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# Ask a Mathematics(Class IX-X Tuition) Question

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P

Prantosh 11/11/2016 in  Mathematics(Class IX-X Tuition),

Mukesh invested in Rs 25 shares of a company paying 12% dividend. If he received 10% per annum on his investment, at what price did he buy each share?

Devajyoti replied | 09/12/2016

Dividend paid by company =12% of Rs 25 = 25 x 12/100 = Rs. 3/-.
10% of investment = Rs. 3/-.
100% investment = Rs 3 x 100/10 = Rs. 30/-.

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G

Ganesh 11/11/2016 in  Mathematics(Class IX-X Tuition),

Which is better investment: 7% Rs 100 shares at Rs 120 or 8% Rs 10 shares at Rs 13.50?

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S

Sujit 11/11/2016 in  Mathematics(Class IX-X Tuition),

A man invests Rs 45000 in 15% Rs 100 shares quoted at Rs 125. When the market value of these shares rose to Rs 140, he sold some shares, just enough to raise Rs 8400. Calculate the dividend due to him on these value and also the number of shares he still holds?

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R

Rituparna 11/11/2016 in  Mathematics(Class IX-X Tuition),

A man invest Rs 1,680 in buying shares of nominal value Rs 24 and selling at 12% premium. The dividend on the shares is 15% per annum. Calculate the number of shares he buys?

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H

Hardik 11/11/2016 in  Mathematics(Class IX-X Tuition),

What do you mean by share certificate?

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N

Nikhil 11/11/2016 in  Mathematics(Class IX-X Tuition),

A man buys 600, Rs 20 shares at a discount of 205 and receives a return of 10% on his money. Calculate the amount invested by him and also calculate the rate od dividend paid by the company?

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A

Aparna 10/11/2016 in  Mathematics(Class IX-X Tuition),

R

Rose replied | 03/12/2016

The equations represent Coincident lines.
Hence, a1/a2=b1/b2=c1/c2
i.e. 3/6=1/k=8/16
k=2.

Saurabh replied | 04/12/2016

For two lines to be coincident, the condition is that their ratios of coefficients should be 1. This satisfies only when k=2.

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S

Sensoph 10/11/2016 in  Mathematics(Class IX-X Tuition),

Kumar replied | 21/11/2016

x/y=5/6...(1)
(x-8)/(y-8)=4/5
On solving x=40, y=48

Urwashi replied | 23/11/2016

40 and 48.

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S

Subhradeep 10/11/2016 in  Mathematics(Class IX-X Tuition),

Muraleedharan replied | 20/11/2016

log 27 = 3 log 3 = 1.431 ; log 3 = 0.477 ; log 9 = 2 log 3 = 2 x 0.477 = 0.954

Rahul replied | 20/11/2016

0.954

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S

Sarvepalli 09/11/2016 in  Mathematics(Class IX-X Tuition),

Manan replied | 19/11/2016

350 / 100 = Rs. 3.50 per orange.
1 dozen = 12 pieces.
Therefore, sale price is 48/12 = Rs. 4.
Therefore, profit = 4 - 3.5 = Rs. 0.50 per orange.

S

Shiv replied | 19/11/2016

Each orange cost 350/100=3.5 Rs;
selling cost of each orange is=48/12=4 Rs;
100 oranges selling cost=4*100=400;
profit%=(selling price-cost price)/cost price*100;
=(400-350)/350*100;
=(50/350)*100;
=14.26%

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J

Jagabandhu 09/11/2016 in  Mathematics(Class IX-X Tuition),

Sarvajeet replied | 19/11/2016

(m+p)*(n+q)

Muraleedharan replied | 20/11/2016

= n (m + p) + q (m +p)
= (m + p) ( n + q)

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R

Rajeswary 10/11/2016 in  Mathematics(Class IX-X Tuition),

S

Shiv replied | 19/11/2016

x=320,y=-155

Daljit Kaur replied | 22/11/2016

X=340, y=-165

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V

Vishal 10/11/2016 in  Mathematics(Class IX-X Tuition),

P

Priti replied | 19/11/2016

=log? 6/ log? 6^3+{log42-log? 6} /log?7^2
=log6/3log 6+{log(42/7)}/2 log7
=3/2 is ans you got it

Muraleedharan replied | 20/11/2016

216 = 6^3; log 42 - log 6 = log (42/6) = log 7 ; [log (42) - log (6)] / log (49) = log 7 / 2 log 7 = 1/2 ; Answer = 3 + 1/2 = 3.5

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S

Santosh 09/11/2016 in  Mathematics(Class IX-X Tuition),

The marked price of a ceiling fan is Rs.1350 and the shopkeeper allows a discount of 6% on it. Find the selling price of the fan?

0 0 4

Daljit Kaur replied | 27/11/2016

SP=Rs 1269

Rajendran replied | 27/11/2016

MP of the fan Rs. 1350
Discount = 6%
= 1350 * 6/100
= 13.5 * 6
= 81
SP of the fan 1350 - 81 = Rs. 1269.

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B

Bijoy 09/11/2016 in  Mathematics(Class IX-X Tuition),

When a plot is sold for Rs 18,700 the owner loses 15%. At what price must that plot be sold in order to gain 15%?

