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

What do you mean by share certificate?

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

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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Kumar replied | 21/11/2016

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

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Urwashi replied | 23/11/2016

40 and 48.

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

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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

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Rahul replied | 20/11/2016

0.954

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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.

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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  Class IX-X Tuition, Mathematics(Class IX-X Tuition), Factorization

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Sarvajeet replied | 19/11/2016

(m+p)*(n+q)

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Muraleedharan replied | 20/11/2016

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

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R

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S

Shiv replied | 19/11/2016

x=320,y=-155

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Daljit Kaur replied | 22/11/2016

X=340, y=-165

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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

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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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Santosh 09/11/2016 in  Class IX-X Tuition, Mathematics(Class IX-X Tuition), Profit, Loss and Discount

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?

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Daljit Kaur replied | 27/11/2016

SP=Rs 1269

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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  Class IX-X Tuition, Mathematics(Class IX-X Tuition), Profit, Loss and Discount

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%?

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Megha replied | 19/11/2016

Rs. 25300.

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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  Class IX-X Tuition, Mathematics(Class IX-X Tuition), Profit, Loss and Discount

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

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Ram Mohan replied | 21/11/2016

Profit percent = 50

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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  Class IX-X Tuition, Mathematics(Class IX-X Tuition), Expansions

Evaluate the product without multiplying directly: 103 x 107?

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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.

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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  Class IX-X Tuition, Mathematics(Class IX-X Tuition), Linear Equations

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

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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)

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

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?

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Sarvajeet replied | 19/11/2016

10, 15

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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  Class IX-X Tuition, Mathematics(Class IX-X Tuition), Logarithms

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

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Martha Umesh replied | 03/12/2016

-1 power 1/3.

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Lakshmi replied | 04/12/2016

-1

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K

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

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

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Sreedevi replied | 07/12/2016

Time period gets doubled and Rate become half.

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Shambhavi replied | 07/12/2016

Time period gets doubled in this case.

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

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 %?

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Sarvajeet replied | 26/11/2016

(10/3)%

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R

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

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?

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

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?

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Ashwini replied | 26/11/2016

Rs. 493175.

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

What is the formula for finding amount?

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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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Supriyo 09/11/2016 in  Class IX-X Tuition, Mathematics(Class IX-X Tuition), Compound Interest

The difference between simple interest and compound interest compounded annually on a certain sum of money for 2 years at 4% per annum is Rs 1. What is the sum of rupees?

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Ram Mohan replied | 18/11/2016

If 'P' is the Principal Sum , then Compound Interest (CI) @4%, after 2 years is:
CI = P{ (1+4/100) (1+4/100) -1} =P { 1+16/10000 +2*4/100 -1} = P{16/10000+2*4/100}
Simple Interest SI, @ 4% after 2 years is:
SI = P*2*4/100
It is given that, CI - SI= Rs.1, Substituting the values of CI and SI in the above eqn, we get:
P{16/10000 + 2*4/100} - P*2*4/100 =1,...  more»
If 'P' is the Principal Sum , then Compound Interest (CI) @4%, after 2 years is:
CI = P{ (1+4/100) (1+4/100) -1} =P { 1+16/10000 +2*4/100 -1} = P{16/10000+2*4/100}
Simple Interest SI, @ 4% after 2 years is:
SI = P*2*4/100
It is given that, CI - SI= Rs.1, Substituting the values of CI and SI in the above eqn, we get:
P{16/10000 + 2*4/100} - P*2*4/100 =1, ==> P*16/10000 = 1, ==> P = 10000/16, ==> p= 625
Therefore, the principal Sum lent was, Rs.625/-
One Can use the Formula
P*(r/100)*(r/100) = d , where, 'P' is Principal Sum, 'r' is rate of interest for a prescribed period, and 'd' is the difference between CI and SI, for two such periods; can be used in this case. «less

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Suyash replied | 19/11/2016

Given,
Compound Interest -Simple Interest = 1
P { 1+ ( 4/100) } ^ 2 - P - ( P×4×2 / 100 ) = 1
On solving,
P = 625

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S

Srinu 09/11/2016 in  Class IX-X Tuition, Mathematics(Class IX-X Tuition), Compound Interest

What is the difference between the compound interest on Rs. 5000 for 1 year at the rate 4% per annum compounded yearly and half yearly?

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Shikha replied | 22/11/2016

Compound interest = Amount +Principal. One and a half year can be calculated as 3 and half years.

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Geeta 09/11/2016 in  Class IX-X Tuition, Mathematics(Class IX-X Tuition), Compound Interest

The compound interest on Rs 30000 at the 7% per annum is Rs 4347. What is the period (in years)?

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Ram Mohan replied | 21/11/2016

The period is 2 years. Very simple solution
30000[ (1+7/100) power n] - 30000 = 4347 = = > 30000 [ (1+7/100) power n] = 34347 ==>
[ (1+7/100) power n] = 34347/30000 ==> [ (100+7)/100) power n] = [ (100+7)/100 ) power 2 ==> n=2

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Kiran 09/11/2016 in  Class IX-X Tuition, Mathematics(Class IX-X Tuition), Compound Interest

What will be the compound interest on a sum of Rs. 25000 after 3 years at the rate of 12 p.a?

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Shikha replied | 23/11/2016

Amount =[25000(1+12/100)3]
=[25000*28*28*28]/25*25*25
=Rs 35123.20
CI = Rs 35123.2+ Rs 25000= Rs 10,123.20.

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A

Apollo 09/11/2016 in  Class IX-X Tuition, Mathematics(Class IX-X Tuition), Compound Interest

Define compound interest?

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Shikha replied | 23/11/2016

Compound interest is the interest to the principal sum i.e., Compound interest = Principal of last year + Amount
Amount = (1+r%) to the power n, number of years.

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