Homework Help > CBSE Class 10 > Mathematics
Ask questions which are clear, concise and easy to understand.
the sum of intiger and their reciprocal is 145/12. Find the intiger ?

Answers:

Solution:
Let the integer = x
Then reciprocal = 1/x
According to Question
x + 1/x = 145/12
=> (x^{2}+1)/x = 145/12
=> 12x^{2} + 12 = 145x
=> 12x^{2 } 145x + 12 = 0
=> 12x^{2} 144x  x + 12 = 0
=> 12x(x  12)  1(x  12) = 0
=> (x  12)(12x  1) = 0
=> x = 12 and x = 1/12
as X is integer x can't be 1/12
Therefore the required integer is 12
Thanks (0) 
Let the integer be x.
x + 1/x = 145/12
12x^{2}  145x + 12 = 0
12x^{2}  144x  x + 12 = 0
12x(x  12)  1(x  12) = 0
(x  12)(12x  1) = 0
x = 12 and x = 1/12
Therefore the required integer is 12.
Thanks (0)
how to prove the sum of first terms of ap is 42 10th term to its 30th terms is 1:3 find the first term

i think questions is : The sum of first six terms of an ap is 42 .the ratio of its 10th term is and its 30th term is 1:3. calculate the first and thirteenth term
 Add Comment

Answers:

Ans.
Let first term of AP is a and comman difference is d.
as we know
${a}_{n}=a+(n1)d\phantom{\rule{0ex}{0ex}}where{a}_{n}is{n}^{th}term$
=> ${a}_{10}=a+(101)d$
=> ${a}_{10}=a+9d$
And ${a}_{30}=a+(301)d$
=> ${a}_{30}=a+29d$
According to Question
=> $\frac{a+9d}{a+29d}=\frac{1}{3}$
=> 3a + 27d = a + 29d
=> 2a = 2d
=> a = d ............. (1)
Sum of first 6 terms = 42
We know
${S}_{n}=\frac{n}{2}\times \left(2a+(n1)d\right)$
=>${S}_{6}=\frac{6}{2}\times \left(2a+(61)d\right)$
=> $42=3\left(2a+5a\right)$
=> $42=21a$
=> a = 2
First term is 2. Common difference is 2.
=> ${a}_{13}=2+(131)2$
= > ${a}_{13}=2+24$
=> ${a}_{13}=26$
Thanks (0)
Which term of AP 121,117,113.................is its first negative term

Answers:

a = 121
d = 4
let a_{n}= 0
then
a_{n }= a + (n1)d
=> 0 = 121 + (n1)(4)
=> 0 = 121  4n + 4
=> 4n = 125
=> n = 125/4
=> n = 31.25
as n must be natural number then n= 32
32^{nd} term is First negative number.
a_{32} = 121 + (31)4
= 121  124 = 3
Thanks (0)
AB is a straight road leading to C, the foot of a tower, A being at a distance of 120m from C and B being 75m nearer. If the angle of elevation of the tower at B is the double of the angle of elevation of the tower at A, find the height of the tower.

Answers:

Just consider the image for Reference.
Gievn : AB is straigt Road and C is the foot of tower. Let DC is tower of height h, and AC is 120cm and BC is 45 cm as it is 75cm nearer than A.
Solution : In Triangle BCD
$\mathrm{tan}60\xb0=\frac{h}{45}$
=> $h=45\mathrm{tan}2x\phantom{\rule{0ex}{0ex}}=h=45\frac{2\mathrm{tan}x}{1{\mathrm{tan}}^{2}x}..............\left(1\right)\mathrm{tan}2x=\frac{2\mathrm{tan}x}{1{\mathrm{tan}}^{2}x}$
In Triangle, ACD
$\mathrm{tan}x=\frac{h}{120}$
=> $h=120\mathrm{tan}x....................\left(2\right)$
From (1) and (2)
$120\mathrm{tan}x=\frac{45\times 2\mathrm{tan}x}{1{\mathrm{tan}}^{2}x}$
=> $1{\mathrm{tan}}^{2}x=\frac{3}{4}$
=> ${\mathrm{tan}}^{2}x=\frac{1}{4}$
=> $\mathrm{tan}x=\frac{1}{2}$
Put value of tan x in (2)
we get h = 60
so height of tower is 60cm
Thanks (0) 
Are you sure this is the question? Please check it once.
Thanks (0)
Converse of Pythagoras theorem

