Lets assume the CP of watch = Rs x
and sunglasses = Rs y.
2600=96x/100 + 104y/100
2700= 104x/100 + 96y/100
By solving,
y=700
x=1960.
Lets assume quantity of zinc to added = x kg.
zinc quantity in alloy = 5/8*16=10 Kg.
and copper quantity = 3/8*16=6 kg.
Alloy new ratio of zinc and copper = 3:1
Zinc quantity in alloy => (10+x)/6=3/1 => 10 + x = 18 => x = 8 kg.
Lets assume the speed of slower train = x miles/hrs.
So, speed of faster train is = (x+6) miles/hrs.
Given, passed each other at 1:30 PM i.e after 3 1/2 hrs.
Both train travelling towards each other so total relative speed = x+(x+6) = (2x+6)
So, 287/(2x+6) = 7/2 => 574 = 14x + 42
=> 14x = 542 => x = ~38
So spee of faster train = (x+6) miles/hrs. = 44 miles/hrs.
Lets assume the capacity of the vessel = x litres.
x (1- 10/x)^2 = 32
x^2 +100-20x = 32x
x^2 +100 - 52x= 0
x^2 - 50x - 2x + 100 = 0
=> (x-2) (x-50) = 0
=> x=2 & x=50
As 10 litres is withdrawn so vessel capacity will be 50 litres.
Lets assume Ritu present age = x years
so sheema age = (x-6)/6 years
As per question:
sheema age [(x-6)/6 + 4] = 12
(x-6)/6 = 12 - 4
=> (x-6)/6 = 8
=> x - 6 = 48 => x = 54
So Ritu age = 54 years.
Lets assume the radius of 1st circle = x cm.
so, the radius of 2nd circle = (14-x)cm.
(pi*x*x)+(pi*(14-x)*(14-x))=130
By solving above eq.
x = 11 or 3.
Let the three integers be 2x + 1, 2x + 3 , 2x + 5.
Therefore, 6x + 9 = 117,
x = 18
The three integers are 37, 39, 41.
Time=50/60 hr=5/6hr
Speed=48mph
distance=S*T=48*5/6=40km
time=40/60hr=2/3hr
New speed = 40* 3/2 kmph= 60kmph.
Lets assume distance of race = x mtrs.
Then when A finishes x m , B has run (x- 12)mtrs and C has run (x-18) mtrs.
=> at this point B is 6 m ahead of C. Now to finish race b needs to run another 12 m,
=> he runs another 12 m. when B finishes race he is 8 m ahead of C.
so last 12 m B has run, C has run 10 m.as speeds are constant,
=> x-12/ x-18 = 12/10 => x = 48 mtrs.
Lets assume second grandson = x year.
So, first grandson age = (x+2) year ,
third grandson = (x-2) year
So, x+2+x+x-2=3x
i.e grandfather age is 3 times as older as his second grandson.
Amount of work done by 20 men = 24 women = 40 boys
i.e 5man = 6 woman = 10 boys. -----------------(i)
According to the 1st condition, 5 men can do a job in 12 days working for 8 hours a day.
Required :- how many more men are required to work with 6 women and 2 boys.
Let the required number of men be M.
According to the given information,
=>(5 x 12 x 8 )/1=[(M + 5 + 1) x 5 x 12 ]/@( 6 women= 5men & 2 boys =1 men)
=>32= M + 6.
=>M=26.
Hence, 26 men are working with 6 women and 2 boys.
Total late in three day = 16*3 = 48 min.
Total late in one day =16 min.
so late in 1 min. = 16/(24*60) = 0.01111;
Hence in 5 hour (5:00 AM to 10:00 AM on 4th day) =.0111*5*60 = 3.333;
So, Exact time will be = 10:52 AM.
Lets assume speed of boat in still water = x km/hr
So boat downstream speed = (x+2) km/hr.
boat upstream speed = (x-2) km/hr.
6/(x-2)+6/(x+2)=33/60 => 6/(x-2)+6/(x+2)=11/20
=> 120*[(x-2) + (x+2)] = 11*x^2 -44
=> 240 x = 11*x^2-44
=> 11* x^2 -240x -44 = 0 ---------(1)
By solving eq. (1)
x = 22 km/hr.
