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Nachiket Deshmukh
@ThinkAce
11,961

the remainder when x^200 -2*x^199 + x^50 - 2*x^49 + x^2 + x + 1 is divided by (x-1)(x-2) is :

a)7 b) 2*x-3 c)6*x-5 d)0

Let f(x) = x^200 -2*x^199 + x^50 - 2*x^49 + x^2 + x + 1

= (x-1)(x-2) * g(x) + (ax+b) ..... ax+b = Remainder

Put x = 1

=> a+b = 1 - 2 + 1 - 2 + 1 + 1 + 1 = 1

Put x = 2

=> 2a+b = 2^200 - 2^200 + 2^50 - 2^50 + 4 + 2 + 1 = 7

(a+b = 1, 2a+b = 7)

=> a = 6 and b = -5

=> Remainder = 6x - 5

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the remainder when x^200 -2*x^199 + x^50 - 2*x^49 + x^2 + x + 1 is divided by (x-1)(x-2) is :

a)7 b) 2*x-3 c)6*x-5 d)0

sol: my take- 6*x-5

f(x) when divided by x-1, rem =f(1)

putting x=1, f(1)=1

putting x=2; f(2)=7

now, f(x)= k(x-1)(x-2)+ (ax+b)

=> using above results, we see that ax+b= 6*x-5 satisfies...

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a)7 b) 2*x-3 c)6*x-5 d)0

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Nachiket Deshmukh
@ThinkAce
11,961

What is the remainder when (81)^21 + (27)^21 + (9)^21 + (3)^21 + 1 is divided by 3^20 + 1?

a)0 b)1 c)61 d)121

It should be 61

81^21 + 27^21 + 9^21 + 3^21 + 1

= + + + + 1

...... (a^2 - b^2) has a factor (a+b) .... To be used for even powers of 3^20

...... (a^3 + b^3) has a factor (a+b) .... To be used for odd powers of 3^20

=> Remainder = 81 - 27 + 9 - 3 + 1 = 61

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What is the remainder when (81)^21 + (27)^21 + (9)^21 + (3)^21 + 1 is divided by 3^20 + 1?

a)0 b)1 c)61 d)121

61 is d rem.

let 3^20+1=x

(x-1)^3 *27+(x-1)^4*81+9*(x-1)^2+3x-3+1/x

-27+81+9-2 =61

rest all are in multiples of x

- 2 Likes

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a)0 b)1 c)61 d)121

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@HeavenSent
1,409

A train met with an accident 3 hours after starting,which detains it for one hour,after which it proceeds at 75% of its original speed.It arrives at the destination 4 hours late.Had the accident taken plae 150km further along the railway line,the train would have arrived only 3 1/2 hours late.Find the length of the trip and the original speed of the train.

Options

1) 900 km,110 km/hr

2) 970 km,70 km/hr3) 1200 km,100 km/hr

4) 1500 km,120 km/hr

:cheers:

So, when speed got reduced by 75% ,time must have increased by 4/3 t.

In second case , train travels 150 km with normal speed and time saving 30 min. Hence

4t/3 - t = 30

t/3 = 30 => t = 90 min

So train travels 150 km in 90 min , hence the speed of train is 100 km/hr .

So option (3)

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Options

1) 900 km,110 km/hr

2) 970 km,70 km/hr

3) 1200 km,100 km/hr

4) 1500 km,120 km/hr

:cheers:

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Up Pande
@upudit0
1,640

A tourist takes a trip through a city in stages. Each stage consists of three segments of length 100 meters separated by right turns

of 60 degree.Between the last segment of one stage and the first segment of the next stage, the tourist makes a left turn of 60 degree.

At what distance will the tourist be from his initial position after 1997 stages?

1. 100m

2. 200m

3. 300m

4. 400m

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Up Pande
@upudit0
1,640

bhaut der se koi question nahi one from my side............

A city has two perfectly circular and concentric ring roads, the outer ring road (OR) being twice as long as the inner ring road (IR).

There are also four (straight line) chord roads from E1, the east end point of OR to N2, the north end point of IR; from N1, the north

end point of OR to W2, the west end point of IR; from W1, the west end point of OR, to S2, the south end point of IR; and from S1, the

south end point of OR to E2, the east end point of IR. Traffic moves at a constant speed of 30 ? km/hr on the OR road, 20 ? km/hr on the

IR road, and 15 5 km/hr on all the chord roads.

247. Amit wants to reach N2 from S1. It would take him 90 minutes if he goes on minor arc S1 E1 on OR, and then on the chord

road E1 N2. What is the radius of the outer ring road in kms?

60 40 30 20 none of the foregoing

248. Amit wants to reach E2 from N1 using first the chord N1 W2 and then the inner ring road. What will be his travel time in minutes

on the basis of information given in the above question?

60 45 90 105 none of the foregoing

249. The ratio of the sum of the lengths of all chord roads to the length of the outer ring road is

5: 2 5: 2? 5: ? None of the above. **varun bhai__________________/\_______________**

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