CLASS XI KVPY EXAMINATION 2010 CAREER POINT
PART -A
One-Mark Question
MATHEMATICS
1. A student notices that the roots of the equation x2 + bx + a = 0 are each 1 less than the roots of the equation
x2 + ax + b = 0. Then a + b is.
(A) Possibly any real number (B) – 2
(C) – 4 (D) – 5
Ans. (C)
Sol. α + β = – b and αβ = a
α+1+β+1=–a ⇒ α+β+2=–a ⇒ –b+2=–a
b–a=2 …..(1)
(α + 1) (β + 1) = b ⇒ αβ + α + β + 1 = b ⇒ a + (– b) + 1 = b
2b – a = 1 .…(2)
from (1) & (2), b = – 1, a = – 3, a + b = –4
x x
+1 −1
2. If x, y are real numbers such that 3 y − 3 y = 24, then the value of (x + y) / (x – y) is
(A) 0 (B) 1 (C) 2 (D) 3
Ans. (D)
x x x x
+1 −1 3y
Sol. 3 y
−3 y
= 24 ⇒ 3 × 3 −
y
= 24
3
3
3 ( )
8 x/ y
= 24 ⇒
x
y
=2
x+ y
=3
x− y
3. The number of positive integers n in the set {1, 2, 3,…….100} for which the number
12 + 2 2 + 32 + ....... + n 2
is an integer is
1 + 2 + 3 + ...... + n
(A) 33 (B) 34 (C) 50 (D) 100
Ans. (D)
a n + b n + c n + .......
Sol. is always an integer if n = odd
a + b + c + ......
n = 7 odd here
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4. The three different face diagonals of a cuboid (rectangular parallelopiped) have lengths 39, 40, 41. The
length of the main diagonal of the cuboid which joins a pair of opposite corners is -
(A) 49 (B) 49 2 (C) 60 (D) 60 2
Ans. (A)
Sol. a 2 + b 2 = 39 ….(i)
b 2 + c 2 = 40 ….(ii)
c 2 + a 2 = 41 …(iii)
a2 + b2 + c2 = ?
Square and add (i), (ii) & (iii)
2(a2 + b2 + c2) = 392 + 402 + 412 = 4802
a2 + b2 + c2 = 2401 ⇒ a 2 + b 2 + c 2 = 49
5. The sides of a triangle ABC are positive integers. The smallest side has length l. Which of the following
statement is true ?
(A) The area of ABC is always a rational number
(B) The area of ABC is always an irrational number
(C) The perimeter of ABC is an even integer
(D) The information provided is not sufficient to conclude any of the statements A, B or C above
Ans. (B)
a=1 b
Sol.
c
1+b>c ⇒ c–b<1
1+c>b ⇒ b–c<1
–1 < b – c < 1
b, c are integers so b – c = 0 ⇒ b = c
2b + 1 1
semi perimeter S = = b+
2 2
1 1 1 b 1 1 2 1
area A = b + b + − b b + − c + − 1 = b − = Irrational
2 2 2 2 2 2 4
6. Consider a square ABCD of side 12 and let M, N be the midpoints of AB, CD respectively. Take a point P
on MN and let AP = r, PC = s. Then the area of the triangle whose sides are r, s, 12 is-
rs rs
(A) 72 (B) 36 (C) (D)
2 7
Ans. (B)
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, CLASS XI KVPY EXAMINATION 2010 CAREER POINT
Sol.
D N C
s
P 6
r
r
A 6 M 6 B
PA = r, PC = s
So PB = r
Triangle with sides r, s & 12 is ∆PCB
P
r 6
B C
12
1 1
area = base × height = × 6 × 12 = 36
2 2
7. A cow is tied to a corner (vertex) of a regular hexagonal fenced area of side a metres by a rope of length 5a/2
metres in a grass field. (The cow cannot graze inside the fenced area.) What is the maximum possible area
of the grass field to which the cow has access to graze ?
5 2
(A) 5πa2 (B) πa (C) 6πa2 (D) 3πa2
2
Ans. (A)
120 5a
2
60 3a
2
60 a
2
Sol. Area = 2 × π + × π + × π
360 2 360 2 360 2
2π 25a 2 9a 2 a 2
= + +
3 4 8 8
2π 30a 2
= × = 5πa2
3 4
60º
a/2
a/2
a
120º 60º
60º 60º 3a/2
a a
120º
5a/2
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PART -A
One-Mark Question
MATHEMATICS
1. A student notices that the roots of the equation x2 + bx + a = 0 are each 1 less than the roots of the equation
x2 + ax + b = 0. Then a + b is.
