M4maths Previous Puzzles

05 Oct 2016 Cost price of the bicycle

A bicycle is sold at a gain of 27%. If it had been sold for Rs. 45 more, 31% would have been gained. The cost price of the bicycle is:

OPtion

1) Rs.1050
2) Rs.1200
3) Rs.1500
4) Rs.1800
5) Rs.1250
6) Rs.1125
7) Rs.1850
8) Rs.1525
9) Rs.1825
10)None of these

Solution

let the C.P. be Rs.x then,
131% of x - 127% of x = 45
=> 4% of x = 45
=> x = 45*100/4 = 1125.

Option 6)

Correct Option: 6 Best Solution (42)

Sudhakarshona   8 years AGO

From the given data, 4% gain is equal to Rs.45..
Therefore, 1% gain will be Rs.11.25..
Therefore, the cost price will be equal to 100*11.25=Rs.1125

Like? Yes (4) | No (3)  

Lakshmipathi   8 years AGO

31-27=4%
100%=100/4*45
=1125

Like? Yes (3) | No (2)  

Hemanathan.R.K   8 years AGO

The difference in profit is 4% which is because of raising Rs.45/-
So 4%(CP)=45
So, the real cost price will amount to 100%(45) which is Rs.1125/-

Like? Yes (2) | No (2)  

Lakshmi Narayanan.k   8 years AGO

gain%=(sp-cp)100/cp.
cp=100sp/127......(1)
if 45 is added to sp
131cp=100sp+4500.....(2)
by solving 2 eqns . we get cp=1125(opt 6)

Like? Yes (2) | No (2)  

G.S.Kantharao   8 years AGO

Difference of gain by raising Rs.45 = 31%-27%=4%
The cost price is= 45*(100/4) =Rs.1125
Right choice is 6.

Like? Yes (2) | No (2)  

shivani shukla   8 years AGO

in this question it is given that
31%-27%=45
100%=1125

Like? Yes (1) | No (1)  

manik garg   8 years AGO

let y be the cost price and x be the selling price
according to que(27/100)=(x=y)/y
and (31/100)=((x+45)-y)/y
solving we get cost price=y=1125

Like? Yes | No   

sanjeet kumar   8 years AGO

Since 4% of total is 45.
So cost would be 1125 that us 100%

Like? Yes (1) | No (1)  

Nishu Kumari   8 years AGO

Difference in assumption 4 out of 100 when profso in 1125 different 45 nd profit is 31%

Like? Yes | No   

Vishal Athwani   8 years AGO

27x/100 + 45 = 31x/100
27x + 4500 = 31x
x=1125

Like? Yes | No   

Priyanka devi E   8 years AGO

4%(31-27) of x=45;
(4/100)*x= 45;
x=(45*100)/4;
x=1125;
therefore the cost price is Rs.1125/-;
correct answer is option 6.

Like? Yes | No   

Dinesh Sharma   8 years AGO

difference between profits = 31% - 27% = 4%

4% of x = 45
0.04x=45
x = 45/0.04=1125

Like? Yes (1) | No (2)  

chetan   8 years AGO

1.27*x + 45 = 1.31*x
then x = 1125.
so option 6 is correct

Like? Yes (1) | No (2)  

nimisha gupta   8 years AGO

Let sp be x
When profit=27%
Then cp= (100*x)/127
When sp is 45 more the profit=31%
So now the cp=(100/131)*(x+45)
Solving the 2 equation we get x=127*45/4
Using this to solve cp,
We get cp=1125

Like? Yes (1) | No (2)  

shaik jasmin   8 years AGO

let cost price be x
its selling price=(127/100)*x
new selling price is 45 more than old selling price
i.e (131/100)*x=45+(127/100)*x
(131x)/100=(4500+127x)/100
131x=4500+127x
4x=4500
x=1125

Like? Yes | No (1)  

Surendra Naraga   8 years AGO

Let x be the cost of the bicycle and
y=the profit obtained when the bicycle sold at gain of 27%
y=x*27/100 -----(1)
When the bicycle sold at a gain of 31% then the profit is y+45
y+45=x*31/100 ------(2)
By solving eqn(1) and eqn(2)
we can get x as rs.1125/-

Like? Yes | No (2)  

Dinesh kumar sah   8 years AGO

Very simple,
Let the CP is Rs.100.
at1st it is sold at profit of 27?
then SP=Rs.127
if it is sold Rs.45 more then profit is 31? with original CP.
now SP=Rs.172
so difference in both SPs is profit of 4?.
Therefore 4? is 45
then 100%=(45/4)*100=Rs.1125 Ans
Hence option (6) is the right answer.

Like? Yes | No (2)  

gubba ranjith kumar   8 years AGO

X+45:X::131:127=> X=1425.75
Bicycle sold for 1425.75
Real price=1425.75*100/127 =1125

Like? Yes | No (2)  

prakash sharma   8 years AGO

let suppose we buy a bicycle in price x
then after sold at 27% gain
selling price=127*x/100
if we sold it in 45rupee more then it will bi 31% gain means
((127*x)/100)+45=((131*x)/100)
=>4x=4500
=>x=1125(option 6)

Like? Yes | No (2)  

Abhishek Divakar   8 years AGO

127% of x + 45 = 131% of x (where x is cost price)
>> 127x +4500 = 131x
x = 1125

Like? Yes | No (2)  

JAYA TRIPATHI   8 years AGO

let the cp of bicycle is x and sp be y
then
y = x+27x/100
y = 127x/100
now again,
y+45 = x+31x/100
y+45 = 131x/100
127x+45 = 131x/100
131x/100+127x/100 = 45
4x/100 = 45
x = 45*25
x = 1125
therefore the cp of bicycle is = 1125 , option(6).

Like? Yes | No (2)  

Ramsb   8 years AGO

131x/100 - 127x/100=45
4x=4500
x=1125

Like? Yes | No (2)  

khushbu   8 years AGO

1.27x=y
1.31x=y+45
where x =cost of bicycle
y=price when 27% gain

Like? Yes | No (2)  

MD TAHERA sultana   8 years AGO

difference between the percentages is eqaul to 45 .equate and solve them
131c.p/100-127c.p/100=45
c.p= 1125

Like? Yes | No (2)  

SHUBHAM GUPTA   8 years AGO

let selling price be SP and cost price be CP
the SP=(127/100)*CP .....(1)
SP+45=(131/100)*CP ....(2)
from (1) and (2)
SP=(127*45)/4.
then CP=1125.
correct option is 6) 1125

Like? Yes | No (2)  

Yahia boudah   8 years AGO

31-27= 4%
4 times 25 is equal to a hundred percent.
thus: 45x25= 1125
since 4% is Rs. 45
the price of this bicycle is Rs. 1125
(without considering the gain)

Like? Yes | No (2)  

SaiKumar   8 years AGO

let x be the cost of the cycle
then with the gain the cost let y
then
(y/x)*100=27
(y+45)*100=31
by solving the both equations you will get the value of x will be 1125

Like? Yes | No (2)  

KARTIK CHOUDHARY   8 years AGO

let x be the cost price then according to question
we have x+(27/100)*x+45=x+(31/100)*x
solve for x we get x=1125
so option 6

Like? Yes | No (2)  

Debashish Dash   8 years AGO

Let the cost price of the cycle is x.
(127x/100) + 45 = (131x/100)
By solving the above equation, x=1125/-

Like? Yes | No (2)  

Nandini JB   8 years AGO

let cost price be x
27% of x=.21
31% of x = 0.31
.31x -.27x = 45
x(.31-.27) = 45
x = 45/0.04
x=1125

Like? Yes | No (2)  

imran khan   8 years AGO

let original cost price is x
selling price is =127x/100
New selling price =131x/100
131x/100-127x/100=1125
option =6

Like? Yes | No (2)  

PINTU PRASAD GUPTA   8 years AGO

let Cost price cycle be x
therefor 131*x/100 - 127x/100 = 45
4*x= 45*100
x= 45*25=1125 ans
So, Option 6 is correct ans

Like? Yes | No (2)  

m@h€sh ®edd¥   8 years AGO

difference b/w the gain percentage is =45
31%-27%=45
4%=45
thrn 100% is 45*100/4
1125

Like? Yes | No (2)  

ankita sachdev   8 years AGO

letcp=x
sp=1.27x;
now sp=1.27x+45
therefore
%p=[(sp-cp)/cp]*100
31=[(1.27x+45-x)/x]*100
31x=27x+4500
4x=4500
x=1125
cp of bicyce=x=1125

Like? Yes | No (2)  

Rahul Ranjan   8 years AGO

Let assume c.p=100
If 27%gain then s.p=127
And if 31% gain then s.p=131
So difference in gain % = 31-27=4%
And gain amount =45

Using unitary method
If 4%------45
1%------45/4
100%---45*100/4=1125

C.p=Rs. 1125

Like? Yes | No (2)  

Nitish kesarwani   8 years AGO

let cp be x
sp=127x/100
now,more 45
sp=127x/100+45=131x/100
4x/100=45
x=45*25=1125(optoin 6)

Like? Yes | No (2)  

Sonu Kumar   8 years AGO

Here 31% - 27% = 4%.
the Value of 4% =45
so 1%= 11.25 So Cost Price is 1125
bicycle sold at a gain 27%= 1125+ 303.75 and at a gain of 31%= 1125+348.75 .
difference between 27% and 31% are Rs. 45 . So the cost price is Rs.1125.

