# TCS off campus latest questions - 30

Star mark question:

### 1. 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
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 ${(26)_{10}} = {(10)_{26}}$
So we put 0 and 1 carry over. But 0 in this system is A.
Now O + O + 1 = 14 + 14 + 1 = 29
Therefore, ${(29)_{10}} = {(13)_{26}}$
But 1 = B and 3 = D in that system. So ONE + ONE = BDAI

### 2. 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
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.

### 3.  What is in the 200th position of 1234 12344 123444 1234444....?

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 = $3n + \dfrac{{n(n + 1)}}{2}$
So $3n + \dfrac{{n(n + 1)}}{2} \le 200$
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 $123\underbrace {444......4}_{17{\rm{ times}}}$.  So last digit is 4 and last two digits are 44.

### 4. 2345 23455 234555 234555........... what was last 2 numbers at 200th digit?

Explanation:
Proceed as above.  The last two digits in the 200th place is 55.

### 5. 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?

Explanation:
Let the boys = b and girls = g
Given $\dfrac{b}{{g - 12}} = \dfrac{2}{1}$
Substitute b = g in the above equation. g = 24. So total students = 24 + 24 = 48

### 6. a bb ccc dddd eeeee .........What is the 120th letter?

Explanation:
Number of letters in each term are in AP. 1, 2, 3, ...
So $\dfrac{{n(n + 1)}}{2} \le 120$
For n = 15, we get LHS = 120. So 15th letter in the alphabet is O. So 15th term contains 15 O's.

### 7. There are 120 male and 100 female in a society. Out of 25% male and 20% female are rural. 20% of male and 25% of female rural people passed in the exam. What % of rural students have passed the exam?

Explanation:
From the above data, Rural male = 25%(120) = 30, Rural female = 20%(100) = 20.
Passed students from rural: male = 20%(30) = 6, female = 25%(20) = 5
Required percentage = $\dfrac{{11}}{{50}} \times 100 = 22\%$

### 8. 1/7 th of the tank contains fuel. If 22 litres of fuel is poured into the tank the indicator rests at 1/5th mark. What is the quantity of the tank?

Explanation:
Let the tank capacity = $v$ liters.
Given, $\dfrac{v}{7} + 22 = \dfrac{v}{5}$
$\dfrac{v}{5} - \dfrac{v}{7} = 22 \Rightarrow v = 385$

### 9. What is the probability of getting sum 3 or 4 when 2 dice are rolled

Total ways = ${6^2} = 36$
Probability = $\dfrac{5}{{36}}$