Question

There are three laces A, B, C in a straight line as shown below.If distance between place A and B is $(2.4\times {10}^6)$ m and distance between B and C is $(5.2\times {10}^5)m$, then find the distance between place A and C is standard form.

• A. $(292\times {10}^6)m$
• B. $(2.92\times {10}^6)m$
• C. $(292\times {10}^5)m$
• D. $(2.92\times {10}^4)m$

Answer 1

Answer: (B)

$(2.92\times {10}^6)m$

Distance between place A and B $=2.4\times {10}^{6}m=2400000m$
Distance between place B and C $=5.2\times {10}^{5}m=5.2\times 100000m=520000m$
Distance between place A and C $=(2.4\times {10}^6+5.2\times {10}^5)m=(2400000+520000)m=292\times {10}^{4}m$
Standard form of $292\times {10}^{4}m=2.92\times {10}^{6}m$