Quantitative Aptitude – Modern Maths – P and C – How many two-digit numbers

Quantitative Aptitude – Number System – How many two-digit numbers

Slot -2 – Quantitative Aptitude – Modern Maths – P & C – How many two-digit numbers

 How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?
a) 6

b) 8

c) 7

d) 5

Answer: 6

Solution:

Let the number is ab so
10a+b >3(10b+a)

7a>29b

a/b>29/7

So the possible pair of (a,b) are (5,1), (6,1), (7,1) , (8,1) , (9,1) & (9,2)
Thus total 6 such numbers are possible.

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Quantitative Aptitude – Modern Maths – Permutation and Combination – In a tournament, there are 43

Quantitative Aptitude – Modern Maths - Permutation and Combination – In a tournament, there are 43

Slot -2 – Quantitative Aptitude – Modern Maths – Permutation and Combination – In a tournament, there are 43 junior

In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is?

Answer: 1098

Solution: total number of matches = 43C2 + 51C2 = 2178
Girl vs girl matches at junior level = 153 = (17×18)/2

Thus number of girls = 18
So number of boys at junior level = 43 – 18 = 25

the number of boy versus boy matches in senior level = 276 =(23×24)/2

Number of boys at senior level = 24
Number of girls at senior level = 51 – 24 = 27

So total number of girl vs boys matches =(18×25+27×24)=1098

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Quantitative Aptitude – Modern Maths – Permutation and Combination – How many numbers with

Quantitative Aptitude – Modern Maths - Permutation and Combination – How many numbers with

Slot -1 – Quantitative Aptitude – Modern Maths – Permutation and Combination – How many numbers with two or more

How many numbers with two or more digits can be formed with the digits 1,2,3,4,5,6,7,8,9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?

Answer: 510

Solution :

As all given digits are different natural numbers and we have to use them at most once, ways of arrangements for each set of digits will be exactly once.
Thus Total such numbers = 9(baseC2) + 9(baseC3) +9(baseC4) +9(baseC5) +9(baseC6) + 9(baseC7) +9(baseC8) +9(baseC9) = 2^9 – 2 = 510

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