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Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes SECTION – A {Mathematics} 1. Find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 3 (a) 866 (b) 867 (c) 868 (d) 869 Solution: First natural no. between 250 & 1000 divisible by 3 = 252 Last Natural No. between 250 & 1000 Divisible by 3 = 999 252 + 255 + …………999 Let no. of terms = n 999 = 252 + (n – 1) 3 999 – 252 = (n – 1) 3 747 3 n 1 249 = n – 1 S 250 [2 2 252 n 250 250 1 3] = 125[504 + 249 × 3] = 125 [504 + 747] = 125 × 1251 = 156375 There is no option in the given question (in order to minimize fluke and guessing probability ) 2. Find the value of k so that: 2k + 1, k2 +k + 1, 3k2 –3k +3 are in A.P. (a) 2,3 (b) 1,-2 (c) 1,2 (d) 3,4 Solution: 2k + 1, k2 + k + 1, 3k2 – 3k + 3 are in A.P (k2 + k + 1) – (2k + 1) = (3k2 – 3k + 3) – (k2 + k + 1) www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 1 Pioneer Education {The Best Way To Success} k2 + k + 1 – 2k – 1 = 3k2 Medical and Non - Medical Classes – 3k + 3 – k2 –k–1 k2 – k = 2k2 – 4k + 2 0 = 2k2 – 4k + 2 + k – k2 k2 – 3k + 2 = 0 k2 – 2k – k + 2 = 0 k(k – 2) – 1 (k – 2) = 0 k = 1, 2 Ans. (c) 3. If the first, second and last term of an A.P are a, b and 2a respectively, then its sum is (a) ab 2(b a) (b) ab (b a) (c) 3ab 2(b a) (d) none of these Solution: First term of A.P. =a 2nd term of A.P. = b common difference = 2nd term – first term d b a last term of A.P. = 2a a n 1 d 2a a n 1 b a n 1 a b a n 1 Sn 2a n 2a 2 a b a n b a a b a n b b a n 1 d b 2a 2 b a b b a 1 b a www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 2 Pioneer Education {The Best Way To Success} b 2a 2 b a b b a b a b 2a a 2 b a Medical and Non - Medical Classes b a 3ab a b a Ans. (c) 4. Find the value of x of the given quadratic equation (5 2 6)x (a) 6, 2 3 (5 2 6)x 2 3 (b) 5, 6 =10 (c) 3, 5 (d) 2, 3 2 Solution: 5 2 6 x2 3 5 2 6 Let 5 2 6 y2 1 y 10 52 2 6 5 2 6 2 25 24 5 2 6 1 5 2 6 y x2 3 5 2 6 1 y x2 3 x2 3 5 2 6 5 2 6 5 2 6 Let 5 2 6 y 5 2 6 x2 3 1 5 2 6 x2 3 10 y 10 10 y 2 1 10y y 2 10y 1 0 y 10 100 4 2 www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 3 Pioneer Education {The Best Way To Success} y 10 96 5 2 6 2 5 2 6 Medical and Non - Medical Classes x2 3 5 2 6 x2 – 3 = 1 x2 = 4 x= 5 2 6 2 1 5 2 6 x2 3 x2 – 3 = – 1 x2 = 2 x 2 Ans. (d) 5. Solve the given quadratic equation 6 6 (a) 1 (c) 3 (b) 2 6 .............. (d) 4 Solution: Let x = 6 x 6 6 .............. 6 x x2 = 6 + x x2 – x – 6 = 0 x2 – 3x + 2x – 6 = 0 x (x – 3) + 2 (x – 3) = 0 x = - 2, 3 (Negative value is rejected Ans. x = 3 6. Form a quadratic equation whose roots are given as: 3 + 5 , 3 – 5 (a) x2 + 6x - 4 = 0 (b) x2 – 6x + 4 = 0 (c) x2 – 6x – 4 = 0 (d) x2 +6x + 4 = 0 Solution: x2 – 5x + P = 0 www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 4 Pioneer Education {The Best Way To Success} S 3 5 3 Medical and Non - Medical Classes 5 =6 P 3 5 3 5 9 5 4 x2 – 6x + 4 = 0 Ans. (b) 7. If the root of the equation x2 px q 0 and x2 qx p 0 differ by the same quantity, then p q k .find the value of k (a) 1 (b)-1 (c)4 (d) -4 Solution: x2 px q 0 , x2 qx p 0 square both the sides 2 2 2 2 4 4 p2 4q q2 4p p2 q2 4q4p p q p q p q 4 p q 4 Ans. (d) 8. Determine the ratio in which the line y –x + 2 = 0 divides the line segment joining the points (3, –1) and (8,9) (a) 1:3 (b)1:2 (c)1:4 (d) 2:3 Solution: www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 5 Pioneer Education {The Best Way To Success} P Medical and Non - Medical Classes 