Tag: writing and expanding numbers

Questions Related to writing and expanding numbers

The number of zeroes that comes after $1$ for $10$ millions are

  1. $8$

  2. $7$

  3. $6$

  4. $9$


Correct Option: B
Explanation:

Million $=1,000,000$ ($6$ zeros)
$10$   million $=10,000,000$
Hence the number of zeros after $1$ in $10$ million is $7.$

Match the following.

Column I Column II 
(i). 100 crores (P) 1 thousand
(ii) 10 lakhs (Q) 1 lakh
(iii) 100 thousands (R) 1 billion
(iv) 100 tens (S) 1 million
  1. (i)>(P), (ii) > (Q), (iii) > (S), (iv) > (R)

  2. (i)>(R), (ii) > (S), (iii) > (P), (iv) > (Q)

  3. (i)>(R), (ii) > (S), (iii) > (Q), (iv) > (P)

  4. (i)>(P), (ii) > (S), (iii) > (Q), (iv) > (R)


Correct Option: C
Explanation:

(i) $100$ crores $= 1,000,000,000 = 1$ billion (R)

(ii) $10$ lakhs $= 1,000,000 =  1$ million (S)
(iii) $100$ thousands $= 1,00,000 = 1$ lakh (Q)
(iv) $100$ tens $= 1,000 = 1$ thousand (P)

A train overtakes two persons who are walking in the same direction in which the train is going at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. The length of the train is:

  1. 20 m

  2. 30 m

  3. 40 m

  4. 50 m


Correct Option: D
Explanation:
Let the length of the train be x metres and speed  be y $m/s$

Speed of the first person$=2kmph$

                                          $=2\times\dfrac{5}{18}$

                                            $=\dfrac{5}{9}m/s$

Speed of the second train $=4kmph$

                                              $=4\times\dfrac{5}{18}$

                                               $=\dfrac{10}{9}m/s$

$\dfrac{x}{y-\dfrac{5}{9}}=9$

$9y-5=x$   
                                       
$90y-50=10x$...........................................(1)

                                            
$\dfrac{x}{y-\dfrac{10}{9}}=10$

$90y-100=9x$...........................................(2)

Substracting $1$ and $2$

$90y-50-90y+100=10x-9x$

$x=50$

So, the length of the train$=50m$



Express in Indian system of numeration.
Three million four hundred thirty five thousand two hundred twenty five.

  1. $345,225$

  2. $34,352,25$

  3. $3,43,52,25$

  4. $34,35,225$


Correct Option: D
Explanation:

Let us convert the given number in words into numerals as follows:


Three million$=30,00,000$
Four hundred thirty five thousand$=435,000$
Two hunderd$=200$
Twenty$=20$
Five$=5$

Now, add the numerals as shown below:

$3000000+435000+200+20+5=34,35,225$

Therefore, the required number is $34,35,225$.

Hence, the Indian system of numeration is $34,35,225$.

Express in Indian number system:
$35,987,123$

  1. $35,987,123$

  2. $3,59,87,123$

  3. $3,987,123$

  4. $3,59,8,71,23$


Correct Option: B
Explanation:
Let us convert the given numerals into words as follows:

$3,00,00,000=$Three crores
$59,00,000=$Fifty nine lakh
$87,000=$Eighty seven thousand
$100=$One hundred
$20=$Twenty
$3=$Three

Therefore, the number can be written in words as Three crores fifty nine lakh eighty seven thousand one hundred and twenty three.

Now, add the above numerals as shown below:

$3,00,00,000+59,00,000+87,000+100+20+3=3,59,87,123$ 
where $3$ is at one's place, $2$ is at tens place, $1$ is at hundredth place, $7$ is at thousandth place and so on. 

Hence, the Indian number system is $3,59,87,123$.

Express in Indian number system.
$4,456,765$

  1. Four crore forty five lakh seven hundred sixty five,

  2. Forty four lakh fifty six thousand seven hundred sixty five,

  3. Forty four crore forty five lakh seven hundred sixty five,

  4. Forty five lakh seven hundred sixty five,


Correct Option: B
Explanation:
Let us convert the given numerals into words as follows:

$44,00,000=$Forty four lakh
$56,000=$Fifty six thousand
$700=$Seven hundred
$60=$Sixty
$5=$Five

Now, combine the numbers in words as shown below:

Fourty four lakh fifty six thousand seven hundred sixty five.

