Test of Mathematics at 10+2 Level TOMATO OBJECTIVE SOLUTIONS Mail : [email protected] 1. A worker suffers a 20% cut in wages. He regains his original pay by obtaining a rise of (a) 20% (b) 22.5% (c)25% (d) 27.5% Solution : Let his wage is x. After 20% cut his wage is x – 20x/100 = 80x/100 In 80x/100 he needs to regain 20x/100 In 1 he needs to regain (20x/100)/(80x/100) = ¼ In100 he needs to regain 100*(1/4) = 25 Therefore, he needs a rise of 25% to regain his wage. Option (c) is correct. 2. If m men can do a job in d days, then the number of days in which m + r men can do the job is (a) d + r (b) (d/m)(m + r) (c)d/(m + r) (d) md/(m + r) Solution : m men can do a job in d days 1 man can do the job in md days m + r men can do the job in md/(m + r) days. Option (d) is correct. 3. A boy walks from his home to school at 6 km per hour (kmph). He walks back at 2 kmph. His average speed, in kmph, is (a) 3 (b) 4 (c)5 (d) √12 Solution : Let the distance from his home to school is x km. Therefore, total distance covered = x + x = 2x. Time taken to go to school = x/6 hours. Time taken to come home from school = x/2 hours. Total time = x/6 + x/2 So, average speed = total distance/total time = 2x/(x/6 + x/2) = 2/(1/6 + ½) = 3 kmph Option (a) is correct. 4. A car travels from P to Q at 30 kilometres per hour (kmph) and returns from Q to P at 40 kmph by the same route. Its average speed, in kmph, is nearest to (a) 33 (b) 34 (c)35 (d) 36 Solution : Let the distance between P and Q is x km. Total distance covered = x + x = 2x Time taken to go from P to Q = x/30 hours. Time taken to go from Q to P = x/40 hours. Total time = x/30 + x/40 Average speed = total distance/total time = 2x/(x/40 + x/30) = 2/(1/40 + 1/30) = 2*40*30/70 = 240/7 = 34.285 (approx.) Option (b) is correct. 5. A man invests Rs. 10000 for a year. Of this Rs. 4000 is invested at the interest rate of 5% per year, Rs. 3500 at 4% per year and the rest at α% per year. His total interest for the year is Rs. 500. Then α equals (a) 6.2 (b) 6.3 (c)6.4 (d) 6.5 Solution : Interest from Rs. 4000 = 5*4000/100 = Rs. 200 Interest from Rs. 3500 = 4*3500/100 = Rs. 140 Rest money = 10000 – (4000 + 3500) = Rs. 2500 Interest from Rs. 2500 = α*2500/100 = 25α As per the question, 200 + 140 + 25α = 500 ðα = 160/25 = 6.4 Option (c) is correct. 6. Let x , x , ….., x be positive integers such that x + x = k for all i, 1 2 100 i i+1 where k is constant. If x = 1, then the value of x is 10 1 (a) k (b) k – 1 (c)k + 1 (d) 1 Solution : Clearly, x = k – 1, x = 1, x = k – 1, ….., x = k – 1 Option (b) is 9 8 7 1 correct. 7. If a = 1, a = 1 and a = a a + 1 for n > 1, then (a) a is odd and a 0 1 n n-1 n-2 465 466 is even (b) a is odd and a is odd (c)a is even and a is even (d) a 465 466 465 466 465 is even and a is odd. 466 Solution : As, a and a both odd so, a = 2 = even. 0 1 2 As a is even, both a and a will be odd because between a and a one is 2 3 4 n-1 n-2 even and hence added to 1 becomes odd. Then a will be even as a and a are both odd. So, the sequence will go in the 5 3 4 way, a , a odd, a even, a , a odd, a even, a , a odd, a even and so on.. 0 1 2 3 4 5 6 7 8 So, the numbers which are congruent to 2 modulus 3 are even and rest are odd. Now, 465 ≡ 0 (mod 3) and 466 ≡ 1 (mod 3) ð a and a are both odd. 465 466 Option (b) is correct. 