NMTC Paper_Class IX And X_2013
PART -A (2013)
Note : • Only one of the choices A,B,C,D is correct for each question. Shake that alphabet of your
choice in teh response sheet. (If you have any doubt in the method of answering seek the guidance of your supervisor).
• For each correct response you get I mark : for each incorrect response you lose 12 mark.
1. Two sides of a triangle are 10cm and 5cm in length and the length of the median to the third side is 6
2 1
cm. The area of the triangle is 6 x cm2. The value of x is
(A) 12 (B) 13 (C) 14 (D) 15 1. 10 5 y y 6 2 1 By opollonius theorem 102 + 52 = 2 2 y 2 13 12 y = 4 169 2 125 = 2 9 so side of is 10, 5, 9 s = 2 9 5 10 = 12 = 12(2)(7)(3) = 6 14 = 6 x x = 14
2. x and y are real numbers such that 7x
– 16y = 0 and 4x– 49 y = 0, then the value of (y – x) is
(A) 2 5 (B) 5 19 (C) 2013 4115 (D) 784 1569 2. 7X – 16y = 0 4X– 49y = 0 y = 16 7X y = 49 4X 16 7X = 49 4X x = – 2 y = 16 49 1 y – x = 16 49 1 + 2 = 16 49 32 49 1 = 784 1569
3. a, b are positive integers such that (A) the sum of their square is S
(B) the sum of their cubes is C times the sum of the numbers itself. (C) S – C = 28. The numbers of such pairs (a, b) is
(A) 1 (B) 2 (C) 3 (D) 6 3. S = a2 + b2 a3 + b3 = C(a + b) C = b a b a3 3 = a2 + b2 – ab S – C = 28 a2 + b2 – (a2 + b2– ab) = 28 ab = 28 = 1 28 or 28 1 = 2 14 or 14 2 = 7 4 or 7 4
so six pairs (a, b) exist
4. The number of numbers of the form 30a0b03 where a, b are digits which are divisible by 13 is
(A) 5 (B) 6 (C) 7 (D) 0
Sol. 30a0b03 = 3000003 + 10000a + 100b
= 13k1 + 6 + 13k2 + 3a + 13k3 + 9b = 13 (k1 + k2 + k3) + (6 + 3a + 9b) = 13 (k1 + k2 + k3) + 3(2 + a + 3b) = 2 + a + 3b should be divisible by 13 as a, b are the digits
2 + a + 3b = 13, 26
when 2 + a +3b = 13
then the possible values b are 1, 2, 3 so 3 cases are possible when 2 + a +3b = 26
then the possible values b are 5, 6, 7, 8 so 4 cases are possible so total cases are 7
5. The number of positive integeral values of (x, y) which satisfy the equation 3x + 3y = 4 ; x + y = 28
simultaneously is (A) 1 (B) 2 (C) 0 (D) 3 5. 3 x + 3y = 4, x + y = 28
3 3
3 y x = 43 x + y + 3 3xy (4) = 64 28 + 12 3xy = 64 xy = 27 as x + y = 28 xy = 27 if x = 1 then y = 27 if x = 27 then y = 1 2 no. of positive integral value of (x, y) which satisfy the above equation.
