M4maths Previous Puzzles

16 Aug 2017 Standard Deviation

Age of 4 members in a family are: 2 years, 6 years, 8 years and 16 years.
What is the population standard deviation of their ages?

OPtion

1) 4.00
2) 4.39
3) 5.00
4) 5.89
5) 5.39
6) 5.09
7) 4.81
8) 5.99
9) 5.81
10)None of these

Solution

Mean of 4 ages = (2+6+8+16)/32 = 8
Sum of squared difference from mean for each number = (8-2)^2 + (8-6)^2 + (8-8)^2 + (8-16)^2
= 36+4+0+64 = 104

Average of Squared difference (Variance) = 104/4 = 26
Population Standard Deviation = Square root of variance = 26^1/2 = 5.09

Correct Option: 6 Best Solution (14)

G.S.Kantharao   7 years AGO

Mean of avea=(2+6+8+16)/4=8
Sum of the square=(2-8)^2+(6-8)^2+(8-8)^2+(16-8)^2/4
=36+4+0+64/4=26
Popular ion SD=√26=5.09
Hence the Right option is 6👈

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Mukund Manohar   7 years AGO

Mean of 4 ages = (2+6+8+16)/4 = 8
Sum of squared difference from mean for each number = (8-2)^2 + (8-6)^2 + (8-8)^2 + (8-16)^2 = 36 + 4 + 0 + 64 = 104
Average of Squared Difference(Variance) = 104/4 = 26
Population Standard Deviation = Square Root of Variance = 26^(1/2) = 5.09

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SAYAN CHATTERJEE   7 years AGO

mean= 1/4*(6+2+8+16)
= 8
sd=(1/4*((8-6)^2+(8-2)^2+(8-8)^2+(16-8)^2)
= 5.09.
Option 6.

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Ravi   7 years AGO

2+6+8+16/4=8 now 8-2=6,8-6=2,8-8=0 and 8-16=-8 so (6)^2+2^2+0^2+8^2/4
which will be 26 now take square root of 26
√26=5.09

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Varun   7 years AGO

Standard deviation is how much the value as been deviated from the nominal value(average value), so every value is subtracted with the average value, but by doing so, it will give both positive deviations. But we are interested only in deviation,whether positive(to the right) or negative(to the left), so, what we do is square the differences and find its mean, and then take out the applied square back.

Here, it is :
Mean initially = (2+6+8+16)/4 = 8
Therefore deviations = -6 -2 0 +8
Squaring : 36 4 0 64
Summation = 104
Mean now =104/4 = 26
Taking back the applied square = Root ( 26 ) = 5 .099

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Bodla Anupriya   7 years AGO

variance=E(x^2)-E(x)^2=(4+6=36+64+256)/4-(2+6+8+16/4)^2=
90-64=26
standard deviation=√26=5.09

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Vishwas   7 years AGO

firstly calculate the mean=(2+6+8+16)/4=8
subtract each of them with their mean and square the result
result will be 36,4,0,64
Again find the mean of them
Mean=26
Standard Deviation=square_root(26)=5.09

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Divya Batra   7 years AGO

we will use simple formula to calculate population standard deviation.

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AYAN PAUL   7 years AGO

Mean of those ages is = (2+6+8+16)/4 = 8
Now we have to find the squared difference of mean and the all values i.e. (8-2)^2 = 36 , (8-6)^2 = 4, (8-8)^2 = 0, (16-8)^2 = 64
now mean value of these squared difference is = (36+64+4)/4 = 26
now square root of 26 will be equal to = 5.09

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prasanna   7 years AGO

We know that to calculate standard deviation
1.find mean of all numbers
2.substract the mean from number and find square of it
3.find mean of these squared numbers
4.take square root of that number
Thats all we get population standard deviation of ages

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shail   7 years AGO

sd is nothing but deviation to mean of each data point , take values 2,6,8,16 and daviation is -6,-2,0,8
for stdd deviation take squares and add then decide by total no of data points .that's called variance
take it's sqr root to obtain sttdd deviation

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AMAN RASTOGI   7 years AGO

mean 8
mean minus each number square divide by n and square rooting it ..will provide the same

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palak Jain   7 years AGO

1.Mean =2+6+8+16/4=8
2. 6^2+2^2+0^2+8^2/4=104/4=26
3 under root of 26=5.099

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shreya   7 years AGO

mean= 8
2-8=-6.......16-8=8
36+4+64/4=26
√26=5.09

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15 Aug 2017 Happy Independence Day!

Find Rank of the word SWATANTRA in the dictionary of words formed by the letters of the word SWATANTRA

OPtion

1) 19472
2) 19772
3) 19172
4) 19272
5) 19372
6) 19572
7) 19672
8) 19872
9) 19972
10) None of these

Solution

Lets arrange the letters alphabetically = AAANRSTTW

Words starting with A = 8!/(2!*2!) = 10080
Words starting with N = 8!/(3!*2!) = 3360
Words starting with R = 8!/(3!*2!) = 3360
Words starting with SA = 7!/(2!*2!) = 1260
Words starting with SN = 7!/(3!*2!) = 420
Words starting with SR = 7!/(3!*2!) = 420
Words starting with ST = 7!/(3!) = 840
Words starting with SWAA = 5!/(2!) = 60
Words starting with SWAN = 5!/(2!*2!) = 30
Words starting with SWAR = 5!/(2!*2!) = 30
Words starting with SWATAA = 3! = 6
Words starting with SWATANA = 2! = 2
Words starting with SWATANR = 2! = 2
Words starting with SWATANTA = 1 = 1
And next word is SWATANTRA so + 1

Sum of all = 19872
Option 8)

Correct Option: 8 Best Solution (6)

G.S.Kantharao   7 years AGO

Word is SWATANTRA
Order is AAANRSTTW
Words with A=8!/2!*2!=10080
With N = 8!/3!*2! =3360
With R =8!/3!*2!=3360
With SA =7!/2!*2!=1260
With SN = 7!/3!*2!= 420
WithSR 7!/3!*2!=420
With ST=7!/3!=840
With SWAA=5!/2!=60
With SWAN=5!/2!*2!=30
With SWAR = 5!/2!*2!=30
With SWATA=3!=6
With SWATANA =2!=2
With SWATANR = 2!=2
With SWATANTA =1
With SWATANTRA=1
Total rank =10080+3360+3360+1260+420+420+840+60+30+30+6+2+2+1+1=19872
Hence the right option is 8👈

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Varun   7 years AGO

5*8!/(3!2!) + 7*7!/(3!2!)+0*6!/(3!*2!)+4*4!/(2!*2!)+...+1*1!/(1)+0*0!/(0) + 1 = 19872

