The sum of two numbers is 31.021 if one of them is 11.56 then what should be other number

Recalling adding and subtracting decimals - class-VII

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A train leaves the station at $8:15$ pm and reach destination at $6:25$ am . The duration of the journey is 

1. $10$ hours $15$ minutes

2. $9$ hours $25$ minutes

3. $10$ hours $10$ minutes

4. $9$ hours $40$ minutes

Correct Option: C

Explanation:

The time interval between $8:15$ pm and $6:25 am$ = $10\,\, hrs\,\, 10\,\, min$

Therefore, the duration of the journey of the train is $10\,\, hr \,\,10\,\, min.$

Therefore, option C is correct.

Simplify : $[ 0.9 - { 2.3 - 3.2 - 3.2 - ( 7.1- 5.4 - 3.5 ) } ]$

Correct Option: C

Explanation:

$\displaystyle \left [ 0.9-\left { 2.3-3.2-\left ( 7.1-5.4-3.5 \right ) \right } \right ]$

=$\displaystyle \left [ 0.9-\left { 2.3-3.2-\left ( 7.1-8.9 \right ) \right } \right ]$

= $\displaystyle \left [ 0.9-\left { 2.3-3.2-\left ( -1.8 \right ) \right } \right ]$

= $\displaystyle \left [ 0.9-\left { 2.3-3.2+1.8 \right } \right ]$

= $\displaystyle \left [ 0.9-\left { 4.1-3.2 \right } \right ]$

= $\displaystyle \left [ 0.9-0.9 \right ]=0 $

The value of $\displaystyle 0.\overline{2}+0.\overline{3}+0.\overline{4}+0.\overline{9}+0.\overline{39}$ is 

1. $\displaystyle 0.\overline{57}$

2. $\displaystyle 1\frac{20}{33}$

3. $\displaystyle 2\frac{1}{3}$

4. $\displaystyle 2\frac{13}{33}$

Correct Option: D

Explanation:

$\displaystyle 0.\overline{2}+0.\overline{3}+0.\overline{4}+0.\overline{9}+0.\overline{39}$

= $\displaystyle \frac{2}{9}+\frac{3}{9}+\frac{4}{9}+\frac{9}{9}+\frac{39}{99}$

= $\displaystyle \frac{22+33+44+99+39}{99}$

= $\displaystyle \frac{237}{99}=2\frac{13}{33}$

Simplify $\displaystyle :0.\overline{4}+0.\overline{61}+0.\overline{11}-0.\overline{36}$

1. $\displaystyle 0.\overline{83}$

2. $\displaystyle 0.\overline{87}$

3. $\displaystyle 0.\overline{80}$

4. $\displaystyle 0.\overline{85}$

Correct Option: C

Explanation:

$\displaystyle :0.\overline{4}+0.\overline{61}+0.\overline{11}-0.\overline{36}=\frac{4}{9}+\frac{61}{99}+\frac{11}{99}-\frac{36}{99}=\frac{4}{9}+\frac{72}{99}-\frac{36}{99}$

$\displaystyle=\frac{4}{9}+\frac{36}{99}=\frac{44}{99}+\frac{36}{99}=\frac{80}{99}=0.\overline{80}$

Sum of $1.8, 16.3$ and $72.985$ is _____ 

1. $91.85$

2. $9108.5$

3. $91.085$

4. $9.1085$

Correct Option: C

Explanation:

For obtaining the sum of decimal numbers with different decimal values we expand the decimal values of all the numbers with respect to the number with the highest decimal value. 

Because the number with the largest decimal places is $72.985$ with $3$ decimal places, we expand all numbers up to $3$ decimal places.

$1.800+16.300+72.985=91.085$

So option C is the correct answer.

$3.\overline { 87 } -2.\overline { 59 } =$?

1. $1.20$

2. $1.\overline { 2 } $

3. $1.\overline { 27 } $

4. $1.\overline { 28 } $

Correct Option: D

Explanation:

$3.\overline { 87 } -2.\overline { 59 } =\left( 3+0.\overline { 87 }  \right) -\left( 2+0.\overline { 59 }  \right) $

$=\left( 3+\dfrac { 87 }{ 99 }  \right) -\left( 2+\dfrac { 59 }{ 99 }  \right) $

$=1+\left( \dfrac { 87 }{ 99 } -\dfrac { 59 }{ 99 }  \right) $

$=1+\dfrac { 28 }{ 99 } $

$=1.\overline { 28 } $.

