Dear Friends,
Last Saturday we posted the first quiz on CSAT as part of our Mission 2016 . Click below to view the quiz
Solutions to Saturday CSAT Quiz #1
Nine professors – A, B, C, D, E, F, G, H and I are to appear on a series of three panels. Each panel will consist of three professors and each professor will appear exactly once. The panel must be arranged according to the following conditions.
I. C and H must be on the same panel.
II. E and F must be on the same panel.
III. I and D cannot be on the same panel.
IV. G must appear on the second panel.
V. Either D or G or both must appear on the panel with B.
Question 1
Which of the following professors could appear on a panel together?
(A) A, F, I
(B) A, D, G
(C) E, C, G
(D) H, C, D
This questions can be solved only by elimination method. If you tried solving it through any other method, it would be a huge wastage of time.
Option (A) violates the rule E and F must be on the same panel.
Option ( B) violates the rule Either D or G or both must appear on the panel with B.
Option ( C) violates the rule E and F must be on the same panel.
Option ( D) is the correct answer
Question 2
Which of the following could be true?
(A) C appears on the 2nd panel
(B) B appears on the 3rd panel
(C) I appears on the 3rd panel
(D) D and I appear on the 1st panel
Since, G must appear on the second panel (condition IV), D will have to appear along with B on 3rd panel.
Therefore, both D and B cannot appear on different panels.
Question 3
The 3rd panel could consist of which the following?
(A) A, C, D
(B) E, C, D
(C) A, B, D
(D) E, F, I
Combination of A, D and C violates condition 1, which requires the combination of C with H
Question 4
All the following could be on the same panel as E, except
(A) A
(B) I
(C) D
(D) G
If one of the panelists is E, other would be F as given in condition II,
C cannot appear with E and F because C always appears with H [refer condition I] and there cannot be more than three panelists
We hope you like the question. UPSC does not set difficult questions in CSAT. This question required solving by elimination.
Q5 – B
Q6 – B
Q7 – C
Q8 – A
Q9 – A
Explanations
5. The total length of roads in the year 1990 = 30,00,000 kms.
Urban roads are 7% of total roads. The length of urban roads = 7 X 30,00,000/100 = 2,10,000.
6. The maximum increases are 2 00,000 kms in both 1996-97 and 1997-98.
Percentage increase in 1996-97 = 2,00,000/25,00,000 * 100 = 8%.
Percentage increase in 1997-98 = 2,00,000/27,00,000 * 100 = 7.4%.
In other years, the increase is only 1,00,000 kms. So they obviously won’t be anywhere near to the above two years. Overall, the highest percentage increase is for 1996-97.
7. The length of highways in the year 1996 = 75% of 2500 = 75 X 2500/100 = 1875 thousand kms.
The length of highways in 1999 = 86% of 3000 = 86 X 3000/100 = 2580 thousand kms.
The increase in length of highways = 7,05 thousand kms.
8. The percentage share of highways in 2000 = 86 + 4 = 90%.
The increase in total length = 5% = 5 X 3000 / 100 = 150 thousand kms. The total length in the year 2000 = 3150 thousand kms.
The length of highways = 90 X 3150 / 100 = 2835 thousand kms
9. Length of project roads in 1995 = 0 km.
Length of project roads in 1999 = 7% of 3000 thousand km = 210 thousand kms.
Increase over these four years is therefore = 210,000 kms.
So, increase per year = 210,000/4 = 52,500 km.
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