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kasarla swathi   7 years AGO

provide alphabetical order numbering:
s w a t a n t r a
4 6 15 1 2 5 3 1
now find the number of digits less than 4,6,... in forward direction like
5 7 0 4 0 1 2 1 0
now divide each one with the number of repetatives including them we get
5/3!2! 7/3!2! 0/3!2! 4/2!2! 0/2! 1/1! 2/1! 1/1! 0/1!
now multiply each term with their factorials starting from 0! right we get
16800 2940 120 6 4 1
now add them we get 19871

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pooja   7 years AGO

S W A T A N T R A
4 6 1 5 1 2 5 3 1
alphabetical arrangement
((5/(3!2!))*8!)+((7/(3!2!))*7!)+(0*6!)+((4/(2!2!))*5!)+(0*4!)+((1/0!)*3!)+((2/0!)*2!)+((1/0!)*1!)+((0/0!)*0!)=19871
19871+1=19872

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himadhar   7 years AGO

A-10080
N-3360
R-3360
SA-1260
SN-420
SR-420
ST-840
SWN-30
SWAN-30
SWAR-2
SWATANTRA-1

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Mukund Manohar   7 years AGO

Rank of the word SWATANTRA
= 5/(3!*2!) + 7/(3!*2!) + 0/(3!*2!) + 4/(2!*2!) + 0/2! + 1/0! + 2/0! + 1/0! + 0/0!
= 16800+2940+0+120+0+6+4+1+0+1 = 19871 + 1 = 19872

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14 Aug 2017 Bankers Discount

The banker's discount on a bill due 8 months hence at 15% is Rs. 1947. What is the true discount?

OPtion

OPTION
1) 1770
2) 1857
3) 1950
4) 1750
5) 1942
6) 1870
7) 1707
8) 1907
9) 1850
10) None of these

Solution

T.D.=B.D. x 100 /(100 + (R x T))
=> 1947*100/(100+15*8/12)
=> 1947*100/110
=> 1770

Correct Option: 1 Best Solution (6)

Ravi   7 years AGO

T.D.
= B.D. x 100
100 + (R x T)
= Rs. 1947 x 100 /
(100 + 15 x 2/3)
= Rs. 1947 x 100 /110
= Rs. 1770

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AYAN PAUL   7 years AGO

We know the formula for true discount is = (Amount * Rate * time) / 100+(Rate * Time )
Now putting the value from above question we can get that True Discount = 177
So the discounted value will be 1947 - 177 = Rs. 1770

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Poulami Banik   7 years AGO

15% pa means 10% in 8 months.
So 1947 is 110% of the principle amount.. Therefore 1947/1.1=1770

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Mukund Manohar   7 years AGO

TD = BD*100 / 100+(R*T)
= 1947*100 / 100+(15* 2/3)
= 194700/110 = Rs 1770

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Madhuvani v   7 years AGO

T.D = (B.G*100/100+(R*T))
= (1947*100/100+(15*2/3))
= 19470/11 = 1770
option = 1

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Janani   7 years AGO

TD=1947*100/100+(15*8/12)
on solving this we get ,
and:Rs.1770

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11 Aug 2017 Trains

Two trains are running in opposite directions on parallel tracks. Train running with speed 60 Km/hr has 240 meter length and the second train running with speed of 40km/hr has 290 meter length.
What is the time taken by the faster train to completely pass through the length of slower train (i.e from front to rear)?

OPtion

1) 17.50 seconds
2) 10.44 seconds
3) 19.80 seconds
4) 19.02 seconds
5) 10.80 seconds
6) 19.50 seconds
7) 18.80 seconds
8) 19.90 seconds
9) 19.08 seconds
10) None of these

Solution

Relative speed= 60+40 = 100 km/hr = 100 * 5/18 = 250/9 m/sec

We are calculating the time taken by the faster train to completely pass the length of slower one.
Hence, distance = length of the slower train + length of faster train = 240+290 = 530 meter

Time = 530/250/9 = 19.08 seconds

Correct Option: 9 Best Solution (6)

SAYAN CHATTERJEE   7 years AGO

relative velocity = (60+40) = 100 kmph
total distance to be travelled= (290+240) = 530
So, time reqd. = (530/1000)/ 100= 0.0053 hr.
= 0.0053*3600 secs.
= 19.08 secs.
Option 9.

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Ravi   7 years AGO

t=D1+D2/S1+S2
t=240+290/(60+40)/5/18=477/25
so t= 19.08sec

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G.S.Kantharao   7 years AGO

Relative speed=60+40=100kmph
Speed Per Second=100*5/18=500/18
Relative length =240+290=530
Time taken to cross=530*18/500=19.08 sec.
Hence my option is 9👈

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D Annapoorna   7 years AGO

time to be taken=530×18÷100×5=19.08

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Madhuvani v   7 years AGO

Relative speed=(60+40) km/hr = 100 km/hr
= (100*5/18)m/sec = 250/9 m/sec
distance covered = (240+290) m = 530 m
required time = (530*9/250) sec = 19.08 sec

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Mukund Manohar   7 years AGO

Relative Speed of the train = (60+40) = 100 km/hr = 500/18 m/s
Required Time = Sum of length of trains/Relative Speed
= (240+290) / (500/18) = 530/ (500/18) = 19.08 seconds

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10 Aug 2017 Stocks & Shares

The market value of a 10.5% stock, in which an income of Rs. 1020 is derived by investing Rs. 17000, brokerage being 3/4 %, is:

OPtion

1) 176.75
2) 174.25
3) 175.00
4) 175.50
5) 175.75
6) 175.25
7) 176.50
8) 174.00
9) 174.75
10) None of these

Solution

For an income of Rs. 1020, investment = Rs. 17000.

For an income of Rs. 10.5, investment = Rs. (17000/1020) * 10.5 = 175
For a Rs. 100 stock, investment = Rs. 175

Market value of Rs. 100 stock = Rs. 175 - 0.75 = Rs. 174.25

Correct Option: 2 Best Solution (7)

Mukund Manohar   7 years AGO

For an income of Rs 1020, investment = Rs 17000
For an income of Rs 21/2, investment = 17000/1020 * 21/2
= Rs 175
For a Rs 100 stock, investment = Rs 175
Market Value of Rs 100 stock = (175 - 3/4) = Rs 174.25

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Ravi   7 years AGO

For an income of Rs. 1020, investment = Rs. 17000.

