Explore topic-wise InterviewSolutions in .

Sample space in throwing a dice = { 1,2,3,4,5,6}

Total possibilities = 6

i)A = Getting greater than 4

A = { 5,6 }

P(A) = 2/6 = 1/3

ii)B = less than or equal to 4

B = { 1,2,3,4}

P(B) = 4/6 = 2/3

1

2

Let be the angles 2x+4x+2x+1x+1x=10xTotal angle=54010x=540x=540/10=54Each angle 2x=108 4x=216 2x=108 1x=54 1x=54

Let be the angles 2x+4x+2x+1x+1x=10xTotal angle=54010x=540x=540/10=54Each angle 2x=108 4x=216 2x=108 1x=54 1x=54

From the given figure, we can see that there are 2 large shaded semicircle and 2 small shaded semicircle

Let R be the radius of large semicircle and r be the radius of small semicircle.

Area of large semicircle

Area = (piR^2)/2

There are two large semi circle,

therefore total area = 2 x (piR^2)/2 = piR^2

Area of small semicircle

Area = (pir^2)/2

There are two large semi circle,

therefore total area = 2 x (pir^2)/2 = pir^2

Area of shaded region

Area = [piR^2 -pir^2] - pir^2

join all the shaded region we get a circle with radius 42

Area of circle with radius 42 = πR^2=22/7 x 42 x 42 = 5544 cm^2

if the diameter of large circle=28so radius is 141starea of large semi circle =1/2×pie×r×r1/2×22/7×14×14=3082edarea of two circle =2×1/2×pie×r×rr=14/2 =7area=2×1/2×22/7×7×7=154total area =area of large semi circle +area of two small semi circle=308+154=462

Let the length of the room is represented by 'l' and breadth by 'b'.

Given that length and breadth of the room is increased by 1m.

= > (l + 1) * (b + 1).

We know that Area of floor = l * b.

Given that area of the floor is increased by 21 m^2.

= > (l + 1) * (b + 1) = lb + 21

= > (l + 1) * (b + 1) - lb = 21

= > lb + l + b + 1 - lb = 21

= > l + b + 1 = 21

= > l + b = 20 -------- (1)

-----------------------------------------------------------------------------------------------------------

Given that length is increased by 1m and breadth is decreased by 1m.

= > length = l + 1

= > breadth = b + 1.

Given that area is decreased by 5m.

= > (l + 1) * (b - 1) = lb - 5

= > (l + 1) * (b - 1) - lb = -5

= > lb - l + b - 1 - lb = -5

= > -(l - b) = -4

= > l - b = 4 ------- (2)

----------------------------------------------------------------------------------------------------------

On solving (1) & (2), we get

= > l + b = 20

= > l - b = 4

---------------

2b = 24

b = 12.

Substitute b = 12 in (1), we get

= > l + b = 20

= > l + 12 = 20

= > l = 8.

Now,

We know that perimeter of the floor = 2(l + b)

= > 2(8 + 12)

= > 2(20)

= > 40m.

error in question replace 3 by 5

40x8-320 he read 320 pages in 8days

10 days to finish a book and 40 pages everyday so total pages = 40 × 10 = 400 pages now to complete in 8 days every day oages must be = 400 / 8 = 50 pages per day

no. x×8= 40×10 i.e X=50

ans is 50

increasing cost of entry ticket to a fair in the ratio= 10:13,decrease in the number of visitors to the fair =6:5

Let original Tickets cost $10 and and number of visitors is 600

Original revenue: 10 x 600 = $6000then,Tickets cost= $13 number of visitors = 500New Revenue: 13 x 500 = $6500

Increase of total collection ratio= 6500/6000reduction= 13:12

Question you have submitted is incomplete. Please post a complete question.

hope it helps u

दोनों का योग 1 होगा

TQ

Let the third pipe empty the cistern in x minutes.Part of cistren filled in 1 minute when all three pipes are opened simultaneously

1/60+1/75-1/x= 1/50x= 300/3= 100minOption A

angle B = 180 - 90 - 65 = 25°

Angle 180-90-65=25. It is the best and right answer

Three languages

If you like the solution, Please give it a 👍

3x+1+4x+5+90 =180°7x+96°=180°7x=180-967x=84°x=84°\7x=22°

Let A = Event that A speaks the truthB = Event that B speaks the truth

Then P(A) = 75/100 = 3/4P(B) = 80/100 = 4/5

P(A-lie) = 1-3/4 = 1/4P(B-lie) = 1-4/5 = 1/5

NowA and B contradict each other =[A lies and B true] or [B true and B lies]= P(A).P(B-lie) + P(A-lie).P(B)[Please note that we are adding at the place of OR]= (3/5*1/5) + (1/4*4/5) = 7/20= (7/20 * 100) % = 35%

Like my answer if you find it useful!

