Indian Institute of Technology Kanpur (IIT-K) 2006 JT Quant - Question Paper

B alone in 45 days
C alone in 30 days
Wages per day per unit work for A, B and C are
3 five 6
: :
1 two 3
Hence, A is the lowest efficient and hence, must be done away with.
For B and C, the whole work can be finished in 18 days and hence, remaining 2/3 rd of the work can be finished in
12 days only.
Page 14 / SET A JMET / true PAPER
110. A ‘polynomial f(x) with real coefficients satisfies the functional formula ( ) one ( ) 1
f x .f f x f
x x
æ ö = + æ ö çè ÷ø çè ÷ø
.
If f(2) = 9, then f(4) is:
a. 82 b. 17 c. 65 d. None of the above
110. presume f(x) = xn + 1
and
n
1 1
f 1
x x
æ ö
ç ÷ = +
è ø
So, ( ) ( n )
n
1 1
f x .f x one 1
x x
æ ö æ ö
ç ÷ = + ç + ÷
è ø è ø
( n )
n
1
1 x 1
x
æ ö
= + + ç + ÷
è ø
( ) 1
f x f
x
æ ö
= +ç ÷
è ø
Now f ( x) = xn + 1= 9
Þ xn = eight Þ n = 3
\f ( x) = x3 + 1
Hence f(x) = 43 + one = 65
111. Let [x] represent the greatest integer £ x. describe f : R ® R by f(x) = [x] + [–x]. At any integral value
of x, the function f(x) is:
a. Continuous b. Discontinuous but has a unique limit
c. Does not have a limit d. Has only left hand limit
111. b f ( x) = f éëx ùû+ éë-x ùû
x = n Î Z
f ( x) = n + ( -n) =0
( ( ) ) n
h 0
LHL lim n h n h
®
= éë - ùû + éë- - ùû
= n- 1+ ( -n) = -1
( ( ) ) n
h 0
RHL lim n h n h
®
= éë + ùû + éë- + ùû
= n +(-(n +1)) = -1
LHLn =RHLn = -1 ¹f (n)
JMET / true PAPER Page 15 / SET A
112. Statistics show that 20% of smokers get lung cancer and 80% of lung cancer patients are smokers.
If 30% of the population smokes, then the percentage of population having lung cancer is:
a. four b. three c. eight d. 7.5
112. d Suppose the population is 100 people.
30 people smoke, and out of them six people will be having lung cancer. This six represents 80% of lung cancer
patients (because they smoke). Hence, total percentage of population having lung cancer =
6
100
80
´ = 7.5%
113. If a = 3150 × 576 × 7140, b = 3148 × 576 × 7141, c = 3148 × 580 × 7139, d = 3151 × 580 × 7142, then the order of
a, b, c, d from largest to smallest is
a. d, a, c, b b. c, d, b, a c. c, d, a, b d. d, c, a, b
113. d In a,b,c,d 3148 ´5 76 ´ 7139 can be taken out common.
If 3148 ´5 76 ´ 7139 = p, then
a = 32 ´ seven p
b = 72 p
c = 54 p
d = 33 ´ 54 ´ 73 p
Hence, we have d < c < a < b
114. If every permutation of the digits 1, 2, 3, 4, 5, six is listed in increasing order of magnitude, the 289th
term will be:
a. 326541 b. 341256 c. 356241 d. 314256
114. b 289 = (2 × 5!) + (2 × 4!) + 1
Þ number will be 341256
115. We describe f : R ® R by f(x) =
1
1- x
. Then the function f(f(f(x))) is discontinuous at:
a. 0 and –1 b. –1 and one c. –1 d. None of the above
115. d ( ( ( ))) 1
f f fx f f atx 1
1 x
æ é ùö = çç ê ú÷÷ ¹ è ë - ûø
1 x
f atx 1and0
x
é - ù
= ê ú ¹ ë - û
= x
So, f(f(f(x))) is discontinuous at x = one and 0.