0 0 4

Megha replied | 19/11/2016

Rs. 25300.

S

Shiv replied | 19/11/2016

sellingprice=((100-loss%)/100)*costprice;
18700=((100-15)/100)*costprice;
costprice=22,000.
sellingprice=((100+gain%)/100)*costprice;
sellingprice=((100+15)/100)*22000;
=25,300 ans

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S

Sammeyka 09/11/2016 in  Mathematics(Class IX-X Tuition),

Joy bought pens at Rs 120 a dozen. He sold it for Rs 15 each. What is his profit percent?

0 0 3

Ram Mohan replied | 21/11/2016

Profit percent = 50

Chandan replied | 05/12/2016

CP of 1 pen = 120/12= 10
SP of 1 pen = 15
Profit = 5
P % = (p/CP)*100=(5/10) x 100
= 50%.

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S

Shubham 09/11/2016 in  Mathematics(Class IX-X Tuition),

Evaluate the product without multiplying directly: 103 x 107?

0 0 4

Ram Mohan replied | 21/11/2016

Their product is 11021. First add 7 to 103 OR add 3 to 107 we get 110 either way. take this 110 as left portion of our answer 110, now multiply 3 and 7, take this product 21 as right portion of the answer. 21 putting the two portions together, we get the product of 103 and 107 as 11021.

Mamata replied | 29/11/2016

103 +3
107 +7
....................
11021
103 is 3 above base 100 and 107 is 7 above base 100
103+7=107+3=110 and multiply the deficiencies i.e 3 x 7=21.
So, 103 107=11021.

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J

Joydip 10/11/2016 in  Mathematics(Class IX-X Tuition),

What will be the condition of lines if a pair of linear equations is consistent?

0 0 4

Gourav Khandelwal replied | 19/11/2016

For two lines to be coincident the two equations say, a1x+b1y+c1= 0 = a2x+b2y+c2 must follow the the following property: (a1/a2) = (b1/b2) = (c1/c2)

V

Vimala replied | 19/11/2016

There are three conditions for a pair of lines:
1. The two lines intersect at a point, then there is a unique solution and hence the pair of linear equations is consistent
2. If the lines coincide and hence there are infinite number of solutions. And the equations are dependent
3. If the lines are parallel, there is no solution, because parallel lines never intersect.
If...  more»
There are three conditions for a pair of lines:
1. The two lines intersect at a point, then there is a unique solution and hence the pair of linear equations is consistent
2. If the lines coincide and hence there are infinite number of solutions. And the equations are dependent
3. If the lines are parallel, there is no solution, because parallel lines never intersect.
If the pair of equations is having at least one solution, then the system is said to be consistent
Therefore, from the above cases, 1 and 2 are said to be consistent. «less

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S

Shabana 10/11/2016 in  Mathematics(Class IX-X Tuition),

The cost of 4 pens and 4 pencil boxes is Rs 100. Three times the cost of a pen is Rs 15 more than the cost of a pencil box. From the pair of linear equations for the above situation. Find the cost of a pen and a pencil box?

0 0 3

Sarvajeet replied | 19/11/2016

10, 15

Suyash replied | 19/11/2016

Given ,
4 Pens + 4 Pencil box = 100
3 Pens = Pencil box + 15
On solving ,
Pen = 10
Pencil box = 15

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A

Amritangshu 10/11/2016 in  Mathematics(Class IX-X Tuition),

Simplify: {[(3 - 9) / 6] power 1 / 3}?

0 0 4

Martha Umesh replied | 03/12/2016

-1 power 1/3.

Lakshmi replied | 04/12/2016

-1

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K

Kartika 09/11/2016 in  Mathematics(Class IX-X Tuition),

What happen to the time period if the interest is compounded half yearly?

Sreedevi replied | 07/12/2016

Time period gets doubled and Rate become half.

Shambhavi replied | 07/12/2016

Time period gets doubled in this case.

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R

Rawinder 09/11/2016 in  Mathematics(Class IX-X Tuition),

The SI on a sum of money is 1/9 of the principle, and the number of years is equal to the rate % per annum. Find the rate %?

Sarvajeet replied | 26/11/2016

(10/3)%

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R

Rituraj 09/11/2016 in  Mathematics(Class IX-X Tuition),

Ramesh invests Rs. 12800 for three years at the rate of 10% per annum compound interest. Find the sum due to Ramesh at the end of the first year?

Sreedevi replied | 21/11/2016

The amount at the end of Year I is 10% of the principle added to the principle itself. So it would be 12800+(10% of 12800)=14080

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V

Vijay 09/11/2016 in  Mathematics(Class IX-X Tuition),

The cost of car, purchased 2 years ago, depreciates at the rate of 20% every year. If its present worth is Rs 315600, find its purchase value?

A

Ashwini replied | 26/11/2016

Rs. 493175.

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V

Vinay 09/11/2016 in  Mathematics(Class IX-X Tuition),

What is the formula for finding amount?

Kumar replied | 27/11/2016

A=P(1+r/100)^n. Where, A is amount, P is Principal , r is rate p.a., n is nos. of years.

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