For an income of Rs. 21/2, investment = Rs. 17000/1020 x 21/2 = Rs. 175.
For a Rs. 100 stock, investment = Rs. 175.
Market value of Rs. 100 stock = Rs. 175 - 3/4 = Rs. 174.25

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SAYAN CHATTERJEE   7 years AGO

For an income of 10.5, the investment is 17000/1020*10.5
= 175
Brokerage is 0.75%
So, market value of Rs 100 stock= (175-0.75) = 174.25.
Option 2.

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Varun   7 years AGO

Gain = 1020
Put = 17000
Profit ratio = 17000/1020 = 16.66
Ratio has = 10.5%
Therefore 16.66*10.5 = 175
Brokerage = 3/4 = 0.75
So market = 175-0.75 = 174.25

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Tavva Deepthi   7 years AGO

assume that face value=rs 100
dividend per share=rs 10.5(as it is 10.5% stock)
by investing rs 17000,earning=rs 1020
to get an earning of rs 10.5 investment required =17000*10.5/1020=175
market value of rs 100 stock=rs 175-3/4
=rs 174.25

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Himaja   7 years AGO

for an income of rs.1020,investment=rs.17000.
for an income of rs.21/2,investment=rs.(17000/1020)*(21/2)=16.66*(21/2)=rs.175.
for a rs.100 stock ,investment=rs.175.
market value of rs.100 stock=rs.(175-(3/4))=rs.174.25

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KRISHNAKUMAR   7 years AGO

dividend per share=10.5%
by investing rs 17000,earnings=1020
the investment is=17000*10.5/1020=175
175-3/4=174.25

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09 Aug 2017 Distance

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 60° and 30° respectively. If the lighthouse is 144 m high, the distance between the two ships is:

OPtion

1) 312.55 m
2) 322.55 m
3) 362.55 m
4) 302.55 m
5) 342.55 m
6) 352.55 m
7) 332.55 m
8) 372.55 m
9) 382.55 m
10)None of these

Solution

Let AB be the lighthouse where A is the bottom point of light house and C and D be the positions of the ships.
Then, AB = 144 m, Angle ACB = 60° and angle ADB = 30°.
AB/AC = tan 60° = sq root(3)
=> AC = AB / sq root(3) = 144/3^(1/2) = 48*3^(1/2)

AB/AD = tan 30° = 1/sq root(3)
=> AD = AB*sq root(3) = 144*3^(1/2)

CD = (AC + AD) = 3^(1/2) (48+144)
= 192*3^(1/2)
= 332.55

Correct Option: 7 Best Solution (22)

SAYAN CHATTERJEE   7 years AGO

Distance between two ships is 144*(1/tan60+a1tan30)
= 332.55
Option 7.

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Mukund Manohar   7 years AGO

Required Distance = 144/tan 60 + 144*tan 30
= 144/sq.root(3) + 144*sq.root(3)
= 332.55 m

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shreya   7 years AGO

144/x = tan 30
144/y =tan 60
x+y= 144√3 + (144/√3) =332.55

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G.S.Kantharao   7 years AGO

Angle of elevations are 60,30 degrees
Height of the light house =144 mts.
Tan 60= 144/x=
144/x =√3
X=144/√3
Tan 30 =144/y=1/√3
Y=144√3
Distance between the ship's is =x + y
X+y=144/√3+144√3=(144+144*3)/√3=576/√3=332.5 mts.
Hence the right option is 7 👈.

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mandeep   7 years AGO

tan 60=144/x
tan 30=144/y
find x and y and then add both to get 332.55

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Divakar   7 years AGO

easy
tan60 = 144/x
x= 144/3^1/2. similarily, tan30 = 144/y
y = 144* 3^1/2
x+Y = 332.55

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Ravi   7 years AGO

let AB=144 so angle Angle ACB=60° AND ANGLE ADB=30° SO AB/AC=tan60° and AB/AD=tan30° so after solving we will get AC and AD value after adding we get 332.55m

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tumati koushik   7 years AGO

x1=144/tan 60 ,x2 =144/tan30
x= x1+x2=144(4/sqrt3)
x=332.5538

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Madhuvani v   7 years AGO

Let AB be the lighthouse and C and D be the positions of the ships.
AB=144m,∠ACB=60,∠ADB=30∘
AB/AC=tan60=√3
AC=144/√3
AB/AD= tan30 =1/√3
AD=144√3
CD=AC+AD=( 144/√3+144√3)=144(1/√3+√3)=332.55 m
ans=7

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sangita kundu   7 years AGO

let the distance between the lighthouse and one ship is x1 and another one is x2 and the angle is 60°and 30°.
sin30°= x1/144 and sin60°=x2/144.
from these two equation we get the values of xand x2.then we add these values.i.e. the required answer

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Vignesh   7 years AGO

Angle of elevation from lighthouse for first boat 60° and for second 30°.
Height = 144 m.

For first boat:


cos 60° = 144/ hyp
Hyp= 288
Sin 60° = X / 288
X = 249.4 m

For second boat :

Cos 30° = 144/ hyp
Hyp= 166.2
Sin 30° = y / 166.2
Y = 83.138 m

Distance = x + y
= 332.55 m

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umend   7 years AGO

Let ship be at a distance of x and y from lighthouse
Then tan60=144/x
tan30=144/y
We get x=83.14. y=249.41
Distance between two ships=x+y=83.14+249.41
=332.55
Ans:332.55

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Bodla Anupriya   7 years AGO

144(tan30+tan60)=144(1/√3 +√3 )
=332.55

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palak Jain   7 years AGO

angle of elevation of first ship=60 degree
So,tan 60=p/b1 =144/b1
so,b1=144/1.732
tan 30=p/b2
so,b2=144*1.732
b1+b2=332.55 m

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Himaja   7 years AGO

let a triangle is bcd .there is lighthouse of height 144m i.e ab.ab/ac=tan60,ab/ad=tan30,cd=ac+ad.finally ac=144/1.73 and ad=144*1.73.so,cd=332.55m

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Sethu Bargavi M   7 years AGO

tan 60 = 144/B
Ship1 distance from light house = 144/1.732 (sq.root of 3) = 83.14
tan 30 = 144/B
Ship2 distance from light house = 144*1.732 = 249.40

Add both 332.54

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kanimozhi kalimuthu   7 years AGO

consider distance between ship 1 and ship 2 be x,y respectively
tan 60=opp/adj=144/x
x=83.13
tan 30=144/y
y=249.41
distance=(x+y)=83.13+249.41=332.55m

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Sahil Mehta   7 years AGO

tan 60=144/base
base=√3*144
base=249.12 m
tan30=144/base2
base2=83.2 m
the total distance will be =base 1 +base2
249.12+83.2 =332.55 m