Answers:

If the square of one side of triangle is equal to the sum of square of other two sides,then the triangle is right triangle.
Thanks (1)
Why volume of cone is 1/3 of cylinder volume

Answers:

Because cone is 1/3 part of a cylinder
Thanks (0)
Why volume of cone is1/3 of cylinder volume

Answers:
Find the probability that a non leap year chosen at random has
 52 Sundays
 53 Sundays

Answers:

2) A nonleap year has 365 days
A year has 52 weeks. Hence there will be 52 Sundays for sure.
52 weeks = 52 x 7 = 364 days .
365– 364 = 1day extra.
In a nonleap year there will be 52 Sundays and 1day will be left.
This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday,friday,Saturday, Sunday.
Of these total 7 outcomes, the favourable outcomes are 1.
Hence the probability of getting 53 sundays = 1 / 7.
1)
∴ probability of getting 52 sundays = 1  1/ 7 = 6 / 7.Thanks (2)
The vertices of a triangle are A(1,3), B(1,1), C(5,1). Find the length of the median through the vertex C.

Answers:

Let the points A(1, 3), B(1, 1) and C(5, 1) be the vertices of a triangle ABC.
We know median is the line segment joining the mid point of line to opposite vertex.
Let D be the mid point of AB. Then the coordinate of D
$\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1+}{y}_{2}}{2}\right)\phantom{\rule{0ex}{0ex}}\left(\frac{1+1}{2},\frac{31}{2}\right)\phantom{\rule{0ex}{0ex}}\left(0,1\right)\phantom{\rule{0ex}{0ex}}$
Coorditane of D( 0,1)
Lenght of Median CD=
Using distance formula
$\sqrt{{\left({x}_{2}{x}_{1}\right)}^{2}+{\left({y}_{2}{y}_{1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\sqrt{{\left(50\right)}^{2}+(11{)}^{2}}\phantom{\rule{0ex}{0ex}}\sqrt{25+0}\phantom{\rule{0ex}{0ex}}\sqrt{25}\phantom{\rule{0ex}{0ex}}5$
Thanks (1)
A train covers a distance of 90kms at a uniform speed. It would have taken 90 minutes less if the speed had been 15km/hr more. Calculate the original duration of the journey.

Answers:

The Questions is it would have taken 30 Minutes less instead of 90 Minutes
Solution:
Let “x” be the usual speed of the train
Let T_{1} be the time taken to cover the distance 90 km in the speed x km/hr
Let T_{2 }be the time taken to cover the distance 90 km in the speed (x + 15) km/hr
Time = Distance/Speed
T_{1} = $\frac{90}{x}$
T_{2} = $\frac{90}{x+15}$
By using the given condition
T_{1}  T_{2} = $\frac{30}{60}\phantom{\rule{0ex}{0ex}}$
$\frac{90}{x}$ $\frac{90}{x+15}$ = $\frac{1}{2}$
Taking 90 commonly from two fractions
$\left(90\left(\frac{1}{x}\frac{1}{x+15}\right)\right)=\frac{1}{2}$
$\frac{x+15x}{\left(x\right)\left(x+15\right)}=\frac{1}{2*90}$
$\frac{15}{x\left(x+15\right)}=\frac{1}{180}$
${x}^{2}+15x=2700$
${x}^{2}+15x2700=0\phantom{\rule{0ex}{0ex}}{x}^{2}+60x45x2700=0\phantom{\rule{0ex}{0ex}}x(x+60)45(x+60)=0\phantom{\rule{0ex}{0ex}}(x+60)(x45)=0\phantom{\rule{0ex}{0ex}}x+60=0orx45=0\phantom{\rule{0ex}{0ex}}x=45,60\phantom{\rule{0ex}{0ex}}$
Speed can't be negative so,
Speed is 45km/h
Thanks (1)
3x^{2}23x110=0 by using quadratic equation

Answers:

3x^{2}23x110=0
3x^{2}33x+10x110=0
3x(x11)+10(x11)=0
(x11)(3x+10)=0
x11=0, 3x+10=0
x = 11, 10/3
Thanks (0)
For what value of K are the roots of the quadratic equation kx(x2)+6=0