8k 3 9k 1 , k 1 k 1 Since point P lies on the line y–x+2=0 9k 1 k 1 8k 3 k 1 9k 1 8k 3 k 1 2 0 2 k 1 0 9k – 1 – 8k – 3 + 2k + 2 = 0 3k – 2 = 0 k= 2 3 Ratio 2 : 3 Ans. (d) 9. What point on x-axis is equidistant from the points A (1, 3) and B (2, -5)? (a) 16 ,0 2 (b) 19 ,0 2 (c) 18 ,0 2 (d) 21 ,0 2 Solution: Let the required point on A(1, 3) x-axis is P(x, 0) B(2, – 5) PA = PB x 1 2 0 3 2 x 2 2 x 5 2 Square both sides x 1 2 9 x 2 2 25 www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 6 Pioneer Education {The Best Way To Success} x2 + 1 – 2x + 9 = x2 Medical and Non - Medical Classes + 4 – 4 x + 25 4x – 2x + 10 = 29 2x = 19 19 2 x Ans. (b) 10. The coordinate of centroid of a triangle are ( 3,2 ) and two of its vertices are (2 3, 1) and (2 3,5) Find the third vertex of the triangle. 3, 2 (a) (b) 3,2 (c) 3, 2 (d) 3,2 Solution: x1 x 2 x3 y 1 y 2 y 3 , Centroid = 3 3 Let the coordinates of the third vertex = (x, y) x 2 3 2 3 y 1 5 , 3 3 3,2 x 4 3 3 y 4 2 3 3, x 4 3 3 3, y 4 6 x 3, y Ans. 3, 2 2 Ans. (a) 11. Find the third vertex of a triangle whose centroid is (1, 4) and two vertices are (–1, 4) and (3, –1). (a)(1,9) (b) (1,6) (c) (1,5) (d) (1,8) Solution: Two vertices (–1, 4) & (3, –1) centroid = (3, –1) www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 7 Pioneer Education {The Best Way To Success} when the three vertices of the are (x1, y2), (x2, y2) & (x3, y3) x1 x 2 x3 y 1 y 2 y 3 , 3 3 Coordinates of centroid are = Let the third vertices of the 1,4 Medical and Non - Medical Classes be (x, y). x 1 3 y 4 1 , 3 3 x 1 3 y 4 1 1 & 4 3 3 x 2 3 & x 1 y 3 4 3 y 3 12 y 9 Ans.(a) (1, 9) 12. Find the value of x such that PQ = QR, where P, Q and R are (6,1), (1,3) and (x, 8) respectively. (a) –1 (b) –2 (c) –3 (d) –4 Solution: PQ = QR 6 1 P(6, 1) Q(1, 3) 2 1 3 2 1 x R(x, 8) 2 3 8 2 square both the sides 25 + 4 = (1 – x)2 + 25 4 = 1 + x2 – 2x x2 – 2x = 3 x2 – 2x – 3 =0 x2 – 3x + x – 3 =0 x(x –3) + 1(x–3)=0 x = –1, 3 Ans. (a) 13. If pth of an AP is q and qth term is p, then its nth term is (a) (p +q –n) (b) (p –q +n) (c) (p +q +n) (d) (p –q –n) www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 8 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes Solution: A + (p – 1) D = q ..(1) A + (q – 1) D = p ..(2) (1) – (2) (p – q) D = q – p D=–1 A + (p – 1) D = q A + (p – 1) (– 1) = q A–p+1=q A=p+q–1 Now, nth term = A + (n – 1) D = (p + q – 1) + (n – 1) (– 1) =p+q–1–n+1 =p+q–n Ans. (a) 14. The value of a such that x2 – 11x + a =0 and x2–14x + 2a = 0 may have a common root is (a) 0 (b) 12 (c) 24 (d) 32 Solution: Let the common root is 2 11 a 0 ............ 1 2 14 2a 0 ........ 2 Solving by cross multiplication method 2 22a 14a a 2a 1 14 11 www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 9 Pioneer Education {The Best Way To Success} 2 8a Medical and Non - Medical Classes 1 3 a 2 a , 3 8a a 2 8 2 a 3 a2 9 8 2 8 a 3 8a 3 a2 3 8a a2 8a 0 3 a a 8 3 a a 24 3 a a 24 0 0 0 a 0, 24 Ans. (a), (c) 15. The points (a2, 0) (0, b2) and (1, 1) are collinear if (a) 1 (b)2 (c)3 1 a2 1 b2 k . find the value of k (d)4 Solution: a2 ,0 0,b2 1,1 Since the three points are collinear x1 y 2 y3 a2 b2 1 x2 y 3 y 1 01 0 x3 y 1 y 2 1 0 b2 0 www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 10 Pioneer Education {The Best Way To Success} 2 2 ab a 2 2 0 b Medical and Non - Medical Classes 0 a2 b2 a2b2 Dividing both sides by a b2 a2 a2b2 1 a2 b2 a2b2 1 b2 a2b2 a2b2 1 k 1 Ans. (a) 16. If the pth, qth and rth terms of an A.P are a, b, and c