E1 and E2 are mutually exclusive P(E1) = 1/3 P(E2) = 3/5 P(E1 U E2) = 1/3 + 3/5 - 1/3*3/5 = 1/3 + 3/5 - 1/5 = 1/3 + 2/5 = (5+6)/(5×3) = 11/15

On dividing a polynomial F(x) by x - a, the remainder will be F(a).

Proof:

Let F(x) be a polynomial divided by (x - a).

Let Q(x) be the quotient and R be the remainder.

By division algorithm,

F(x) = Q(x)(x - a) + R.......................(i)

[Dividend= (Divisorx quotient) + Remainder]

Substituting x = a inequation(i), we have

F(a) = Q(a)(a - a) + R

⇒⇒F(a) = R

Hence, the remainder is F(a).

1

2

Whenkept in copperorbrass vessels, curd(and other acidic food items) reacts withcopperand forms toxic compounds which may be unfit for human consumption, and may cause food poisoning. Thus it isnot advisable to keepcurdand other acidic or sour food items incopperor brassvessels

1. distance of op is 9 cm because op is radius of given circle

2.Q does lie inside of circle

tnx

CosA-sinA+1/cosA+sinA-1=(cosA-sinA+1)(cosA+sinA+1)/(cosA+sinA-1)(cosA+sinA+1)=(cos²A-cosAsinA+cosA+cosAsinA-sin²A+sinA+cosA-sinA+1)/{(cosA+sinA)²-(1)²}=(cos²A-sin²A+2cosA+1)/(cos²A+2cosAsinA+sin²A-1)={cos²A+2cosA+(1-sin²A)}/(1+2cosAsinA-1) [∵, sin²A+cos²A=1]=(cos²A+2cosA+cos²A)/2cosAsinA=(2cos²A+2cosA)/2cosAsinA=2cosA(cosA+1)/2cosAsinA=(cosA+1)/sinA=cosA/sinA+1/sinA=cotA+cosecA=cosecA+cotA (Proved)

long method

It is given that,

Five years ago sagar was twice as old as tiru.

Ten years later sagar age will be ten year more than tiru age

Let x be the age of tiru five years ago.

Then the age of sager 5 years ago = 2x

Present age of tiru = x + 5

Present age of sagar = 2x + 5

After 10 years,

age of tiru = (x + 5 )+ 10 = x + 15

age of sagar = (2x + 5)+10 = 2x + 15

To find x

Ten years later sagar age will be ten year more than tiru age

we can write,

(x + 15) + 10 = 2x + 15

x + 25 = 2x + 15

x = 10

To find present age of tiru and sagar

Present age of tiru = x + 5 = 10 + 5 = 15

Present age of sagar = 2x + 5 = 2*10 + 5 = 25

To find the age of sagar when tiru was born

Present age of tiru = 15

Present age of sagar = 25

The age of sagar when tiru was born = Present age of sagar - present age of tiru = 25 - 15 = 10 years

Therefore the age of sagar when tiru was born = 10 years

Question you have submitted is incomplete. Please post a complete question.

Question you have submitted is incomplete. Please post a complete question.

🤔

i) 10/2(2(2)+9(5))=5(4+45)=5(49)=245ii) 12/2(2(-37)+11(4))=6(-74+44)=6(-30)=-180iii) 100/2(2(0.6)+99(1.1))=50(1.2+108.9)=50(110.1)=5505iv)11/2(2(1/15)+10(1/12-1/15))=11/2(2/15+10(3/(15×12)))=11/2(2/15+10(1/60))=11/2(2/15+1/6)=11/2(27/90)=11*3/(10*2)=33/20

so the option is D)none of these.

cosec^2-1=cot^2AHence cot^2A*tan^2A=1 as cotA=1/tanA