- Reasoning Test Preparation
- Reasoning - Home
- Reasoning - Overview
- Reasoning

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

Following quiz provides Multiple Choice Questions (MCQs) related to **Cube and Cuboid**. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using **Show Answer** button. You can use **Next Quiz** button to check new set of questions in the quiz.

Here x = 6. So x - 2 = 6 - 2 = 4. 4 × 4 × 4 = 64. Hence option D is the answer.

Here x = 10. So x - 2 = 10 - 2 = 8. So 8 × 8 × 8 = 512.

Q 3 - A big cube whose all the corners are named as K, L, M, N, O, P, Q and R. Its each portion is of 36 cm in length. The cube is segmented into tiny cubes and length of the portion of each tiny cube is 4 cm. Then how many such cubes are possible?

To find the number of tiny cubes, first we have to find x. Here x = (36/4) = 9. So number of tiny cubes M = 9 × 9 × 9 = 729. Hence option B is the answer.

Q 4 - What will be length of the portion of tiny cubes, if each portion of the original big cube is 14 cm and the cube is segmented into 343 tiny ones?

Here number of tiny cubes = 343. Cube root of 343 = 7. So x = 7 cm.

2 = (14/portion of tiny cube) or portion of tiny cube = 14/7 = 2 cm.

Q 5 - A big cube whose all the corners are named as A, B, C, D, E, F, G and H. Its each portion is of 20 cm length. The cube is segmented into tiny cubes and length of the portion of each tiny cube is 4 cm. Then how many such cubes are possible?

To find the number of tiny cubes, first we have to find x. So x = (20/4) = 5. So number of tiny cubes = 5 × 5 × 5 = 125. Hence option C.

Q 6 - A big cube is having 9 cm portion and the tiny cubes cut out of it is having 3 cm for each portion. How many tiny cubes will be formed such that each face of these cubes is surrounded by other cubes?

Here x = 9/3 = 3. Such cubes can be found by following method. x – 2 = 3 - 2 = 1. 1 × 1 × 1 = 1. So, number of cubes that will be formed such that each face of these cubes is surrounded by other cubes is only one.

Q 7 - What will be the length of the portion of tiny cubes, if the original big cube having each portion of 36 cm is segmented into 216 tiny ones?

Total number of tiny cubes = 216. Cube root of 216 = 6. So x = 6 cm.

6 = (36/ portion of tiny cube) or portion of tiny cube = 36/6 = 6 cm.

The answer is the number of corners available which is 8. Hence option C is the correct answer.

x = 8. So x - 2 = 8 - 2 = 6. 6 × 6 × 6 = 216.

Q 10 - A big cube is having 80 cm portion and the tiny cubes cut out of it is having 8 cm each portion. Then how many tiny cubes will be formed such that each face of these cubes is surrounded by other cubes?

Here x = 80/8 = 10. Such cubes can be found by following method. X – 2 = 10 - 2 = 8. So 8 × 8 × 8 = 512. So number of cubes will be formed such that each face of these cubes is surrounded by other cubes is 512.

reasoning_cube_and_cuboid.htm

Advertisements