respectively, then: (b–c) p + (c–a)q +(a–b)r = k. Find the value of k (a) 0 (b) 1 (c) 2 (d) 3 Solution : A + (p –1) D = a A + (q –1) D = b A + (r –1) D = c (b – c)p + (c – a) q + (a – b)r = k [{A + (q –1) D} – {A + (p –1) }] p + [{ A + (r –1) D} – { A + (p –1) D}] r = [(q –r)D] p + [(r –p) D]q + [(p –q) D] r = D[pq –pr +rq –pq + pr –qr] =0 Ans. (A) 17. If the sum of p, q and r terms of an A.P be a, b and c respectively. Then: a b c (q r) (r p) (p q) k p q r (a) 0 (b) 1 (c) 2 (d) 3 Solution : www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 11 Pioneer Education {The Best Way To Success} p 2A 2 p 1 D a q 2A 2 q 1 D b r 2A 2 r 1 D a q r p Now, Medical and Non - Medical Classes c b r p q c p q r = p 2A 2p = 1 2A 2 = 1 2A q r r p p q 2 p 1 D p r q 2A 2q 1 2A 2 p 1 D q r q 1 D r p q 1 D r p r 2A 2r 1 2A 2 r 1 D p q r 1 D p q D pq pr q r qr pq r p rp rq p q =0 Ans. (A) 18. The digits of a positive integer, having three digits are in A.P and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number. (a) 850 (b)851 (c) 852 (d) 853 Solution : Look at the options only 3rd option is satisfying all (C) the condition as per question 19. Find the value of x in the given quadratic equation 51+x +51–x = 26. (a) 1 (b) 2 (c) 3 (d) 4 Solution : 55+x + 51–x = 26 51 . 5x 51 + x 5 26 Let 5x = y www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 12 Pioneer Education {The Best Way To Success} 5y + Medical and Non - Medical Classes 5 26 y 5y2 – 26y + 5 = 0 5y2 – 25y +–y + 5 =0 5y(y–5) –1(y –5) =0 y= 5x = 1 ,5 5 1 5 x = –1 5x =5 x=1 Ans. (a) 20. The co-ordinates of A, B, C are (6, 3), (3, 5) and (4, –2) respectively and P is a point (X, Y), Then area PBC area ABC (a) 6 X Y 2 . The value of k is : k (b) 7 (c) 8 (d) 9 Solution : A(6, 3), B(3, 5), C(4, –2) area of PBC areaof ABC P (x, y) x y 2 k 1 3 2 y 4 y 5 x 5 2 2 1 6 7 3 2 3 4 3 5 2 1 6 3y 4y 20 7x 42 15 81 7x y 26 19 x y 2 k x y 2 k x y 2 k We cannot find the value of k from this sequence. There is no option in the given question (in order to minimize fluke and guessing probability ) www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 13 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes 21. The coordinates of two points A and B are (–1, 4) and (5, 1) respectively Find the coordinates of the point P which lie on extended line AB such that it is three times as far from B as from A. (a) (–4,–11/2) (b) (4,11/2) (c) (4,–11/2) (d) (-4,11/2) Solution : PA = 3P B PA PB 3 1 AB PB 3 PB 1 AB PB 3PB AB 2PB AB 2 PB 1 5= 2X 1 2 1 2X 1 3 5 15 1 2x 2x 16 x 8 1= 2y 4 2 1 3 2y 4 3 – 4 =2y y= 1 2 www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 14 Pioneer Education {The Best Way To Success} 8, Medical and Non - Medical Classes 1 2 PB = 3PB 1 2xx 1 5 2 1 3 2x 5 8 2x x 4= 4 2y 1 1 2 1 12 2y 1 11 2y y 11 2 4, 11 2 Ans. (d) 22. Find the area of triangle formed by joining the mid points of the sides of triangle whose vertices are (0,–1), (2, 1) and (0, 3). Find the ratio of this triangle with given triangle. (a) 1:4 (b)4:1 (c) 2:1 (d)1:2 Solution : www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 15 Pioneer Education {The Best Way To Success} Area of DEF Medical and Non - Medical Classes 1 12 1 2 11 0 00 2 1 01 3 2 23 1 0 1 1 1 1 1 1 2 Area of ABC = 1 8 4 2 Areaof DEF Areaof ABC 1 4 Ans.(a) 23. Form a quadratic equation whose roots are given as: 3, (a) 3x2 4x 3 0 3 0 (c) 3x2 4x Solution : 1 3 1 1 5= 3 3 3 3 1 1 P= 3. 