Answers:

For Equal roots D = 0
kx(x2) + 6 = 0
kx^{2} 2kx + 6 = 0
comparing with ax^{2}+ bx + c = 0, we get
a = k, b = 2k , c = 6
D = b^{2}  4ac
Put value of a b c
(2k)^{2}  4*6*k = 0
4k^{2}  24k = 0
4k(k6) =0
4k = 0 or k6=0
k = 6
Thanks (0)
Two tangents are drawn from an external point P to the circle such that angle oba 10 then find the value of bpa

Answers:
The height of a cone is 24 cm.A small cone is cut off at the top by a plane parallel to the base.If the volume of the remaining part be 7/8of the given cone, at what height above the base is the section made?

Let ABC be a cone of radius R, height 24cm. let this cone be cut by a place A`B` parallel to AB. O` is the center of base of the cut out cone A`B`C`. let h be the height, r be the redius of cone A`B`C`. Cleary ΔA`B`C` ~ Δ AOC
 Add Comment

Answers:

Ans.
So h/24 = r/R
h = 24r/R ..... (1)
Volume of Cone ABC = 24πR^{2}/3 = 8πR^{2 }  (2)
Remaining Portion is Know as Frustum as its volume is 7/8 of Cone ABC.
So Volume of Smaller Cone A`B`C` = 1/8 of Cone ABC
Volume of Cone A`B`C` = π r^{2}h / 3
from (1),
π r^{2}24r/3R = 8π r^{3}/R  (3)
Accroding to questions
8π r^{3}/R = 8πR^{2}/8
8π r^{3}/R = πR^{2}
r^{3}/R^{3} = 1/8
r/R = 1/2
put this value in (1)
We get h = 24*1/2 = 12cm
So cone is cut from the height 2412 = 12 cm
Thanks (0)
The diffrence of the ages of sohrab &his father is 30 years .if the diffrence of the square of their ages is 1560 then find their ages

Answers:

let age of Sohrab = x
then age of his father = x + 30
Given : (x+30)^{2}  (x)^{2} = 1560
x^{2} + 900 + 60x  x^{2} = 1560
=> 60x = 660
=> x = 11
Age of Sourab = 11
age of his father = 41
Thanks (1)
Find the diffrence between the points (a,b) (a,b)

Answers:

$Differencebetween(a,b)and(a,b)=\sqrt{{\left(aa\right)}^{2}+{\left(bb\right)}^{2}}$
$=\sqrt{{\left(2a\right)}^{2}+{\left(2b\right)}^{2}}=\sqrt{4{a}^{2}+4{b}^{2}}$
$=\sqrt{4({a}^{2}+{b}^{2})}$
$=2\sqrt{{a}^{2}+{b}^{2}}$
Thanks (0)
The distance between the points (2,2)&(1,x) is 5 then one of the value of x is

Answers:

Given : A (2,2) and B(1,x) and Distance b/w these two is 5
according to distance formula
(x_{2}x_{1})^{2} + (y_{2}y_{1})^{2} = d^{2}
=> (12)^{2} + (x+2)^{2} = 25
=> 9 + x^{2} + 4 +4x = 25
=> x^{2} + 4x  12 = 0
=> x^{2} + 6x 2x  12 = 0
=> x (x+6) 2(x+6) = 0
=> (x+6) (x2) = 0
either x + 6 = 0 or x2=0
so x = 6, 2
Thanks (0)
Show that the points (7,3) (3,0) (0,4) (4,1)are vertices of rhombus

Answers:

Let A(7,3) , B(3,0) , C(0,4) and D(4,1) represent the points.
To prove that these points are vertices of a rhombus we have to show that all sides are equal. We may calculate it using distance formula.
AB= $\sqrt{(37{)}^{2}+(03{)}^{2}}=\sqrt{16+9}=\sqrt{25}=5units$
BC=$\sqrt{(03{)}^{2}+(40{)}^{2}}=\sqrt{9+16}=\sqrt{25}=5units$
CD=$\sqrt{(40{)}^{2}+(1+4{)}^{2}}=\sqrt{16+9}=\sqrt{25}=5units$
DA=$\sqrt{(74{)}^{2}+(3+1{)}^{2}}=\sqrt{9+16}=\sqrt{25}=5units$
The diagonals are not equal.
AC= $\sqrt{(70{)}^{2}+(3+4{)}^{2}}=\sqrt{49+49}=\sqrt{98}=7\sqrt{2}units\phantom{\rule{0ex}{0ex}}$
BD=$\sqrt{(43{)}^{2}+(10{)}^{2}}=\sqrt{1+1}=\sqrt{2}units$
This shows that these points are vertices of rhombus.
Thanks (1)
Find the sum of all 3 digit no. Which leave a remainder 3 when divided by 5 .

