CAREER POINT THE ASSOCIATION OF MATHEMATICS TEACHERS OF INDIA

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1 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 and the institution through which you appear in the specified places.. iagrams are only visual aids; they are NOT drawn to scale. 3. You are free to do rough work on separate sheets. 4. uration of the test : pm to 4 pm hours. PRT - Note Only one of the choices, B,, 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 1/ mark. Q.1 Which one of the following numbers is NOT the sum of two prime numbers?. 4 B Sol. Option () 4 ( ) Q 4 and 30 even number (B) 30 ( ) sum of two prime no. is always even (excluding two) () 1 ( + 19) () 67 odd no ( + 65 ) It is not an prime no. 67 can't be represented as sum of two prime numbers Q. B is a square and PB = P. The perimeter of the rectangle PQ is 80 cm. The perimeter of B in cms is. 100 B P B Q REER POINT Ltd., P Tower, IPI, Road No.1, Kota (Raj.), Ph:

2 Sol. Option B Let B = B = = = 3x Given Perimeter of PQ = 80 cm or (x + 3x) = 80 8x = 80 x = 10 cm Perimeter of square B = (3x + 3x) = 1x = 10 cm 3x x P x Q B Q.3 Saket added up all the even numbers from 1 to 101. Then, from the total he obtained, he subtracted all odd numbers between 0 and 100. The answer he would have obtained is. 0 B Sol. Option Sum of even no. from 1 to 101 Let S 1 = This is an.p with common difference = T n = 100 = + (n 1) 98 = n 1 = n 1 = 49 n = S 1 = ( + 100) = S 1 = 550 (1) and Let S = sum of odd no. between 0 & 100 S = P with ommon difference = T n = 99 = 1 + (n 1) n = S = (1 + 99) = = 500 () ccording to question S 1 S = = Q.4 The value of 4 8 is B REER POINT Ltd., P Tower, IPI, Road No.1, Kota (Raj.), Ph:

3 Sol. Option = = 16 1 ns. Q.5 B is a rectangle. B = 8 cm and B = 6 cm. Q is the midpoint of B. P, R are on and B respectively such that P = cm, R = 1 cm. rea of the shaded triangle in square cms is. 1 B Q B P R Sol. Option B = = 8 cm B = 6 cm Q is midpoint 4cm Q 4cm B Q = QB = 8 = 4 cm cm P 5cm P = cm P = 6 = 4 cm R = 1 cm BR = 6 1 = 5 cm rea of shaded ΔPQR = area of rectangle B area of (ΔPQ + ΔBQR) area of trap. PR 4cm 8cm R 1cm = (4 + 1) 8 = = = 14 cm Q.6 The Rishimoolam of a number is defined as follows. onsider the number 34. By multiplying its digits, 3 and 4, we obtain 3 4 = 4. gain, multiplying the digits of 4, we get 4 = 8. We say 8 is the Rishimoolam of the number 34. If 0 is the Rishimoolam, we say the number has no Rishimoolam. Which one of the following has no Rishimoolam?. 736 B Sol. Option () = 16 again 1 6 = 1 again 1 = Rishimoolam of 736 = REER POINT Ltd., P Tower, IPI, Road No.1, Kota (Raj.), Ph:

4 (B) = 168 again = 48 again 4 8 = 3 again 3 = 6 Rishimoolam of 647 = 6 () = 4 again 4 = 8 Rishimoolam of 831 = 8 () = 54 again 5 4 = 0 again 0 = 0 cc. To question if zero is the Rishimoolam Number has no Rishimoolam 619 has no Rishimoolam Q.7 Two circles touch two parallel lines as shown in the diagram. The radius of each circle is 1 cm. The distance between the centres of the circles is 5 cm. The area of the shaded region in square cms is. 5π B. 10π. 10 π π Sol. Option Radius of circle = 1 cm iameter of circle = cm = B = cm area of shaded region = area of rectangle B area of two semi circle π(1) = (B B) = 5 π = 10 π nswer 1 5 cm B Q.8 Samrud wrote two consecutive integers, one of which ends in a 5. He multiplied both. He squared the answer. The last two digits of his answer is.50 B Sol. Option When we multiply two consecutive number in which one of them ends with 5, then the resulting multiplication ends up with zero and when we square the result, the squared number ends up with 00. last two digit of his answer is 00. e.g : 14, = 10 & (10) = REER POINT Ltd., P Tower, IPI, Road No.1, Kota (Raj.), Ph:

5 Q.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, 59 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.6 B Sol. Option We know that (even number) + (prime number) = odd number Here we do not use prime number = 44 + prime number odd number Because highest number given = 59 & 38 + prime number odd number nd if we add to 44 We get 46 & 46 < 59 odd number + prime number When prime No. = even number odd number cc. To explanation in case of even no, we require sum of number on each card is odd number In case of 59, prime number = sum = 61 In case of x = 61 x = 17 Prime no. = 17 In case of y = 61 y = 3 Prime no. = 3 ifference both largest & second largest prime no. = 3 17 = 6 Q.10 The number of three digit numbers abc such that a b c = 15 is. B Sol. Option B a b c = = No. of these digits no. = 3! = 6 & numbers are Total = 6 REER POINT Ltd., P Tower, IPI, Road No.1, Kota (Raj.), Ph:

