Intially it was 35 gms and price was 12250 (35^2=1225)prices varies with square of the weight
Later 2 : 5=10gms and 25 gms so their price will be 1000+6250=7250
so loss=12250-7250
oh sorry answer given in Arun sharma is 6000..
oh sorry answer given in Arun sharma is 6000..
EagleMenace Saysoh sorry answer given in Arun sharma is 6000..
Answers in Arun Sharma are often wrong.
Hi All is this the place to discuss DI from Arun Sharma too. Please let me know ASAP. If not which one will be the right thread to post queries for Arun Sharma DI.
hey puys, I have a doubt in Arun sharma`s profit and loss level 2 difficulty problem...
A driver of a autorickshaw makes a profit of 20% on every trip when he carries 3 passengers and the price of petrol is Rs30 a litre. Find the percentage profit for the same journey if he goes for 4 passengers/trip and the price petrol is reduced to 24 a trip? (Assume that revenue/passenger is the same in both cases)..
1)33.33
2)65.66
3)100
4)data inadequate
Profit is 20%. Therefore earnings are 1.2*30 = 36. Therefore revenue per passenger is 36/3= 12.
New earnings = 12*4= 48
New cost of fuel= 24.
Therefore new profit is (48-24)/24 = 24, which is 100% profit. 😃
arijitprepares SaysHi All is this the place to discuss DI from Arun Sharma too. Please let me know ASAP. If not which one will be the right thread to post queries for Arun Sharma DI.
yes go ahead and post frd
Hi puys, I was just going through the Geometry theory in Arun Sharma. Got a couple of doubts, would be glad if someone helped me out.
1. It says that in an equilateral triangle
r=h/3=a/2*sqrt(3) where r is the inradius, a is the side and h is the height of the equilateral triangle. But using the earlier generic formula for any triangle i.e
Area = r*s where r is the inradius and a is the side
I get the following derivation i.e.
r=h/2=sqrt(3)*a/4. Could someone clarify?
2. It says that ' The incentre and circumcentre lies at a point that divides the height in the ratio 2:1' What does this mean?
3. For an equilateral triangle it says
Area= r*s
Then it says that s=(a+b+c)/2 where a,b,c are sides of a triangle. How did this happen?
Regards,
Intellageek
hii puys i was doing NUMBER SYSTEM from ARUN SHARMA...gt a cpl of doubts wish if any1 could help
1 Three runners running around a circular track, can complete one revolution in 2,4 and 5.5 hrs resp. When will they meet at the starting pt
a.22 b.33 c.11 d.44
2. Find the greatest no. which will divide 215, 167, and 135 so as to leave the same remainder in each case.
a.64 b.32 c.24 d.16
Plzz explain the logic also
thx
hii puys i was doing NUMBER SYSTEM from ARUN SHARMA...gt a cpl of doubts wish if any1 could help
1 Three runners running around a circular track, can complete one revolution in 2,4 and 5.5 hrs resp. When will they meet at the starting pt
a.22 b.33 c.11 d.44
2. Find the greatest no. which will divide 215, 167, and 135 so as to leave the same remainder in each case.
a.64 b.32 c.24 d.16
Plzz explain the logic also
thx
hii puys i was doing NUMBER SYSTEM from ARUN SHARMA...gt a cpl of doubts wish if any1 could help
1 Three runners running around a circular track, can complete one revolution in 2,4 and 5.5 hrs resp. When will they meet at the starting pt
a.22 b.33 c.11 d.44
Time at which A completes one round are 2,4,6,8,....
Time at which B completes one round are 4,8,12,16....
Time at which C completes one round are 5.5,11,16.5,22....
So the option satisfying all the three are d.44
hii puys i was doing NUMBER SYSTEM from ARUN SHARMA...gt a cpl of doubts wish if any1 could help
1 Three runners running around a circular track, can complete one revolution in 2,4 and 5.5 hrs resp. When will they meet at the starting pt
a.22 b.33 c.11 d.44
2. Find the greatest no. which will divide 215, 167, and 135 so as to leave the same remainder in each case.
a.64 b.32 c.24 d.16
Plzz explain the logic also
thx
1. In such case take LCM
LCM(2,4,5.5) = Lcm(120,240,330) = 11*10*8*3 mint
In hours = 11*10*8*3 /60 =44
2. Consider r be the reminder; then the number (215-r),(167-r),and (135-r) are exactly divisible by the required number.Its obvious that if two numbers be divisible bt a certain number,then their difference is also disvisible by the number.Hence
(215-r)-(167-r) ,(167-r)-(135-r) , and (215-r)-(135-r) are divisible by required number.