Hence, the Indian number system is fourty four lakh fifty six thousand seven hundred sixty five.

A data has highest value $120$ and the lowest value $71.A$ frequency distribution in descending order with seven classes is to be constructed. The limits of the second class interval shall be 

  1. $77$ and $78$

  2. $78$ and $85$

  3. $85$ and $113$

  4. $113$ and $120$


Correct Option: D
Explanation:
Range of Frequency distribution=Highest Value-Lowest value
 $=120-71=49$

Dividing this into Seven $(7)$ equal classes.

$\Rightarrow \dfrac{49}{7}=7$

Thus the class width should be 7

Now  arranging  in descending order

Class interval $1 \rightarrow (120-7) to\space 120 \rightarrow 113-120$

Class interval $2 \rightarrow (113-7) to \space 113 \rightarrow 106-113$

Hence class interval $1$ and $2$ is $113$ and $120$ 

Arrange the following fractions is ascending order :
$\dfrac{7}{10},\dfrac{3}{8},\dfrac{4}{5}$

  1. $\dfrac{3}{8},\dfrac{7}{10},\dfrac{4}{5}$

  2. $\dfrac{3}{8},\dfrac{4}{5},\dfrac{7}{10}$

  3. $\dfrac{4}{5},\dfrac{3}{8},\dfrac{7}{10}$

  4. $\dfrac{7}{10},\dfrac{3}{8},\dfrac{4}{5}$


Correct Option: A
Explanation:
$7/10= 0.7$
$3/8 = 0.375$
$4/5=0.8$

So ascending order, = $\dfrac 3 8, \dfrac 7 {10}, \dfrac 4 5$

Arrange the following rational number in ascending order $\displaystyle \frac{3}{7},\frac{4}{5},\frac{7}{9},\frac{1}{2}$

  1. $\displaystyle \frac{4}{5},\frac{7}{5},\frac{3}{9},\frac{1}{2}$

  2. $\displaystyle \frac{3}{7},\frac{1}{2},\frac{7}{9},\frac{4}{5}$

  3. $\displaystyle \frac{4}{5},\frac{7}{9},\frac{1}{2},\frac{3}{7}$

  4. $\displaystyle \frac{1}{2},\frac{3}{7},\frac{7}{9},\frac{4}{5}$


Correct Option: B
Explanation:

The LCM of 2, 5, 7 and 9 is 630.
$\frac{3}{7} = \frac{270}{630}$
$\frac{4}{5} = \frac{504}{630}$
$\frac{7}{9}=\frac{490}{630}$
$\frac{1}{2}=\frac{315}{630}$
The ascending order will be $\frac{3}{7},\frac{1}{2},\frac{7}{9},\frac{4}{5}$

Arrange in ascending order of magnitude $\sqrt 3, \sqrt [5]{15}, \sqrt [10]{227}$

  1. $\sqrt [5]{15} < \sqrt [10]{227} < \sqrt 3$

  2. $\sqrt 3 < \sqrt [5]{15} < \sqrt [10]{227}$

  3. $\sqrt [10]{227} < \sqrt 3 < \sqrt [5]{15}$

  4. None of these


Correct Option: A
Explanation:

$\sqrt 3, \sqrt [5]{15}, \sqrt [10]{227}$


LCM of $2, 5$ and $10=10$

$\sqrt 3=\sqrt [2\times 5]{3^5}=\sqrt [10]{3\times 3\times 3\times 3\times 3}=\sqrt [10]{243}$

$\sqrt [5]{15}=\sqrt [5\times 2]{15^2}=\sqrt [10]{15\times 15}=\sqrt [10]{225}$

$\sqrt [10]{227}=\sqrt [10]{227}$

$\therefore \sqrt [5]{15} < \sqrt [10]{227} < \sqrt 3$