(A) Possibly any real number (B) – 2
(C) – 4 (D) – 5
Ans. (C)
Sol. α + β = – b and αβ = a
α+1+β+1=–a ⇒ α+β+2=–a ⇒ –b+2=–a
b–a=2 …..(1)
(α + 1) (β + 1) = b ⇒ αβ + α + β + 1 = b ⇒ a + (– b) + 1 = b
2b – a = 1 .…(2)
from (1) & (2), b = – 1, a = – 3, a + b = –4
x x
+1 −1
2. If x, y are real numbers such that 3 y − 3 y = 24, then the value of (x + y) / (x – y) is
(A) 0 (B) 1 (C) 2 (D) 3
Ans. (D)
x x x x
+1 −1 3y
Sol. 3 y
−3 y
= 24 ⇒ 3 × 3 −
y
= 24
3
3
3 ( )
8 x/ y
= 24 ⇒
x
y
=2
x+ y
=3
x− y
3. The number of positive integers n in the set {1, 2, 3,…….100} for which the number
12 + 2 2 + 32 + ....... + n 2
is an integer is
1 + 2 + 3 + ...... + n
(A) 33 (B) 34 (C) 50 (D) 100
Ans. (D)
a n + b n + c n + .......
Sol. is always an integer if n = odd
a + b + c + ......
n = 7 odd here
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, CLASS XI KVPY EXAMINATION 2010 CAREER POINT
4. The three different face diagonals of a cuboid (rectangular parallelopiped) have lengths 39, 40, 41. The
length of the main diagonal of the cuboid which joins a pair of opposite corners is -
(A) 49 (B) 49 2 (C) 60 (D) 60 2
Ans. (A)
Sol. a 2 + b 2 = 39 ….(i)
b 2 + c 2 = 40 ….(ii)
c 2 + a 2 = 41 …(iii)
a2 + b2 + c2 = ?
Square and add (i), (ii) & (iii)
2(a2 + b2 + c2) = 392 + 402 + 412 = 4802
a2 + b2 + c2 = 2401 ⇒ a 2 + b 2 + c 2 = 49
5. The sides of a triangle ABC are positive integers. The smallest side has length l. Which of the following
statement is true ?
(A) The area of ABC is always a rational number
(B) The area of ABC is always an irrational number
(C) The perimeter of ABC is an even integer
(D) The information provided is not sufficient to conclude any of the statements A, B or C above
Ans. (B)
a=1 b
Sol.
c
1+b>c ⇒ c–b<1
1+c>b ⇒ b–c<1
–1 < b – c < 1
b, c are integers so b – c = 0 ⇒ b = c
2b + 1 1
semi perimeter S = = b+
2 2
1 1 1 b 1 1 2 1
area A = b + b + − b b + − c + − 1 = b − = Irrational
2 2 2 2 2 2 4
6. Consider a square ABCD of side 12 and let M, N be the midpoints of AB, CD respectively. Take a point P
on MN and let AP = r, PC = s. Then the area of the triangle whose sides are r, s, 12 is-
rs rs
(A) 72 (B) 36 (C) (D)
2 7
Ans. (B)
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, CLASS XI KVPY EXAMINATION 2010 CAREER POINT
Sol.
D N C
s
P 6
r
r
A 6 M 6 B
PA = r, PC = s
So PB = r
Triangle with sides r, s & 12 is ∆PCB
P
r 6
B C
12
1 1
area = base × height = × 6 × 12 = 36
2 2
7. A cow is tied to a corner (vertex) of a regular hexagonal fenced area of side a metres by a rope of length 5a/2
metres in a grass field. (The cow cannot graze inside the fenced area.) What is the maximum possible area
of the grass field to which the cow has access to graze ?
5 2
(A) 5πa2 (B) πa (C) 6πa2 (D) 3πa2
2
Ans. (A)
120 5a
2
60 3a
2
60 a
2
Sol. Area = 2 × π + × π + × π
360 2 360 2 360 2
2π 25a 2 9a 2 a 2
= + +
3 4 8 8
2π 30a 2
= × = 5πa2
3 4
60º
a/2
a/2
a
120º 60º
60º 60º 3a/2
a a
120º
5a/2
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