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Two digit numbers

11, 12, 13, 21, 22, 23, 31, 32, 33.

i. P (both digits being same) = \(\frac{3}{9}=\frac{1}{3}\)

ii. P(sum of digits being 4) = \(\frac{3}{9}=\frac{1}{3}\)

23 days = 3 weeks + 2 days Wednesday comes on Tuesday + Wednesday, Wednesday + Thursday when two days are taken. 

Total probabilities (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday) 

∴ Total probabilities = 7 

∴ Probabilities = 2/7

Total 3 digit numbers = 900

a. Numbers with first and last digits are equal

101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393

There are 90 such numbers.

∴ Probability = \(\frac{90}{900}=\frac{1}{10}\)

b. The numbers with last two digits zero are 100, 200, 300,……… 900 total numbers 9.

∴ Probability = \(\frac{90}{900}=\frac{1}{100}\)

c. The numbers with the last digit greater than the first digit is 36

∴ probability = \(\frac{36}{90}=\frac{1}{5}\)

If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.Now the area of ∆APQ = 1/2 × AP × QN (Since, area of a triangle= 1/2× Base × Height)

Similarly, area of ∆PBQ= 1/2 × PB × QN

area of ∆APQ = 1/2 × AQ × PM

Also,area of ∆QCP = 1/2 × QC × PM ………… (1)

Now, if we find the ratio of the area of triangles ∆APQand ∆PBQ, we have 

Now the area of ∆APQ = 1/2 × AP × QN (Since, area of a triangle= 1/2× Base × Height)

Similarly, area of ∆PBQ= 1/2 × PB × QN

area of ∆APQ = 1/2 × AQ × PM

Also,area of ∆QCP = 1/2 × QC × PM ………… (1)

Now, if we find the ratio of the area of triangles ∆APQand ∆PBQ, we have

=

=

Similarly,

=

=

………..(2)

According to the property of triangles, the triangles drawn between the same parallel lines and on the same base have equal areas.

Therefore, we can say that ∆PBQ and QCP have the same area.

area of ∆PBQ = area of ∆QCP …………..(3)

Therefore, from the equations (1), (2) and (3) we can say that,

AP/PB = AQ/QC

Also, ∆ABC and ∆APQ fulfil the conditions for similar triangles, as stated above. Thus, we can say that ∆ABC ~∆APQ.

The MidPoint theorem is a special case of the basic proportionality theorem.

According to mid-point theorem, a line drawn joining the midpoints of the two sides of a triangle is parallel to the third side.

According to the property of triangles, the triangles drawn between the same parallel lines and on the same base have equal areas.

\begin{array}{l} \frac {area ~of~ ∆APQ}{area~ of~ ∆PBQ}\end{array}

Therefore, we can say that ∆PBQ and QCP have the same area.

area of ∆PBQ = area of ∆QCP …………..(3)

Therefore, from the equations (1), (2) and (3) we can say that,

AP/PB = AQ/QC

Also, ∆ABC and ∆APQ fulfil the conditions for similar triangles, as stated above. Thus, we can say that ∆ABC ~∆APQ.

The MidPoint theorem is a special case of the basic proportionality theorem.

According to mid-point theorem, a line drawn joining the midpoints of the two sides of a triangle is parallel to the third side.

Total number of two digit number : 90 

Total number of two digit perfect squares: 6

Number of two digit numbers which are not perfeet squares : 90 – 6 = 84,

Probability = \(\frac{84}{90}=\frac{42}{45}\)

i. Probability of two digits being same = \(\frac{9}{90}=\frac{1}{10}\)

(11, 22, 33, 44, 55, 66, 77, 88, 99)

ii. Numbers in which first digit is greater than second digit are 10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65, 70, 71, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 96, 97, 98,

45 outcomes, Required Probability = \(\frac{45}{90}=\frac{1}{2}\)

iii. Numbers in which first digit is smaller than second digit are

12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89, 36 total no.of favourable outcomes = 36

Probability = \(\frac{36}{90}=\frac{18}{45}=\frac{6}{15}=\frac{2}{5}\)

A journal entry is a record of the business transactions in the accounting books of a business. A properly documented journal entry consists of the correct date, amounts to be debited and credited, description of the transaction and a unique reference number. A journal entry is the first step in the accounting cycle.

(A) A and B

A liquidity ratio is an indicator of whether a company's current assets will be sufficient to meet the company's obligations when they become due. The liquidity ratios include the current ratio and the acid test or quick ratio. The current ratio and quick ratio are also referred to as solvency ratios.

If the radius of the shaded semicircle is r,

Area of all semicircles = \(4\times\frac{\pi r^2}{2}=2\pi r^2\)

Area of the bigger circle = \(\pi(4r)^2=16\pi r^2\)

Probability = \(\frac{2 r^2}{16 r^2}=\frac{1}{8}\)

Distribution function, mathematical expression that describes the probability that a system will take on a specific value or set of values. The classic examples are associated with games of chance. The binomial distribution gives the probabilities that heads will come up a times and tails n − a times (for 0 ≤ a ≤ n), when a fair coin is tossed n times. 

The closure property of the whole number states that addition and multiplication of two whole numbers is always a whole number. 

For example, consider whole numbers 7 and 8, 7 + 8 = 15 and 7 × 8 = 56. Here 15 and 56 are whole numbers as well. This property is not applicable on subtraction and division.

a. Number of two-digit natural numbers = 90

b. Numbers whose sum of digits will be 10 is 19, 28, 37, 46, 55, 64, 73, 82, 91

Probability that the sum of digits of that number will be 10 = \(\frac{9}{90}=\frac{1}{10}\)

E =sin(π/14) sin(3π/14) sin(5π/14)

Put : x = π/14.

Then : 14x = π.

∴ E = sin x. sin 3x, sin 5x.

∴ E = sin (π/14). sin (3π/14). sin (5π/14) ......... (1)

Now,

sin π/14 = sin (π/2 - 6π/14) = cos 6π/14 = cos (π - 8π/14) = - cos 8π/14 = - cos 8x

Similarly, sin 3π/14 = cos 4π/14 = cos 4x and

sin 5π/14 = cos 2π/14 = cos 2x.

∴ from (1),

E = - cos 8x. cos 4x. cos 2x

Multiply and divide by 2 sin2x,

=  -1/ (2 sin2x ). [( 2 sin 2x. cos 2x ). cos 4x. cos 8x]

= (-1/ 4 sin 2x ). ( 2 sin 4x. cos 4x ). cos 8x

= (-1/ 8 sin 2x ). ( 2 sin 8x. cos 8x )

= (-1/ 8 sin 2x ). sin 16x

= (-1/8). ( 1/ (sin π/7)). sin 8π/7

= (-1/8). ( 1/ (sin π/7)). sin ( π + π/7 )

= - (1/8). ( 1/ sin π/7 ). ( - sin π/7 )

= 1/8.

∴ E = 1/8

∵ School fee of one student in a month = ₹1050

∴ School fees collected from 36 students in a month = ₹1050 x 36 = 37800

Answer is 90 two-digit even numbers

Answer is 10 prime numbers

Solution:  

 395/1050 = 79/210 = 79/3*7*10
Since the denominator has 3 as its factor, 

Therefore, 395/1050 has a non-terminating decimal expansion.

Pairs of numbers in tokens are (1.1) , (1,2), (1,3), (1,4) (2.1), (2, 2), (2, 3), (2, 4) (3, 1), (3,2), (3; 3), (3,4) (4, 1), (4, 2), (4, 3), (4,4).

Pairs getting sum as prime numbers (1.1) , (1,2), (1,4), (2, 1), (2, 3), (3, 2), (3, 4), (4, 1), (4, 3) 

Probability of getting sum of the face numbers a prime number = \(\frac{9}{16}\)

a. Total outcomes

(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 = 36

b. Sum of two numbers is to be 8 = 5 (2,6), (3,5), (4,4), (5,3), (6,2), Probability = 5/36

c. Probability of being same number =(1,1), (2,2), (3,3), (4,4), (5,5), (6,6),

Total = 6

Probability = \(\frac{6}{36}=\frac{1}{6}\)

Hey! We can surely say that the sum will be infinity.

But if we go on fun experiment then we will get the sum as -1/12. How?

Let's take sum be S

S=1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + …= ?

