Mathematics Questions and Answers – Comparing Very Small and Large Number

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Comparing Very Small and Large Number”.

1. Mass of object A is 120000 and that of B is 0.240. Compare their masses.
a) Mass of Object B is approximately is 105 times larger than A
b) Mass of Object A is approximately is 105 times larger than B
c) Mass of Object A is approximately is 106 times larger than B
d) Mass of Object B is approximately is 106 times larger than A
View Answer

Answer: c
Explanation: \(\frac{mass\, of\, object\, A}{mass\, of\, object\, B} = \frac{1.2 × 10^5}{2.4 × 10^{-1}} = \frac{1.2 × 10^6}{2.4}\)
⇒ Mass of Object A is approximately is 106 times larger than B.

2. Compare the radius of circle P (r1 = 10.2 × 10-6) with the radius of circle B (r1 = 0.2 × 10-5).
a) r1 = 6.1r2
b) r2 = 5r1
c) r1 = 5r2
d) r1 = 5.1r2
View Answer

Answer: d
Explanation: \(\frac{r_1}{r_2} = \frac{10.2 × 10^{-6}}{0.2 × 10^{-5}} = \frac{10.2 × 10^{-1}}{0.2}\) = \(\frac{1.02}{0.2}\) = 5.1.
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3. Mass of a neutron is 1.675 × 10-27 and mass of proton is 1.672 × 10-27. Find the total mass.
a) 3.347 × 10-27
b) 3.337 × 10-27
c) 3.447 × 10-27
d) 3.357 × 10-27
View Answer

Answer: a
Explanation: Total mass = mass of neutron + mass of proton = 1.675 × 10-27 + 1.672 × 10-27 = (1.675 + 1.672) × 10-27 = 3.347 × 10-27.
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4. Distance between A to C is 2.1 × 1011 and that of A to B is 3.12 × 109. Find the distance between B and C.

a) 2.1312 × 1011
b) 2.0412 × 1011
c) 2.1312 × 1011
d) 2.1312 × 1010
View Answer

Answer: c
Explanation: Distance between B and C = AC – AB = 2.1 × 1011 – 3.12 × 109 = (2.1 + 0.0312) × 1011 = 2.1312 × 1011.

5. Compare volumes of cube A having side 2 × 102 m and cube B having side 1 × 101 m.
a) V1 = 800V2
b) V1 = 8000V2
c) V2 = 8000V1
d) V2 = 800V1
View Answer

Answer: b
Explanation: volume of cube A = (2 × 102)3 = 8 × 106 m3.
⇒ Volume of cube B = (101)3 = 103
⇒ \(\frac{V_1}{V_2} = \frac{8 × 10^6}{10^3}\) = 8000.
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Sanfoundry Global Education & Learning Series – Mathematics – Class 8.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter