2 easy questions: Probability/Percentages and Digits

• Sep 13th 2011, 02:37 AM
2 easy questions: Probability/Percentages and Digits
In the life sciences department, 65% of the total number of students watch the news on
television, 40% read a newspaper and 25% both read a newspaper and watch the news on
television. What percentage of the department's students do not watch the news on
television and do not read a newspaper?
(1) 0%
(2) 10%
(3) 20%
(4) 25%

My solution:
100 students.
40-25 = 15 only newspaper
65- 25 = 40 only TV

not sure..
guess:
65+25 = 90?
100-90 is 10?

not sure...

DIGITS question:

25. A, B, C and D are letters representing different digits, so that ABCD is a four-digit
number.
Given: A·AB = CD
If we divide ABCD by the number AB, what three-digit number will be obtained?
(1) 101
(2) 1AB
(3) 11A
(4) 10A

AB into ABCD..

CD is A^2B
and AB into A^2B is...? 0A
AB/AAB leave you with.. 1/A

not sure ..

• Sep 13th 2011, 07:34 AM
BobP
Re: 2 easy questions: Probability/Percentages and Digits
For the first question, there are just four groups. Those that just watch TV, those that just read, those that do both and those that do neither. You've worked out numbers for the first two and you are given the number for the third group. If all four total 100 you should be able to deduce the fourth. (It's easy to see what's going on if you draw a Venn diagram).

For the second question, get what's going on straight in your mind by working through a long division using actual numbers. Try for example 1236/12, 12's into 1 don't go, 12's into 12 go once etc.
• Sep 13th 2011, 08:31 AM
Soroban
Re: 2 easy questions: Probability/Percentages and Digits

Quote:

25. $\displaystyle A, B, C, D$ are letters representing different digits,
. . .so that $\displaystyle ABCD$ is a four-digit number.

Given: .$\displaystyle A\cdot AB \,=\, CD$
If we divide $\displaystyle ABCD$ by the number $\displaystyle AB$, what three-digit number will be obtained?

. . $\displaystyle (1)\;101 \qquad (2)\;1AB \qquad (3)\;11A \qquad (4)\;10A$

Simply "do" the long division . . .

. . $\displaystyle \begin{array}{cccccc}&&&1 & 0 & A \\ && -- & -- & -- & -- \\ A\:\;\;B & | & A & B & C & D \\ && A & B \\ && -- & -- \\ &&&& C \\ &&&& 0 \\ &&&& -- \\ &&&& C & D \\ &&&& C & D \\ &&&& -- & -- \end{array}$

• Sep 13th 2011, 10:34 PM
Re: 2 easy questions: Probability/Percentages and Digits
hello..

how did you pull that off with those awesome graphics/image? im impressed, i would like to knw..

q1
15+40+25 = 80
100-80 = 20

Quote:

Originally Posted by Soroban
. . $\displaystyle \begin{array}{cccccc}&&&1 & 0 & A \\ && -- & -- & -- & -- \\ A\:\;\;B & | & A & B & C & D \\ && A & B \\ && -- & -- \\ &&&& C \\ &&&& 0 \\ &&&& -- \\ &&&& C & D \\ &&&& C & D \\ &&&& -- & -- \end{array}$