Ravikant Singh:
6480/3== 2160
6370/2160==637/216
216==6^3
X= 7/6
Rate=1/6== 16.66%
SI===8640×1/6×3== 4320✓✓
EXPLANATION=========👇👇👇👇👇👇👇👇👇👇👇👇👇
1 + r/100 = k
p*r/100 = 6480/3
p*(k - 1) = 2160

p*k^4-pk^3+pk^3-pk^2===
p*k⁴ - p*k² = 6370
pk² * ( k² - 1) = 6370
p*k² *(k + 1) * (k - 1) = 6370
k²*(k + 1) * 2160 = 6370
k = 7/6
rate = 16.66%

8640*1/6*3 = 4320✓✓

. A B C
Sp 4x 5x 6x
P% 5p 11p 6p
Cp Z Z 150

4x/100+5p = 5x/100+11p
P = 100/19%

6y*100/100+600/19%*100 = 150
Y = 625/19

Total sp = 15*625/19

5*625/19 / 100+1100/19 *100
Cp= 625/6 (A)
15*625/19 - 2*625/6+150
= 135 rs Approx✔️

(100+5x)==4a
(100+11x)==5a
Divide..

X==100/19

500/19 : 1100/19 : 600/19

Cp ratio: 76/24 : 76/24 : 114/25
SP ratio: 4 : 5 : 6

Profit==616/150

114/25==150
616/150==135 approx

In short plzzzzz✍
Ans. 500

> (m+6w)10=2mp=8w(p+5)=k

> w = k/8 * 1/(p+5)

> m = k/2 * 1/p

> (m+6w)10 = k

> 1/2p + 3/4 * 1/(p+5) = 1/10


> 2p² - 15p - 50 = 0

> 2p²-20p+5p -50 = 0

> 2p(p-10)+5(p-10) = 0

> p => 10 dys


> m+6w = 2m
> m = 6w

> 5/1 => 500%

2 pipes X and Y tegether can fill a tank in 80/9 minutes. If the efficiency of pipe X is increased such that it canX now fill the tank (in 2 minutes less than before. Also, the efficiency of pipe Y is decreased such that it now takes 1 mjhute more than before to fill the tank, then with their changed efficiencies they together can fill it m 42/5 minutes. How much time X alone will take to fill the same tank?

Let tw=80
Efficiency x=a,y=9-a
Time after change=
X=80/a -2,y=80/9-a +1
Efficiency=
X=80a/80-2a
Y=80(9-x)/89-a
ATQ====
80
--------------------------------------=42/5
80*(a/80-2a)+(9-a)/89-a
Solving
a^2-129a+3560
------------------------=21/5
a^2-9a+720
Solving
2a^2+57a-335=0
a=5✓✓
80/5====16 hours✓✓

2 pipes X and Y tegether can fill a tank in 80/9 minutes. If the efficiency of pipe X is increased such that it canX now fill the tank (in 2 minutes less than before. Also, the efficiency of pipe Y is decreased such that it now takes 1 mjhute more than before to fill the tank, then with their changed efficiencies they together can fill it m 42/5 minutes. How much time X alone will take to fill the same tank?

Let tw=80
Efficiency x=a,y=9-a
Time after change=
X=80/a -2,y=80/9-a +1
Efficiency=
X=80a/80-2a
Y=80(9-x)/89-a
ATQ====
80
--------------------------------------=42/5
80*(a/80-2a)+(9-a)/89-a
Solving
a^2-129a+3560
------------------------=21/5
a^2-9a+720
Solving
2a^2+57a-335=0
a=5✓✓
80/5====16 hours✓✓

. P Q R
Mp% 25% 50% 40%
P% a% b% c%

For (a)
4x^2-43x-255 = 0
4x^2-60x-17x-255 = 0
4x(x-15) +17(x-15)
x=15%
a=15%
For (b)
3y^2-46y-144 = 0
3y^2-54y+8y-144=0
y(y-18) +8(y-18)
y=18%
b=18%
For (c)
5z^2-64z-84 = 0
5z(z-14)+6(z-14)
z=14%
c=14%
(Q)
114 = 798 sp
100 = 700 cp
140 = 980 cp
(R)
50-18 = 272
100 = 850 cp
118 = 1003 sp
150 = 1275
(P)
115 =1275-102 sp
100 = 1020 cp
150 = 1530 mp
98+153+153=404rs✔️

80%=14 days
100%=35/2 days

24. 56. 60. 35/2
840
35. 15. 14. 48
Sita eff=35
Lokesh=15
Minu=14
Jitu=48*7/12=28
Kishore=48*5/12=20
Ritu=35*60%=21
840*20%=168
840*30%=252
168/56=3 days
252/42=6 days
A=6-3=3 days
B=840/20=42 days

35*6=210
210/840 *100=25%=c

35*12=420
420/840=1/2 part=d

840/63=40/3 days=e

Sita ,lokesh, minu
24,56,60
Lcm=840..tw
Sita=35
Lokesh=15
Minu=14
Ritu=0.6*35=21
Jitu=1.4kishor

Minu and ritu total eff=35

35*6/840=1/4~25%=C✓✓
2.4kishor*14=84*8
Kishor=20
Jitu=28


Kishor ritu lokes time 20% =84*2/56=3 days
Minu and jitu 30% work=84*3/42=6
6-3=3=A✓✓

Kishore alon=840/20=42 days=B✓✓

Lokesh and kishor in 12 days=35*12
35*12/840=1/2=50%=D✓✓

Ritu minu jeetu total work=840/63=40/3= E

Let total imported=100a
Xiomi===
Male=14a
Female=21a
Oppo===
M=5.25a+45
F=21a+180
Vivo===
M=8.4a+72
F=228-8.4a
Realme===
M=100
F=570-21a
ATQ====
5.25a+1195=65a
a=20✓
Total=2000✓
Vivo=72+228=300✓
300/20=15%✓✓