Like? Yes | No (2)  

Akshay Pillai   8 years AGO

31%-27%=4%
4%=Rs 100
100%=Rs(45/4)*100
=1125

Like? Yes | No (2)  

sachin   8 years AGO

initial CP=100
first item SP=127
second item SP=131
4 unit =45
1unit =45/4
CP=45*100/4=1125

Like? Yes | No (2)  

Raj   8 years AGO

let CP=x
acc. to problem
SP=(1+(27/100))of CP................................(1)
SP+45=(1+(31/100))ofCP..............................(2)

Solving (1) and (2) we get
CP=1125

Like? Yes | No (2)  

sainath   8 years AGO

let initial s.p be x
so c.p.=(100/127)x...........(1)
and final s.p. =x + 45
then c.p. = (100/131)x+45.....(2)
solving above two equations we get x = s.p.= Rs.1433/-
and Finally substitute it into (1) eqn
we get c.p.=(100/127)1433
= 1125

Like? Yes | No (2)  

Nwobi Chibuzor    8 years AGO

let the cost price be z.
price at 27% gain is 1.27z
price at 31% gain is 1.31z
Therefore;
1.27z + 45 = 1.31z
The cost price of the bicycle, z is 1125

Like? Yes | No (3)  

04 Oct 2016 Coded language

If in coded language
DINESH = 12
GOPESH = 15
SANDEEP = 35
Then
ROHAN = ?

OPtion

1) 16
2) 32
3) 24
4) 31
5) 38
6) 26
7) 42
8) 28
9) 18
10)23

Solution

A(1), B(2), C(3), D(4), E(5), F(6), G(7), H(8), I(9), J(10), K(11), L(12), M(13), N(14), O(15), P(16), Q(17), R(18), S(19), T(20), U(21), V(22), W(23), X(24), Y(25), Z(26)
DINESH = (D+H) = 4+8 = 12 (First character + Last character)
GOPESH = (G+H) = 7+8 = 15
SANDEEP = (S+P) = 19+16 = 35
Similarly
ROHAN = (R+N) = 18+14 = 32

Option 2)

Correct Option: 2 Best Solution (48)

Lakshmipathi   8 years AGO

first letter + last letter ;
DINESH=D+H, 4+8=12
GOPESH=G+H , 7+8=15
Like wise

ROHAN=, R+N=18+14=32
ans=32
option (2):

Like? Yes (4) | No (2)  

Aakash Beniwal   8 years AGO

Use A=1,B=2,C=3, ... ETC
And add first and last alphabet like
DINESH > D=4,H = 8
4+8 = 12
SIMILARLY
ROHAN > R =18,N=14
18+14 =32 ANSWER

Like? Yes (1) | No (1)  

Lakshmi Narayanan.k   8 years AGO

DHINESH = (D+H) = 4+8 =12.
ROHAN = (R+N) = 18+14 =32 (OPT 2)

Like? Yes (1) | No (1)  

RISHABH   8 years AGO

In DINESH = D+H=12
Similarly, G+H=15
S+P=35
So, for ROHAN= R+N=32

Like? Yes | No   

saigoutham   8 years AGO

DINESH=d=4,h=12 d+h=4+8=12
gopesh=g=7,h=8 g+h=7+8=15
sandeep=s=19,p=16 s+p=35
rohan=r+n=32

Like? Yes | No   

Ankita Sharma   8 years AGO

the first and last alphabets of the names when added according to their numeral positions gives the result
D=4+H=8==12
G=7+H=8==15
S=19+P=16==35
similarly R=18+N=14==32

Like? Yes | No   

manik garg   8 years AGO

D+H=(4+8)=12
G+H=(7+8)=15
S+P=(19+16)=35
R+N=(18+14)=32

Like? Yes | No   

Nwobi Chibuzor    8 years AGO

According to the alphabet numbering A-Z = 1-26;
DINESH =D (4) + H (8) = 12
GOPESH =G (7) + H (8) = 15
SANDEEP = S (19) + P (16) = 35
ROHAN = R (18) + N (14) = 32

Like? Yes (1) | No (1)  

A.P.Singh   8 years AGO

ROHAN
R..18 the letter
N.. 14 th letter
18+14=32

Like? Yes (1) | No (1)  

renuka.o75@gmail.com    8 years AGO

in dinesh we add d+h=4+8=12
same way for gopesh and sandeep
ROHAN=R+N=18+14=32

Like? Yes | No   

Janki asati   8 years AGO

Dinesh=12=d+h
Gopesh=15=g+h
Sandeep=35=s+p
Rohan=32=r+n=18+14

Like? Yes | No   

Ajay Reddy Peddinti   8 years AGO

The main logic in this is adding the first and last alphabets number.
In DINESH
D=4 and H=8
4+8=12
Likewise in GOPESH, G(7)+H(8)=15 and
in SANDEEP, S(19)+P(16)=35.
So finally in ROHAN, R=18 and N=14
Therefore 18+14=32 is the answer

Like? Yes | No   

NAVEENKUMAR HV   8 years AGO

D=4
H=8
So D+H=12 similarly
R=18
N=14
so R+N=32

Like? Yes | No   

Abhinay   8 years AGO

The words are coded as the sum of their place for the first and last letter.
DINESH= D+H= 4+8=12
GOPESH= G+H=7+8=15
SANDEEP= S+P= 19+16=35

THEREFORE,
ROHAN=R+N= 18+14=32

Like? Yes | No   

Sowjanya a v   8 years AGO

A to Z are numbered 1 to 26.
ROHAN=First letter + last letter=R+H=18+14=32
Ans-32

Like? Yes | No   

jayasree   8 years AGO

Write alphabets numbering 1 to 26
For DINESH we get
4 9 14 5 19 8 first term n last number i.e 4 plus 8 equal to 12
For GOPESH we get
7 15 16 5 19 8 adding 7 and 8 we get 15
For SANDEEP we get
19 1 14 4 5 5 16 adding 19 and 16 we get 35
For ROHAN we get
18 15 8 1 14 adding 18 and 14 we get 32
So answer is 32

Like? Yes | No   

chaitanya muttineni   8 years AGO

DINESH=
FIRST LETTER D=4, LAST LETTER H=8,so,D+H=4+8=12
GOPESH=
G=7,H=8,SO,G+H=7+8=15
SIMILLARLY
ROHAN---->R=18,N=14 SO,R+N=32
SO,32(2)IS THE ANSWER

Like? Yes | No   

Sourav Mondal   8 years AGO

rank of 1st latter + rank of last latter
Dinesh= 4(d)+8(h)=15
Rohan= 18(r)+14(n)=32

Like? Yes | No   

priyanshuchhawchharia   8 years AGO

Sum of first and last word like as
D + H = 12
G + H = 15
S + P = 35
HENCE
R + N = 32

Like? Yes | No   

nimisha gupta   8 years AGO

A=1,B=2,C=3........and so on till Z=26
In DINESH,
D=4 & H=8
4+8=12
Similarly,
G+H=15
S+P=35
So in ROHAN
R+H=32

Like? Yes | No   

Anamika singh   8 years AGO

Dinesh=D=4 Nd H=8 so 4+8=12
Gopesh=G=7nd H=8so 7+8=15
Sandeep=S=19+ P=16=35
Rohan=R=18+N=14=32
option no.(2)👍

Like? Yes | No   

Sonu Kumar   8 years AGO

DINESH D=4 ,H=8, 4+8=12
GOPAL G=7,H=8 7+8=15
SANDEEP S=19,P=16, 19+16=35 THEN
ROHAN R=18,N=14, 18+14=32

Like? Yes | No   

Samjukta Das   8 years AGO

Here in DINESH, D+H=4+8=12
GOPESH, G+H=7+8=15
SANDEEP=S+P=19+16=35

So, ROHAN, R+N=18+14=32

Like? Yes | No   

gubba ranjith kumar   8 years AGO

DINESH---D+H=4+8=12
GOPESH--G+H=7+8=15
SANDEEP-S+P=19+16=35
lly ROHAN-R+N=18+14=32

Like? Yes | No (1)  

prakash sharma   8 years AGO

According to weighted code of these alphabets answer is addition of first and last character's weighted code
eg-DINESH=12
means Weighted code of ( D + H)
=4+8
and GOPESH=15
means Weghted code (G+H)
=7+8
so same as
ROHAN=(Weighted code of R) + (weighted code of N)
=>18+14
=>32(option 2)

Like? Yes | No (1)  

G.S.Kantharao   8 years AGO

The first and last letters sum is the given Numerical
D(4)INESH(8) = 4+8=12
G(7)OPESH(8) = 7+8=15
S(19)ANDEEP(16) = 19+16 =35
Then R(18)OHAN(14) = 18+14=32
Hence the right option is 2

Like? Yes (2) | No (3)  

SaiKumar   8 years AGO

convert last letter into numbers and add them you will get the answer as 32 for ROHAN
R=18 N=14 so the ans is 32

Like? Yes | No (1)  

Dinesh kumar sah   8 years AGO

Very simple,
To find the answer add 1st and last number of the corresponding alphabet.
ie DINESH=D+H=4+8=12
similarly,
GOPESH=G+H=7+8=15 and
SANDEEP=19+16=35
finally
ROHAN=R+N=18+14=32 Ans
Hence option (2) is the right answer.

Like? Yes | No (1)  

chetan   8 years AGO

D+H = 4+8 = 12
G+H = 7+8 = 15
S+P =19+16=35
R+N=18+14=32
so option 2 is correct.

Like? Yes | No (1)  

Aparna S   8 years AGO

D+H=> 4+8 => 12
G+H=>7+8=>15

So,
R+N=> 18+14 => 32

Like? Yes | No (1)  

JAYA TRIPATHI   8 years AGO

DINESH = 4+8 = 12 ( SUM 0F 1ST LEETER & LAST LETTER COAD ACCOURDING TO ALPHABETICAL SERIES)
GOPESH = 7+8 = 15
SANDEEP = 19+16 = 35

ROHAN = 18+14 = 32
OPTION (2).

Like? Yes | No (1)  

amit dixit   8 years AGO

D+H=12
G+H=15
S+P=35
SO ANS IS=
R+N=32

Like? Yes | No (1)  

Maha Ganesh   8 years AGO

first letter + last letter
R(18)+ N(14)
32

Like? Yes | No (1)  

Sudhakarshona   8 years AGO

Let A,B,C,D,... be coded as 1,2,3,4,...
In DINESH, the first letter D can be coded as 4 and the last letter H can be coded as 8, so the result is the sum of the code numbers of the first letter and the last letter of the word..