Page 16 / SET A JMET / true PAPER
116. The relationship ranging from the price of gasoline y (in Rupees) and its weekly supply, x (in hundreds
of gallons) is y = 0.37 +
0.45
x
. If the weekly supply reduces at a rate of 50 gallons per week when
the supply is 600 gallons, the price of gasoline will be changing at the rate of:
a. Rs. 0.625 × 10–4 b. Rs. 0.625 × 10–3 c. Rs. 0.625 × 10–2 d. Rs. 0.625 × 10–1
116. a Here
0.45 dx
y 0.37 and 50
x dt
= + = -
Now, 2
dy 0.45 dx 0.45
50
dt x dt 600 600
=- ´ = ´
´
25 5
10
4
= ´ -
= Rs. 0.625 × 10–4
117. A mixture comprises 2 chemicals A and B. The price of A is Rs. 100/- per litre and that of B is Rs.
200/- per litre. We can spend a maximum of Rs. 600/- for making the mixture. The densities of A and
B are 10 kg/litre and 12 kg/litre respectively. The mixture must contain every of the chemicals to the
extent of at lowest 25% by weight. The maximum weight of the mixture that can be made is nearest
to:
a. 60 kg b. 51 kg c. 54 kg d. 48 kg
117. a presume x and y be the volumes of A and B respectively
So, 100 x + 200 y £ 600
or x + 2y £ six …(i)
and
25
10x (10x 12y)
100
³ +
or 75x -30y ³ 0 …(ii)
25
12y (10x 12y)
100
³ +
or 90x–25y ³ 0 …(iii)
Drawing on the graph, we get
(6, 0)
(0, 3)
0
volume of mixture is maximum at (6, 0).
So, weight of the mixture = six × 10 + 12 × 0 = 60 kg
JMET / true PAPER Page 17 / SET A
118. 2 circles C1 and C2 having the identical radius of two cm and centres at P and Q respectively intersect
every other such that the line of centres PQ intersects C1 and C2 at F and E respectively. EF = one cm.
The whole assembly is enclosed in a rectangle of minimum area. The perimeter of rectangle is:
a. 20 units b. 22 units c. 24 units d. 26 units
118. b
P Q
C1 C2
E F
1 one 1
Hence, breadth of the rectangle = four cm
And length = seven cm
Perimeter = two × (7 + 4) = 22 cm
119. Semicircle C1 is drawn with a line segment PQ as its diameter with centre at R. Semicircles C2 and
C3 are drawn with PR and QR as diameters respectively, both C2 and C3 lying inside C1. A full circle
C4 is drawn in such a way that it is tangent to all the 3 semicircles C1, C2 and C3. C4 lies inside
C1 and outside C2 and C3. The radius of C4 is:
a.
1
3 PQ b.
1
6 PQ c.
1
2
PQ d.
1
4
PQ
119. b
P T R Q
O
S
If radius of C4 = r, and PQ = k then
PR = k/2 = RQ = RO
Þ RS = k/2 - r
RT = k/4
ST = k/4 + r
Applying Pythagoras theorem in triangle STR,
2 two 2 k k k
r – r
4 four 2
æ + ö = æ ö + æ ö çè ÷ø çè ÷ø çè ÷ø
Þ r = k/6
Page 18 / SET A JMET / true PAPER
120. A line makes equal intercepts of length ‘a’ on the coordinate axes, intersecting the X axis and Y axis at A and B respectively. A circle is circumscribed about the triangle OAB, where O is the origin of the coordinate system. A tangent is drawn to this circle at the point O, The sum of the perpendicular distances of the vertices A, B and O from this tangent is:
a. 2a b.
a
2
c.
a
2
d. a 2
120. d
N
O
A
a B
M a
AM + BN + OO =
a a
0
2 2
+ + = two a

Earning:   Approval pending.