6. ABCD is a rectangle. Through C a variable line is drawn so as to cut AB at X and DA produced at Y. Then BX.DY is
(A) twice the area of the rectangle ABCD (B) equal to the area of the rectangle ABCD (C) a variable quantity which lies between the area of rectangle ABCD and twice the area of the rectangle ABCD
6. D C B A b b bk bk a Y x XAY ~ XBC XB XA = BC AY = XC XY XB XA = b a = XC XY XA = aK XB = bK = (aK + bK) = K (a + b) BX.DY = bK (a + b) = b[K(a + b)] = bl = area of rectangle
7. The number of ordered triples (x, y, z) such that x, y, z are primes and xy + 1 = z is
(A) 0 (B) 1 (C) 2 (D) infinitely many
7. xy + 1 = Z
x cannot be odd because if it is odd then xy + 1 even so z is even
but z is prime. so that why x cannot be odd
x is even x = 2
2y + 1 = z
x = 2, y = 2, 3 = 5
but when y is odd then 2y + 1 is always be divisible by 2 + 1 = 3. so
2y + 1 = z is not prime only 1 triplet is possible (2, 3, 5)
8. In the adjoining figure O is the centre of the circle. ACOB is a square with A on the circle. Through B a line parallel to OA is drawn to cut the circle at D nearer to A. Then BOD =
D B O A C (A) 20° (B) 18° (C) 15° (D) 22 2 1 °
8. D B O A C E x y a a a a b b DBA = BAO = 45° OBE = BOA = 45° OBD = 90 + 45 = 135° BEO BE = OE = b b2 + b2 = a2 b = 2 a OED siny = a 2 2 / a = 2 1 y = 30° DOE = 60° x = DOE – EOB = 60° – 45° x =15
9. The number of real solutions of the equation x + x2 x3 1
= 1 is (A) 1 (B) 2 (C) 3 (D) 0 9. x + x2 x3 1 = 1 1 x x2 3 = 1 – x x2 + 1 x3 = 1 + x 2 – 2x x3 + 1 = 1 + 4x2 – 4x x3 – 4x2 + 4x = 0 x(x2 – 4x + 4) = 0 x(x – 2)2 = 0 x = 0,2
Put x = 2 not satisfy of the equation
x = 0 is the only solution of the above equation.
ans = 1
10. In the adjoining figure PQRS is a square of side 2 units PTR and QTS are quadrants of circles of radius 2 units. With SR as diameter a semicircle is drawn. A, B denote the areas of the portions shaded. Then (A – B) = P Q R S A B T (A*) 2 3 – 4 (B) – 3 1 (C) 2 3 – 4 1 (D)
(2)2 = 4 ) 2 ( 2 + 4 ) 2 ( 2 – [(A) + 2 ) 1 ( 2 ] + (B) 4 = 2– A – 2 + B A – B = 2– 2 – 4 A – B = 2 3 – 4
11. a, b, c are digits of a 3-digit number such that 64a + 8b + c = 403, then the value of a + b + c + 2013 is (A) 2024 (B) 2025 (C) 2034 (D) 2035 11. 64a + 8b + c = 403 64a + 8b = 403 – C as LHS is multiple of 8. RHS should be a multiple of 8 C = 3 64a + 8b = 403 – 3 = 400 8a + b = 50 8a = 50 – b LHS is multiple of 8 RHS is multiple of 8 b = 2 8a = 48 a = 6 a + b + c + 2013 = 6 + 2 + 3 + 2013 = 2024
12. What is the sum of the digits of (9999999999)3
(A) 99 (B) 108 (C) 180 (D) 199 12. 93 = 729 993 = 997002999 (9999999999)3 = (10 – 1)nine 700 ...2(10) times 9 sum of digit = 9 9 + 7 + 10 9 + 2 = 81 + 7 + 90 = 180
13. The number of three digits number which are divisible by 3 and have the additional property that the sum of their digits is 4 times their middle digit is
(A) 7 (B) 4 (C) 11 (D) 10
13. a + b + c = 4b
a + c = 3b
sum of two digit is not more than 18 as a + c is equal to multiple of 3 also
a + c = 3, 6, 9, 12, 15, 18 b = 1, 2, 3, 4, 5, 6 as no. is div. by 3 a + b + c = 3k 3b + b = 3k 4b = 3k b is multiple of 3 If b = 3 then a + c = 9 If b = 6 then a + c = 18 a = 1, c = 8 a = 9, c = 8
a = 2 , c = 7 so only 1 cases is form
and so on a = 9, c = 0 so 9 cases are form so total cases are 10
14. In the adjoining figure AB is a diameter of a circle. AB is produced to P such that BP = radius of the circle. PC is a tangent to the circle. The tangent at B and AC produced cut at E. Then CDE is
A
C E
D
B
P
(A) isosceles with EC = ED (B) isosceles with EC = CD
(C) equilateral (D) a scalene triangle
