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( c ) 15 litres
Mohammad Gauri
I am looking for a qualified chartered accountant in my profile who is known for his excellence in the field.
30
a = 2
d = 5 – 2 = 3
Then 12th term is :
This in arithmetic progression
L = a + (12 – 1) d
L = 2 + 11 * 3
L = 2 + 33 = 35
20000
x and y can be equal to 1..
If so,then he gets 1/1 of Rs.10 which is equal to 10 and
again 1/1 of Rs.10=Rs.10..so,he gets total of 20..and
returns 20..so,no gain and no loss..
If x=1,y=2,he gets, 5+20=25..and returns 20..so he may not
lose..
So,whatever be the values of x and y,only these two answers
are possible..
so its a)He never loses..
wake up guys and see modulo % can return only integer value.
So answer will be 1 and all other answers are wrong. You can
run your equation in C code and check the output.
A. Rs 140
7+8+11=26
B: 7/26*520= 140
8 cubes
Total number of pairs is NC2^{N}C_2NC2. Number of pairs standing next to each other = N. Therefore, number of pairs in question = NC2^{N}C_2NC2 – N = 28/2 = 14. If N = 7,
7C2 – 7 = 21 – 7 = 14….
N =7
2