Let the capacity of 1 bucket = $ x $.
Then, the capacity of tank = 25$ x $.
New capacity of bucket =$ \dfrac{2}{5} x $
$\therefore$ Required number of buckets =$ \dfrac{25x}{(2x/5)} $
=$ \left( 25x\times\dfrac{5}{2x} \right) $
=$ \dfrac{125}{2} $
= 62.5
Let the cost of a chair and that of a table be Rs. $ x $ and Rs. $ y $ respectively.
Then, 10$ x $ = 4$ y $ or $ y $ =$ \dfrac{5}{2} x $.
$\therefore$ 15$ x $ + 2$ y $ = 4000
$\Rightarrow$ 15$ x $ + 2$ \times \dfrac{5}{2} x $ = 4000
$\Rightarrow$ 20$ x $ = 4000
$\therefore x $ = 200.
So, $ y $ =$ \left(\dfrac{5}{2} \times 200\right) $= 500.
Hence, the cost of 12 chairs and 3 tables = 12$ x $ + 3$ y $
= Rs. $\left(2400 + 1500\right)$
= Rs. 3900.
Suppose their paths cross after $ x $ minutes.
Then, 11 + 57$ x $ = 51 - 63$ x $ $\Leftrightarrow$ 120$ x $ = 40
x = $ \dfrac{1}{3} $
Number of floors covered by David in $\left(1/3\right)$ min. =$ \left(\dfrac{1}{3} \times 57\right) $= 19.
So, their paths cross at $\left(11 +19\right)$i.e., 30th floor.
Let savings in N.S.C and P.P.F. be Rs. $ x $ and Rs. $\left(150000 - x \right)$ respectively. Then,
$ \dfrac{1}{3} $x =$ \dfrac{1}{2} \left(150000 - x \right)$
$\Rightarrow \dfrac{x}{3} $+$ \dfrac{x}{2} $= 75000
$\Rightarrow \dfrac{5x}{6} $= 75000
$\Rightarrow x $ =$ \dfrac{75000 \times 6}{5} $= 90000
$\therefore$ Savings in Public Provident Fund = Rs. $\left(150000 - 90000\right)$ = Rs. 60000
Let the number of ducks be d
and number of cows be cThen, total number of legs = 2d + 4c = 2$\left(d + 2c\right)$
total number of heads = c + d
Given that total number of legs are 28 more than twice the number of heads
=> 2(d + 2c) = 28 + 2(c + d)
=> d + 2c = 14 + c + d
=> 2c = 14 + c
=> c = 14
i.e., total number of cows = 14
Let the total number of shots be $ x $. Then,
Shots fired by A =$ \dfrac{5}{8} x $
Shots fired by B =$ \dfrac{3}{8} x $
Killing shots by A =$ \dfrac{1}{3} $of$ \dfrac{5}{8} x $=$ \dfrac{5}{24} x $
Shots missed by B =$ \dfrac{1}{2} $of$ \dfrac{3}{8} x $=$ \dfrac{3}{16} x $
$\therefore \dfrac{3x}{16} $= 27 or $ x $ =$ \left(\dfrac{27 \times 16}{3} \right) $= 144.
Birds killed by A =$ \dfrac{5x}{24} $=$ \left(\dfrac{5}{24} \times 144\right) $= 30.
240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (11 × 24 ÷ 8 × 3 – 69) × 2}
= 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (264 ÷ 8 × 3 – 69) × 2}
= 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (33 × 3 – 69) × 2}
= 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (99 – 69) × 2}
= 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ 30 x 2}
= 240 ÷ 8 × 512 ÷ 4 + ½ of {60 x 2}
= 240 ÷ 8 × 512 ÷ 4 + ½ of 120
= 240 ÷ 8 × 512 ÷ 4 + ½ × 120
= 240 ÷ 8 × 512 ÷ 4 + 60
= 30 × 512 ÷ 4 + 60
= 15360 ÷ 4 + 60
= 3840 + 60
= 3900
25 – 1/2{5 + 4 – (3 + 2 – 1 + 3)}
= 25 – 1/2{5 + 4 – 7}
= 25 – 1/2*2
= 25 –1
= 24
27 – [38 – {46 – (15 – 13 – 2)}]
= 27 – [38 – {46 – 0}]
=27 – [38 – 46]
=27 – [–8]
=27 + 8
=35
2550 – [510 – {270 – (90 – 80 + 70)}]
= 2550 – [510 – {270 – (10 + 70)}]
= 2550 – [510 – {270 – 80}]
= 2550 – [510 – 190]
= 2550 – 320
= 2230
4 + (1/5) [{–10 x (25 – 13 – 3)} ÷ (–5)]
= 4 + (1/5) [{–10 x 9} ÷ (–5)]
= 4 + (1/5) [–90 ÷ (–5)]
= 4 + (1/5) [18]
= 4 + 1 ÷ 5 × 18
= 4 + .2 × 18
= 4 + 3.6
= 7.6
22 – (1/4) {–5 – (– 48) ÷ (–16)}
= 22 – (1/4) {–5 – 3}
= 22 – (1/4) × –8
= 22 – 1 × –2
= 22 + 2
= 24
63 – (–3) {–2 – 8 – 3} ÷ 3{5 + (–2) (–1)}
= 63 – (–3) {–2 – 8 – 3} ÷ 3{5 + 2}
= 63 – (–3) {–2 – 8 – 3} ÷ 3 × 7
= 63 – (–3) × (–13) ÷ 3 × 7
= 63 – 39 ÷ 3 × 7
= 63 – 13 × 7
= 63 – 91
= –28
[29 – (–2) {6 – (7 – 3)}] ÷ [3 x {5 + (–3) x (–2)}]
= [29 – (–2) {6 – 4}] ÷ [3 x {5 + 6}]
= [29 – (–2) × 2] ÷ [3 x 11]
= [29 – (–4)] ÷ 33
= [29 + 4] ÷ 33
= 33 ÷ 33
= 1
1/(3+1/(3+1/(3–1/3)))
= 1/(3 + 1/(3 + 1/(8/3)))
= 1/(3 + 1/(3 + 3/8))
= 1/(3 + 8/27)
= 1/(89/27)
= 27/89