1)What is the product of the irrational roots of the equation (2x-1)(2x-3)(2x-5)(2x-7)=9? (a)3/2 (b)4 (c)3 (d)3/4 Ans. (a)
2)In an examination, 35% candidates failed in one subject and 42% failed in another subject while 15% failed in both the subjects. If 2500 candidates appeared in the examination, how many passed in either subject but not in both? a) 325 b) 1075 c) 1175 d) 2125 e) 2250
In A subject 35% of 2500 failed i.e. 875. In B subject 42% of 2500 failed i.e. 1050. Failed in both the subject 15%of 2500 i.e. 375. => Student failed in A alone=875-375=500. => Student failed in B alone=1050-375=675. Therefore, Student failed in either of the two subjects = 500+675 = 1175.
1)A ladder is lying against a wall of length 25 feet. Distance between bottm of ladder and wall is 7 ft. Now ladder got slide down by 4 feet. So wts the new distance betwen there botom ends.
2)A train length is 100 mtr cross a man in 8 seconds (Same direction), man speed is 5km/hr and after 15 minutes train reached at station then find the time in wich the person reach at station
3)A man want to paint roofs of two houses. He have a tin of paint. When he paints first floor he remained with a amount of paint by which he can paint 33 sqr. Feet more but when he paints the 2nd floor he left 162 sqr feet area unpainted. Whats the amount of paint in tin
A man sells an article at 5% abpve its cost price. If he had bought it at 5% less than what he paid for it and sold it for Rs 2 less, he would have gained 10%. Find the cost price of the article a)500 b)360 c)425 d)400
A man sells an article at 5% abpve its cost price. If he had bought it at 5% less than what he paid for it and sold it for Rs 2 less, he would have gained 10%. Find the cost price of the article a)500 b)360 c)425 d)400
A man sells an article at 5% abpve its cost price. If he had bought it at 5% less than what he paid for it and sold it for Rs 2 less, he would have gained 10%. Find the cost price of the article a)500 b)360 c)425 d)400
My take is 400.
CP = 100x and SP = 105x. Now, CP' = 95x and SP' = 105x - 2. => (105x - 2 - 95x)/95x = 0.10 => x = 4. So, CP = 400.
1.P(x) is a degree 2 polynomial in x. It leaves remainder 5 on division by (x - 1) and remainder 2 on division by (x + 2). When P(x) is divided by (x - 1) (x + 2) then find the remainder.
Ans- Cant be determined.take two cases by trial and error method for P(x)=47 and x=7 and p(x)=102 and x=8 the remainders will be diff in diff cases. Please guys let me know if you have some other way.
1.P(x) is a degree 2 polynomial in x. It leaves remainder 5 on division by (x - 1) and remainder 2 on division by (x + 2). When P(x) is divided by (x - 1) (x + 2) then find the remainder.
A man sells an article at 5% abpve its cost price. If he had bought it at 5% less than what he paid for it and sold it for Rs 2 less, he would have gained 10%. Find the cost price of the article a)500 b)360 c)425 d)400
1)The clock is gaining 8 min per two hours, if it is set at 10.10,then what is the correct time if it shows 4.34 on the same day
2)Two trains started from A and B towards each other at a speed of 16 and 21 kmph. What is the distance between the stations if one of the trains is 60 kmph more than the other when they met?
3)a person has to travel 48 km in 5 hours at a fixed speed.but he covers first 24 kms at a speed 20% less than the fixed speed and the rest 24 km at 2km more than the fixed speed.what is his fixed speed.
1)The clock is gaining 8 min per two hours, if it is set at 10.10,then what is the correct time if it shows 4.34 on the same day
2)Two trains started from A and B towards each other at a speed of 16 and 21 kmph. What is the distance between the stations if one of the trains is 60 kmph more than the other when they met?
3)a person has to travel 48 km in 5 hours at a fixed speed.but he covers first 24 kms at a speed 20% less than the fixed speed and the rest 24 km at 2km more than the fixed speed.what is his fixed speed.
pls explain the proceedings in detail
1. My Answer is 4.10pm Initially both the clocks are set to 10.10(fake clock and original-time clock)
after every two hours time procceds as follow:- FAKE ORIGINAL 10.10 10.10 12.18 12.10 14.26 14.10 16.34(4.34) 16.10(4.10)...
1)The clock is gaining 8 min per two hours, if it is set at 10.10,then what is the correct time if it shows 4.34 on the same day
2)Two trains started from A and B towards each other at a speed of 16 and 21 kmph. What is the distance between the stations if one of the trains is 60 kmph more than the other when they met?
3)a person has to travel 48 km in 5 hours at a fixed speed.but he covers first 24 kms at a speed 20% less than the fixed speed and the rest 24 km at 2km more than the fixed speed.what is his fixed speed.
pls explain the proceedings in detail
3. my answer is that fixed speed is 10kmph.
Let fixed speed is x. then time taken for first 24kms = 24/0.8x time taken for next 24kms = 24/(x+2) total time taken is 5hrs. => 24/0.8x + 24/(x+2) = 5 => 30/x + 24/(x+2) = 5 after solving above equation we get x=-6/5 and 10. Hence, 10kmph is the answer....
50^51^52 divided by 11 what will be the remainder?
Lets Break down the ques by checking Mod at each step
50 Mod 11 leaves rem = 6
(6^51) / 11 would leave rem as 6
Euler's num = 11 (1-1/11) =10 (Since 6 and 11 are co-primes) So 6^10 would leave rem = 1 and hence 6^51 can be broken down to (6^10)^5 * 6^1 which would leave rem as 1*6 = 6
Now 6^52 can be broken down using the same Euler's num as (6^10)^5 * 6^2 which would leave rem as (1*36 Mod 11) = 3
Hence 3
another is 2^100 divided by 101 what will be remainder?
plz help me out....
Is it 1?
Using Euler's num again as 2 and 101 are co-primes. So, Euler's num = 101 * (1-1/101) = 100 So, 2^100 leaves rem = 1
Hence 1
PS: I know I have solved it in a very unorganized manner. So, feel free to ask whatever you dont understand.
1)A ladder is lying against a wall of length 25 feet. Distance between bottm of ladder and wall is 7 ft. Now ladder got slide down by 4 feet. So wts the new distance betwen there botom ends.
Length of ladder 24 (7,24,25)
Now slide down by 4 => 25-4 =21
So distance bet bottom end = root (24^2-21^2) = _/ 3*3*3*5 = 3_/15
Lets Break down the ques by checking Mod at each step
50 Mod 11 leaves rem = 6
(6^51) / 11 would leave rem as 6
Euler's num = 11 (1-1/11) =10 (Since 6 and 11 are co-primes) So 6^10 would leave rem = 1 and hence 6^51 can be broken down to (6^10)^5 * 6^1 which would leave rem as 1*6 = 6
Now 6^52 can be broken down using the same Euler's num as (6^10)^5 * 6^2 which would leave rem as (1*36 Mod 11) = 3
Hence 3
Utsav, (6^51)^52 is not same as 6^(51^52). Pls check.