3 x2 – 5x + P =0 x2 – 1 3 (b) 3x2 4x 3 0 (d) 3x2 4x 3 0 4 x 1 0 3 www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 16 Pioneer Education {The Best Way To Success} 3x 2 4x Medical and Non - Medical Classes 3 0 Ans. (c) 24. x 12 12 12........ (a)1 (b) 2 (c)3 (d)4 Solution : x 12 12 12 x 12 x square both the sides x2 = 12 +x x2 – x –12 =0 x2 –4x +3x –12 =0 x(x–4) +3(x –4) =0 (x–4) (x +3)=0 x = –3, 4 Ans. (d) 25. If Sr denotes the sum of the first r terms of an A.P. Then, S3n : (S2n – Sn) is (a) n (b) 3n (c) 3 (d) none of these Solution : Sn n 2a 2 Sn S2n Sn n 1 d 3n 2a 2 2n 2a 2 3n 2a 2 3n 1 d 3n 1 d n 2a 2 n 1 d 3n 1 d n 4a 2d 2n 1 2 2a n 1 d www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 17 Pioneer Education {The Best Way To Success} 3n 2a 2 3n 1 d n 4a 2d 2n 1 2 3 2a Medical and Non - Medical Classes 2a n 1 d 3n 1 d 2a 2d 2n 1 n 1 d 6a 3d 3n 1 2a d 4n 2 n 1 6a 9nd 3d 2a d 3n 1 6a 9nd 3d 2a 3nd d 3 2a 3nd d 2a 3nd d 3 Ans.(c) 26. If 7 times of 7th term of an AP is equal to 11 times the 11th term. then find the value of 18th term (a)1 (b)0 (c) 2 (d)3 Solution : 7[A + (7 –1) D] = 11[ A + (11 – 1) D] 7A +42 D = 11A + 110 D 42D –110D = 4A –68D = 4A 4A +68D=0 2A + 34D =0 A +17D =0 A + (8 –1)D=0 18th Ans. (b) www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 18 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes 27. If , are roots of the equation ax2 + bx + c = 0 find the value of : 3 + 3 3abc b3 3abc b3 3abc b3 3abc b3 (a) (b) (c) (d) a3 a3 a3 a3 Solution : ax2 + bx + c =0 b c , a a 3 3 2 = 3 b = a 2 b a b b2 = a a2 3 2 2 c a 3c a b b2 3ac = a a2 b3 3acb 3abc b3 = a3 a3 Ans. (d) 28. If the roots of the equation p(q r)x2 q(r p)x r(p q) 0 are equal, then 1 1 p r N . q then the value of N is: (a) 0 (b) 1 (c) 2 (d) 3 Solution : p(q –r)x2 + q(r –p) x +r(p –q) =0 D =0 q r p 2 4pr q r p q 0 q2 (r –p)2 – 4pr(pq –q2 –pr + qr) =0 q2(r 2 + p2 –2pr) – 4p2rq + 4q2pr + 4p2r2 – 4p2rq + 4q2pr +4p2r2 – 4pqr2 =0 q2r2 + p2q2 + 2prq2 +(2pr)2 – 4p2rq –4pqr2 =0 www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 19 Pioneer Education {The Best Way To Success} (qr + pq – 2pr)2 Medical and Non - Medical Classes =0 qr + pq =2pr 1 1 p r 2 q N 2 Ans. (c) 29. If the point (a, 0) (0,b) and (1,1) are collinear then (a)1 (b) 2 1 1 k find the value of k a b (c)3 (d)4 Solution : (a, 0), (0, b) (1, 1) x 1(y2 –y3) + x2(y3 – y1) x3(y1 – y2) =0 a(b –1) + 0(1 –0) + 1(0 –b) =0 ab –a –b =0 a + b =ab 1 1 1 a b Ans. (a) 30. In an A.P if Sn = n2p and Sm = m2 p , m≠ n, show that Sp = k p2 (a)1 (b) 2 (c) 3 (d) 4 Solution : Sn = n2p n 2A 2 n 1 D 2A +(n –1)D = 2np n2p …..(1) ……(2) (1) –(2) (n –m) D =2np –2mp D= 2p n m n m D = 2p www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 20 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes 2A +(n–1) D = 2np 2A + (n –1)2p = 2np A + np –p = np = A p Sp = = p 2A 2 p 2p 2 p 1 D p 1 2p p = ,2p 1 p 1 2 =p2 . p = p3 = p . p2 k p Ans. There is no option in the given question (in order to minimize fluke and guessing probability) www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 21 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes SECTION – B {PHYSICS} 31. Velocity of light in medium 1 is 2.4×107 m/s and velocity of light in medium 2 is 1.8× 107 m/s then the refractive index of medium 2 with respect to medium 1 is (a) 3 (b) 4 4 (c) 1 3 3 (d) 1 4 Solution: (b) 1 n2 V1 V2 2.4 107 4 = 1.8 107 3 Ans. (b) 32. If the a convex lens of refractive index n1 is placed in a medium of refractive index n2 such that n2 > n1 the lens will behave as a (a) convex lens itself (b) diverging