A student E scored 30 marks more than that by C whose total marks is more than that by A but less th...........

Quantitative Aptitude (SBI-PO) Case Studies

Directions (): Study the following information carefully and answer the given questions.
In a school 4 students A, B, C and D appeared in an English examination which is consist of 15 short and 4 long questions are asked. Each of short question carry 5 marks and each of long questions carry 10 mark and for a spelling error 0.5 mark is deducted. For not attempting a question, zero marks is awarded. In the English exam, A answered (x + 6) short questions, x long questions and did (2x + 10) spelling errors. Ratio of number of spelling errors made by A and B is 8 : 3. Total marks scored by D is 86. Ratio of short and long questions answered by B In the English exam is 5 : 1 and he scored total 102 marks. C scored 35 marks more in short questions than long questions except spelling errors and the number of spelling errors made by C is equal to the difference of the number of spelling errors made by A and B. Number of short questions answered by D is average of that answered by A and B. Percentage of number of long questions answered among all the question answered by D is 20%. Average of the number of spelling errors made by all four students A, B, C and D is 10. Each student attempts integral number of short as well as long questions.

A student E, scored 30 marks more than that by C, whose total marks is more than that by A but less than that by D. If the number of spelling errors made by E are 10 more than that by C, then the total marks scored by C is what percent less than that by E?

37.5%

33(1/3)%

25%

30%

None of these

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Explanation:

Let the number of long questions answered by B and C be y and z respectively, then
the number of short questions answered by B = 5y
Number of spelling errors made by A = (2x + 10), then
Number of spelling errors made by B =

For C, number of short questions answered = (Number of long questions answered by B × 10 + 35)/5 = (2z + 7)
Number of short questions answered by D is average of that answered by A and B

Total marks scored by D = 86
5(x + 5y + 6)/2 +10(x + 5y + 6)/8 – (20 – 4x)/2 = 86
⇒15(x + 5y + 6)/4 – (10 – 2x) = 86
⇒23x + 75y = 294 …. (i)
Also, 5 × 5y + 10 × y – 3(x + 5)/8 = 102
⇒280y – 3x = 831 …. (ii)
On solving equation (i) and (ii), we get
x = y = 3

∴ The number of question short and long questions answered, number of spelling errors and total marks scored will be as follow

It is given that
Total marks of A < Total marks of C < Total marks of D
68 < [5(2z + 7) + 10z – 5] < 86
⇒68 < (20z + 30) < 86
⇒38 < 20z < 56
∴Possible value of z = 2
So, total marks scored by C = (20z + 30) = 70
Total marks scored by E = Total marks scored by C + 30 = 70 + 30 = 100
Hence, the required percentage = (100-70)/100× 100 = 30%.