Extramarks-background-image
Q:

The distance of a point (2, 5) from X-axis is

A:

5 units.

Q:

If the distance between points (p, –7),(9, –7) is 15 units, then p is

A:

– 6 or 24.

Q:

If 7 is the mean of 5, 3, .5, 4.5, b, 8.5, 9.5, then the value of b is

A:

18

Q:

Which of the following is the HCF of two consecutive natural numbers?

A:

1

Q:

A:

sec+ tan.

Q:

The value of sin 15° =

A:

Q:

The greatest number which divides 258 and 323 leaving remainders 2 and 3 respectively is

A:

64

Q:

The coordinates of the centre of a circle Which is passing through (1, 2),(3, –4) and (5, –6) is

A:

(11, 2)

Q:

In the given triangle the value of sin A is

A:
5/13.
Q:

The lines y = 4x – 10 and y = 7x + 11 intersect at

A:

(-7, -38).

Q:

If the roots of quadratic equation 3x2 – 2x – k = 0, are real and equal, then value of k is _____.

A:
The given quadratic equation is: 3x 2 −2x−k=0 Since the roots of the given equation are real and equal. Therefore, D=0⇒ b 2 −4ac=0 ⇒ ( −2 ) 2 −4( 3 )( −k )=0 ⇒       4+12k=0 ⇒                    k=− 4 12 ⇒                    k=− 1 3 ∴ If the roots of quadratic equation 3x 2 −2x−k=0,are real and equal, then value of k is − 1 3 ¯ . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@5F53@
Q:

If x2 – 3x – 18 = 0, then the roots of quadratic equation are 6 and ____.

A:
The given quadratic equation is:               x 2 −3x−18=0     x 2 −6x+3x−18=0 x( x−6 )+3( x−6 )=0 ⇒    ( x−6 )( x+3 )=0 ⇒          x=6,−3 ∴ If x 2 –3x–18=0, then the roots of quadratic equation are 6 and −3 ¯ . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@EC75@
Q:

If the first term of an AP is ‘x’ and the common difference is ‘y’, then the nth term of the series is _________.

A:

If the first term and common difference of an AP be ‘a’ and ‘d’ respectively, then nth term of AP is:
Tn = a + (n – 1)d
Here, a = x and d = y
Then, for the given series, the nth term is:
Tn = x + (n – 1)y
Therefore, if the first term of an AP is ‘x’ and the common difference is ‘y’,

then the nth term of the series is x + (n – 1)y.

Q:

If the sides of two similar triangles are in the ratio 3:4, then the areas of these triangles are in the ratio_____.

A:
By area ratio thereom, Area of first triangle Area of second triangle = ( side of first triangle ) 2 ( side of second triangle ) 2     = ( 3 4 ) 2     = 9 16 Therefore, the required ratio is 9:16.MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@DBB7@
Q:

The radii of the ends of a frustum of a cone 35 cm high are 7 cm and 4 cm. The volume of the frustum is______.

A:
The volume of frustum of a cone= 1 3 πh( r 1 2 + r 2 2 + r 1 r 2 )     = 1 3 × 22 7 ×35( 7 2 + 4 2 +7×4 )     = 1 3 ×22×5( 49+16+28 )    = 1 3 ×110×( 93 )     =3410 Therefore, the volume of the frustum of the cone is 3410 cm 3 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@F517@
Q:

A number is chosen at random from the numbers –8, –7, –6, –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5, 6, 7, and 8. If the selected number is a positive multiple of 3, then the probability of the selected number is ________.

A:
Total number of the given numbers = 17 The positive multiples of 3 are 3 and 6. The probability of selected number= Number of multiples of 3 Total numbers = 2 17 Therefore, the required probabilty is 2 17 .
Q:

what is the cojugate number of 2+5?

A:
The conjugate number of 2+5 is 2 -5.
Q:

Solve the equation 3x2 - 2x - 1 = 0.

A:
3x2 - 2x - 1 = 0
3x2 - 3x + x - 1 = 0
3x(x - 1) + (x - 1) = 0
(x - 1)(3x + 1) = 0
x = 1 and x = -1/3
Q:

The sum of three numbers of an AP is 15. Find its first term.

