Test Index

Problems based on Numbers Test 6

© examsnet.com

Question : 4 of 40

Marks: +1, -0

If

1^3 + 2^3 + \ldots + 9^3 = 2,025

, then the approximate value of

(0.11)^3 + (0.22)^3 + \ldots + (0.99)^3

is:

0.2695
0.3695
2.695
3.695

Solution:

Concept:

Factor out the common cube from each term.

Explanation:

Each term can be written as

(0.11k)^3 = (0.11)^3 k^3

.

So,

(0.11)^3 + (0.22)^3 + \ldots + (0.99)^3 = (0.11)^3(1^3 + 2^3 + \ldots + 9^3)

.

Given

1^3 + 2^3 + \ldots + 9^3 = 2025

.

Thus, value

= (0.11)^3 \times 2025

.

(0.11)^3 = 0.001331

.

0.001331 \times 2025 = 2.695275

.

Approximate value

= 2.695

.

Answer:

2.695

© examsnet.com

Go to Question:

Prev Question

Next Question