Answer: A Check the digits A =1,2,3,4,......9 Note that for a perfect square number with ten's place odd, unit's place of the number must be 6. 66*66 = 4356 Then D=4.
Q. No. 20:
A certain number when divided by 222 leaves a remainder 35, another number when divided by 407 leaves a remainder 47. What is the remainder when the sum of these two numbers is divided by 37?
Answer: C Number of zeroes will be decided by the power of 2 and 5 in the product. Since, the power of 5 is less than the power of 2 hence number of zero will be equal to power of 5. Power of 5 = 55*1010*1515.......*100100 =>(5+10+15+20+ (25*2)+ 30+40+.....) =>(5+10+15+....100) + (25+50+75+100) => 20/2 [2*5 + 19*5] + 250 => 1050+250 = 1300
Q. No. 22:
If a person makes a row of toys of 20 each, there would be 15 toys left. If they made to stand in rows of 25 each, there would be 20 toys left, if they made to stand in rows of 38 each, there would be 33 toys left and if they are made to stand in rows of 40 each, there would be 35 toys left. What is the minimum number of toys the person have?
Answer: D Required number of toys = LCM (20,25,28,38 and 40) - 5 => 3,800-5 = 3,795.
Q. No. 23:
Each of X alarm tolls at regular intervals. All of them tolls together twelve times a day. No two alarm at equal intervals of time. If each alarm tolls after a whole number of minutes, what is the maximum possible value of X?
Answer: A The alarm tolls together twelves times a day. Therefore, they toll together once every 2 hours (or 120 minutes). Since no two alarms toll at equal intervals of time, the total number of distinct factors of 120, including 1 and 120 itself = 23(3)(5) The number of factors = (3+1)(2)(2) = 16. The maximum value of X is 16.