Similarly, In GOPESH, G-7 and H-8, it's summation is equal to 15..

Similarly, In SANDEEP, S-19 and P-16, it's summation is equal to 35..

Likewise, In ROHAN, the first letter R is coded as 18 and the last letter N is coded as 14, and hence it's summation will be equal to 32..

Like? Yes | No (1)  

SIVA KUMAR   8 years AGO

first +last =32 which means alaphet of 1st number and last alphabet number

Like? Yes | No (1)  

Dinesh Sharma   8 years AGO

Sum of First Character Value and Last Character Value i.e.

D+H=4+8=12
G+H=7+8=15
S+P=19+16=35

So, for ROHAN: R+N=18+14=32.

Like? Yes | No (1)  

ROZINA KHATUN   8 years AGO

adding the code of first and last letter I.e, R=18 and N=14,R+N=18+14=32

Like? Yes | No (1)  

Manisha gouri   8 years AGO

given that,
he added first and last alphabets
dinesh =12 = d+h=4+8=12
gopesh=>g+h=7+8=15
sandeep=s+p=19+16=35
then rohan=r+n=18+14=32
so, option 2 is ans

Like? Yes | No (1)  

Nitish kesarwani   8 years AGO

since A=1 To Z=26 so each Name written in question first and last sum=32 so option 2 is correct. such as Gopesh=15 ,g=7 & h=8 their sum=15 ,r=18 & n=14 sum=32

Like? Yes | No (1)  

pooja nagar   8 years AGO

because coded digit is sum of first and last latter of respected word
.

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Mukund Manohar   8 years AGO

DINESH = D+H = 4+8 = 12
GOPESH = G+H = 7+8 = 15
SANDEEP = S+P = 19+16 = 35
ROHAN = R+N = 18+14 = 32

Like? Yes | No (1)  

ADITI GOEL   8 years AGO

assign number to english alphabets from 1 t0 26.
LOGIC:sum of first and last letter's number is the answer.
then in DINESH,D+H i.e 4+8=12
in GOPESH, G+H i.e 7+8=15
in SANDEEP, S+P i.e 19+16=35
so, in ROHAN, R+N i.e 18+14=32

Like? Yes (1) | No (2)  

Ayyappa swamy   8 years AGO

Let us letters we coaded as A is ,B is 2.....Z is 26 like that
we can add first and last letters of numbers we can get the result

Like? Yes | No (1)  

gulshan   8 years AGO

DINESH =12 => D(4)+H(8) = 12
GOPESH = 15 => G(7)+H(8) = 15
SANDEEP = 35 => S(19)+P(16) =35
Then
ROHAN = ?(32) => R(18)+N(14)=32

Like? Yes | No (1)  

SATHEESH V   8 years AGO

(A,1),(B,2),(C,3),............(Y,25),(Z,26)

Sum of FIRST and LAST letter equivalent numbers is the solution

D+H=4+8=12
G+H=7+8=15
S+P=19+16=35
similarly
R+N=18+14=32

Like? Yes | No (1)  

shivani shukla   8 years AGO

A=1,B=2,C=3,D=4,E=5.F=6,G=7,H=8,I=9,J=10,K=11,L=12,M=13,N=14,O=15,P=16,Q=17,R=18,S=19,T=20,U=21,V=22,W=23,X=24,Y=25,Z=26
in the question it is taking the sum of first and last numeric of the alphabet i.e.
In DINESH=12
D=4 and H=8 so D+H=4+8=12
In GOPESH=15
G=7 and H=8
G+H=7+8=15
In SANDEEP=35
S=19 and P=16
S+P=19+16=35
So in ROHAN
R=18 and N=14
R+N=18+14=32
hence ROHAN=32

Like? Yes | No (1)  

ayushi mishra   8 years AGO

DINESH=D+H=4+8=12
GOPESH=G+H=7+8=15
SANDEEP=S+P=19+16=35
THEN
ROHAN=R+N=18+14=32

Like? Yes | No (1)  

Veerabhadra rao   8 years AGO

in dinesh d+h is 4+8=12
in gopesh g+h is 7+8=15
in sandeep s+p is 19+16=35
in rohan r+n is 18+14=32

Like? Yes | No (2)  

03 Oct 2016 Sequence

What is the missing number of the following sequence
5, 39, 8, 9, 40, 11, 13, 27, 14, 15, ......, 17

OPtion

1) 16
2) 28
3) 68
4) 42
5) 64
6) 36
7) 49
8) 52
9) 38
10)72

Solution

5, 39, 8, 9, 40, 11, 13, 27, 14, 15, ......, 17
(5, 39, 8), (9, 40, 11), (13, 27, 14), (15, ......, 17)
a b c
b = c^2 - a^2
8^2-5^2 = 64-25 = 39
11^2-9^2 = 121-81 = 40
14^2-13^2 = 196-169 = 27
Similarly
17^2-15^2 = 289-225 = 64

Correct Option: 5 Best Solution (15)

prakash sharma   8 years AGO

we divided this series into 3-3 pairs like
(5,39,8),(9,40,11),(13,27,14),(15,_,17)
we assume this series as (A,B,C)
then
B=(C-A)*(A+C)
eg-(5,39,8)
means 39=(8-5)*(5+8)
so same as in last pair of this series
B=(17-15)*(15+17)
B=64(Option 5 in answer)

Like? Yes (3) | No (1)  

chetan   8 years AGO

(8+5)*(8-5) =39
(11+9)*(11-9) =40
(14+13)*(14-13) =27
(17+15)*(17-15) =64
so option 5 is correct

Like? Yes (3) | No (2)  

G.S.Kantharao   8 years AGO

Series is5,39,8,940,11,13,27,14,15,...,17
(5,39,8),(9,40,11),(13,27,14),(15,......,17)
[5,{(8+5)(8-5)},8],[9,{(11+9)(11-9)},11],[13,{(14+13)(14-13)},14],[15,{(17+15)(17-15)},17]
Hence the number missing is 64 and right option is 5.

Like? Yes (2) | No (1)  

Lakshmi Narayanan.k   8 years AGO

(5,39,8), (9,40,11), (13,27,14), (15,-,17).
5+8=13*3=39.
11+9=20*2=40.
13+14=27*1=27.
15+17=32*2=64(OPT 5)

Like? Yes (1) | No   

rrr.j   8 years AGO

5+8=13 so 13*3=39; (since difference of 8-5=3 we multiply with 13)
9+11=20 so 20*2=40; (same 11-9=2 so multiply with 20)
13+14=27 so 27*1;
15+17=32 so 32*2=64;
option 5)64

Like? Yes | No   

Mukund Manohar   8 years AGO

(5,39,8),(9,40,11),(13,27,14),(15,_,17)
8+5=13 & 8-5=3
13*3=39
So,(5,39,8)
Applying the same logic,
11+9=20 & 11-9=2
20*2=40
So, (9,40,11)
14+13=27 & 14-13=1
27*1=27
So, (13,27,14)
17+15=32 & 17-15=2
32*2=64
So,(15,64,17)
64 is the missing number in the given sequence.

Like? Yes | No   

manik garg   8 years AGO

taking 3 number at a time we get
(5+8)*(8-5)=39
similarly for all and at last
(17+15)*(17-15)=64

Like? Yes | No   

Dinesh Sharma   8 years AGO

In general, take 3 consecutive number, says n1,n2,n3 and apply, (n3+n1)*(n3-n1)=n2

(8+5)*(8-5)=39
(11+9)*(11-9)=40
(14+13)*(14-13)=27
(17+15)*(17-15)=64

Like? Yes | No   

Siva   8 years AGO

(5+8)*(8-5)=39,
(9+11)*(11-9)=40,
(13+14)*(14-13)=27,
(15+17)*(17-15)=64.

Like? Yes | No   

gnanesh mannam   8 years AGO

8+5=13,8-5=3,,13×3=39,,
11+9=20,11-9=2,20×2=40
14+13=27,14-13=1,27×1=27,
17+15=32,17-15=2,32×2=64
answer is 64

Like? Yes | No   

Lalit Kumar   8 years AGO

(5+8)*3=39
(11+9)*2=40
(13+14)=27
(15+17)*2=64

Like? Yes | No   

nagarjuna   8 years AGO

5,39,8 :- 8+5=13 ,8-5=3 :- 13*3=39
9,40,11 :- 9+11=20 ,11-9=2 :- 20*2=40
......
15+17=32, 17-15=2 :- 32*2=64

Like? Yes | No   

imran khan   8 years AGO

5+8=13*3=39
9+11=20*2=40
13+14=27*1=27
similarly 15+17=32*2=64

Like? Yes | No   

SATHEESH V   8 years AGO

(5+8)*(8-5)=39
(9+11)*(11-9)=40
(13+14)*(14-13)=27
similarly
(15+17)*(17-15)=64

Like? Yes (1) | No (2)  

KARTIK CHOUDHARY   8 years AGO

here we take the 3 numbers series and
5+8=13 and (8-5)*13=39
similarly(17+15)=32 and (17-15)*32=64

Like? Yes | No (1)  

30 Sep 2016 Number of coin

A bag contains Rs.1800 in form of one rupee, 50-paise and 25 paise coins in the ratio 3:4:12. The number of 25 paise coin is:

OPtion

1) 600
2) 900
3) 1200
4) 1376
5) 2700
6) 2100
7) 3000
8) 675
9) 1500
10)None of these

Solution

Ratio of values of coin = 3/1: 4/2: 12/4 = 3: 2: 3.
Value of 25 paise coin = Rs.(1800*3/8) = Rs.675
Number of these coins = (675*4) = 2700

Option 5)

Correct Option: 5 Best Solution (54)

G.S.Kantharao   8 years AGO

Let 3x,4x,12x are the coins of 1Rs,50-paise,25-paise coins
3x*1+4x*1/2+12x*1/4 = 1800
3x+2x+3x =1800
8x=1800
x=225
25-paise coins=12*225=2700
Hence the right option is 5

Like? Yes (23) | No (2)  

gubba ranjith kumar   8 years AGO

1:0.5:0.25=3:4:12=3x:4x:12x
Let a be cost made using 0.25 rs
3(4a)+4(2a)+12a=1800
32a=1800
a=56.25
But no.of 0.25 coins=56.25*12*4=2700

Like? Yes (7) | No (2)  

Kunal nemu   8 years AGO

ratio of one rupee,50 paise,and 25 paise coins 3:4:12
= 100*3 : 4*50 : 12*25 =300 : 200 : 300 = 3 : 2 :3
sum of the ratio = 3+2+3 =8
number of one rupee coins = 3/8 *1800 = 675
number of 50 paise coins = (2/8 *1800)*2 =900
number of 25 paise coins = (3/8 * 1800)*4 = 2700

Like? Yes (4) | No   

JAYA TRIPATHI   8 years AGO

given ratio = 3:4:12
therefore, 3x+4x/2+12x/4 = 1800
12x+8x+12x = 1800*4
32x = 7200
x = 7200/32 =225
therefore the no. of 25 paise coin is = 225*12 = 2700

Like? Yes (5) | No (2)  

ANAND MOHAN RAI   8 years AGO

let the number of 100 paise,50 paise and 25 paise coins are 3x ,4x,12x respectively.
by condition,
3x*100+4x*50+12x*25=1800*100
x=225
hence number of 25 paise coins=12x=12*225=2700
option 5)

Like? Yes (2) | No   

sai sreenivas   8 years AGO

given ratio is 3:4:12.
let the number of coins be x
one rupee consists of 100p so (3x*100)/100
50p coins=(4x*50)/100
25p coins=(12*25x)/100
adding all of them
(3x*100)/100+(4x*50)/100+(12x*25)/100=1800
3x+2x+3x=1800
8x=1800
x=225
To get the number of 25p coins. 225*12=2700.

Like? Yes (1) | No   

MAHESHKUMAR KOTAMBRI   8 years AGO

let x
(300x)+(200x)+(300x)=180000
after solving, x=225
hence The no of 25 paise coins are 12*225=2700

Like? Yes (1) | No   

amit dixit   8 years AGO

3x+4x(.5)+12x(.25)=1800
8x=1800
x=225
225*12=2700

Like? Yes (1) | No   

dokku vasu yadav   8 years AGO

in this one we given a sum of 1800RS in the form of 1rs,50p,25p here and the ratio is 3:4:12 and we have convert these ratios into convertable ones and by means of it we get
1rs=100p*3=300&50p*4=200&25p*12=300
300:200:300
the is said to be of 3:2:3
we have to calucalte the number of coins in 25p so
3/8*1800=675
we have to find number of coins so 675*4=2700 coins paisa

Like? Yes (1) | No   

Nwobi Chibuzor    8 years AGO

For every 3 (1 rupee) there are 4 (50 paise coins) and 12 (25 paise coins) which sums up to 3rupees + 2rupees + 3rupees = 8 rupees. There are 1800 rupees in the bag so that means there are 1800/8=225 portions of 3 (1rupee), 4 (50 paise coins) and 12 (25 paise coins).
Therefore there are 225 * 12 =2700 (25 paise coins)

Like? Yes (1) | No   

Prashant Kumar   8 years AGO

Let x be the number of coins/rupees, then

[3*x+{4x*0.5}+{12x*0.25}]=1800

From here , we get x as 225,

so 1 rupee note = 3*225 = 675
50 paise coins = 4*225 = 900
25 paise coins = 12*225 = 2700
So, the correct answer is 5) 2700

Like? Yes | No   

Abhilash Reddy   8 years AGO

let's take individual coins as 3 rupee coins 4 50 MP coins 12 25 paise coins
total AMT leads to RS 8
1800/8=225
225*3=2700

Like? Yes (2) | No (2)  

Horendra Pal Singh   8 years AGO

x{3+(4/2)+(12/4)}=1800
x(3+2+3)=1800
8x=1800
x=225
225*12=2700

Like? Yes | No   

kondapuram tirumalreddy   8 years AGO

total amount=Rs.1800

then 3*Rs.1 + 4*Rs.0.50 + 12*Rs.0.25= Rs.8

no. of Rs0.25 coins are 1800*[12/8]=2700

Like? Yes | No   

PINTU PRASAD GUPTA   8 years AGO

let ration be x
there for one rupees number = 3*x
50 paisa number = 4*x
25 paisa number = 12*x
1*3*x + 1/2*(4*x)+1/4*(12*x) = 1800
8x =1800
x= 225
there for 25 paisa Number = 12*225= 2700
Option 5 will be correct

Like? Yes | No   

phanikumar   8 years AGO

given 1rupe=100pisa and 50 pisa and 25 pisa
the ratio given 3:4:12
so finla ration is 300:200:300
so final is 3:2:3

no of 25 pisa coins is (3/8)*1800
equals to 675

Like? Yes | No   

gyan   8 years AGO

3x+4x/2+12x/4=1800
=> x=225
no of 25 paise coins =x*12=225*12=2700

Like? Yes | No   

manik garg   8 years AGO

3X+(.5*4X)+(.25*12X)=1800
ON SOLVING X=225
so number of 25 coins=225*12=2700

Like? Yes | No   

Priyanka Tukaram Dudalkar   8 years AGO

Let the number of 1 Rupee(100 paise), 50 paise and 25 paise coin be 3x,4x,12x(3:4:12) respectively.
Then their sum = Rs. (100*3x)/100 + (50*4x)/100 + (25*12x)/100

=Rs. 8x
I.e. 8x=1800
X=225

No.of 25 paise=12x=12*225=2700

Like? Yes | No   

vanshika agarwal   8 years AGO

Let the ratio be 3x:4x:12x

Then 3x+(4x/2)+(12x/4)=1800
X=225
Hence 12x= 225*12=2700

Like? Yes | No   

SHALINI   8 years AGO

let x be one rs
then 3:4:12 is
3x+4x(.5)+2x(.25)=1800
8x=1800
=225
then 25 paise coins is 225*12=2700

Like? Yes | No   

Sachin Upreti   8 years AGO

The number of 25 paise coins is = 12*(1800/(3*1+4*0.5+12*0.25))
=2700

Like? Yes | No   

sachin   8 years AGO

(1800*100)/(100*3 + 50*4 + 25*12)= 1 unit
1 unit =225
no of 25 p coins= 12*225=2700

Like? Yes | No   

shivakumar   8 years AGO

X(3(1)+4(0.5)+12(0.25))=1800
X(8)=1800
X=225
so number of 25 paise coins is 12 X=12(225)=2700

Like? Yes | No   

shunpriya   8 years AGO

3x+4x*1/2+12x*1/4=1800
8x=1800
x=225
12x=2700

Like? Yes | No   

Dinesh   8 years AGO

3x+1/2*4x+1/4*12x=1800
x=225
12*225=2700

Like? Yes | No   

naveen singh   8 years AGO

2700*.25=675
2700 is 12th portion so 1 portion is 225
acc 3 portion is 675 and 4th portion is 900
675 + 675*1+900*.5=1800

Like? Yes | No   

sathish   8 years AGO

For example take 2700/12=225, 225 is single a ratio
1 ruppe cion 3*225=675*1 = 675
50 paise cion 4*225=900*0.5 = 450
25 paise cion 12*225=2700*0.25 = 675
Total = 1800

Like? Yes | No   

sanjeet kumar   8 years AGO

Let the coins of 1 rupee , 50 paise and 25 paise be 3x ,4x and25x.
Then 3x + 2x + 3x =1800
x=225
so number of 25 paise coin is 2700

Like? Yes | No   

Kandhimalla   8 years AGO

3(x)+(0.5*4)x+(0.25*12)x=1800
8x=1800
X=225
25 paise coins=225*12=2700

Like? Yes | No   

Vidyalakshmi G   8 years AGO

3x+4x+12x=1800, as 50p n 25p the eqn becomes 3x+2x+3x=1800, x = 225, 12x = 2700

Like? Yes | No   

anupama   8 years AGO

from the given data
3*1x+4*1/2x+12*1/4x=1800
x=225
to find 25 paise
x=225*12=2700

Like? Yes | No   

SAPNA KUMARI   8 years AGO

3x+1/2*4x+1/4*12x=1800
x=225
then the no of coin is 25 paise is
12*225=2700

Like? Yes | No   

Raju Sampath   8 years AGO

here convert the ratio into one rupee coins
i.e, 3:2:3
here total parts are 8 .
1800/8 is equal to 225 ..
to get 25 paise count multiply 225 with 4 will get 2700

Like? Yes | No   

shivani shanishchara   8 years AGO

3x+(4x/2)+(12x/4)=1800
evaluate x
x=225
for 25 paisa(3:4:12)
12*225=2700 which is the number of 25 paisa coin

Like? Yes | No   

JENSI KARUNYA.S   8 years AGO

the coins in the ratio is 3:4:12
(3*100)X+(4*50)X+(12*25)X=180000
(300+200+300)X=180000
X=225
the value of the coins, Re.1=675, 50 paise=450, 25 paise=675
number of 25 paise coins =67500/25=2700
option 5) 2700 is the correct answer.

