Solution:
5 greater than 4.
Answer: (a)
If fares for cities L and M from K are x and y respectively, then
2x + 3y = 77 or 6x + 9y = 231...(1)
3x + 2y = 73 or 6x + 4y = 146...(2)
subtracting 2 from 1, we get
5y = 85
y = Rs. 17 => x = Rs. 13
4, 14, 24, 34, 44, 54, 64, 74, 84, 94 => has 11 number of (4's)
40, 41, 42, 43, 45, 46, 47, 48, 49 => has 9 number of (4's)
So, totally (11 + 9 = 20) number of 4's in 1 to 100.
In case (A), x and y could, for example, both be equal fractions and x/y would be an integer.
You can find the integers which when divided by 5 have a remainder 2 by adding 2 to all multiples of 5. So we have n = 7 , 12, 17, 22 etc.
From this series we can see that n does not have to be odd.
Also n + 1 can be a prime because, for example, 12 + 1 = 13
And (n + 2) / 7 has a remainder 2 in some cases but not all.
Remember the question asks us for what MUST be true, and we see that none of the statements are true in all cases. However, adding 3 to any of the values of n will always give a multiple of 5.
Given exp . is like this
so ans = 854 - 276 = 578
Ans.
4.5 = 7 x 0.5 + 1
5.5 = 4.5 x 1 + 1
12 = 5.5 x 2 + 1
49 = 12 x 4 + 1
? = 49 x 8 + 1, i.e. ? = 393
Let the number be x
When 10 is added, number = x + 10
When 7 is multiplied, number = 7 X (x + 10)
According to question,
?[7 X (x+10) / 5] ?5 = 44
?7x + 70 = 49 X 5
?x + 10 = 35,
?x = 25
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