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AYAN PAUL   7 years AGO

From the question we can write the distance between a ship and the lighthouse is = 144*1.732 = 249.408
and the distance from the lighthouse to another ship is 144/1.732 = 83.14
Now combining these two distance we can write the total distance between two ships will be = 249.408 + 83.14 = 332.55 m (approximately)

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prasanna   7 years AGO

Let's take The distance between lighthouse and ship1 is x and distance between lighthouse and ship2 is d-x than by giving that angles from lighthouse to ships are 30 and 60
By solving trigonometric equations we get the distance between 2 ships is 332.55m

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ANANDA PRIYA T.R   7 years AGO

On one end of the light house the
Distance is (144/tan 60)=83.138
On the other side of the light house the
Distance is (144 /tan 30) =249.41
83.138 +249.41 =332.548

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shashank kumar singh   7 years AGO

Let base be x.
tan(30)=144
x=144×√3
tan(60)=144
x=144÷√3
By adding both the x we get 332.548

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08 Aug 2017 Radius and Volume

3 solid spheres of radius 7 m, 5 solid cylinder of radius 5 m, height 3 meter and 7 solid cones of radius 3 m, height 7 m are melted to form a bigger solid sphere. What is the radius of this bigger solid sphere in meters?

OPtion

1) 10.22
2) 11.14
3) 12.09
4) 11.24
5) 10.89
6) 11.09
7) 12.09
8) 10.94
9) 12.04
10)None of these

Solution

Volume of 3 smaller sphere's = 3*(4/3)*Pi*r1^3
Volume of 5 Cylinder = 5*(Pi*r2^2*h1)
Volume of 7 Cones = 7*(1/3)*Pi*r3^2*h2
Volume of Larger Sphere = (4/3)*Pi*R^3

Here r1=7m, r2=5m, r3=3m, h1=3m, h2=7m, and R=?
Volume of Larger Sphere = Volume of 3 smaller sphere's + Volume of 5 Cylinder +Volume of 7 Cones
=> (4/3)*Pi*R^3 = 3*(4/3)*Pi*r1^3 + 5*P1*r2^2*h1 + 7*(1/3)*Pi*r3^2*h2

=> 4/3*R^3 = 4*343 + 5*75 + 7*21 = 1894
=> R^3 = 1420.5
R = 1420.5^(1/3) = 11.24 m

Option 4)

Correct Option: 4 Best Solution (14)

G.S.Kantharao   7 years AGO

Radii are 7m,5m,3m
Heights are 3m,7m
No of solids are 3,5,7
Total volume= 3*4*π*(7^3)/3 + 5* π*(5^2)*3+7*π*(3^2)*7/3
=(1372π+375π+147π)m^3
4*πR^3/3=1894πm^3
R=(1894*3/4)^(1/3)=11.24m
Hence the right option is 4👈

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Mukund Manohar   7 years AGO

Volume of bigger solid sphere = (3*1436.76)+(5*235.62)+(7*65.97)
= 4310.28 + 1178.1 + 461.79
= 5950.17 m^3
Radius of the bigger solid sphere = cube root(1419.93)
= 11.24 m

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SAYAN CHATTERJEE   7 years AGO

(3*4/3*3.14*7^3+5*5^2*3+7*1/3*3.14*3^2*7)=4/3*3.14*R^3
=> R= 11.24.
Option 4.

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Harshal Sapkade   7 years AGO

by melting all the solids we get there volumes. by applying the formulas for particular cubes we can get the answer.

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shreya   7 years AGO

vol of sphere×3 + vol of cylinder×5 + vol of cone×7 = vol of sphere
1894π = 4π r ^3 / 3
r=11.24

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Ravi   7 years AGO

add all the volumes of sphere, cylinder and come and put equal to the big solid sphere volume which is 4/3πr^3 so the r will be equal to 11.24

Like? Yes | No   

KRISHNAKUMAR   7 years AGO

the volume of 3 sphere is = 4310.28
the volume of 5 cylinder is =1178.1
the volume of 7 cone is=461.79
the volume of new cone is=4310.28+1178.1+461.79
=5950.74
from above value radius is=11.2397

Like? Yes | No   

Madhuvani v   7 years AGO

Volume of 3 smaller sphere's = 3*(4/3)*Pi*r1^3
Volume of 5 Cylinder = 5*(Pi*r2^2*h1)
Volume of 7 Cones = 7*(1/3)*Pi*r3^2*h2
Volume of Larger Sphere = (4/3)*Pi*R^3

Here r1=7m, r2=5m, r3=3m, h1=3m, h2=7m, and R=?
Volume of Larger Sphere = Volume of 3 smaller sphere's + Volume of 5 Cylinder +Volume of 7 Cones
=> (4/3)*Pi*R^3 = 3*(4/3)*Pi*r1^3 + 5*P1*r2^2*h1 + 7*(1/3)*Pi*r3^2*h2
(4/3)*pi*R^3 = 1372*pi+375*pi+147*pi
=> 4/3*R^3 = 1894pi
=> R^3 = 1420.5
R =cube root(1420.5) = 11.24 m

Option 4)

Like? Yes | No   

Samiksha Sonam   7 years AGO

Area of bigger sphere=3*[(4/3)*(22/7)*343]+5*[(22/7)*25*3]+[7*(1/3)*(22/7)*9*7]=5952.6
Radius of bigger sphere=11.24

Like? Yes | No   

Harshit Gupta   7 years AGO

The basic concept for this is that Total Volume remains the same when all the solid spheres, solid cylinders and solid cones are added up to form a single solid sphere.

Like? Yes | No   

Ramkumar saini   7 years AGO

3*1436.76=4310.08
5*235.62=1178.1
7*65.97=461.79
r^3=1419.16
r=11.2376

Like? Yes | No   

AYAN PAUL   7 years AGO

Here the equation become (4/3)*22/7*r^3 = 3*(4/3)*22/7*343 + 5*(22/7)*25*3 + 7*(1/3)*(22/7)*9*7
Solving this equation we get that r = 11.24 m

Like? Yes | No   

palak Jain   7 years AGO

volume of 3 spheres=4/3(pie)*r^3 where r=7m=1372 pie m^3
volume of 5 cylinders=pie*r^2*h where r=5m,h=3m=375pie m ^3
volume of 7 solid cones=4/3 pie *r^2*h where r=3m ,h=7m=147 piem^3
total volume=1894 pie m^3
1894 pie m^3=4/3 pie r^3
r=11.24 m

Like? Yes | No   

Abinesh.R   7 years AGO

volume of sphere + volume of cylinders+volume of cone=volume of sphere..

Like? Yes | No   

07 Aug 2017 Decode

If DEEPAWALI = 76 and DURGAPUJA 99
Then RAKSHABANDHAN = ?