Like? Yes | No   

ALOK RANJAN DAS   8 years AGO

the ratio of coins is 3:4:12
therefore 3 coins of 1=Rs3
4 coins of .50=rs2
12 coins of .25=rs3
total amount with the ratio=8
1800/8=225
therefore225 sets of coins with ratio of 3:4:12 are present
from here we can calculate no of 25 paise coins will be 225*12=2700

Like? Yes | No   

gobinath   8 years AGO

3x+4#.5x+12#.25x =1800
3+2+3=8. 1800/8=225
225#12=2700

Like? Yes | No   

KEERTHI SUDHA JAVVAJI   8 years AGO

3x(1)+4x(1/2)+12x(1//4)=1800
=>x=450/2
We have to find 12x
=> 12x=12*450/2=2700

Like? Yes | No   

Adarsh dubey    8 years AGO

3:4:12
by using same no. multiplication in all the tree numbers , we get
675 coins of 1 rs
900 coins of 50 paise
2700 coins of 25 paise

Like? Yes | No   

arijit   8 years AGO

3x+ 1/2 *4x+ 1/4*12x = 1800
x=225
no. of 25 paisa coin = 12*225= 2700

Like? Yes | No   

Chitrala Venkatesh Babu   8 years AGO

given 3x+(4/2)x+(12/4)x=1800
=>8x=1800
=>x=225
therfore no.of 25paise coins is 12*x=12*225=2700

Like? Yes | No   

Rohit   8 years AGO

2700 is the answer.

explanation:
1rupee coin : 50p coin : 25paise coin=3:4:12
means in terms of money
there are 3x*1 rupee=3xrupee
4x*50paisa=2x rupee
12x*25paisa=3x rupee
so the ratio is 3:2:3
now combining of 3x,2x &3x is 1800.
So,8x=1800
=>x=1800/8
=225
so 3x=225*3=675
so 675 rs are in terms of 25 paise coins.
4 coins make 1 rupee
so,4*675 coins make 675 rupee
so answer is 675*4=2700.

Alternative way:
so 25 paisa contains 12x coins
=12*225
=2700

Like? Yes | No   

Rupesh verma   8 years AGO

total rs=1800
ratio of 25 p is 12x
rupees ratio of 25p is 3.(12/4)
so total sum of ratio =3+2+3=8.
so 1800/8=2.25
so total sum of rupees of 25p is=2.25*3=6.75
so total coins of 25p is=6.75*4=2700

Like? Yes | No   

sainath   8 years AGO

let us take 3x,4x,12x then
((100*3x)+(50*4x)+(25*12x))/100
=8x/100
x=225.so 12x=12*225=2700

Like? Yes | No   

Dileepkumar Palli   8 years AGO

3x(1)+4x(1/2)+12x(1/4)=1800
X=225
number of 25 paise coins = 225×12=2700

Like? Yes | No   

komal kujur   8 years AGO

3x/1+4x/2+12x/4=1800
=>x=225
no 25 paise coin=12*225=2700

Like? Yes | No   

Snehasish Debnath   8 years AGO

let, the no. of one rupee(100 paise), 50 paise, 25 paise are 3x, 4x, 12x respectively.
Now their value will be 300x paise, 200x paise, 300x paise respectively. [as 12x coins of 25 paise values = (25 * 12x) = 300x paise]
now total value = (300x + 200x + 300x) = 800x paise
now, 800x = 1800 * 100 [as 1800 rupee = 1800*100 paise]
or, x = 225
so, the no. of coins of 25 is = 12x = 12*225 = 2700 (Ans.)

Like? Yes | No   

Samjukta Das   8 years AGO

3x+4x*0.5+12x*0.25=1800
3x+2x+3x=1800
x=225

225*12=2700

Like? Yes | No   

SATHEESH V   8 years AGO

19........8
?.........1800

1800*19/8=4275

4275*12/19=2700

Like? Yes | No (1)  

Abhishek Dubey   8 years AGO

3x+(4/2)x+(12/4)x=1800 as 2 50 paise coin is 1 rs and 4 25 paise coin is 1 rs and total amount is 1800

Like? Yes (1) | No (2)  

MD TAHERA sultana   8 years AGO

converting coins into rupee
1 rupee -3 rs 50 paise - 2 rs
25paise -3 rs total 8rs in rupees 25 paise coins are 12 so in total 1800 it is 2700

Like? Yes | No (1)  

KARTIK CHOUDHARY   8 years AGO

3x+(0.5*4x)+(0.25*12x)=1800
on solving x we get
x=225
so number of 25 paise coins is 225*12=2700

Like? Yes | No (1)  

manish sharma   8 years AGO

Given Ratio for total 1800rs is
rs 1 : 50ps : 25ps
3 : 4 : 12
if there are 2700 coins of 25ps than this will fulfill above cnditiion

than the total no of coins for respective amunt will be
675: 900: 2700
1rs coin 675 = 675
50 ps coin 900 = 0.50*900=450
25ps coin 2700 = 0.25*2700= 675
total = 675 + 450 +675 = 1800
So number of 25ps coins will be 2700.

Like? Yes | No (1)  

29 Sep 2016 Average Weight

The average weight of 19 students was 50 kg. By the admission of a new student the average weight is reduced to 49.6 kg. The weight of the new student is:

OPtion

1) 48.8kg
2) 40.8kg
3) 40.6kg
4) 42.2kg
5) 44.8kg
6) 46.6kg
7) 42kg
8) 48kg
9) 44kg
10)None of these

Solution

Weight of the new student is = (20*49.6-19*50)
= 42kg.

Option 7)

Correct Option: 7 Best Solution (32)

phanikumar   8 years AGO

x/19=50 and y/20=49.6
x=950 and y=992
so new age is y-x=992-950=42

Like? Yes (4) | No (2)  

Lakshmi Narayanan.k   8 years AGO

the avg of 19 students is 50kg.
sum 0f 19 students=19*50=950.
sum of 20 students=20*49.6=992.(992-950=42) opt 7

Like? Yes (3) | No (1)  

dokku vasu yadav   8 years AGO

intially the given average of 19 students is 50 by means of calculating the total weight of all students is 950 and by means of adding one extra student the average to 49.6
adding the extra weight of 42 kg to the 950 we get
992/20=49.6

Like? Yes (1) | No   

Nrupesh A N   8 years AGO

total weight of 20 people is 992 .and total weight of 19 people 950..so 992-950=42

Like? Yes (1) | No   

shipra bharti   8 years AGO

sum of weight of 19 student is 950kg
and sum of 20 student is 992kg
so now weight of new student is 992-950

Like? Yes (1) | No   

ANAND MOHAN RAI   8 years AGO

weight of new student=49.6*20-50*19=42 kg
hence option 7)

Like? Yes | No   

Sandip Das   8 years AGO

weight of new student=(total weight of 20 student)-(total weight of 19 student)
=(20*49.6)-(19*50)
=992-950
=42 (option 7)

Like? Yes (1) | No (1)  

Y M ANIEF   8 years AGO

Total weight of 19 students=19*50
950
Including 20th student=20*49.6=992
20 students - 19 students weight=new student weight
992-950=42

Like? Yes | No   

Dinesh kumar sah   8 years AGO

Sum of weight of 19 students=950 kg
Let the age of new student is x kg.
then according to the question,
(950+x)/20=49.6
so x=42kg Ans
Hence option (7) is the right answer.

Like? Yes (1) | No (1)  

Vignesh.S   8 years AGO

average weight of 19 student is 50 therefore the sum of weight of 50 student is 950

now (950+x)/20 =49.6

by solving this we get x as 42

therefor the answer is 42

option(7)

Like? Yes | No   

RAMPALA JYOTHI   8 years AGO

let weight of 19 students is x;
and weight of new student is y;
avg weight of 19 students (x/19)=50;
then x=950;
with new student avg weight is
((x+y)/20)=49.6;
y=(49.6*20)-(950);
y=42.00.
so weight of new student is 42kg.

Like? Yes | No   

vasantha kumar kalli   8 years AGO

total weight=19*50=950.
total weight after adding person=20*49.6=992
so the difference =992-950=42

Like? Yes | No   

Dinesh Sharma   8 years AGO

let total weight of 19 students = x
Therefore, x/19 = 50 i.e. x= 950

Now one more student take admission, it means total students = 20

now assume total weight of 20 students = y
so, y/20 = 49.6 i.e. y = 992

Therefore weight of new student = 992 - 950 = 42 kg , so option(7) is correct answer.

Like? Yes | No   

naveenkumar   8 years AGO

19*50=950
49.6*50=992
992-950=42
ans=42

Like? Yes | No   

amit dixit   8 years AGO

50*19=950
49.6*20=992

992-950=42

Like? Yes | No   

Anamika singh   8 years AGO

reduce in average=50-49.6=0.4kg
reduce in average total=20×0.4=8kg
so the weight of new student=50-8=42kg👍

Like? Yes | No   

Govind Sharma   8 years AGO

old total weight = 19*avg.
old total weight = 19*50
old total weight = 950

new total weight = 20*49.6
new total weight = 992kg

therefore weight of newly added guy is = 992-950
= 42kg

Like? Yes | No   

Uneet Kumar Singh   8 years AGO

Total weight of the students = 19*50
let weight of the new student be x then new average is (19*50 + x)/20
Equating it to to 49.6 and solving it for x we get x=42

Like? Yes | No   

vanshika agarwal   8 years AGO

Let the sum of wieght of 19 people is X kg then
X/19=50
X=950kg

Now,
Let the weight of next added person be Y kg then,

(X+Y)/20 = 49.6

Hence by putting X= 950
Y= 42 kg..which is the required answer..

Like? Yes | No   

sowmya   8 years AGO

avg(19 students)= 50 kgs
sum(19 students)= 50*19= 950 kgs
avg(20 students)= 49.6kgs
sum of 20 students= 992kgs
weight of new student= sum of 20 students- sum of 19 students
=992-950
=42 kgs

Like? Yes | No   

SAI VISWANADH KINTALI   8 years AGO

(a1+a2+......+a19)/19=50
a1+a2+....+a19=50*19=950 -------> eq1

(a1+a2+a3+........+a19+a20)/20=49.6
a1+a2+.....+a19+a20=49.6*20=992
950+a20=992 (from eq1)
a20=992-950=42

Like? Yes | No   

sai sreenivas   8 years AGO

The total weight of 19 students is 19*50=950
The weight of 20 students after admission of a new student is 20*49.6=992.
The weight of the new student is 992-950=42.

Like? Yes | No   

gulshan   8 years AGO

(19*50+x)/20=49.6
(950+x)/20=49.6
950+x=49.6*20
950+x=992
x=42

Like? Yes | No   

Samjukta Das   8 years AGO

Sum of ages of 19 students= 50*19=950
Now there are 20 students after adding a new student

Now 20*49.6=992

Weight of student added=992-950=42kg

Like? Yes | No   

NATARAJ   8 years AGO

the total of 19 students is 950 and next after new one enter the total of 20 students is 992 so the diff is the new student weight. the weight of new student is 42.

Like? Yes | No   

Nwobi Chibuzor    8 years AGO

Let the sum of the weight of all students be Z.
Therefore, Z÷19=50 gives Z=950.
Let the weight of the new student be Y.
Therefore, (950+Y)÷20=49.6 which makes Y=42

Like? Yes | No   

RIYA GUPTA   8 years AGO

Average weight*Number of students=50*19=950
New average weight*Number of students=49.6*20=992.0
Now 992-950=42kg
Weight of new student=42kg

Like? Yes | No   

Sahil dandawate   8 years AGO

(W1+w2+.....w19)/19=50
W1+w2+.....w19=950
Now , (w1+w2+w3+.....w20)/20=49.6
W20 =492-450=42

Like? Yes | No   

Divya   8 years AGO

sum of weights of students/no. of students=average weight
=>sum of weights of 19 students= 19*50
=950.
now,
let the weight of the new student be x.
=>(950+x)/20=49.6
=>950+x=992
=>x=42kg.

Like? Yes | No   

Divya   8 years AGO

first taken tha avg of 19 mem
x/19=50=>950
next avg of 20 mem
x/20=49.6=>992
age of new student=992-950=42

Like? Yes | No   

Siva   8 years AGO

Total weight of 19 students is 950, after add a student total weight is 992, so the weight of added student is 42

Like? Yes | No (1)  

JAYA TRIPATHI   8 years AGO

let the weight of 20 students is x1, x2, x3........x20
now,given that
x1+x2+x3+......+x19/19 = 50kg
=> x1+x2+x3+....+x19 = 50*19
= x1+x2+x3+....+x19 = 950kg
again, given
x1+x2+x3+.....+x20/20 = 49.6kg
=> x1+x2+x3+.....+x20 = 49.6*20
=x1+x2+x3+.....+x20 = 992kg
=950+x20 = 992
=> x20 = 992-950= 42kg
therefore the weight of new student is 42kg.

Like? Yes | No (1)  

28 Sep 2016 Interest

A sum of Rs.2500 is lent out in two parts one at 12% and another one at 12.5%. If the total annual income is Rs.306, the money lent at 12% is

OPtion

1) Rs.1200
2) Rs.2400
3) Rs.1300
4) Rs.1700
5) Rs.1240
6) Rs.2600
7) Rs.1340
8) Rs.2480
9) Rs.2680
10)None of these

Solution

Let the money lent at 12.5% be x
Then, money at 12.5% = 2500-x
=> x*12*1/100 + (2500-x) * 25/2 * 1/100 = 306
or
24x + 25(2500-x) = 61200
or x = 62500 - 61200 = 1300
Option 3)

Correct Option: 3 Best Solution (10)

JAYA TRIPATHI   8 years AGO

let the amount lent out on 12% be x then,
the amount lent out on 12.5% be (2500-x)
therefore,
12x/100+(2500-x)*12.5/100 = 306
12x+31250-12.5x = 30600
0.5x = 31250-30600
0.5x = 650
=> x = 6500/5 = 1300
therefore, the amount lent out on 12% is 1300.

Like? Yes (19) | No (6)  

Lakshmi Narayanan.k   8 years AGO

x+y= 2500
12x+12.5y= 30600.by solving these qns we get x=1300,y=1200.

Like? Yes (12) | No (8)  

Pooja Priti   8 years AGO

let the sum for 12% be x,then for 12.5% be 2500-x.
so as we know si=(p*r*t)/100
then
(x*12*1)/100+((2500-x)*12.5*1)/100=306
On solving this equation we get x=1300
So the amount for 12% is Rs.1300

Like? Yes (4) | No (2)  

ishaan khan   8 years AGO

let amnt at 12% be x then amnt at 12.5% be 2500-x
hence
(x*12*1/100)+((2500-x)*12.5*1/100)=306
on solving it
we get x=1300

Like? Yes (5) | No (3)  

Siva   8 years AGO

12x/100+(2500-x)×12.5/100
=306
-.5x+31250=30600
.5x= 650
x = 1300.

Like? Yes (3) | No (3)  

Ravi kumar   8 years AGO

Taking the amount at 12% as (x)
amount at 12.5% as (2500-x)

now....

(x*12*1)/100 + ((2500-x)*12.5*1)/100 = 306

12x + 31250 - 12.5x = 30600

x = 1300

answer = 1300 (3)

Like? Yes (3) | No (3)  

sai sreenivas   8 years AGO

let the two amounts be x and 2500-x.
let the r.o.i on x is 12 and the r.o.i on 2500-x is 12.5.
(x*12*1)/100+((2500-x)*12.5*1)/100=306.
on solving we get 1300 for the value of x which is the r.o.i of 12%.

Like? Yes (1) | No (2)  

Dinesh kumar sah   8 years AGO

Very simple,
Let the money lent on 12? is Rs.x.
Then according to the question,
(x*12*1)/100 +((2500-x)*12.5*1)/100=306
if we solve the above expression
then we will get x=Rs.1300 Ans
Hence option (3) is the right answer.

Like? Yes (1) | No (4)  

shivani shukla   8 years AGO

let the money which 12% have to lent will be=
[306-1200*(0.125)]/0.12=1300

Like? Yes (2) | No (5)  

venkatesh   8 years AGO

P1+P2=2500
12P1+12.5P2=30000
IF SOLVE IT THEN GET
P1=1300;
ANS P1=1300

Like? Yes (2) | No (6)  

27 Sep 2016 Coded language

If in coded language
RAMESH = 235
SHIVAM = 363
MANISH = 291
Then
SANJAY = ?

OPtion

1) 184
2) 426
3) 192
4) 356
5) 426
6) 291
7) 176
8) 364
9) 375
10)198

Solution

A(1), B(2), C(3), D(4), E(5), F(6), G(7), H(8), I(9), J(10), K(11), L(12), M(13), N(14), O(15), P(16), Q(17), R(18), S(19), T(20), U(21), V(22), W(23), X(24), Y(25), Z(26)
RAMESH = R*A + M*E + S*H = 18*1 + 13*5 + 19*8 = 18+65+152 = 235
SHIVAM = S*H + I*V + A*M = 19*8 + 9*22 + 1*13 = 152+198+13 = 363
MANISH = M*A + N*I + S*H = 13*1 + 14*9 + 19*8 = 13+126+152 = 291
Similarly
SANJAY = S*A + N*J + A*Y = 19*1 + 14*10 + 1*25 = 19+140+25 = 184

Option 1)

Correct Option: 1 Best Solution (16)

nimisha gupta   8 years AGO

S=19
A=1
N=14
J=10
A=1
And y=25
So 19*1+14*10+1*25
=184

Like? Yes (1) | No   

manish sharma   8 years AGO

ramesh =(18*1)+(13*5)+(19*8)=235
shiVam = (19*8)+(9*22)+(1*13)=363
manish = (13*1)+(14*9)+(19*8)= 291
sanjay= (19*1)+(14*10)+(1*25)=184

Like? Yes (1) | No   

prakash sharma   8 years AGO

according to Weighted code like
RAMESH (18,1,13,5,19,8)=>18*1+13*5+19*8=235
so same as
SANJAY(19,1,14,10,1,25)=>19*1+14*10+1*25=184(option 1)

Like? Yes | No   

Siva   8 years AGO

18×1+13×5+19×8=235,
19×8+9×22+1×13=363,
13×1+14×9+19×8=291,
As like this
19×1+14×10+1×25=184.

Like? Yes | No   

Geetha   8 years AGO

A(1)B(2)....Z(26).
S(19)A(1)N(14)J(10)A(1)Y(25)
Split into three sets,perform mulitipication with in the set and then perform addition with the sets
SA,NJ,AY
(19*1)+(14*10)+(1*25)=184

Like? Yes | No   

MAHESHKUMAR KOTAMBRI   8 years AGO

s=19, a=1,n=14,j=10,y=25
SA+Nj+ay=19+140+25=184

Like? Yes | No   

y.nikhil   8 years AGO

s=19, a=1, n=14, j=10, a=1, y=25
s*a=19*1=19
n*j=14*10=140
a*y=1*25
19+140+25=184

Like? Yes | No   

Kunal nemu   8 years AGO

we get in the position of word...
RAMESH = 18*1+13*5+19*8 =235
same MANISH =13*1+14*9+19*8=291
same SANJAY = 19*1+14*10+1*25= 184
so the answer is 184

Like? Yes | No   

dattatrayareddy   8 years AGO

first to multiply and next two multiply and next two multiply and add the three values we get the answr

Like? Yes | No   

rajshree dadhich   8 years AGO

consecutive multiplication of the numbers and then add all the numbers

Like? Yes | No   

chary   8 years AGO

Multiply each two letter"s place values and add them to get the answer

Like? Yes | No   

Akhila pathivada   8 years AGO

RAJESH= 18*1+13*5+9*8
=235
Likewise
SANJAY=19*1+14*10+1*25
=184

Like? Yes | No   

SATHEESH V   8 years AGO

(18×1)+(13×5)+(19×8)=235
(19×8)+(9×22)+(1×13)=363
Similarly
(19×1)+(14×10)+(1×25)=184

Like? Yes | No   

Satishbabu   8 years AGO

RAMESH -> 18 1 13 5 19 8 -> 18*1+13*5+19*8 = 235
SHIVAM -> 19 8 9 22 1 13 -> 19*8+9*22+1*13 = 363
MANISH -> 13 1 14 9 19 8 -> 13*1+14*9+19*8 = 291

=> SANJAY = 19 1 14 10 1 25 -> 19*1+14*10+1*25 = 184

Like? Yes | No   

Aakash Beniwal   8 years AGO

logic
take A=1, B=2 ,C=3, ... R=18,S=19, ... SIMILARLY
for RAMESH
18, 1 , 13 ,5 , 19 ,8
take two at once and multiple them and add all result like:>
(18 *1) + (13*5) +(19*8) = 18+65+152 = 235
similarly solve for SANJAY
my answer is 184

Like? Yes | No (1)  

Kshitij   8 years AGO

RAMESH = (18*1)+(13*5)+(19*8) = 235
SHIVAM = (19*8)+(9*22)+(1*13) = 363
MANISH = (13*1)+(14*9)+(19*8) = 291
So, SANJAY = (19*1)+(14*10)+(1*25) = 184

Like? Yes | No (1)  

26 Sep 2016 Sequence

What is the next number of the following sequence
246, 20, 689, 66, 479, 55, 369, ....

