A potter asked his two sons to sell some pots inthe market. The amount received for each potwas same as the number of pots sold. The twobrothers spent the entire amount on somepackets of potato chips and one packet of bananachips. One brother had the packet of banana chipsalong with some packets of potato chips, whilethe other brother just had potato chips. Eachpacket of potato chips costs 10/- and the packetof banana chips costs less than 10/-. Thepackets of chips were divided between the twobrothers so each brother received equal numberof packets. How much money should one brothergive to the other to make the division financiallyequitable?
A potter asked his two sons to sell some pots inthe market. The amount received for each potwas same as the number of pots sold. The twobrothers spent the entire amount on somepackets of potato chips and one packet of bananachips. One brother had the packet of banana chipsalong with some packets of potato chips, whilethe other brother just had potato chips. Eachpacket of potato chips costs 10/- and the packetof banana chips costs less than 10/-. Thepackets of chips were divided between the twobrothers so each brother received equal numberof packets. How much money should one brothergive to the other to make the division financiallyequitable?
A potter asked his two sons to sell some pots inthe market. The amount received for each potwas same as the number of pots sold. The twobrothers spent the entire amount on somepackets of potato chips and one packet of bananachips. One brother had the packet of banana chipsalong with some packets of potato chips, whilethe other brother just had potato chips. Eachpacket of potato chips costs 10/- and the packetof banana chips costs less than 10/-. Thepackets of chips were divided between the twobrothers so each brother received equal numberof packets. How much money should one brothergive to the other to make the division financiallyequitable?
A potter asked his two sons to sell some pots inthe market. The amount received for each potwas same as the number of pots sold. The twobrothers spent the entire amount on somepackets of potato chips and one packet of bananachips. One brother had the packet of banana chipsalong with some packets of potato chips, whilethe other brother just had potato chips. Eachpacket of potato chips costs 10/- and the packetof banana chips costs less than 10/-. Thepackets of chips were divided between the twobrothers so each brother received equal numberof packets. How much money should one brothergive to the other to make the division financiallyequitable?
A potter asked his two sons to sell some pots inthe market. The amount received for each potwas same as the number of pots sold. The twobrothers spent the entire amount on somepackets of potato chips and one packet of bananachips. One brother had the packet of banana chipsalong with some packets of potato chips, whilethe other brother just had potato chips. Eachpacket of potato chips costs 10/- and the packetof banana chips costs less than 10/-. Thepackets of chips were divided between the twobrothers so each brother received equal numberof packets. How much money should one brothergive to the other to make the division financiallyequitable?(5) 7
Number of pots = n
Amount received for each pot = n
Packets of potato = p , Cost of potato = 10, Banana cost = b ( b
p + 1 was equally divided among brothers [ p has to be odd]
so, A has (1 + (p-1)/2) [1 banana + rest potato] and B has (p + 1)/2 [all potatoes]
Money with A = b + 5*(p-1) = 5p + (b-5)
Money with B = 5*(p + 1) = 5p + 5
Difference between the amount = (10 - b)
Sum of the money = 10*p + b = n^2
p = odd only
p = 1, b = 6 => n = 4
p = 3, b = 6 => n = 6
p = 19, b= 6 => n = 14
p = 25, b= 6 => n = 16
b is always 6 somehow .. (Need to prove it => since p is odd -> second last digit is odd, and since second last digit is odd, the perfect square must end in 6 :), hence b = 6 )
A potter asked his two sons to sell some pots inthe market. The amount received for each potwas same as the number of pots sold. The twobrothers spent the entire amount on somepackets of potato chips and one packet of bananachips. One brother had the packet of banana chipsalong with some packets of potato chips, whilethe other brother just had potato chips. Eachpacket of potato chips costs 10/- and the packetof banana chips costs less than 10/-. Thepackets of chips were divided between the twobrothers so each brother received equal numberof packets. How much money should one brothergive to the other to make the division financiallyequitable?
