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Distance moved = 40mCircumference of the garden roller = 4m

Hence, no. of revolutions it makes in moving 40m = 40m/4m = 10.

Diameter = 1.4 m r = d/2 r = 1.4/2r = 0.7m

Curved surface area of cyclinder= 2 * 22/7 * r^2

= 2 * 22/7 * 0.7 * 0.7

= 8.8 m^2

No of revolution = 50

Therefore = 8.8 * 50

= 440 m2

Diameter of the roller = d = 1.4 m

Radius ( r ) = d/2 = 1.4/2 = 0.7m

Length of the roller = ( h ) = 2 m

If the roller complete one revolution

Then the area covered = curved surface area of the roller

= 2πrh

Area covered in 5 revolutions

= 5 × 2πrh

= 10 × 22/7 × 0.7 × 2

= 44 square meters

Loss = 742-700/742= 42/742 x 1005 .66

price before discount=700discount=30%price after discount=700*0.7=490

Diameter = 1.4 m r = d/2 r = 1.4/2r = 0.7m curved surface area of cyclinder = 2 * 22/7 * r^2 = 2 * 22/7 * 0.7 * 0.7= 8.8 m^2 No of revolution = 5Therefore = 8.8 * 5 = 44 m2

the

A)x^2+y^2=(3K)^2.....this is the right answer

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expression under square root extends infinitely then expression in brackets would equal toxxitself so we can rewrite given expression asx=2+x−−−−−√x=2+x. Square both sidesx²=2+x-->after solving this equationx^2-x-2= 0x=2orx=−1 As given thatx>1then only one solution is valid:x=2

25% of 5000 = 25×5000/100 = 1250 Rs.

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let X be the common multiplenow5x,4x,3x are sidesdifference in longest and smallest5x-3x=2X=1hence sides are 5,4,3

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a = 2 and b = 1 is the correct answer of the given question

az+ by=7____(1) bx+ay=3___(2) 3x+b=7___(3), a+2b=3____(4); a=2;: b=1,

first take cost of 1 kg sugar by dividing and then multiply 877.50 from the answer that you have taken.

27is the correct answer of the given question

sugar = 260/8 = 32.5, 877.50/32.5=27 kg

first cost of 1kg summary dividing and then multiply 877.50 from the answer that you have taken

27 kg is the correct answer

the correct answer is 27

27 is the right answer

Thanks

=(sinA+cosecA)²+(cosA+secA)²

=sin²A+cosec²A+2sinAcosecA+cos²A+sec²A+2cosAsecA

=sin²A+cos²A+cosec²A+sec²A+2sinA×1/sinA+2cosA×1/cosA

=1+cosec²A+sec²A+2+2

=5+(1+cot²A)+(1+tan²A)

=7+tan²A+cot²A

Identities used:1+tan²A=sec²A

1+cot²A=cosec²A

sin²A+cos²A=1

cosecA=1/sinA

secA=1/cosA

distance=speed×time7×48=336km is the distancenow speed=28km/hournow time=distance/time=336/28=12hourso the car will take 12 hours to cover the same distance

Given equations are : ax + by = 7..........(1)bx + ay = 5 .........(2)

according to question, (3,1) is the point of intersection of given equations. so, (3,1) will satisfy both of given equations.

put (3,1) in equation (1), 3a + b = 7 ........(3)

put (3,1) in equation (2), 3b + a = 5......... (4)

multiplying 3 with equation (3) and then subtracting equation (4),

3(3a + b) - (3b + a) = 3 × 7 - 5

9a + 3b - 3b - a = 21 - 5

8a = 16 => a = 2 , put it in equation (3)

b = 7 - 3a = 7 - 6 = 1

hence, a = 2 and b = 1

Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)

Let’s consider the ratio these two AP’s m th terms as am: a’m→(1)Recall the nth term of AP formula, an= a + (n – 1)d

Hence equation (1) becomes,am: a’m= a + (m – 1)d : a’ + (m – 1)d’

On multiplying by 2, we getam: a’m= [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]

= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]

= S2m – 1: S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23]

Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].

8 kg= rs 2601 rs =8/260 kg877.50 rs = 8*877.5/260=7020/260=27kg

260rs=8kg sugar1rs=8/260 sugar877.50 rs=8*877.5/260= 27kg

P=Rs.16,000, T=2yrs, R=10% and 12% p.a

For 1st yr:

16000 X 1 X10 ÷ 100 = 1600

Amount = 16000 + 1600 = 17,600

He repays 5600 at the end of 1st yr

Therefore, 17,600 - 5600 = 12000

New Principal = 12000

For 2nd yr,

Interest = 12000 X 1 X 12 ÷ 100 = 1440

Amount = 12000 + 1440 = 13440

Therefore, the amount outstanding for the 2nd yr is = Rs. 13,440

27 kg of sugar can bought for Rs 877.50

principle -40000amount- 44100rate- 5%

amount=p (1+r/100)n44100=40000 (1+5/100)n44100/40000=(1 + 1/20)n44100/40000= (21/20)n(21/20)^2 = ( 21/20)n2 years = time

a=15d=8

Therefore the AP becomes-15,23,31,39,47.........

Let the journey covered by train, bus and car = ‘4a’, ‘3a’ and ‘2a’ respectively.

Let fare of train, bus & car = ‘b’, ‘2b’ and ‘4b’ per km.

Then, fare paid for train, bus & car = 4ab, 6ab, 8ab respectively.

∴ Total fare = 4ab + 6ab + 8ab = 18ab

Given that,

⇒ 18ab = 720

Or, ab = 40

∴ Fare spent on train = 4ab

= 4 × 40 = Rs. 160.

Fare for 1st km = Rs. 15

Fare for 2nd km=Rs.15+8=Rs23=Rs.15+8=Rs23

Fare for 3rd km=Rs.23+8=31=Rs.23+8=31

Here, each subsequent term is obtained by adding a fixed number (8) to the previous term.

Hence, it is an AP

the list of fare = {15,15+8,15+2*8,......}= {15,23,31,39,......}

These terms make an arithmetic progression as the difference between two consecutive terms is 8.

There is an arithmetic progression in the fare after each kilometre increment. The fare goes like this: 20, 28,36,44,..where the common difference is 8.

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what is your question

(-316)×(-1)=-316×-1=316

correct answer is 316

316 is the right answer

316 is the correct answer of the given question

+ 316 is your answer

(-316)×(-1)=-316×-1=316

316 is correct answer

316 is the right answer

correct answer this question is 316

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this is an example of Arithmetic series because a comman fare is being applied just the comman ratio in arithmetic progressionso the series will be15,15+8,23+8,31+8+..