8. Two trains of equal length L, travelling at speeds V and V miles per hour 1 2 in opposite directions, take T seconds to cross each other. Then L in feet (1 mile 5280 feet) is (a) 11T/15(V + V ) 1 2 (b) 15T/11(V + V ) 1 2 (c)11(V + V )T/15 1 2 (d) 11(V + V )/15T 1 2 Solution : Speed = V miles per hour = V *5280/3600 feet/second = 22V /15 1 1 1 feet/second Relative velocity = 22V /15 + 22V /15 = 22(V + V )/15 feet/second Total 1 2 1 2 distance covered = sum of train lengths = L + L = 2L Therefore, 2L = {22(V + V )/15}*T ð L = 11(V + V )T/15 1 2 1 2 Option (c) is correct. 9. A salesman sold two pipes at Rs. 12 each. His profit on one was 20% and the loss on the other was 20%. Then on the whole, he (a) Lost Re. 1 (b) Gained Re. 1 (c)Neither gained nor lost (d) Lost Rs. 2 Solution : Let the cost price of the pipe on which he made profit = x. ð x + 20x/100 = 12 ð 120x/100 = 12 ð x = 12*100/120 ð x = 10 Let the cost price of the pipe on which he lost = y. ð y – 20y/100 = 12 ð 80y/100 = 12 ð y = 12*100/80 ð y = 15 Therefore, total cost price = 10 + 15 = 25 Total selling price = 2*12 = 24 So, he lost (25 – 24) = Re. 1 Option (a) is correct. 10. The value of (256)0.16(16)0.18 is (a) 4 (b) 16 (c)64 (d) 256.25 Solution : 0.16(16)0.18 = 28*0.16*24*0.18 = 28*0.16 + 4*0.18 = 24(2*0.16 + 0.18) = 24*0.5 = 22(256) = 4 Option (a) is correct. 11. The digit in the unit place of the integer 1! + 2! + 3! + ….. + 99! Is (a) 3 (b) 0 (c)1 (d) 7 Solution : Now, after 5! = 120 all the terms end with 0 i.e. unit place digit is 0. So, the unit place digit of the given integer is the unit place digit of the integer 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 i.e. 3 Option (a) is correct. 12. July 3, 1977 was a SUNDAY. Then July 3, 1970 was a (a) Wednesday (b) Friday (c)Sunday (d) Tuesday Solution : In 1970 after July 3 there are = 28 + 31 + 30 + 31 + 30 + 31 = 181 days. In 1971 there are 365 days. In 1972 there are 366 days. In 1973, 1974, 1975 there are 3*365 = 1095 days. In 1976 there are 366 days. In 1977 up to July 3 there are 31 + 28 + 31 + 30 + 31 + 30 + 3 = 184 days. Therefore total number of days up to July 3, 1970 from July 3, 1977 is 181 + 365 + 366 + 1095 + 366 + 184 = 2557 2557 ≡ 2 (mod 7) ð July 3, 1970 was a Friday. (Sunday – 2) Option (b) is correct. 13. June 10, 1979 was a SUNDAY. Then May 10, 1972, was a (a) Wednesday (b) Thursday (c)Tuesday (d) Friday Solution : After May 10, 1972 there are 21 + 30 + 31 + 31 + 30 + 31 + 30 + 31 = 235 days. In 1973, 1974, 1975 there are 3*365 = 1095 days. In 1976 there are 366 days. In 1977, 1978 there are 365*2 = 730 days. In 1979 up to June 10, there are 31 + 28 + 31 + 30 + 31 + 10 = 161 days. Therefore, total number of days from May 10, 1972 to June 10, 1979 is 235 + 1095 + 366 + 730 + 161 = 2587 days. Now, 2587 ≡ 4 (mod 7) ð May 10, 1972 was a Wednesday. (Sunday – 4) Option (a) is correct. 