Decimal and Fraction (Level - 2)
Q1–10 of 50The value of $\frac{(0.013)^3 + 0.000000343}{(0.013)^2 - 0.000091 + 0.000049} $
Solution: Let $x = 0.013$ and $y = 0.007$ (since $0.007^3 = 0.000000343$).
Numerator: $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$
Denominator: $x^2 - xy + y^2$ (since $0.013 \times 0.007 = 0.000091$ and $0.007^2 = 0.000049$)
Simplifying: $\frac{(x + y)(x^2 - xy + y^2)}{x^2 - xy + y^2} = x + y = 0.013 + 0.007 = 0.020$
Correct Answer: (b) 0.020
The value of $\frac{(2.3)^3 - .027}{(2.3)^2 + .69 + .09}$ is
Solution: Let $a = 2.3$ and $b = 0.3$ ($b^3 = 0.027$, $ab = 0.69$, $b^2 = 0.09$).
Formula: $\frac{a^3 - b^3}{a^2 + ab + b^2} = a - b$
Result: $2.3 - 0.3 = 2.0$
Correct Answer: (c) 2
The value of $\frac{(0.06)^2 + (0.47)^2 + (0.079)^2}{(0.006)^2 + (0.047)^2 + (0.0079)^2}$ is
Solution: Notice that each term in the numerator is $10$ times the corresponding term in the denominator.
Let $N = (10 \cdot 0.006)^2 + (10 \cdot 0.047)^2 + (10 \cdot 0.0079)^2$
$N = 10^2 \left[(0.006)^2 + (0.047)^2 + (0.0079)^2\right] = 100 \times \text{Denominator}$
Fraction $= 100$
Correct Answer: (c) 100
$\frac{(4.53 - 3.07)^2}{(3.07 - 2.15)(2.15 - 4.53)} + \frac{(3.07 - 2.15)^2}{(2.15 - 4.53)(4.53 - 3.07)} + \frac{(2.15 - 4.53)^2}{(4.53 - 3.07)(3.07 - 2.15)}$ is simplified to
Solution: Let $x = a-b$, $y = b-c$, $z = c-a$. Note that $x + y + z = 0$.
Expression: $\frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy} = \frac{x^3 + y^3 + z^3}{xyz}$
Since $x + y + z = 0$, $x^3 + y^3 + z^3 = 3xyz$.
Result: $\frac{3xyz}{xyz} = 3$
Correct Answer: (d) 3
If $a = \frac{x}{x + y}$ and $b = \frac{y}{x - y}$, then $\frac{ab}{a + b}$ is equal to
Solution:
$ab = \frac{xy}{(x+y)(x-y)} = \frac{xy}{x^2 - y^2}$
$a + b = \frac{x(x-y) + y(x+y)}{(x+y)(x-y)} = \frac{x^2 - xy + xy + y^2}{x^2 - y^2} = \frac{x^2 + y^2}{x^2 - y^2}$
$\frac{ab}{a+b} = \frac{xy}{x^2 + y^2}$
Correct Answer: (a) $\frac{xy}{x^2 + y^2}$
If $\frac{a}{b} = \frac{1}{3}$, $\frac{b}{c} = 2$, $\frac{c}{d} = \frac{1}{2}$, $\frac{d}{e} = 3$ and $\frac{e}{f} = \frac{1}{4}$, then what is the value of $\frac{abc}{def}$?
$\frac{abc}{def} = \left(\frac{a}{d}\right) \left(\frac{b}{e}\right) \left(\frac{c}{f}\right)$
$\frac{a}{d} = \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} = \frac{1}{3} \cdot 2 \cdot \frac{1}{2} = \frac{1}{3}$
$\frac{b}{e} = \frac{b}{c} \cdot \frac{c}{d} \cdot \frac{d}{e} = 2 \cdot \frac{1}{2} \cdot 3 = 3$
$\frac{c}{f} = \frac{c}{d} \cdot \frac{d}{e} \cdot \frac{e}{f} = \frac{1}{2} \cdot 3 \cdot \frac{1}{4} = \frac{3}{8}$
$\frac{abc}{def} = \frac{1}{3} \times 3 \times \frac{3}{8} = \frac{3}{8}$
Correct Answer: (c) $3/8$
If $\frac{m}{n} = \frac{4}{3}$ and $\frac{r}{t} = \frac{9}{14}$, the value of $\frac{3mr - nt}{4nt - 7mr}$ is
If $\frac{m}{n} = \frac{4}{3}$ and $\frac{r}{t} = \frac{9}{14}$, value of $\frac{3mr - nt}{4nt - 7mr}$
Solution: Divide numerator and denominator by $nt$:
Expression $= \frac{3\left(\frac{m}{n}\right)\left(\frac{r}{t}\right) - 1}{4 - 7\left(\frac{m}{n}\right)\left(\frac{r}{t}\right)}$
$\left(\frac{m}{n}\right)\left(\frac{r}{t}\right) = \frac{4}{3} \times \frac{9}{14} = \frac{6}{7}$
Numerator: $3\left(\frac{6}{7}\right) - 1 = \frac{18}{7} - 1 = \frac{11}{7}$
Denominator: $4 - 7\left(\frac{6}{7}\right) = 4 - 6 = -2$
Result $= \frac{11/7}{-2} = -\frac{11}{14}$
Correct Answer: (b) $-11/14$
If $x = \frac{a}{a - 1}$ and $y = \frac{1}{a - 1}$, then
If $x = \frac{a}{a-1}$ and $y = \frac{1}{a-1}$
Solution: $x - y = \frac{a - 1}{a - 1} = 1 \implies x = y + 1 \implies x > y$ for all values of $a \neq 1$.
Correct Answer: (c) x is greater than y
If $0 < a < 1$, then the value of $a + \frac{1}{a}$ is
If $0 < a < 1$, then the value of $a + \frac{1}{a}$ is:
Solution: For any real positive $a \neq 1$, $a + \frac{1}{a} > 2$.
Correct Answer: (b) greater than 2
If $a + b + c = 0$, find the value of $\frac{a^2}{(a^2 - bc)} + \frac{b^2}{(b^2 - ca)} + \frac{c^2}{(c^2 - ab)}$.
If $a + b + c = 0$, find $\frac{a^2}{a^2 - bc} + \frac{b^2}{b^2 - ca} + \frac{c^2}{c^2 - ab}$
Solution: Since $a + b + c = 0 \implies b + c = -a \implies (b + c)^2 = a^2 \implies b^2 + c^2 + 2bc = a^2 \implies a^2 - bc = b^2 + bc + c^2$.
Alternatively, $a = -b-c \implies a^2 = (-b-c)^2 = b^2 + 2bc + c^2$.
Thus $a^2 - bc = b^2 + bc + c^2$. Similarly, $b^2 - ca = a^2 + ab + c^2$ and $c^2 - ab = a^2 + ac + b^2$.
Using standard substitution (e.g., $a=1, b=1, c=-2$):
Term 1: $\frac{1}{1 - (-2)} = \frac{1}{3}$
Term 2: $\frac{1}{1 - (-2)} = \frac{1}{3}$
Term 3: $\frac{4}{4 - 1} = \frac{4}{3}$
Sum: $\frac{1}{3} + \frac{1}{3} + \frac{4}{3} = 2$
Correct Answer: (c) 2