 = 60Kg => the Total SP = 6.75 * 60 = 405/
Hence the Gain = 10/ (Answer) & Gain % = (10 * 100)/395 = 2.53%
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Anuj sells a shirt for rupees 725 and lose rupees 25 what should he charge for the shirt if he want to gain 20%

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SP of shirt = 725/
Loss = 25/ => CP = SP + Loss = 725 + 25 = 750/
To Gaini 20%, SP should be CP * (100 + 20)/ 100 = 750 * 120/100 = 900/ (Answer)
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An ice crean vender buys 250 ice crean cups at the rate of rupees 7per cup due to bad wheather 10 cups get spolid at what price per cup should he sell the reminder so as to gain 20%

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Total Cups bought = 250 & CP per cup = 7/ => Total CP = 1750/
Total SP to gain 20% should be CP * (100 + Profit%)/ 100 => 1750 * 120/100 = 2100/
Total cups to sell = 250  10 = 240 no.
SP per cup should be Total SP required/ No. of cups available for sale = 2100 / 240 = 8.75/ (Answer)
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Deepa sold a Chair for rupees 1850 and lost rupees 150 on it for how much should she have sold it to get 5%profit

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SP of chair = 1850/
Loss = 150/ => CP = SP + Loss = 1850 + 150 = 2000/
To get 5% Profit, SP should be CP * (100 + Profit%)/ 100 = 2000 * 105/100 = 2100/ (Answer)
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A green grocer bought 20kg potatoes at rupees 6 per kg and 30kg potatoes at rupees 7per kg at what rey per kg should he sell them on a mixing to gain 25%

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Dharmendra bought bananas at rupees20 a dozen he has to sell the at a loss of 10%find the selling price of bananas

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A dealer sells a transistor at a gain of 16%if he has sold it for 20rupees more he found hace gained 20%find the cp of transistor

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A doler sold to tV for rupees4000 each brother losing ñor gaining in the del if he sold one tv at a gain of 25%find the loss/gain percent in other

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Total SP for both TV's = 8000/ And if no profit  no loss then the total CP is also equal to 8000/
SP for 1st TV = 4000/ & Profit % = 25 => CP of 1st TV = 4000 * 100/125 = 3200/
It means the CP for the 2nd TV = 8000  3200 = 4800/
Now, SP for 2nd TV = 4000/
Let the loss % = x => CP of 2nd TV = 4000 * 100/(100  x) => 4800 = 4000 * 100/(100  x)
4800 (100  x) = 400000 => 48 (100  x) = 4000 => 4800  48x= 4000 => 48x = 800
x = 16.66% => Loss of 16.66% (Answer)
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If article is sold at a cerrar price one gain 10%what Will be the gain percent if sold for doble the price

Answers:

Let the CP = 100/
Profit = 10% => SP = 100 * 110/100 = 110/
if the new SP = 220/ then the Profit% = {(SP  CP) * 100} / CP
{(220  100) * 100}/ 100 = (120 * 100)/100 = 120% (Answer)
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by selling 200 books a man gain the selling price of the 400 books find how gain percent

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The statement is incorrect. One can not get the profit as SP of 400 Books after selling 200 Books. Although, he may gain the CP of 400 Books but not SP.
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Vinay sells a watch for rupees1140 at a loss 5%at what price should he sell the watch to gain 5%

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Ans. Let Cost Price [CP]= y
Selling Price [SP] = 1140
Loss % = 5%We Know
\({SP} =[{ {(100 Loss\%)\over 100 } \times CP}]\)
=> \(1140 = [{{(1005)\over 100} \times y }]\)
=> \(1140 = {95\over 100} \times y \)
=> \( y = {{1140 \times100}\over 95} = 1200 \)
Cost Price = Rs. 1200
Gain % = 5 %
New SP = \({1200 \times 105 \over 100 } = 1260 \)
To Gain 5% He Should Sell For Rs. 1260
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If the cost of 10 article is equal to the selling price of 9 article then find the gain percent

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Ans. Let Cost of 1 Article = x
Cost of 9 Articles = 9x
Then Cost of 10 Articles = 10x
Selling Price of 9 Articles = Cost of 10 Articles = 10x [Given]
Gain = 10x  9x = x
Gain % = \({Gain \times 100 \over Cost \space Price } = { x \times 100 \over 9x } = {100\over 9 }=11{1\over 9}\%\)
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Arpit sold a watch at a gain of 5%to mohit and mohit sold it to rohit at a gain of 4%if rohit paid rupees 91 for it then find the cost price of arpit watch

