CBSE math
CBSE maths CBSE Sample Guess Question Papers CBSE hindi social science results X X11 syllabus
Question16. I need CBSE maths new paper pattern. by Ashmita class X answered on Jan 9, 2007 at 8:00 a.m. (+ 5.5 G.M.T.)
Answer16. You can download it from here click here
Question17. Can i get full information about project work of 10th board exam? by sunny kejriwal answered on Jan 10, 2007 at 9:08 p.m. I.S.T.
Answer17. Sure, you can download Maths Lab Manual from here. click here
The Project work in maths carries 5 marks and out of 5 project you have to make one project. If after going through the Lab manual if you have any doubt feel free to contact us again.
Question18. Find the value of p for which the points (-5,1),(1,p),(4,-2) are collinear. by Chandini class x answered on Jan 11, 2007 at 10:15 I.S.T.
Answer18. Let (1,p) divides the join of (-5,1) and (4,-2) in ratio k:1. Using section formula 1 = (4k - 5)/(k+1)
k = 2
p = (- 2k+1)/(k+1)
p = - 1
Question19. Can you please tell what is Siddhacharyas formula? by ravikant class X answered on Jan 12, 2007 at 7:40 p.m. I.S.T.
Answer19. Quadratic formula is called Siddhacharyas formula.
Question20. Find probability of 53 Sundays in a leap year. by Simran class x, answered on Jan 13, 2007 at 8:15 a.m.
Answer20. Total days in a leap year is 366. (366/7) = 52 2/7. So there are 52 sundays the remaining two days can be any two consecutive days for example sunday and monday. Out of the total outcomes thus obtained 2 contain sunday.So the Probability of 53 sundays in a leap is 2/7.
Answer16. You can download it from here click here
Question17. Can i get full information about project work of 10th board exam? by sunny kejriwal answered on Jan 10, 2007 at 9:08 p.m. I.S.T.
Answer17. Sure, you can download Maths Lab Manual from here. click here
The Project work in maths carries 5 marks and out of 5 project you have to make one project. If after going through the Lab manual if you have any doubt feel free to contact us again.
Question18. Find the value of p for which the points (-5,1),(1,p),(4,-2) are collinear. by Chandini class x answered on Jan 11, 2007 at 10:15 I.S.T.
Answer18. Let (1,p) divides the join of (-5,1) and (4,-2) in ratio k:1. Using section formula 1 = (4k - 5)/(k+1)
k = 2
p = (- 2k+1)/(k+1)
p = - 1
Question19. Can you please tell what is Siddhacharyas formula? by ravikant class X answered on Jan 12, 2007 at 7:40 p.m. I.S.T.
Answer19. Quadratic formula is called Siddhacharyas formula.
Question20. Find probability of 53 Sundays in a leap year. by Simran class x, answered on Jan 13, 2007 at 8:15 a.m.
Answer20. Total days in a leap year is 366. (366/7) = 52 2/7. So there are 52 sundays the remaining two days can be any two consecutive days for example sunday and monday. Out of the total outcomes thus obtained 2 contain sunday.So the Probability of 53 sundays in a leap is 2/7.
Question21. I have not recieved the answer of my 9th class question. i had asked the question in last week.by Hitesh Kumar class 1x, answered on Jan 14, 2007 at 9:00 p.m. I.S.T.
Answer21. At this time we are answering only one question per day, so we are not able to answer all questions we receive . These questions are more than 50 per day. You have second time so we will try to answer your question at the earliest.
Question22. Sum of 3 terms of an A.P. is 3 and product is - 24. Find the three of A.P. by Aakansha class x, answered on Jan 17, 2007 at 10:00 p.m. I.S.T.
Answer22. Let the 3 terms be a + d, a, a + d
Sum = a + d + a + a + d = 3
3a = 3
a = 1
Product = (a + d)a(a - d) = - 24
(1 + d)1(1 - d) = - 24 (since a = 1)
1 - d^2 = - 24
d = +- 5
Terms of A.P. are -4, 1, 6 or 6, 1, - 4
Question23. I want your suggestion for my son about maths for comming exam cbse final 2007 by Arup Kr, Das, answered on Jan 19, 2007 at 5:31 p.m. I.S.T.
