AP ICET 2020 Question Paper Shift-2 (10th Sep)
The following Pie chart depicts the distribution of Annual expenditure of a company under 6 broad heads. Based on this data answer the questions.
AP ICET 2020 Shift-2 (10th Sep) - Question 51
What is the ratio of the expenditure under the heads B, C and D put together and that of the heads A and F put together?
AP ICET 2020 Shift-2 (10th Sep) - Question 52
If the company cut down its annual expenditure by 5% and 20% under the heads E and F respectively the new degrees in the section E and F in the pie chart will be respectively.
AP ICET 2020 Shift-2 (10th Sep) - Question 53
If the company decides to create an extra head of account G by transferring 10% and 20% of its expenditure under the heads E and F respectively. the sectorial angle in the pie chart for G will be
In a class of 80 students, each student chooses to play at least one of the games A, B, C. Out of them 45 play game A, 35 play game B, 30 play game C. Also 13 play both A and B, 10 play both B and C and 5 play all the 3 games. Based on this information answer questions.
AP ICET 2020 Shift-2 (10th Sep) - Question 54
The number of students who play exactly one game is
AP ICET 2020 Shift-2 (10th Sep) - Question 55
The number of students who play games A and C but not B is
The 26 letters of English are given ranks 1,2, ...., 26, in the alphabetical order. Then for $$1 \leq r \leq 26, r^{th}$$ letter is coded as $$k^{th}$$ letter where $$7r + 3 = 26M + k$$ where $$0 \leq k \leq 25$$ and $$M ∈ Z$$. When $$k = O, k^{th}$$ letter is coded as 'z'. For example the letter 'E' is coded as 'L' because when $$r = 5$$ and $$7r + 3 = 38$$, then $$k = 12$$ and $$12^{th}$$ letter is 'L'. For decoding the reverse process is used. Based on this answer the questions.
AP ICET 2020 Shift-2 (10th Sep) - Question 57
The number of letters that remain unchanged in this code is
AP ICET 2020 Shift-2 (10th Sep) - Question 58
The number of alphabets that are coded to their next letters is