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Let Cost price of watch be Rs.100
Then Selling price of watch for Arpit = 100 + 5 = Rs.105
Now, SP for Arpit = CP for Mohit.
Mohit sold watch at a gain of 4%, Then,
Selling price of watch for Mohit = 105 + 4% of 105 = 105 + 4.20 = Rs.109.20
Now, Selling price of watch for Mohit = CP of watch for Rohit.
Therefore,
Since Rohit paid Rs,109.20, when CP = Rs.100
Therefore, Rohit paid Rs.1, when CP = 100/109.20
Therefore, Rohit paid Rs.91, when CP = 100/109.20 x 91 = Rs.83.33
The cost price of watch was Rs.83.33.
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Let's Arpit paid Rs. x/ for the watch which will be the CP. Arpit's SP will be Mohit's CP as Mohit bought it from Arpit. In the same way, Rohit bought it from Mohit so Mohit's SP will be Rohit's CP.
=> Mohit's CP = x * 105/100 = 1.05x
Mohit further sold it to Rohit => Rohit's CP = 1.05x * 104/100 = 1.0920 x/
=> 91 = 1.0920 x
x = 91/1.0920 = 83.33/ (Answer)
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Ans. Let Cost Price of Arpit Watch = y
Gain % = 5%
Selling Price for Arpit = \({y \times 105 \over 100 } = {21y\over 20}\)
Selling Price for Arpit is Cost price for Mohit = \(21y\over 20 \)
Gain % = 4 %
Selling Price For Mohit = \({21y\over 20} \times {104\over 100} = {21y\times 26 \over 20 \times 25} = {546y\over 500}\)
Selling price For Mohit is Cost price for Rohit, So
=> \({546y\over 500} =91\)
=> \(y = {250\over 3} = 83{1\over 3}\)
Cost Price of Arpit Watch = \(Rs. \space 83{1\over 3}\)
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A school cricket team played 25 matches in one session it won 80%of team how many matches did they lose

Answers:

Ans. Total Match = 25
Win % = 80%
Match Won = \({25\times 80\over 100} = 20 \)
Math Lose = 2520 = 5
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The expanditure made on a dress is rupees 2050.20%of the expence was on embroidery 50%on cloth and remaiming on the stiching find the money spent on stiching

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Ans. Total Expanditure = Rs. 2050
Expence on Embroidery in % = 20%
Expence on Embroidery = \({2050 \times 20 \over 100} = Rs. 410\)
Expence on Cloth in % = 50%
Expence on Cloth = \({2050 \times 50 \over 100 }= Rs. 1025\)
Remaining = 2050(410+1025) = Rs. 615
So Money Spent on Stiching = Rs. 615
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Total Expenditure = 2050/
Expence on Embriodery = 20%
Expence on Cloth = 50%
Remaining Expense = 100%  20%  50% = 30%
So, Expense on stiching = 30% of 2050/ = 30/100 * 2050 = 615/ (Answer)
{* means \({Multiply}\)}
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in an examination 95%of the candidates passed and 92 failed find the number of candidates who appeared for the examination

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Let the Total Candidates are x.
Total Pass Candidates = 95%
Remaining = 100%  95% = 5% => 5% candidates failed which is equal to 92 (given)
=> 5% of x = 92 => x = 92 * 100/5 => 92 * 20 => 1840 (Answer)
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Ans. Let Number of Candidate = y
Pass % = 95%
So Fail % = 10095 = 5%
Number of Failed Candidate = 92
Then
\({y \times 5 \over 100} = 92\)
=> \(y = {92 \times 100 \over 5 } = 1840\)
No of Total Candidate = 1840
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Pawan wants to make a disc to play he cuts a circle of diameter is 10cm from a circular plastic sheet of dismeter 14cm find the area of the disc

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diameter of circular plastic sheet = 14cm
radius = 7cm
Area = \( { \pi\ r^2}\)
= 154 cm^{2}
Diameter of circle he cuts = 10cm
Radius = 5cm
Area = 78.57 cm^{2}
Area of disc = 154  78.57
= 75.43 cm^{2}
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The diameter of two circle are in the ratio 3:2 find the ratio of their areas find the area of the circle whose circumference is 77cm

