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the sum of intiger and their reciprocal is 145/12. Find the intiger ?

Posted by Pankaj Asery (Jan 18, 2017) (Question ID: 1452)

• Solution:

Let the integer = x

Then reciprocal = 1/x

According to Question

x + 1/x = 145/12

=> (x2+1)/x = 145/12

=> 12x2 + 12 = 145x

=> 12x- 145x + 12 = 0

=> 12x2- 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

Answered by Naveen Sharma (Jan 19, 2017)
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• Let the integer be x.

x + 1/x = 145/12

12x2 - 145x + 12 = 0

12x2 - 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.

Answered by Rashmi Bajpayee (Jan 18, 2017)
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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

Posted by rohit singh (Jan 18, 2017) (Question ID: 1444)

• 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

Posted by Naveen Sharma (Jan 18, 2017)
• Ans.

Let first term of AP is a and comman difference is d.

as we know

=>

=>

And

=>

According to Question

=>

=> 3a + 27d = a + 29d

=> 2a = 2d

=> a = d       ............. (1)

Sum of first 6 terms = 42

We know

=>

=>

=>

=> a = 2

First term is 2. Common difference is 2.

=>

= >

=>

Answered by Naveen Sharma (Jan 18, 2017)
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Which term of AP 121,117,113.................is its first negative term

Posted by Akash A.S (Jan 16, 2017) (Question ID: 1419)

• a = 121

d = -4

let an= 0

then

a= a + (n-1)d

=> 0 = 121 + (n-1)(-4)

=> 0 = 121 - 4n + 4

=> 4n = 125

=> n = 125/4

=> n = 31.25

as n must be natural number then n= 32

32nd term is First negative number.

a32 = 121 + (31)-4

= 121 - 124 = -3

Answered by Naveen Sharma (Jan 16, 2017)
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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.

Posted by Shruti Sharma (Jan 16, 2017) (Question ID: 1415)

• 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

=>

In Triangle, ACD

=>

From (1) and (2)

=>

=>

=>

Put value of tan x in (2)

we get h = 60

so height of tower is 60cm

Answered by Naveen Sharma (Jan 17, 2017)
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• Are you sure this is the question? Please check it once.

Answered by Shweta Gulati (Jan 17, 2017)
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Converse of Pythagoras theorem

Posted by Israr Khan (Jan 16, 2017) (Question ID: 1394)

• 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

Posted by raman SINGH (Jan 15, 2017) (Question ID: 1391)

• Because cone is 1/3 part of a cylinder

Answered by himani sah (Jan 15, 2017)
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Why volume of cone is1/3 of cylinder volume

Posted by raman SINGH (Jan 15, 2017) (Question ID: 1390)

Find the probability that a non leap year chosen  at random has

1. 52 Sundays
2. 53 Sundays
Posted by Rohit R Mahatungade (Jan 14, 2017) (Question ID: 1367)

• 2) A non-leap 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 non-leap 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.

Answered by Naveen Sharma (Jan 14, 2017)
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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.

Posted by Arpit Singla (Jan 14, 2017) (Question ID: 1361)

• 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

Coorditane of D( 0,1)

Lenght of Median CD=

Using distance formula

Answered by Naveen Sharma (Jan 14, 2017)
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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.

Posted by Arpit Singla (Jan 14, 2017) (Question ID: 1360)

• 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 T1 be the time taken to cover the distance 90 km in the speed x km/hr

Let Tbe the time taken to cover the distance 90 km in the speed (x + 15) km/hr

Time = Distance/Speed

T1 = $\frac{90}{x}$

T2 = $\frac{90}{x+15}$

By using the given condition

T1 - T2 = $\frac{30}{60}\phantom{\rule{0ex}{0ex}}$

$\frac{90}{x}$- $\frac{90}{x+15}$ = $\frac{1}{2}$

Taking 90 commonly from two fractions

Speed can't be negative so,

Speed is 45km/h

Answered by Naveen Sharma (Jan 14, 2017)
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Posted by Yash Partap Singh (Jan 12, 2017) (Question ID: 1302)

• 3x2-23x-110=0

3x2-33x+10x-110=0

3x(x-11)+10(x-11)=0

(x-11)(3x+10)=0

x-11=0, 3x+10=0

x = 11, -10/3

Answered by Naveen Sharma (Jan 12, 2017)
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For what value of  K are the  roots of the  quadratic equation kx(x-2)+6=0