6 PRT - B Note Write the correct answer in the space provided in the response sheet. For each correct response you get 1 mark. For each incorrect response you lose 1/4 mark. Q.11 Five chairs cost as much as 1 desks, 7 desks cost as much as tables and 3 tables cost as much as sofas. If the cost of 5 sofas is Rs 550, then the cost of a chair (in Rs) is Sol. cost of 5 sofas = Rs cost of 1 sofa = Rs. = Rs Now, cost of 3 table = cost of sofas = 1050 = Rs cost of 1 table = Rs. = Rs cost of 7 desk = cost of table = 700 = Rs cost of 1 desk = = Rs & cost of 5 chair = cost of 1 desk = 00 1 = Rs cost of 1 chair = Rs. 5 = Rs. 480 Q.1 The average age of a class of 0 children is 1.6 years. 5 new children joined with an average age of 1. years. The new average of the class (to one decimal place) Sol. Let 0 children are x 1, x, x 3,., x 0 cc to question x1 + x x 0 = x 1 + x + x x 0 = = 5 (1) Let 5 new children are y 1, y, y 3, y 4, y 5 given y1 + y + y3 + y4 + y5 = 1. 5 y 1 + y + + y 5 = 1. 5 = 61 () New average of class x1 + x x 0 + y1 + y + y3 + y4 + y5 = 5 = = 5 = 1.5 nswer REER POINT Ltd., P Tower, IPI, Road No.1, Kota (Raj.), Ph:

7 Q is a two digit prime and when we reverse its digits, the number 31 obtained is also a prime number. The number of two digit numbers having this property is Sol. Such nos are 13, 31 71, 17 17, 71 73, 37 31, 13 11, 11 37, 73 79, 97 97, 79 Total = 9 numbers Q.14 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 5 cm in every 6 months. The number of months after which the two plants will have equal height is Sol. & Now, 3 cm in every months x cm 5 cm in 6 months 44 cm 80 cm Plant -1 Plant - Let equal height = x cm Plant 1 have to grow (x 44) cm Plant have to grow (x 80) cm For plant 1, It grows 3 cm in months It grows (x 44) cm in = 3 (x 44) months.(1) For plant, It grows 5 cm in 6 months It grows (x 80) cm in = 5 6 (x 80) months.() cc. to question 6 (x 44) = (x 80) 3 5 = 10x 440 = 18x x = x = equal height = = 15 cm 8 x 44 = = 81 cm or plant 1 grow 81 cm From (1), growth time = 3 81 = 54 Months nswer REER POINT Ltd., P Tower, IPI, Road No.1, Kota (Raj.), Ph:

8 Q.15 In 5 days a man walked a total of 85 KM. Every day he walked 4 KM less than the previous day. The number of KM he walked on the last day is Sol. Let first day he walk x km cc. to question x + (x 4) + (x 8) + (x 1) + (x 16) = 85 5x 40 = 85 5x = 15 x = 5 He walked on last day = (x 16) km = 5 16 = 9 km Q.16 In the adjoining figure, B is parallel to. The value of x is B 35 Sol P B x BP = = BP = P = 35 (alternate angles) & P + P = 180 (linear pair) P = = 131 Now, x is exterior angle for ΔP x = = 166 nswer x Q.17 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 13 wheels. The number of bicycles is more than tricycles. The number of unicycles in the shop is Sol. Let there are x unicycles y bicycles z tricycles cc. To question x + y + z = 7 (1) nd, x + y + 3z = 13 () Eq. () Eq. (1) y + z = 6 Q y > z (given) Only possible condition REER POINT Ltd., P Tower, IPI, Road No.1, Kota (Raj.), Ph:

9 z = 1 & y = 4 From eqn. (1) x = 7 x = No. of unicycles in shop = Q.18 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 Sol. No. of branches = b and No. of parrots = p Given, When 6 parrots sat on each branch, all branch are occupied but three parrots were left over Total no. of parrots p = 6b + 3.(1) and also, when 9 parrots sat on each branch, all parrots seated but two branch were empty Total no. of parrots are p = 9 (b ).() From (1) & () 6b + 3 = 9 (b ) 6b + 3 = 9b 18 3b = 1 b = 7 From (1) p = 6 (7) + 3 = 45 p + b = = 5 nswer Q.19 The income of and B are in the ratio 3 :. Their expenditures are in the ratio 5 : 3. If each saves Rs 10,000, then s income is (in Rs) Sol. Let income of = 3x income of B = x & Expenditure of = 5y Expenditure of B = 3y cc. to question 3x 5y = 10, 000.(1) & x 3y = 10, 000.() 3 6x 10y = 0000.(3) 6x 9y = (4) Eq. (4) Eq. (3) y = From (1) 3x 5 (10000) = x = or income of = 3x = Rs nswer REER POINT Ltd., P Tower, IPI, Road No.1, Kota (Raj.), Ph:

10 Q.0 The radius of a circle is increased so that its circumference is increased by 5%. The area of the circle will increase by % Sol. 1 = πr = 1 + 5% of 1 = = 1.05 πr = π 1.05r When circumference increased by 5% radius increased by 5% 1 = πr 105 = π r % increase in area = π r πr 100 = 100 πr = ( ) ( ) ( 100) = ( )( ) = 100 = 10.5 % increase nswer REER POINT Ltd., P Tower, IPI, Road No.1, Kota (Raj.), Ph:

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