## A chair is sold for Rs.825. gain is 10%. find C.P.

S.P. = 825 gain = 10% C.P. = 825 $\times $ $\frac{100}{100 + 10}$ 825 $\times $ $\frac{100}{110}$ = 750 C.P. = 750

» Read moreS.P. = 825 gain = 10% C.P. = 825 $\times $ $\frac{100}{100 + 10}$ 825 $\times $ $\frac{100}{110}$ = 750 C.P. = 750

» Read moreC.P. = 750 loss = 4% S.P. = 750 $\times $ $\frac{100 – 4}{100}$ 750 $\times $ $\frac{96}{100}$ = 360

» Read moreC.P. = 750 gain = 10% S.P. = 750 $\times $ $\frac{100 + 10}{100}$ 750 $\times $ $\frac{110}{100}$ = 825 $\therefore$ S.P. = 825

» Read moreLoss = 450 – 405 = 45 Loss% = $\frac{45}{450}$ $\times $ 100 = 10 $\therefore$ loss% = 10

» Read moreS.P. = 495 C.P. = 450 Gain = 495 – 450 = 45 Gain % = $\frac{45}{450}$ $\times $ 100 = 10 $\therefore$ gain% = 10

» Read moreAshok’s speed = 20 km/hr = $(20\times \frac{5}{18})$m/sec $=$ $(\frac{50}{9})$ m/sec Time taken to cover 400 meters = $(400\times \frac{9}{50})$sec = 72 sec = 1$\frac{1}{5}$ minutes

» Read moreSpeed = $\frac{Distance}{Time}$ = $\frac{180}{6}$ = 30 meters 30 $\times $ $\frac{18}{5}$ = 108 km/hr

» Read moreTime = 12 minutes = $\frac{12}{60}$ hours Speed = 3.5 km/hr Distance = speed $\times $ time 3.5 $\times $ $\frac{12}{60}$ = $\frac{7}{10}$ km Distance in meters = $\frac{7}{10}$ $\times

» Read moreTo convert m/sec into km/hr multiply with $\frac{18}{5}$ 22 $\times $ $\frac{18}{5}$ = $\frac{396}{5}$ = 79.2 km/hr

» Read moreTo convert km/hr into m/sec multiply with $\frac{5}{18}$ 90 $\times $ $\frac{5}{18}$ = 25 m/sec

» Read moreA’ s one day’s work = $\frac{1}{10}$ B’ s one day’s work = $\frac{1}{12}$ Ratio of (A+B)’s one day work = $\frac{1}{10} : \frac{1}{12}$ = 6 : 5 Amount should

» Read more(A+B)’s 5 days work = 5$\lbrack \frac{1}{25}+\frac{1}{20}\rbrack $ = 5$ \lbrack \frac{4+5}{100}\rbrack $ = 5 $\lbrack \frac{9}{100}\rbrack $ = $\frac{9}{20}$ Remaining work = 1 – $\frac{9}{20}$ = $\frac{11}{20}$ B can

» Read more(A+B)’s one day’s work = $\frac{1}{12}$ B’s one day work = $\frac{1}{28}$ $\lbrack (A+B)- (B) = A\rbrack $ A’s one day work = $\frac{1}{12}$ – $\frac{1}{28}$ = $\frac{7-3}{84}$ = $\frac{4}{84}$

» Read moreA’s one day’s work = $\frac{1}{6}$ B’s one day’s work = $\frac{1}{8}$ C’s one day’s work = $\frac{1}{12}$ (A+B+C)’s one day work = $\frac{1}{6}$ + $\frac{1}{8}+\frac{1}{12}$ = $\frac{4+3+2}{24}$ = $\frac{9}{24}$

» Read moreA’s one day’s work = $\frac{1}{20}$ B’s one day’s work = $\frac{1}{30}$ (A + B)’s one day work = $\frac{1}{20}$ + $\frac{1}{30}$ = $\frac{3 + 2}{60}$ = $\frac{5}{60}$ = $\frac{1}{12}$

» Read moreLet P’s age and Q’s age be 6$x$ years and 7$x$ years respectively. Then, 7$x$ – 6$x$ = 4 $x$ = 4 Required ratio = $(6x+4)$ : $(7x+4)$ = 28

» Read moreLet Rahul’s age be $x$ years. Then, Sachin’s age = $(x -7)$ years $\frac{x – 7}{x}$ = $\frac{7}{9}$ 9$x$ – 63 = 7$x$ 2$x$ = 63 $x$ = 31.5 Hence,

» Read moreLet the ages of Suneel and his father 10 years ago be $x$ and 5$x$ years respectively. Suneel’s age after 6 years = $(x+10)$ + 6 = $(x+16)$ years Father’s

» Read moreLet Gaurav’s and Sachin’s ages one year ago be 6$x$ and 7$x$ years respectively. Gaurav’s age 4 years hence = $(6x+1)$ + 4 = $(6x+5)$ years Sachin’s age 4 years

» Read moreLet the son’s present age be $x$ years. Then father’s present age = $\left(3x+3\right)$ years. $\left(3x + 3 + 3\right) = 2\left(x+3\right) + 10$ $3x + 6 = 2x +

» Read moreLet Ankit’s age be $x$ years. Then, Nikita’s age = $\frac{240}{x}$ years. $2 \times \frac{240}{x} – x = 4$ $480 – x^2 = 4x$ $x^2 + 4x – 480 =

» Read moreLet Raghu’s present age be $x$ years. Then, Raghu’s age after 15 years = ($x$ + 15) years. Raghu’s age 5 years back = ($x$ – 5) years. ($x$+ 15)

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