Posted by abhishek yadav (Jan 11, 2017) (Question ID: 1273)

• For Equal roots D = 0

kx(x-2) + 6 = 0

kx2 -2kx + 6 = 0

comparing with ax2+ bx + c = 0, we get

a = k,  b = -2k , c = 6

D = b2 - 4ac

Put value of a b c

(-2k)2 - 4*6*k = 0

4k2 - 24k = 0

4k(k-6) =0

4k = 0 or k-6=0

k = 6

Answered by Naveen Sharma (Jan 11, 2017)
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Two tangents are drawn from an external point P to the circle such that angle oba 10 then find the value of bpa

Posted by Shashant Ahuja (Jan 10, 2017) (Question ID: 1267)

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?

Posted by Rajeev Kumar (Jan 09, 2017) (Question ID: 1230)

• Let ABC be a cone of radius R, height 24cm. let this cone be cut by a place AB parallel to AB. O is the center of base of the cut out cone ABC. let h be the height, r be the redius of cone ABC. Cleary ΔABC ~ Δ AOC

Posted by Naveen Sharma (Jan 12, 2017)
• Ans.

So h/24 = r/R

h = 24r/R ..... (1)

Volume of Cone ABC =  24πR2/3  = 8πR2      ----- (2)

Remaining Portion is Know as Frustum as its volume is 7/8 of Cone ABC.

So Volume of Smaller Cone ABC = 1/8 of Cone ABC

Volume of Cone ABC  =  π r2h / 3

from (1),

π r224r/3R  =  8π r3/R ------ (3)

Accroding to questions

8π r3/R  = 8πR2/8

8π r3/R = πR2

r3/R3 = 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 24-12 = 12 cm

Answered by Naveen Sharma (Jan 12, 2017)
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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

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1218)

• let age of Sohrab = x

then age of his father = x + 30

Given : (x+30)2 - (x)2 = 1560

x2 + 900 + 60x - x2 = 1560

=> 60x = 660

=> x = 11

Age of Sourab = 11

age of his father = 41

Answered by Naveen Sharma (Jan 10, 2017)
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Find the diffrence between the points (a,b) (-a,-b)

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1217)

• $=\sqrt{{\left(-2a\right)}^{2}+{\left(-2b\right)}^{2}}=\sqrt{4{a}^{2}+4{b}^{2}}$

$=\sqrt{4\left({a}^{2}+{b}^{2}\right)}$

$=2\sqrt{{a}^{2}+{b}^{2}}$

Answered by Neeraj Sharma (Jan 14, 2017)
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The distance between the points (2,-2)&(-1,x) is 5 then one of the value of x is

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1216)

• Given : A (2,-2) and B(-1,x) and Distance b/w these two is 5

according to distance formula

(x2-x1)2 + (y2-y1)2 = d2

=> (-1-2)2 + (x+2)2 = 25

=> 9 + x2 + 4 +4x = 25

=> x2 + 4x - 12 = 0

=> x2 + 6x -2x - 12 = 0

=> x (x+6) -2(x+6) = 0

=> (x+6) (x-2) = 0

either x + 6 = 0 or x-2=0

so x = -6, 2

Answered by Naveen Sharma (Jan 10, 2017)
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Show that the points (7,3) (3,0) (0,-4) (4,-1)are vertices of rhombus

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1215)

• 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=

BC=

CD=

DA=

The diagonals are not equal.

AC= $\sqrt{\left(7-0{\right)}^{2}+\left(3+4{\right)}^{2}}=\sqrt{49+49}=\sqrt{98}=7\sqrt{2}units\phantom{\rule{0ex}{0ex}}$

BD=

This shows that these points are vertices of rhombus.

Answered by Shweta Gulati (Jan 10, 2017)
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Find the sum of all 3 digit no. Which leave a remainder 3 when divided by 5 .