S₁ = 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + …

Take a negitive S₁

-S₁ = − 1 + 1 − 1 + 1 − 1 + 1 − 1 + …

Now 1 - S₁ = 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + …

So , 1 - S₁ = S₁

Hence S₁ = 1/2

S₂ =1 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + …

S₂ + S₂

=1 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + …

+ 0+ 1 − 2 + 3 − 4 + 5 − 6 + 7 + …

=1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + …

= S₁

So,

2S₂ = S₁

2S₂ = 1/2

S₂ = 1/4

S − S₂

=1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + …

− 1 + 2 − 3 + 4 − 5 + 6 − 7 + 8 + …

=0 + 4 + 0 + 8 + 0 + 12 + 0 + 16 + …

= 4( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + … )

= 4S

So , S - S₂ = 4S

Finally , S – 1/4 = 4S

⇒ S = – 1/12
 Which means

S = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + …= – 1/12

Hence proved. But this answer is absolutely wrong as it breaks the sum law.

Answer is (a) Remainder.

If remainder is not zero, then the divisor is not a factor of dividend.

Refers to the below link

http://bit.ly/2DxInlE

Given that,

Two dice are thrown at same time. 

Therefore,

Total possible outcomes = n(s) 

= 6 × 6 = 36. 

Let event E be the event that sum of two numbers appearing on the top is more than 9. 

Hence,

Favorable outcomes of event E = {(4, 6), (5, 5), (5, 6), (6, 5), (6, 6), (6, 4)} 

Hence, 

Total outcomes favorable to event E is n(E) = 6.

∴ P(E) = \(\frac{Total\,outcomes\,favourable\,to\,event\,E}{Total\,possible\,outcomes}\) 

= \(\frac{6}{36}\) = \(\frac{1}{6}\).

Hence,

Probability of event E is \(\frac{1}{6}\)

Answer is (c) P(A/B) = 1

Answer is (b) 5/8

Answer is (a) ) Both the statements are true and Statement. It is the correct explanation of Statement I.

Answer is (d) Statement I is false, but Statement II is true.

tan^-1(\(\frac{cos\theta+sin\theta}{cos\theta-sin\theta}\)) = tan^-1\(\left(\cfrac{1+\frac{sin\theta}{cos\theta}}{1-\frac{sin\theta}{cos\theta}}\right)\)(By dividing both numerator and denominator by cos θ)

\(=tan^{-1}(\frac{1+tan\theta}{1-tan\theta})\) \((\because\frac{sin\theta}{cos\theta}=\tan\theta)\)

\(=tan^{-1}\left(\cfrac{tan\frac{\pi}4+tan\theta}{1-tan\frac{\pi}4tan\theta}\right)\)\((\because tan\frac{\pi}4=1)\) 

\(=tan^{-1}(tan(\frac{\pi}4+\theta))\) \((\because\frac{tanA+tanB}{1+tanAtanB}=tan(A+B))\)

\(=\frac{\pi}4+\theta\) \((\because tan^{-1}(tan\theta)=\theta)\)

Total number of cards in the deck = 52. 

a). Total number of king card in the deck = 4. 

The probability that the drawn card is a king = \(\frac{Total\,number \,of\,red\,8\,card}{Total\,number\,of\,cards\,in\,deck}\) = \(\frac{4}{52} = \frac{1}{13}\) . 

Hence, option (ii) is correct. 

b). Total number of cards which shown 8 in the deck is 4 in which 2 are black and 2 are red suited. 

Therefore total number of red 8 cards in the deck = 2 

Now, the probability that the draw card is a red eight = \(\frac{Total\,number \,of\,red\,8\,card}{Total\,number\,of\,cards\,in\,deck}\) = \(\frac{2}{52} = \frac{1}{26}\) . 

Hence, option (iv) is correct. 

c). The face cards are jack, queen and king cards, each of them have 4 quantities in the deck. 

Therefore, total number of face cards in the deck = 4 × 3 = 12. 

Hence, the probability that the drawn card is a face card = \(\frac{Total\,number \,of\,red\,8\,card}{Total\,number\,of\,cards\,in\,deck}\) = \(\frac{2}{52} = \frac{13}{13}\).