14. E P A B r r 2r O D C PC2 = r 3r PC = 3.r tan = r 3 r = 3 1 = 30° BDP = 180 – (90 + 30) = 60° COB = 60° CAO = 2 1 (60°) = 30° In EBA E = 180 – (90 + 30) = 60° In CDE C = D = E it is an equilateral triangle
15. Nine numbers are written in asscending order. The middle number is the average of the nine numbers. The average of the five largest number is 68 and the average of the five smallest numbers is 44. The sum of all numbers is
(A) 560 (B) 504 (C) 112 (D) 122 15. x1, x2,....,x9 x5 = 9 x .... x x1 2 9 x1 + x2 + ... + x9 = 9x5 ..(i) 68 = 5 x ... x x5 6 9 x5 + x6 + ... + x9 = 68 5 = 340 ..(ii) 44 = 5 x ... x x1 2 5 x1 + x2 + ... + x5 = 44 5 = 220 ..(iii)
equation (ii) + (iii)
x5 + (x1 + x2 + ...+x9) = 340 + 220 x5 + 9x5 = 560
10x5 = 560 x5 = 56
PART - B (2012)
Note : • Write the correct answer in the space provided in the response sheet.
• For each correct response you get I mark ; for each incorrect response you lose 1 mark.4
16. The least value of the positive integer ‘n’ such that (n + 20) + (n + 21) + (n + 22) + ...+ (n + 100) is a
perfect square is ___________ 16. (n + 20) + (n + 21) + ... + (n + 100) = m2 81n + (20 + 21 + ...+ 100) = m2 81n + 2 81 [20 + 100] = m2 81[n + 60] = m2 n = 4
17. In the adjoining figure two equal circles of radii 2 units each touch. AB is the common diameter. The tangent at B meets the tangent from A to circle at C as shown. If BC = K 2 then the value of K is ___________ C B A 17. A B C D 4 2 2 O BC = K 2 AD = 2 2 2 6 = 32 = 4 2 BC = DC = K 2 AC2 = AB2 + BC2 (4 2 + K 2 )2 = 82 + (K 2 )2 32 + 2K2 + 16K = 64 + 2K2 16K = 32 K = 32/16 = 2
18. When the number 333332 + 22222 is written as a single decimal number, the sum of its digits is ___________ 18. 333332 + 22222 = (3 11111)2 + 2 11111 = (11111) [99999 + 2] = 11111 [100001] = 1111111111 so sum of digit is 10
19. The number of three digits numbers such that the product of their digits is a prime number is
19. 112 113 117 115
211 311 711 511
so 12, three digit no such that product of digit is prime.
20. The number of real values (x, y) for which 2x+1 + 3y = 3y+2
– 2x is ___________ 20. 2x+1 + 3y = 3y+2 – 2x 2x(2+1) = 3y (9 – 1) 2x.3 = 3y.8 2x.31 = 3y.23 x = 3, y = 1
21. In the adjoining figure. ABC is equilateral AD, BE and CE are respectively perpendicular to AB, BC and AC. Then AreaAreaofof ABCDEF___________
B A C D F E 21. F C E B D A 60° 30° 60° 60° 30°
ABC is also an equilateral
as ABC = BCA = CAB = 60°
cos60° = DB AD 2 1 = DB x DB = 2x tan60° = AD AB 3 = x AB AB = 3 x DE = DB + BE = 2x + x = 3x as ABC ~ DEF ABC area DEF area = 2 x 3 x 3 = 3 : 1 22. If f(x) = ax + b and f(f(f(x))) = 27x + 26 then (a + b) = __________ 22. f(x) = ax + b f(f(x)) = a(ax + b) + b
f(f(f(x))) = a[a (ax + b) + b] + b = 27x + 26 a3x + a2b + ab + b = 27x + 26 a3 = 27 a = 3 a2b + ab + b = 26 9b + 3b + b = 26 13b = 26 b = 2 a + b = 3 + 2 = 5
23. The eight digits 6, 5, 5, 4, 4, 3, 2 and 1 are used to form two 3-digit numbers and one 2-digit number. The largest possible sum of three number is ____________
23. H T U 6 5 3 5 4 2 4 1 _________________ 12 3 6
First we fill hundred place with bigger digit i.e. 6, 5 than we fill tens digits 5, 4, 4 then we fill unit digit 3, 2, 1 so the largest sum is 1236
24. a 0, b 0 The number of real number pair (a, b) which satisfy the equation a4 + b4 = (a + b)4 is
24. a4 + b4 = (a + b)4
a4 + b4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 2ab (2a2 + 3ab + 2b2) = 0
2ab = 0 or 2a2 + 3ab + 2b2 = 0 but 2ab 0 because as a 0, b 0
2a2 + 3ab + 2b2 = 0
Let consider it as Q.E. with variable a & b as constant.
