# KVPY-SA 2017 Mathematics Paper with Solutions

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### KVPY SA 2017 - Maths

Question 1: A quadrilateral has distinct integer side lengths. If the second-largest side has length 10, then the maximum possible length of the largest side is

1. a. 25
2. b. 26
3. c. 27
4. d. 28

Solution:

Let a, b, c, d are the four sides of a quadrilateral

Where a > b > c > d are distinct positive integers and b = 10 (given)

Now, we know that in a quadrilateral

Sum of any three sides > 4th side

⇒ a < b + c + d

Since b = 10 and b > c > d, so the largest possible values of c and d are 9 and 8 respectively.

⇒ a < 10 + 9 + 8

⇒ a < 27

Therefore, the largest possible value of a is 26.

Question 2: The largest power of 2 that divides 200!/100!

1. a. 98
2. b. 99
3. c. 100
4. d. 101

Solution:

The exponent of 2 in 200!

= (200/2) + (200/22) + ……… + (200/27)

= 100 + 50 + 12 + 6 + 3 + 1

= 197.

The exponent of 2 in 100!

= (100/2)+(100/22)+…(100/26)

= 50+12+6+3+1

= 97

Therefore, the exponent of 2 in 200!/100! = 197 – 97 = 100.

Question 3: Let a1, a2, a3, a4 be real numbers such that a1+ a2 +a3 + a4 = 0 and a12+ a22 +a32 + a42 = 1. Then the smallest possible value of the expression (a1 – a2)2 + (a2 – a3)2 + (a3 – a4)2 + (a4 – a1)2 lies in the interval

1. a. (0, 1.5)
2. b. (1.5, 2.5)
3. c. (2.5, 3)
4. d. (3, 3.5)

Solution:

Given, a12+ a22 +a32 + a42 = 1.

Therefore (a1 – a2)2 + (a2 – a3)2 + (a3 – a4)2 + (a4 – a1)2 = 0

When a1 = a2 = a3 = a4 = +1/2 or -1/2.

Question 4: Let S be the set of all ordered pairs (x, y) of positive integers satisfying the condition x2 – y2 = 12345678. Then

1. a. S is an infinite set
2. b. S is the empty set
3. c. S has exactly one element
4. d. S is a finite set and has at least two elements

Solution:

x2 – y2 = 12345678

OR

(x – y) (x + y) = 12345678 x, y belongs to z+

Case I: x, y belongs to odd

Difference of squares of two odd integers is multiple of 8 (or divisible by 8)

⇒ x2 – y2 = 8k, k belongs to I

But R.H.S is not divisible by ‘8’

Therefore, equality can’t hold.

Case II:x, y belongs to even

⇒ x – y belongs to even, x + y belongs to even

L.H.S = (x – y) (x + y)

= 2n×2m (n, m belongs to I)

= 4 mn (multiple of 4)

But R.H.S is not divisible by ‘4’

Again not possible.

Case III:x belongs to odd, y belongs to even

⇒ x – y belongs to odd, x + y belongs to odd

⇒ L.H.S = (x – y) (x + y)

= odd × odd

= odd

But R.H.S is even.

So not possible.

Therefore, S is an empty Set.

Question 5: Let A1A2A3....A9 be a nine-sided regular polygon with side length 2 units. The difference between the lengths of the diagonals A1A5 and A2A4 equals

1. a. 2+ √12
2. b. √12-2
3. c. 6
4. d. 2

Solution:

In ΔA1OA2

∠ A1OA2 = 2 π/9 = 400

⇒ cos 40° = (x2+x2-22)/2.x.x {using cosine rule}

⇒ 2x2 (cos 400–1) = –4

⇒ x2 = 2/(1-cos 400) = 2/2sin2200

⇒ x = 1/sin 200 ……(1)

In Δ A1OA5

Now, ∠ A1OA5 = 8π/9

⇒ cos 8π/9 = (x2+x2-(A1A5)2)/2x2 {using cosine rule}

⇒ A1A52 = 2x2 (1 – cos 1600)

⇒ A1A5 = 2 x sin800 ……. (ii)

Similarly, In Δ A2OA4

A2A4 = 2 x sin400

Therefore A1A5 – A2A4 = 2 x (sin 800 – sin400)

= 2(1/sin 200)2.sin 200cos 600 (sin C – sin D = 2 cos(C+D)/2 sin (C-D)/2)

= 2

Question 6: Let a1, a2,.... an be n nonzero real numbers, of which p are positive and remaining are negative. The number of ordered pairs (j, k), j < k, for which aj ak is positive, is 55. Similarly, the number of ordered pairs (j, k), j < k, for which aj ak is negative is 50. Then the value of p2 + (n – p)2 is

1. a. 629
2. b. 325
3. c. 125
4. d. 221

Solution:

Given, a1, a2, .... an be n nonzero real numbers, among which p are positive and n – p are negative.