Amount received = x²
Potato chips bought = 2p - 1
Amount spent = 20p + a - 10, where a
So, one brother paid 10p and other one paid 10p - 10 + a
=> One will pay (10 - a)/2 to other one
x² = 20p - 10 + a
Unit digit of x² is 'a' (so unit digit can not be 0) and tens digit is always odd
Now, all the perfect square having tens digit as odd have unit digit as 6
Find the complex number having the least possible argument and satisfying z-5i............I dont have the oa.......so plz post your solutions with approach
Number of pots = nAmount received for each pot = nPackets of potato = p , Cost of potato = 10, Banana cost = b ( bp + 1 was equally divided among brothers [ p has to be odd]so, A has (1 + (p-1)/2) [1 banana + rest potato] and B has (p + 1)/2 [all potatoes]Money with A = b + 5*(p-1) = 5p + (b-5)Money with B = 5*(p + 1) = 5p + 5 Difference between the amount = (10 - b)Sum of the money = 10*p + b = n^2p = odd only p = 1, b = 6 => n = 4p = 3, b = 6 => n = 6p = 19, b= 6 => n = 14p = 25, b= 6 => n = 16b is always 6 somehow .. (Need to prove it => since p is odd -> second last digit is odd, and since second last digit is odd, the perfect square must end in 6 , hence b = 6 ) => money to be given = 2 ?
Amount received = x²Potato chips bought = 2p - 1Amount spent = 20p + a - 10, where a So, one brother paid 10p and other one paid 10p - 10 + a=> One will pay (10 - a)/2 to other onex² = 20p - 10 + aUnit digit of x² is 'a' (so unit digit can not be 0) and tens digit is always oddNow, all the perfect square having tens digit as odd have unit digit as 6=> a = 6So, money to be paid = (10 - 6)/2 = 2
Given that product of a two-digit number AB and 74 is a three-digit number EEE, where A, B and E are distinct non zero digits. What is the value of €˜E €™? A. 9 B. 7 C. 6 D. 8 E. None of these
Given that product of a two-digit number AB and 74 is a three-digit number EEE, where A, B and E aredistinct non zero digits. What is the value of €˜E €™?A. 9B. 7C. 6D. 8E. None of these
Given that product of a two-digit number AB and 74 is a three-digit number EEE, where A, B and E aredistinct non zero digits. What is the value of €˜E €™?A. 9B. 7C. 6D. 8E. None of these
Total number of bananas with three friends €“ Moti, Sumit and Manky €“ is 10. The sum of the reciprocals of the number of bananas with the three friends is 1. If the number of bananas with each of the three friends is an integer, which of the following could be the difference between the number of bananas with Moti and Sumit? A. 2 B. 3 C. 0 D. 1 E. Either (A) or (C) (2
Total number of bananas with three friends €“ Moti, Sumit and Manky €“ is 10. The sum of the reciprocals ofthe number of bananas with the three friends is 1. If the number of bananas with each of the three friends isan integer, which of the following could be the difference between the number of bananas with Moti andSumit?A. 2B. 3C. 0D. 1E. Either (A) or (C) (2
Given that product of a two-digit number AB and 74 is a three-digit number EEE, where A, B and E aredistinct non zero digits. What is the value of €˜E €™?A. 9B. 7C. 6D. 8E. None of these
If 'a' is an odd natural number and 'b' is an even natural number, what is the total number of solutions of the equation ab = 2a + b + 598? A. 16 B. 12 C. 5 D. 6 E. 8 (2
If 'a' is an odd natural number and 'b' is an even natural number, what is the total number of solutions of theequation ab = 2a + b + 598?A. 16B. 12C. 5D. 6E. 8 (2
If 'a' is an odd natural number and 'b' is an even natural number, what is the total number of solutions of theequation ab = 2a + b + 598?A. 16B. 12C. 5D. 6E. 8 (2
B. 12?
ab = 2a + b + 598 (2k-1)(2p) = 4k-2 + 2p + 598 4kp - 2p = 4k -2 + 2p + 598 4k(p-1) = 4p + 596 k(p-1) = p + 149 k = (p+149)/(p-1) k = 1 + 150/(p-1) Find p to get positive integral values for k 150 = 2*3*5^2 Total factors = 2*2*3 = 12
If 'a' is an odd natural number and 'b' is an even natural number, what is the total number of solutions of theequation ab = 2a + b + 598?A. 16B. 12C. 5D. 6E. 8 (2
.. (Need to prove it 
=> since p is odd -> second last digit is odd, and since second last digit is odd, the perfect square must end in 6 :), hence b = 6 ) 