1. Fill in the response sheet with your Name, Class and the institution through which you appear in the specified places.

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1 Note: THE ASSOIATION OF MATHEMATIS TEAHERS OF INDIA Screening Test - Kaprekar ontest NMT at SUB JUNIOR LEVEL - VII & VIII Standards Saturday, 6th August, 07. Fill in the response sheet with your Name, lass and the institution through which you appear in the specified places.. Diagrams are only visual aids; they are NOT drawn to scale. 3. You are free to do rough work on separate sheets.. Duration of the test: pm to pm -- hours. PART A Note Only one of the choices A. B,, 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 mark. For each incorrect response you lose mark.. The fraction is written in the decimal form 0.a a a 3... The value of a 07 is 37 (A) 8 (B) 0 () (D) is divided by 3 The remainder is So The a 07 digit is. Option (). The number of integers x satisfying the equation (x 3x + ) x+ = is (A) (B) 3 () (D) 5 For (x 3x + ) x + = ase - I x + = 0 ase - II x 3x + = and powers can be any integer ase - III x 3x + = and power must be odd. ase I x + = 0 x = ase II x 3x + x 3x = 0 x(x 3) = 0 x = 0, 3 ase III x 3x + = x 3 3x + = 0 = (x ) (x ) = 0 x =, x = is discarded So, no. of solution = (, 0, 3, )

2 3. The number of two digit numbers ab such that the number ab ba is a prime number is (A) 0 (B) () (D) 3 ab ba = 00 ab 0 b a = 9a 9b 9 (a b) 9 is always the factor of ab ba so, no such number is present. If A , then (A) < A < (B) < A < 3 () 3 < A < (D) A < A = = = 66 < A < 3 Option (B) 5. What is the 07 th letter in ABRAADABRAABRAADABRA..., where the word ABRAADABRA is repeatedly written? (A) A (B) B () (D) R ABRAADABRA Total digits are = Remainder when 07 is divided by So the 07 th letter = (A) Option (A) 6. How many of the following statements are true? (a) A 0% increase followed by another 5% increase is equivalent to a 5% increase (b) If the radius of a circle is doubled then the ratio of the area of the circle to the circumference is doubled (c) If a positive fraction is subtracted from and the resulting fraction is again subtracted from we get the original fraction. (A) 0 (B) () (D) 3 (a) Is false (b) Is true because After doubling it A r r. () r A' r ' r r So, ratio of area and circumference is doubled (c) Is true Let n m is a positive fraction ATQ, m n Option () is correct m n

3 7. In the adjoining figure the breadth of the rectangle is 0 units. Two semicircles are drawn on the breadth as diameter. The area of the shaded region is 00 sq units. The shortest distance between the semicircles is 5 (A) 5 (B) 5 () 3 x 3 (D) Let the length be x Area of shaded region = Area of rectangle [Area of semi circles] = 0x [5] = 00 0x = x 0 5 x 0 0 Shortest distance = x 0 5 = Option (A) 8. When you arrange the following in descending order A. 5% of 30 B. 8% of 5. 0% of 0 D. 6% of 0 E. 9% of 5 the middle one is (A) 5% of 30 (B) 8% of 5 () 0% of 0 (D) 6% of 0 A. 5% of 30 =.5 B. 8% of 5 =.. 0% of 0 = D. 6% of 0 =.6 E. 9% of 5 =.5 When arranged in descending order the middle one is, Option (D)

4 9. A boy aims at a target shown in the figure. When he hits the center circle he gets 7 points, first annular region 5 points and second annular region 3 points. He shoots six times. Which one of the following is a possible score? (A) 6 (B) 6 () 9 (D) 6 is the possible score 7, 5, 5, 3, 3, 3 0. After simplifying the fraction we get a term independent of b a a ab a b ab x y y xy xy a yx y (A) a,y (B) b,x () a,b (D) x,y b a x y a y ab xy b a x y ab a yx y a y ab xy = bx (independent of a, y) Option (A). If 7 Rasagullas are distributed to each boy of a group, 0 rasagullas would be left. If 8 are given to each boy then 5 rasagullas would be left. So the person who distributes the rasagullas brought 5 more rasagullas and distributed the same number (x) rasagullas to each. There is no rasagulla left. Then x is (A) 0 (B) () (D) Let us suppose, there are n boys in group and there and total r rasgullas are there, ATQ. r 7n = 0 () r 8n = 5 () Subtract equation () and () n = 5 r = 5 Now, if x rasgullas are distributed to each boy, such that no rasgullas are left. And 5 more rasgullas are brought Total rasgullas = 5 +5 = Each boy should x 5 Option ()

5 . In the adjoining diagram all squares are of the same size. The total area of the figure is 88 square cms. The perimeter of the figure (in cm) is (A) 86 (B) 96 () 06 (D) 9 Total square are = Total area = 88 Let the side of a square = x So, 8x = 88 x = 36 x = 6 E F G H I B x K J A P O L N M Perimeter = AB + B + D + DE + EF + FG + GH + HI + IJ + JK + KL + LM + MN + NO + OP + PA = 6 x = 6 6 = 96 cm 3. When Newton was a primary school student he had to multiply a number by 5. But by mistake he divided the number by 5. The percentage error he committed is (A) 95% (B) 96% () 50% (D) 75% Let the no. be x Desired result = 5x Obtained result = 5 x x Error = 5x x 5 5 x % error = 00 96% 5 5x 3. AB is an isosceles triangle with sides AB A 3x x 3. The area of the equilateral triangle with side length x is (A) 3 3 (B) 36 3 () 5 3 (D) 6 3 3x AB = A = 3x = 3 3 9x 3x x x = 6