OPtion

1) 20
2) 64
3) 66
4) 92
5) 55
6) 85
7) 45
8) 74
9) 36
10)35

Solution

(246, 20), (689, 66), (479, 55), (369, ....)
abc = (a*b + b*c)-(a*c)
246 = (2*4 + 4*6)-(2*6) = (8+24)-12 = 32-12 = 20
689 = (6*8 + 8*9)-(6*9) = (48+72)-54 = 120-54 = 66
479 = (4*7 + 7*9)-(4*9) = (28+63)-36 = 91-36 = 55
Similarly
369 = (3*6 + 6*9)-(3*9) = (18+54)-27 = 72-27 = 45

Option 7)

Correct Option: 7 Best Solution (3)

Ankit Sharma   8 years AGO

Take pairs... i.e
246,20 689,66 479,55
In every pair..... Let it be abc
So 2nd term is as
b*c-a*(c-b)
Therefore it will be
6*9-3(9-6)=54-9= 45

Like? Yes (2) | No (1)  

Lakshmi Narayanan.k   8 years AGO

246,20,689,66,479,55,369,_
246=2*4+(4-2)6=20.
689=6*8+(8-6)9=66.
479=4*7+(7-4)9=55.
369=3*6+(6-3)9=45. (opt 7)

Like? Yes (4) | No (4)  

gulshan   8 years AGO

246= 2*6+4+4=20
689 =6*9+8+4=66
479 =4*9+7+12=55
369 = 3*9+6+12=45
that my ans is 45

Like? Yes (3) | No (5)  

23 Sep 2016 Amount of the bill

The true discount on a bill due 10 months hence at 6% per annum is Rs.26.25. Find the amount of the bill.

OPtion

1) Rs. 750.25
2) Rs. 855.5
3) Rs. 1250.25
4) Rs. 1150.25
5) Rs. 551.25
6) Rs. 645.25
7) Rs. 945.5
8) Rs. 1653.25
9) Rs. 1102.5
10)None of these

Solution

S.I. on Rs.100 for 10 months at 6% per annum
= Rs.(1000 * 10/12 * 6/100) = Rs.5
=> Amount = Rs.(100+5) = Rs.105
So, T.D. = Rs.(105-100) = Rs.5
If T.D. is Rs.5, sum due = Rs.105
If T.D. is Rs.26.25, sum due = Rs.(105/5 * 26.25) = Rs.551.25
Option 5)

Correct Option: 5 Best Solution (15)

Raju Sampath   8 years AGO

given that for 12 months discount is 6%,
hence for 10 months the discount is 5%.
Rs.26.25 is 5% of bill then , amount is (26.25*100/5)=525
total bill is 525+26.25=551.25

Like? Yes (25) | No (6)  

Indhrajeeth   8 years AGO

S.I=P*N*R/100
26.25=P*(10/12)*(6/100)
P= Rs.525
A= P+S.I = 525+26.25 = 551.25

Like? Yes (12) | No (5)  

karthi   8 years AGO

Given,R=6%,T=10/12yrs,TD=Rs.26.25,A=?
TD=A*R*T/100+R*T
26.25=A*6*10/12/100+(6*10/12)
26.25=A*5/105
A=>Rs.551.25

option 5)

Like? Yes (1) | No   

PINTU PRASAD GUPTA   8 years AGO

x be the amount bill after discount
x*10/12*6/100 = 26.25
x= 525
hence actual bill = 525+26.25= 551.25 ans
Hence option is 5.

Like? Yes (4) | No (4)  

ramyaraj   8 years AGO

TD= (amount�R�T)/(100+(R�T)
26.25=(x�6�(10/12)) / (100+(6�(10/12)))
5x=26.25(105)
x=551.25

Like? Yes | No   

Abhash Kumar   8 years AGO

SI on rupees 100 for 10 months at 6% per annum = (100 x 10 x 6)/(12 x 100) = Rs. 5.
∴ Amount = 100 + 5 = 105.
So true discount = 105 - 100 = Rs. 5

So If TD = 5 then sum due = 100 
So, when TD = 26.25 the sum = (100 x 26.25) / 5 = Rs. 551.25

Like? Yes (2) | No (3)  

Maholleshwari Nekkanti   8 years AGO

Truediscount=PTR/100;
26.25=P*10*6/(12*100);
p=525
amount=P+I
amount=26.25+525=551.25

Like? Yes (2) | No (3)  

phanikumar   8 years AGO

si on rupees 100 for 10 months at 6% for annum =(100*10/12*6/100)=5
amount =100+5=105
iff TD is 5 then sum is 100
so TD OF 26.25 is=(100*26.25)/5=551.25

Like? Yes (2) | No (3)  

sanjeev   8 years AGO

let amount be Rs X.

TRUE DISCOUNT = (X*RATE*TIME)/(100+(RATE*TIME))

26.25 = (X*6*(10/12)) / (100+(6*10/12))
After solving you will get
26.25*105 = 5X

X= 551.25Rs

Like? Yes (2) | No (3)  

Kshitij   8 years AGO

12 months or 1year discount = 6%
So, 10 months discount = 6/12*10 % = 5%
Now, 10 months dicount on Rs x = Rs 26.25
or 5/100* x =26.25
or x = 26.25*20 = 525
so total bill = bill + discount = 525 + 26.25 = Rs 551.25

Like? Yes (2) | No (4)  

Dinesh kumar sah   8 years AGO

Simply, given that T.D=Rs. 26.25,T=10 months=5/6 year,R=6? then Amount=?

We know that
T.D=(Amount �R�T)/(100+R�T)
or,Amount=T.D�(100+R�T)/R�T

if we put all values and then solve then we will get Amount=Rs. 551.25 Ans
Hence option (5) is the right answer.

Like? Yes (1) | No (4)  

Anamika singh   8 years AGO

Amount=100+5=105
true discount=105-100=5
if TD=5then sum due =100
TD=26.25 then sum=(100×26.25)/5
=551.25answer👍

Like? Yes (1) | No (4)  

Sandip Das   8 years AGO

TD=26.25
r=6%
t=10/12
=5/6
TD=(amt*6*(5/6))/(100+(6*(5/6))
amt=(26.25*105)/5
=551.25 (option 5)

Like? Yes | No (4)  

imran khan   8 years AGO

let present worth be x
SI of X × 6% × 10 months= 26.25
X * 6/100 × 10/12 = 26.25
X =525
AMOUNT =true discount + present worth
=525 +26.25=551.25
option =5

Like? Yes | No (4)  

Venkatesh kadali   8 years AGO

x*6*(10/12)*(1/100)=26.25
x=525
bil amount=525+26.25=551.25
so option is 5.

Like? Yes (1) | No (8)  

22 Sep 2016 Remaining area

A paper is in the form of a square of side 20 m. Semi-circular are drawn inside the square paper on two opposite sides as diameter. The semi-circular portions are cut off. the area of the remaining paper is

OPtion

1) (400-2*Pi) m^2
2) (400-100*Pi) m^2
3) (400-200*Pi) m^2
4) 200*Pi m^2
5) 400*Pi m^2
6) (1600-2*Pi) m^2
7) (1600-200*Pi) m^2
8) (1600-100*Pi) m^2
9) 800*Pi m^2
10)None of these

Solution

Required Area = 20*20 - Pi*10^2 = (400-100*Pi) m^2

Option 2)

Correct Option: 2 Best Solution (48)

sruthi pujari   8 years AGO

20*20 area of square=400
Area of 2 semi circles =2*pi*10^2
Area remaining:(400-100*pi)m^2

Like? Yes (10) | No (4)  

shivani shukla   8 years AGO

let the side of square be a then,
area of square =a^2
as the semi circle are drawn on the square on opposite side the area of semi circle =(pi/2)*r^2
and the diameter of circle is side of a square so the formula becomes (pi/2)*a^2
the both the semi circle are cut the the remaining area is equal to = area of square -2(area of semi circle)
= a^2-(2*(pi/2*a^2))
where a=20 and r=(diameter)/2=20/2=10
=(20)^2-(2*(pi/2*(10)^2))m^2
=(400-100*pi)m^2

Like? Yes (6) | No (3)  

Lakshmi Narayanan.k   8 years AGO

area of square= 400m^2
area of two semicircles= (100pi+100pi)/2
area of remaining portion=400-100pi. (opt 2)

Like? Yes (4) | No (2)  

naveenkumar   8 years AGO

diameter of circle is 20
area of circle is =pi* r^2
sub the area of square
we get remaining area

Like? Yes (2) | No   

ravi   8 years AGO

total area of square =400
radius of each circle is 10
area of both circle is pi*100
free space is = 400-100*pi

Like? Yes (3) | No (1)  

G.S.Kantharao   8 years AGO

Side of a Square = 20m
Area of Square = 400 m^2
Radius of Semi - Circle = 10m
Area of 2 Semi - Circles = 2(pi*r^2)/2 = 100 pi m^2.
Remaining Piece Area = (400 - 100 pi) m^2.
Hence the answer is (400 - 100pi) m^2
The right choice is 2.

