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

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1 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 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. 4. Duration of the test: pm to 4 pm - hours. PART A Note Only one of the choices A,,, 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.. Which one of the following numbers is NOT the sum of two prime numbers? (A) 4 () 30 () 67 (D) Sum of two prime numbers excluding two is always an even number, so option A and are not true. Option D is also not true because 9 + = So option () 67 is true. AD is a square and P = AP. The perimeter of the rectangle APQD is 80 cm. The perimeter of AD in cm is (A) 00 () 0 () 40 (D) 60 A P 90º D Q Let AP = x So P = x

2 A x x 3x 3x D x Q x Perimeter of rectangle A PQD = 80 (x + 3x) = 80 8x = 80 x = 0 cm So perimeter of square = AD 4 side = 4 3x = x or 0 = 0 cm Option () 3. Saket added up all the even numbers from to 0. Then, from the total he obtained, he subtracted all odd numbers between 0 and 00. The answer he would have obtained is (A) 0 () 0 () 30 (D) 0 Sum of first n even number is n (n + ) Here n = 0 So 0() = 0 Now sum of first n odd number is n Odd number from 0 to 00 are 0 So 0 = 00 Difference = 0 00 = 0 Option (D) The value of is 4 8 (A) 6 () 4 () 4 (D) Option (D) AD is a rectangle. A = 8 cm and = 6 cm. Q is the midpoint of A. P,R are on AD and respectively such that AP = cm, R = cm. Area of the shaded triangle in square cms is (A) () 3 () 4 (D) 6 Q P D R

3 4 Q 4 4 P 6 Area of shaded portion = D 8 R Area of rectangle AD [area of AQP + area of QR + area of Trapezium PRD] Area of rectangle AD = 48 cm Area of AQP = 4 4cm Area of QR = 4 0cm Area of trapezium PRD = 4 8 0cm 48 [ ] = = 4 cm Option () 6. The Rishimoolam of a number is defined as follows. onsider the number 34. y multiplying its digits,3 and 4, we obtain 3 4 = 4. Again, multiplying digits of 4, we get 4 = 8. We say 8 is the Rishimoolam of the number is the Rishimoolam, we say the number has no Rishimoolam. Which one of the following has no Rishimoolam? (A) 736 () 647 () 83 (D) 69 (A) 736 = = 6 6 = = = () = 68 = 6 8 = = 3 3 = 6 () 8 3 = 4 = 4 = 8 (D) 6 9 = 4 = 4 = 0 = 0 = 0 Option D has Rishimoolam 7. Two circles touch two parallel lines as shown in the diagram. The radius of each circle is cm. The distance between the centres of the circles is cm. The area of the shaded region in square cms is (A) () 0 () 0 (D) 0 + P Q S R Area shaded region is [Area of Rectangle PQRS Area of two semicircles ] 0 Option ()

4 8. Samrud wrote two consecutive integers, one of which ends in a. He multiplied both. He squared the answer. The last two digits of his answer is (A) 0 () 40 () 0 (D) 00 Product of any no. which ends with and its consecutive always ends up with zero, hence when use square it, then last two digit of resultant will give 00. Option (D) 9. Vishwa wrote a number on each side of 3 cards. In each card, the numbers written on the sides are different. One side of each card is a prime number and the other sides had 44, 9 and 38 respectively. Given that the sum of the numbers on each card is the same, the difference between the largest and the second largest of the prime numbers on the cards is (A) 6 () 7 () 9 (D) 4 All prime numbers greater than 9 are odd, so on finding their difference result will be even & there is only one even prime no. i.e. So next prime to 9 is 6. Hence on this card is written Now for finding others 6 44 = 7 and 6 38 = 3 Difference between largest and second largest is 3 7 = 6 Option (A) 0. The numbers of three digit numbers abc such that a b c = is (A) () 6 () 8 (D) 9 a b c = Total = 6 Option () PART 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 4 mark.. Five chairs cost as much as desks, 7 desks cost as much as tables and 3 tables cost a as much as sofas. If the cost of sofas is Rs.0, then the cost of a chair (in Rs.) is Let = D 7D = T 3T = S ost of sofas = 0 0 ost of sofas = 00

5 3T = Table lost = D = T = 700 Desk cost = = D = So cost of chair = Rs.480. The average age of a class of 0 children is.6 years. new children joined with; an average age of. years. The new average of the class (to one decimal place) x x x3...x0 6 0 new children joined y y y3 y4 y. y + y + y 3 + y 4 + y = 6 x x... x0 y y...y New average = is a two digit prime and when we reverse its digits, the number 3 obtained is also a prime number. The number of two digit numbers having this property is Total = In a garden there are two plants. One plant is 44 cm tall and the other is 80 cm tall. The first plant grows 3 cm in every months and the second cm in every 6 months.the number of months after which the two plants will have equal height is 3 44 x 80 x 6 3 9x x x 36 6 x x x = 36 6 x = 9 6 = 4

6 . In days a man walked a total of 8 KM. Every day he walked 4 KM less than the previous day. The number of KM he walked on the last day is Let a man walk x km on first day x + x 4 + x 8 + x + x 6 = 8 x 40 = 8 x = x = So on last day he walked 6 = 9kms 6. In the adjoining figure, A is parallel to D. The value of x is 3º 49º D xº 3º xº 49º t y = = 3º now y = z (Alternate angles) z = 3º also 49 + t = 80 D t = = 3 x = t + z (Exterior angle) x = = 66º 7. In Mahadevans cycle shop for children, there are unicycles, having only one wheel, bicycles, having two wheels and tricycles, having three wheels. Samrud counts the seats and wheels and finds that there are totally 7 seats and 3 wheels. The number of bicycles is more than tricycles. The number of unicycles in the shop is Let x be unicycles y be bicycles z be triangles So epuation for wheels is x + y + 3z = 3 () For seats x + y + z = 7 () Subtracting () & () y + z = 6 Now bicycles is more than tricycles, y have to be z = 6 z = z = So unicycle = x = 7 x = 7 x =

7 8. There is a tree with several branches. Many parrots came to rest on the tree. When 6 parrots sat on each branch of the tree, all the branches were occupied but three parrots were left over. When 9 parrots sat on each branch, all parrots were seated but two branches were empty. If b is the number of branches and p is the number of parrots, the value of b + p is Let ranches be b and parrots be p A.T.P. 6b = p 3 () 9(b ) = p () Solving the above equations p = 6b + 3 p = 9b 8 6b + 3 = 9b 8 = 3b b 3 7 = b Putting in equation () 6 7 = p = p 4 = p b + p = = 9. The incomes of A and are in the ratio 3:. Their expenditures are in the ratio :3. If each saves Rs 0,000, then As income is (in Rs) Let incomes of A and are 3x and x & Expenditures be y and 3y Income = Expenditure + saving (3x y = 0000) x 3y = x 0 y = x 9y = y = 0000 When y = 0000 x = 0000 So A s income = 3x = Rs. 0. The radius of a circle is increased so that its circumference is increased by %. The area of the circle will increase by % ircumference increased by % 0r New Radius = 00 Old area = r New area = 0 00 r 0 r r 0 r 0 r % change = % r r 0000 r.

CAREER POINT THE ASSOCIATION OF MATHEMATICS TEACHERS OF INDIA

CAREER POINT THE ASSOCIATION OF MATHEMATICS TEACHERS OF INDIA REER POINT THE SSOITION OF MTHEMTIS TEHERS OF INI Screening Test Gauss ontest NMT at PRIMRY LEVEL V & VI Standards Saturday, 6 th ugust, 017 P Note : 1. Fill in the response sheet with your Name, lass