6 Area of an equilateral having x side B Two distinct numbers a and b are selected from,, 3, 60. The maximum value of (A) 6750 (B) 570 () 850 (D) 350 a = 60, b = 59 a b a b a b is a b PART B Note Write the correct answer in the space provided in the response sheet. For each correct response you get mark. For each incorrect response you lose mark. 6. Two cogged wheels of which one has 6 cogs and the other 7 cogs, mesh into each other. If the latter turns 80 times in three quarters of a minute, the number of turns made by the other in 8 seconds is. Logs Seconds Turns 6 8 x Turns second Turns cog x 5 6 x =. Ans. 7. If n is a positive integer such that a n =, then a 6n 6 is a n = (a n ) 3 = 3 a 6n = 8 So, (a 6n ) 6 (8) 6 = 0 Ans. 8. The least number of children in a family such that every child has at least one sister and one brother is Family should have atleast four children ( male and female)

7 9. A water tank is 5 full. When 0 liters of water is removed, it becomes 3 full. The capacity of the tank in liters is Let the capacity of the tank = x ATQ 3 x 0 x 5 x 3x 6x 5x x 0 x AB is an equilateral triangle. Squares are described on the sides AB and A as shown. The value of x is A B F xº E A G 60º D 60º 60º AB is an equilateral Let side = y FGBDE is a hexagon. Angle sum = 70. [(90 + x) ] = x = 360 x = x = 30º Ans. B. ABD is a trapezium with AB = 6 cm, AD = 8 cm and D = 8 cm. The sides AB and D are parallel and AD is perpendicular to AB. P is the point of intersection of A and BD. The difference between the areas of the triangles PD and PAB in square cms is A 6 cm 8 cm P 90º D

8 A 6 cm h 8 cm P 90º h D 8 PAB ~ PD h + h = 8 () h h 6 8 h 3 h 3 () h + 3h = 8 h = 8 h = h = 6 Ar (PB) Ar(PAB) = 8 h 6 h = 9 (6) 3 () = 5 6 = 8 cm Ans.. The price of cooking oil has increased by 5%. The percentage of reduction that a family should effect in the use of oil so as not to increase the expenditure is Rate of oil is x per and consumption is y Expenditure = Rs.xy. () ATQ x 5x New, Rate of oil = x Rs per Let consumption = y 5x Expenditure = y' Expenditure must be same 5x y' xy y' y 5 % Reduction in consumption = y y y y ' % Ans. y y 5 3. The number of natural numbers between 99 and 999 which contains exactly one zero is From 0 09 = 9 numbers 0, 0, 30, 0, 50, 60, 70, 80, 90 9 numbers Total no. s from 0 99, which are having exactly one zero 8 numbers similarly from numbers and so on. So, Total nos = 9 8 = 6 Ans.

9 . In the adjoining figure we have semicircles and AB = B = D. The ratio of the unshaded area to the shaded area is B D d B d d D d x d Area of shaded position Area of unshaded position = 3d x 3d x () () 9d x 9d 9d Area of unshaded position 3 : Ans. Area shadedposition x x 3d 5. Gold is 9 times as heavy as water and copper is 9 times as heavy as water. The ratio in which these two metals be mixed so that the mixture is 5 times as heavy as water is Let the weight of water is g Wt. of unit of gold = 9 g Wt. of unit of copper = 9 g Let is suppose, x units are taken from cold and y units are taken from copper 9(x) + 9y = 5(x + y) x 3 x = 6y Ans. y 6. Five angles of a heptagon (seven sided polygon) are 60, 35, 85, 5 and 5 " If the other two angles are both equal to x, then x is - Angle sum of heptagon = (7 ) 80 = 5 80 = 900 = x = 900 x = 50 X = 75º Ans.

10 7. ABD is a trapezium with AB parallel to D and AD perpendicular to AB. If AB = 3 cm, D = 35 cm and AD = 5 cm. The perimeter of the given trapezium in cms is A D F 3 35 F = 35 3 = In a rt. BF, B = 5 + B = 69 B = 3 cm Perimeter = = 76 cm Ans. 8. The number of three digit numbers which are multiples of is No. of numbers divisible by ( to 999) = 90 No. of numbers divisible by (to 99) = 9 No. of numbers divisible by (00 999) = 90 9 = 8 Ans. 9. If a, b are digits, ab denotes the number l0a + b. Similarly, when a, b, c are digits. abc denotes the number 00a + 0b + c. If X, Y, Z are digits such that XX+YY+ZZ = XYZ, then XX YY ZZ is Given: XX + YY + ZZ = XYZ XX digit no. X + Y + Z = 00 X + 0 Y + Z 0Z + Y = 89 X X =, Y = 9, Z = 8 So, XX, YY, ZZ = = 9583 Ans. 30. The positive integer n has, 5 and 6 as its factors and the positive integer m has,8, as its factors. The smallest value of m + n is N has factors =,5,6 N can be = 3 5 = 30 M has factors =,8, M can be = 3 = M + N = 30 + = 5 Ans.

THE ASSOCIATION OF MATHEMATICS TEACHERS OF INDIA. Screening Test - Gauss Contest NMTC at PRIMARY LEVEL -V & VI Standards Saturday, 26th August, 2017

THE ASSOCIATION OF MATHEMATICS TEACHERS OF INDIA. Screening Test - Gauss Contest NMTC at PRIMARY LEVEL -V & VI Standards Saturday, 26th August, 2017 THE ASSOIATION OF MATHEMATIS TEAHERS OF INDIA Note: Screening Test - Gauss ontest NMT at PRIMARY LEVEL -V & VI Standards Saturday, 6th August, 07. Fill in the response sheet with your Name, lass and the