Answer23. For maths exam suggestion refer this page click here . If you have any other question write to us once again through the same form.
Question24. Find the sum of first hundred even positive integers divisible by 5 by Debanik Bhattacharjee answered on Jan 20, 2007 at 8:05 a.m. I.S.T. ( +5.5 G.M.T.)
Answer24. A.P. of reqd. numbers is 10, 20, 30, 40, .........
Here first term = a = 10, common difference = d = 10
100th term = a + 99 d
= 1000
Sum of 100 terms n/2(a + l) where l is last term
= 100/2(10 + 1000)
= 50500
Question25. Sum of 3terms of A.P. is 36 and product is 1296. Find A.P? by mohammed noufal answered on Jan 21, 2007 at 9:43 p.m. I.S.T.
Answer25. This type of question has been answered (see Q 22)
Answer21. At this time we are answering only one question per day, so we are not able to answer all questions we receive . These questions are more than 50 per day. You have second time so we will try to answer your question at the earliest.
Question22. Sum of 3 terms of an A.P. is 3 and product is - 24. Find the three of A.P. by Aakansha class x, answered on Jan 17, 2007 at 10:00 p.m. I.S.T.
Answer22. Let the 3 terms be a + d, a, a + d
Sum = a + d + a + a + d = 3
3a = 3
a = 1
Product = (a + d)a(a - d) = - 24
(1 + d)1(1 - d) = - 24 (since a = 1)
1 - d^2 = - 24
d = +- 5
Terms of A.P. are -4, 1, 6 or 6, 1, - 4
Question23. I want your suggestion for my son about maths for comming exam cbse final 2007 by Arup Kr, Das, answered on Jan 19, 2007 at 5:31 p.m. I.S.T.
Answer23. For maths exam suggestion refer this page click here . If you have any other question write to us once again through the same form.
Question24. Find the sum of first hundred even positive integers divisible by 5 by Debanik Bhattacharjee answered on Jan 20, 2007 at 8:05 a.m. I.S.T. ( +5.5 G.M.T.)
Answer24. A.P. of reqd. numbers is 10, 20, 30, 40, .........
Here first term = a = 10, common difference = d = 10
100th term = a + 99 d
= 1000
Sum of 100 terms n/2(a + l) where l is last term
= 100/2(10 + 1000)
= 50500
Question25. Sum of 3terms of A.P. is 36 and product is 1296. Find A.P? by mohammed noufal answered on Jan 21, 2007 at 9:43 p.m. I.S.T.
Answer25. This type of question has been answered (see Q 22)
Question26. solve for x:(x-3)(x+9)(x-7)(x+5)=208 by kanika khanna Email kanikakhanna............@hotmail.com answered on 22.1.07 at 8:43 p.m.
Answer 26. These type of questions do not form part of the current class x syllabus. But still we are giving the solution.
Step 1 rearrange (x+9)(x-7)(x-3)(x+5) = 208
Step 2 (x^2 + 2x - 63)(x^2 + 2x - 15) = 208
Step 3 Put x^2 + 2x = y
Step 4 (y - 63)(y - 15) = 208
Step 5 y^2 - 78y + 945 = 208
Step 6 y^2 - 78y + 737 = 0
Step 6 (y - 7)(y - 71) = 0
Step 7 y = 7, y = 71
Step 8 putting value of y we get x^2 + 2x = 7 or x^2 + 2x = 71 now we solve for x
We are back after Republic day break and will start answering questions from today. Posted on Jan 30,2007 at 9:04 I.S.T.
Question27. find the weight of a lead pipe 3.5m iong if the external diameter of the pipe is 2.4m and the thickness is 2mm and 1 cubiic cm of lead weighs 11g by chaitanya kaul class x, Email add chaitanya................@gmail.com, answered on Jan 31, 2007 at 7:32 I.S.T.
Answer 27. Height = 3.5m = 3500 mm, outer radius (R) = 1.2 m = 1200 mm, Inner radius(r) = 1200 - 2 = 1198 mm
Volume = outer volume - inner volume = Pie h (R^2 - r^2) = (22/7) 3500 (1200^2 - 1198^2)
= 22*500(1200 - 1198)(1200 + 1198) using identity
= 22*500*2*2398
= 22*1000*2398
= 52756000 mm^3
= 52756 cm^3
Density = 11g/cm^3
Mass = Volume * density = 52756*11
= 580316 grams or 580.316 k.g.
