44365.An urn contains 5 red and 3 blue balls.In how many different ways,2 red and 1 blue balls can be drawn?
20
10
30
5
Explanation:
The urn contains 5 red and we want 2 red balls.So ways of selecting red balls=$^5C_{2}$= $\dfrac{5\times4}{2\times1}$=10
Similarly the ways of selecting 1 blue ball from 3 blue balls=$^3C_{1}$=$\dfrac{3}{1}$=3
So totoal ways to select 2 red and 1 blue ball will be=$10\times3$=30
44387.2.008 = ?
$\dfrac{251}{125}$
$\dfrac{125}{251}$
Explanation:
2.008 = $\dfrac{2008}{1000}$ = $\dfrac{251}{125}$
2.008 = $\dfrac{2008}{1000}$ = $\dfrac{251}{125}$
44390.To find the product (.68 x .79)
0.5372
0.005372
5.372
53.72
Explanation:
Now, 68 x 79 = 5372.
Sum of decimal places = (2 + 2 ) = 4.
$\therefore$ .68 x .79 = 0.5372
Now, 68 x 79 = 5372.
Sum of decimal places = (2 + 2 ) = 4.
$\therefore$ .68 x .79 = 0.5372
44391.0.63 ÷ 9 = ?
0.07
0.007
Explanation:
63 / 9 = 7
Decimal places in dividend = 2
∴ 0.63 / 9 = 0.07
63 / 9 = 7
Decimal places in dividend = 2
∴ 0.63 / 9 = 0.07
44392.$\dfrac{0.00042}{0.06}$ = ?
0.007
0.7
Explanation:
$\dfrac{ 0.00042}{0.06}$ = $\dfrac{(0.00042 x 100 )}{ (0.06 x 100)}$
= $\dfrac{0.042 }{6}$
= 0.007
$\dfrac{ 0.00042}{0.06}$ = $\dfrac{(0.00042 x 100 )}{ (0.06 x 100)}$
= $\dfrac{0.042 }{6}$
= 0.007
44393.Arrange the following fractions $\dfrac {1}{2}$, $\dfrac {3}{4}$, $\dfrac {7}{8}$, $\dfrac {5}{12}$ in descending order
$\dfrac {7}{8}$ > $\dfrac {1}{2}$ > $\dfrac {5}{12}$ > $\dfrac {3}{4}$
$\dfrac {7}{8}$ > $\dfrac {3}{4}$ > $\dfrac {1}{2}$ > $\dfrac {5}{12}$
Explanation:
Now, $\dfrac {1}{2}$ = 0.5, $\dfrac {3}{4}$ = 0.75, $\dfrac {7}{8}$ = 0.875 , $\dfrac {5}{12}$ = 0.416...
Since, 0.875 > 0.75 > 0.5 > 0.416... . So $\dfrac {7}{8}$ > $\dfrac {3}{4}$ > $\dfrac {1}{2}$ > $\dfrac {5}{12}$
44394.$\dfrac{22}{7}$ = ?
$3.\overline{142857}$
0$.\overline{3}$
Explanation:
$\dfrac{22}{7}$ = 3.142857142857.... = $3.\overline{142857}$
$\dfrac{22}{7}$ = 3.142857142857.... = $3.\overline{142857}$
44395.0.$\overline{53}$ = ?
$\dfrac{3}{99}$
$\dfrac{53}{99}$
Explanation:
0.$\overline{53}$ = $\dfrac{53}{99}$
0.$\overline{53}$ = $\dfrac{53}{99}$
44396.0.22$\overline{73}$ = ?
