Quantitative Aptitude - Numbers and Algebra
- Option : A
- Explanation : let average of the seven men be x ∴ x = (10x6+(x+3)) / 7 => 60 + x +3 = 7x => x = 63/6 = 10.5 Total amount = 10.5 x 7 = 73.5
- Option : B
- Explanation :
6 - 4 = 2, 7 - 5 = 2, 8 - 6 = 2, 9 - 7 = 2, 10 -8 = 2
L.C.M. of 6, 7, 8, 9, 10 = 2520;
Greatest number of 6 digits = 999999
2520 x 396 + 2079 = 999999
Remainder = 2079.
Subtract 2079 from 999999, then we get 999999- 2079 = 997920.
Subtract 2 from this number to get required number, which is 997918 and which will give the remainders 4, 5, 6, 7, 8 when divided by 5, 6, 7, 8, 9 respectively.
- Option : D
- Explanation : Number of students who answered 1 or more questions wrongly = 2N-1.
Number of students who answered 2 or more questions wrongly = 2N-2
Hence number of students who answered 1 question wrongly = 2N-1- 2N-2 = 2N-2
Similarly it can be shown that number of students who answered 2 questions wrongly = 2N-2 - 2N-3 = 2N-3
Similarly we can find number of students who answered K questions wrongly where K ≥ 3
Hence total number of questions attempted wrongly
s = 2N-2+2(2N-3)+3(2N-4 )+...+(N-1)(20) + N(1) ........(1)
∴ s/2 = 2N-3+ 2(2n-4) +.......+ (N-2)(20) + (N-1)/2 + N/2.......... (2)
Substracting equations (1) from (2)
s/2 = 2N-2 + 2N-3 +...+20+1/2
=> s = 2N + 2N-2+2N-3+...+1 = 2N -1
= 8191
=> N = 13
- Option : A
- Explanation :
Worst scenario is when other four get equal number of votes.
Let the winning candidate get x votes.
∴ x > (261-x) / 4
=> x > 52
x = 53
30. If 7x + 6y = 420, x and y are natural numbers, then what can be said about x?
- Option : B
- Explanation :
7x + 6y = 420
Equation is of the form:
7x + even number = even number.
∴7x has to be even
Hence x has to be even.
