Title 2-Magic Circles _ Squares.pdf ✅

2 CHAPTER CONTENTS  Magic Circle  Magic Square  Miscellaneous Pattern Magic Circle Problems This unit is based on numerical calculations. Usually these are circles, the first two of which have four numbers at four points on the circle and one inside the circle. These numbers are placed according to some rules or sequence. The third cycle has any four numbers with fifth missing. We are required to find this number from the given choice, according to the same rule that holds good for other two circles. EXAMPLES  Ex.1 12 3 2 4 5 18 6 3 5 2 ? 2 8 5 2 (A) 12 (B) 14 (C) 16 (D) 18 Sol. The answer to above question is (C) i.e. 16, because the numbers inside the first two circles are obtained by multiplying the outside numbers and dividing by 10, i.e. 10 53 4 2 = 10 120 = 12 The same follows for the second and third circles. Ex.2 26 36 25 49 64 21 9 16 81 25 ? 25 36 64 144 (A) 24 (B) 25 (C) 23 (D) 31 Sol. The answer is (D) i.e. 31, because the numbers inside the first two circles are obtained by taking the sum of the square roots of the four numbers outside the circles, e.g. 49 + 64 + 25 + 36 = 7 + 8 + 5 + 6 = 26 (Ist Circle) and 16 + 25 + 9 + 81 = 4 + 5 + 3 + 9 = 21 (IInd Circle) Ex.3 14 3 2 4 5 18 4 3 6 5 ? 6 5 8 7 (A) 24 (B) 26 (C) 28 (D) 22 Sol. The answer is (B) i.e. 26, because (52 – 4 2) + (32 – 2 2) = 14 (Ist circle) and (62 – 5 2) + (42 – 3 2) = 18 (IInd circle) Ex.4 8 16 12 9 5 4 15 13 11 9 ? 18 15 9 7 (A) 12 (B) 8 (C) 4 (D) 5 Sol. The answer is (D) i.e. 5, because (9 – 5) + (16 – 12) = 8 (Ist Circle) (11 – 9) + (15 – 13) = 4 (IInd Circle) so, (9 – 7) + (18 – 15) = 5 (IIIrd Circle) Ex.5 13 5 3 7 4 5 9 3 8 4 ? 8 3 9 4 (A) 4 (B) 8 (C) 12 (D) 15 Sol. The answer is (C) i.e. 12 because (7 × 4) – (5 × 3) = 28 – 15 = 13 (Ist Circle) (8 × 4) – (9 × 3) = 32 – 27 = 05 (IInd Circle) So, (9 × 4) – (8 × 3) = 36 – 24 = 12 (IIIrd Circle) MAGIC CIRCLES & SQUARES
Ex.6 29 7 2 3 5 39 6 4 5 3 ? 6 5 7 3 (A) 49 (B) 51 (C) 59 (D) 21 Sol. Answer is (B) i.e. 51, because (3 × 5) + (7 × 2) = 29 and (5 × 3) + (6 × 4) = 39 Ex.7 13 27 3 4 16 11 42 7 13 65 ? 27 9 8 72 (A) 9 (B) 12 (C) 15 (D) 18 Sol. Answer is (B) i.e. 12, because (16 ÷ 4) + (27 ÷ 3) = 13 (Ist Circle) and (65 ÷ 13) + (42 ÷ 7) = 5 + 6 = 11 (IInd Circle) So, (72 ÷ 8) + (27 ÷ 9) = 9 + 3 = 12 (IIIrd Circle) Ex.8 2 2 6 3 8 1 3 8 4 6 ? 6 4 12 8 (A) 3 (B) 4 (C) 5 (D) 6 Sol. Answer is (B) i.e. 4 because (3 × 8) ÷ (2 × 6) = 24 ÷ 12 = 2 (Ist Circle) (4 × 6) ÷ (8 × 3) = 24 ÷ 24 = 1 (IInd Circle) So, (12 × 8) ÷ (6 × 4) = 96 ÷ 24 = 4 (IIIrd Circle) Ex.9 14 3 2 5 4 18 4 3 6 5 ? 