The value of $(1.02)^4+(0.98)^4$ upto three places of decimal is

1. $2.004$

2. $2.003$

3. $2.04$

4. $2.0004$

Correct Option: A

Explanation:

${\left( 1.02 \right)}^{4} + {\left( 0.98 \right)}^{4}$

$= {\left( 1 + 0.02 \right)}^{4} + {\left( 1 - 0.02 \right)}^{4}$

$= \left[ {^{4}{C} _{0}} {\left( 1 \right)}^{4} + {^{4}{C} _{1}} {\left( 1 \right)}^{4-1} {\left( 0.02 \right)}^{1} + {^{4}{C} _{2}} {\left( 1 \right)}^{4-2} {\left( 0.02 \right)}^{2} + {^{4}{C} _{3}} {\left( 1 \right)}^{4-3} {\left( 0.02 \right)}^{3} + {^{4}{C} _{4}} {\left( 1 \right)}^{4-4} {\left( 0.02 \right)}^{4} \right] + \ \left[ {^{4}{C} _{0}} {\left( 1 \right)}^{4} - {^{4}{C} _{1}} {\left( 1 \right)}^{4-1} {\left( 0.02 \right)}^{1} + {^{4}{C} _{2}} {\left( 1 \right)}^{4-2} {\left( 0.02 \right)}^{2} - {^{4}{C} _{3}} {\left( 1 \right)}^{4-3} {\left( 0.02 \right)}^{3} + {^{4}{C} _{4}} {\left( 1 \right)}^{4-4} {\left( 0.02 \right)}^{4} \right]$

$= 2 \left[ {^{4}{C} _{0}} + {^{4}{C} _{2}} \left( 0.0004 \right) + {^{4}{C} _{4}} \left( 0.00000016 \right) \right]$

$= 2 \times \left( 1 + 0.0024 + 0.00000016 \right)$

$= 2 \times 1.002$

$= 2.004$

Hence the value of given expression upto three places of decimal is $2.004$.

The value of $0.\overline{1}+0.0\overline{1}+0.00\overline{1}$ is equal to

1. $\dfrac{343}{900}$

2. $\dfrac{37}{300}$

3. $\dfrac{4}{33}$

4. $\dfrac{1343}{10989}$

Correct Option: B

Explanation:

Let $  x = 0.\bar{1} \Rightarrow 10x = 1.\bar{1}$

$ 10x -x = 1 \Rightarrow  x = 1/9 ...(I)$

$ y = 0.0\bar{1} = 0.01+0.001+0.0001+...$

$ = \dfrac{1}{100}+\dfrac{1}{1000}+\dfrac{1}{10000}+...$

$ = \dfrac{1}{100}(1+\dfrac{1}{10}+\dfrac{1}{100}+...)$

$ = \dfrac{1}{100}\times \dfrac{1}{1-\dfrac{1}{10}} = \dfrac{1}{90}...(II)$

$ z = 0.00\bar{1} = 0.001+0.0001 + 0.00001+...$

$=0.001(1+\dfrac{1}{10}+\dfrac{1}{100}+...)$

$ =\dfrac{1}{1000}\times \dfrac{1}{1-\frac{1}{10}}  = \dfrac{1}{1000} \times \dfrac{10}{9} = \dfrac{1}{900}...(III)$

From (1),(2),(3) $ 0.\bar{1}+0.0\bar{1}+0.00\bar{1} = \dfrac{1}{9}+\dfrac{1}{90}+\dfrac{1}{900}= \dfrac{111}{900} = \dfrac{37}{300}$

$ \therefore $ option B is correct

Simplify the following :

$0.4 \times \displaystyle \frac{7}{3} \div \frac{15}{8}  $ of $  \left ( \dfrac{7}{5} - \dfrac{4}{3} \right )$.

1. $\displaystyle 5 \frac{8}{3375}$

2. $\displaystyle 7 \frac{8}{15}$

3. $\displaystyle 7 \frac{7}{15}$

4. $\displaystyle 5 \frac{7}{3375}$

Correct Option: C

Explanation:

The given expression is

$0.4\times \dfrac { 7 }{ 3 } \div \dfrac { 15 }{ 8 } $ of $\left( \dfrac { 7 }{ 5 } -\dfrac { 4 }{ 3 }  \right) $

$=\dfrac { 4 }{ 10 } \times \dfrac { 7 }{ 3 } \div \dfrac { 15 }{ 8 } $ of $\dfrac { 1 }{ 15 } $

$ =\dfrac { 2 }{ 5 } \times \dfrac { 7 }{ 3 } \div \dfrac { 1 }{ 8 } $

$ =\dfrac { 2 }{ 5 } \times \dfrac { 7 }{ 3 } \times \dfrac { 8 }{ 1 } $

$ =\dfrac { 112 }{ 15 } $

$=7\dfrac { 7 }{ 15 } $

The value of $\displaystyle \frac{0.25\times0.25-0.24\times0.24}{0.49}=?$

1. $0.0006$

2. $0.49$

3. $0.01$

4. $0.1$

Correct Option: C

Explanation:

Given, $\displaystyle \frac {0.25\times0.25-0.24\times0.24}{0.49}$Using the identity $a^2 - b^2 = (a+b)(a-b)$$\therefore \displaystyle 0.25\times 0.25 - 0.24\times 0.24 = (0.25+0.24)(0.252-0.24) = (0.49)(0.01)$

