NATIONAL TALENT SEARCH EXAMINATION (NTSE-2019) STAGE -1 [PAPER CODE : X ] BENGALURU, RNATAKA STATE : TAMIL NADU PAPER : SAT Dte: 04-11-2018 Max. Mrks: 100 SOLUTIONS Tim lowed: 120 nutes 1 éùæö4 1 1 êú9ç+ 1253 ÷ 101. The value of êúç-1 ÷ is : ëûèø64 3 (1) 9 (2) 3 (3) 81 (4) 994 A. (2) 1 éù4 éù1 1 êú+ 3 S. 9êú- 1 125 ëûêúëû64 3 1 =+ëûéù945[]4 = 3 102. If m+- n p0=, then the value of ()mnp+- 2 is (1) mn (2) –mn (3) 2 mn (4) 4 mn A. (4) S. m+- n p0= Þ mnp+= ÞSquaring both sides: mn2mnp++= Þ m+ n - p =- 2 mn Þ ()m+- n p2= 4mn 1 1 103. If x2 += 14 , then the value of x 3 + is : x2 x3 (1) 52 (2) 42 (3) 24 (4) 25 A. (1) 1 1 S. x2 += 14 x2 æö1 2 Þ çx+=÷ 16 èøx 1 Þx4+ =± x Taking + 4, 1 x4+= x Cubing both sides: æö1 3 çx+=÷ 64 èøx 11æö Þx3+ + 3ç x +÷ ×= 1 64 xx3èø 1 Þx3+ = 64- 12= 52 x3 104. The polynomials ax4x3x432+ +- and x3 - 4xa+ leave the same remainder when divided by x – 3, then the value of a is : (1) –1 (2) –4 (3) 4 (4) 1 A. (1) S. 1 : ax3 + 4x2 + 3x – 4 Acc. to question: 27a + 36 + 9 – 4 = 27 – 12 + a Þ 26a = – 26 a = – 1 105. The bisectors of Ð B and Ð C of a triangle ABC meet at a point O. If Ð A=60°, then Ð BOC is: (1) 30° (2) 60° (3) 90° (4) 120° A. (4) A 60° O S. x y x y B C 2 2x + 2y = 120° x + y = 60° Þ ÐBOC = 120° 106. The side BC of D ABC is produced to a point D. The bisectors of Ð ABC and Ð ACD meet at point E. If Ð BAC = 60°, then Ð BEC is: (1) 15° (2) 30° (3) 60° (4) 120° A. (2) S. A E 60° y x y x D B C 2y = 60 + 2x ---------- (i) ÐACB =- 180 2y In DBEC ÐBEC =-+- 180( x 180 y) = y – x From equation (i) y – x = 30° Þ ÐBEC=° 30 107. D ABC is an isosceles triangle in which AB = AC. If the side BA is produced to D such that AD = AB, then Ð BCD is : (1) 30° (2) 45° (3) 60° (4) 90° A. (4) D y A 2x S. y x x B C 3 Now 2y + 2x = 180° Þ y + x = 90° Þ ÐBCD=° 90 108. In an A.P., the sum of m terms is equal to n and the sum of n terms is equal to m, then the sum of (m+n) terms is : (1) m + n (2) -(mn+) (3) (mn- ) (4) (nm- ) A. (2) S. Given Sm = n Sn = m m Þ ëûé2am1dn +() - ù= 2 n Also: ëûé2an1dm +() - ù= 2 2n Þ2a+-() m 1d= m 2m 2a+-() n 1d= n 2n 2m Subtracting : ()m- nd =- mn é+(nmnm)( -ù) dmn2()-Þêú ëûmn -2nm( +) Þd = ----------------- (i) mn 2n Also : 2a + (m – 1)d = m 2n Þ 2a + (m + n – 1)d = + nd m 2n é-+ù2nm( ) = + n êú from (i) mëû mn = –2 Thus, sum of m + n terms 4 æömn+ =ç÷ éëû2a++() m n-ù 1d èø2 =-(mn+) 109. If the roots of the equation (abx2bacxbc02+ 22) -( +) ++=( 22) are equal, then : 2ac (1) 2bac=+ (2) b = (3) 2 = (4) b = ac ac+ b ac A. (3) S. (a2 + b2) x2 – 2b (a + c)x + b2 + c2 = 0 Þ D = 0 Þ 4bac2()+=++2 4abbc( 2222)( ) 22éù+2+=22+22 ++ 4 22 Þbaëûc 2ac ab ac b bc Þ2acb2 = a2c2 + b4 Þ (b2 – ac)2 = 0 Þ b2 = ac 110. A right triangle has hypotenuse of length p cm and one side of length q cm. If (p – q) = 1, then the length of third side is : (1) 2q1+ (2) 2q1+ (3) 2p1+ (4) 2p1+ A. (2) A p q S. C B x xqp222+= x=- p22 q=+=+ pq 2q1 111. The longest pole that can be kept in a room of dimentions 5m´´ 4m 2m is (1) 9 5m (2) 6 5m (3) 3 5m (4) 5 3m A. (3) S. 5 E D F G H 2 C 4 A 5 B Longest pole joint A & D AC = 25+= 16 41 AD = 414+ = 45 = 35 m 112. If the volume of a sphere is equal to its surface area, then the circumference of a cross sectional circle whose centre coincides with the sphere is : (1) 2p (2) 4p (3) 6p (4) 8p A. (3) S. Given 4 pR32 =p 4R 3 R = 3 Circumference = 2r6p=p 113. A circle is inscribed in a triangle ABC with right angle