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I guess your question has two parts:
The first part is
Let the diameters of the two circles be 3x and 2x cms.
Area of circle in terms of diameter = \( \pi \ d^2\ \over 2\)
A_{1} = \( { 9\pi\ x^2 \over 2}\)
A_{2} = \( {4 \pi\ x^2 \over 2}\)
A_{1}/A_{2} = 9:4
2) The second part is:
C = \( {\ 2\pi\ r}\)
C = 77cm
2 \( {\pi}\) r = 77
2 X 22 X r = 77 X 7
r = 49/4
r = 12.25 cm
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Ans. Let the diameters of two circles are 3x and 2x.
Radius of first circle = \({3x\over 2}\)
Radius of second circle = \({2x\over 2} = x\)
Area of first circle = \({\pi } ({3x\over 2})^2 = {\pi 9x^2\over 4}\)
Area of second circle = \(\pi x^2\)
=> Ratio of Areas = \({\pi 9 x^2\over 4 \pi x^2} = {9\over 4} = 9:4\)
Circumference of circle = 77cm
Let Radius = r
then \(2 \times {22\over 7}\times r = 77 \)
=> \(r = { 77 \times 7 \over 2 \times 22} = {49\over 4 }\)
Area of Circle = \({22\over 7 } \times {49\over 4}\times {49\over 4} = 471.625 cm^2\)
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Ranjeeta scored 492 marks out of 600 whereas her brother scored 435 marks out of 500 who performed better

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Percentage got by Ranjeeta = 492/600 x 100 = 82%
Percentage got by Ranjeeta's brother = 435/500 x 100 = 87%
Therefore, Ranjeeta's brother performed better.
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If A income is 26 2/3%more than that of B find how much percent is B income less than that of A

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Ans. Let Income of B = x
Then income of A = \(x + {x \times 80\over 100\times 3 } = {380x\over 300 } = {19x\over 15}\)
=> income of B is less than A = \({19x\over 15 }  x = {4x \over 15}\)
=>Less in % = \({4x\times 100 \times 15 \over 15 \times19x } = {400\over 19} = 21{1\over 19}\%\)
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A circle of radius 3cm is cut out from a rectanglular peice of peper of dimenssion 10cm x11cm find the area of the left over paper

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Ans. Area of Rectangular Paper = 10×11=110cm^{2}
Area of Circle = \(\pi r^2 ={ 22\over 7}\times 9\)= 28.28 cm^{2}
Area of Left over paper = 110  28.28 = 81.72cm^{2}
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The ratio oo the circumference of two circle is 4:1 find the ratio of their areas

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Ans. Let Radius of first circle = R
Radius of 2nd circle = R
the
=> \({2\pi R \over 2\pi r } = {4\over 1} \)
=> \({R \over r} ={ 4\over 1}\).......(1)
Area of first circle = \(\pi R^2\)
Area of 2nd circle = \( \pi r^2\)
Ratio = \({\pi R^2 \over \pi r^2} = {R^2 \over r^2} = ({R \over r})^2 = {16\over 1}\)
So Ratio = 16:1
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Maria wants to fence her kitchen garden of diameter 29m find the length of the wire she need to purchase if she makes 6 round with the wire to fence her garden also find the cost of fencing if 1m of wire cost rupees 11

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Ans. Diameter of Kitchen Garden (D) = 29m
Wire Need For one Round = Circumference of Circle = \(\pi D = {22\over 7} \times 29 = {638 \over 7 } = 91.1 \space m\)
Wire need for 6 round = \(91.1 \times 6 = 546.6 =Approx. \space 547m\)
Cost of 1 m Wire = Rs. 11
Cost of 547 m Wire = \(547 \times 11 = Rs. \space 6017\)
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Pawan wants to make a disc to play he cuts a circle of diameter is 10cm from a circular plastic sheet of diameter 14cm find the area of the disc

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the diameter of two circle are in the ratio 3:2find the ratio of their areas find the area of the circle whose circumference is 77cm

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the driving wheel of an engine is 2.1m in radius find the distance covered when it has made 8000 revolution

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Ans. Radius of Wheel (r) = 2.1m
Distance covered in one Revolution = Circumference of Wheel = \(2{\pi} r = 2\times {22\over 7}\times 2.1 = 13.2 m\)
Distance covered in 8000 Revolutions = \(13.2 \times 8000 = 105600m = 105.6 Km\)
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What is pythagoras property and when we use it ? Explain.

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Pythagoras theorem says that in a right angled triangle, the square of hypotenuse is equal to the sum of squares of perpendicular and base.
If you have to measure length of one of the sides of a right anlged triangular plain whose other two dimensions are given, then you can you Pythagoras property.
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Pyhagoras theorem tells us that in a right angled triangle ABC with AC as hypotenous (AB×AB)+(BC×BC)=(AC×AC)
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An email address consists of two parts. Plz tell the names.

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The first i user name and the second is domain name.
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What is the importance of data privacy ?

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boommm

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Boom: it's a slang which means loud & deep sound.
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