Question28. An isosceles triangle ABC is inscribed in a circle. If AB = AC = 13cm and BC = 10cm,find the radius of the circle. by Francis Thomas class x/1x, Email add francis..............@yahoo.co.in answered on Feb 1, 2007 at 11:08 p.m. I.S.T.
Answer 28. Draw AD perpendicular to BC intersecting BC at D.
AD bisects BC , So BD = ½ BC = ½(10) = 5cm(since altitude to base in an isosceles triangle is also median)
So AD becomes perpendicular bisector of BC and hence passes through centre of circle let it be O
(since perpendicular bisector of a chord passes through centre of circle.)
In right triangle ADB, AB^2 = AD^2 + BD^2 (pythagoras theorem)
13^2 = AD^2 + 5^2
169 = AD^2 + 25
AD^2 = 169 - 25
AD = 12 cm.
Let radius OA = x cm.
OD = (12 - x) cm.
OB = OA = x cm.(radii of same circle)
In right triangle BDO, OB^2 = OD^2 + BD^2
x^2 = (12 - x)^2 + 5^2
x^2 = 144 + x^2 - 24x + 25
24x = 169
x = 169/24
Radius = 169/24 cm.
Question29. find the weight of a lead pipe 3.5m iong if the external diameter of the pipe is 2.4m and the thickness is 2mm and 1 cubiic cm of lead weighs 11g by chaitanya kaul class x, Email add chaitanya................@gmail.com, answered on Jan 31, 2007 at 7:32 I.S.T.
Answer29. Height = 3.5m = 3500 mm, outer radius (R) = 1.2 m = 1200 mm, Inner radius(r) = 1200 - 2 = 1198 mm
Volume = outer volume - inner volume = Pie h (R^2 - r^2) = (22/7) 3500 (1200^2 - 1198^2)
= 22*500(1200 - 1198)(1200 + 1198) using identity
= 22*500*2*2398
= 22*1000*2398
= 52756000 mm^3
= 52756 cm^3
Density = 11g/cm^3
Mass = Volume * density = 52756*11
= 580316 grams or 580.316 k.g.
Question30. An isosceles triangle ABC is inscribed in a circle. If AB = AC = 13cm and BC = 10cm,find the radius of the circle. by Francis Thomas class x/1x, Email add francis..............@yahoo.co.in answered on Feb 1, 2007 at 11:08 p.m. I.S.T.
Answer30. Draw AD perpendicular to BC intersecting BC at D.
AD bisects BC , So BD = ½ BC = ½(10) = 5cm(since altitude to base in an isosceles triangle is also median)
So AD becomes perpendicular bisector of BC and hence passes through centre of circle let it be O (since perpendicular
bisector of a chord passes through centre of circle.)
In right triangle ADB, AB^2 = AD^2 + BD^2 (pythagoras theorem)
13^2 = AD^2 + 5^2
169 = AD^2 + 25
AD^2 = 169 - 25
AD = 12 cm.
Let radius OA = x cm.
OD = (12 - x) cm.
OB = OA = x cm.(radii of same circle)
In right triangle BDO, OB^2 = OD^2 + BD^2
x^2 = (12 - x)^2 + 5^2
x^2 = 144 + x^2 - 24x + 25
24x = 169
x = 169/24
Radius = 169/24 cm.
Answer 26. These type of questions do not form part of the current class x syllabus. But still we are giving the solution.
Step 1 rearrange (x+9)(x-7)(x-3)(x+5) = 208
Step 2 (x^2 + 2x - 63)(x^2 + 2x - 15) = 208
Step 3 Put x^2 + 2x = y
Step 4 (y - 63)(y - 15) = 208
Step 5 y^2 - 78y + 945 = 208
Step 6 y^2 - 78y + 737 = 0
Step 6 (y - 7)(y - 71) = 0
Step 7 y = 7, y = 71
Step 8 putting value of y we get x^2 + 2x = 7 or x^2 + 2x = 71 now we solve for x
We are back after Republic day break and will start answering questions from today. Posted on Jan 30,2007 at 9:04 I.S.T.
