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^{3}-4x^{2}+ax+b is exactly divisible by (x-1) and (x+2).

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x

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1. multiple of 7.

2. multiple of 3 and 5.

3. multiple of 3 or 5.

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Total numbers = 1 to 35 = 35

1. Multiples of 7 (7, 14, 21, 28, 35) = 5

Probability of selecting a multiple of 7 = 5/35 = 1/7

2. Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33

Multiples of 5 = 5, 10, 15, 20, 25, 30, 35

Multiples of 3 and 5( 15, 30 )= 2

Probability of selecting a multiple of 3 and 5 = 2/35

3. Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33

Multiples of 5 = 5, 10, 15, 20, 25, 30, 35

Multiples of 3 or 5 = 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35

Multiples of 3 or 5 = 16

Probability of selecting a multiple of 3 or 5 = 16/35

1. Multiples of 7 (7, 14, 21, 28, 35) = 5

Probability of selecting a multiple of 7 = 5/35 = 1/7

2. Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33

Multiples of 5 = 5, 10, 15, 20, 25, 30, 35

Multiples of 3 and 5( 15, 30 )= 2

Probability of selecting a multiple of 3 and 5 = 2/35

3. Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33

Multiples of 5 = 5, 10, 15, 20, 25, 30, 35

Multiples of 3 or 5 = 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35

Multiples of 3 or 5 = 16

Probability of selecting a multiple of 3 or 5 = 16/35

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First term, a = 12

Common difference, d = 6

Last term, l = 252

Let total terms be n

l = a + (n-1)d

252 = 12 + (n - 1)6

240 = 6n - 6

6n = 246

n = 246/6

n = 41

Sum of all terms

S_{n} = n(a+l)/2

S_{n} = 41(12 + 252)/2

S_{n }= 41(264)/2

S_{n} = 5412

Common difference, d = 6

Last term, l = 252

Let total terms be n

l = a + (n-1)d

252 = 12 + (n - 1)6

240 = 6n - 6

6n = 246

n = 246/6

n = 41

Sum of all terms

S

S

S

S

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Let point *C*(-3, *k*) divides the line segment joining the points (-5, -4) and (-2, 3) in ratio *m* : 1.

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^{2}-PR^{2})= ^{2}-4^{2})=3 cm.

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If the mean of the following distribution is 27, find the value of p:

Class-interval 0-10 10-20 20-30 30-40 40-50 |

No. of Workers 8 p 12 13 10 |

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Show that a, c, b, d are in proportion, if

(ma^{2} + nb^{2}) : (mc^{2} + nd^{2}) : : (ma^{2} - nb^{2}) : (mc^{2} - nd^{2}).

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Let one number be x and other number be 16 - x

Their reciprocal will become 1/x and 1/(16 - x).

A.T.Q

(1/x) + [1/(16-x)] = (1/3)

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The ordered pairs for ‘less than Ogive’ are

(30, 10), (40, 18), (50, 30), (60, 54), (70, 60), (80, 85), (90, 100).

__Case 2__:

The ordered pairs for ‘more than Ogive’ are

(20, 100), (30, 90), (40, 82), (50, 70), (60, 46), (70, 40), (80, 15).

And the graph is shown below:

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Two tangents *TP* and *TQ* are drawn to a circle with centre *O* from an external point *T*. Prove that *PTQ* = 2*OPQ*.

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a. Mr. Abhinav took term insurance policy for him and paid ₹2700 as SGST. What is the total annual premium paid by him for this policy given that the rate of GST is 18%?

b. Mr. John went to a trip to Jaipur. He took a room in hotel for three days at the rate of ₹3000 per day. On second day, his friend Sam also joined him. Hotel provided an extra bedding at ₹1000 per day for the bed. How much GST at the rate of 18% is charged by the hotel in the bill to Mr. John for the three days?

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b.

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Let AB be a building of height 20 m and BC be the transmission tower of height x m and D be any point on the ground.

Here, BDA=45° and ADC=60°

Now in ADC, we have

tan60°=AC/AD 3=(x+20)/AD

AD=(x+20)/ 3 …(1)

Again in ADB, we have

tan45°=AB/AD1=20/AD

AD=20 m.

Putting the value in equation (1), we have

20=(x+20)/ 3 203=x+20

x=203 – 20

=20(3 – 1)

=20(1.732 – 1)

=14.64 m.

Thus, the height of tower is 14.64 metres.

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Steps of construction:

1) Join *O* to *P.*

2) Draw a perpendicular bisector of *OP*.

3) Take *O*’ as a center and *O*’*O* as radius draw a circle it intersect the previous circle at point *A* and *B*.

4) Join *P* to *A* and *B.*

5) *PA* and *PB* are required tangents of the circle from the point *P*.

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