OPtion

1) 112
2) 100
3) 109
4) 122
5) 101
6) 102
7) 103
8) 99
9) 98
10) None of these

Solution

A=1, B=2, C=2 and so on till Z= 26
DEEPAWALI = Sum of Positions of each alphabet = 4+5+5+16+1+23+1+12+9

DURGAPUJA = Sum of Positions of each alphabet = 764+21+18+7+1+16+21+10+1 = 99

Similarly RAKSHABANDHAN = Sum of Positions of each alphabet = 18+1+11+19+8+1+2+1+14+4+8+1+14 = 102

Correct Option: 6 Best Solution (15)

Shilpi   7 years AGO

Deepawali=4+5+5+16+1+23+1+12+9=76.similarly others.

Like? Yes (2) | No   

Mukund Manohar   7 years AGO

A=1,B=2,C=3,D=4,E=5..........Y=25,Z=26
Logic :- Sum of position of alphabets
DEEPAWALI = D+E+E+P+A+W+A+L+I = 4+5+5+16+1+23+1+12+9 = 76
Similarly, DURGAPUJA = 99
RAKSHABANDHAN = 18+1+11+19+8+1+2+1+14+4+8+1+14 = 102

Like? Yes | No   

Aakash Beniwal   7 years AGO

add numeric value of alphabet of RAKSHABANDHAN i.e. A=1, B=2 .....

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Pritam Paul   7 years AGO

RAKSHABANDHAN
According to Alphabet series...
R= 18
A= 1
K= 11
S= 19
H= 8
A= 1
B= 2
A= 1
N= 14
D= 4
H= 8
A= 1
N= 14
Sum of these letters = 102

Like? Yes | No   

KRISHNAKUMAR   7 years AGO

Assume a=1 ,b=2 ,..... Z=26
Deepawali=4+5+5+16+1+23+1+12+9=76
Like that
Rakshabandhan = 18+1+11+19+8+1+2+1+14+4+8+1+14=102

Like? Yes | No   

anusha mathur   7 years AGO

R=18 + A=1 + K=11 + S=19 + H=8 + A=1 + B=2 + A=1 + N=14 + D=4 + H=8 + A=1 + N=14 = 102

Like? Yes | No   

KOWSALYA   7 years AGO

DEEPAWALI=4+5+5+16+1+23+1+12+9=76
LIKE WISE
RAKSHABANDHAN=18+1+11+19+8+1+2+1+14+4+8+1+14=102
option 6) 102

Like? Yes | No   

SAYAN CHATTERJEE   7 years AGO

we we put no.s from 1 to 26 from A to Z
RAKSHABANDHAN comes out to be
(18+1+11+19+8+1+2+1+14+4+8+1+14)
= 102.
Option 6.

Like? Yes | No   

jyothi chhabria   7 years AGO

rakshabandan as per numbering of alphabets r=18 a=1 and so on by adding we get 102

Like? Yes | No   

Ravi   7 years AGO

add all the alphabet no. come in series

Like? Yes | No   

prasanna   7 years AGO

give numbering from 1 to 26 to the alphabets A to Z.
than add the respective number of the alphabet than we get 102 for RAKSHABANDHAN

Like? Yes | No   

Samiksha Sonam   7 years AGO

R+A+K+S+H+A+B+A+N+D+H+A+N=18+1+11+19+8+1+2+1+14+4+8+1+14=102

Like? Yes | No   

Macharla Ramya   7 years AGO

alphabetical order
R+A+K+S+H+A+B+A+N+D+H+A+N=18+1+11+19+8+1+2+1+14+4+8+1+14=102

Like? Yes | No   

AYAN PAUL   7 years AGO

If we decode the code DEEPAWALI with their alphabetic value then the summation of those values will be 76
Similarly if we decode RAKSHABANDHAN along their alphabetic value then the answer will be 102

Like? Yes | No   

swathi   7 years AGO

becoz they are sum of numbers of alphabets

Like? Yes | No (1)  

04 Aug 2017 Age

A is 4 years older than B and B is twice as old as C. The total of the ages of A, B and C is 44 years. What is the product of their ages?

OPtion

1) 2360
2) 2380
3) 3036
4) 2760
5) 2690
6) 2380
7) 2630
8) 2560
9) 3618
10)None of these

Solution

Let’s assume Age of B be B years
Age of A = B+4
Age of C = B/2

Total age = B + 4 + B + B/2 = 44
Solving 5B/2 = 40
B = 16
C = B/2 = 16/2 = 8 years
A = 16+4 = 20 years
Product = 20*8*16 = 2560

Option 8)

Correct Option: 8 Best Solution (54)

Mukund Manohar   7 years AGO

A = B+4 and B/2 = C
A+B+C = 44
=> B+4+B+B/2 = 44
=> 5B+8 = 88
=> B = 16 years
So, A = 20 years and C = 8 years
Product of their ages = 20*16*8 = 2560

Like? Yes | No   

AYAN PAUL   7 years AGO

Let the Age of B is x so Age of A will be x+4 and Age of C will be x/2
So we can write x+x/2+x+4 = 44
from here we get x = 16
So A = 20 B = 16 and C = 8
Now product of ages of A B & C will be 16*20*8 = 2560

Like? Yes | No   

SAYAN CHATTERJEE   7 years AGO

B=2C
A=B+4
=2C+4
SO, C+2C+2C+4= 44
SO, C= 8
B= 16
A= 20
SO, A*B*C= 20*16*8
2560.
Option 8.

Like? Yes | No   

Ravi   7 years AGO

let c's age is x then B's will be 2x and c's will be 2x+4 so total age of A+B+C=44
then 2x+4+2x+x=44 ,then x= 8 age of A=20
B=16 and C=8 product of their ages will be
20×16×8=2560

Like? Yes | No   

Varun   7 years AGO

Let C's age be x
Then B's age = 2x
A's age = 2x + 4
Sum of ages = 5x + 4 = 44
Therefore x = 8

That gives present ages as :
C = 8
B = 16
A = 20

So, it gives product as 2560.

Like? Yes | No   

abhishek   7 years AGO

Let age of C is x.
Then age of B is 2x.
And age of A is 2x + 4.
Now x + 2x + 2x + 4 = 144.
5x + 4 = 44
x = 8.
Then age of A is 20, B is 16 and C is 8.
Therefore there product is 8 * 16 * 20 = 2560.

Like? Yes | No   

Deepesh Mittal   7 years AGO

C age x,
then B age is 2x.
so A age is 2x+4.
(2x+4)+2x+x=44
5x=40
x=8
So, B is 2x=2(8)=16
A is (2x+4)=20
C is 8.
Hence, 20*16*8 = 2560.

