Tag: rounding off digits

Questions Related to rounding off digits

On a recent trip, Cindy drove her car 290 miles, rounded to the nearest 10 miles, and used 12 gallons of gasoline, rounded to the nearest gallon. The actual number of miles per gallon that Cindy's car got on this trip must have been between.

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $\displaystyle \frac { 290 }{ 12.5 } $ and $\displaystyle \frac { 290 }{ 11.5 } $

  2. $\displaystyle \frac { 295 }{ 12 } $ and $\displaystyle \frac { 285 }{ 11.5 } $

  3. $\displaystyle \frac { 285 }{ 12 } $ and $\displaystyle \frac { 295 }{ 12 } $

  4. $\displaystyle \frac { 285 }{ 12.5 } $ and $\displaystyle \frac { 295 }{ 11.5 } $

  5. $\displaystyle \frac { 295 }{ 12.5 } $ and $\displaystyle \frac { 285 }{ 11.5 } $

Show answer
Correct Option: D
Explanation:
Here, the rounding off of numbers is tested.
Distance rounded to nearest 10 miles = 290. 
So actual distance covered may be b/w 285 and 295. 
Gasoline used rounded to nearest gallon = 12.
So the actual gas used may be b/w 11.5 and 12.5. 
Now to get range of of miles/gallon, 
least value = the least of distance/the max of gas = 285/12.5
highest value = max of distance/least of gas = 295/11.5 
Option D is the correct choice.

The radius of the sun is $7\times 10^8 m$ and its mass is $2\times 10^{30} kg$. What is the order of magnitude of density of the sun?

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $1.4\times 10^3 kg/m^3$

  2. $ 10^7 kg/m^3$

  3. $1.5\times 10^3 kg/m^3$

  4. $10^3 kg/m^3$

Show answer
Correct Option: D
Explanation:

Given :  $R = 7\times 10^8 \ m$     and    $M = 2\times 10^{30} \ kg$
Volume of sun  $V = \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi\times (7\times 10^8)^3 = 1.4\times 10^{27} \ m^3$
Density of sun   $\rho = \dfrac{M}{V} = \dfrac{2\times 10^{30}}{1.4\times 10^{27}} = 1.43\times 10^3 \ kg/m^3$
Thus order of magnitude of density of sun is $10^3 \ kg/m^3$.

The radius of the earth is given $6.4\times 10^6 m$. What is the order of magnitude of the size of the earth? 

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $10^6 m$

  2. $6 \times 10^6 m$

  3. $10^7 m$

  4. $5 \times 10^6 m$

Show answer
Correct Option: C
Explanation:

When a number rounded to the nearest power of 10, it is called order of magnitude. Here a number which is less than 5 is taken as 1 and a number greater than 5 is taken as 10. 
As $6.4>10, $ so it would be takes as 10. 
The order of magnitude of earth's size $=10\times 10^6=10^7 m$  

What is the order of magnitude of the distance of a quasar from us if light takes 2.9 billion years to reach us ?  

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $2.7\times 10^{25} m$

  2. $ 10^{25} m$

  3. $10^24 m$

  4. $3\times 10^{25} m$

Show answer
Correct Option: B
Explanation:

Here, time taken $t=2.9$ billion years $=2.9\times 10^9$ years $ =2.9\times 10^9\times (365\times 24\times 3600) s=9.14\times 10^{16} s$
Distance $=$ velocity $\times $ time
or $d=(3\times 10^8)\times (9.14\times 10^{16})=2.74\times 10^{25} m$
As $2.74 <5$ so it will be taken as 1. 
Thus, the order of magnitude of distance $d=1\times 10^{25} m=10^{25} m$

Match the following:

Column - I Column - II
A Deca p $10^3$
B Hecto q $10^6$
C Kilo r $10^1$
D Mega s $10^2$
E Giga t $10^8$
u $10^9$


physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. A-r, B-s, C-p, D-q, E-u

  2. A-s, B-r, C-p, D-q, E-u

  3. A-r, B-s, C-u, D-q, E-p

  4. A-q, B-s, C-p, D-r, E-u

Show answer
Correct Option: A
Explanation:

Here column-I represents the some prefixes and column-II represents the power of ten corresponding prefixes. 

Deca $=10^1$, hecto =$10^2$, kilo =$10^3$, mega $=10^6$, giga $=10^9$ 
Thus, A-r, B-s, C-p, D-q, E-u

$ 5\times 10^7 \mu s$ is equivalent to ____

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. 0.5 s

  2. 5 s

  3. 50 s

  4. 500 s

Show answer
Correct Option: C
Explanation:

We know that  $1\mu s = 10^{-6} \ s$
Thus,  $5\times 10^{7}\mu s = 5\times 10^{7}\mu s\times \dfrac{10^{-6} \ s}{\mu s} = 50 \ s$

The order of magnitude of $379$ is

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: B
Explanation:

Order of magnitude of $379$ is $\dfrac { 379 }{ { 10 }^{ 2 } } =3.79<{ 5 }^{ -1 }$    i.e, $379\approx { 10 }^{ 2 }$

order of magnitude $=2$.

With due regard to significant figures, add the following:
a. 953 and 0.324
b. 953 and 0.625
c. 953.0 and 0.324
d. 953.0 and 0.374

physics units and measurement: error analysis rounding off digits rounding of digits standard form

  1. a. $953$; b. $954$ c. $953.3$ d. $953.4$

  2. a. $953$; b. $955$ c. $953.3$ d. $953.4$

  3. a. $952$; b. $954$ c. $953.3$ d. $953.4$

  4. a. $953$; b. $954$ c. $953.3$ d. $952.4$

Show answer
Correct Option: A
Explanation:
a.  $953+0.324=953.324\approx \boxed { 953 } $
b.  $953+0.625=953.625\approx \boxed { 954 } $
c.  $953.0+0.324=953.324\approx \boxed { 953.3 } $
d.  $953.0+0.374=953.374\approx \boxed { 953.4 } $