$\dfrac{2251}{9900}$
$\dfrac{2273}{9900}$
Explanation:
0.22$\overline{73}$ = $\dfrac{2273 - 22}{9900}$ = $\dfrac{2251}{9900}$
0.22$\overline{73}$ = $\dfrac{2273 - 22}{9900}$ = $\dfrac{2251}{9900}$
44449.simplify $\sqrt{125x^{3}}$
5x$\sqrt{5x}$
25x$\sqrt{5x}$
5x$\sqrt{25x}$
15x$\sqrt{5x}$
Explanation:
$\sqrt{125x^{3}}=\left(\sqrt{(25x^{2})} \times \sqrt{(5x)}\right)$ [using the rule $\sqrt{(a \times b)}=\sqrt{a} \times \sqrt{b}$]
=5x$\sqrt{5x}$
$\sqrt{125x^{3}}=\left(\sqrt{(25x^{2})} \times \sqrt{(5x)}\right)$ [using the rule $\sqrt{(a \times b)}=\sqrt{a} \times \sqrt{b}$]
=5x$\sqrt{5x}$
44450.Write $\sqrt{\dfrac{32}{144}}$ in the simplified form $a\sqrt{n}$
$\dfrac{\sqrt{32}}{3}$
$\dfrac{\sqrt{2}}{13}$
$\dfrac{\sqrt{2}}{3}$
$\dfrac{\sqrt{3}}{2}$
Explanation:
$\sqrt{\dfrac{32}{144}}=\dfrac{\sqrt{32}}{\sqrt{144}}$[using the rule $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$]
=$\dfrac{(\sqrt{16}\times \sqrt{2})}{12}$ [using the rule $\sqrt{(a \times b)}=\sqrt{a} \times \sqrt{b}$]
=$\dfrac{4 \sqrt{2}}{12}$ [simplify $\dfrac{4}{12}$]
=$\dfrac{\sqrt{2}}{3}$
$\sqrt{\dfrac{32}{144}}=\dfrac{\sqrt{32}}{\sqrt{144}}$[using the rule $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$]
=$\dfrac{(\sqrt{16}\times \sqrt{2})}{12}$ [using the rule $\sqrt{(a \times b)}=\sqrt{a} \times \sqrt{b}$]
=$\dfrac{4 \sqrt{2}}{12}$ [simplify $\dfrac{4}{12}$]
=$\dfrac{\sqrt{2}}{3}$
44451.simplify $\dfrac{10\sqrt{3}}{\sqrt{5}}$
4$\sqrt{15}$
2$\sqrt{15}$
2$\sqrt{5}$
4$\sqrt{5}$
Explanation:
$\dfrac{10\sqrt{3}}{\sqrt{5}}=\dfrac{10\sqrt{3}}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}}$ [using the rule $\dfrac{b}{\sqrt{a}}=\dfrac{b}{\sqrt{a}} \times \dfrac{\sqrt{a}}{\sqrt{a}}=\dfrac{b \sqrt{a}}{a}$]
=$\dfrac{10(\sqrt{15})}{5}$ [using the rule $\sqrt{a} \times \sqrt{b}=\sqrt{(a \times b)}$]
=2$\sqrt{15}$ [since $\dfrac{10}{5}=2$]
$\dfrac{10\sqrt{3}}{\sqrt{5}}=\dfrac{10\sqrt{3}}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}}$ [using the rule $\dfrac{b}{\sqrt{a}}=\dfrac{b}{\sqrt{a}} \times \dfrac{\sqrt{a}}{\sqrt{a}}=\dfrac{b \sqrt{a}}{a}$]
=$\dfrac{10(\sqrt{15})}{5}$ [using the rule $\sqrt{a} \times \sqrt{b}=\sqrt{(a \times b)}$]
=2$\sqrt{15}$ [since $\dfrac{10}{5}=2$]
44452.Simplify $\dfrac{2\sqrt{3}}{5}+\sqrt{108}$
$\dfrac{3 \sqrt{3}}{5}$
$\dfrac{32 \sqrt{3}}{5}$
$\dfrac{32 \sqrt{2}}{5}$
$\dfrac{32 \sqrt{3}}{15}$
Explanation:
$\dfrac{2\sqrt{3}}{5}+\sqrt{108}=\dfrac{2\sqrt{3}+5\sqrt{108}}{5}$ [find the LCD to add]
