Latest Elitmus Aptitude Question SOLUTION: The table below, Gives the marks of obtained by six students P,Q,R,S,T in five subjects A,B,C,D,E. a, b, c, d, e denote the minimum marks scored in the A,B,C,D,E respectively

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(#M40018178) ELITMUS Logical Reasoning question Keep an EYE

The table below, Gives the marks of obtained by six students P,Q,R,S,T in five subjects A,B,C,D,E.
a, b, c, d, e denote the minimum marks scored in the A,B,C,D,E respectively
Max marks =100

A B C D E
P a+15 b+5 c d+4 e+1
Q a+14 b+8 c+5 d+2 e+2
R a+13 b+2 c+6 d+5 e
S a b+1 c+7 d+6 e+3
T a+8 b+4 c+3 d+3 e+6
U a+5 b c+9 d e+9

Also the following information is known
T scored equal marks in A and C
R scored one mark less in B than T scored in D
P scored 2 marks more in E then Q scored in B
The sum of of all the marks scored all the students in all subjects was 2600.
No one score less than 75 in any subject.

Questions:-
1)If U’s score in B was not less than P’s score in C, then how much did S score in E?
a)91 b)93 c)87 d)90
2)Referring to the previous question, who scored the max in which subject?
a)P,A b)Q,B c)U,E d)U,C
3)If P’s marks in A more than the marks anyone got in other subject, Then how much could T have scored In D?
a)64 b)83 c)82 d) 79
4)Which of the following cannot be R’s score in C?
a)94 b)90 c)93 d)96

Asked In Elitmus sravan (11 years ago)

Unsolved Read Solution (12)

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Read Solution (12) :

Sumit Rawal  (11 years ago)

1.a+8=c+3
c-a=5
2.d+3-(b+2)=1
d-b=0
d=b
3.(e+1)-(b+8)=2
e-b=9
e-d=9
4.(6a+55)+(6b+20)+(6c+30)+(6d+20)+(6E+21)=2600
6(a+B+C+D+E)+146=2600
6(a+B+C+D+E)=2600-146
a+b+c+d+e=2454/6=409
Min marks by each student is 75
if a =75 then c=80 also b=80 d=80 then e=89
but their sum is 404
so adding 1 to each quantity we get sum
a =76,b=c=d=81,e=90

question 1st
U's score in B= P's score in C
i.b=c
then S's score in e = e+3=90+3=93
thus answer is b 93
question 2nd
from the above calculations we observed that e has scored max i.e 90 and U score in E is e+9 =90+9=99(max)
Thus anwer is cU,E
question 3rd
it is given that p got max marks in A thus min marks is 75 sum of marks 409
taking e=90 so max value e+9=90+9=99 which must be less than p+15(since it is biggest)
p+15>99
p+15=100
so solving equation for t
100+8+D+4+D+3+D+3+90+6=409
-214+3D=409
-d=82*3=246
d=82
answer is c
Question 4
r's score in c cannot be 94 is score is 94 c=88,if c=88, a =83,e=97
d+3

Is this solution Helpfull? Yes (19) | No (11)  

pavan  (11 years ago)

sravan can u plz send ur mobile no.. i want to talk to u about elitmus plzzzz. help me

Is this solution Helpfull? Yes (6) | No (2)  

Ajitesh rai  (8 years ago)

With the given information----------

a = c - 5 ----------------- (1)
d = b --------------------(2)
e = b + 9 --------------------(3)
and
a + b + c + d + e = 409 ------------------(4) { adding all the marks }
also
a,b,c,d,e > = 75
Now coming to the first question------------
Relation between 'b' and 'c' is given , Thus convert equation (4) in 'b' and 'c' using (1) (2) (3)---

We get , 3b + 2c = 405 ----------(5)

Now either b = c OR b > c ( according to question )
--------------Now friends , here comes the main logic ----------------

a = c - 5 , We know that 'a' must be least 75 , hence c must be least 80

In (5)
CASE 1 , if c = 80 then b = 81. 6 ( b > c , but decimal hence reject }

CASE 2 if c = 81 then b = 81 ( b = c }

CASE 3 if c = 82 then b = 80.3 { b < c reject}

Hence , only possible solution is b = c = d = 81

a = 76 and e = 90

ANS 1 e + 3 = 93

ANS 2 we can easily see that e + 9 is maximum , hence ( U , E }

ANS 3 -----We have to check all the options to check for invalid options ----

Now maximum marks are 100 , hence a + 15 < 100 ,---------- 75 < a < 85 ---------

T Score is d + 3

(a) # Cannot be 64 ( INVALID )

(b) 83 , i.e d = 80 , Now using (1) (2) (3) (5) , We get d = b = 80 , e = 89 c = 82.5 ( INVALID )

(c) 82 , i.e d = 79 , ( d = b = 79 , e = 88 and c = 84 , a = 79 ) -------- ( VALID )

(d) 79 , i.e d = 76 , ( d = b = 76 , e = 85 and c = 83.5 ( INVALID )

Hence (c) 82 is correct

ANS 4 ---Do it yourself-----

Is this solution Helpfull? Yes (4) | No   

Bankim  (11 years ago)

a=76
b=c=d= 81
e= 90

Is this solution Helpfull? Yes (2) | No (2)  

taashi  (11 years ago)

1.b
2.c
3.c
4.a

Is this solution Helpfull? Yes (1) | No   

sravan  (11 years ago)

Table:

A B C D E

P a+15 b+5 c d+4 e+1

Q a+14 b+8 c+5 d+2 e+2

R a+13 b+2 c+6 d+5 e

S a b+1 c+7 d+6 e+3

T a+8 b+4 c+3 d+3 e+6

U a+5 b c+9 d e+9

Is this solution Helpfull? Yes | No (2)  

sravan  (11 years ago)

BANKIM can u explain how?

Is this solution Helpfull? Yes | No (2)  

sravan  (11 years ago)

9618476235

Is this solution Helpfull? Yes | No (3)  

vishav bansal  (11 years ago)

please anybody explain its proper solution

Is this solution Helpfull? Yes | No (1)  

Azhar Hussain  (10 years ago)

@sumit what if we take the value of a=75, b=80,c =84,d=80, e=89
therefore , by adding a+b+c+d+e = 408 . hence we will add 1 to c to make the total 409 i.e c=84+1=85
A/Q e+3 =? ; e+3= 89+3=92 .

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MR. X  (9 years ago)

@sumit how b=c??

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Satvik  (7 years ago)

@Ajitesh,
in http://www.m4maths.com/18178-The-table-below-Gives-the-marks-of-obtained-by-six-students-P-Q-R-S-T-in-five-subjects-A.html#
question 3, we can't assume that the condition given in the 1st question (b>=c), will be valid, unless stated.

then why did we rule out case of

a=82 b=77 c=87 d=77 e=86
a=85 b=75 c=90 d=75 e=84

Moreover, the solution you assumed for ques 3, a=79 b=79 c=84 d=79 e=88, it negates the statement that P scored the highest in A (a+15=94) than any other student in any other subject, as then T will have e+9=97 in E.

please clarify?

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