A:

Let three numbers in an AP are a –d, a, a + d.

Therefore, a – d + a + a + d = 15
3a=15

a = 5

Q:

What is the necessary condition for two triangles to be similar?

A:

Two triangles are similar if:
1) their corresponding angles are equal or
2) their corresponding sides are in same ratio.

Q:

In figure, AQ and AR are tangents from A to the circle with centre O. P is a point on the circle. Prove that AB + BP = AC + CP.

A:

Q:

In figure, DABC is circumscribing circle. Find the length of BC.

A:
Since the tangents from an external point to a circle

are equal in length, therefore

AR=AQ=3cm, CP=CQ=5 cm

BR=BA-AR=10-3=7 cm, BP=BR=7 cm

Now, BC=BP+CP=7+5=12 cm

Thus, the length of BC= 12 cm.

Q:

The sum of two numbers is 36 and their difference is 14. Find the numbers.

A:
Q:

Find the values of k for which the system of linear equations has unique solution.
2x + 3y = 7, kx + 9y = 28.

A:
Q:

Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.

A:
Since the given points are collinear, the area of the triangle formed by them must be 0,
i.e.,
Q:

In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)?

A:
Let (– 4, 6) divides AB internally in the ratio k : 1.
Using the section formula, we get


So, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.
Q:

Prove that if a and b are odd positive integers then a2+b2 is even integer.

A:
Q:

ABC is right angled at B and A = C. Is cosA = cosC?

A:

Q:

A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.

A:
Q:

The side of a square is 10 cm. Find the area of circumscribed and inscribed circles.

A:
Q:

In triangle ABC, C = 3B = 2(A + B). Find the three angles.

A:

Q:

In a simultaneous throw of a pair of dice, find the probability of getting 8 as a sum.

A:
Elementary events associated to the random experiment of throwing two dice.

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Total events = 6 x 6 = 36

Favourable event is when the sum is 8.

(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)

Favourable events = 5

Hence, probability of getting 8 as a sum = 5/36

Q:

A bag contains 48 red balls and some blue balls. If the probability of drawing a blue ball is half that of a red ball, determine the number of blue balls in the bag.

A:

Let number of blue balls in the bag=x

Number of red balls in the bag=48

Total balls in the bag=(x+48)

P(red ball)=48/( x+48)

P(blue ball)=x/( x+48)

According to given condition,

P(blue ball)=1/2 {P(red ball)}

x/( x+48)=1/2{48/( x+48)}

x=24.

Thus, the number of blue balls in the bag is 24.

Q:

Express 32760 as the product of its prime factors.

A:
Using the factor tree for prime factorisation, we have

So, we have 32760 = 2 2 2 3 3 5 7 13

Q:

Divide -x3 + 3x2 - 3x + 5 by -x2 + x - 1 and verify the division algorithm.

A:
Dividend = -x3 + 3x2 - 3x + 5
Divisor = -x2 + x - 1

So quotient = x - 2, remainder = 3

Divisor Quotient + Remainder

= (-x2 + x - 1) (x - 2) + 3

= -x3 + 3x2 - 3x + 5 = Dividend

Q:

A:
Q:

If the sum of first n terms of an AP is Sn=4n-n2, Find AP and nth term.

A:
Q:

How many terms of the AP 3, 5, 7, 9, … must be added to get the sum 120?

A:

Q:

ABC is a right triangle right-angled at C. Let BC = a, CA = b, AB = c and let P be the length of perpendicular form C on AB prove that

(i) cp = ab
(ii)

A:

(i) Let CD AB. Then, CD = p

Area of ABC = (AB CD)/2 = cp/2 ...(1)

Also,
Area of ABC = (BC AC)/2 = ab/2 ...(ii)


From (i) and (ii) we have,
cp = ab.

(ii) Since ABC is a right triangle right–angled at C

Q:

A:

Q:

If 7 cosecA-3cotA = 7, prove that 7cotA - 3cosecA = 3.

