THE ASSOCIATION OF MATHEMATICS TEACHERS OF INDIA Screening Test - Bhaskara Contest NMTC at JUNIOR LEVEL - IX & X Standards Saturday, 1 September, 2018 Fill in the response sheet with your Name, Class and the institution through which you appear in the specified places. 2. Diagrams are only visual aids; they are NOT drawn to scale. 3. You are free to do rough work on separate sheets. 4 Duration of the test: 2 pm to 4 pm -- 2 hours. PART—A Note . Only one of the choices A. B, C, D is correct for each question. Shade the alphabet of your choice in the response sheet. If you have any doubt in the method of answering. seek the guidance of the supervisor. + For each correct response you get 1 mark. For each incorrect response you lose ; mark. 3+v6 1. The value of ————._-—_—— is 33-212 - a2 + (60-27 we @) 3 ©) V6 (©) ie Ans. (8) ol 346 36 (5+) "e843 -4V2+5V2-3v3 V3 +2 V3 +2 2 A train moving with a constant speed crosses a stationary pole in 4 seconds and a platform 75 m long in 9 seconds, The length of the train is (in meters) (A) 56 (B) 58 (C) 60 (0) 62 Ans. (C) Sol. _ Let the train have length (m and speed s m/sec. C se4 “ (475 8 o_ 78 4 gr = 40 +300 s= (By using (i) 5¢= 300 ¢= 60m.4 Ans. Ans. Sol. Sol. (One of the factors of &x® — 42" - 24xy + 16y* + 20y — 18x + 10is (Bonus) (A) 3x — 4y — 22 (B) 3x + 4y - 2z (C) 3x + 4y +2z (D) 3x — 4y + 22 ‘The natural number which is subtracted from each of the four numbers 17,31,25,47 to give four ‘numbers in proportion is at (8)2 (cy3 (4 (©) Let x be subtracted so that 17, 31, 25, 47 are in proportion. 17-x _ 25-x Bix” 47—x (17 x) (47 -x) = (25x) (31 - x) 799 + x’ — 64x = 775 + x" — 56x 24 = 8x x=3, The solution to the equation 5(3") + 3(5") = 510is (Bonus) (a2 (4 (5 (0) No solution («+ 1)?=x, the value of 11x° + 8x" + 8x-2is (at (2 ©3 (4 (a) (c+ 1)? =x exe t=0 11x° + 8x7 + Bx —2 C4 x41) (1x3) 41 = (0) (14x-3) +124 ‘There are two values of m for which the equation 4x* + mx + 8x + 9 = 0 has only one solution for x. The sum of these two value of mis (Bonus) yt (B)2 (3 (4 D=0 (m+8)-449 m+8=212 m=4,-20 sum = ~20=~16. ‘The number of zeros in the product of the first 100 natural numbers is (A) 12 (B) 15, (c) 18 (D) 24 (©) [22] of] [2] « 5] lel ls 204440404 corer 4 ‘So number of zeros is 24. The length of each side of a triangle in increased by 20% then the percentage increase of area is (A) 60% (8) 120% (C) 80% (0) 44% 0) Let side of A are a, b,c (a) area = (sa) (© 5) —e) When each side increased by 20% a’ =1.2a1. Ans. Sol. 12. b 2b c= 12¢ = 12 (arb+e) _, s (a") new area = /f-2s(7.28-1.2a) (1.28-1.2b) (1.28-1.20) 2s a, 15> The number of pais of eatvely prime postive integers (ab) such that 2+ "52 js an integers wt 2 3 4 © Given , bate relative prime postive integer such hat 2. "© = where ks an integer) = ask 15" 4a Nowas HGF. (a,b) =1 so. alts > alt,3,5,15) Pu a=i Similarly only b = 2 simplify Put a=3 =b=2 ass Sb=2 =15 Sb=2 (a, b)=(1,2),(3,2),(6.2),(15, 2) 4 Ans. ‘The four digit number 8ab9 is a perfect square. The value of a” + b” is (a) 52 (8) 62 (C)54 (0) 68 a 93° = 8649 =6,b= ab + b= 6+ 4? = 52, 1 a, b are positive real numbers such that ++ °~1. The smallest value of a+ bis a (a) 15 16 ov (0) 18 @)atb>10+6 