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1213)

• 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 + ( n-1)d

998 = 103 + ( n-1)5

895/ 5 = n -1

179 = n -1

n = 180

now find sum of 180 terms of AP

S180 = 180/ 2 ( 103 + 998 )

=> S180 = 90 * 1001

=> S180 = 900090

Answered by Naveen Sharma (Jan 09, 2017)
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If mth term of an A.P. is 1/n and nth term is 1/m .show that its (mn)th term is 1

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1212)

• Let a and d are first term and common difference of AP

Answered by Naveen Sharma (Jan 09, 2017)
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Find k so that  15,k,-1are in A.P. series

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1211)

• As they are in AP

so 2b = a + c

2k = 15 - 1

2k = 14

k = 7

Answered by Naveen Sharma (Jan 09, 2017)
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Solve 4x2-2(a2+b2)x+a2b2=0  find the value of x

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1210)

• a= 4

b= -2(a2+b2)

c= a2b2

Using Discriminant formula,

D= b2-4ac

[-2(a2+b2)]2-4 x4xa2b2

4(a4+b4+2a2b2)-16a2b2

(2a2-2b2)2

x=

x= [2(a2+b2)+2(a2-b2)]/8 = 4a2/8 = a2/2

x= [2(a2+b2)-2(a2-b2)]/8 = 4b2/8 = b2/2

Answered by Shweta Gulati (Jan 10, 2017)
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If ax2+bx+c=0has equal roots then proove that c=b2/4a

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1208)

• For equal roots,  discriminant D=0

i.e. b2-4ac=0

b2=4ac

So, c= b2/4a

Answered by Shweta Gulati (Jan 12, 2017)
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If-5 is root of quadratic equation 2x^2+2px-15=0and the quadratic equation   p(x^2+x)+k=0has equal roots .find the value of k

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1202)

• Thanks for the answer but can u  answers some more questions

Posted by Kanishk Goyal (Jan 09, 2017)
• as -5 is root of equation 2x2+2px-15=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(x2+x) + k = 0 have equal roots.

px2+px + k = 0

3.5x2+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

Answered by Naveen Sharma (Jan 09, 2017)
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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)

Posted by Himesh kaushik (Jan 07, 2017) (Question ID: 1136)

• V =

5390 = 22/7(196-r2)15

196-r2 = 114.33

r2= 196-114.33

r2=81.67

r = 9.03

Answered by Devanshu Agarwal (Jan 08, 2017)
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If P(2,4) is equidistant from Q(7,0) and R (X,9), find the value of X. Also find the distance PQ.

Posted by Himesh kaushik (Jan 07, 2017) (Question ID: 1134)

• Given: P(2,4) is equidistnat from Q(7,0) and R(X,9).

then PQ = PR

Using Distance Formula : =

PQ =  = $\sqrt{41}\phantom{\rule{0ex}{0ex}}$

PR =

PQ = PR

Sqauring Both Side, We get

x2 + 4 -4x + 25 = 41

=> x- 4x -12 = 0

=> x- 6x +2x -12 = 0

=> x(x-6) +2(x-6) = 0

=> (x-6) (x+2) = 0

=> x = -2 , 6

Answered by Naveen Sharma (Jan 12, 2017)
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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.

Posted by Himesh kaushik (Jan 07, 2017) (Question ID: 1133)

• 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

Answered by Naveen Sharma (Jan 12, 2017)
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x-a/ x-b +x -b/ x-a = a/b+ b/a

Posted by Sudhanshu Yadav (Jan 07, 2017) (Question ID: 1126)

• $\frac{x-a}{x-b}+\frac{x-b}{x-a}=\frac{a}{b}+\frac{b}{a}$

$⇒y+\frac{1}{y}=\frac{{a}^{2}+{b}^{2}}{ab}$

$⇒\frac{{y}^{2}+1}{y}=\frac{{a}^{2}+{b}^{2}}{ab}⇒ab{y}^{2}-{a}^{2}y-{b}^{2}y-ab=0$

$⇒ay\left(by-a\right)-b\left(by-a\right)=0⇒\left(by-a\right)\left(ay-b\right)=0$

$case-1$

$bx-ax=0⇒x\left(b-a\right)=0⇒x=0$

$Case-2$

$⇒ax-bx={a}^{2}-{b}^{2}⇒x\left(a-b\right)=\left(a-b\right)\left(a+b\right)$

$x=\left(a+b\right)$

Answered by Neeraj Sharma (Jan 14, 2017)
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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.......

Posted by Pawan Goswami (Jan 07, 2017) (Question ID: 1125)

• 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

(AB2) = (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.

Answered by Naveen Sharma (Jan 12, 2017)
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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 ?

Posted by Lakshmi kumari (Jan 06, 2017) (Question ID: 1115)