Reasoning: (A = 2/3 B and B = 1/4 C) = A/B = 2/3 and B/C = 1/4
A:B = 2:3 and B:C = 1:4 = 3:12
A:B:C = 2:3:12
A;s share = 510 * 2/17 = Rs. 60
B's share = 510 * 3/17 = Rs. 90
C's share = 510 * 12/17 = Rs. 360.
Ans: (B)
Question: If Rs. 510 be divided among A, B, C in such a way that A gets 2/3 of what B gets and B gets 1/4 of what C gets, then their shares are respectively:
Options:
(A) Rs. 60, Rs. 90, Rs. 389
(B) Rs. 60, Rs. 90, Rs. 360
(C) Rs. 60, Rs. 90, Rs. 368
(D) Rs. 60, Rs. 90, Rs. 329
(E) Rs. 60, Rs. 30, Rs. 329
Rationale: First even minus first odd = 2-1 = 1;
The sum of first 2 even integers minus the sum of first 2 odd integers = (2+4)-(1+3) = 2;
The sum of first 3 even integers minus the sum of first 3 odd integers = (2+4+6)-(1+3+5) = 3;
We can see the patterns here, so the sum of first 10 positive even integers minus the sum of first 10 positive odd integers will be 10.
The answer: (A)
Question: If X is the sum of first 10 positive even integers and Y is the sum of first 10 positive odd integers, what is the value of x-y?
Please try to explain your answers
Options:
(A) 10
(B) 25
(C) 50
(D) 75
(E) 100
Reasoning: {Total} = {finance} + {marketing} - {both} + {neither}
100 = 25 + 54 - {both} + 40
{both} = 19%.
Ans: (D)
Question: Walking across campus, a student interviewed a group of students. 25% of the students took a finance class last semester, 54% took a marketing class last semester, and 40% took neither a finance nor a marketing class last semester. What percent of the students in the group took both a finance and a marketing class?
Options:
(A) 60%
(B) 50%
(C) 25%
(D) 19%
(E) 10%
Rationale: The differences between two successive terms from the beginning are 7, 5, 7, 5, 7, 5. So, 40 is wrong.
The answer: (C)
Question: Find the invalid no.from the following series 13, 18, 25, 30, 37, 40
Options:
(A) 23
(B) 26
(C) 40
(D) 42
(E) 49
Reasoning: {Total} = {M} + {N} - {Both}
{Total} = 47 + 28 - 12 = 63.
Ans: (E)
Question: There are 47 distinct numbers in set M, there are 28 distinct numbers in set N, and there are 12 distinct numbers that are in both sets M and N. Set H is the set containing the elements that are in at least one of sets M and N. How many elements are in set H?
Options:
(A) 39
(B) 40
(C) 51
(D) 58
(E) 63
Reasoning: Number of ways of selecting 2 fiction books = 7C2
Number of ways of selecting 2 non fiction books = 6C2
7C2 * 6C2 = 21 * 15 = 315
Ans: (C)
Question: There are 7 fictions and 6 non-fictions. How many cases are there such that 2 fictions and 2 non-fictions are selected from them?
Options:
(A) 90
(B) 120
(C) 315
(D) 180
(E) 200