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180-110=70°(linear pair)

70+80=x(Exterior angle)x=150

Option d is correct

option (2) is correct.

<ACD+<ACB=180°( rethink yugm angle) <ACD= 130°( given) to 130°+ <ACB= 180°<ACB= 180°- 130°<ACB= 50°<A+<B+<ACB=180°(∆ ke tino angle ka addition) x+60°+50°=180°x+110°=180°x=70°

he travel 60 km by scooter

he travels 60 kms by scooter

total Distance covered by man = 300 kmdistance travelled by car = 1/5 of 300 = 60 kmdistance travelled by train = 3/5 of 300 = 180 kmdistance travelled by scooter = 300 - (60+180) = 300 - 240 = 60 km ( ANS )

answer will be the 60 kilometres

he travel 60 km by scooter

He travel 60 km by scooter

in the first questionangle ACB = 180-125=55

in triangle ABC50+55+x=180x= 75°

I need all answers

Let's assume

the number of children in each row of the square is x

Since, the number of students in each row is equal to the number of rows

so, the number of children in each column of the square is x

6250 students of different schools

But 9 children are left out

so, total number of student in rows and columns =6250-9=6241

total number of student in rows and columns=(the number of children in each row of the square)*(the number of children in each column of the square)

now, we can plug values

6241 =x × xx^2 =6241

now, we can solve for xx=√6421 =79So,

the number of children in each row of the square is 79...........Answer

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Let number of rows is x and number of students is y in a row.Then total number of students = xy

A/C to question,(x - 2)(y + 1) = xy ⇒xy + x - 2y - 2 = xy ⇒x - 2y = 2 -------(1)

Again, (x + 3)(y - 1) = xy ⇒xy - x + 3y - 3= xy ⇒ -x + 3y = 3 -------(1)

Solve equations (1) and (2), y = 5 and put it equation (1) x = 12

Hence, total number of students = xy = 12 × 5 = 60

Let her new weight be x70:x=5:470*4=5x280=5xx=56Reena weight = 56kg

thnks

what

Earning of Amit = Rs 32000

Fraction of income spent on food = 1/4

Amount of money spent on food = Rs (32000 × 1/4) = Rs 8000

Remaining money = Rs (32000 - 8000) = Rs 24000

Money spent on house rent = 3/10 of Rs 24000

= Rs (24000 × 3/10)

= Rs 7200

Money remaining = Rs (24000 - 7200) = Rs 16800

Money spent on education of children = 5/21 of Rs 16800

= Rs (16800 × 5/21)

= Rs 4000

Therefore,-----------------

Money left = Rs {32000 - (8000 + 7200 + 4000)}

= Rs (24000 - 19200)

= Rs 4800

33 years is the correct answer

20 children avg age = 1220children total age = 12×20= 240 20-1=19student total age = 240-10= 230year19 student total age = 23019 student avg age = 230/19= 12.105 years appox

average of 20 students=12 years, total age of 20 students= 20 x 12 = 240 years. total age 20 students and teachers = 13 x 21=26+13=273 years, age of teachers = 273-240=33 years

33 is correct answer

Circumference of a circular park= 352 mWidth of a road = 7 mCircumference = 2πr352 = 2 ×( 22/7)× r352×7 = 44rr = (352×7)/44 = 8 × 7 = 56 mInner radius of the park (r) = 56 mOuter radius of the Park including the road (R )= width + rR = 7 + 56 = 63 mR = 63 mArea of the road = π(R² - r²)= π(R+r) (R - r) [a² - b² = (a-b)(a-b)]= 22/7 (63+56)(63-56)= 22/7 × 119 × 7= 22 × 119Area of the road= 2618 m²Hence, the Area of the road is 2618 m².

Area of garden = 125*60 = 7500 m^2Area of outer region = (125+6)(60+6) = (131)*(66)= 8646m^2 Area of road = 8646 - 7500 = 1146 m^2

Cost of repairing road = 1146*150 = Rs 171900

x-y=0 ------1y+3=0 ------2From 2, y=-3Substitute it in 1x-(-3)=0x+3=0x=-3

x^3 + y^3 = (x+y)(x^2 - xy + y^2)

LHS: x^3 + y^3

RHS: (x+y)(x^2 - xy + y^2)=(x^3 - x^2y + xy^2) + (x^2y - xy^2 + y^3)= (x^3 + - x^2y + x^2y+ xy^2 - xy^2 + y^3)= x^3 + y^3

LHS = RHS

Hence proved

ABC is a right angled at B and D is the midpoint of BC.

In ∆ABD

AD²= AB²+BD²…………(1)[By Pythagoras theorem]

In ∆ ABCAC²= AB²+BC²…………(2)

From equation (1)

AD²= AB²+ (BC/2)²[D is the midpoint of BC]

AD²= AB²+ BC²/4AD²= AB²/1+ BC²/44AD²= 4AB²+ BC²BC²= 4AD²-4AB²…………….(3)Using the value of BC² in eq. 2

AC²=AB²+4AD²-4AB²

AC²=AB²-4AB+4AD²AC²= 4AD²-3AB²

Solution

ABC is a triangle right angled at B, and D and E are points of trisection of BC.Let BD = DE = EC= xThen BE = 2x and BC = 3xInΔ ABD,AD² = AB²+BD² AD² = AB²+x²InΔ ABE,AE² = AB² + BE²AE² = AB² + (2x)²AE² = AB² + 4x²InΔ ABC,AC² = AB² + BC² AC² =AB + (3x)² AC² =AB²+ 9x²

Now,3AC²+5AD² = 3(AB² + 9x²) + 5(AB² + x²)8AB² + 32x²8(AB² + 4x²)= 8AE²⇒ 8AE² = 3AC² + 5AD²Hence proved.

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All the optional are verified and correct

10²=6²+8²100=36+64100=100hence, (6,8,10) is a set of pythagoras triplets.

a is the correct answer

triplet a(6,8,10) is the true pythagoras triplet.

a is the correct Pythagorean triplet

option a Is the correct answer of the given question

AB = BCAB - BF = BC - BDand AF = CD say 1BD = DE = EF = BF say 2

so the area ofΔAEF = 1/2 x AF x EFarea of triangle CDE = 1/2 CD x DEor 1/2 AF x EF from equation 1 and 2

so the area ofΔAEF = area ofΔ CDE

Its Option Das √50= 5√2and√5^2+5^2= 5√2so its a right anglr too

answer is option D. but i want to know "how to find it is right angled triangle?"

What is motion . classify motion as uniform and non uniform motion

2(x²+y²+z²+2yz) is the correct answer of the given question

Radius = 7cmSum of interior angles of a quadrilateral = 360°Therefore​ theta = 360Area grazed = πr^2 x theta/360= 22/7 x 7 x 7 x 360/360= 22 x 7= 154 cm^2Area grazed = 154 cm^2

thanks

3/4 = 0.75 6/8 = 3/4 = 0.75 so yes both are equivalent

option 3 is right because:1 dozen = 12 items,so 20 dozens=240 items.so 1 item cost =250 rs.240 items cost=240×250=60000 which is greater than 50000.