A shopkeeper makes a profit of Q% by selling an object for rs.24/-. Had the cost price and selling price been interchanged ,it would have led to a loss of 62.5Q%.With the latter cost price , what would be the new selling price to get a profit of Q%nthis is q.16 Lod 2 profit and losscan someone help me understand how to get a solution to this.
let the cost price be "x" and the selling price by 24.. so profit% will be 24-x/x = Q Also if the CP and SP are interchanged then 24 will be the cost price and x will be the selling price. Loss% is given by 24-x/24 = 62.5Q/100
But from the first condition 24-x= xQ/100 substitute for second case xQ/24= 62.5Q/100 ==> x= 1500/100 = 15 So cost price is 15 selling price is 24 therefore profit percentage is 60% Now when the CP is 24 profit percentage 60% substitute and get the answer as 38.4
Well the answer is 5. Can't a number like 2*7*1 be possible? Why only prime squares?
sorry i read the question wrongly.. according to the question your logic is correct So the possible numbers will be (1 is not a prime number ) 2*3*5,2*3*7,2*3*11,2*3*13,2*5*7 Hence 5
sorry i read the question wrongly.. according to the question your logic is correctSo the possible numbers will be (1 is not a prime number )2*3*5,2*3*7,2*3*11,2*3*13,2*5*7Hence 5
Ah right. Thanks man.:)
One more question:
How many divisors/factors of 1200 end with only one €˜zero €™?
Ah right. Thanks man.One more question:How many divisors/factors of 1200 end with only one €˜zero €™?1614121011
answer is 10...look when you factorize it you will get 1200=2^4*3*5^2 So to get a single zero any multiplication of 2 with 5 wil give you single zero. so total 4 possibility next 5^2*2 = 50 5*3*2^x here x can be any of the four values 1,2,3,4. Hence 4 posssibilities next 5^2*3*2 = 150 hence 1 possibility.
answer is 10...look when you factorize it you will get 1200=2^4*3*5^2So to get a single zero any multiplication of 2 with 5 wil give you single zero. so total 4 possibilitynext 5^2*2 = 50 5*3*2^x here x can be any of the four values 1,2,3,4. Hence 4 posssibilitiesnext 5^2*3*2 = 150 hence 1 possibility.is the answer right?
The answer is apparently 12.
Tell me if this is right: Like you said, 1200=2^4*3*5^2
So to make the divisors end in one zero we can write this: (2^3*3*5)*10
So, now we cannot allow 2 and 5 to multiply inside the bracket. So, factors possible with 2 &3 will be (3+1)*(1+1)=8
And factors possible with 3 & 5 will be (1+1)*(1+1)= 4
So 8+4= 12. I think this is right. Not very sure though.
The answer is apparently 12.Tell me if this is right: Like you said, 1200=2^4*3*5^2So to make the divisors end in one zero we can write this: (2^3*3*5)*10So, now we cannot allow 2 and 5 to multiply inside the bracket. So, factors possible with 2 &3 will be (3+1)*(1+1)=8And factors possible with 3 & 5 will be (1+1)*(1+1)= 4So 8+4= 12. I think this is right. Not very sure though.
Lovely. A small permutation doubt:There are 6 pups and 4 cats.In how many ways can they be seated in a row so that no cats sit together?Answer: 6!*P(7, 4)Why is it P(7,4) and not C(7,4)? The cats seem the same, why do we bother with order here?
in that case all the pups are also same why 6! we shud consider all the pups and cat as distinct
Number Systems LOD3.43. The remainder when 2^2 + 22^2 + 222^2 + .......... (222.....49 two's)^2 is divided by 9 is?????a.2b.5c.6d.744. The sum of the digits in the multiplication N = 202 x 20002 x 2000002 x .....2(31 zeros) is???a. 112b.160c.144d.cannot be determined
43) 6 44) 160 both have been discussed in our quant section 2-3 days back.
Can neone help me wid this problem pls..!!! time and distance pg 299. Two cyclist start simulataneously towars each other from aurangabad and ellora which are 28 km apart. an hour later they meet up and keep peadaling with the same speed without stopping. The second cyclist arrives at ellora 35 minutes later than the first arrives at aurangabad.Find the spped of the cyclist who started from ellora.
This is question no. 14 in LOD 2 of Time Speed Distance chapter in Arun Sharma's Quant book: Two indian tourists in the US cycled towards each other, one from point A and the other from point B. The first tourist left point A 6 hrs later than the second left point B, and it turned that he travelled 12 km less than the second tourist. After their meeting, they kept cycling with the same speed, and the first tourist arrived at B 8 hours later and the second arrived at A 9 hours later. Find the speed of the faster tourist. (a) 4 km/hr (b) 6 km/hr (c) 9 km/hr (d) 2 km/hr (e) 5 km/hr
Bhai can u pls xplain how to solve this problem.. gettin stuck all d tym... 😐
