In a pocket of A, the ratio of Re1 coins, 50p coins and 25p coins can be expressed by three consecutive odd prime numbers that are in ascending order. The total value of coins in the bag is Rs 58. If the number of Re 1, 50p, 25p coins are reversed, find the new total value of coins in the pocket of A?
Answer: D Since the ratio of the number of Re1 , 50p and 25p coins can be represented by 3 consecutive odd numbers that are prime in ascending order, the only possibility for the ratio is 3:5:7. Let the number of Re1, 50p and 25p coins be 3k, 5k and 7k respectively. Hence, total value of coins in paise => 100(3k) + 50(5k) + 25(7k) = 725k = 5800 => k= 8. If the number of coins of Re1 ,50p and 25p is reversed, the total value of coins in the bag (in paise) = 100(7k) + 50(5k) + 25(3k) = 1025k (In above we find the value of k). => 8200 = Rs 82.
Q. No. 2:
Manish, Rahul and Bharti have some stones with each of the. Five times the number of stones with Rahul equals seven times the number of stones with Manish while five times the number of stones with Manish equals seven times the number of stones with Bharti. What is the minimum number of stones that can be there with all three of them put together?
Answer: B Let the stones with Manish, rahul and Bharti be m,r and b respectively. Given, 5r=7m and 5m =7b => 25r=35m and 35m = 49b => 25r=35m=49b=k => r/49 = m/35 = b/25 The least possible integral values for r,m,b will be r= 49, m= 35 and b=25 => Total = 49+35+25 = 109.
There are N numbers of gold biscuits in the house, in which four people are lived. If the first men woke up and divided the biscuits into 5 equal piles and found one extra biscuit. He took one of those piles along with the extra biscuit and hid them. He then gathered the 4 remaining piles into a big pile, woke up the second person and went to sleep. Each of the other 3 persons did the same one by one i.e divided the big pile into 5 equal piles and found one extra biscuit. Each hid one of the piles along with the extra biscuit and gathered the remaining 4 piles into a big pile.
Answer: D Suppose N= 5x+1 A took x+1 biscuit. Now 4x is of the form 5y+1 then x must be in the form 5z+4 => 4(5z+4) = 5y+1 => y= 4z+3 and x= 5z+4 The ratio of number of biscuits that A and B took is [(5z+4)+1] : [(4z+3)+1] = 5:4 So, we can say that any two successive persons A,B,C and D takes coins in the ratio of 5:4 Let the number of biscuits that A,B,C and D took be a,b,c and d respectively. a:b = b:c = c:d = 5:4 a:b:c:d = 125:100:80:64 => a = 125k => x= 125k-1 and N= 5x +1 = 625k-4 As, N>1000, the least value of N is when k =2 => N= 1246
Q. No. 2:
If N<1000, how many biscuits were left after the fourth man took his share?
Answer: C Suppose N= 5x+1
A took x+1 biscuit.
Now 4x is of the form 5y+1 then x must be in the form 5z+4
=> 4(5z+4) = 5y+1
=> y= 4z+3 and x= 5z+4
The ratio of number of biscuits that A and B took is
[(5z+4)+1] : [(4z+3)+1] = 5:4
So, we can say that any two successive persons A,B,C and D takes coins in the ratio of 5:4
Let the number of biscuits that A,B,C and D took be a,b,c and d respectively.
a:b = b:c = c:d = 5:4
a:b:c:d = 125:100:80:64
=> a = 125k
=> x= 125k-1 and N= 5x +1 = 625k-4 => N<1000, then k=1 => N= 621 => 621 = 5(124)+3 4(124)= 5(99)+1 4(99) = 5(79) +1 4(79) =5(63)+1 After the fourth man took his share 5(63)+1, the biscuits lefts is 4(63) =252
Q. No. 3:
In the previous question, what is the sum of the number of biscuits hidden by the last 2 men.
Answer: B Suppose N= 5x+1
A took x+1 biscuit.
Now 4x is of the form 5y+1 then x must be in the form 5z+4
=> 4(5z+4) = 5y+1
=> y= 4z+3 and x= 5z+4
The ratio of number of biscuits that A and B took is
[(5z+4)+1] : [(4z+3)+1] = 5:4
So, we can say that any two successive persons A,B,C and D takes coins in the ratio of 5:4
Let the number of biscuits that A,B,C and D took be a,b,c and d respectively.
a:b = b:c = c:d = 5:4
a:b:c:d = 125:100:80:64
=> a = 125k
=> x= 125k-1 and N= 5x +1 = 625k-4 => N<1000, then k=1 => N= 621 => 621 = 5(124)+3 4(124)= 5(99)+1 4(99) = 5(79) +1 4(79) =5(63)+1 The number of biscuits hidden by 3rd and the 4th men is 79+1 = 80 and 63+1=64 i.e a total of 144.
Q. No. 4:
A bus and a truck are available to cross a jungle. The speed of the truck is thrice that of the bus. The capacity of the truck is 50 persons and that of bus is 30 persons. The average occupancy of the bus is twice that of the truck. The tickets for the bus and the truck cost Re 1 and Re 1.50 respectively. What is the ratio of the average rupee collection of the truck to that of the bus in a day?Assume there is no wastage time between trips and the occupancy of the bus/truck is defined as the ratio of the actual number of persons boarding it and its capacity.
Answer: A Average Rupee collection = Speed* capacity * Occupancy * Ticket rate. ratio of average Rupee collection of truck to that of boat= product of above rates. => (3*50*1*1.5) : (1*30*2*1) = 15:4