Answer: B) Father = 40; Son =24 Explanation: Let the present age of the man be 'P' and son be 'Q',Given, P + Q = 64 or Q = (64 - P)Now the man says "I am five times as old as you were, when I was as old as you are",So, P = 5[B - (P - Q)]We get 6P = 10Q,Substitute value for Q,6P = 10(64 - P),Therefore P = 40, Q = 24.

Answer: C) 1001 Explanation: Here by trial and error method, we can obseve that 404404 = 404 x 1001; 415415 = 415 x 1001, etc. So, any number of this form is divisible by 1001.

Answer: D) 4 Explanation: On dividing 627349 by 15, we get remainder = 4.Therefore, the obtained remainder is the least number to be subtracted from the given number so that the the mnumber is divisible by 15.Here 4 is the least number to be subtracted from 627349 so that it is divisible by 15.

Answer: D) None Explanation: As given Kapil's 8th birthday was celebrated 2 years ago, Kapil’s present age = 8 + 2 = 10 years. Father's age is twice of Kapil’s age after 10 years. => F + 10 = 2 (K + 10) = 2 (10+ 10)F = 30years, Avanthi's age =1/6(F) = 30/6 = 5 years.

Answer: D) 555681 Explanation: Given number is 987 = 3 x 7 x 47. So, required number must be divisible by each one of 3, 7, 47. None of the numbers in 553681 and 555181 are divisible by 3. While 556581 is not divisible by 7.Correct answer is 555681.

Answer: C) 668, 4023 Explanation: Let the smaller number be k.Then, larger number = (3355 + k)Therefore 3355 + k = (6k + 15)‹=› 5k = 3340‹=› k = 668. Therefore, the smaller number k = 668 and now the larger number = 3355+668 = 4023.

Answer: A) 15 days Explanation: In one day work done by K = 1/20 Work done by K in 2 days = 1/10   Work done by K,L,M on 3rd day =1/20 + 1/30 + 1/60 = 1/10 Therefore, workdone in 3 days is = 1/10 + 1/10 = 1/5   1/5th of the work will be completed at the end of 3 days. The remaining work = 1 - 1/5 = 4/5 1/5 of work is completed in 3 days Total work wiil be done in 3 x 5 = 15 days.

Answer: A) 3 times Explanation: Let us assume A is the faster and B is slower, A takes 10s to finish a lap of 50 m, B takes 17s(approx) to finish a lap of 50 m.So at 11th sec A starts his 2nd lap by the time B will be in between so that is the first time they will meet each other, at 21st sec A starts his 3rd lap by the time B will be coming back to finish his 2nd lap. So that is the second time they meet in opposite. Then at 31st sec A will start his 4th lap by the time B would not even finish 2nd lap as he take 34(2*17) sec to finish 2 laps. So at 35th sec B will start his 3rd lap by the time A will be coming back to finish his 4th lap in the half way so they will meet 3rd time. At 41st sec A will start its fifth lap by the time B will be going to finish 3rd lap in same direction. At 50th second both will complete A's 5th lap and B's 3rd lap. So they will start next lap(A's 6th and B's 4th) from the same point in same direction. And A will finish the race after completing 300m(6*50) total distance and then they wont meet opposite to each other. So totally they met 3 times in opposite direction.