Like? Yes (4) | No (2)  

Shweta Mishra   8 years AGO

Since area of the square = 20*20 = 400 m^2
and radius of the semi circular portion = 20/2 = 10 m
so, area of the portions cut off = 2*((Pi*10*10)/2) m^2 = (100*Pi) m^2
and, area of remaining portion = (400-100*Pi) m^2
Hence, correct ans is option 2.

Like? Yes (2) | No (1)  

suba   8 years AGO

area of square - area of semicircles=(20)^2-pi*(10)^2=400-100pi

Like? Yes (1) | No   

Sabarisan   8 years AGO

area of the square=400 m^2
area of two semicircles=100 * pi m^2
so, area of rremaining area= 400-100*pi m^2

Like? Yes (2) | No (1)  

Shubham pandey   8 years AGO

area of square is 400
area of circle is pie/4*400
so, result is = (400-pie *100)m^2

Like? Yes (1) | No   

W J Krishna Teja   8 years AGO

required area= area of square-(area of two semi circles)
radius of semicircle=20/2=10
so required area= s^2-pi*r^2
s=20;r=10
ans=(20^2-pi*10^2) m^2
=(400-100*Pi) m^2

Like? Yes (1) | No   

karthikeyan   8 years AGO

area of aquare = 400
area of 2 semi circles combine can be taken as area of a circle with dia 20 which is equal to (pi*20^2 )/4 . so remaining area = (400- 100*pi)m^2

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khushboo goenka   8 years AGO

area of 2 semicircle= pi*100
area of paper= 20*20=400m^2

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odugu naveen varma   8 years AGO

square has area of 400 sq.m.
then area of two semi circulars having diameter 20m is ((pi/4)*sqr(20)) = 100* pi
then area of square other than semi circles = (400-(100*pi)).

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ALOK RANJAN DAS   8 years AGO

total area of paper = 20*20=400 m^2
area of semi circle= pi*10^2 /2
total area of 2 semi circle= pi*10^2
the remaining area of paper will be (400-100*pi)m^2
hence ans is option 2

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Tanmoy Majumder   8 years AGO

area of the half circle is (pi*10^2)/2.
The area we have to subtract is equal pi*10^2, since there are two half portion.
The remaining area will be (400-pi*100) sq. m

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Vignesh.S   8 years AGO

=areaof square - area of circle ( two semicircle forms 1)

=20^2 - pi*10^2
= 400-100pi (answer)

option (2)

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Madhuri Gupta   8 years AGO

Area of Square = 400 m^2
Total Area of semi-circular on two opposite sides is = Pi * r^2 = Pi*10*10 = Pi*100 m^2
expected area = (400-Pi*100) m^2

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Raju Sampath   8 years AGO

area of total paper is (20 m)^2
two semi-circular portions forms a circle with radius 10 m.
area of circle is pi*r^2=(100*pi) m^2
remaining is equal to difference between area of square and area of circle.

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Ankit Sharma   8 years AGO

remaining area = (area of square- area of 2 semi-circles)
area of square= 20*20=400 m^2
as 2 semi-circles have diameter 20, so radius wil be 10m
hence 2 semi-cirle will make a complete circle of radius 10m
hence,
area of 2 semi_circles=pi*r*r=pi*100=100pi
so remaining area is
= (400-100pi) m^2
i.e option 2

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Rajdip Dey   8 years AGO

area of the square= 20*20 = 400 m^2
area of 2 semi circles= (2*0.5*pi*10*10) m^2
hence, area of remaining part = (400-100*pi) m^2

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Samruddha   8 years AGO

area of square=(side)^2 (20)^2 =400
area of cricle=pir^2 pi(10)^2 = 100*pi
area of remaining part=area of square-area of cricle
(400-100*pi)m^2

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nimisha gupta   8 years AGO

Area of square=20*20
=400
Area of semi circle =(pi*10*10)/2
Area of 2 semi circle =area of circle
=pi*10*10
=100*pi
Area of remaining =area of square-area of 2 semi circle
=400-100*pi

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JENSI KARUNYA.S   8 years AGO

area of the square=20^2=400
side of the square is taken as diameter of semicircles
so r=20/2=10
area of the semicircles=pi*(50+50)=100*pi
the remaining area=400-(pi*100) m^2
option 2) 400-(100*pi) m^2 is the correct answer.

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ARCHANENDRA   8 years AGO

area of square =a^2=20*20=400
area of two semicircle=2(Pi*r^2/2)=pi*(10*10)=100Pi
remaining area=area of square-area of two semicircles
=400-100Pi

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jothi lakshmi   8 years AGO

Area of the square=20*20=400
Area of the circle formed by 2 semi circles in opposite side =Pi*(10)^2
Since diameter=20
Remaining required area =Area of square-Area of circle.
(i.e) Area = (400-100*Pi)m^2

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Chitrala Venkatesh Babu   8 years AGO

two half semi circles make one circle of radius half of side of square .therefore remaining area is
=square area-circle area
=(20)^2-pi*(10)^2
=400-100*pi

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vivek Thakur   8 years AGO

(area of sq.) - (area of circle)
20^2-?(10)^2
=400-100?

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Sruthi S Nair   8 years AGO

area of sqaure 400meter sqr
radius of the semicircle 10 m
area of 2 semicircles =2*(10*10*pi)
remaining area = area of square - area of 2 semicircles
= (400 -100pi) meter sqr

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haritha reddy   8 years AGO

Area of square = 20 * 20 = 400 m^2
two semicircles of same diameter = one complete cycle.
As diameter=20, Radius = 10
Area of circle = (100*pi)m^2
So area of remaining paper
= (400-100*pi)m^2. i.ee option 2

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prashant   8 years AGO

(Area of square - Area of two semi circle of radius of 10 m )=[(20*20)-(10*10*pi)m^2]

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imran khan   8 years AGO

area of square is = a^2=20^2=400
area of circle is = pi r^2 =pi 10×10=100 pi
area of remaining paper is =area of square - area of circle= 400-100*pi =(400-100*pi)m^2
option = 2

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Avinash Singh   8 years AGO

Area of remaining part=area of square- area of circle
= (20*20)-(pi*100)=(400-100*Pi)m^2.

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Anant Mishra   8 years AGO

rem area=area of square -area of circle
=20^2-pi*10*10
=(400-100*pi)m^2

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Indrajit Bhandari   8 years AGO

Total area of the square is = 20m*20m
Now, the semi circle just touch each other as there radius is 10 and drawn in opposite sides of the square as diameter.
So, total area cut off is = pi*10m*10m
= 100*pi*m*m
So the area remain = 400m*m - 100*pi*m*m
= (400 - 100*pi)m^2

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PINTU PRASAD GUPTA   8 years AGO

area remain = area of square - area of circle having diameter 20 m
So , Radius of Circle = 10 m
Remaining area = side*side -PI * r *r =20*20 - pi*10*10
= 400 -100*pi
So option 2 will be answer

Like? Yes | No (1)  

Gayathri Devi G.R   8 years AGO

Area of square=20^2=400m^2
total area of semicircles=2*(pi*(10^2))/2=pi*(10^2)=100*pi
So,remaining area= (400-100*pi)m^2

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Dhaval Vekariya   8 years AGO

area of square=20*20=400
area of circle=pi*10^2=100*pi
remaining area=(400-100*pi) m^2

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shunpriya   8 years AGO

area of square=pi*r^2
area of circle=2*2pi*r^2
so,(400-100*pi)m^2

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shaik jasmin   8 years AGO

Required area = [ (20 * 20) - (pi * (10*10)) ]
=[ 400 - 100 * pi ]m^2

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sani musa tsoho   8 years AGO

Area of Square= L^2 =20^2=400
Area of semi-Circle=Pi*r^2
but diameter =20m
r=d/2=20/2
r=10
Area of semi-circle=Pi*10^2=100*Pi
therefore, the Area of the remaining paper=[ (Area of Square)-(Area of semi-circle)]m^2
= (400-100*Pi)m^2
hence Option No. 2 is correct answer....#@smtsoho

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phanikumar   8 years AGO

required area =((20*20-pi*(10)^2)
that is (400-100*pi)m^2

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Prakash   8 years AGO

area of remaing paper= area of square-2(area of semi circle)
ans=(20^2)+2((pi*10^2)/2)
ans=400-100pi

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GIFTY SHAJU   8 years AGO

area of square - sum of areas of two semicircles with radius as half the side of the square,since each had a diameter equal to one of the sides of the square.
ie, (20^2-pi*10^2)m^2

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ayushi mishra   8 years AGO

area of square = a^2
area of semi circle=pi/2*r^2
the semi circle are drawn on the square and then cuted from the square so,
the remaining area =area of square-(2*(area of semi circle))
=a^2-(2(pi/2*r^2))
where a=20 and r=d/2=20/2=10
=20^2-(2*(pi/2*(10)^2))m^2
=(400-pi*100)m^2

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prakash sharma   8 years AGO

Area remaining=area of square-2*(Area of semi-circle)
=>(20*20-(pi*10^2))m^2
=>(400-100*pi)m^2

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Dinesh kumar sah   8 years AGO

Let the side of square is a m and radius of semi circle is r m.
Simply area of the remaining portion=area of complete square- area of two semi circle
=a^2-2�((pi�r^2)/2)
=20^2-pi�100
=(400-100pi) m^2

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pugazh gandhi   8 years AGO

area of square is 400m^2
area of 2 semicircle = 2[pi*100/2]
=pi*100
remaining area = [400-pi*100]m^2
therefore correct opt is 2

Like? Yes | No (2)