or (215- 167) , (167-135) , and (215-135)
the required number = HCF((215- 167) , (167-135) , and (215-135)) = HCF(48,32,80) = 16
hii puys i was doing NUMBER SYSTEM from ARUN SHARMA...gt a cpl of doubts wish if any1 could help
1 Three runners running around a circular track, can complete one revolution in 2,4 and 5.5 hrs resp. When will they meet at the starting pt
a.22 b.33 c.11 d.44
2. Find the greatest no. which will divide 215, 167, and 135 so as to leave the same remainder in each case.
a.64 b.32 c.24 d.16
Plzz explain the logic also
thx
1.d->44 For them to meet together, they need to be at the same place at the same time. for that to happen, the number has to be a common multiple, the first time they meet will be the lowest common multiple.
2.d->16
Never use logic when you can use options and never use options without logic. (just divide the numbers and see)
The process by which to actually solve it is almost immaterial. The funda is X-r, Y-r, Z-r (r=common remainder). When you take cat/xat, what matters is the answer and not how you get it.
200! is divisible by (x!)^n...whats the max. possible value of n when x = 70....plz give ans with explanation...i'm gettin 2,bt dat's not d ans
arnabsuave Says200! is divisible by (x!)^n...whats the max. possible value of n when x = 70....plz give ans with explanation...i'm gettin 2,bt dat's not d ans
my answer is : 31
Basically you have to find the highest power of 70 in 200!
for that prime factors of 70= 2*5*7
So we just need to calculate the power of 7 in 200
so 200/7=28/7=3 ... So answer =28+3=31
my answer is : 31
Basically you have to find the highest power of 70 in 200!
for that prime factors of 70= 2*5*7
So we just need to calculate the power of 7 in 200
so 200/7=28/7=3 ... So answer =28+3=31
Should be 32, you just missed out on an easy calculation :)
200/7 = 28
28/7 = 4
4/7 = 0
28+4+0 = 32
hii puys i was doing NUMBER SYSTEM from ARUN SHARMA...gt a cpl of doubts wish if any1 could help
1 Three runners running around a circular track, can complete one revolution in 2,4 and 5.5 hrs resp. When will they meet at the starting pt
a.22 b.33 c.11 d.44
2. Find the greatest no. which will divide 215, 167, and 135 so as to leave the same remainder in each case.
a.64 b.32 c.24 d.16
Plzz explain the logic also
thx
For 2nd question the method is :
when the remainder is same in each case and we have to find the greatest no that can divide the three numbers, you just need to take the HCF of the difference between the numbers.
like 215-167=48 and 215-135=80
So HCF(48,160)= 16
Concept: N1=q*a +b;
N2=q*c+b
So N2-N1= q(c-a) ... clearly q divides n2-n1.. so in every case take the HCF of the differences
Should be 32, you just missed out on an easy calculation :)
200/7 = 28
28/7 = 4
4/7 = 0
28+4+0 = 32
these kind of mistakes cost very dearly in CAT. :banghead:
Should be 32, you just missed out on an easy calculation :)
200/7 = 28
28/7 = 4
4/7 = 0
28+4+0 = 32
but d answer is 3...d constrain where d denominator(let the denominator b x!) is a factorial is often d highest prime no. below d denominator no.(ie highest prime no. below x).. so basically here 67 shud b d constrain(highest prime no. below 70)..bt 200! has max. 2 67's...so,i got d ans as 2...
i learned dis concept in a thread in pagalguy.com only...bt d ans given to dis question dere is 3...
plzzz help guys!!!
my answer is : 31
Basically you have to find the highest power of 70 in 200!
for that prime factors of 70= 2*5*7
So we just need to calculate the power of 7 in 200
so 200/7=28/7=3 ... So answer =28+3=31
it's not 70 here,remember,it's 70!...so calculation here has lots of prime no.,eg 23,29,31,37,41 etc....
no idea though how ans has been given as 3....if v apply d logic dat 67 will b d constrain v shud get 2 as ans!!!!
arnabsuave Says200! is divisible by (x!)^n...whats the max. possible value of n when x = 70....plz give ans with explanation...i'm gettin 2,bt dat's not d ans
200!=2^197*3^97*5^49*7^32*11^19*13^16*17^11*19^10*23^8*29^6*31^6*37^5*41^4*43^4*47^4*53^3*59^3*61^3*67*2......
70!=2^67*3^32*5^16*7^11*11^6*13^5*17^4*19^3*23^3....67^1
So,it is clear that the answer is 2.
Now,Arun Sharma has some misprints so you can never be sure if the answer is correct or not.Just the basics are important.If your approach is right,then answer should not be a problem 😃