A:
7cosecA-3cotA=7
 7cosecA-7=3cotA
 7(cosecA-1)=3cotA
Multiplying by (cosecA+1) both sides, we get-
7(cosecA-1) (cosecA+1)=3cotA (cosecA+1)
7(cosec2A-1)=3CotA (cosecA+1)
7cot2A=3 cotA (cosecA+1)
7cotA= 3(cosecA+1)
7cotA-3 cosecA=3
Q:

Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle.

A:
Step 1: Taking any point O as centre, draw a circle of 6 cm radius. Locate a point P, 10 cm away from O. Join OP.

Step 2: Bisect OP. Let M be the mid-point of PO.

Step 3: Taking M as centre and MO as radius, draw a circle.

Step 4: Let this circle intersect the previous circle at point Q and R.

Step 5: Join PQ and PR. PQ and PR are the required tangents.

Q:

If the median of the following frequency distribution is 46. Find the missing frequency.

Variable 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Total

F: 12 30 P 65 q 25 18 229

A:

Q:

If the roots of the equation
(b-c) x2 +(c-a) x+ (a-b)=0 are equal, then prove that 2b= a+c.

A:
Q:

Some students arranged a picnic. The budget for food was Rs.240. Because four students of the group failed to go, the cost of food to each student got increased by Rs 5. How many students went for the picnic?

A:

Q:

A Point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D prove that OB2 + OD2 = OC2 + OA2.

A:

Let ABCD be the given rectangle and let O be a point within it. Join OA, OB, OC and OD.

Through O, draw EOF || AB. Then, ABFE is a rectangle

In right triangles OEA and OFC, we have


Now, in right triangles OFB and ODE, we have

Q:

A toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 6 cm and its height is 4 cm. Find the cost of painting the toy at the rate of Rs. 5 per 1000 sq cm.

A:


Q:

A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface of the remainder is 8/9 of the curved surface of the whole cone, find the ratio of the line segments into which the altitude of the cone is divided by the plane.

A:


Let VAB be a hollow cone of height H, slant height L and the base radius R.

Suppose this cone is cut by a plane parallel to the base such that O' is the centre of the circular section of the cone.

Let h be the height, l be the slant height and r be the base radius of the smaller cone VA'B'.

In VO’A’ and VOA

A = A’

O = O’

Therefore by AA postulate

VO’A’ VOA

Q:

Determine the height of a mountain if the elevation of its top at an unknown distance from the base is 30° and at a distance 10 km further off the mountain, along the same line, the angle of elevation is 15°. [Use tan15°=0.27]

A:


Let AB be a mountain of height h metres and angle of elevation of its top is 30° from a point at x km away from mountain and that of 15° from (x+10)km away from the base of the mountain.

InAOQ,

tan30°=h/x(1/3)=h/x

x=3h … (1)

In AOP,

tan15°=h/(x+10)

0.27= h/(x+10)

h=0.27(x+10)

h=0.27(3h+10)

=0.273h+2.7

h - 0.273h=2.7

h(1-0.271.732)=2.7

h(0.54)=2.7

h=2.7/0.54

h=270/54

=5 km.

Thus, height of the mountain is 5 km.

Q:

The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 15003 m, find the speed of the jet plane.

A:

Let P and Q be the two positions of the plane and let A be the point of observation. Let ABC be the horizontal line through A. It is given that angles of elevation of the plane in two positions P and Q from a point A are 60° and 30° respectively.

Then, PAB=60°, QAB=30°

Here PB=15003 metres

In ABP, we have

tan60°=BP/AB 3=15003/AB

AB=1500 m

In ACQ, we have

tan30°=CQ/AC(1/3)= 15003/AC

AC=4500 m

So, PQ=BC=AC - AB

=4500 - 1500

=3000m

Thus, the plane travels 3000 m in 15 seconds.

Hence, the speed of plane=3000/15=200 m/sec.

Q:

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm an 6 cm respectively. Find the sides AB and AC.

A:


Download App

Download Extramarks – The Learning App

Take your CBSE board preparation to another level with AI based and rich media animation on Extramarks - The Learning App.

Features of Learning App

  • Learn with studybot “Alex”
  • Personalized Learning journey
  • Unlimited doubt Solving
  • Quiz, challenge friends
  • Leaderboard Ranking
  • Academic Guru

Download App Download App