6 5 8 7 (A) 49 (B) 26 (C) 59 (D) 21 Sol. Answer is (B) i.e. 26 because (5 × 4) – (3 × 2) = 14 (5 × 6) – (3 × 4) = 18 So, (8 × 7) – (5 × 6) = 26 This example has been solved already (see example 3), so it can be concluded that there are many method to solve the same problem. Ex.10 3.5 8 9 5 13 3.6 9 8 7 12 ? 7 12 9 11 (A) 3.7 (B) 3.8 (C) 3.9 (D) 3.1 Sol. Answer is 3.9 because 10 (5 13  9  8) = 35 ÷ 10 = 3.5 So, 10 (7  9 1112) = 39 ÷ 10 = 3.9 Ex.11 1 35 5 3 18 2 42 7 4 32 ? 40 5 6 54 (A) 0 (B) 1 (C) 3 (D) 5 Sol. Answer is (B) i.e. 1 because pattern is (35 ÷ 5) – (18 ÷ 3) = 7 – 6 = 1 (32 ÷ 4) – (42 ÷ 7) = 8 – 6 = 2 So, (54 ÷ 6) – (40 ÷ 5) = 9 – 8 = 1 Ans. Ex.12 5 7 9 4 5 12 14 20 75 35 ? 9 50 65 45 (A) 7 (B) 9 (C) 13 (D) 15 Sol. Answer is (C) i.e. 13 because 4  7  5  9 = 25 = 5 65  45  9  50 = 169 = 13 Ans. Ex.13 20 25 11 7 57 10 7 4 6 8 ? 21 11 19 13 (A) 16 (B) 25 (C) 36 (D) 49 Sol. Answer is (A) i.e. 16 because 2 × 25  57 11 7 = 2 100 = 20 2 × 7  8  4  6 = 2 25 = 10 So, 2 × 2113 19 11 = 2 64 = 16 Ans. Ex.14 8 16 12 9 5 4 15 13 11 9 ? 18 15 9 7 (A) 24 (B) 5 (C) 28 (D) 27 Sol. Answer is (B) i.e. 5 because (16 + 9) – (12 + 5) = 8 (11 + 15) – (9 + 13) = 4 So, (9 + 18) – (15 + 7) = 5 Ans
Ex.15 8 16 36 1 25 17 64 100 49 81 ? 9 25 4 16 (A) 6 (B) 7 (C) 8 (D) 9 Sol. Answer is (B) i.e. 7 because ( 1 + 16 + 25 + 36 ) ÷ 2 = 16 ÷ 2 = 8 ( 49 + 64 + 81 + 100 ) ÷ 2 = 34 ÷ 2 = 17 So, ( 4 + 9 + 16 + 25 ) ÷ 2 = 14 ÷ 2 = 7 Ans. Magic Squares Problems In these types of questions, numbers are arranged according to some rule in the cells made into a square. One cell is left empty, and we have to find the way the numbers are arranged and mark the answer from the choice given below. Let us consider some examples :– EXAMPLES  Ex.16 9 3 ? 21 7 27 5 9 (A) 12 (B) 15 (C) 30 (D) 28 Sol. Answer is (B) i.e. 15 because the numbers in opposite cells are 3-times or 1/3 each other. e.g. 7 × 3 = 21, 3 × 3 = 9, 9 × 3 = 27, then 5 × 3 = 15 Ex.17 108 2 ? 18 6 3 (A) 1 (B) 36 (C) 216 (D) 1944 Sol. The answer is 1944 as the numbers are arranged in the following way, 2 × 3 = 6, 3 × 6 = 18, 6 × 18 = 108, 18 × 108 = 1944 Ex.18 7 6 5 3 3 4 2 3 ? (A) 12 (B) 3 (C) 4 (D) 5 Sol. The answer is 3 because the sum of the numbers in each column is 12. Ex.19 32 35 39 42 46 51 3 8 ? (A) 11 (B) 90 (C) 60 (D) 14 Sol. The answer is 14 because the numbers are increasing by 3 and 4 in the first row, 4 and 5 in the second row and 5 and 6 in the third row. Ex.20 7 9 16 4 15 ? 