at A. The length of the t*° sides containing the right angle are 6 cm and 8 cm respectively. The radius of the circle is: (1) 2 cm (2) 6 cm (3) 8 cm (4) 10 cm A. (1) 10 6 S. 8 1 Area of D=´´68 2 = 24 Semi perimeter = 12 D 24 r=== 2cm 8 12 6 114. A fair die is thrown once. The probability of getting neither a prime nor a composite number is : 5 1 (1) 1 (2) 0 (3) (4) 6 6 A. (4) 1 S. 6 115. If the product of two zeroes of the polynomial x32-+- 6x11x6 is 2, then the third zero is : (1) 1 (2) 2 (3) 3 (4) 4 A. (3) S. x3 – 6x2 + 11x – 6 = 0 abg=6 Given ab=2 Þ g=3 116. If the HCF of 55 and 22 is expressed in the form of 55m-´ 22 2 then the value of m is : (1) 2 (2) 1 (3) 11 (4) 22 A. (2) S. H.C.F of 55 & 22 = 11 Given: 55m – 2 ´ 22 = 111 Þ m = 1 117. The graphs of the linear system xy1;2x2y2+= += gives : (1) no solution (2) unique solution (3) infinitely many solutions (4) two solutions A. (3) S. Conceptual a x 118. If tan q= , then the value of is : x ax22+ (1) cos q (2) sin q (3) cos ecq (4) secq A. (1) S. Conceptual 119. The remainder when xnn + divided by x1- is : (1) n (2) cannot be deterrnined (3) n + 1 (4) 0 A. (3) S. Conceptual 7 120. Which of the following statements are not true ? (a) sum of two irrational numbers always irrational (b) difference between two irrational numbers is irrational (c) product of two irrational numbers irrational (d) quotient of two irrational numbers is irrational (1) (a) and (b) only (2) (a), (b), (c) and (d) (3) (a), (b) and (c) only (4) none of the above A. Conceptual Sl. (2) 121. A car travels from Chennai to Bengaluru with a speed of 60 km/hr and returns back along the same path with a speed of 40 km/hr- The average speed of the car is given by: (1) 50 km/1hr (2) 13-8 m/s (3) 48 km/hr (4) 172.8 m/s A. (3) 2vv 2´´ 60 40 2 ´´ 60 40 km V====12 48 S. av ++ v12 v 60 40 100 hr 122. What will be the percentage change in momentum of a body when both its mass and velocity are doubled ? (1) 400 (2) 75 (3) 500 (4) 300 A. (4) S. p1== m 11 v( 2m)( 2v) == 4 mv 4p (4pp-) % change = ´100= 300% p 123. The displacement-time graph for two particles are shown in the figure. The ratio of velocity of A to velocity of B is : ) m B ( t n e m e c A a l 30° p s i D 30° Time (S) 1 (1) 1 (2) 3 (3) (4) 3 3 A. (3) 1 V tan301° S. A = ==3 ° Vtan603B 3 8 124. The velocity - time graph of a body moving along a staright line is shown below. The acceleration of the body along OA, AB and BC is : (1) 5m/s,0,22- 5m/s (2) -5m / s22 ,0,+ 5m /s (3) 5m / s2 ,1.6m / s 22 ,- 5m / s (4) -5m/s2 ,1.6m/s 22 ,5m/s A. (1) D-v 100 S. a=== 5m/s2 OA D-t 20 O a== 0 m/s2 AB Dt 0- 10 a= =- 5 m/s2 BC 108- 125. Two bodies A and B having masses 2 kg and 4 kg respectively are separated by 2m. Where should a body of mass 1 kg be placed so that the gravitational force on this body due to A and B is zero? (1) 8.3 m (2) 0.83 m (3) 3.8 m (4) 0.38 m A. (2) A B S. 1 kg 2 kg 4 kg x (2–x) 2 G ()2()1G()4 (1) = x2 (2x-)2 9 12= x2()2x-2 12 Þ= Þ-()2 x= 2x x() 2x- 2=+ x 2x 2 Þx== 0.83m 12+ 126. A ship of mass 3´ 107 kg initially at rest is pulled by force of 5´ 104 N through a distance of 3 m. Assuming that the resistance due to water is negligible the speed of the ship is: (1) 1.5 m/s (2) 60 m/s (3) 0.1 m/s (4) 5 m / s A. (3) F 5´ 104 S. a == m 3´ 107 45 æö2´´ 5 10 10- 1 V222=+=+ u 2as 0ç´ 3÷=== 10 èø3´ 1077 10 100 11 V2= Þ V== 0.1m/s 100 10 127. An electric bulb is rated 220 V, 110 W.