CSAT DAILY MISSION #62

Examine the information given below and answer the questions that follow:

Question 1

How many students have scored the lowest marks in two or more subjects?

(A) 2

(B) 3

(D) 0

Question 2

Who has the highest marks overall?

(A) D

(B) C

(D) E

Question 3

If 75% marks are required overall for distinction, how many people will get a distinction?

(A) 1

(B) 2

(D) 4

Question 4

What per cent are the marks obtained by C in Hindi of the marks obtained by A in the same subject?

(A) 75.9%

(B) 78.4%

(D) 75.3%

Question 5

What is the average marks obtained by all the students in Social Studies?

(A) 55.7

(B) 57.5

(D) 59.5

SOLUTIONS TO DAILY CSAT MISSION # 63

1. (D) 2. (B) 3. (B) 4. (D) 5. (D)

Explanations
Let us first try to find the arrangement of the letters. We start off by arranging seven spots in ascending order for the letters and placing B at the centre (fourth position):

Next, it is given that A – D = 3. The only spots that can be found which satisfy this are:

Next, it is given that B – F = C – D. This means F is less than B. Let’s try to find the position of F.
In Case 1, there are two positions (first and second) where F could be. If F is at the first position, then B – F = 3. In that case, C – D would have to be 3 as well. But A – D = 3 (given). So, the first position is not possible for F. If F is at the second position, then B – F = 2. In that case, C – D would have to be 2 as well. This means C will be at the fifth position. F at second position remains a possibility.
In Case 2, there are two positions (first and third) where F could be. If F is at the first position, then B – F = 3. In that case, C – D would have to be 3 as well. But A – D = 3 (given). So, the first position is not possible for F. If F is at the third position, then B – F = 1. In that case, C – D would have to be 1 as well. This leaves C at the third position. But F is already at the third position. So, F at third position is also not a possibility.

Only one scenario remains that satisfies all conditions i.e. F at second and C at fifth positions:

Since G > E, G is obviously at the seventh and E is at the first position. The final arrangement is:

1. Any alphabet will have its greatest value when E = 4 and G = 10. And, any alphabet will have its lowest value when E = 1 and G = 7. Value of C when E = 4 is 8. Value of D when E = 1 is 3. Difference is 5.

2. B will have its highest value when E = 4. B will be 7 in this case.
B will have its lowest value when E = 1. B will be 4 in this case. Difference is 3.

3. If A = 7, E = 2 and G = 8.

4. C is 4 greater than E. So, X is 4 greater than C.
When E = 1: C = 5 and X = 9. (A + E) in this case is 7, which is not equal to X. So, E = 1 is not the case.
When E = 2: C = 6 and X = 10. (A + E) in this case is 9, which is not equal to X. So, E = 2 is not the case.
When E = 3: C = 7 and X = 11. (A + E) in this case is 11, which is equal to X. So, E = 3 is a possible case.
When E = 4: C = 8 and X = 12. (A + E) in this case is 13, which is not equal to X. So, E = 4 is not the case.
When E = 3, D = 5.

5. A is 4 greater than F, and D is 4 less than G.