Question27. find the weight of a lead pipe 3.5m iong if the external diameter of the pipe is 2.4m and the thickness is 2mm and 1 cubiic cm of lead weighs 11g by chaitanya kaul class x, Email add chaitanya................@gmail.com, answered on Jan 31, 2007 at 7:32 I.S.T.
Answer 27. Height = 3.5m = 3500 mm, outer radius (R) = 1.2 m = 1200 mm, Inner radius(r) = 1200 - 2 = 1198 mm
Volume = outer volume - inner volume = Pie h (R^2 - r^2) = (22/7) 3500 (1200^2 - 1198^2)
= 22*500(1200 - 1198)(1200 + 1198) using identity
= 22*500*2*2398
= 22*1000*2398
= 52756000 mm^3
= 52756 cm^3
Density = 11g/cm^3
Mass = Volume * density = 52756*11
= 580316 grams or 580.316 k.g.
Question28. An isosceles triangle ABC is inscribed in a circle. If AB = AC = 13cm and BC = 10cm,find the radius of the circle. by Francis Thomas class x/1x, Email add francis..............@yahoo.co.in answered on Feb 1, 2007 at 11:08 p.m. I.S.T.
Answer 28. Draw AD perpendicular to BC intersecting BC at D.
AD bisects BC , So BD = ½ BC = ½(10) = 5cm(since altitude to base in an isosceles triangle is also median)
So AD becomes perpendicular bisector of BC and hence passes through centre of circle let it be O
(since perpendicular bisector of a chord passes through centre of circle.)
In right triangle ADB, AB^2 = AD^2 + BD^2 (pythagoras theorem)
13^2 = AD^2 + 5^2
169 = AD^2 + 25
AD^2 = 169 - 25
AD = 12 cm.
Let radius OA = x cm.
OD = (12 - x) cm.
OB = OA = x cm.(radii of same circle)
In right triangle BDO, OB^2 = OD^2 + BD^2
x^2 = (12 - x)^2 + 5^2
x^2 = 144 + x^2 - 24x + 25
24x = 169
x = 169/24
Radius = 169/24 cm.
Question29. find the weight of a lead pipe 3.5m iong if the external diameter of the pipe is 2.4m and the thickness is 2mm and 1 cubiic cm of lead weighs 11g by chaitanya kaul class x, Email add chaitanya................@gmail.com, answered on Jan 31, 2007 at 7:32 I.S.T.
Answer29. Height = 3.5m = 3500 mm, outer radius (R) = 1.2 m = 1200 mm, Inner radius(r) = 1200 - 2 = 1198 mm
Volume = outer volume - inner volume = Pie h (R^2 - r^2) = (22/7) 3500 (1200^2 - 1198^2)
= 22*500(1200 - 1198)(1200 + 1198) using identity
= 22*500*2*2398
= 22*1000*2398
= 52756000 mm^3
= 52756 cm^3
Density = 11g/cm^3
Mass = Volume * density = 52756*11
= 580316 grams or 580.316 k.g.
Question30. An isosceles triangle ABC is inscribed in a circle. If AB = AC = 13cm and BC = 10cm,find the radius of the circle. by Francis Thomas class x/1x, Email add francis..............@yahoo.co.in answered on Feb 1, 2007 at 11:08 p.m. I.S.T.
Answer30. Draw AD perpendicular to BC intersecting BC at D.
AD bisects BC , So BD = ½ BC = ½(10) = 5cm(since altitude to base in an isosceles triangle is also median)
So AD becomes perpendicular bisector of BC and hence passes through centre of circle let it be O (since perpendicular
bisector of a chord passes through centre of circle.)
In right triangle ADB, AB^2 = AD^2 + BD^2 (pythagoras theorem)
13^2 = AD^2 + 5^2
169 = AD^2 + 25
AD^2 = 169 - 25
AD = 12 cm.
Let radius OA = x cm.
OD = (12 - x) cm.
OB = OA = x cm.(radii of same circle)
In right triangle BDO, OB^2 = OD^2 + BD^2
x^2 = (12 - x)^2 + 5^2
x^2 = 144 + x^2 - 24x + 25
24x = 169
x = 169/24
Radius = 169/24 cm.
