Answer: A) Rs. 404.80 Explanation: Let C1 be the cost price of the first article and C2 be the cost price of the second article.Let the first article be sold at a profit of 22%, while the second one be sold at a loss of 8%. We know, C1 + C2 = 600.The first article was sold at a profit of 22%. Therefore, the selling price of the first article = C1 + (22/100)C1 = 1.22C1The second article was sold at a loss of 8%. Therefore, the selling price of the second article = C2 - (8/100)C2 = 0.92C2. The total selling price of the first and second article = 1.22C1 + 0.92C2. As the merchant did not make any profit or loss in the entire transaction, his combined selling price of article 1 and 2 is the same as the cost price of article 1 and 2. Therefore, 1.22C1 + 0.92C2 = C1+C2 = 600As C1 + C2 = 600, C2 = 600 - C1. Substituting this in 1.22C1 + 0.92C2 = 600, we get1.22C1 + 0.92(600 - C1) = 600or 1.22C1 - 0.92C1 = 600 - 0.92*600or 0.3C1 = 0.08*600 = 48or C1 = 48/(0.3) = 160.If C1 = 160, then C2 = 600 - 160 = 440.The item that is sold at loss is article 2. The selling price of article 2 = 0.92*C2 = 0.92*440 = 404.80.

Answer: B) Rs.12,000 Explanation: We have P = 8748, R = -10 and n = 3Therefore, the purchase price of the machine= P1+R100n= 87481-101003= 8748 x (100/90) x (100/90) x (100/90)= 12,000.   Therefore, the purchase price of the machine was Rs. 12,000.

Answer: C) 175 Explanation: Given that,  working days = 5  working hours = 8  A man get rupess per hour is Rs.2.40  So in one day the man get total rupees is 2.40 x 8 = 19.2  So in 5 days week the man get total rupees is 19.2 x 5 = 96  So in 4 week the man get total rupees is 96 x 4 = 384  So the man worked for = 160hours in 4 weeks    But given that the man earned Rs.432    Hence remaning money is (432-384 = 48)which is earn by doing overtime work   Overtime hours  = 48/3.20 = 15 So total worked hours is = 15 + 160 = 175.

Answer: B) 13 + 1/3 days Explanation: C alone can finish the work in 40 days.As given C does half as much work as A and B together => (A + B) can do it in 20 days(A + B)s 1 days wok = 1/20.A's 1 days work : B's 1 days Work = 1/2 : 1 = 1:2(given)A's 1 day’s work = (1/20) x (1/3) = (1/60) [Divide 1/20 in the raio 1:2]B's 1 days work = (1/20) x (2/3) = 1/30(A+B+C)'s 1 day's work = (1/60) + (1/30) + (1/40) = 9/120 = 3/40 All the three together will finish it in 40/3 = 13 and 1/3 days.

Answer: A) 45 metric tonnes Explanation: 2 engines of former type for one hour consumes (2 x 24)/(6 x8) = 1 metric ton  i.e. 3 engines of latter type consumes 1 ton for one hour Hence 9 engines consumes 3 tons for one hour For 15 hours it is 15 x 3 = 45 metric tonnes.

Answer: B) A = 30000 B = 40000 C = 48000 Explanation: Compound ratio of A:B:CA:B = 3:4B:C = 5:6----------A:B:C = 15:20:24 We can divide Rs.118000 in this ratio. Share of A = 15/59 x 118000 = 30000 Share of B = 20/59 x 118000 = 40000 Share of C = 24/59 x 118000 = 48000

Answer: D) 72 Explanation: If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are k, l, m, n & o respectively, thenk + l + m + n = 58 x 4 = 232 ---- (i) & l + m + n + o = 60 x 4 = 240 ---- (ii)Subtracting (i) from (ii), we get o-k = 8 ---(iii)Given, k/o = 8/9 ----(iv)So from (iii) & (iv), we geta = 64, e = 72Therefore, number of visitors on 5th day is 72.

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.