Quiz – 1

As 3x – 2 = 2x + 4 ..........(i)

=> x = 6

Now,
\frac{a+2}{3}=6
a=16

\frac{2+3\div 2\times (-1)\times 2+3\div 2\times 3}{2-3\div 2\times (1)-3+2\times 3\div 2-2}

\frac{2+\frac{3}{2}\times (-1)\times 2+\frac{3}{2}\times 3}{2-\frac{3}{2}\times (1)-3+2\times \frac{3}{2}-2}

\frac{2-3+\frac{9}{2}}{2-\frac{3}{2}-3+3-2}=\frac{-1+\frac{9}{2}}{\frac{3}{2}}=\frac{\frac{7}{2}}{\frac{-3}{2}}=\frac{-7}{3}

Let one part be x and the other part be (300 – x). It is given that half of one part is 48 less than the other
\frac{1}{2}x=(300-x)-48

\frac{1}{2}x + x= 252

\frac{3}{2}x = 252
x=168
Hence, one part is 168 and the other part is = 300 – 168 = 132.

Solution:

\frac{1}{x-1}+\frac{1}{x-2}=\frac{3}{x-3}

\frac{x-2+x-1}{x^{2}-2x-x+2}=\frac{3}{x-3}

\frac{2x-3}{x^{2}-3x+2}=\frac{3}{x-3}

3x^{2}-9x+6=2x^{2}-6x-3x+9

x^{2}=3

x=\pm \sqrt{3}

27.33\times \left ( \frac{567.33-483.79}{58.67} \right )=27.33(\frac{128.54}{58.67})

27.33 x 2.19=59.85

Let the number of members in the party be x.

As, per the given condition:

Each member has given twice as many rupees as number of members,

Therefore, x\times 2x = 9248

2x^{2} = 9248

x^{2} = 4624

Taking square root of both the sides,

x = 68.

Solution:

Let the required number be x.
3x-\frac{3x}{5}=60

x=25

Let f(x)=x^{3}-3x^{2}+kx-10
Now, f(x) at x=5, will be zero
So 40+5k=0
k=-8

Solution:

\frac{4}{7}\times\frac{14}{4}\div 2+6\frac{1}{3}=\frac{4}{7}\times \frac{14}{4}\div 2+\frac{19}{3}
\frac{4}{7}\times \frac{7}{4}+\frac{19}{3}=1+\frac{19}{3}=\frac{22}{3}

(x+\frac{1}{x})^{2}=x^{2}+2+\frac{1}{x^{2}}
(x+\frac{1}{x})^{2}=79+2=81
(x+\frac{1}{x})=\pm 9