Q: Find the odd man out. 2, 5, 10, 17, 26, 37, 50, 64
Solution: (1*1)+1 , (2*2)+1 , (3*3)+1 , (4*4)+1 , (5*5)+1 , (6*6)+1 , (7*7)+1 , (8*8)+1 But, 64 is out of pattern. The correct answer is: D) 64
Q: What is the probability of getting a sum 9 from two throws of a dice? . P(E) =n(E)/n(S)=4/36=1/9. }
Solution: In two throws of a die, n(S) = (6 x 6) = 36. Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)
The correct answer is: C) 1/9Q: What percent of 7.2 kg is 18 gms ?
Solution: Required percentage = (18/7200 * 100)% = 1/4% = 0.25% The correct answer is: A) .25%
Q: In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
Solution: Required number of ways = (7C5*3C2) = (7C2*3C1) = 63 The correct answer is: B) 63
Q: Find the Odd One Out? 4, 6, 12, 30, 81, 315.
Solution: Given the number series is 4, 6, 12, 30, 81, 315. Here it follows a pattern that 4 4 x 1.5 = 6 6 x 2 = 12 12 x 2.5 = 30 30 x 3 = 90 not equals to 81 90 x 3.5 = 315 Hence the odd one in the series is 81. The correct answer is: C) 81
Q: The difference between compound interest and simple interest on a sum for two years at 8% per annum, where the interest is compounded annually is Rs.16. if the interest were compounded half yearly , the difference in two interests would be nearly
Solution: For 1st year S.I =C.I. Thus, Rs.16 is the S.I. on S.I. for 1 year, which at 8% is thus Rs.200 i.e S.I on the principal for 1 year is Rs.200 Principle = Rs.100*2008*1 = Rs.2500 Amount for 2 years, compounded half-yearly Rs.2500*1+41004=Rs.2924.4 C.I = Rs.424.64 Also, S.I=Rs.2500*8*2100=Rs.400 Hence, [(C.I) - (S.I)] = Rs. (424.64 - 400) = Rs.24.64 The correct answer is: A) Rs.24.64
Q: The total age of A and B is 12 years more than the total age of B and C. C is how many years younger than A ?
Solution: (A+B) - (B+C) = 12 A - C = 12. C is younger than A by 12 years. The correct answer is: A) 12
Q: In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
Solution: Let , S - sample space E - event of selecting 1 girl and 2 boys. Then, n(S) = Number ways of selecting 3 students out of 25 = 25C3 = 2300. n(E) = 10C1×15C2 = 1050. P(E) = n(E)/n(s) = 1050/2300 = 21/46 The correct answer is: A) 21/46
Q: A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?
Solution: ns=C230 Let A be the event of getting two oranges and B be the event of getting two non-defective fruits. and A∩B be the event of getting two non-defective oranges ∴ PA=C220C230, PB=C222C230 and PA∩B=C215C230 ∴PA∪B=PA+PB-PA∩B = C220C230+C222C230-C215C230=316435 The correct answer is: C) 316/435
Q: Insert the missing number. 7, 26, 63, 124, 215, 342, (....)
Solution: Numbers are (23 - 1), (33 - 1), (43 - 1), (53 - 1), (63 - 1), (73 - 1) etc. So, the next number is (83 - 1) = (512 - 1) = 511. The correct answer is: D) 511
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