Hence, option (ii) is correct. 

d). Total queen cards is 4 in the deck in which 2 are black suited and 2 are red suited.

Therefore, total number of queen cards of black suit = 2.

The probability of getting a queen of black suit = \(\frac{Total\,number \,of\,red\,8\,card}{Total\,number\,of\,cards\,in\,deck}\) = \(\frac{2}{52} = \frac{1}{26}\).

Hence, option (i) is correct.

e). Total number of ace cards in the deck = 4 

Total number of spade cards in the deck = 13. 

Hence, total number of cards which are either an ace or a card of spades = 4 + 13 – 1 = 16. (∵ One ace card is shade card ) 

Therefore, the probability that the drawn card is a card of spades or an ace card = 
\(\frac{Therefore\,of\,cards\,which\,are\,either\,an\,ace\,or\,acard\,of\,spades}{Total\,number\,of\,cards\,in\,the\,deck}\) = \(\frac{16}{52} = \frac{4}{13}\).

Hence, option (ii) is correct.

∫(1 − x)√x dx =

 ∫(√x − x√x) dx 

= ∫ (x ^{\(\frac{1}{2}\)} − x ^{\(\frac{3}{2}\)}) dx 

= ∫ x ^{\(\frac{1}{2}\)} − ∫ x ^{\(\frac{3}{2}\)} dx

=\(\frac{2}{3} x ^{\frac{3}{2}}\) − \(\frac{2}{5} x^{\frac{5}{2}}\) + c, where C is an integral constant.

(\(\because\) ∫ xⁿdx = \(\big(\frac{x^{n+1}}{n+1}\big)\))

Hence, ∫(1-x)√x dx = \(\frac{2}{3} x ^{\frac{3}{2}}\) - \(\frac{2}{5} x^{\frac{5}{2}}\) + c.

Answer is (a) Both the statements are true and Statement. It is the correct explanation of Statement I.

Answer is (a) Both the statements are true and Statement. It is the correct explanation of Statement I.

Answer is (d) Statement I is false, but Statement II is true.

Answer is (a) 0

Answer is (c) |vector a|² - |vector b|²

Answer is (d) 25

Answer is 

I. (d) z = 0

II. (a) (0,1,0)

III. (b) 8/5

IV. (c) parallel

We have,

sinθ = 1/2

⇒ cosecθ = \(\frac{1}{sin\theta}\) 

= 2

Now, 

3cot²θ+3 = 3(1+cot²θ) 

= 3cosec²θ

= 3×2² 

= 3×4 

= 12.

Hence,

The value of (3cot²θ+3) is 12.

Option:(a) 1/√1-x²

Option: (a) sec²x

Option: (b) π/4

MB = 1/ 2 × AB = 1 / 2 × 12 = 6 cm (perpendicular drawn from the centre of the circle to the chord bisects the chord)

OB² = OM² + MB² .....................(Pythagoras theorem) 

= 8² + 6² 

= 64 + 36 

= 100 ∴ OB 

= 10 cm

Considering the given series

31, 102, 165, 256, 345, 446, 562

The series follows the following pattern:

∴ Wrong term in given number series is 165.

Slip rings in AC motor are used to change the direction of current in the coil continuously.

The melting point of fuse wire is lower than that of other metals

(i) Starts appear to be twinkling due to atmospheric refraction. 

(ii) Star light reaches the surface of the earth through many layers of atmosphere. 

(iii) These layers have different optical densities due to which they offer different refractive index values to the incoming light. 

(iv) So, the light bends many times giving different apparent positions of the star which we see as twinkling. 

The flow of current in the circuit stops simultaneously with the melting of fuse wire.

The given convex lens is madeup of five different materials.So they have different refractive indices/ different focal lengths. Hence they form five different images.

The current in the circuit may increase due to reasons such as short circuit, overload, excess flow of current or any problems in the insulation. As the higher temperature produced in the circuit due to these reasons makes the fuse wire to melt and the flow of current stops. Thus the circuit and the appliances are protected.

• The ends of the fuse wire must be connected firmly at appropriate points.
• The fuse wire should not project out of the carrier base.
• Use the fuse wire of proper amperage.