Two people A and B start from P and Q(distance= D) at the same time towards each other. They meet at a point R, which is at a distance of 0.4D from P. They continue to move two and fro between the two points. Find the distance from point P at which the fourth meeting takes place.
a) 0.8D b) 0.6D c) 0.3D d) 0.4D
Let the distance is between P and Q 120 km
they first meet .4d means 48 km from p
from this we can say that P has walked 48 km and q has walked 72 km
So their ratio is 2:3
at first time they will walked 120 km for meeting
and now for every time they meet after first meeting they will walked a total distance of 120*2=240 km
so check the distance walked by A in next three meeting=240/5*2*3=288 km
288+48 (first meeting)=336 km is the distance walked by a from starting point to fourth meeting 120+120+96=336
means 96 km from P they will meet at fourth time 96/120=.8D ans
The series of numbers(1, 1/2, 1/3, .. 1/9172)is taken. Now 2 numbers are taken from this series say (x,y). Then the operation x+y+x.y is performed to get the consolidated number. The process is repeated. What will be the value of the set after all the numbers are consolidated in to one number ? A]1970 b]1971 c]1972 d]None of these
M is a 2 digit number which has the property that: The product of factorials of its digitis > sum of factorials How many such number exist ? A]56 b]64 c]63 d]None of these
== > In thest type of questions Focus on what we need to get ( e.g. 150 )
7200 = 25 X 32 X 52 Sumation form = ( 20 + 21 + 22 + 23 + 24 + 25 ) ( 30 + 31 + 32 ) ( 50 + 51 + 52 ) We want to get all factors which were divisible by 150. So minimum number we need here is 150. i.e. 52 * 3 * 2. IN 5s all numbers below 52 will be excluded. IN 3s all numbers below 31 will be excluded Because any number multiplied above this will be a Multiple of 150 IN 2s all numbers below 21 will be excluded. So the number boils down to ( 21 + 22 + 23 + 24 + 25 ) ( 31 + 32 ) ( 52 ) Number of factors will be -- > FIVE 2s * TWO 3s * ONE 5s 5 * 3 * 1 == > 15
100 million bacteria can completely decompose a garbage dump in 15 days and 60 million bacteria can do so in 30 days.If same quantity of garbage is added to the initial quantity of the dump everyday,how many bacteria will be required to completely decompose the dump in 10 days?
a)150mn b)180mn c)140mn d)120mn e)125mn the qn. is nt much clear to.kindly explain with a detailed sol.
100 million bacteria can completely decompose a garbage dump in 15 days and 60 million bacteria can do so in 30 days.If same quantity of garbage is added to the initial quantity of the dump everyday,how many bacteria will be required to completely decompose the dump in 10 days?
a)150mn b)180mn c)140mn d)120mn e)125mn the qn. is nt much clear to me.kindly explain with a detailed sol.
The series of numbers(1, 1/2, 1/3, .. 1/9172)is taken. Now 2 numbers are taken from this series say (x,y). Then the operation x+y+x.y is performed to get the consolidated number. The process is repeated. What will be the value of the set after all the numbers are consolidated in to one number ? A]1970 b]1971 c]1972 d]None of these
M is a 2 digit number which has the property that: The product of factorials of its digitis > sum of factorials How many such number exist ? A]56 b]64 c]63 d]None of these
hi kambarish, take first two numbers 1 and 1/2 will give 2 then 2 and 1/3 will give 3 so trend is that last denominator comes as answer thereby conclude that at the end 1972 will come..ans=1972
for the next one..ans=63 check manually u will get!!!
Two friends Shayam and kailash own two versions of a car. Shayam owns the diesel version of the car, while kailash owns the petrol version. Kailashs car gives an average that is 20% higher than Shayams (in terms of litres per kilometre). It is known that petrol costs 60% of its price higher than diesel. The ratio of the cost per kilometer of Kailashs car to Shayams car is (a) 3:1 (b) 1:3(c) 1.92:1(d)Cant be determined
answer should be (c) for this but it's given as (a) in Arun Sharma..
can u pls temme the approach for the below qns also... pls
Kailashs car gives an average that is 20% higher than Shayams (in terms of litres per kilometre). It is known that petrol costs 60% of its price higher than diesel.
2. If shyams car gives an avg of 20km/lit, then the difference in the cost of travel per km between the two cars is: a. Rs4.3 b. Rs3.5 c. Rs 2.5 d. cant be determined.
3. For the above qn, the ratio of the cost per km of shyams travel to kailash travel is a. 3:1 b. 1:3 c. 1:1.92 d. 2:1 e. cant be determined
4. If diesel costs Rs.12.5 per litre, then the difference in the cost of travel per km b/w Kailashs and shyams is (assume an avg of 20km/lit for shyams car and also assume that petrol is 50% of its own price higher than diesel) a. Rs 1.75 b. Rs 0.875 c. Rs. 1.25 d. Rs 1.125 e. None of the above
can u pls temme the approach for the below qns also... pls
Kailashs car gives an average that is 20% higher than Shayams (in terms of litres per kilometre). It is known that petrol costs 60% of its price higher than diesel.
2. If shyams car gives an avg of 20km/lit, then the difference in the cost of travel per km between the two cars is: a. Rs4.3 b. Rs3.5 c. Rs 2.5 d. cant be determined.
3. For the above qn, the ratio of the cost per km of shyams travel to kailash travel is a. 3:1 b. 1:3 c. 1:1.92 d. 2:1 e. cant be determined
4. If diesel costs Rs.12.5 per litre, then the difference in the cost of travel per km b/w Kailashs and shyams is (assume an avg of 20km/lit for shyams car and also assume that petrol is 50% of its own price higher than diesel) a. Rs 1.75 b. Rs 0.875 c. Rs. 1.25 d. Rs 1.125 e. None of the above