a+b>16, a-2bis a, b real numbers. The least value of a” + ab +b (at (0 ()-1 (2 (c) fa, b) =a? + ab +b*-a-2b at b—1 +2b-2 fa = 1 {or stationary points we need f, = 1, = 0 which gives 2a+ a+2b=2 a=0,b=1 conly one point (a,b) Now check fas few — fan” (2) (2)- (17 =3>0 Now at (a, b) = (0, 1) faa > O and fi, > 0 So, (a, b) = (0, 1) will given minimum value ‘so f(0, 1) = 0+ 0+ 1-0-2 —1is the minimum value. lis the incenter of a triangle ABC in which 2A = 80°. ZBIC = (a) 120° (8) 10 (C) 125° (0) 130° (0) A 80 ‘ 2BIC-= 90+ 5 = 90+ > = 190°. In the adjoining figure ABCD is a square and DFEB is a rhombus “CDF = 5 a B D: e (a) 15° (8) 18° (C) 20° (0) 30° (a)Sol. Let side of square DG =6C Apply sine rule ACDF vz sin o= “3-1 and we knowsin 15° = ¥3=1 22 22 0-15 PART -B Note + Write the correct answer in the space provided in the response sheet + Foreach correct response you get 1 mark. For each incorect response you lose + marks 46 ABCD isa square EF are point on BC, CD respectively and EAF = 45°. The value of aeue is Ans. So : it ve18. 20. By ASA AABE = AADG AE = AG BE=GD — (CPCT) By SAS AGAF = AEAF ‘The average of 5 consecutive natural numbers is 10. The sum of the second and fourth of these numbers is, : 20 XPT EZ EES HE ag 5 5x +10 =50 80 the number are 8, 9, 10, 11, 12 9+11=20, ‘The number of natural number n for which n? + 96 is a perfect square is, n= 23,10,2,5 ‘so number of values of nis 4. ‘nis an integer and is also an integer. The sum of all such nis -6 Van n+1 noted nt1=-8 > ‘So sum of two value of So that = is @ traction where a, b have no common factors other 1. b exceeds a by 3. Ifthe numerator is increased by 7, the fraction is increased by unity. The value of a+b "Ans. Sol. 23. Ans. Sol. I inf 2 = 5 + 6x 2x* — 6x = 5. ‘The angle of a heptagon are 160°, 135%, 185°, 140°, 125°, x°, x°. The value of xis__. nt 2 160 + 135 + 185+ 140 + 125 + 2x = 900° 745° + 2x = 900" 2x = 155° (Sy-7 2 2° 3(AB? - AC?) ABCs a tangle and AD sits attude, i 8D = SDC, then the value of “=> is 2 ra a S{AB*— AC?) _ 3fth? + 25x) - (he +x) _ 3 [24x"] BC? (ex 36x24, Ans. Sol. ‘As sphere is inscribed in a cube that has surface area of 24 cm?. A second cube is then inscribed within the sphere. The surface area of the inner cube (in cm’) is 8 A positive integer n is multiple of 7. It Vf lies between 15 and 16, the number of possible values (s) of nis 4 15< yh <16 225 << 256 as nis multiple of 7 are 231, 238, 245, 252 so total 4 numbers. 2. M man do a work in m days. I there had been N men more, the work would have been finished n days eater then the value of Mis 1 Men | Day | Work Mm [m [Mm M+N | m—n | (M+N)(m—n) Mm = (M+N) (m—n)Mim Mm Mn+ Nn Nm —Mn= Nn oo | mM mN-Mn No, noN aN Nn 28. The sum of the digit of a two number is 15. Ifthe digits of the given number are reversed, the number is increased by the square of 3. The original number is 78 N=10a+b N= t0b+a atb=15 N =N+3? 10b + a=10a+b+9 9b -9a=9 b-a=t a=7,b=8 N=78 29. When expanded the units place of (3127)' "is % Ans. 7 Sol. Cyclicty of 7is 4 173 =4x43+1 50 unit digit is 7” 30. 1:3andc:(a+b)=5:7, then: (c+a)is Ans. Sol. Sa+5b-7o=0 by (i) x5 ~ (il) x3, we get 15a-5b-5c=0 154+ 15b -21c=0 - - + — 20b+ 16c=0 16, ~ 20 Pubs Sc in@)