The number of ordered pair (j,k), j < k such that the product aj . ak is positive is 55.

Which is possible when either both aj, ak are positive or both are negative.

pC2 + n–p C2 = 55 (No. of ways of selecting two positive and two negative numbers)

p(p-1)/2 +(n-p)(n-p-1)/2 = 55

p2 + (n – p)2 – n = 55 ….(i)

The number of ordered pair (j,k), j < k such that the product aj . ak is negative is 50.

Which is possible when one of aj, ak is positive and other negative.

Required number of ways = Total number of ways of selecting two numbers - No. of

ways of selecting two positive and two negative numbers

nC2 – 55 = 50

n(n-1)/2 = 105

n2 – n – 210 = 0

n = 15, –14 (n can’t be negative therefore – 14 is rejected)

Now, put n = 15 in equation (i)

So p2 + (n – p)2 = 125

Question 7: If a, b, c, d are four distinct numbers chosen from the set {1,2,3 ....,9}, then the minimum value of is

1. a. 3/8
2. b. 1/3
3. c. 13/36
4. d. 25/72

Solution:

a, b, c, d ∈ {1, 2, 3, ………9}

Now, (a/b)+(c/d) is minimum if (a/b) and (c/d) individually become minimum.

a/b -> min, when a -> min and b -> max.

c/d -> min, when c -> min and d -> max.

Also a, b, c, d are distinct

So minimum (a/b)+(c/d) = (1/9)+(2/8) Or (1/8)+(2/9)

= 26/72 or 25/72

Question 8: If 72x. 48y = 6xy, where x and y are non zero rational numbers, then x + y equals

1. a. 3
2. b. 10/3
3. c. –3
4. d. –10/3

Solution:

72x. 48y = 6 x. y

(23.32)x.(24.31)y = (2.3)x.y

23x+ 4y. 32x + y = 2x. y . 3xy

Equating the powers of 2 and 3, we get

3x + 4y = xy …………(i)

2x + y = xy ……………(ii)

Subtract (ii) from (i)

x = – 3y { put in (i) }

3(–3y)+ 4y = (–3y).y

– 5y = – 3y2

y = 5/3, x = – 5

Hence x + y = (5/3)-5

= -10/3

Question 9: Let AB be a line segment of length 2. Construct a semicircle S with AB as diameter. Let C be the midpoint of the arc AB. Construct another semicircle T external to the triangle ABC with chord AC as diameter. The area of the region inside the semicircle T but outside S is

1. a. π/2
2. b. 1/2
3. c. π/√2
4. d. 1/√2

Solution:

C being midpoint of arc ACB ⇒ AX = BX ⇒ ∠CAB = ∠CBA

Also ∠ ACB = 90 (Angle in semicircle)

OC AB (Construction)

⇒ ∠OAC = ∠OCA = 45

OA =OX = 1

In ΔOAC

AC2 = OA2 + OC2

⇒ AC = √2

Now, Required Area = Area of semicircle on AC as diameter – Area of segment A.M.C.A

= (1/2) ×π (√2/2)2-((π /4)-(1/2) )

= ((π /4)-(π /4)+(1/2) )

= 1/2

Question 10: Let r(x) be the remainder when the polynomial x135 + x125 – x115 + x5 + 1 is divided by x3– x. Then.

1. a. r(x) is the zero polynomial
2. b. r(x) is a nonzero constant
3. c. degree of r(x) is one
4. d. degree of r(x) is two

Solution:

x135+ x125 – x115 + x5 + 1 = q(x) (x3 – x) + Ax2 +Bx + C

Where q(x) is quotient and r(x) = Ax2 +Bx + C is remainder

Now, putting x = 0, 1, –1

We get

⇒ 1 = C

⇒ 3 = A + B + C

⇒ A+B = 2 ….(i)

⇒ –1 = A – B + C

⇒ -1 = A-B+C

⇒ A - B = -2

⇒ 2A = 0

⇒ A = 0 and B = 2

So r(x) = 2x + 1 a polynomial of degree 1.