14. A man started from home at 14:30 hours and drove to a village, arriving there when the village clock indicated 15:15 hours. After staying for 25 minutes (min), he drove back by a different route of length (5/4) times the first route at a rate twice as fast, reaching home at 16:00 hours. As compared to the clock at home, the village clock is (a) 10 min slow (b) 5 min slow (c)5 min fast (d) 20 min fast Solution : Let the distance from home to the village is x and he drove to the village by v speed. Therefore, time taken to reach village is x/v. Now, time taken to come back home = (5x/4)/2v = 5x/8v Total time = x/v + 5x/8v + 25 = (13/8)(x/v) + 25 ð (13/8)(x/v) + 25 = (16:00 – 14:30)*60 = 90 ð (x/v) = 65*8/13 = 40 So, he should reach village at 14:30 + 40 min = 15:10 hours. Therefore, the village clock is (15:15 – 15:10) = 5 min fast. Option (c) is correct. 15. If (a + b)/(b + c) = (c + d)/(d + a), then (a) a = c (b) either a = c or a + b + c + d = 0 (c)a + b + c + d = 0 (d) a = c and b = d Solution : (a + b)/(b + c) = (c + d)/(d + a) ð (a + b)/(b + c) – 1 = (c + d)/(d + a) – 1 ð (a – c)/(b + c) = - (a – c)/(d + a) ð (a – c)/(b + c) + (a – c)/(d + a) = 0 ð (a – c){1/(b + c) + 1/(d + a)} = 0 ð (a – c)(a + b + c + d)/{(b + c)(d + a)} = 0 ð (a – c)(a + b + c + d) = 0 ð Either a = c or a + b + c + d = 0 Option (b) is correct. 16. The expression (1 + q)(1 + q2)(1 + q4)(1 + q8)(1 + q16)(1 + q32)(1 + q64), q ≠ 1, equals (a) (1 – q128)/(1 – q) (b) (1 – q64)/(1 – q) (c){1 – q^(21 + 2 + …. + 6)}/(1 – q) (d) None of the foregoing expressions. Solution : Let E = (1 + q)(1 + q2)(1 + q4)(1 + q8)(1 + q16)(1 + q32)(1 + q64) ð (1 – q)*E – (1 – q)(1 + q)(1 + q2)(1 + q4)(1 + q8)(1 + q16)(1 + q32)(1 + q64) (q ≠ 1) ð (1 – q)*E = (1 – q2)(1 + q)(1 + q2)(1 + q4)(1 + q8)(1 + q16)(1 + q32)(1 + q64) ð (1 – q)*E = (1 – q4)(1 + q4)(1 + q8)(1 + q16)(1 + q32)(1 + q64) ð (1 – q)*E = (1 – q8)(1 + q8)(1 + q16)(1 + q32)(1 + q64) ð (1 – q)*E = (1 – q16)(1 + q16)(1 + q32)1 + q64) ð (1 – q)*E = (1 – q32)(1 + q32)(1 + q64) ð (1 – q)*E = (1 – q64)(1 + q64) ð (1 – q)*E = (1 – q128) ð E = (1 – q128)/(1 – q) (q ≠ 1) Option (a) is correct. 17. In an election 10% of the voters on the voters‟ list did not cast their votes and 60 voters cast their ballot paper blank. There were only two candidates. The winner was supported by 47% of all voters in the list and he got 308 votes more than his rival. The number of voters on the list was (a) 3600 (b) 6200 (c)4575 (d) 6028 Solution : Let number of voters on the voters‟ list is x. Did not cast their vote = 10x/100 Therefore, total number of voters who cast their vote in favor of a candidate = x – 10x/100 – 60 = 90x/100 – 60 Winner got 47x/100 votes. Rival got 90x/100 – 60 – 47x/100 = 43x/100 – 60 According to question, 47x/100 – (43x/100 – 60) = 308 ð 4x/100 + 60 = 308 ð x/25 = 308 – 60 ð x/25 = 248 ð x = 248*25 ð x = 6200 Option (b) is correct. 18. A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. He obtained (3/5) part of the total full marks. Then the number of papers in which he got more than 50% marks is (a) 2 (b) 3 (c)4 (d) 5 Solution : Let the full mark of each paper is x. Total full marks = 5x He got total 5x*(3/5) = 3x marks In first paper he got 3x*6/(6 + 7 + 8 + 9 + 10) = 9x/20 In second paper he got 3x*7/(6 + 7 + 8 + 9 + 10) = 21x/40 In third paper he got 3x*8/(6 + 7 + 8 + 9 + 10) = 3x/5 In fourth paper he got 3x*9/(6 + 7 + 8 + 9 + 10) = 27x/40 In fifth paper he got 3x*10/(6 + 7 + 8 + 9 + 10) = 3x/4 In first paper percentage of marks = (9x/20)*100/x = 45% In second paper percentage of marks = (21x/40)*100/x = 52.5%