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as we can see that 9 ^2+12^2 = 144+81= 225that is equal to 15^2hence it is a right angle triangle

3

(B) SSS

But how?

Total CP = 3000+3000 = 6000 Profit on one is 500 so SP1 = 3000+500 = 3500 Profit on other is 5% so SP2 = 3000*1.05 = 3150 total SP = 3500+3150 = 6650 overall profit = 6650-6000 = 650

the remainder should be 10

By remainder theoremwe can find that3(-1)^4+2(-1)+5= 10

the remainder is 10 🤔

the remainder should be 10

by dividing 3x⁴+2x²+5 by x+1 answer (remainder) should be 10.....

answer A is the correct ansew

XC is right answer of this question

100-(15+25+20+10)=100-70=30%leftif X is the amount of total moneyX*30/100=240X=800rupeeshence 20% on education20/100*800=160rupees

56 p as a per cent of Rs.2.8056 p as a per cent of 280 p=(56/280)*100=20

We know that

Sum of first n natural numbers

= n ( n + 1 )/2

Here ,

n = 30,

S30 = [ 30 ( 30 + 1 ) ] /2

= ( 30 × 31 ) / 2

= 930 / 2

= 465

Therefore ,

Sum of the first 30 natural numbers

= S30 = 465

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i) 3.478 km, ii) 25.5 km, iii) 0.465 km, iv) 490.5 km

hfsobc Sri on visitors

-17/56 is correct answer

thanks kal mara paper ha math ka

P = Rs. 24000Time or n= 6 months = 2 periods of 3 months each or n= 2.Rate = 20 paisa per onerupee per annum= 20 paisa per 100 paisaR = 20 % per annumThere are 4 quarters in a year so, R = 20/4 = 5 % for 3 monthsA = P(1 + R/100)ⁿA = 24000(1 + 5/100)²A = 24000× 105/100× 105/100A = 12× 105× 21A = 26460 Rs.Compound Interest = A - PCompound Interest = 26460 - 24000Compound Interest = 2460 Rs.

465 coins consist of 1 rupee, 50 paise and 25 paise coins.Their values are in the ratio 5 : 3 : 1.The number of each type of coins respectively is

Let 1 re coin = x, 50p coin =y, 25p coin = z,x + y + z = 465 and 4x : 2y : z = 5 : 3 : 1 This is the value ratio.4x : 2y = 5 : 312x = 10yx/y = 5/62y : z = 3 : 12y = 3zy/z = 3/2x : y : z = 5 : 6 : 4 = 31(5 : 6 : 4)Solving x = 155, y = 186, z = 124 coins

Answer:155,186,124 numbers of 1 re, 50 paise and 25 paise coins.

Thanks

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rong Answar

for option c6x-2y=32(3x-y)=33x-y=3/2so the graph of 6x-2y=3 os parallel to the graph of 3x-y=2

A=P(1+R/300)^2 =24000(1+20/300)^2=24000(32/30)^2 so A=27306.67so interest=3306.67

(x-y)^3 = x^3 -y^3 -3xy(x-y) 5^3 = x^3 -y^3 -3×84×51260+125 = x^3 -y^3

Principal amount (P) = 4000 rupeesTime (t) = 4/3 yearRate = 2 paiseper rupee(100 paise) per month = 2% monthely = 2×12 = 24% annual

S.I = P×R×t/100 = 4000*24*4/100*3 = 16*80 = 1280

Interest = 1280 rupeesso, total Amount after 4/3 years = 4000+1280= 5280 rupees

Soin decimal= 15/56= 0.267857

a2+b2+2ab is formula ka answer hoga

a²+b²+2ab is the answer

(a^2 + b^2) = a^2 + b^2 + 2ab

In ∆ ABC,∠A + ∠ABC + ∠ACB = 180°∠A = 180° - 100° = 80°

∆ABC and ∆ DBC have same base.∠B = ∠A∠B = 80°

(6x+1/5)(6x-1/5)= (6x)²- (1/5)²36x²-b = 36x²- 1/25-b = 36x²-1/25 -36x²b = 1/25

36x^2-b=36x^2-6x/5+6x/5-1/25-b=-1/25b=1/25

A = (√3-√2)/(√3+√2) =(√3-√2)/(√3+√2)×(√3-√2)/(√3-√2) [ rationalizing it] =(√3 -√2)²/1 =(√3 -√2)² = 5- 2√6b=(√3+√2)/(√3-√2)×(√3+√2)/(√3+√2) [ rationalizing it]=(√3+√2)²/1 =(√3+√2)² = 5+2√6now, a²+b² - 5ab= (5-2√6)² + (5+2√6)² - 5(5-2√6)(5+2√6)=25 +24 -20√6 +25 +24 + 20√6 - 5(25-24)=98 -5 = 93

given z=2-3iZ̅=2+3iz+z̅=(2-3i)+(2+3i) =4

f(z)=sin zf(π/4)= sin π/4=1/√2

Given,a = 2b = 4ca = 4c......(1)b = 4/2 c = 2c..........(2)

So, (a + b + c)^2/a^2 + b^2 + c^2 Put value of a and b from eq(1) and eq(2)= (4c + 2c + c)^2 / 16c^2 + 4c^2 + c^2 = 49c^2/21c^2 = 7/3

that's wrong

simple interest =p×r×t. now simple interest =400×20×6. =48000

simple interest= P*R*T = 400*20*6= 48000

480 is the correct answer

Put the formula of S. I that is P×R×T/100

if it is per annum the time must be in years

Interest =p×r×t/100 so, 400×20×6/100= 480

simpl interest - 48000

40 is correct answer

3600 रूपए छः महीने के

is a right answer480

छः महीने के तिन हजार छः सो रूपए

40 is the correct answer

480 it is right answer

P- 400R-20T-6SI=P×R×T400×20×6=48,000

simple internet= principle×rate×time=400×20×6=48000

48000 is the best answer

480 is the correct answer of this question

480 is right answer that si=p*r*t/100

Here,P=400R=20T=6S.I=PRT/100400×20×6/1004×20×6=480Rs

principal = 400 rupeesrate = 20 paise per annum time = 6 monthsby the PRT/100 rule(400*20*6/12)/100= 40 rupees therefore simple interest = 40 rupees

simple interest =P*R*T=400*20*6=48000

1)printer2)hard disk3)port4)motherboard5)CD-RW(read write)

name the oldest computer

name the printer used for printing text and graphics with high speed but at low cost

1) Printer

2) Hard disk

3)Port

4)Motherboard

5)cd - rw (read write

1 . Hardware and software

s n national public school .

The ____ is often referred to as the mother of all components attached to it, which often include peripherals, interface cards, sound cards, video cards and so on