Question 11: It is given that the number 43361 can be written as a product of two distinct prime numbers p1 , p2. Further, assume that there are 42900 numbers which are less than 43361 and are co-prime to it. Then, p1+p2 is

1. a. 462
2. b. 464
3. c. 400
4. d. 402

Solution:

The prime factors given number are,

43361 = 131 × 331

p1 + p2 = 131 + 331

= 462

Question 12: Let ABC be a triangle with ∠ C = 900. Draw CD perpendicular to AB. Choose points M and N on sides AC and BC respectively such that DM is parallel to BC and DN is parallel to AC. If DM = 5, DN = 4, then AC and BC are respectively equal to

1. a. 41/4, 41/5
2. b. 39/4, 39/5
3. c. 38/4, 38/5
4. d. 37/4, 37/5

Solution:

CM || ND and CN || MD

Therefore CMDN is a rectangle

⇒ CM = ND = 4 and CN = MD = 5

In Δ CMD

By Pythagoras theorem

CD2 = CM2 + MD2

CD = √41

Also, Δ AMD ~ Δ DNB

⇒ y/5 = 4/x = z/t …….(i) {Sides of similar triangles are in proportion}

By Pythagoras Theorem

(x + 4)2 = 41 + t2 ……..(ii)

In Δ BDC

By Pythagoras Theorem

(y + 5)2 = 41 +z2 ………(iii)

From (ii) & (iii)

((x+4)2-41)/((y+5)2-41) = (t/z)2

((x+4)2-41)/((20/x)+5)2-41) = (x/4)2 {using (i)}

16 (x2 + 16 + 8x – 41) = 400 +25x2 + 200x – 41x2

4x2 – 9x – 100 = 0

x = 25/4 , y = 16/5 {using (i) again }

Hence AC = x + 4 = 41/4 and BC = 5 + y = 41/5.

Question 13: Let A, G and H be the arithmetic mean, geometric mean and harmonic mean, respectively of two distinct positive real numbers. If α is the smallest of the two roots of the equation A(G – H)x2 + G (H – A) x + H(A –G) = 0, then

1. a. –2 < α < –1
2. b. 0 < α < 1
3. c. –1 < α < 0
4. d. 1 < α < 2

Solution:

Given, A(G–H) x2 + G(H–A) x + H(A–G) = 0

Put x = 1 in above equation

A(G–H) + G(H–A) + H(A–G) = 0

0 = 0

⇒ x = 1 is one root (let it be ‘α’ and other root be ‘β’)

Now using product of roots,

α. β = A(A-G)/A(G-H)

1. β = (H.A-H.G)/A(G-H)

β = (G2-H.G)/A(G-H) {A.H = G2}

β = G(G-H)/A(G-H)

β = G/A < 1{ as A > G}

Question 14: In the figure, ABCD is a unit square. A circle is drawn with centre O on the extended line CD and passing through A. If the diagonal AC is tangent to the circle, then the area of the shaded region is

1. a. (9-π )/6
2. b. (8-π )/6
3. c. (7-π )/4
4. d. (6-π )/4

Solution:

AC is tangent to circle with radius OA

Therefore ∠ OAC = 900

Also, AC is diagonal of square

⇒ OA = AD cosec(450) = √2

Now, Required Area = Area of square – (area of sector OAX – area of Δ OAD)

= 1-( π(√2)2. ((π/4)/2 π)-(1/2)(1).(1))

= 1-((π/4)-(1/2))

= (3/2)-( π/4)

= (6-π )/4

Question 15: The sum of all non-integer roots of the equation x5 – 6x4 + 11x3 – 5x2 – 3x + 2 = 0 is

1. a. 6
2. b. –11
3. c. –5
4. d. 3

Solution:

x5 – 6x4 + 11x3 – 5x2 – 3x + 2 = 0

(x – 1) (x – 2) (x3 – 3x + 1) = 0 {factorizing}

Now, x3 – 3x + 1 has non integral roots whose sum is 3.

Question 16: Let S be the circle in xy-plane which touches the x-axis at point A, the y-axis at point B and the unit circle x2 + y2 = 1 at point C externally. If O denotes the origin, then the angle OCA equals

1. a. 5π/8
2. b. π/2
3. c. 3π/4
4. d. 3π/5

Solution:

Circle S touches both the axes, therefore its center lies on y = x, let it be O’(a,a).

So A(a,0) and B(0,a)

⇒ ∠ O’OA = 45

‘C’ lies on x2 + y2 = 1 & on OO’

Therefore C (1/√2, 1/√2)

OO’ = OC + CO’

√2a = 1+a

a = 1/√2-1 = √2+1

Now, slope of AC = (0-1/√2)/(1+√2-(1/√2)

= – 1/(√2+1)

= – (√2-1)

= – tan π/8

= tan (π –π /8)

= tan (7π /8)

So ∠ CAX = 7π/8

Also ∠ CAX = ∠ OCA + ∠ COA {Exterior angle property}

⇒ 7 π /8 = ∠ OCA + π /4

⇒ ∠ OCA = 5 π /8.