lens (c) concave mirror (d)convex mirror Solution: (b) 33. A concave mirror forms an erect image twice the size of an object. The object distance from the mirror is: (a) f/2 (b) 2f (c) 3f/2 (d) 2f/3 Solution: (a) Since h2 h1 v u Since h2 = 2h1 Therefore v = 2u According to mirror formula 1 2u 1 u 1 v 1 u 1 f 1 f Therefore u = f 2 34. A point source of light B is placed at a distance L in front of the centre of a mirror of width d hung vertically on a wall. A man walks in front of the mirror Along a line www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 22 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes parallel to a mirror at a distance 2L from it, as shown in fig. find the greatest distance over which he can see the image of the light source from the mirror. (a) d (b) d/2 (c) 2d (d) 3d Solution: From geometry, CD = d = C’D’ BD = DP = d/2; CB = QC = d/2 P’Q’ = C’D’ + P’D’ + Q’C’ ND DP ND' D'P' L d/2 2L D'P' D'P' d Similarly, Q’C’ = d Hence, distance through which the man can observe the image = Q’C’ + C’D’ + D’P’ = d +d + d = 3d www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 23 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes 35. Two plane mirrors AB and CD each of length 2m are arranged parallel to each other 3 m apart and a ray of light is incident at A as shown in figure. How much refraction does the ray of light undergo? (a) 4 (b) 2 (c) 3 (d) 1 Solution: (c) DC AD tan300 DC DC AD tan 300 3 1 3 DC = 1 m Now no. of images = 2 + 1 =3 36. A glass lens has a focal length 5 cm in air. In water its focal length would be (a) 5 cm (b) More than 5 cm but finite (c) infinite (d) Less than 5 cm Solution:(b) www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 24 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes 37. In the above equation what is the angle of deviation of the incident ray. (a) 300 (b) 400 (c)900 (d) 1200 Solution: Since the instant ray came out with same angle (450) Therefore angle of deviation is 900. Ans.(c) 38. Consider two convex lens of same focal length f separated by a distance 2f. If a parallel rays of light falls on one convex lens as shown in figure then the final image will be formed at (a) at F (b) at infinity (c) at 2f (d) none of these www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 25 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes Solution : (b) Since the distance between the two lenses is 2f their for the focus of the two lenses will coincide each other thus first of all ray will focus at point f and then it will move to infinity when passes through second lens. 39. A air bubble in water will act like a (a) Concave mirror (b) Convex lens (c) Convex mirror (d) Concave lens Solution:(d) 40. If camera a has f= 4.5 m lens and camera B has f =2.8m lens, and the diameters of both the lenses are equal, then (a) A is better for photographing fast moving objects (b) Pictures taken by B will always be sharper (c) Pictures taken by A will always be sharper (d) B is better for photographing fast moving objects Solution:(c) www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 26 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes SECTION – C {Chemistry} 41. In CH3 CH2 OH the bond that undergoes heterolytic cleavage most readily is (a)C–C (b)C–O (c)C–H (d)O–H Solution: (b) 42. 'Drinking alcohol' is very harmful and it ruins the health 'Drinking alcohol' stands for(a) drinking methyl alcohol (b) drinking ethyl alcohol (c) drinking propyl alcohol (d) drinking isopropyl alcohol Solution: (b) drinking ethyl alcohol 43. Which of the following is a soap (a) Sodium alkylsulphonate (b) Sodium chloride (c) Sodium sulphonate (d) Sodium stearate Solution: (d) Sodium stearate (C17H35 COONa) 44. Acidic hydrogen is present in (a) Ethyne (b) Ethene (c) Ethane (d) Benzene Solution: (a) Ethyne (CH CH) RC CH RC C– + H+ 45. An organic compound 'A' with molecular formula C2H4O2 turns blue litmus to red and gives brisk effervescence with sodium hydrogen carbonate. Identify the compound (a) Methanoic acid (b) Ethanoic acid (c) Propanoic acid (d) Butanoic acid Solution: (b) Ethanoic acid (CH3COOH) 46. Power alcohol contains (a) 50% petrol and 50 % ethanol (b) 75% petrol and 25 % ethane (c) 25% petrol and 75% ethane (d) 70% petrol and 30% ethane Solution:(b) www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 27 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes 47. Glacial acetic acid is a(a) frozen acetic acid (b) 5-8% of solution of acetic acid in water (c) mixture of acetic acid and alcohol (d) None of these Solution:(a) 48. A & B both compounds give H2 gas with solution. If A & B react in presence of acid catalyst then they from ethyl acetate. Thus, A & B would be (a) CH3COOH, CH3OH (b) HCOOH, CH3COOH (c) CH3COOH, C2H5OH (d) C3H7COOH, C3H7OH Solution :(a) 49. The IUPAC name for (a) 1, 1-Dimethyl-1-1,3-butanediol (b) 2-methyl-2,4-pentanediol (c) 4-methyl-2,4-pentanediol (d) 1,3,3-Trimethyl1-1,3-propanediol Solution :(b) 50. The number of 40 carbon atoms in 2, 3, 4, 4-tetramethyl pentane is (a) 1 (b) 2 (c) 3 (d)4 Solution:(b) www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 28 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes SECTION – D {Biology} 51. Mucosal lining of uterus is (a) Mesometrium (b) Endometrium (c) Epimetrium (d) Epidermis Solution: (b) 52. In spirogyra, asexual reproduction takes place by (a) breaking up the filaments into smaller bits (b) division of a cell into two cells (c) division of a cell into many cells (d) formation of young cells from older cells Solution: (a) 53. Jasmine is multiplied vegetatively through (a) Stem cutting (b) Leaves (c) root cutting (d) Layering Solution: (d) 54. Out of the four sides I, II, III, IV whose details are shown, which one should be focused under the microscope for showing budding in yeast? (a) I (b) II (c) III (d) IV Solution: (c) 55. Seed is formed from (a) Unfertilized ovary (c) Fertilized ovule (b) Fertilized ovary (d) Unfertilized ovule Solution: (c) 56. Characters transmitted from parents to offspring are present in (a) Cytoplasm (b) Ribosome (c) Golgi bodies (d) Genes Solution :(d) www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 29 Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes 57. Gestation period in humans is (a) 270 days (b) 290 days (c) 200 days (d) 245 days Solution :(a) 58. The slide of Amoeba showing binary fission is characteristic by (a) Rounded Amoeba with elongated nucleus (b) Amoeba having pseudopodia and rounded nucleus (c) Amoeba constructed in the middle with round nucleus (d) Amoeba constructed in the middle with elongated nucleus Solution :(d) 59. Which of the following diagrams shows Yeast cell undergoing budding (a)I (b) II (c) III (d) IV Solution :(d) 60. A surgical technique of reversible but permanent contraception in females is (a) Vasectomy (b) Tubectomy (c) Norplant (d) IUCD Solution :(b) ALL THE BEST www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721 30