Tag: maths

$ Given\quad that\quad x\quad and\quad y\quad are\quad positive\quad number\quad and\quad x>y.\ We\quad may\quad calculate\ (a)\quad on\quad the\quad positive\quad side\quad of\quad number\quad line\quad x\quad will\quad be\ \quad \quad to\quad the\quad right\quad of\quad y\quad and\quad \ (b)\quad on\quad the\quad negative\quad side\quad of\quad the\quad number\quad line\quad x\quad will\ \quad \quad be\quad to\quad the\quad left\quad of\quad y.\ Statement\quad A\longrightarrow -x>-y\quad i.e\quad -x\quad to\quad the\quad right\quad side\quad of\quad -y\ which\quad is\quad not\quad complying\quad with\quad (b).\quad So\quad it\quad is\quad not\quad true.\ Statement\quad B\longrightarrow -x<-y\quad i.e\quad -x\quad lies\quad to\quad the\quad left\quad of\quad -y.\ This\quad complies\quad with\quad (b).\quad So\quad the\quad statement\quad is\quad true.\ Statement\quad C\longrightarrow \frac { 1 }{ x } >\frac { 1 }{ y } .\quad Here\quad the\quad numerators\quad of\quad the\ two\quad fractions\quad are\quad equal.\quad But\quad the\quad denominator\quad of\quad \frac { 1 }{ x } \ is\quad greater\quad than\quad that\quad of\quad \frac { 1 }{ y } .\ So\quad this\quad statement\quad is\quad not\quad true.\ Statement\quad D\longrightarrow -\frac { 1 }{ x } >\frac { 1 }{ y } ,\ -\frac { 1 }{ x } is\quad a\quad negative\quad number\quad so\quad it\quad will\quad always\quad be\quad smaller\ than\quad a\quad positive\quad number.\ \therefore \quad -\frac { 1 }{ x } \ngtr \frac { 1 }{ y } .\quad So\quad this\quad statement\quad is\quad not\quad true.\ Answer-\quad Statement\quad B\quad is\quad correct.\  $

Every integer can be expressed as a rational number by dividing itself by 1.
Now, it becomes a rational number with the numerator the number itself and denominator 1.

$\begin{array}{l} x=0.566\ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _  \ 10x=5.666\ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ (i) \ 100x=56.6666\ _ \ _ \ _ \ _ \ _ \ _ (ii) \ (ii)-(i) \ 90x=51 \ x=\frac { { 51 } }{ { 90 } }  \end{array}$

Here,$ 2^{111}$ = $2^{3\times27}$
which is divisible by $2^{3} = 8$
Hence, the digit in the unit place is 0

chart of place value
thous hund  tens  ones
1        0         0      0

Hundred is $\dfrac{1}{10}$ less than thousand
Similarly, tens is $\dfrac{1}{10}$ less than hundred
So, the required answer is $\dfrac{1}{10}$

The place value of 6 here is 600.

1 lakh $ = 100 $ thousand

Since $ 1,00,000 = 100 \times 1,000 $

Commas are inserted in a number after each period.
Option C is correct.

Numeral for sixteen crore four lakh fifty thousand three hundred ten is 16,04,50,310
Option B is correct.