Like? Yes | No   

G.S.Kantharao   7 years AGO

Let the age of C =x
Then B's age=2x
A'sage is =2x+4
Their sum =(2x +4)+ (2x)+ (x)=5x+4=44
5x=40
X=8
Then their ages are=20,16,8
Product=20*16*8=2560
Hence the right option is 8👈

Like? Yes | No   

KRISHNAKUMAR   7 years AGO

Let age of c is =x
The age of b is =2x
The age of a is =2x+4
Sum of age is =5x+4=44
So x= 8
Product is =8×16×20=2560

Like? Yes | No   

Pritam Paul   7 years AGO

Let the age of C is X.
Age of B is 2X.
Age of A is 2X+4

X+2X+(2X+4)=44
X= 8
Age of C is 8, B is 16 and A is 20
The product of A,B and C is (8×16×20)= 2560

Like? Yes | No   

Tarun varshney   7 years AGO

let b age = x
and a age =x+4
then c age = x/2
age ( a+b+c)=44
x.=16
age multiple =20*16*8

Like? Yes | No   

mandeep   7 years AGO

a=b+4. b=2c
a+b+c=44
solve and get a=20 b=16 c=8
and the product is 2560

Like? Yes | No   

g lokesh   7 years AGO

Let us assume age of C = x yrs
so , B =2x
A = 2x+4
total age is x+2x+2x+4=44
on solving X= 8
ages of a,b and c is 20,16 and 8
product of them is 2560

Like? Yes | No   

Nidhi kumari   7 years AGO

A= B+4 years
B= 2C years
C = B/2

A+B+C= 44years
B+4+B+B/2 =44years
B = 16years
A= B+4= 16+4= 20years
C= B/2 = 16/2= 8years

Like? Yes | No   

umend   7 years AGO

Let A,B,C be the ages of A,B,C respectively
Then A=4+B
B=2C
A+B+C=44
Solving abv 3 equation gives
A=20. B=16. C=8
ABC=20*16*8=2560
Ans:2560

Like? Yes | No   

Madhuvani v   7 years AGO

Let C=X years, B=2X years , A=(2X+4) years
A+B+C=44
2X+4+2X+X=44
X=8
The product of the ages is = (2X+4)(2X)(X)
= 4X^3+8X^2 = 2048+512 = 2560

Like? Yes | No   

Arijit sarkar   7 years AGO

let c age be x.
then B age will be 2x.
A will be 2x+4.
5x =40
since sum is given . so x=8.
after multiplying we get 2560

Like? Yes | No   

ManikumarSoruvu   7 years AGO

c=x,b=2x,A=2x+4
2x+4+2x+x=44
x=8
Therefore,product=8*16*20=2560

Like? Yes | No   

pooja khannawalia   7 years AGO

Let the age of C be x.
then the age of B = 2x.
Now the A is four years older than B i.e., 2x + 4.
A.T.Q.
x + 2x + 2x+4 = 44
x= 8
2x = 16
2x + 4 = 20
8*16*20 = 2560 (Ans)

Like? Yes | No   

D Ramya   7 years AGO

first we will find B age by using given information..then remaining ...so the product will be 2560

Like? Yes | No   

Akash   7 years AGO

A=B+4
B=2C => A=2C+4
WKT A+B+C = 44
2C+4+2C+C=44
5C+4=44
5C=40
C=8
A=20
B=16
A*B*C=2560

Like? Yes | No   

Shaswata Bisi   7 years AGO

Let the age of A, B and C be a,b and c respectively
b=2c
a=4+b=4+2c
a+b+c=44
4+2c+2c+c=44
c=8
b=16
a=20
product of ages= a*b*c=20*16*8=2560

Like? Yes | No   

babita maurya   7 years AGO

let C=x years old.Then according to the question
B=2x,
A=4+2x
now it is given that sum of the ages of A B C is 44 years
i.e. x+2x+(4+2x)=44
5x=40
x=8
It means A=20
B=16
C=8
so product of their ages=20*16*8
=2560
hence option 8 is correct.

Like? Yes | No   

krishna reddy   7 years AGO

A= B+4 and B = 2C so we can write that
A=2C+4
and we also know that A+B+C=44
substituting all the equations we get
C=8.
now B=16. And A=20
ABC=8×20×16=2560.

Like? Yes | No   

Vijeta Budania   7 years AGO

let age of c=x
so, age of b and a are 2x and 2x+4 respectively.
given, a+b+c=44
2x+X+2x+4=44
x=8
therefore, abc= 2560

Like? Yes | No   

KOWSALYA   7 years AGO

a=4+b;b=2c
a+b+c=44
c=8;b=16;a=20
ABC=2560

Like? Yes | No   

Sai Charan   7 years AGO

Age of A = B + 4
Age of B = 2C
i.e, Age of C = B/2

Given that,
A + B + C = 44
(B + 4) + B + (B/2) = 44
B = 16
A = 20
C = 8
A X B X C = 2560

Like? Yes | No   

SARATCHANDRAN   7 years AGO

B=2C
A=(B+4)=2C+4
Given A+B+C= (2C+4)+2C+C=44
5C=40
C=8
therefore B=16 and A=20
A×B×C=20×16×8=20×128 =2560

Like? Yes | No   

shubham kumar   7 years AGO

A=2x+4
b=2x
c=x
2x+4+2x+x=44
x=8
20*16*8=2560

Like? Yes | No   

prasanna   7 years AGO

a+4=b,b=2c
a+b+c=44
solving above equations we get a=20,b=16,c=8
so product of these a,b,c is 2560

Like? Yes | No   

shreya   7 years AGO

A=B+4
B=2C
A+B+C=44
(B+4)+B+(B/2)=44
B=16
A=20
C=8
AXBXC=2560

Like? Yes | No   

Gayatri   7 years AGO

4x=40
x=8
A=2x+4=20
B=2x=16
C=x=8

Like? Yes | No   

S. Subramanyam   7 years AGO

C=x then B=2x and A=2x+4
A+B+C=44
so x=8
then A*B*C=8×16×20=2560

Like? Yes | No   

B.Sri Pravallika   7 years AGO

a=b+4;b=2c;c
a+b+c=44
2c+4+2c+c=44
c=8
a*b*c=2560

Like? Yes | No   

Anandhakumar   7 years AGO

A=B+4
B=2C
A=2C+4
A+B+C=5C=44
So C=8,B=16,A=20,
Product of ABC=2560

Like? Yes | No   

Anjali M Verma   7 years AGO

2x+4+2x+x=44
x is the age of C
so solving we get
x=8
so A=8
B=16
C=20
A×B×C=2560

Like? Yes | No   

Macharla Ramya   7 years AGO

let age of c be 'x'
b=2x,a=2x+4
a+b+c=44
x+2x+2x+4=44
x=8
a=8,b=16,c=20
abc=8×16×20==2560