Question 17: In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the non-parallel sides are all equal to 30. In order to maximize the area of the trapezium, the smallest angle should be

1. a. π/6
2. b. π/4
3. c. π/3
4. d. π/2

Solution:

x = 30 cos θ; y = 30 sin θ

Now, Area of trapezium

A = ½ (30+30+2x)×(y) { Area =1/2 (sum of parallel sides)height }

= (30+30cos θ) 30 sin θ

= 900 sin θ (1+cos θ)

Differentiating both sides with respect to θ

dA/d θ = 900 {sin θ. (– sin θ) +(1 + cos θ). cos θ } = 0 {for critical points}

⇒ cos2 θ – sin2 θ + cos θ = 0

⇒ 2cos2 θ + cos θ – 1 =0

⇒ (2cos θ – 1) (cos θ +1) = 0

⇒ cos θ = ½ or –1

θ = π /3 OR π (not possible).

Question 18: Let A1, A2, A3 be regions in the xy-plane defined by

A1 = {(x, y) : x2 + 2y2 ≤1},

A2 = {(x, y) : |x|3 + 2√2 |y|3 ≤ 1},

A3 = {(x, y) : max (|x|, √2 |y|) ≤1. Then

1. a. A1 ⊃A2 ⊃A3
2. b. A3 ⊃A1 ⊃A2
3. c. A2 ⊃A3 ⊃A1
4. d. A3 ⊃A2 ⊃A1

Solution:

$A_{1}=4\int_{0}^{1}\frac{{\sqrt{1-x^{2}}}}{\sqrt{2}}dx$ {ellipse}

= $\frac{4}{\sqrt{2}}\left \{ (\frac{x^{2}}{2}\sqrt{1-x^{2}}+\frac{1}{2}\sin ^{-1}x)_{0}^{1} \right \}$

= (4/√2)(0+π/4 -0)

= π/√2 ..(1)

Now for x(0,1)

x2> x3

1-x2<1-x3

$\sqrt[3]{1-x^{3}}> \sqrt{1-x^{2}}$

Therefore A2 = $4\int_{0}^{1}\frac{\sqrt[3]{1-x^{3}}}{\sqrt{2}}dx> 4\int_{0}^{1}\frac{\sqrt{1-x^{2}}}{\sqrt{2}}dx$ …(2)

= 2√2 ……..(3)

So from (1), (2) and (3)

A3 ⊃A2 ⊃A1

Question 19: Let ABCD be a square and E be a point outside ABCD such that E, A, C are collinear in that order. Suppose EB = ED = √130 and the areas of triangle EAB and square ABCD are equal. Then the area of square ABCD is

1. a. 8
2. b. 10
3. c. √120
4. d. √125

Solution:

Let side of square be ‘a’

⇒ Diagonal BD = a ⇒ OB = BD/2 = a/2

Now, In Δ OBE

BE2 = OE2+OB2 (using Pythagoras theorem)

⇒ OE = √130-a2/2

Now, ar (Δ EAB) = ar (Δ EOB) – ar (ΔAOB)

= (½)(a/√2) √(130-a2/2)-(1/2)(a/√2)2

= (a/4) √(130-a2/2)-a2/4

ar (square ABCD) = a2

Given, ar (Δ EAB) = ar (square ABCD)

⇒ (a2/4)√(260-a2)-a2/4 = a2

a2 (260 – a2) = 25a4

260 – a2 = 25a2

a2 = 10 (area of square).

Question 20: Consider the set A = {1, 2, 3, ..., 30}. The number of ways in which one can choose three distinct numbers from A so that the product of the chosen numbers is divisible by 9 is

1. a. 1590
2. b. 1505
3. c. 1110
4. d. 1025

Solution:

A = {1, 2, 3 …..….30}

Let X, Y , Z be subsets of A :

X = {9, 18, 27}

Y = {3, 6, 12, 15, 21, 24, 30}

Z = {1, 2, 4, 5, …….}

n(X) = 3, n(Y) = 7, n(Z) = 20

Now, the product of three selected numbers is divisible by 9, if

1) One from X, two from remaining

2) Two from X, one from remaining

3) All three from X

4) Two from Y, one from Z

5) All three from Y

So required no. of cases

= 3C1 ×27C2 +3C2 ×27C1 +3C3 + 7C2×20C1 +7C3

= 1590