Tag: maths

Questions Related to maths

If the decimal o.d25d25d25 ................ is expressible in the form n/27, then d+n must be

  1. 9

  2. 28

  3. 30

  4. 34


Correct Option: D
Explanation:

$x = 0.d25 d25d25 -----$
$x = 0.\overline{d25}$
$1000 x = d25. \overline{d25}$
$999 x = d 25$
$x = \displaystyle \frac{d 25}{999}$
$x = \displaystyle \frac{d25}{37.27}$
take d = 9 then $x = \displaystyle \frac{25}{27}$
$d = 9          n = 25$
$d + n = 34$

Express $ \dfrac {5}{13} $ correct to $3$ significant figures.

  1. $1.26$

  2. $0.385$

  3. $0.00385$

  4. $0.103$


Correct Option: B
Explanation:

$\dfrac {5}{13} = 0.3846$


Rounding off to $3$ places to nearest decimal, we get $0.385$.
So, option $B$ is correct.

A $3$ digit id a $3$ digit number (not starting with zero) which reads the same backwards as forwards. For example $171$. The sum of all even $3$ digit palindromes, is 

  1. $22380$

  2. $25700$

  3. $22000$

  4. $22400$


Correct Option: A

The number of significant digits in the measurement of the side of a square whose computed area is $1.1025$ square inches to the nearest tenthousandth of a square cm is

  1. $2$

  2. $3$

  3. $4$

  4. $5$

  5. $1$


Correct Option: D
Explanation:

(d) is the correct choice.

Which of the following statements is incorrect regarding significant figures?

  1. All the non-zero digits are significant.

  2. All the zeros between two non-zero digits are significant.

  3. Greater the number of significant figures in a measurement, smaller is the percentage error.

  4. The power of 10 is counted while counting the number of significant figures.


Correct Option: D
Explanation:

The term significant figures are referring to the number of important digits (0 through 9 inclusive) in the coefficient of some expression in the scientific notation. The number of significant figures in any expression indicate the confidence or precision with which we can state a quantity.

Some rules for significant figures are:

1. All non-zero numbers are significant.

2. Zeros located between non-zero digits are significant.

3. Trailing zeros at the end will be significant only if the number contains a decimal point; otherwise, they are insignificant.

4. Zeros to the left of the first nonzero digit are insignificant.

5. Number in exponents (for example power of 10) is insignificant.

Thus option D is correct. 

A shopkeeper purchased $346$ kg $500$ g of orange. Later on, he found that $109$ kg $300$ g of oranges were rotten. Find the quantity of oranges in good condition.

  1. $150$ kg $670$ g

  2. $204$ kg $600$ g

  3. $237$ kg $200$ g

  4. $250$ kg $940$ g


Correct Option: C
Explanation:

We know that $gram$  is abbreviated as $gm$

$1$  $kg =1000$  $gram$
Total  weight of  Oranges purchased $ = B = $   $346$ $kg$  $500$  $gm$ 
Total weight of rotten oranges $=A = $   $109$ $kg$  $300$  $gm$ 
(i) We have to subtract $109\ kg$ $300\ gm$ from $346$ $kg$  $500$  $gm$ to get the weight of good conditioned oranges

$109$ $kg$  $300$  $gm$ $ = 109kg + 300gm$  $=A $   ...................(1)

$346$ $kg$  $500$  $gm$ $ = 346kg + 500gm$  $=B $   ...................(2)


$B-A = [346\ kg+500\ gm]-[109\ kg +300\ gm]$
$ = [346\ kg-109\ kg] +  $ $[500$ $gm$ $-$ $300$  $gm$]
$= 237  $  $kg $ $+  $  $200$ $gm$ 

Therefore, $237  $  $kg $  $200$ $gm$   of oranges are in good condition.

How much heavier is the toffee packet which has mass $1$ kg $345$ g in comparison to $1$ kg of chocolates?

  1. $1$ kg $345$ g

  2. $345$ g

  3. $1$ kg

  4. $1$ kg $234$ g


Correct Option: B
Explanation:

Weight of first toffee packet $=1.345\ kg$ 


Weight of second toffee packet $=1\ kg$ 

Required difference
$=1.345-1$

$=0.345\ kg$

$=345\ gm$

Hence, this is the answer.

If $15$ bananas measure $1.2$ kg, then find the weight of one banana. (Give yous answer in grams)

  1. $80$ gms

  2. $70$ gms

  3. $60$ gms

  4. $50$ gms


Correct Option: A
Explanation:

Total number of Bananas $=15$


Total mass of 15 Bananas  $=1.2 kg$

Total mass of $1$ banana $=\dfrac{1.2}{15} $   $kg$

We know that

$1$  $kg =1000$  $gram$

$1$  $gm =\dfrac1{1000}$  $kg$

we have to convert $\dfrac{1.2}{15}$  $kg$  to   $grams$


$\dfrac{1.2}{15}$  $kg =\dfrac{1.2}{15} \times 1000$  $gm$

                $=80$  $gm$

So, Option $A $ is correct 

A hollow iron pipe is $21\,cm$ long and its external diameter is $8\,cm$. If the thickness of the pipe is $1\,cm$ and iron weighs $8\,g/c{m^3}$, then the weight of pipe is :

  1. $3.6\,kg$

  2. $3.696\,kg$

  3. $36\,kg$

  4. $36.9\,kg$


Correct Option: B
Explanation:

Volume of iron $=\pi(R^2-r^2)\times h$


$=\pi(4^2-3^2)\times 21$

$=\cfrac{22}{7}\times(16-9)\times 21$

$=462cm^2$

Weight $=\cfrac{8\times 462}{1000}kg=3.696kg$

Working together, pipes $A$ and $B$ can fill an empty tank in $10\ hours$. they worked together for $4$ hours and then $B$ stopped and $A$ continued filling the tank till was full. It took a total of $13\ hours$ to fill the tank. How long would it take $A$  to fill the empty tank alone?

  1. $13\ hours$

  2. $15\ hours$

  3. $17\ hours$

  4. $18\ hours$


Correct Option: B