Thus, $\displaystyle \frac {0.25\times0.25-0.24\times0.24}{0.49} = \frac {0.49\times 0.01}{0.49}$ = $0.01$

$\displaystyle \frac{42.31-26.43}{42.31+26.43}\div \frac{423.1-264.3}{4.231+2.643}$  is equal to 

1. $\displaystyle 10^{-2}$

2. $\displaystyle 10^{-1}$

3. 10

4. $\displaystyle 10^{2}$

Correct Option: A

Explanation:

$\displaystyle \frac{42.31-26.43}{42.31+26.43}\div \frac{423.1-264.3}{4.231+2.643}$

$=\displaystyle \frac{15.88}{68.74}\div \frac{158.8}{6.874}$

$=\displaystyle \frac{15.88}{68.74}\times \frac{6.874}{158.8}=\frac{1588}{6874}\times \frac{68.74}{1588}$

= $\displaystyle \frac{68.74}{6874}=\frac{6874}{687400}=\frac{1}{100}=0.01$

Which pair of operations will make the equation below true when inserted into the blank spaces in the order shown? $\displaystyle 2\frac{3}{10} $      $1.5$  _  $2=1.8$

1. $-$ and $+$

2. $\displaystyle \times $ and $+$

3. $+$ and $-$

4. $\displaystyle \times $and $-$

Correct Option: C

Explanation:

Let us first write the mixed fraction $2\dfrac {3}{10}$ in decimals as follows:

$2\dfrac { 3 }{ 10 } =\dfrac { (2\times 10)+3 }{ 10 } =\dfrac { 20+3 }{ 10 } =\dfrac { 23 }{ 10 } =2.3$

Therefore, $2\dfrac {3}{10}=2.3$

Now, add $1.5$ to $2.3$ then we get,

$2.3+1.5=3.8$

Now subtract $2$ from $3.8$ as follows:

$3.8-2=1.8$

Therefore, we have $2.3+1.5-2=1.8$

Hence, $2\dfrac { 3 }{ 10 } +1.5-2=1.8$

In the expression $24 - [ 2.4 - { 0.24 - (0.024 - x)}] = 21.8184$, the value of x is 

1. $0.0024$

2. $0.024$

3. $0.24$

4. $2.4$

Correct Option: A

Explanation:

We solve the given expression as follows:

$24−\left[ 2.4−{ \left{ 0.24−\left( 0.024−x \right)  \right}  } \right] =21.8184\ \Rightarrow 24−\left[ 2.4−{ \left{ 0.24−0.024+x \right}  } \right] =21.8184\ \Rightarrow 24−\left[ 2.4−{ \left{ 0.216+x \right}  } \right] =21.8184\ \Rightarrow 24−\left[ 2.4−{ 0.216-x } \right] =21.8184$

$\Rightarrow 24−\left[ 2.184-x \right] =21.8184\ \Rightarrow 24−2.184+x=21.8184\ \Rightarrow 21.816+x=21.8184\ \Rightarrow x=21.8184-21.816\ \Rightarrow x=0.0024$

Hence, $x=0.0024$

The simplification of $\displaystyle3\overline{36}-2.\overline{05}+1\overline{33}$ is equal to

1. 2.6

2. 2.64

3. $\displaystyle 2.\overline{61}$

4. $\displaystyle 2.\overline{64}$

Correct Option: D

Explanation:

$\displaystyle 3.\overline{36}-2.\overline{05}+1.\overline{33}$

= $\displaystyle 3+0.\overline{36}-\left ( 2+0.\overline{05} \right )+1+0.\overline{33}$

= $\displaystyle 3+\frac{36}{99}-2-\frac{5}{99}+1+\frac{33}{99}$

= $\displaystyle 2+\frac{64}{99}=2+0.\overline{64}=2.\overline{64}$

Match the following.

1. (i)$\rightarrow$ (Q), (ii) $\rightarrow$ (R), (iii)$\rightarrow$ (S), (iv) $\rightarrow$ (P)

2. (i)$\rightarrow$ (R), (ii) $\rightarrow$ (P), (iii) $\rightarrow$(S), (iv) $\rightarrow$(Q)

3. (i)$\rightarrow$ (R), (ii)$\rightarrow$(S), (iii) $\rightarrow$ (P), (iv) $\rightarrow$ (Q)

4. (i)$\rightarrow$ (Q), (ii) $\rightarrow$(S), (iii) $\rightarrow$ (P), (iv)$\rightarrow$(R)

Correct Option: B

Explanation:

(I) $715+12.59+685.35=1412.94$
(II) $518-(216.80-115.96)=518-100.84=417.16$
(III) $4.090+0.050+6.500=10.640$
(IV) $36.050+198.05-21.03=213.07$.