=$\dfrac{2\sqrt{3}+5(\sqrt{36}\times \sqrt{3})}{5}$ [using the rule $\sqrt{(a \times b)}=\sqrt{a} \times \sqrt{b}$]
=$\dfrac{2\sqrt{3}+30 \sqrt{3}}{5}$ [Evaluate \sqrt{36} and multiply to 5]
=$\dfrac{(2+30) \sqrt{3}}{5}$ [using the rule $a \sqrt{c}\pm b\sqrt{c}=(a \pm b) \sqrt{c}$]
=$\dfrac{32 \sqrt{3}}{5}$
$\dfrac{2\sqrt{3}}{5}+\sqrt{108}=\dfrac{2\sqrt{3}+5\sqrt{108}}{5}$ [find the LCD to add]
=$\dfrac{2\sqrt{3}+5(\sqrt{36}\times \sqrt{3})}{5}$ [using the rule $\sqrt{(a \times b)}=\sqrt{a} \times \sqrt{b}$]
=$\dfrac{2\sqrt{3}+30 \sqrt{3}}{5}$ [Evaluate \sqrt{36} and multiply to 5]
=$\dfrac{(2+30) \sqrt{3}}{5}$ [using the rule $a \sqrt{c}\pm b\sqrt{c}=(a \pm b) \sqrt{c}$]
=$\dfrac{32 \sqrt{3}}{5}$
44453.Rationalise the denominaor in $\dfrac{7}{\sqrt{3}+2}$
$-7\sqrt{3}+14$
$7\sqrt{3}+14$
$-7\sqrt{3}-14$
$7\sqrt{3}-14$
Explanation:
$\dfrac{7}{\sqrt{3}+2}=\dfrac{7}{\sqrt{3}+2} \times \dfrac{\sqrt{3}-2}{\sqrt{3}-2}$ [multiply the numerator and denominator by $\sqrt{3}-2$]
=$\dfrac{7\sqrt{3}-14}{3-4}$
=$\dfrac{7\sqrt{3}-14}{-1}$
=$-7\sqrt{3}+14$
$\dfrac{7}{\sqrt{3}+2}=\dfrac{7}{\sqrt{3}+2} \times \dfrac{\sqrt{3}-2}{\sqrt{3}-2}$ [multiply the numerator and denominator by $\sqrt{3}-2$]
=$\dfrac{7\sqrt{3}-14}{3-4}$
=$\dfrac{7\sqrt{3}-14}{-1}$
=$-7\sqrt{3}+14$
44454.Rationalise the denominator in $\dfrac{2}{1-\sqrt{2}}$
$2+2\sqrt{2}$
$2-2\sqrt{2}$
$-2+2\sqrt{2}$
$-2-2\sqrt{2}$
Explanation:
$\dfrac{2}{1-\sqrt{2}} = \dfrac{2}{1-\sqrt{2}} \times \dfrac{1+\sqrt{2}}{1+\sqrt{2}}$ [multiply the numerator and denominator by $1+\sqrt{2}$]
=$\dfrac{2+2\sqrt{2}}{1-2}$
=$\dfrac{2+2\sqrt{2}}{-1}$
=$-2-2\sqrt{2}$
$\dfrac{2}{1-\sqrt{2}} = \dfrac{2}{1-\sqrt{2}} \times \dfrac{1+\sqrt{2}}{1+\sqrt{2}}$ [multiply the numerator and denominator by $1+\sqrt{2}$]
=$\dfrac{2+2\sqrt{2}}{1-2}$
=$\dfrac{2+2\sqrt{2}}{-1}$
=$-2-2\sqrt{2}$
44455.Simplify 3m8n3 $\div$ (3m8n3)0
3m8n3
0
1
None of these
Explanation:
3m8n3 $\div$ (3m8n3)0=3m8n3 $\div$ 1 [using the rule a0=1]
=3m8n3
3m8n3 $\div$ (3m8n3)0=3m8n3 $\div$ 1 [using the rule a0=1]
=3m8n3
44456.Simplify : (3a)-2
$\dfrac{1}{9a^{-2}}$
$\dfrac{1}{3a^{2}}$
$\dfrac{1}{9a^{2}}$
$\dfrac{-1}{9a^{2}}$
Explanation:
(3a)-2=$\dfrac{1}{(3a)^{2}}$ [Using Rule a-m =$\dfrac{1}{a^{m}}$]
=$\dfrac{1}{3^{2}a^{2}}$ [Using Rule (am)n=amn]
=$\dfrac{1}{9a^{2}}$ [Evaluate 32]
(3a)-2=$\dfrac{1}{(3a)^{2}}$ [Using Rule a-m =$\dfrac{1}{a^{m}}$]
=$\dfrac{1}{3^{2}a^{2}}$ [Using Rule (am)n=amn]
=$\dfrac{1}{9a^{2}}$ [Evaluate 32]