Start learning 50% faster. Sign in now
Maximum ten boxes are placed one above the other. There are four boxes between Box E and Box B, which at the topmost position. Two boxes are placed between Box E and Box C. One box is placed between Box C and Box D, which has 12 balls. One box is placed between Box F and Box D. Inferences: From above statements, Given, maximum 10 boxes are placed one above the other. Let the position of the box in lowermost is numbered as 1 and topmost is 10 (note: We start with 10 boxes, if necessary it can be changed with respect to given statements) Given, Box B is at topmost position. Then, Box E is fifth to below Box B (4 boxes between them) Box C is placed either 3rd to above Box E or 3rd to below Box E (2 boxes between them). Thus we get two possibilities. Case-1: Here, Box C is placed 3rd to above Box E. Then, Box D (12 balls) is placed 2nd to below Box C (1 box is between them). Box F is 2nd to below Box D (1 box is between them, only possibility) Case-2: Here, Box C is placed 3rd to below Box E. Then, Box D (12 balls) is placed 2nd to above Box C (1 box is between them, only possibility). Box F is 2nd to above Box D (1 box is between them, only possibility) By using above information, we get the following table as shown
Practice Next
I. p2+ 2p – 8 = 0 II. q2 – 5q + 6 = 0
I. 3x2 – 17x + 10 = 0
II. y2 – 17y + 52 = 0
I. 2 x ² + x – 1 = 0
II. 2 y ² - 3 y + 1 = 0
...I. 2y2 + 31y + 99 = 0
II. 4x2 + 8x – 45 = 0
I. 4x2+ 3√7 x-7 =0
II. 7y2+ 4√7 y-5=0
I. x² - 33x + 270 = 0
II. y² - 41y + 414 = 0
I. x2 – 18x + 81 = 0
II. y2 – 3y - 28 = 0
I. x2 – 13x + 40 = 0
II. 2y2 – 15y + 13 = 0
I. 8x² + 2x – 3 = 0
II. 6y² + 11y + 4 = 0
I. 96y² - 76y – 77 = 0
II. 6x² - 19x + 15 = 0