Given Expression =$\dfrac{(a^2 - b^2)}{(a+b)(a-b)}$=$\dfrac{(a^2 - b^2)}{(a^2 - b^2)}$=1
Given Expression =$ \dfrac{8 - 2.8}{1.3} $=$ \dfrac{5.2}{1.3} $=$ \dfrac{52}{13} $= 4.
Part filled in 4 minutes = 4$ \left(\dfrac{1}{15} +\dfrac{1}{20} \right) $=$ \dfrac{7}{15} $.
Remaining part =$ \left(1 -\dfrac{7}{15} \right) $=$ \dfrac{8}{15} $.
Part filled by B in 1 minute =$ \dfrac{1}{20} $
$\therefore \dfrac{1}{20} $:$ \dfrac{8}{15} $:: 1 : $ x $
$ x $ =$ \left(\dfrac{8}{15} \times 1 \times 20\right) $= 10$ \dfrac{2}{3} $min = 10 min. 40 sec.
$\therefore$ The tank will be full in $\left(4 min. + 10 min. + 40 sec.\right)$ = 14 min. 40 sec.
Let B be turned off after $ x $ minutes. Then,
Part filled by $\left(A + B\right)$ in $ x $ min. + Part filled by A in $\left(30 - x \right)$ min. = 1.
$\therefore x \left(\dfrac{2}{75} +\dfrac{1}{45} \right) $+ $\left(30 - x \right)$.$ \dfrac{2}{75} $= 1
$\Rightarrow \dfrac{11x}{225} $+$ \dfrac{(60 -2x)}{75} $= 1
$\Rightarrow$ 11$ x $ + 180 - 6$ x $ = 225.
$\Rightarrow x $ = 9.
Earlier dimensions of the wall = 100 × 3 × 0.30.
New dimensions = L × 1.5 × 0.3.
∴ As men, women and children are given to be equally efficient, so in the first case, the total number of persons is 100 (i.e. 30 + 20 + 50)
and the same in the second case is 75 (15 + 25 + 35).
Length of wall = L = (75x100) x (2x9) x (15x20) x (100 x 3 x 3 x 0.3)/(1.5 x 0.3) ⇒ L = 25 m.
The word 'RUMOUR' has 6 letters.
In these 6 letters, 'R' occurs 2 times, 'U' occurs 2 times and rest of the letters are different.
Hence, number of ways to arrange these letters
=$\dfrac{6!}{\left(2!\right)\left(2!\right)}$
=$\dfrac{6\times5\times4\times3\times2\times1}{\left(2\times1\right)\left(2\times1\right)}$
=180
Given that any number of flags can be hoisted at a time. Hence we need to find out number of signals that can be made using 1 flag, 2 flags, 3 flags, 4 flags, 5 flags and 6 flags and then add all these.
Number of signals that can be made using 1 flag
= $^6P_{1}$ =6
Number of signals that can be made using 2 flags
=$ ^6P_{2} $
=6×5=30
Number of signals that can be made using 3 flags
=$ ^6P_{3}$
=6×5×4=120
Number of signals that can be made using 4 flags
=$ ^6P_{4} $
=6×5×4×3=360
Number of signals that can be made using 5 flags
= $^6P_{5}$
=6×5×4×3×2=720
Number of signals that can be made using 6 flags
= $^6P_{6 }$
=6×5×4×3×2×1=720
Therefore, required number of signals
=6+30+120+360+720+720=1956
If B receives Rs.9600 as his share of the total profit of Rs.19,100 for the year, how much did B invest in the company?
The total profit for the year is 19100. Of this B gets Rs.9600. Therefore, A would get (19100−9600) =Rs. 9500
The partners split their profits in the ratio of their investments.
Therefore, the ratio of the investments of,
A:B=9500:9600=95:96.
A invested Rs.10000 initially for a period of 4 months. Then, he withdrew Rs.2000.
Hence, his investment has reduced to Rs.8000 (for the next 5 months).
Then he withdraws another Rs.3000. Hence, his investment will stand reduced to Rs.5000 during the last three months.
So, the amount of money that he had invested in the company on a money-month basis will be,
=4×10000+5×8000+3×5000
=40000+40000+15000
=95000
If A had 95000 money months invested in the company, B would have had 96,000 money months invested in the company (as the ratio of their investments is 95:96).
If B had 96,000 money-months invested in the company, he has essentially invested $\dfrac{96000}{12}$=Rs. 8000
Since the ratio of the number of Rs. 1, 50p and 25p coins can be represented by 3 consecutive odd numbers that are prime in ascending order, the only possibility for the ratio is 3:5:7.
Let the number of Re1, 50p and 25p coins be 3k,5k and 7k respectively.
Hence, total value of coins in paise
⇒100×3k+50×5k+25×7k
=725k
=5800
⇒k=8.
If the number of coins of Rs. 1,50p and 25p is reversed, the total value of coins in the Bag (in paise)
=100×7k+50×5k+25×3k=1025k (In above we find the value of k).
⇒8200p=Rs. 82.
$log_{4}x+log_{2}x$=12
=>$\dfrac{log\:x}{log\:4}+\dfrac{log\:x}{log\:2}$=12
=>$\dfrac{log\:x}{log\:2^2}+\dfrac{log\:x}{log\:2}$=12
=>$\dfrac{log\:x}{2\:log\:2}+\dfrac{log\:x}{log\:2}$=12
=>$\dfrac{log\:x+2\:log\:x}{2\:log\:2}$=12
=>$\dfrac{3\:log\:x}{2\:log\:2}$=12
Therefore,
logx=$\dfrac{12\times2\:log\:2}{3}$
=8 log 2
=$log(2^8)$
=log(256)
x=256