Answers:

First 3 Digit Number that leaves remainder 3 when divided by 5 is 103.
and last will be 998
A.P. = 103, 108, 113, ..................., 998
first term a = 103
Common Difference d = 5
nth term = 998
n = ?
as we know
nth term = a + ( n1)d
998 = 103 + ( n1)5
895/ 5 = n 1
179 = n 1
n = 180
now find sum of 180 terms of AP
S_{180 = }180/ 2 ( 103 + 998 )
=> S_{180 = }90 * 1001
=> S_{180 = }900090
Thanks (0)
If m^{th} term of an A.P. is 1/n and n^{th }term is 1/m .show that its (mn)^{th} term is 1

Answers:

Let a and d are first term and common difference of AP
Thanks (0)
Find k so that 15,k,1are in A.P. series

Answers:

As they are in AP
so 2b = a + c
2k = 15  1
2k = 14
k = 7
Thanks (1)
Solve 4x^{2}2(a^{2}+b^{2})x+a^{2}b^{2}=0 find the value of x

Answers:

a= 4
b= 2(a^{2}+b^{2})
c= a^{2}b^{2}
Using Discriminant formula,
D= b^{2}4ac
[2(a^{2}+b^{2})]^{2}4 x4xa^{2}b^{2}
4(a^{4}+b^{4}+2a^{2}b^{2})16a^{2}b^{2}
(2a^{2}2b^{2})^{2}
x= $(b\pm \sqrt{D})/2a$
x= [2(a^{2}+b^{2})+2(a^{2}b^{2})]/8 = 4a^{2}/8 = a^{2}/2
x= [2(a^{2}+b^{2})2(a^{2}b^{2})]/8 = 4b^{2}/8 = b^{2}/2
Thanks (1)
If ax^{2}+bx+c=0has equal roots then proove that c=b^{2}/4a

Answers:

For equal roots, discriminant D=0
i.e. b^{2}4ac=0
b^{2}=4ac
So, c= b^{2}/4a
Thanks (0)
If5 is root of quadratic equation 2x^2+2px15=0and the quadratic equation p(x^2+x)+k=0has equal roots .find the value of k

Thanks for the answer but can u answers some more questions
 Add Comment

Answers:

as 5 is root of equation 2x^{2}+2px15=0
then it ll satisfy this equation, Put x = 1
2(5)^{2} + 2p(5)  15 = 0
=> 50  10p 15 = 0
=> 35 = 10p
p = 3.5
And it is given that eq. p(x^{2}+x) + k = 0 have equal roots.
px^{2}+px + k = 0
3.5x^{2}+3.5x + k = 0
a = 3.5, b = 3.5, c = k
so d = 0
(3.5)^{2}  4 (3.5)k = 0
=> 14 k = 12.25
k = 3.0625
Thanks (2)
The radius of the cirular ends of a bucket of height 15cm are 14cm and rcm (r<14). If the volume of bucket is 5390cm*3 then the value of r.(π=22÷7)

Answers:

V = $\pi ({{r}_{2}}^{2}{{r}_{1}}^{2})h\phantom{\rule{0ex}{0ex}}$
5390 = 22/7(196r^{2})15
196r^{2} = 114.33
r^{2}= 196114.33
r^{2}=81.67
r = 9.03
Thanks (0)
If P(2,4) is equidistant from Q(7,0) and R (X,9), find the value of X. Also find the distance PQ.