a = ) 2 ( 2 ) b 2 )( 2 ( 4 ) b 3 ( b 3 2 2 a = 4 b 7 b 3 2
for any real value b 0 the 2
b 7
is an img quantity
a is an img no. for any real value of b, b 0
there in no real value of (a, b) which satisfy the above equation.
25. The number of integers greater than and less than 70 that can be written as ab (where b >1 is)____ 25. N = (3, 4, ..., 6, 9)
b > 1 N = ab
so we have to search no between which are perfect square, perfect cube etc. 4 = 22 8 = 23 9 = 33 16 = 42 or 24 25 = 52 27 = 33 36 = 62 49 = 72 64 = 82or 26
26. ABCD is parallelogram P is a point on AD such that AD AP = 2013 1
. Q is the point of intersection of ACAC
and BP. Then AC AQ __________ 26. A B C D Q P Let AD = 2013x AP = x PD = AD – AP = 2013 x – x = 2012x AD AP = 2013 1 AP PD = 1 2012 AD = BC = 2013 x By AA PQA ~ BQC BC PA = QC AQ x 2013 x = QC AQ 2013 1 = QC AQ AC AQ = 2014 1
27. ABCD is a square. E and F are respectively point on BC and CD such that EAF = 45°. AE and AF cut
the diagonal BD at P, Q respectively. Then
APQ of Area AEF of Area = _________ 27. E C F D A B 45° 45° 45° Q P
28. m, n are natural numbers. The number of pairs (m, n) for which m2 + n2 + 2mn
– 2013 m – 2013n – 2014 = 0 is __________ 28. m2 + n2 + 2mn – 2013m – 2013n – 2014 = 0 (m + n)2 – 2013 (m + n) – 2014 = 0 (m + n)2 – 2014 (m + n) + (m + n) – 2014 = 0 (m + n)[m + n – 2014] + 1[m + n – 2014] = 0 (m + n + 1) (m + n – 2014) = 0 m + n = – 1 or m + n = 2014 as m, n are natural m + n = –1 is not possible m + n = 2014
(1, 2013), (2, 2012) ... (2013, 1)
2013 pair are possible
29. In the adjoining figure BAC is a 30° – 60° – 90° triangle. D is the midpoint of AC. The perpendicular at
D to AC meets the line parallel to AB through C at E. The line through E perpendicular to DE meets BA produced at F. If DF = 5 x the x = __________ 60° 20 F E C D A B 29. 60° 30° 20 F E C D A B 1 3 4 In ABC sin30 = AB AC = 20 AC 2 1 = 20 AC AC = 10 AD = DC = 2 AC = 5 1 = 60° AF || EC 3 = 4 = 90° AC || EF FACE is parallogram EF = CA = 10 In EDC tan60 = DC ED 3 = ED10 ED = 10 3 In FED FD = 2 2 ED FE = 2 2 ) 3 10 ( 10 = 100(13) = 20
30. PR and PQ are tangent to a circle and QS is a diameter. Then RQS QPR = ____________ 30. R A O S Q x y P QPR = x ; RQS = y In quad OQPR O = 180 – x In OQR