Like? Yes | No   

Rohan shenoy   7 years AGO

C age is taken as 'x' then B will be '2x' and A will be'2x+4'
2x+4+2x+x=44
x=8
(20)(16)(8)=2560

Like? Yes | No   

Venkatesh V   7 years AGO

by solving with the equation
we get the ages 8,16,20
product = 2560

Like? Yes | No   

Raji Baskar   7 years AGO

B=2C
A=2C+4
A+B+C=44
=>2C+4+2C+C=44
On Solving,
C=8
A*B*C=>20*16*8==>2560

Like? Yes | No   

Rex   7 years AGO

a=b+4;b=2c and a+b+c=44 then converting all in terms of c we have a=2c+4; b=2c; then putting them in the equation a+b+c=44 we have
->(2c+4)+(2c)+(c)=44
->5c+4=44
->5c=40
->c=8
Therefore,
b=2c=2(8)=16 and a=b+4=16+4=20.
Hence the product of their ages = 20*16*8=320*8=2560 answer.

Like? Yes | No   

Navini Kulurkar   7 years AGO

5c=44-4
C=8
B=16
A=4+16=20
A*B*C=20*16*8=2560

Like? Yes | No   

tushar   7 years AGO

(C)x+(B)2x+(A)(2x+4)=44
5x=40
x=8
So 8*16*20=2560

Like? Yes | No   

manisha   7 years AGO

Let b=x, a=x+4, c=x/2
x+x+4+x/2=44
x=16=b , a=16+4=20, c=8
Abc= 16*8*20= 2560

Like? Yes | No   

Raj   7 years AGO

A=B+4
B=2c
A+B+C=44
B=16
A=20
C=8
A*B*C=2560

Like? Yes | No   

Saleena Qureshi   7 years AGO

A= B+4
B = 2C
A+B+C=44

(B+4)+B+(B/2) = 88

B=16
Therefore, C=8, A= 20

Like? Yes | No   

d bose   7 years AGO

let C BE X
SO, B=2X
A=2X+4
SO,A+B+C=44
2X+4+2X+X=44
X=8
SO, C=8
A=20
B=16
SO, ABC=2560

Like? Yes | No   

Riya malaiya   7 years AGO

let c age is x than b age will be 2x and a age will be 2x+4........now.......x+2x+2x+4=44...........we get x=8...now..............8*16*20=2560=ans.

Like? Yes | No   

Divakar   7 years AGO

its simple
x+2x+2x+4=44
x=8
now product of their ages = 8*16*20= 2560

Like? Yes | No   

Sanju Ambiger   7 years AGO

A=4+B
B=2C
4+2C+2C+C=44. c=8
A=20
B=16

Like? Yes | No   

Raghu kln   7 years AGO

a=2x+4
b=2x
c=x
2x+4+2x+x=44
5x+4=44
x=8
20*16*8=2560

Like? Yes | No   

Prabhanshu   7 years AGO

Age of C = 8 yrs.
Age of B = 16yrs.
Age of A= 16+4 = 20yrs
total = 8+16+20=44 yrs..
product = 2560..

Like? Yes | No   

Priti Dhanawade   7 years AGO

A=20
B=16
C=8


20*16*8=2560

Like? Yes | No   

aishwarya pandey   7 years AGO

A=x+4
B=x
C=x/2
as B = 2y
x=2y
y=x÷2
So,
A=x+4
B=x
C=x÷2
SO,
x+4+x+(x÷2)=88
(5*x)+8=80
x=16

A=20
B=16
C=8
20*16*8=2560

Like? Yes | No   

03 Aug 2017 Time and Work

P can do a work in 17 days. Q can do the same work in 8.5 days and R can do the same in 25.5 days. Q and R start the work and leave after 3 days. P finishes the remaining work in X days. What is the value of X?

OPtion

1) 9.0 days
2) 8.0 days
3) 8.5 days
4) 10.5 days
5) 10.0 days
6) 11.0 days
7) 12 days
8) 12.5 days
9) 7.0 days
10) None of these

Solution

Amount of work done by Q and R in 3 days = 3*(1/8.5 + 1/25.5) = 3*2*4/51 = 8/17

Amount of work remaining after 3 days = 9/17

Amount of work done by P in 1 day = 1/17
So for 1/17 work P takes 1 day
Hence for 9/17 work P would take (1/(1/17))*8/17 = 9 days

Correct Option: 1 Best Solution (31)

SAYAN CHATTERJEE   7 years AGO

Tasks completed in 3 days by Q & R is 3*(1/8.5+1/25.5)
so, works left= 1- 3*(1/8.5+1/25.5)
So, P does the same in (1-3*(1/8.5+1/25.5))
So, X=9.
Option 1.

Like? Yes | No   

Mukund Manohar   7 years AGO

Amount of work done by Q and R in 3 days = 3*(1/8.5 + 1/25.5) = 3*8/51 = 8/17
Remaining amount of work = 9/17
Now, Amount of work done by P in 1 day = 1/17
If 1/17 work is done by P in 1 day
So 9/17 work is done by P in 9 days
Value of X = 9 days

Like? Yes | No   

Ravi   7 years AGO

P's one day work is 30,q's is 60 and R's is 20 and total work will be 510
so p+q work for 3 days which will be 240 and remaining 270 will be done by p in
270/30= 9 days

Like? Yes | No   

Naimish   7 years AGO

Let 1 be the total work. In 3 days, Q and R complete 0.47 part of work. Therefore 0.53 work was left for P to do. P takes 17 days to complete entire work so for completing 0.53 work he would take 0.53*17 i.e. 9.01 days approx 9.0 days

Like? Yes | No   

Pritam Paul   7 years AGO

Q does a work in 3 days = 3/8.5
R does the same in 3 days = 3/25.5
Both Q & R does this togetherly = (3/8.5 + 3/25.5)= 120/255
Work rests for P = 1-120/255 = 135/255
So, P does the remaining portion in (17×135/255)= 9 days.

Like? Yes | No   

shreya   7 years AGO

17 days -1work by P
Q and R together 6.375 days 1 work...therefore in 3 days .47 work
remaining .53 work to be done by P so
17×.53=9

Like? Yes | No   

ManikumarSoruvu   7 years AGO

Ans:9.0 days
3 days Work of Q and R=(3/8.5)+(3/25.5)=0.470588235
Remaining work Carried By P=1-0.470588235=0.529411765
Therefore, P complete the Remaining work in 17*0.529411765=9 days

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Gokul Krishnan   7 years AGO

P do in 17 days, so capacity to do in a day is 1.5 days.
Q do in 8.5 days,so capacity to do in a day is 3 days.
R do in 25.5 days,so capacity to do in a day is 1 day.
so,P do in 9 days.

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g lokesh   7 years AGO

The part of work done by R and Q in 1 day is 1/8.5+1/25.5
=2/17+2/51
=8/51
part of work done in 3 days is 8*3/51
=8/17
remaining part of work is 9/17
P can do a part of work in 17, 9/17 part of work can be in done 9 days

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Nidhi kumari   7 years AGO

p can do a work in 17days = 1/17
Q can do a work in 8.5 days = 1/ 8.5
R can do a work in 25.5 days= 1/ 25.5
Q & R leave the work after 3days= 3(1/8.5+1/25.5) = 8/17
Remaining work = 1-8/17 = 9/ 17
p finishes the remaining work in x days = 9/17*17 = 9days
p finishes the remaining work in x days= 9 days

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G.S.Kantharao   7 years AGO

P,Q,R, can do in 17days,8.5days,25.5days respectively
In 1 day they each can do 1/17,2/17,2/51
Q+R can do in 1 day =2/17+2/51=(6+2)/51=8*3/51=8/17
Remaining =9/17
P can complete in 9days.

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Sasirekha   7 years AGO

p can do 3 parts of work in a day. totally 51 parts available.
Q and R have already done 24 parts of work in 3 days.
So remaining 27 parts can be done in 9 days.

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Abinesh.R   7 years AGO

total work to be done = 51
Q AND R FOR 3 days =24
remaining work = 27
p take 9 days to finish remaining work

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Anandhakumar   7 years AGO

Take total work 25.5
P can do a 25.5 work in 17 days so1.5 work /day
Q can do a 25.5 work in 8.5 days so 3 work per day
R can do a25.5 work in 25.5. Days so 1work per day first 3 days Q and R finished 12 works so P finish the remaining 13.5 Works in 9 days(1.5 works per day)

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Venkatesh V   7 years AGO

total work =51
p does 3 work per day
q does 6 work per day
r does 2 work per day
q+r=8
3*8=24
51-24=27
27/3=9
9 days

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umend   7 years AGO

Q one day work=1/8.5
R one day work=1/25.5
Work completed when Q and R works for 3 days
=(1/8.5+1/25.5)*3=0.47
Left over work=1-0.47=0.53
P takes X days to complete remaining 0.53 work,
P one day work=1/17
For X days=(1/17)*X=0.53
X=9
Ans:9 days

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Shaswata Bisi   7 years AGO

P can do 1/17 of the work in 1 day
Q can do 1/8.5 of the work in 1 day
R can do 1/22.5 of the work in 1 day.
In 3 days R and Q finish 3/8.5 + 3/22.5 of the work =24/51.
Amount of work left= 1-24/51 = 27/51.
Days required for P to finish the work= (27/51)/(1/17)
=(27/51)*17)= 9 days

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P.jagadeesh   7 years AGO

The efficiencies are p-1.5
Q-3
R-1
In 3days Q+R=12/25.5 th work
So the remaining 13.5/25.5 will be completed by p in 9 days

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Muraleeswaran   7 years AGO

to find the total works... and find induvual work, then calculate the day

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KRISHNAKUMAR   7 years AGO

p can do work per day=1/17
q can do work per day=1/8.5
r can do work per day=1/25.5
q and r do work in 3 days=3(1/8.5+1/25.5)=8/17
balance work=1-8/17=9/17
days need for p=(9/17)/(1/17)=9

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babita maurya   7 years AGO

p 1 day's work=1/17
similarly q i day's work=2/17,
and q=2/51
now, according to the question q and r work together for 3 days so, work completed by p and q in 3 days=3*2/17+3*2/51=24/51
hence remaining work that need to be completed by p is=1-24/51=27/51
now p can completes the work alone in 17 days.
so 27/51 work can be completed by p =27/51*17=9 days
so correct option is 1

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Ramkumar saini   7 years AGO

p=17=2x
q=8.5=x
r=25.5=3x
so we can find now

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Varun   7 years AGO

12/25.5 done by them, remaimning 13.5/25.5 he does in 9 days as he does 1/17 per day.

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VARUN   7 years AGO

-->Suppose P works for "p" hours in a day.Similarly Q for "q"
hours ,R for "r" hours.
-->Assume the works takes total "K" hours to complete.
-->By given condition
17p=8.5q=25.5r=K
3q+3r+pX=K
-->Solving above to equations for X
we get X=9

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Raji Baskar   7 years AGO

P's one day work => 1/17
Q's one day work => 1/8.5
R's one day work => 1/25.5

Q+R's 3 days work => 3*(1/8.5+1/25.5)=> 8/17
Remaining work =>1-8/17 =>9/17
If P's 1 day work is 1/17 then his x day work is 9/17,
=>x = 9/17*17/1 ==> 9days

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Mounika   7 years AGO

P's 1 day's work=1/17
Q's 1 day's work=1/8.5=2/17
R's 1 day's work=1/25.5=2/51
Q and R together 1day's work=2/17 +2/51 =8/51
then (Q+R)'s 3 days work=3*8/51=8/17
remaining work =1-8/17=9/17
9/17th remaining work is done by A in=17*9/17=9days

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jyothi chhabria   7 years AGO

let total unit of work is 102 units
p=1/17=6 units
q=12 units
r=4 u
q+r worked for 3 days I.e 3*(12+4)=48
remaining work is done by p =102-48=54
p can do in 1 day 6 units therefore 54 units can be done in 9 days

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NABA KISHORE PADHY   7 years AGO

1-3*(2/17+2/51)=1-24/51=27/51 part done by A
required day is 17*27/51=9 days

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meena   7 years AGO

find individual unit and find Q and r completion of work . total unit /p 's individual

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Deepak   7 years AGO

let total unit of work=25.5
eff P=1.5
eff Q=3
eff R=1
so eff Q and R =3+1=4 worked for 3 days so 4*3 unit is finished
left unit of work=25.5-12=13.5
required days by p =13.5/1.5 =9.0

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saikamaltejpavuluri   7 years AGO

LCM of 17,8.5 and 25.5 =51
P one day work=3
Q one day work=6
R one day work=2
Work done by Q and R in 3 days= 3*(6+2) i.e 24
Remaining work (27) will be completed by P in (27/3)=9 days

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