# Logical Reasoning Capsules - Edwin Jose, CAT DILR 100 Percentiler - Set 8

• Directions: Answer the following questions on the basis of information given below:
Five actresses Careena, Catrina, Carishma, Coena and Cashmira are disguised as U, V, X, Y and Z, not necessarily in this order. Director Bhakti Kapoor interviewed the actresses in the order Z, Y, X, U and V for his film “Bhoot Aunty”.

1. Catrina was interviewed after Cashmira.
2. Careena was interviewed before Carishma.
3. The fee (in Rs.) demanded by X, U, V were 2397 lakhs, 2379 lakhs and 1213 lakhs respectively.
4. One of the remaining two actress demanded Rs. 1123 lakhs and the other demanded a sum (in Rs.) between 1201 lakhs and 2288 lakhs.
5. The sum of the fees demanded by Careena and Carishma is the same as the sum of the fees demanded by Cashmira and Catrina.

(1) Who is disguised as Y?
(a) Coena or Cashmira
(b) Coena or Careena
(c) Coena or Cashmira or Careena
(d) Coena or Cashmira or Catrina

(2) What is the fee (in Rs.) demanded by Coena?
(a) 2397 or 1213 lakhs
(b) 1213 lakhs
(c) 1213 or 2379 lakhs
(d) 1123 lakhs

(3) Who was interviewed last ?
(a) Coena or Catrina
(b) Coena or Carishma
(c) Carishma or Catrina
(d) None of these

Solutions:

Let the fee (in Rs. lakhs) demanded by Careena, Carishma, Cashmira and Catrina be a, b, c and d (not necessarily in that order).
Let the fee (in Rs. lakhs) demanded by Coena be e.
Also, either Z or Y must have demanded Rs. 1123 lakhs as the prices demanded by X, U and V are already given.
Let e = 1123.
a + b = c + d.
It can be seen that 2397 – 2379 = 18
and 18 + 1213 = 1231.
So, 1231 + 2379 = 2397 + 1213.
Also, 1201 < 1231 < 2288.
So, 1231 is one such acceptable value.
Coena must have demanded the fee of Rs. 1123 lakhs and so she must be either Z or Y (as concluded above).

The following cases are possible: Note:
e cannot take any value other than 1123.
E.g. let e = 2397.
In such a case no equation of the type ‘a + b = c + d’ is possible, which satisfies the given conditions. A possible equation is
‘1123 + 2289 = 1213 + 2379’ but since 2289 > 2288, it cannot be taken

(1) Option C
(2) Option D
(3) Option C

DIRECTIONS: Answer the questions on the basis of the information given below. Four machines A, B, C and D can produce four items E, F, G and H. The efficiency (in units/hr) of the machines for each product (while working alone on that product) is given in the table below. (1) If 5000 units each of E, F, G and H are required then which machine will take the least time if used alone?
(a) C
(b) D
(c) B
(d) A

(2) 1680 units each of E, F, G and H are required. Each machine can produce only one item and no two machines can work simultaneously. How should the items be assigned to different machines so that the total time taken is minimum?
(a) A – H, B – E, C – G, D – F
(b) A – H, B – G, C – E, D – F
(c) A – F, B – E, C – G, D – H
(d) A – F, B – E, C – H, D – G

(3) Machines A and B can be operated only between 10 a.m. and 2 p.m. on a given day. They have to produce equal number of units of each item e.g. If machine A produces 100 units then it has to be 25 units each of E, F, G and H. What is the ratio of the maximum number of units that A and B can produce respectively?
(a) 31 : 36
(b) 83 : 96
(c) 31 : 37
(d) None of these

Solutions:

(1) Option A
Total time taken (in hrs.) to produce 5000 units each of E, F, G and H:
For Machine A:
=5000/75+5000/150+5000/125+5000/50
=5000×(1/75+1/150+1/50+1/125)
=240
For Machine B:
=5000/125+5000/75+5000/100+5000/100
=200×(1/5+1/3+1/4+1/4)
=206.67
For Machine C:
=5000/100+5000/125+5000/125+5000/75
=200×(1/4+1/5+1/5+1/3)
=196.67
For Machine D :
=5000/125+5000/75+5000/75+5000/125
=200×(1/3+1/5+1/5+1/3)
=213.33

(2) Option C
Observation of the table tells us that the answer should be
A-F, B-E, C-G, D-H.

(3) Option B
Let n units of each item be produced by machine A:
Hence, n/75+n/150+n/125+n/50 = 4
n(1/75+1/150+1/125+1/50)=4
n = 83.33
Since the machine, at maximum, can work for four hours, we will have to discard decimal part of 83.33. At maximum, 4 × 83 = 332 units can be produced by machine A.
Hence, 4n = 332.
Let k units of each item be produced by machine B:
Hence, k/125+k/75+n/100+n/100 = 4
k(1/75+1/100+1/125+1/100)=4
k = 96.77
At maximum, 4 × 96 = 384 units can be produced by machine B.
Hence, 4k = 384
Ratio = 332: 384 = 83: 96.

1

1

3

3

8

1

1

1