Practice Model question set 1 - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) What is the solution of the equation x $log_{10} ({10}/3) + log_{10} 3 = log_{10} (2 + 3^x)$ + x ?
(a)
(b)
(c)
(d)
$x log_{10}({10}/3) + log_{10} 3 = log_{10} (2 + 3^x) + x$
$xlog_{10} 10 - x log_{10} 3 + log_{10} 3 = log_{10} (2 + 3^x)$ + x
x - $log_{10^3^x} + log_{10^3} = log_{10} (2 + 3^x)$ + x
$log_{10[3/{3^x}]} log_{10} (2 + 3^x)$
⇒$3^{1–x} = 2 + 3^x$
⇒$3^{1–x} –3^x = 3^1 – 3^0$
x = 0
Q-2) What is the value of $2 log(5/8) + log({128}/{125}) + log (5/2)?$
(a)
(b)
(c)
(d)
$2 log (5/8) + log ({128}/{125}) + log (5/2)$
= $log(5/8)^2 + log({128}/{125}) + log(5/2)$
= $log {5^2 × 128 × 5}/{8^2 × 125 × 2} = log {5^2 × 2^7 × 5}/{(2^3)^2 × 5^3 × 2}$
= $log {2^7 × 5^3}/{2^6 × 5^3 × 2} = log {2^7 × 5^3}/{2^7 × 5^3}$ = log 1 = 0
Q-3) What is the value of $({1/3} \text"log"_{10} 125 -2\text"log"_{10} 4 + \text"log"_{10} 32 + \text"log"_{10} 1)$
(a)
(b)
(c)
(d)
(e)
${1/3} \text"log"_{10} 125 – 2{\text"log"_{10}} 4 + {\text"log"_{10}} 32 + {\text"log"_{10}}$ 1
=${1/3} \text"log"_{10} (5)^3 – 2{\text"log"_{10}}(2)^2 + {\text"log"_{10}}(2)^5$ + 0
=$\text"log"_{10} 5 – 4{\text"log"_{10}} 2 + 5{\text"log"_{10}2}$
= $\text"log"_{10}5 +\text"log"_{10}2 = \text"log"_{10}10 = 1$
Q-4) If ${\text"log"_7}{\text"log"_5} (√x + 5 + √x)$ = 0 , find the value of x.
(a)
(b)
(c)
(d)
(e)
$\text"log"_7 \text"log"_5(√x + 5 + √x)$ = 0
use $\text"log"_a$ x = b⇒$a^b$=x
∴ $\text"log"_5 (√x + 5 + √x) = 7^0$ = 1
$√x + 5 + √ x = 5^1 = 5⇒2√x$ = 0
∴ x = 0
Q-5) What is the value of ${[log_{13}(10)]}/{[log_{169}(10)]}$
(a)
(b)
(c)
(d)
${log_{13}(10)}/{log_{169}(10)} = {log_{13}(10)}/{log_{13^2}(10)}$
$(∵ log_{a^b} c = 1/b log_a c)$
= ${log_{13}10}/{1/2 log_{13}10} = 1/{1/2}$ = 2
Q-6) If ${\text"log"_a} b = 1/2, {\text"log"_b}$ c = $1/3$ and ${\text"log"_c}$ a = $k/5$, then the value of k is
(a)
(b)
(c)
(d)
(e)
${\text"log"_a} b = {1/2}, {\text"log"_b}c= {1/3}$, and ${\text"log"_c} a = k/5$
⇒${\text"log" b}/{\text"log" a} = {1/2}, {\text"log" c}/{\text"log" b} = {1/3}, {\text"log" a}/{\text"log" c} = k/5$
⇒${1/2} × {1/3} × {k/5}$ = 1⇒k = 30
Q-7) If ${\text"log"_8} x + {\text"log"_8} {1/6} = 1/3$, then find the value of x.
(a)
(b)
(c)
(d)
(e)
${\text"log"_8}x +{ \text"log"_8} {1/6} = 1/3$
or ${\text"log"_8} (x × {1/6}) = 1/3 or {\text"log"_8} (x/6) = 1/3$
or $x/6 = (8)^{1/3}$ {∵ $\text"log"_a$ b = x⇔(a)x=b}
or ${x/6} = (2^3)^{1/3}$ or x = 12
Q-8) It is given that $log_{10} 2$ = 0.301 and $log_{10} 3 = 0.477.$ How many digits are there in $(108)^{10}$ ?
(a)
(b)
(c)
(d)
$log (108)^{10} = 10 log 108 = 10 log (2^2 × 3^3)$ = 10 (2log2 + 3log3)
= 10 (2 × 0.301 + 3 × 0.477 ) = 10 (.602 + 1.431)
= 10 × 2.033 = 20.33
integral part = 20
No. of digits = 20 + 1 = 21
Q-9) There are n zeroes appearing immediately after the decimal point in the value of $(0.2)^{25}$. It is given that the value of $log_{10}2$ = 0.30103. The value of n is
(a)
(b)
(c)
(d)
Let y = $(0.2)^{25}$
Talking log on both sides, we got
log(y) = $(0.2)^{25}$ = 25. log(0.2)
= 25 $(log (2/{10}))$ = 25 (log2 – log10)
= 25 (0.330103 – 1) = –17.47475
Hence, n = 17
Q-10) If $\text"log" 2$ = 0.30103, then find the number of digits in $2^{56}$ .
(a)
(b)
(c)
(d)
(e)
Let x = $2^{56}$
⇒$\text"log" x = 56 \text"log" 2$ = 56 × 0.30103 = 16.85
∴ Number of digits in $2^{56}$ = 17