13 8 21 (A) 29 (B) 19 (C) 23 (D) 25 Sol. The answer is 19 because the sum of the first two numbers in each row gives the third number, i.e. 7 + 9 = 16, 4 + 15 = 19, 13 + 8 = 21 Ex.21 127 ? 7 63 31 15 (A) 255 (B) 440 (C) 190 (D) 250 Sol. The answer is 255 as the number is doubled and increased by 1 in each cell, starting from 7, i.e. 7 × 2 = 14 + 1 = 15 15 × 2 = 30 + 1 = 31 31 × 2 = 62 + 1 = 63 63 × 2 = 126 + 1 = 127 127 × 2 = 254 + 1 = 255 Ex.22 17 11 19 12 13 16 25 4 ? (A) 36 (B) 9 (C) 25 (D) 64 Sol. In the first column 25 = (17 – 12)2 therefore (19 – 16)2 is 9
Ex.23 21 56 70 45 87 84 115 180 ? (A) 130 (B) 195 (C) 295 (D) 150 Sol. The rule is that in each row the difference of first two numbers is doubled. i.e. (56 – 21) × 2 = 70. Hence the required number (180 – 115) × 2 = 130 Ex.24 17 15 8 99 95 64 36 45 ? (A) 729 (B) 1331 (C) – 729 (D) –343 Sol. The rule is that in a row as (17 – 15)3 = 8. Therefore (36 – 45)3 = (–9)3 = –729 Miscellaneous Pattern Directions : Find the missing character from among the given alternatives. Ex.25 2 10 13 6 3 15 7 12 25 441 ? 289 (A) 625 (B) 25 (C) 125 (D) 156 Sol. Clearly, (3 + 2)2 = 25; (15 + 6)2 = (21)2 = 441 ; (10 + 7)2 = (17)2 = 289. So, missing number = (12 + 13)2 = (25)2 = 625. Hence, the answer is (A). Ex.26 5 19 3 4 7 ? 5 6 (a) (b) 6 29 4 5 (c) (A) 25 (B) 37 (C) 41 (D) 47 Sol. Clearly, in fig. (a), 5 × 3 + 4 = 19. In fig (c) = 6 × 4 + 5 = 29.  In fig.(b), missing number = 7 × 5 + 6 = 35 + 6 = 41. Hence, the answer is (C). Ex.27 492 5 7 8 9 6 4 572 7 6 4 9 5 8 A B ? 4 7 5 5 3 2 C (A) 115 (B) 130 (C) 135 (D) 140 Sol. Clearly, the number inside the circle is equal to the sum of the product of the upper three numbers and the product of the lower three numbers. Thus, In fig. A, (5 × 6 × 8) + (7 × 4 × 9) = 240 + 252 = 492. In fig. B, (7 × 5 × 4) + (6 × 8 × 9) = 140 + 432 = 572.  In fig C, missing number = (4 × 3 × 5) + (7 × 2 × 5) = 60 + 70 = 130. Hence, the answer is (B). Ex.28 ? 1 2 21 22 40 20 23 43 (A) 5 (B) 4 (C) 3 (D) 2 Sol. Clearly, in the second column, 22 + 2 – 23 = 1. In the third column, 40 + 5 – 43 = 2.  In the first column, missing number = 21 + 1 – 20 = 2. Hence, the answer is (D). Ex.29 6 18 15 3 2 5 4 3 ? 8 27 9 (A) 11 (B) 6 (C) 3 (D) 2 Sol. Clearly, in the first column, 3 6 4 = 3 24 = 8. In the second column, 2 18 3 = 2 54 = 27. Let the missing number in the third column be x. Then, 5 15 x = 9 or 15x = 45 or x = 3. Hence, the answer is (C).