Aptitude Question Bank For AIEEE SSC

The sum of three from the four numbers A, B, C, D are 4024, 4087, 4524 and 4573.  What is the largest of the numbers A, B, C, D? a. 1712 b. 1650 c. 1164 d. 1211 Answer: a Explanation: a+b+c=4024 b+c+d= 4087 a+c+d=4524 a+b+d=4573 Combining all we get 3(a+b+c+d) = 17208 ⇒ a + b + c +d  = 3736 Now we find individual values. a = 1649, b = 1212, c = 1163, d = 1712. So maximum value is 1712.

You have been given a physical balance and 7 weights of 52, 50, 48, 44, 45, 46 and 78 kgs. Keeping weights on one pan and object on the other, what is the maximum you can weigh less than 183 kgs. a. 180 b. 181 c. 182 d. 178 Answer: A Explanation: 52+50+78 = 180

  Find the probability that a leap year chosen at random will have 53 Sundays. a. 1/7 b. 2/7 c. 1/49 d. 3/7 Answer: B Explanation: A leap year has 366 day which is 52 full weeks + 2 odd days. Now these two odd days may be (sun + mon), (mon + tue), .... (Sat + sun).  Now there are total 7 ways. Of which Sunday appeared two times. So answer 2/7

Which of the following numbers must be added to 5678 to give a remainder of 35 when divided by 460? a. 955 b. 980 c. 797 d. 618 Answer: C Explanation: 5678 - 35 + (one of the answer option) should be divisible by 460.  Only option C satisfies.

a bb ccc dddd eeeee .........What is the 120th letter? Answer: O Explanation: Number of letters in each term are in AP. 1, 2, 3, ... So  n ( n + 1 ) 2 ≤ 120 n ( n + 1 ) 2 ≤ 120 For n = 15, we get LHS = 120. So 15th letter in the alphabet is O. So 15th term contains 15 O's.

There are equal number of boys and girls in a class. If 12 girls entered out, twice the boys as girls remain. What was the total number of students in a class? Answer: 48 Explanation: Let the boys = b and girls = g Given  b g − 12 = 2 1 b g − 12 = 2 1 Substitute b = g in the above equation. g = 24. So total students = 24 + 24 = 48

2345 23455 234555 234555........... what was last 2 numbers at 200th digit? Answer: 55 Explanation: Proceed as above.  The last two digits in the 200th place is 55.

What is in the 200th position of 1234 12344 123444 1234444....? Answer: 4 Explanation: The given series is 1234, 12344, 123444, 1234444, ..... So the number of digits in each term are 4, 5, 6, ... or (3 + 1), (3 + 2), (3 + 3), .....upto n terms =  3 n + n ( n + 1 ) 2 3 n + n ( n + 1 ) 2 So  3 n + n ( n + 1 ) 2 ≤ 200 3 n + n ( n + 1 ) 2 ≤ 200 For n = 16, We get 184 in the left hand side. So after 16 terms the number of digits equal to 184.  And 16 them contains 16 + 3 = 19 digits. Now 17 term contains 20 digits and  123 444......4          17 t i m e s 123 444......4 ⏟ 17 t i m e s .  So last digit is 4 and last two digits are 44.

Find the number of perfect squares in the given series 2013, 2020, 2027,................, 2300   (Hint 44^2=1936) a. 1   b. 2    c. 3   d. Can’t be determined Answer: a Explanation: The given series is an AP with common difference of 7. So the terms in the above series are in the form of 2013 + 7k.  We have to find the perfect squares in this format in the given series. Given that 44^2 = 1936. Shortcut: To find the next perfect square, add 45th odd number to 44^2. So 45^2 = 1936 + (2 x 45 -1) = 2025 46^2 = 2025 + (2 x 46 - 1) = 2116 47^2 = 2116 + (2 x 47 - 1) = 2209 Now subtract 2013 from the above numbers and divide by 7. Only 2209 is in the format of 2013 + 7k.  One number satisfies.

 In particular language if A=0, B=1, C=2,…….. ..     , Y=24, Z=25 then what is the value of  ONE+ONE (in the form of alphabets only) a. BDAI    b. ABDI    c. DABI    d. CIDA Answer: a Explanation: This problem is based on Base 26 rather than regular base 10 (decimal system) that we normally use.  In base 10 there are 10 digits 0 to 9 exist.  In base 26 there are 26 digits 0 to 25 exist.  To convert any number into base 26, we have to divide the number with 26 and find the remainder. ( Study this Base system chapter ). Here, ONE + ONE = E has value of 4. So E + E = 8 which is equal to I. Now N + N = 13 + 13 = 26.  But in base 26, there is no 26.  So  ( 26 ) 10 = ( 10 ) 26 ( 26 ) 10 = ( 10 ) 26 So we put 0 and 1 carry over. But 0 in this system is A. Now O + O + 1 = 14 + 14 + 1 = 29 Therefore,  ( 29 ) 10 = ( 13 ) 26 ( 29 ) 10 = ( 13 ) 26 But 1 ...