First time distance= H
Second time = 80H/100 = 4H/5
similarly third time 80% of 4H/5 = H(4^2)/(5^2)
and so on..
This will lead to infinite terms of geometric progression
i.e H+2*4H/5+2*16H/25..
So, Sum = H+ 2*4H/(5(1-4/5)) = 9H.
Tax reduced from 3 1/2 or 7/2% to 3 1/3% or 10/3%.
So, the difference in tax = (7/2 - 10/3)% = 1/6%.
=> 1/6 % of Rs. 8400 = 8400*1/6% = 8400* (1/6 * 1/100) = Rs. 14.
Lets assume sum of money = Rs. x ;
A gets Rs. 33 when sum distributed in the ratio of 3:2:5
so, 33=x*3/10
=> x=110.
Lets assume day required to complete the job = x day
Men Day
---------------------
3(u) 4(d)
4 x
x/4 = 3/4
=> x = 3 days.
[(6.23/100)*258.43]-'X'+[(3.11/100)*127]=13.87
16.100189-X+3.9497=13.87
X=6.179889.
Consider a solution = 1 ltr
X is the certain quantity which has to be replaced
Now,
40%of (1-x)+(25%of x)=35/100
=> 40/100 * (1-x) + 25x/100 = 35/100
=> 40/100 - 40x/100 + 25x/100 = 35/100
=> 15x/100 = 5/100
=> x = 1/3.
If the difference of the sum of digits at odd places and the sum of its digits at even places, is either 0 or divisible by 11, then clearly the number is divisible by 11.
[(Sum of digits at odd places) - (Sum of digits at even places)]/11
=> [(3+4+6+2) - (2+5+8)]/11 => (15 - 15)/11 = 0/11 = 0
similarly others number are divisible by 11.
Given Two No, ratio = 3:4 and their HCF = 4
So, No. = 3*4 =12 and 4*4=16
LCM of 12,16 = 48.
Lets assume marks of Ajith, Sachu, Karna, Saheep and Ramesh respectively u, v, w, x, y
So, as per the question:
z+7=x ---(i)
u- 9=z ---(ii)
x+ z=v ---(iii)
v+w=110 ---(iv)
Given, u=32 ---- (v)
By solving eq. (i), (ii), (iii), (iv) and (v)
w=57.
Lets assume additional hours = x hrs.
So total No. hours in journey = (4+x)
[(55*4)+(70*x)]/(4+x)=60 =>
=>x=2
Therefore, Total No. of hrs. in Journey = (x+4) = 6 hrs.
Simply appreciates variety 1 by 26% and depreciates variety 2 by 26% as:
3969(1-(1/26))^n=1369(1+)1/26))^n
For n = 2 we get both values equal.
Variety 2 Variety1
3969.00 1369 Initially
2937.06 1724.94 after I year
2173.42 2173.42 after II year
So the price become same after 2 years.
26-22=4
39-35=4
52-48=4
65-61=4
LCM of (26,39,52,65)=780
So smallest number = (780-4) = 776.
Lets assume distance between P and Q = d km.
Time taken by A in both side = d/10 + d/9 = 19d/90 hrs.
Time taken by B in both side = 2d/12 = d/6 hrs.
B tool 10 min. or 1/6 hrs. less than A.
So, 19d/90 - d/6 = 1/6
=> (19d-15d)/90 = 1/6
=> 4d/90 = 1/6
=> d = 15/4 km = 3.75 km.
C.P. of the article = Rs. 100
So Marked price = (100*100)/80 = Rs. 125
SP after the discount = Rs.(125*88)/100 = Rs. 110
therefore Gain percent = 10.
If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are a, b, c, d & e respectively, then
a+b+c+d=58*4=232 ----(i)
b+c+d+e=60*4=240 ----(ii)
Subtracting eq. (i) from (ii)
e-a=8 ---(iii)
Given, a/e=7/8 ---(iv)
So from eq. (iii) & (iv)
a=56, e=64.
According to baye's theorem:
p(no of way to drawn a ball from 2nd urns is red and ball trferred from 1 urns)=(3c1/5c1)*(3c1/9c1)=9/45
Total=(2c1/5c1)*(4c1/9c1)+(3c1/5c1)*(3c1/9c1)=17/45
so probability=(9/45)/(17/45)=9/17.