Answers:

Given: P(2,4) is equidistnat from Q(7,0) and R(X,9).
then PQ = PR
Using Distance Formula : = $\sqrt{({x}_{2}{x}_{1}{)}^{2}+({y}_{2}{y}_{1}{)}^{2}}$
PQ = $\sqrt{25+16}\phantom{\rule{0ex}{0ex}}$ = $\sqrt{41}\phantom{\rule{0ex}{0ex}}$
PR = $\sqrt{(x2{)}^{2}+(5{)}^{2}}$
PQ = PR
Sqauring Both Side, We get
x^{2} + 4 4x + 25 = 41
=> x^{2 } 4x 12 = 0
=> x^{2 } 6x +2x 12 = 0
=> x(x6) +2(x6) = 0
=> (x6) (x+2) = 0
=> x = 2 , 6
Thanks (0)
PQ perpendicular OQ. The tangent to the circle with centre O at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.

Answers:

Ans.
Given: AP and AQ are tangents to the circle with centre O, AP ⊥ AQ and AP = AQ = 5 cm
we know that radius of a circle is perpendicular to the tangent at the point of contact
⇒ OP ⊥ AP and OQ ⊥ AQ
Also sum of all angles of a quadrilateral is 360° ⇒∠O + ∠P + ∠A + ∠Q = 360°
⇒∠O + 90° + 90° + 90° = 360°
⇒∠O = 360° – 270° = 90°
Thus ∠O = ∠P = ∠A = ∠Q = 90°
⇒ OPAQ is a rectangle but since adjacent sides of OPAQ i.e. AP and AQ are equal.
Thus OPAQ is a square
radius = OP = OQ = AP = AQ = 5 cm
Since diagonals of a square are perpendicular bisector of each other.
Hence PQ and OA are perpendicular bisectors of each other
Thanks (0)
Solve quadratic equation by factorization
xa/ xb +x b/ xa = a/b+ b/a

Answers:

$\frac{xa}{xb}+\frac{xb}{xa}=\frac{a}{b}+\frac{b}{a}$
$let\frac{xa}{xb}=y\Rightarrow \frac{xb}{xa}=\frac{1}{y}$
$\Rightarrow y+\frac{1}{y}=\frac{{a}^{2}+{b}^{2}}{ab}$
$\Rightarrow \frac{{y}^{2}+1}{y}=\frac{{a}^{2}+{b}^{2}}{ab}\Rightarrow ab{y}^{2}{a}^{2}y{b}^{2}yab=0$
$\Rightarrow ay\left(bya\right)b\left(bya\right)=0\Rightarrow \left(bya\right)\left(ayb\right)=0$
$\Rightarrow y=\frac{a}{b}ory=\frac{b}{a}$
$case1$
$ify=\frac{a}{b}\Rightarrow \frac{xa}{xb}=\frac{a}{b}\Rightarrow bxba=axba\Rightarrow bx=ax$
$bxax=0\Rightarrow x(ba)=0\Rightarrow x=0$
$Case2$
$ify=\frac{b}{a}\Rightarrow \frac{xa}{xb}=\frac{b}{a}\Rightarrow ax{a}^{2}=bx{b}^{2}$
$\Rightarrow axbx={a}^{2}{b}^{2}\Rightarrow x(ab)=(ab)(a+b)$
$x=\left(a+b\right)$
$Hencex=0or\left(a+b\right)$
Thanks (0)
A 25 FOOT LADDER IS PLACED AGAINST A VERTICAL WALL OF A BUILDING.THE FOOT OF THE LADDER IS 17 FEET FROM THE BASE OF THE BUILDING,IF THE TOP OF THE LADDER SLIPS 4 FEET,THEN THE FOOT OF THE LADDER WILL SLIP.......

Answers:

I think it is not 17 feet it is 7 feet.
Fiirst we need to find at what height ladder is placed.
Let BC is ladder and AC is Building with foot at C.
Now as given AB = 25
BC = 7
Using Pythagorous Theroem
(AB^{2}) = (BC)^{2} + (AC)^{2}
=> 625 = 49 + (AC)^{2}
=> (AC)^{2} = 576
=> AC = 24
It slips down 4 feet it means its top is at 20 feet height
So Now
AC = 20
AB = 25
then again Using theorem
(BC)^{2} = 625  400
(BC)^{2} = 225
BC = 15
it mean it slips 8 feet.
Thanks (0)
Sir and Madam, In our maths book, we have studied how to construct tangents from any external point without using the center of circle.I have discovered a yet new mode for the same , that is, how to construct tangents by an alternative method...
so, Can I construct the tangents by this method in Exam , If such question is asked ?

Answers: