M4maths Previous Puzzles

26 Aug 2015 Number

In a party of 35 people there are twice as many women as children and twice as many children as men. How many of each are there?

OPtion

1) 20 men, 10 children, 5 women
2) 10 men, 15 children, 10 women
3) 15 men, 5 children, 20 women
4) 9 men, 16 children, 10 women
5) 16 men, 9 children, 10 women
6) 15 men, 15 children, 5 women
7) 5 men, 10 children and 20 women
8) 15 men, 10 children, 10 women
9) 5 men, 5 children, 25 women
10)None of these

Solution

5 men, 10 children and 20 women
If x = the number of men, then x + 2x + 4x =35
therefore 7x = 35
so x = 5

Option 7)

Correct Option: 7 Best Solution (30)

praveendas   9 years AGO

simple, go through verification of options.

Like? Yes (6) | No   

Surit Datta   9 years AGO

IF x = the number of men,
then x + 2x + 4x = 35.
Therefore 7x = 35.
So X = 5.
so option 7
that's it.

Like? Yes (7) | No (3)  

RAHUL KUMAR   9 years AGO

Let men be x.
then children will be 2x and women will 4x
so, x+2x+4x=35
x=5
therefore, men=5, children= 10, and women=20
option 7 is right.

Like? Yes (2) | No   

bhaskar   9 years AGO

m+2m+4m=35
m=5
c=10
w=20

Like? Yes (2) | No (1)  

Rahul kumar   9 years AGO

no of women=2x
no of child=x
no of men=x/2
2x+x+x/2=35
x=10
so women=20,child=10,men=5

Like? Yes (1) | No   

Venkatesh kadali   9 years AGO

the ratio of women and children and men is =4:2:1
7x=35
x=5
so womens are=4*5=20

childrens are=2*5=10

and mens are=1*5=5

so answer is 7

Like? Yes (1) | No   

Srikanthreddy    9 years AGO

let say men=x
from question children=>2*men= 2 x
women=>2*children=2(2 x)
total=>x+2 x+4 x= 35
=>x=5
=>men=5,children=10,women=20

Like? Yes (1) | No   

Nikhil Kumar   9 years AGO

let the number of children=x
then number of men=x/2
number of women=2x
hence 2x+x+x/2=35
solving x=10

Like? Yes (2) | No (1)  

Lakhi kant Roy   9 years AGO

IF x = the number of men,
then x + 2x + 4x = 35.
Therefore 7x = 35.
So X = 5.
Men=5,Children=5 and Women=20

Like? Yes (1) | No   

Shubham Raj   9 years AGO

let take the men=x
children =2x
women = 4x
then from question
x+2x+4x=35
so , x= 5
option- 7

Like? Yes (1) | No   

Prabhat Burnwal   9 years AGO

let the number of children=x
then number of men=x/2
number of women=2x
hence 2x+x+x/2=35
solving x=10

Like? Yes (1) | No   

sudiptaa   9 years AGO

Let C = x
W= 2x
M = x/2

x+x/2+2x= 35
x= 10

C= 10
M= 5 W=20

Like? Yes (1) | No   

manohar kumar singh   9 years AGO

let men be x then children are 2 x and women 2 * 2x= 4x

therefore, x+2x+4x=35
7x=35
x=5
men 5, children 10, women 20

Like? Yes (1) | No   

sahil arora   9 years AGO

let no. of men's be 'x'
no. of women's be 'y'
no. of children's be 'z'
acc.to the question
y=2z;
z=2x;
and as it is given in question total no of people are 35
therefore x+y+z=35
on calculation we get x=5;y=20and z=10

Like? Yes (1) | No   

velmala gangadhar   9 years AGO

let no.of womens x
no.of children y
no.of men z
x=2y,y=2z,x+y+z=35...then x=20,y=10,z=5...option 7

Like? Yes (1) | No   

chindam laxman   9 years AGO

w=2C
C=2M
WE KNOW : W+C+M=35
------> C=10
M=5
W=20.

Like? Yes (1) | No   

Kunal nemu   9 years AGO

if x = then number of men.
then x + 2 x + 4 x = 35
so 7 x = 35
therefore x = 5
the answer is 5 men ,10 children,and 20 women

Like? Yes (1) | No   

Nabarupa Das   9 years AGO

if there will be 5 men and children=5*2=10 and 10*2=20 women

Like? Yes (1) | No   

Hemanth kumar   9 years AGO

IF x = the number of men,
then x + 2x + 4x = 35.
Therefore 7x = 35.
So X = 5.

Like? Yes (1) | No   

manish kumar   9 years AGO

let no of men=x
then no. of child=2x
and no.of women=4x
A/q:-
x+2x+4x=35
so that
x=5
hence no of men =5
no.of children=10
and no. of women=20

Like? Yes (1) | No   

K. Vanisri   9 years AGO

children=2*man
woman=2*children=4*man
men+women+children=35
men+4*men+2*men=35
therefore, men=5
women=10
children=20

Like? Yes (1) | No   

jasbir kaur   9 years AGO

M:2M:4M
7M=35 , M=5
so, Men=5, Children=10, Women=20

Like? Yes (1) | No   

ramakrishnateja   9 years AGO

given that c=2m
w=2c, let men=y then c=2y,w=4y.hence 4y+2y+y=35
7y=35 y=5.hence men=5,children=10,women=20.

Like? Yes (1) | No   

Anurag Sahay Sriavsatav   9 years AGO

Select by option. Only option no 7 satisfied the condition given above.

Like? Yes (1) | No   

Ankan Chakraborty   9 years AGO

let the no of men= x
then children=2x
and women = 4x
so
x+2x+4x=35
x=5
men=5,children=10,women=20

Like? Yes (1) | No   

SAGNIK DATTA   9 years AGO

men=x ,children=2x ,women=4x
so 7x=35 so x=5
so men children women = 5 10 20

Like? Yes (1) | No (1)  

sourav pal   9 years AGO

IF x = the number of men,
then x + 2x + 4x = 35.
Therefore 7x = 35.
So X = 5.
So 5 Men ,10 Children and 20 Women

Like? Yes (1) | No (1)  

Pranati Kundu   9 years AGO

let no. of men be x
then no of children is 2x and women is 4x
4x+2x+x=35
x=5

Like? Yes (1) | No (1)  

ramya   9 years AGO

X is the number of men
so x+2x+4x=95
7x=35
x=5

Like? Yes (1) | No (1)  

Moumita Choudhury   9 years AGO

Twice as many women as children so 20 women and 10 children. Similarly 10 children and 5 men going by the options. All together 35 people.

Like? Yes (1) | No (2)  

25 Aug 2015 Number of tiles

The dimension of the floor of a rectangle hall are 4m*3m. The floor of the hall is to be tiled fully with 8cm*6cm rectangular tiles without breaking tiles to smaller size. The number of tiles required is

OPtion

1) 4800
2) 4500
3) 3600
4) 3200
5) 2600
6) 2500
7) 2400
8) 2200
9) 2000
10) None of these

Solution

Number of tiles = 400*300/8*6 = 2500

Option 6)

Correct Option: 6 Best Solution (43)

Surit Datta   9 years AGO

(400*300) / (8*6) = 2500 is the answer.
so option 6
that's it.

Like? Yes (10) | No (1)  

manohar kumar singh   9 years AGO

area of floor = 4*3
area of one tile= .06*.08=.0048
no of tiles =12/.0048=2500

Like? Yes (4) | No   

priyanka burnwal   9 years AGO

area of rectangle hall=4m*3m=12m^2
area of rectangular tiles=0.08m*0.06m=0.0048m^2
so
the no. of tiles =12m^2/0.0048m^2=2500

Like? Yes (6) | No (4)  

PANDIDURAI T   9 years AGO

4m=400cm
3m=300cm
400*300=120000cm^2
6cm*8cm=48cm^2
120000/48=2500

Like? Yes (2) | No (1)  

ramya   9 years AGO

The area of the floor = 12sq.m => 1200sq.cm
The area of 1 tile = 48sq.cm
To find the number of tiles required:
= Area of floor/ Area of tile
= 1200 / 48
=> 25 per centimeter
Therefore, to cover the area of the floor , we need 2500 tiles.

Like? Yes (2) | No (1)  

Ankita Ghosh   9 years AGO

no. of tiles= (area of the rectangular floor)/(area of each tiles)
=(4m*3m)/(8cm*6cm)=(400cm*300cm)/(8cm*6cm)
=2500

Like? Yes (2) | No (1)  

Rishikesh Ranjan   9 years AGO

no of tiles= (4 m*3 m )/ 0.08 m * 0.06 m
=2500

Like? Yes (1) | No   

nara harikrishna   9 years AGO

4m*3m=400cm*300cm
8cm*6cm each rectagular tile
X=400/8*300/6
X=2500

Like? Yes (1) | No   

VIDIT SINGH   9 years AGO

Area of floor = 120000qcm
area of a tile = 48 sqcm
no of tile = 120000/48=2500

Like? Yes (1) | No   

Sireesha Tumati   9 years AGO

area of rectangle=4*100*3*100 --------->convert each meter dimension into centimeter
area of tile=8*6=48
1200000/48=2500

Like? Yes (2) | No (1)  

kumar   9 years AGO

400*300
/
8*6
= 2500

Like? Yes (1) | No (1)  

PRABIR    9 years AGO

option- 6 ans.

total area of the floor= 3m*4m =400*300 cm^2
area of one tile= 8*6 cm^2
no of tiles required =400*300/8*6 =2500

Like? Yes (1) | No (1)  

Subinraj   9 years AGO

Area of the floor=(400*300)cm^2
Area of 1 tile=(8*6)cm^2
Number of tiles=400*300/8*6=2500

Like? Yes (1) | No (1)  

meghana   9 years AGO

Let dimension of the floor in cm =400*300
Let dimension of the tile in cm =8*6

Area of floor=120000 cm
Area of tile =48 cm

so,how many such tiles(48 cm) we need to cover 120000cm area of the floor
i.e 120000/48 =2500
therefore , The number of tiles required is = 2500

Like? Yes (1) | No (1)  

VEMULA HIMAVARSHA   9 years AGO

hall siza 4m*3m=12m2
tile size =8cm*6cm=48cm2
for 12m2 there will be 120000cm2
then tiles required=120000/48=2500

Like? Yes (1) | No (1)  

Shubham Raj   9 years AGO

area of rectangular hall is = 12 * 100 *100 cm^2
area of tile is =8 * 6 cm^2
so, no of tiles required = (12 * 100 * 100 ) / (8*6)
= 2500

Like? Yes (2) | No (2)  

Aman Arora   9 years AGO

area of the rectangular hall is 400*300=120000cm
area of the tiles is 8*6=48 cm
no of tiles is =120000/48
no of tiles is 2500

Like? Yes (2) | No (2)  

Suganya   9 years AGO

area of a floor=300*400=120000cm/sq
area of a tile =8*6=48cm/sq
no. of tiles required =120000/48=2500

Like? Yes (3) | No (3)  

Amber Singh   9 years AGO

surface area of floor= 4m*3m = 12m^2.
surface area of a tile = (8/100)m*(6/100)m = (48/100)m^2.
now, the no. of tiles required will be 12/(48/100) = 2500.

Like? Yes (1) | No (1)  

TINKU MITTAL   9 years AGO

area of floor=400*300 cm^2
area of one tile= 8*6 cm^2

no. of tiles required= (400*300)/(8*6)
= 2500

Like? Yes (1) | No (1)  

Keshav Kishlay   9 years AGO

1m=100cm
so 4m*3m=120000cm
so no. of tiles required=120000/8*6=2500

Like? Yes (1) | No (1)  

Mahesh Chaurasia   9 years AGO

area of floor=400*300=120000 cm2
area of tile=8*6=48cm2
no. of tiles=120000/48=2500

Like? Yes (1) | No (1)  

shushant singh   9 years AGO

ans is 6)2500
the no of tiles =(120000)/(8*6)
=2500

Like? Yes (1) | No (1)  

Nandkishor jaiswal   9 years AGO

total no. of tiles=(area of floor)/(area of each tiles)
=(4m*3m)/(8cm*6cm)
=(400cm*300cm)/(8cm*6cm)
=2500 pieces required

Like? Yes (1) | No (1)  

Math Lovers   9 years AGO

400/8 = 50 in length
300/6 = 50 in width

so total = 50*50 =2500

Like? Yes (1) | No (1)  

praveendas   9 years AGO

area of the floor is =4*3*100^2 cm^2
area of the one tile=48cm^2
no of tiles required is =area of the floor/area of one tile=2500

Like? Yes (1) | No (1)  

Anup Sahay   9 years AGO

Area of rectangle hall=4*3=12m^2
let no. of tiles be x
area of single tile=0.08*0.06=.0048
so, area of single tile*x=area of hall
0.0048*x=12
x=2500
option 6)

Like? Yes (2) | No (2)  

ROHIT GOYAL   9 years AGO

The total area is=400*300=120000cm^2
the area of one tile=6*8=48cm^2
hence, no of tiles =120000/48=2500

Like? Yes (1) | No (1)  

ritu yadav   9 years AGO

area of floor = 3*4= 12msq
area of tile=8*6= 48cmsq= 0.00048msq
no. of tiles used = area of floor /area of tile
= 12/0.000048
=2500

Like? Yes (1) | No (1)  

Vijeta Sinha   9 years AGO

total area of floor= 400*300=120000cm2
total area of tiles= 8*6= 48 cm2
no of tiles = 120000/48=2500

Like? Yes (1) | No (1)  

inderpreet rooprai   9 years AGO

length of hall is 4, length of tile is 0.08 so (4/0.08)=50 tiles in a row
breadth of hall is 3 , breadth of tile is (3/0.06)=50 tiles in column
so 50*50=2500 tiles

Like? Yes (1) | No (1)  

Tarun Ghosh   9 years AGO

area 12m2 .and 1pleate area .0048m2 ...that divided and get answer....

Like? Yes (1) | No (1)  

vikas kulshrestha   9 years AGO

area of room =12 sq.m
area of tiles =48 sq.cm
no. of tiles =12 *10000/48
= 2500

Like? Yes (1) | No (1)  

dipika   9 years AGO

area of floor= 400*300 =120000 cm. sq
area of a tile=8*6 =48 cm. sq
no. of tiles = 120000/48= 2500

Like? Yes (1) | No (1)  

pratik avinash   9 years AGO

Area of hall (400*300 )cm = 120000 sq. cm
Area of tile (8*6)cm = 48 q. cm
so no tile = 120000/48 = 2500 tiles.

Like? Yes (1) | No (1)  

k sureshbabu   9 years AGO

AREA OF FLOOR IS 120000 SqCM
AREA OF EACH TILE IS 48 Sq CM
NO OF TILES REQUIRED 120000/48=2500

Like? Yes (1) | No (1)  

Ankan Chakraborty   9 years AGO

4m=400cm and 3m=300cm

so the no of tiles needed = Area of the rectangle hall
--------------------------------------
Area of rectangular tiles

(400*300)/8*6 = 2500

Like? Yes (1) | No (1)  

sommya   9 years AGO

area of hall / area if tiles
and with conversion of cm to m

4*3/.80*.60 = 2500

Like? Yes (1) | No (1)  

Rahul   9 years AGO

area of floor would be 400 CM * 300 CM = 120000 cm^2.
area covered by a single tile would be 8 cm * 6 cm = 48 cm^2.

so , 48 cm^2 can be covered by 1 tile
then 120000 cm^2 covered by ?

120000/48= 2500 tiles.

Like? Yes (1) | No (2)  

Belal Ahmed   9 years AGO

each tile covers an area of 8cm*6cm i.e in ratio of 4:3 and total area required is 400*300cm
so should be 50*8:50*6
total tiles are 50"50=2500

Like? Yes (1) | No (2)  

Lakhi kant Roy   9 years AGO

Area of the floor =400*300=120000cm^2.
Area of 1 tile=8*6=48cm^2.
So,No of Tiles=120000/48
=2500
Hence,option (6) is correct.

Like? Yes (1) | No (2)  

avanthika   9 years AGO

4*3=12m^2
0.08*0.06 =0.0048 m^2 for single tile
No. of tiles = 12/0.0048 = 2500

Like? Yes (1) | No (2)  

rahul shaw   9 years AGO

total area of room=400*300=120000
area of 1 tile=8*6
so no of tiles=120000/48=2500
option 6

Like? Yes (1) | No (2)  

24 Aug 2015 Sequence

What is the next number of the following sequence.
18, 82, 86, 122, 606, 642,

OPtion

1) 2242
2) 2442
3) 2142
4) 2106
5) 2406
6) 2476
7) 4106
8) 2642
9) 2672
10)2496

Solution

18
18 + 8^2 = 18+64 = 82
82 + 2^2 = 82+4 = 86
86 + 6^2 = 86+36 = 122
122 + 22^2 = 122+484 = 606
606 + 6^2 = 606+36 = 642
Similarly
642 + 42^2 = 2406
Option 5)

Correct Option: 5 Best Solution (36)

Danda Sindhura   9 years AGO

18=8^2=64 18+64=82
82=2^2=4 82+4=86
86=6^2=36 86+36=122
122=22^2=484 122+484=606
606=6^2=36 606+36=642
642=42^2=1764 642+1764=2406

2406

Like? Yes (2) | No   

SAGNIK DATTA   9 years AGO

next term= previous term+(number excluding the first digit of previous term )^2

Like? Yes (2) | No   

BIBIN RAJ B S   9 years AGO

Last digit of 18 is 8. and 8^2=64.
18+64=82.

Last digit of 82 is 2. and 2^2=4
82+4=86

Last digit of 86 is 6 and 6^2=36
86+36=122

Last digits of 122 is 22. and (22)^2=484
122+484=606

Last digits of 606 is 06 and (6)^2=36
606+36=642.

Last digits of 642 is 42 and (42)^2=1764
642+1764=2406.

Like? Yes (2) | No   

meghana   9 years AGO

The Series is 18, 82, 86, 122, 606, 642,

1) The process of this series is if it is a 2 digit number take last digit of that number and find square of last digit add that to the two digit.
2) if it is a 3 digit number take last 2 digits of that number and find square of last 2 digits add that to the the three digit.
3) .....continue this process for all digits

ex:
1. 18+(8)^2 =82 -----------(Take last number in the 18 i.e '8' and add 8^2 i.e 64 to 18 )
2. 82+(2)^2 =86 -----------(Take last number in the 82 i.e '2' and add 2^2 i.e 4 to 82 )
3. 86+(6)^2 =122 -----------(Take last number in the 86 i.e '6' and add 6^2 (i.e 36) to 86 )
4. ...............................
continue this process then final element will be
5. 642+(42)^2 =2406 this is final answer

Like? Yes (2) | No   

sneha   9 years AGO

18 82 86 122 606 642 x
64 4 36 484 36 1764
we can see that the difference of (82-18)=64 which is the sq of 18's last digit 8..
similarly (86-82)=4 which is the sq of 82's last digit 2.
.................
in case of 642 the sq of 42 is 1764.
the next no of the series is (1764+642)=2406

Like? Yes (2) | No   

Testing Account   9 years AGO

add square of unit digjt into no and get the next numuber. so 42^2=1764
add this into 642
so ans is 2406

Like? Yes (1) | No   

kumar   9 years AGO

ADDING LAST (N-1)DIGITS SQUARE VALUE::
18+8^2= 82
82+2^2=86
86+6^2=122
122+ 22^2=606
606+ 06^2= 642
642+42^2= 2406
option:5

Like? Yes (1) | No   

akshay verma   9 years AGO

difference between every no. is a perfect square then check from option

Like? Yes (1) | No   

Amber Singh   9 years AGO

the difference is the square of the number except the last one.
8^2, 2^2, 6^2, 22^2, 06^2, and similarly 42^2=1764.
so the next no in the series is 642+1764=2406.

Like? Yes (1) | No   

Lakhi kant Roy   9 years AGO

Ans is 2406
series is follow like this--------
(given digit+square of right place digit)
18-->18+8^2=18+64=82
82-->82+2^2=82+4=86
86-->86+6^2=86+36=122
122-->122+22^2=122+484=606
606-->606+6^2=606+36=642
642-->642+42^2=642+1764=2406
so,option 5 is correct.

Like? Yes (1) | No   

Shubham Raj   9 years AGO

18
82 ----> 18+64 = 82( previous no + (unit digit no) ^2)
86-----> 82+4 =86
122 ---> 86+484=606
606----> 122+36=642
642----->606+1764=2406
next sequence is 2406 (5)

Like? Yes (1) | No   

Keshav Kishlay   9 years AGO

difference between no. is like that
64,4,36,484,36
which is calculate by multiplying the last digits which will be gets after discarding the first digit of the no. so for the first no. (18) the difference is 8*8=64 like that for (82):- 2*2=4 and so on for (606):-06*06=36 so for (642):-42*42=1764
so the next no. will be 2406

Like? Yes (1) | No   

Adharsh   9 years AGO

The next number in the sequence is the sum of the previous number and the square of the the number formed by the digits other than the first digit.

Like? Yes (1) | No   

Mahesh Chaurasia   9 years AGO

in 18 ,it add 8^2 i.e. 18+8^2=18+64=82
82+2^2=82+4=86
86+6^2=122
122+22^2=606
606+(06)^2=642
642+42^2=2406

Like? Yes (1) | No   

Deepak kumar kumawat   9 years AGO

adding number with square of number form by accept 1st digit of that number
642+(42)^2=2406

Like? Yes (1) | No   

yamini shrivastava   9 years AGO

if we see the serise 18 ,82 ,86 ,122 ,606 ,642
so diff 64, 4 ,36 , 484 ,36
so we can see the diff of two numbers are the square of the last digit of previous number.
8*8=64,2*2=4,6*6=36,22*22=484,6*6=36
so for next number the diff =42*42=1764
so number will be 1764+642=2406

Like? Yes (1) | No   

Neeraj   9 years AGO

18+8^2=82
82+2^2=86
86+6^2=122
122+22^2=606
606+06^2=642
642+42^2=2406

Like? Yes (1) | No   

Nabarupa Das   9 years AGO

(82-18)=64=8^2 that is 8 is the last digit of 18 and its square
similarly
(86-82)=4=2^2 that is 2 is the last digit of 82 and its square
and so on.
(642-606)=36=(06)^2 , 06 the last two digits of 606.
now last two digits of 642 is 42.
(42)^2=1764
so the next number is=1764+642=2406

Like? Yes (1) | No   

ajay   9 years AGO

every 1st digit e.g.8 in 18 squared and added to itself den nxt one comes like dis,,,642+(42*42)

Like? Yes (1) | No   

Binay Shaw   9 years AGO

82-18=64=8*8
86-82=4=2*2
122-86=36=6*6
606-122=484=22*22
642-606=36=6*6
2046-642=1764=42*42

Like? Yes (1) | No   

Tushar Goyal   9 years AGO

choose all digits except first digit and square that number and add to that number.
suppose in 18 the number except first digit is 8, square it i.e 64 and then add it to 18 we get the next number which is 82. now same for last number, the number after removing first digit is 42, square it, we get 1764 and then add this to 642,we get 2406.

Like? Yes (1) | No   

Naveenhaa B   9 years AGO

taking unit digit and then its square is added to the same number.. for example
18-> unit digit is 8, its square is 64, hence the 2nd term is 64+18 = 82
then in 82, unit digit is 2, squaring 2 and added to 82, 82+2^2=86 and so on.. thus last term is 642, its square is 1464 added to 642 =642+1764= 2406

Like? Yes (1) | No   

gaurav shukla   9 years AGO

18+8*8=82
82+2*2=86
86+6*6=122
.
.
642+42*42=2406

Like? Yes (1) | No   

Chitra.S   9 years AGO

82-8=64...8^2
86-82=4............2^2..........................(1)...which is 2
122-86=36........6^2
606-122=484....22^2................(2)...which is 2+20=22
642-606=36.......6^2
2406-642=1764..42^2................(3)..which is (22+20=42)..(this is the clue..we have to add 20 in even places with the previous even place)

Like? Yes (1) | No   

shri sathan   9 years AGO

Sum of the square of the last digit in a number and the number itself will be the solution

Like? Yes (1) | No   

NAGENDRA P   9 years AGO

18+8^2=82
82+2^2=86
86+6^2=122
122+22^2=606
606+6^2=642
642+42^2=2406

Like? Yes (1) | No   

Belal Ahmed   9 years AGO

add square of each numbers last digit to the number so we get the number.
18+8^2=82 similarly 642+42^2= 2406

Like? Yes (1) | No   

RAVAL   9 years AGO

add square of unit digjt into no and get the next numuber. so 42^2=1764
add this into 642
so ans is 2406

Like? Yes (1) | No   

VIDIT SINGH   9 years AGO

18+(8)^2=82
82+(2)^2=86
86+(6)^2=122
122+(22)^2=606
606+(06)^2=642
642+(42)^2=2406

Like? Yes (1) | No   

Manpreet Kaur   9 years AGO

18 + (8*8) = 82
82 + (2*2) = 86
86 + (6*6) = 122
122 + (22*22) = 606
606 + (6*6) = 642
642 + (42*42) = 2406

Like? Yes (1) | No   

Deepanshu Chaturvedi   9 years AGO

82 = 18 + 8*8
86 = 82 + 2*2
122 = 86 + 6*6
606 = 122 + 22*22
642 = 606 + 42*42
similarly last term is
642 + 42*42 = 2406
answer is 2406

Like? Yes (1) | No   

Swapnil Gupta   9 years AGO

18+8^2=82
82+2^2=86
86+6^2=122
122+22^2=606
606+06^2=642
similarly, 642+42^2=2406.
Hence, option (5) is the correct answer.

Like? Yes (1) | No   

ritu yadav   9 years AGO

in this series every last pair of digit is squared and added to the number to get the next number in series.. like in 18 we squared the 8and add it to 18 then we get 82 again taking 2 squaring it and adding it to 82 gives 86 that is are next number in series... and so on so for 642 ew take 42 and its square willl be 1764 and added to 642 gives 2406

Like? Yes (1) | No   

Anup Sahay   9 years AGO

Logic is to square the last digit in a 2 digit no. or square the last 2 digits if the no. is a 3 digit no. and add it to the no. itself and you can get the the next number.

solution:
18>> last digit is 8
8*8=64
now 18+64=82
carry the similar process for each and check it
similarly
642>>last 2 digits=42
42*42=1764
now, 642+1764=2406 ...ans.

Like? Yes (1) | No   

Sruthi Sreejan   9 years AGO

take the square of the digit at units place if it is a 2 digit number or take the square of the last 2 digits if it is a 3 digit number and finally add the number to it to get the next number.
eg for 82, 82 + (2*2)=86.

similarly, (42*42)+ 642 = 2406

Like? Yes (1) | No   

RAJ KUMAR   9 years AGO

18+8²=82
82+2²=86
86+6²=122
122+22²=406
406+6²=442
442+42²=2406

Like? Yes (1) | No   

23 Aug 2015 Number

If I had one more sister I would have twice as many sisters as brothers. If I had one more brother I would have the same number of each. How many brothers and sisters have I?

OPtion

1) 2 sisters and 3 brothers
2) 1 sister and 4 brothers
3) 3 sisters and 3 brother
4) 3 sisters and 2 brothers
5) 4 sisters and 2 brothers
6) 3 sisters and 1 brother
7) 1 sisters and 2 brothers
8) 1 sister and 1 brothers
9) 1 sister and 3 brothers
10)None of these

Solution

let x be the number of sisters and
y the number of brothers:
x+ 1= 2y
y+ 1= x
Therefore, y + 1+ 1= 2y
so y =2
or x + 1= 2x - 2
so x = 3
Three sisters and two brothers
Option 4)

Correct Option: 4 Best Solution (13)

Shubham Raj   9 years AGO

ans- 4) 3 sisters 2 brothers
because ifwe have 3 sisters then
if we increase 1 sister then sisters are twice as brother.
if we increase 1 brother
the condition is same means no of brother and sister is same.

Like? Yes (3) | No (2)  

sireesha.yammanuru   9 years AGO

varying options by the question
1. 3+1 = 2*2
2. 2+1 = 3

Like? Yes (1) | No   

Krishna Pratap Singh   9 years AGO

according to question if 1 more sister than the number of sisters are twice than that of the brothers
and if 1 more brother than there would be the exat number of them
this condition is satisfied by option 4
3 sisters and 2 btothers
one more sister then there would be 4 sisters and 2 brothers
and if one more brother than there would be 3 sisters and 3 brothers

Like? Yes (1) | No   

ABHINAV PUSHPAM   9 years AGO

Ans= 3 sisters & 2 brothers
If number of sisters=x, brothers=y
x+1=2y, x-2y=-1 ---(i) &
x=y+1, x-y = 1 ---(ii)
Solving (i)&(ii), x=3,y=2

Like? Yes (2) | No (2)  

Anubrindha   9 years AGO

My answer is option (4)
First condition: 3+1=(4/2) (i.e.,) (4 sis =2 bro)
It satifyies when one more sister is added "number of sister=2(number of brother).
Second condition: 3=2+1 (i.e.,) (3sis=3bro)
when one bother is added number of sister is equal to number of brother

Like? Yes (1) | No (1)  

Meghana. G   9 years AGO

if he had 3 sister and addition of 1 will make it to 4 sisters which twice of the no of brothers(I.e 2 bros) according to the question. If he had + 1 bro both will be equal I.e. 3 sis = 3 bro

Like? Yes (2) | No (2)  

Puja Dhar   9 years AGO

1 more sister means 4 sisters i.e twice of brothers. again 1 more brother means 3 brothers which is equal to no of sisters

Like? Yes (2) | No (2)  

Kunal nemu   9 years AGO

Ler brothers be x and sisters be y
now y+1=2x
and x+1=y
solve the two equation and we get x=2,y=3
so brothers be 2 and sisters be 3

Like? Yes (1) | No (1)  

vignesh   9 years AGO

Let no of sis be x
No of bros be y
Eq1 x+1=2y
Eq2 y+1=x
Solve eq 1 & eq2
We get x=3 y=2

Like? Yes (1) | No (1)  

Mugesh sivagnanam Krishnapillai   9 years AGO

Let sister be x and brother be y
x+1=2y
y+1=x
solving we get x=3,y=2
Therefore 3 sisters and 2 brothers

Like? Yes (1) | No (1)  

Priyam   9 years AGO

sis=x
bro=y
1)x+1=2y
2)x=y+1
Solving 1 &2 , we get:
x=3 & y=2.

Like? Yes (1) | No (1)  

Surit Datta   9 years AGO

3 sisters and 2 brothers.
so option 4.
that's it.

Like? Yes (3) | No (4)  

sagar srivastava   9 years AGO

increase one sister then it will become twice ..increase one brother they will become equal

Like? Yes (1) | No (2)  

20 Aug 2015 Four digit number

Total four-digit number formed by using non-zero integers (1 to 9) greater than 1745 are

OPtion

1) 6561
2) 6328
3) 6298
4) 6245
5) 6147
6) 6043
7) 5890
8) 5592
9) 5398
10)5160

Solution

Total number of the format
174x are 1*1*1*4 = 4
17xy are 1*1*5*9 = 45
1xyz are 1*2*9*9 = 162
pxyz are 8*9*9*9 = 5832
Tota = 6043

Option 6)

Correct Option: 6 Best Solution (6)

Bodhisattwa Adak   9 years AGO

i have calculated this by programming in the following way.
#include "stdio.h"

void main()
{
long num,i,n,k,count=0,j;
//printf("nEnter a number : ");
//scanf("%d",&num);
for(j=1745;j

Like? Yes (10) | No (5)  

Shashank Gupta   9 years AGO

case 1:
-->1st fix ,2nd fix & 3rd fix
174 _= 4

-->1st fix & 2nd fix
17_ _=5*9=45

-->1st fix
1 _ _ _=2*9*9=162

case 2:
-->when 1 is not count at 1st place
_ _ _ _=8*9*9*9=5832

Total= 4+45+162+5832=6043

Like? Yes (7) | No (4)  

SAGNIK DATTA   9 years AGO

1746-1799=49
1800-1999=162
2000-9999 = 8*9*9*9 = 5832
Total = 6043

Like? Yes (2) | No (2)  

Nikhil Gajjala   9 years AGO

greater than 1745 i.e, by fixing 174_ last place has 4 ways and by fixing 17__ we have 5*9 here we have 4+45=49
by fixing 1___ we have 1*2*9*9=162
up to here 49+162=211
now above 2000 are 8*9*9*9=5832
from above 211+5832=6043

Like? Yes (2) | No (2)  

Moumita Choudhury   9 years AGO

total four digit numbers greater than 2000 using non zero integers ( 1 to 9) = 8*9*9*9 = 5832.
To find four digit numbers greater than 1745 and less than 2000 -
49+81+81 = 211.
So total numbers = 5832+211 = 6043.

Like? Yes (2) | No (2)  

Arpan Chattopadhyay   9 years AGO

total 4 digit nos. using non zero numbers=6561.
numbers less than or equal to 1745 with non zero elements=518. so 6561-518=6043.

Like? Yes (2) | No (3)  

19 Aug 2015 Mixture

A can contains a mixture of two liquids A and B in proportion 7:5. When 9 litre of mixture are drawn off and the can is filled with B, the proportion of A and B becomes 7:9. How many litres of liquid A was contained by the can initially.

OPtion

1) 25
2) 16
3) 15
4) 20
5) 22
6) 18
7) 10
8) 24
9) 21
10)None of these

Solution

Let the can initially contain 7x litre and 5x litre of mixture A and B respectively. Thus out of 12x litre of total mixture, 9 litre were taken out.
Quantity of A in mix. left
= 7x - 9/12x * 7x
= 28x-21 / 4 litre
Quantity of B in mix. left
= 5x- 9/12x*5x
= 20x-15 / 4 litre
Therefore ratio
7:9
or x = 3

Otion 9)

Correct Option: 9 Best Solution (81)

sudiptaa   9 years AGO

IR = 7:5
FR= 7:9

Cross Product = 63-35 = 28
Common factor of first ratio = 9/12 +9*7/28=3
A = 7*3 =21 litres
B= 5*3= 15 litres

Like? Yes (94) | No (37)  

sanjeev kumar   9 years AGO

Initial
A=7x ,B=5x , Mix.=12x
After removing 9 lit. n filled by B mix.=12x-9
A=7/12(12x-9) ,B=5/12(12x-9)+9
A:B=7:9
solving these x=3
Initial A=7x=7*3=21 Liter

Like? Yes (45) | No (7)  

vivek kr. chatterjee   9 years AGO

since initially ratio was 7:5, so just for hit and trial ,take any option which is multiple of 7, such option is only 21

take that value for now, do the calculation and u will it satisfies the conditions.

Like? Yes (43) | No (19)  

vimal tiwari   9 years AGO

initially 7/12 prts is A

after addition become 7/16

let M be total amount than...

(7/12)/(7/16)=(1-(9/M))
M = 36 litres

so A initially was. 7/12 of 36 litre i.e. 21 litres

Like? Yes (28) | No (12)  

Chandrajeet Yadav   9 years AGO

let A = 7x & B = 5x
total mixture = 7x+5x=12x

after 9 litre of mixture are drawn off remaining misture = 12x-9

after filling can with B:
A = (12x-9) * 7/12= 28x-21
B = {(12x-9)* 5/12}+9= 20x+21

So, (28x-21)/(20x+21)= 7/9
by solving above equation, we get
x=3

So liquid A was contained by the can initially = 7x = 7*3 = 21

So Ans = option (9)

Like? Yes (27) | No (12)  

ABHINAV PUSHPAM   9 years AGO

Suppose the can initially contains 7x and 5x litres of mixtures A and B resp.
Quantity of A in mixture left
= (7x - 7/12 x 9) litres = (7x - 21/4) litres.
Quantity of B in mixture left
= (5x - 5/12 x 9) litres = (5x - 15/4) litres.
(7x - 21/4) / [(5x - 15/4)+9] = 7/9 = › 28x - 21/20x + 21 = 7/9 =› 252x - 189 = 140x + 147
=› 112x = 336 =’ x = 3.
So, the can contained 21 litres of A.
option 9)

Like? Yes (9) | No   

Abhishek Anand   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively
Quantity of A in mixture left= (7x - 7/12 * 9) = (7x - 21/4)ltr
Quantity of B in mixture left = (5x - 5/12 *9) = (5x - 15/4)ltr

so, (7x-21/4) / (5x -15/4) + 9 = 7/9

28x-21/20x+21 = 7/9
on solving for x
we get 112x=336
x=3
So, the can contained 21 litres of A.

Like? Yes (6) | No   

bhaskar   9 years AGO

(7x-(7*9/12))/(5x-(5*9/12)+9)=7/9
x=3
7x=21

Like? Yes (7) | No (1)  

shreya saklecha   9 years AGO

7x/(5x+9)=7/9
X=9/4
=>[[9/4*12]+9]*7/12=21

Like? Yes (25) | No (20)  

kondapuram tirumalreddy   9 years AGO

X/Y=7/5 - RATIO OF A to B
5X-7Y=0......(1)
9 litres would have : 7/12*9=5.25 of A and 9-5.25=3.75 of B

New ratio: (X-5.25)/(Y-3.75+9)=7/9
9X-7Y=16*5.25 ....... (2)

Solve eq 1 and 2 : answer is 21 for x

Like? Yes (4) | No   

Tinku   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = (7x - 7/12 x 9 litres
=( 7x - 21/4) litres.

Quantity of B in mixture left = (5x - 5/12 x 9 ) litres
=( 5x - 15/4 ) litres.


Since ( 7x - 21/4) / ( 5x - 15/4 ) + 9 = 7/ 9

= 28x -21 / 20x + 21 = 7 / 9

=> 252x -189 = 140x + 147

=> 112x 336

=> x =3

So, can contained 21 litres of A.

Like? Yes (4) | No   

sahil arora   9 years AGO

simple but tricky question just chck the options which 1 is dividible by 7

Like? Yes (10) | No (6)  

guru teja   9 years AGO

consider 7x and 5x of mixtures A and B.
A's quantity left 7x-21/4 liters
B's quantity left 5x-15/4 liters
x = 3.
So, the can contained 21 litres of A.

Like? Yes (3) | No   

ritu yadav   9 years AGO

let liquid A be 7x and B be 5x
when 9 litres are drawn off and liquid B is added
then , remaining liquid A = 7x-(7/12*9)=7x-21/4
remaing liquid B = 5x-(5/12*9)+9=5X-21/4
now (7x-21/4):(5x-21/4)= 7:9
then x=3
but liquid A is 7x so it will be 21 litres

Like? Yes (4) | No (1)  

vinusha   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4
Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9
28x - 21 = 7
20x + 21 9
252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (2) | No   

Sangeetha G   9 years AGO

(7x-7*9/12)/(5x-5*9/12)
x=3
A=21 litres

Like? Yes (3) | No (1)  

bishal   9 years AGO

first A is 7x and B is 5x total mixtuer 12x now 9 liter drawn so 12x-9+z=16x so z=4x+9 now cheeking answer it came 21

Like? Yes (2) | No   

Nanthini   9 years AGO

A=(7x-97/12)*9)=(7x-21/4)
B=(5x-(5/12)*9)=(5x-15/4)
A/B+9=7/9
=>252x-189=140x+147
x=3
therefore 7x=21

Like? Yes (2) | No   

Mugesh sivagnanam Krishnapillai   9 years AGO

Let the initial mixture be in ratio 7x and 5x
After taking 9 litres
A mixture left=(7x-((7/12)*9)
B mixture left=(5x-((5/12)*9)
then 9 litre of b mixture is added so B=(5x-((5/12)*9)+9
A/B=(7x-((7/12)*9)/(5x-((5/12)*9)+9=7/9
solving we get x=3
Therefore 7x=21 litres.

Like? Yes (2) | No   

Koteswararao   9 years AGO

let total volume (A+B) be=x ltrs (intial)
then A's concentration = 7*x/12
After removal of 9 ltrs A's concentration=(7*x/12)-(63/12)
therfore 7x-63/12=7x/16
x=36;
intial volume of A=7*36/12=21

Like? Yes (2) | No   

shahida   9 years AGO

let initially a and b have 7x and 5x litres..
when 9ltrs are drawn off,quantity of A in mixture left:
7x-(7/12*9)=7x-21/4ltrs
quantity of b left:
5x-(15/4)ltr
therefore the ratio will be:-
7x-(21/4)/5x-(15/4)+9=7/9
after solving,we get the value of x as 3.
which means the answer is 7*3=21

Like? Yes (2) | No   

Nabarupa Das   9 years AGO

{7x-(7/12)*9}/[{5x-(5/12)*9}+9]=7/9
hence x=3.
so A was=3*7=21

Like? Yes (2) | No   

Surit Datta   9 years AGO

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively
Quantity of A in mixture left
= (7x - 7/12 x 9) litres = (7x - 21/4) litres.
Quantity of B in mixture left
= (5x - 5/12 x 9) litres = (5x - 15/4) litres.
(7x - 21/4) / [(5x - 15/4)+9] = 7/9 = › 28x - 21/20x + 21 = 7/9 =› 252x - 189 = 140x + 147
=› 112x = 336 =’ x = 3.
So, the can contained 21 litres of A.
so option 9
that's it.

Like? Yes (6) | No (5)  

kumar   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4
Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9
28x - 21 = 7
20x + 21 9
252x - 189 = 140x + 147

112x = 336

x = 3. so 21 liters is answer

Like? Yes (1) | No   

Narendra Gopi   9 years AGO

Initially ratio is 7:5
9 liters drawn of from can A is 7x-7/12*9=7x-21/4
9 liters drawn from can B is 5x-5/12*9=5x-15/4
their ratio is and adding 9 liters to B is (7x-21/4)/(5x-15/4)+9
=(28x-21)/(20x+19)
their ratio is equal to 7:9
==>(28x-21)/(20x+19)=7/9
==>252x-189=140x+147
==>112x=336
==>x=336/112
==>x=3
A filled by initially is 7x=7*3=21

Like? Yes (1) | No   

SAGNIK DATTA   9 years AGO

(7x-(63/12))/(5x+(63/12))=7/9
Solve x. x=3
Ans : 7x = 21

Like? Yes (2) | No (1)  

ishan chourishi   9 years AGO

let a be 21 so b becomes 15 by ratio 7:5 nd total becomes 36 litres.
when 9 litre of mixture is drawn, total mixture remains 27 litre.
which is 15.75 nd 11.25 by ratio 7:5.
9 litre is added to 11.25 so it becomes 20.25.
so the new mixture has A & B 15.75 & 20.25 litres respectively which is in the ratio 7:9.
hence option 9.

Like? Yes (2) | No (1)  

Vivek   9 years AGO

The original ratio = 7/5
9 litres of mixture is removed
The amount of A solution = 7x - 7/12*9
similarly
The amount of B solution = 5x - 5/12*9
9 litres of B is added hence solution B becomes
5x - 5/12*9+9
Equating the ratio of A and B solution to the new ratio we have
(7x - 7/12*9)/(5x - 5/12*9+9) = 7/9
solving we get
x=3
Hence the amount of original solution =7*3=21
therefore answer is option 9)

Like? Yes (1) | No   

Akhilesh Pandey   9 years AGO

A:B=7:5
9LT drawn.
A=9*7/12=21/4
B=9*5/12=15/4

so,
(7x-21/4)/(5x+21/4)=7/9
x=3
A=7*3=21
option 9

Like? Yes (1) | No   

jayaram   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4

Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4


7x - 21
4
= 7
5x - 15 + 9
4
9

28x - 21 = 7
20x + 21 9

252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

Arpan Chattopadhyay   9 years AGO

if A=21l,the B=15l. if 9 litres is removed then A becomes 15.75l and B 11.25l. now adding 9litres to B gives 20.25l. So 15.75:20.25 gives 7:9.

Like? Yes (2) | No (1)  

Venkatesh kadali   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left=

(7x - 7/12)x 9 litres =(7x -21/4)litres.

Quantity of B in mixture left =

(5x - 5/12)x 9 litres = (5x - 15/4) litres.

---> (7x-21/4) / (5x-15/4)+9 = 7/9

-----> (28x - 21) /(20x + 21)= 7/9

------> 112x=336

----> X=3
So, the A can contained initially 7*x=7*3= 21 litres of A

Like? Yes (1) | No   

Arun Karthick   9 years AGO

Quantity of A in mixture left = ((7x) – ((7/12) * 9)) => 7x – (21 /4) liters
Quantity of B in mixture left = ((5x)-((5/12) * 9)) => 5x – (15 / 4) liters
ratio of A and B in mixture is 7 : 9,
(7x – (21/4)) / ((5x) – (15/4)) = 7 / 9 => x = 3
initially can contains 7x liters of A = 21 liters

Like? Yes (1) | No   

ramya   9 years AGO

Let the can initially contains 7x and 5x litres of mixtures A and B respectively
Quantity of A in mixture left
= (7x - 7/12 x 9) litres = (7x - 21/4) litres.
Quantity of B in mixture left
= (5x - 5/12 x 9) litres = (5x - 15/4) litres.
(7x - 21/4) / [(5x - 15/4)+9] = 7/9 = › 28x - 21/20x + 21 = 7/9 =› 252x - 189 = 140x + 147
=› 112x = 336 =’ x = 3.
So, the can contained 21 litres of A.

Like? Yes (1) | No   

Ajay Kumar Gupta   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left =(7x -7/12*9) litres =(7x -21/4) litres.
Quantity of B in mixture left =(5x -5/12x 9)litres =(5x -15/4) litres.
(7x -21/4)/(5x-15/4)+9=7/9
so, (28x - 21)/(20x + 21)=7/9
so, 252x - 189 = 140x + 147
so,112x = 336
i.e,x = 3.
So, the can contained 7x=21 litres of A.

Like? Yes (1) | No   

Swati   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4
Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9
28x - 21 = 7
20x + 21 9
252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

Vishal   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4
Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9
28x - 21 = 7
20x + 21 9
252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

Dinni   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4
Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9
28x - 21 = 7
20x + 21 9
252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

sourav pal   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4

Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4


7x - 21
4
= 7
5x - 15 + 9
4
9

28x - 21 = 7
20x + 21 9

252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

Sanjeeb Chatterjee   9 years AGO

X/Y=7/5 - RATIO OF A to B
5X-7Y=0......(1)

9 litres would have : 7/12*9=5.25 of A and 9-5.25=3.75 of B

New ratio: (X-5.25)/(Y-3.75+9)=7/9
9X-7Y=16*5.25 ....... (2)

Solve eq 1 and 2 : answer is 21 for x

Like? Yes (1) | No   

Palak   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4
Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9
28x - 21 = 7
20x + 21 9
252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

SHARMILA BISWAS   9 years AGO

let initially qty of A=7x and B=5x
in 9 l of mixture,qty of A=(7/12)*9=21/4 and B=(5/12)*9=15/4
after 9 l is drawn out,A=7x-(21/4) and B=5x-(15/4)
after the can is filled with B again,B=5x-(15/4)+9
and A/B=7/9
solving we get x=3
so initially A was 7x therefore A=21.

Like? Yes (1) | No   

Santosh Kumar Kasaudhan   9 years AGO

let total mixture=x
then
x*7/12-63/12 : x*5/12-45/12+9 = 7:9
so
x*7/12
=
21

Like? Yes (1) | No   

Kethineni Vinodh   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4

Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4


7x - 21
4
= 7
5x - 15 + 9
4
9

28x - 21 = 7
20x + 21 9

252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

JAHAGIRDAR RUPALI MANOJ    9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - (7 *9)/12litres = 7x - (21 /4) litres.

Quantity of B in mixture left = 5x -( 5 * 9)/12 litres = 5x - (15/4) litres.

[7x - (21 /4) ] /[(5x - (15/4) )+9]=7/9

solving this ,
112x = 336
x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

pranay jaroli   9 years AGO

let x litre is contained initially
amount of a=(7/12)*x
amount of b=(5/12)*x

when 9L is taken
amount of a=21/4
amount of b=15/4

remaining a 7x/12 -21/4
remaining b 5x/12 - 15/4 + 9(after adding 9L)
final ratio is given
(7x-63/5x+63 ) = 7/9
solving x=36

part of a contained initially (7/12)*36=21 ans

Like? Yes (1) | No   

raj   9 years AGO

let 7x and 5x are the initial value.
now, ( 7x- (7/12)*9 ) / 12x =7/16
so x=3
therefore, initial concentration = 7 * 3=21

Like? Yes (1) | No   

siddhi datri   9 years AGO

let initial mixtures of A & B be 7x & 5x.
new proportion of A=7x-7*9/12
new proportion of B=5x-5*9/12+9
therefore,( 7x-7*9/12 )/(5x-5*9/12+9)=7/9
7x=21

Like? Yes (1) | No   

Gaurav Verma   9 years AGO

initially
______________
a=7x, b=5x
after decreasing 9 liter
a=7x- 9(7/12)x b=5x- 9(5/12)x

Adding 9 l in B
b=5x - 45x/12 +9

Dividing a/b and equating it with 7/9
we get 7x=a=21L

Like? Yes (1) | No   

ezhilane   9 years AGO

A:B = 7:5
Total = 5+7=12 units

When 9 litres of mixture is withdrawn , ((7/12) * 9) litres of A is withdrawn, and ((5/12) * 9) litres of B is withdrawn
Then 9 litres of B is added
so that new ratio is 7:9

In new mixture, Total volume of A is 7x - ((7/12) * 9) = 7x - 21/4
And total volume of B is 5x - ((5/12) * 9) + 9 = 5x - 3 3/4 + 9 = 5x + 21/4

So, (7x - 21/4)/(5x + 21/4) = 7/9

Solving for x:
x=3
Original Volume of A = 7x = 21 .
so answer is 21.

Like? Yes (1) | No   

KAMARAJAR   9 years AGO

X/Y=7/5 - RATIO OF A to B
5X-7Y=0......(1)

9 litres would have : 7/12*9=5.25 of A and 9-5.25=3.75 of B

New ratio: (X-5.25)/(Y-3.75+9)=7/9
9X-7Y=16*5.25 ....... (2)

Solve eq 1 and 2 : answer is 21 for x

Like? Yes (1) | No   

Manikanteswar   9 years AGO

Assume that the can initially contains 7x and 5x of mixtures A and B respectively
quantity of A in the mixture = 7x-21/4
quantity of b in the miixture = 5x-15/4
but the given ratio after the alteration is 7:9
so
(7x-21/4)/(5x-(15/4)+9) = 7/9
Solve the equation and you will get the answer
and also its apparent that there is no other option which is a factor of 7.

Like? Yes (1) | No   

lavanya   9 years AGO

Let can contains 7x and 5x litres of mixtures A and B respectively.
Quantity of A in mix. left =(7x-7/12*9) = 7x-21/4
Quantity of B in mix. left =(5x-5/12*9) = 5x-15/4
Therefore
(7x-21/4)/( 5x-15/4) = 7/9
x=3
therefore A contains 21 litres

Like? Yes (1) | No   

tamilmani   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4
Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9
28x - 21 = 7
20x + 21 9
252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

Gomathi P   9 years AGO

initially the A and B is 7x and 5x respectively

quantity of A mixtures is left=7x-(7/12*9)=7x-(21/4)liters
quantity of B mixtures is left=5x-(5/12*9)=5x-(15/4)liters

Therefore,

(7x-(21/4))/(5x-15/4+9)=7:9

28x-21/20x+21=7:9
9(28x-21)=7(20X+21)
252x-189=140x+147
112x=336
x=3

so A contains 7*3=21liters

Like? Yes (1) | No   

apoorv gupta   9 years AGO

Initially can has 7x litre of liquid A, 5x litre of liquid B.
After removing 9 litres of mixture and adding 9 litres of B -
A = 7x - (7/12)*9
B = 5x - (5/12)*9 + 9

(7x - (7/12)*9) : (5x - (5/12)*9 + 9) = 7:9
x = 3
So, initially A = 7x = 7*3 = 21

Like? Yes (1) | No   

kajal singh   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4
Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9
28x - 21 = 7
20x + 21 9
252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

gaurav shukla   9 years AGO

let x be total volume in liter.
then A=7x/12 B=5x/12
(7(x-9)/12)/(5(x-9)/12 + 9)=7/9
x=36
A=21

Like? Yes (1) | No   

Karan Somjhani   9 years AGO

let the initial quantity was-
A = 7x, B= 5x ;
from this 9l of liquid was taken out and replaced by B,
so, new quantities are-
A = 7x-(7/12)9 and
B= 5x -(5/12)9 +9
now, above A:B = 7/9 (i.e the new ratio)
solving this we get x=3
so, initially A was 7x=7*3=21 (ans).

Like? Yes (1) | No   

sweta jha   9 years AGO

(7x-21/4)/(5x-15/4)+9=7/9
=>x=3
so A=7*3=21

Like? Yes (1) | No   

NAGENDRA P   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4
Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9
28x - 21 = 7
20x + 21 9
252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

prakiti maji   9 years AGO

initially can contains 7x and 5x litres of mixtures..
quantity of a in mixture left 7x-21/4.. .............................(1)
quantity of b in mixture left 5x-15/4.. .............................(2)
.....
.....
dividing 1 by 2, we get x =3 and after getting the value of x we put the value of x in 7x so the answer is 21.

Like? Yes (1) | No   

Sireesha Tumati   9 years AGO

//7x-(21/4)/(5x-(15/4)+9)=7/9
by solving
x=3
then A's content is 7x=7*3=21

Like? Yes (1) | No   

Swapnil Gupta   9 years AGO

let,initially can contains x ltrs. of liquid.
so, A will be 7x/12 and B will be 5x/12.
now when 9 ltrs. of liquid is taken out so new composition becomes :-
A= 7(x-9)/12 and B= 5(x-9)/12.
now 9 ltrs. of liquid B is added. So, new ratio becomes:-
A:B=7(x-9)/12:[5(x-9)/12]+9=7:9.
now after solving we get x=36 ltrs.
so A=7x/12=21 ltrs.
hence, option (9) is the correct ans.

Like? Yes (1) | No   

praveen Kumar   9 years AGO

a:b=7:5 A:B=7:9
let no. of litres=x
a=7/12,b=5/12 A=7/16 B=9/16
((7/12)x-((7/12)*9))/((5/12)x-((5/12)*9))=7/9
x=36
a=(7/12)*36
a =21

Like? Yes (1) | No   

Sagar Wadjikar   9 years AGO

7K-7/12 (9) =7a

5k - 5/12 (9) + 9 = 5a

Solve for k
k = 3
7k = 21

Like? Yes (1) | No   

KAVITHA   9 years AGO

the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4

Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4


7x - 21
4
= 7
5x - 15 + 9
4
9

28x - 21 = 7
20x + 21 9

252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

Archie   9 years AGO

Fraction of B in original mixture =5/12
Feaction of B in 2nd mixture= 1
Fraction of B in resulting =9/16
Using alligations we get the mixture in which 2 liquids are mixed as 3:1 (Ratio of ((9/16)-(5/12)) & ((1-9/16))
Hence for 9 ltrs of liquid B i s added after taking out 9ltrs , 27ltrs will be qty. of liquid left.
Hence, total mixture contained=27+9=36ltrs
qty. of A contained =(36*7)/12 =21ltrs
option 9 :)

Like? Yes (1) | No   

anjali singh   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4
Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9
28x - 21 = 7
20x + 21 9
252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No   

rahul shaw   9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.
so acc to question
7x-(7x*9/12)/5x-(5x*9/12)
after solving we get
x=3
so A=21 INITIALLY

Like? Yes (2) | No (1)  

Sushmita Ghanty   9 years AGO

let, initial amount of A and B are 7x and 5x litre
now, A =7x-7/12*9=7x-21/4 litre
B=5x-5/12*9=5x-15/4 litre
(7x-7/12*9)/((5x-5/12*9)+9)=7/9
or, x=3;
so A contained 7*3=21litre

Like? Yes (1) | No (1)  

anand vijay   9 years AGO

Let the can initially contains 7x and 5x of mixtures A and B respectively.

quantity of A left in the mixture=7x-7*9/12 litres= 7x-21/4 litres
quantity of B left in the mixture=5x-5*9/12 litres=5x-15/4
litres
therefore, (7x-21/4)/(5x-15/4)+9=7/9
on solving we get, x=3
hence can contains liquid A 7*3= 21 litres initially
option 9)

Like? Yes (1) | No (1)  

PRABIR    9 years AGO

A-7x
B-5x
In 9L A is 21/4 L, B is 15/4
So (7x-21/4)/(5x-15/4 +9)=7/9
7x=21L

Like? Yes (1) | No (1)  

Kunal nemu   9 years AGO

let ,the can initially contains A=7x,B=5x
quantity of A=(7x-7/12*9)
quantity of B=(5x-5/12*9)

so,(7x-21/4)/(5x-15/4)+9=7/9
or (28x-21)/(20x+21)=7/9
or 252x-189=140x+147
or 112x=336
or x=3
So,the can contained 21lit. of A

Like? Yes (1) | No (1)  

Ayush Kumar Singh   9 years AGO

let a+b=x
and A/B=7/5
then A=7/12 of x initially
final will be,A= 7/12(x-9)
and same procedure with B

Like? Yes (2) | No (2)  

Meganathan.I   9 years AGO

X/Y=7/5 - RATIO OF A to B
5X-7Y=0......(1)

9 litres would have : 7/12*9=5.25 of A and 9-5.25=3.75 of B

New ratio: (X-5.25)/(Y-3.75+9)=7/9
9X-7Y=16*5.25 ....... (2)

Solve eq 1 and 2 : answer is 21 for x

Like? Yes (1) | No (1)  

santosh kumar patel   9 years AGO

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively.
Quantity of A in mixture left
= (7x - 7/12 x 9) litres = (7x - 21/4) litres.
Quantity of B in mixture left
= (5x - 5/12 x 9) litres = (5x - 15/4) litres.
(7x - 21/4) / [(5x - 15/4)+9] = 7/9 = › 28x - 21/20x + 21 = 7/9 =› 252x - 189 = 140x + 147
=› 112x = 336 =’ x = 3.
So, the can contained 21 litres of A.

Like? Yes (1) | No (1)  

vikas kulshrestha   9 years AGO

Solution:

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively.

When 9 litres of mixture are drawn off, quantity of A in mixture left:
7x–(
7
12
)×9=7x−
21
4
litres

Similarly quantity of B in mixture left:
5x−(
5
12
)×9=5x−(
15
4
) litres

Therefore ratio becomes:
7x–(
21
4
)
5x−(
15
4
)+9
=
7
9

⇒
(28x−21)
(20x+21)
=
7
9

⇒(252x−189)=140x+147
⇒112x=336
⇒ x=3

So the can contained: 7×x=7×3= 21 litres of A

Like? Yes (1) | No (1)  

Thirupathi Reddy   9 years AGO

A should be a multiple of 7
only one option is divisible by 7 i.e., 21

Like? Yes (3) | No (3)  

Pranav   9 years AGO

considering the amount to be 7x and 5x the the question can be solved.

Like? Yes (1) | No (1)  

Naveenhaa B   9 years AGO

initially can contains 7x and 5x of mixtures A and B.

Quantity of A in mixture left = (7x-(2/4))
Quantity of B in mixture left = (5x-(15/4))

(7x - (21/4))
divided by (5x-(15?4)) and which is equal to 7 : 9
on solving
x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No (1)  

18 Aug 2015 Average

The average of marks obtained by 120 candidates was 35. If the average of passed candidates was 39 and that of failed candidates was 15, the number of candidates who passed the examination, is

OPtion

1) 150
2) 160
3) 120
4) 100
5) 170
6) 110
7) 240
8) 80
9) 90
10) None of these

Solution

Suppose x passed and (120-x) failed
Then
x*39 + (120-x)*15 = 120*35
or
24x = 2400
x = 100
Option 4)

Correct Option: 4 Best Solution (128)

ritu yadav   9 years AGO

let x be no. of passed students and y be failed students
then x+y=120
and 39x+15y=4200
solving these equations will yield x as 100

Like? Yes (59) | No (23)  

Chandrajeet Yadav   9 years AGO

let assume Number of passed candidates = x
Since the Number of total candidate= 120
Then Number of failed candidates = 120-x

Since Average Marks obtained by 120 Candidates = 35
So, Total Marks obtained by 120 Candidates = 35*120 = 4200

Total marks obtained by passed Candidates = 39*x
Total marks obtained by failed Candidates = 15*(120-x)

So, 39*x + 15*(120-x) = 120*35
So, 39*x + 1800-15*x = 4200
So, 39*x-15*x = 4200-1800
So, 24*x = 2400
So, x = 100

So, Number of passed Candidates = x = 100

Like? Yes (29) | No (1)  

Abhishek Anand   9 years AGO

Go thru the option pick 100 as no. of passed student
Then 39 x 100 = 3900
and for failed student = 15 x20 = 300
3900+300=4200 which is equivalent to avg of 120 students

Like? Yes (5) | No (1)  

Vivek   9 years AGO

Average of 120 students is 35
let x number of students pass the exam
therefore (120 - x) students fail the exam
so we have
39*x+15*(120-x)=35*120
solving this we get
x= 100
Hence the option is 4)

Like? Yes (3) | No   

ABHINAV PUSHPAM   9 years AGO

Let the no of passed candidate be 'x' the the failed candidates be '120-x'
x*39+(120-x)*15=120*35
39x+1800-15x=4200
x=100

Like? Yes (4) | No (1)  

Surit Datta   9 years AGO

Let the no of passed candidate be 'x' the the failed candidates be '120 - x'
x*39 + (120 - x) *15 = 120*35
39x + 1800 - 15x = 4200
x = 100
so option 4
that's it.

Like? Yes (2) | No   

Devendra Marghade   9 years AGO

If passed candidates are x, then failed=120-x
So 39x+(120-x)*15=120*35
On solving,x=100
Option 4)

Like? Yes (2) | No   

ishan chourishi   9 years AGO

total 120 students, average marks=35
total marks=120*35=4200
there are total 120 students so options which have values more than 120 nd also 120 will be ignored.
now we have 4th option 100,
so let no of students passed=100
marks of passed=100*39=3900
no of students failed=120-100=20
marks of failed=20*15=300
hence total marks of passed nd failed studentd=3900+300=4200
which is average of 35 of 120 students hence option 4.

Like? Yes (2) | No   

K. Vanisri   9 years AGO

let number of student passed=x and that of failed=120-x
therefore,
(39*x+15(120-x))/120=35
Answer will come x=100
So 100 candidates are passed.

Like? Yes (3) | No (1)  

Nikhil Gajjala   9 years AGO

let the passed candidates be x
then fail candidates be y
I.e, x+y=120
total marks = 120*35=4200
39*x+15*(120-x)=4200
x=100

Like? Yes (2) | No   

nipun   9 years AGO

let the no. of passed student be x
hence no. of failed student =120-x

x*39 + ((120-x)*15) = 120*35
39x - 15x + (120*15) = 120*35
24x = 120*20
x=100

Like? Yes (3) | No (1)  

sourav pal   9 years AGO

otal no.of candidates=120 and avg is 35
then total marks obtained by total candidates will be=120*35(hint:total/num=average)
let us consider passed candidates will be x
failed candidates will be120-x
so

120*35=x*39+(120-x)15

By solving the equation we will get x=100

So ans:100

Like? Yes (1) | No   

S R Y Farhan   9 years AGO

P+F / 120 = 35.
P/np = 39
F/nf= 15
39np + 15nf /120 =35
24np + 15*120 / 120 =35
np = 100 answer

Like? Yes (1) | No   

Deepak kumar kumawat   9 years AGO

let passed candidates is x
(39x+15*(120-x))/120=35
x=100

Like? Yes (1) | No   

Vishal   9 years AGO

let x be no of passed then 120-x is no of failed ones..
now 35*120=4200 is total marks...
39*x + 15*(120-x) = 4200
getting x = 100.

Like? Yes (1) | No   

Math Lovers   9 years AGO

Let Passed students = x and failed Students = y

=> x + y = 120
=> 39x + 15y = 120*35

Solving these equations, x=100

Like? Yes (1) | No   

SWARNADEEP BISWAS   9 years AGO

passed candidates=x
x*39+15*(120-x)=120*35
x=100

Like? Yes (1) | No   

kumar   9 years AGO

let pass students be x fail be 120-x;
so 39*x + (120-x)*15 /120 =35
24x=2400
x=100 so pass stuents =100

Like? Yes (1) | No   

rajat gupta   9 years AGO

39x+(120-x)*15=120*35;
solve for x which is the no of students who passed the examination is equal to 100

Like? Yes (1) | No   

umesh   9 years AGO

39*x+15*(120-x)=120*35
24x=120*20
x=100

Like? Yes (1) | No   

neha gupta   9 years AGO

Let x be the number of candidates who passed and y be the number of candidates who failed then:
x + y =120 ..................(i)
also (39 * x) + (15 * y) = 120 * 35 ..............(ii)
Solving (i) and (ii) we get
x=100 and y=20
so number of candidates who passed are 100.

Like? Yes (1) | No   

IYTHA VENKATA RAMAIAH   9 years AGO

bothe passed and failed studnets marks equals to toal marks of 120 students .i.e,39X+15Y=4200, and solving from weighted average we get x=5y now substitute this in first we get x=100.

Like? Yes (1) | No   

santosh kumar patel   9 years AGO

no.of pass=p
no.of fail=120-p
total marks=pass+fail
120*35=39p+(120-p)*15
solvng we get p=100

Like? Yes (1) | No   

G Mounika   9 years AGO

The average marks obtained by 120 students = 35.
The sum of marks of 120 students = 120*35 = 4200.
The average of passed candidates was 39.
Let no of passed candidates= a and sum of marks of passed candidates = x.
so x/a=39
x = 39a
The average of failed candidates was 15
so (4200-x)/(120-a) = 15
(4200 - 39a) = 15(120 - a)
4200 - 39a = 1800 - 15a
24a = 2400
a = 100.
The number of candidates passed is a = 100.

Like? Yes (1) | No   

Balaselvaraj M   9 years AGO

average mark obtained by 120 student is 35.
therefore total marks obtained by 120 student is 35*120=4200.
average mark of passed candidate is 39.
average mark of failed candidate is 15.

let us take 'x' is the no of passed candidates.
then '120-x' is the no of failed candidates.
therefore,
39x+15(120-x)=4200,we get x=100 candidates.

Like? Yes (1) | No   

Bairi Sowjanya Reddy   9 years AGO

let T be total marks obtained by all students & Tp, Tf be total marks of 'x' passed and 'y' failed students resp..
.'. x+y=120 ----> eq1
avg=T/120 ==> 35 =T/120 ==> T=4200
Tp = 39 * x and Tf = 15 * y
Tp + Tf = T i.e. 4200 = 39x + 15y --> eq2
solving both the equations we get x=100 and y = 20

Like? Yes (1) | No   

Deven Patidar   9 years AGO

let x is the no of passed student
then
(39x+120*(120-x))/120= 35
39x+1800-15x=4200
24x=2400
x=100

Like? Yes (1) | No   

Rahul gope   9 years AGO

let the number of passed candidate=x,and failed=y
average of 120 candidtes is 35.
then total 120 number got=120*35=4200
we cant write 39x+15y=4200
and x+y=120
by solving we get x=100

Like? Yes (1) | No   

lanka koundinya   9 years AGO

let x be no.of students passed
39x+(120-x)*15=35*120
x=100

Like? Yes (1) | No   

Manikandn   9 years AGO

go through the options to get the answer......
if i take 100 the average marks of passed students is = 3900
then 20 students is failed soavg. marks of failed students is = 300
then add two averages and divide total number of students we get 35.. so 100 is correct option

Like? Yes (1) | No   

ezhilane   9 years AGO

we can try and error method
if i take 100 the average marks of passed students is = 3900
then 20 students is failed soavg. marks of failed students is = 300
then add two averages and divide total number of students we get 35.. so 100 is correct option

Like? Yes (1) | No   

Siddharth Sinha   9 years AGO

Let x be the no. of passed students and 120-x the no. of failed students.

So, ATQ

35*120 = 39*x + 15*(120-x)

Solving it we get x = 100 thus required no. of students.

Like? Yes (1) | No   

Sanjeeb Chatterjee   9 years AGO

let passed candidates be x and failed candidates be y
therfore total no of candidates is x+y=120
and 39x+15y=120*35
solving we get x=100
hence passed candidates is 100

Like? Yes (1) | No   

pankaj daripa   9 years AGO

39*x +(120-x)*15=120*35
or,(39x-15x)=4200-1800
or,24x=2400
or,x=100

Like? Yes (1) | No   

keerthi   9 years AGO

let number of passed candidates is = x
x *39+(120-x)15=35
x=100

Like? Yes (1) | No   

PRABIR    9 years AGO

Let x be total passed candidates,
39*x + (120-x)=120*35
24x=2400
X=100

Like? Yes (1) | No   

Ajay Kumar Gupta   9 years AGO

{x*39+(120-x)*15}/120=35
so, x=100 and 120-x=20
passing candidate=x=100

Like? Yes (1) | No   

siddhi datri   9 years AGO

let the no of passed candidates =x
Number of failed candidates=120-x
total marks=120*35
hence,
x*39+(120-x)*15=120*35
24x=120*20
x=100

Like? Yes (1) | No   

Puja Dhar   9 years AGO

let, sum of marks of 120 students is p. then p/120=35
or,p=4200
now let t no of student passed the xm and the sum of their no is x.. so.
x/t=39 again let sum of the number obtained by filed student is y. so, y/(120-t)=15
now,39t+15*(120-t)=4200(as sum of all the marks obtained by all 120 students is 4200)
or,t=100

Like? Yes (1) | No   

Moumita Choudhury   9 years AGO

Let the no of passed candidate be 'x' the the failed candidates be '120-x'
x*39+(120-x)*15=120*35
39x+1800-15x=4200
x=100

Like? Yes (1) | No   

Sandip Shaw   9 years AGO

let pass- x
failed=120-x
120*35=39*x+(120-x)*15
therefore x=100

Like? Yes (1) | No   

Rupa Kumari   9 years AGO

Let, n is the number of students passed the exam
39 n+ 15 (120-n) = 35*120;
on solving,
n = 100

Like? Yes (1) | No   

Ganesh kumar   9 years AGO

let total students marks be T,
total pass students marks be P,
total failed students marks be F.
so T=P+F;
now what given is ( T/120)=35
so P+F=35*120.
now P/x=39. F/y=15
where x and y are passed and failed students.
also x+y=120 total students
so by calculation we can get 100.

Like? Yes (1) | No   

ganesh babu   9 years AGO

120*35=35x * (120-x)15
4200=24x+1800
x=100

Like? Yes (1) | No   

Suganya   9 years AGO

given average of 120 candidates =35
sum/120=35
sum=35*120=4200
let us take from option and compare
sum/100=39
sum =3900
4200-3900=300 gives the average of failed candidates =15
so option is 4

Like? Yes (1) | No   

venkatasubbareddy   9 years AGO

let the number of students passed=x
no of students failed=120-x

then

120*35=39*x+(120-x)15

solving this we get x=100
no of students passed =100

Like? Yes (1) | No   

Nairita   9 years AGO

Marks obtained by total candidate=120×35 let no. Of passed students be x marks obtained by passed candidates=39x marks obtained by failed candidates=15(120-x) So, 39x+15(120-x)=120×35 Solving x=100.

Like? Yes (1) | No   

RAJ KUMAR   9 years AGO

x=passed candidate
y=failed candidate
x+y=120
39x+15y=4200
so x=100 ans............

Like? Yes (1) | No   

abhinash   9 years AGO

let no of passed student=x thn total marks of passed =39x
so eqn now
39x+(120-x)15=120*35
after solve get x=100

Like? Yes (1) | No   

bhaskar   9 years AGO

p+f=120
39p+15f=35(p+f)
p=5f
from lines 1 and 3
p=100(no. of students passed)

Like? Yes (1) | No   

Rahul kumar   9 years AGO

let x be no of passed student
total marks =120*35..........eq1
total marks of passed student=x*39.......eq2
total marks of failed student(120-x)*15......eq3
eq1=eq2+eq3
solving we ger x=100 ie no of passed student

Like? Yes (1) | No   

sommya   9 years AGO

x/120= 35 x= 4200

39y+ 15(120-y)= 4200

y = 100

Like? Yes (1) | No   

KODEISWARAN ELANGOVAN   9 years AGO

cp+cf=120
sum of all=sum of failed total+sum of passed total
solving both equations we get the ans as cp=100

Like? Yes (1) | No   

Pranav Vankar   9 years AGO

(39*x)+ 15(120-x)=25*120 (x is the number of candidates who passed)
solving above equation will give x=100

Like? Yes (1) | No   

Dinni   9 years AGO

39x+15y=4200
x+y=120
By solving x=100 & y=20
Passed students=100

Like? Yes (1) | No   

SHARMILA BISWAS   9 years AGO

average marks of 120 candidates=35
so total marks of 120 candidates=35*120
=4200
average marks of passed candidates=39
let no of passed candidates is x
so total marks of passed candidates=39*x
so no of failed candidates=120-x
then total marks of failed candidates=(120-x)*15
so,39*x+(120-x)*15=4200
then x=100
therefore option 4

Like? Yes (1) | No   

jothi   9 years AGO

(39x+15y)/120=35
39x+15y=4200
39x+15(120-x)=4200
39x+1800-15x=4200
24x=4200-1800
24x=2400
x=100

Like? Yes (1) | No   

Hari singh mandral   9 years AGO

x+y=4200
39x+15y=1800
after solving the eqautions
x will be 100

Like? Yes (1) | No   

Sagar Wadjikar   9 years AGO

pass + fail =120
39*pass +15*fail = 35*120
so pass = 100

Like? Yes (1) | No   

Santosh Kumar   9 years AGO

let no of pass student be x and no of fail student =(120-x)
now,
39(x)+15(120-x)=120*35
===>24x=2400
hence x=100

Like? Yes (1) | No   

THUMMALA VINEESHA   9 years AGO

let x be the passed candidates
x*39+(120-x)*15=120*35
39x+1800-15x=4200
24x=2400
x=100

Like? Yes (1) | No   

anand vijay   9 years AGO

x = no of passed candidate (120-x)=no. of failed candidate
39*x+(120-x)*15=4200
on solving we get, x=100
no of passed candidate=100
option 4)

Like? Yes (1) | No   

rajesh kumar pal   9 years AGO

total marks=120*35=4200
marks of passed candidate=39x
marks of failed candidate=15(120-x)
therefore,
39x+(120-x)15=4200
solve the equation nd u will get ur answer

Like? Yes (1) | No   

vignesh   9 years AGO

39x + 15(120-x) = 120*35
by solving this equation

Like? Yes (1) | No   

kriti kankariya   9 years AGO

let passed students be x
therfore failed will be 120-x.
(39x+15(120-x))/120=35

Like? Yes (1) | No   

A.P.Singh   9 years AGO

If x number of students passed, then
39x+15*(120-x)= 120*35
x=100

Like? Yes (1) | No   

Mohit Shrivastava   9 years AGO

Let no. of students passes b "x" and failed b "y".
Total students=x+y=120
average marks of 120 students is 35.Hence they have a total of 120*35=4200 marks.
Now average of passed student is 39 and no of students is x,hence total marks of passed students is 39x.Similarly for failed students its 15x.
Now we have two equations.
x+y=120
and
39x+15y=4200.
Solve it and I'll get d answer.

Like? Yes (1) | No   

vikas kulshrestha   9 years AGO

39x+15(120-x)=4200
x=100
x is a passed students.

Like? Yes (1) | No   

KULDEEP KUMAR   9 years AGO

Let the number o f passed candidates x.
Then total marks = 1 2 0 x 3 5 = 39x + ( l 2 0 - x ) x l 5
4200 = 39*+1800-15*
24*x = 2400
x=100

Like? Yes (1) | No   

Archie   9 years AGO

Let number of students who passed be x.Then (120-x) students failed..
----> Total marks obtained=x*39+(120-x)*15
=120*35
Solving we get,
x=100 :) Option 4

Like? Yes (1) | No   

velmala gangadhar   9 years AGO

t.p passed students marks
t.f failed students marks
p.s passed students
f.s failed students...
avrg of 120 students marks is 35...total marks=35*120
i.e t.p+t.f=35*120
avg of passed students marks 39..t.p/p.s==39
and avg of failed students marks 15.... t.f/f.s==15
by solving t.p=3900,t.f=300 ..then no.of students passed in exam is..100

Like? Yes (1) | No   

Trinadh   9 years AGO

let pass students be x and failed students be y. then sum of x and y is 120. Therefore 39*x+15*y=35*120. By solving this we can get x=100.

Like? Yes (1) | No   

vivek   9 years AGO

total marks obtained by students= 35*120=4200
also let no. of passed student will be x and no of failled student be y. then,
x+y=120
39*x + 15*y=4200
By solving two equation we get x =100

Like? Yes (1) | No   

Manikanta   9 years AGO

given the average of 120 students is 35 = (total sum/120)
total sum=35*120=4200
the passed candidates avarege =39
the failed candidates average =15
there fore 39*100=3900 passed students is 100 and
failed students is 15*20=300
the total is 4200
so the answer is 100

Like? Yes (1) | No   

vivek kr. chatterjee   9 years AGO

Average marks = 35 , no. of students = 120, so total = 35*120 = 4200
by hit and trial method,
taking no. of passed students (must be less than 120) = 100, their average marks = 39, so in total = 39* 100 =3900, and remaining failed students =20, their avrg = 15, in total = 15* 20 = 3000

since, 39000 + 3000 = 42000 which satisfies the situation

hence answer = option 4 i.e 100

Like? Yes (1) | No   

tamilmani   9 years AGO

The total mark got by 120 students
in the class= (120*35)=4200
No of candidates passed = x
No of candidates failed = 120 - x
39(x) + (120-x)*15 = 4200
39 x +1800 - 15 x = 4200
24 x = 2400
x = 100
Number of candidates passed in the examination(x)=100

Like? Yes (1) | No   

poojasaisri   9 years AGO

35=(x/120) x=4200 =>sum of all candidates marks
passed candidates => 39=y/z =>y=39z
failed candidates => 15=p/(120-z) =>p=15*(120-z)
w.k.t p+y=x =>p+y=4200
15*(120-z)+39z=4200
=> z=100 =>passed candidates

Like? Yes (1) | No   

guru teja   9 years AGO

Lets take no of passed candidate be X the the failed candidates be 120-X
X*39+(120-X)*15=120*35
39X+1800-15X=4200
X=100

Like? Yes (1) | No   

Sharmila Alajangi   9 years AGO

given the average of 120 students is 35 = (total sum/120)
total sum=35*120=4200
the passed candidates avarege =39
the failed candidates average =15
there fore 39*100=3900 passed students is 100 and
failed students is 15*20=300
the total is 4200
so the answer is 100

Like? Yes (1) | No   

shahida   9 years AGO

as avg of 120 students is 35...
let the no of students who passed be x..
so, the no of students who fail will be..120-x..
so,120*35=39*x+(120-x)*15
x=100

Like? Yes (1) | No   

mithilesh kumar   9 years AGO

Let passed student=x
so total no. of student=x*39+(120-x)*15
120*35=24x+1800 4200
4200-1800=24x
=>x=2400/24
x=100
so total passed student=100

Like? Yes (1) | No   

Tanmay Shah   9 years AGO

Let passed students are x.
So failed student are (120-x).
120*35=39*x+15*(120-x)
so 24*x=2400
so x=100
passed students are 100

Like? Yes (1) | No   

Deepanshu Chaturvedi   9 years AGO

x be no. of students passed
y be no. of students failed
equation formed are
x+y=120
39 x+15 y=4200 (sum=120*35=4200)
solving we get
x=100

Like? Yes (1) | No   

Golluru Lalitha   9 years AGO

Let x= no of students pass in exam,y= no of failure students x+y = 120
39x+15y = 120*35
by solving above equations x=100 , y =20.

Like? Yes (1) | No   

Nizamuddin S   9 years AGO

let number of passed candidates be x = 39*m ,
number of failed candidates be y = 15*(120-3)
given x+y = 4200
on substituting we get m=100.

Like? Yes (1) | No   

KAMARAJAR   9 years AGO

ans=100
120*35=4200
100*39=3900
20*15=300
3900+300=4200

Like? Yes (1) | No   

lavanya   9 years AGO

Let the no. of candidates passed exam =x
therefore
120*35 = 39*x +15(120-x)
x= 100

Like? Yes (1) | No   

GOURANGA SUNDAR MANDAL   9 years AGO

let x student passed,so (120-x) student failed
so,in mathematical problem,39*x +(120-x)*15 =120*35
then, x=100
so option 4 is correct answer

Like? Yes (1) | No   

KUMAR ROHIT   9 years AGO

total marks=120*35
let, total passed=X
i.e, total fail=120-X
so,
120*35=39X + 15(120-X)
=> X=100.

Like? Yes (1) | No   

pappu kumar   9 years AGO

let passed eqn=x
x+y=120, 39x+15y=120*35
39x+15y=4200
solving these two eqn,we get
x=100

Like? Yes (1) | No   

Ashish Kumar Karan   9 years AGO

let passed candidates=x
(39*x) + (120-x)*15=120*35

Like? Yes (1) | No   

apoorv gupta   9 years AGO

Let no. of students passed = x
and no. of students failed = y
so, x + y = 120 ---(1)
(39x + 15y) / 120 = 35 ---(2)
on solving above two equations we get x = 100

Like? Yes (1) | No   

rahul shaw   9 years AGO

let x be the no of passed students
so acc to question,
35*120=39x+(120-x)*15
x=100
option 4

Like? Yes (1) | No   

saicharan   9 years AGO

total score=35*120=4200
let y students passed and no of students failed = 120-y
sum of scores of passed students be 39y
therefore sum of scores of failed students is (120-y)*15=4200-39y
therefore y=100

Like? Yes (1) | No   

Gouri Ajit Ghode   9 years AGO

If we consider X candidates passed the exam then there will be 120-X candidates who failed.
So, 120*35=39*X+15*(120-X)

Like? Yes (1) | No   

Bathula Tejesh   9 years AGO

Total no.of students=120
Let 'x' be the number of passsed students
& their avg is 39.
therefore total=39*x=39x......................(i)
then, (120-x) is the no.of failed students
& their avg is 15
therefore total = (120-x)*15=1800-15x...................(ii)
avg of 120 students=35
therefore total = 120*35=4200.................(iii)
then from the above we have to fid the value of ""x""
the required equation is as follows
(i)+(ii)=(iii)
39x+1800-15x=4200
24x=2400
x=2400/24
x=100
therefore the no.of candidates who passed the examination is """""""""""100"""""""""".

Like? Yes (1) | No   

Mugesh sivagnanam Krishnapillai   9 years AGO

Let the number of passed candidates be x.
then total marks = 120 x 35 = 39x + ( 120 – x ) x 15 or, 4200 = 39x + 1800 – 15x
or, 24x = 2400 So x = 100

Like? Yes (1) | No   

dheeraj ramavatar somani   9 years AGO

total marks=35*120
39*passed std+15*(120-passed std)=total marks
on solving
passed std=100

Like? Yes (1) | No   

Akshit Agarwal   9 years AGO

Answer is 100
Total marks 120*35=4200
Total passing marks 39*100=3900
Total failing marks 15*20=300
Total marks 300+3900=4200
Total students 100+20=120

Like? Yes (1) | No   

khush   9 years AGO

p=sum of marks obtained by passed students
f=sum of marks obtained by failed students
np=no. of passed students
nf= no. of failed students

nf+np=120(given)

(p+f)/120=35 ( given)

p/np=39(given)

f/nf=15 (given)

(p/np) + f(120-np) =35*120--->(1st equation)
p+f=35*120;
p=35*120-f--->(2)

Like? Yes (1) | No   

RAVAL   9 years AGO

suppose passed candidates are is X
& faid candidates are Y then,
35=sum/120
so, sum=4200
X+Y = 120 ......(1)
39X+15Y = 4200....(2)

solve both equation,
X= 100 & Y = 20
ans. = 100

Like? Yes (1) | No   

Sushmita Ghanty   9 years AGO

let x no of students passed in the exam. so,
39x + 15(120-x) = 35*120
or, 24x=2400
or, x=100

Like? Yes (1) | No   

Nanthini   9 years AGO

(x1+x2+....xn)/100=39=>3900
(x1+x2+....xn)/20=15=>300
3900+300=4200/120=35

Like? Yes (1) | No   

Amber Singh   9 years AGO

let no of passed candidates be x and no of failed candidates be y. Then 39x+15y=120*35 and x+y=120.
solving these two linear equations we get x=100.

Like? Yes (1) | No   

kondapuram tirumalreddy   9 years AGO

let the number of passed candidates be x
then the number of failed candidates will be 120-x
so
x*39+(120-x)*15=120*35
39x+1800-15x=4200
x=100

Like? Yes (1) | No   

sonia kalsi   9 years AGO

total marks of passed +total marks of failed=4200
let no. of passed students=x
n no. of failed students=120-x
as total no. of students =120
acc. to ques.
total marks of passed students=39x
total marks of failed students=1800-15x

now equation is:
1800-15x+39x=4200
and
x =100

hence no. of passec students =x=100

Like? Yes (1) | No   

vimal tiwari   9 years AGO

total students = 120*35

let pass be X then fail be 120 - X

pass= 39*X
fail=(120-X )*15

pass + fail =total
(x)*39 + (120 - X)*15 =120*35

X=100

Like? Yes (1) | No   

Anirban Basu   9 years AGO

sum = 35*120
x*39+(120-x)*15=35*120
solve for x

Like? Yes (1) | No   

Harish    9 years AGO

s/120=35 where s=marks of students passed+of those who've failed
p->total marks of passed students
p/x=39 where x-> no of students passed.
similarly s-p/120-x=15 for failed students, s=35*120
solving these 2 we get x=100

Like? Yes (1) | No   

SINJAN   9 years AGO

Total is 120*35
Let x pass and (120-x) fail...
then 39*x+(120-x)*15=120*35

Like? Yes (1) | No   

Anusha Kota   9 years AGO

let no of students who passed= k
let no of students who failed = m
Total number of students = 120= k+m
marks obtained by passed candidates=p
marks obtained by failed candidates=f

avg of all students=( (total marks)/120)=35
=((p+f)/120)=35
passed candidates marks average=(p/k)=39
so, p=39*k
failed candidates marks average=(f/m)=15
so, f=15*m
on substituting above equations, we get
( ( (39*k)+(15*m))/120 )=35
and m=120-k,
we get k=100

Like? Yes (1) | No   

RANJAN KUMAR   9 years AGO

total marks=120*35=4200
suppose number of pass student=x
now,
39*x+15(120-x)=4200
39x+1800-15x=4200
24x=2400
x=100

Like? Yes (1) | No   

arun prasad   9 years AGO

by using trial and error method
i take 100 students are passed and average is 39 so total is 3900
remaining 20 students are fail their average is 15 so total is 300
39x+15y=35*120
3900+300=4200
4200=4200

Like? Yes (4) | No (3)  

ankit rawat   9 years AGO

ratio of fail to pass candidate is 1:5
so..
6x=120
x=20
no. of pass candidate is 5x= 5×20=100

Like? Yes (15) | No (14)  

Ananth Chellooss   9 years AGO

Let the number of passed candidates be x. then total marks = 120 x 35 = 39x + ( 120 – x ) x 15 or, 4200 = 39x + 1800 – 15x
or, 24x = 2400 So x = 100

Like? Yes (1) | No   

yamini shrivastava   9 years AGO

(tm)total marks=marks obtain by pass student(tp)+marks obtain by fail student(tf)
tm/120=35
tm=35*120=4200
tp/x=39 //let x student is pass//
tf/12-x=15
4200=39x+15(120-x)
2400=24x
x=100

Like? Yes (1) | No   

Anurag Sahay Sriavsatav   9 years AGO

let the no of passed candidates=x
let the no of failed candidates=y
so x+y=120
39x+15y=4200
so x=100;
so option (4) is correct.

Like? Yes (1) | No   

Tru Jan   9 years AGO

Total Students = 120
avg marks = 35
ie:- 120*35=4200 (Total marks given)

if pass student = 100
avg marks =39
hence total marks given for passed students = 100*39 =3900

failed students =20
avg marks = 15
hence total marks given for passed students = 20*15 =300

and Total marks given = 4200 = 3900 + 300
ie :- passed + failed student marks....

Like? Yes (1) | No   

S.S.Akhilesh   9 years AGO

sum of marks of total students = average*total number
using that,sum of fail students + sum of passed students = total students

Like? Yes (1) | No   

jonsarath   9 years AGO

total=120*35
pass + fail is 39x+15(120-x)=4200
24x=2400
x=100

Like? Yes (1) | No   

raghvendra   9 years AGO

120*35=4200 total marks
let n be the candidate who pass the so
39 n+(120-n)*15=4200
n=100

Like? Yes (1) | No   

praveen Kumar   9 years AGO

sum of 120 students marks / 120=35
sum of marks of passed students / number of students passed=39
sum of marks of failed students / number of students failed =15
sum of the marks of passed students(mp)+failed students(mf)=sum of 120 student marks(m120).
no. of passed student(p)+failed students(f)=120
m120/120=35

mp/p=39
mf/f=15

(mp+mf)/120=35
(p*39+f*15)/120=35

p*39+(120-p)*15=35*120


p=100

Like? Yes (1) | No   

sahil arora   9 years AGO

sry previous was wrong correct is option 4

Like? Yes (1) | No   

shyamal gayen   9 years AGO

Let the no. of passed candidate is x.so no. of failed candidate is (120-x).
now 39x+15(120-x)=120*35
so x=100

Like? Yes (1) | No   

Aniket Rai   9 years AGO

total marks =120*35;
let n student passed ..
marks obtained by n student = 39*n
marks obtained by failed student = 15*(120-n)
according to the question
120*35 = 39n + 15*(120-n)

solving n = 100

Like? Yes (1) | No   

SAGNIK DATTA   9 years AGO

let passed = x failed =120-x
so 39x + 15(120-x)=120*35
x=100

Like? Yes (1) | No (1)  

Narendra Gopi   9 years AGO

marks obtained by 120 students is 120*35
passed candidates=p
failed candidates-120-p
120*35=39x+(120-x)*15
4200=14x+1800
24x=2400
x=100

Like? Yes (1) | No (1)  

ABID PASHA   9 years AGO

average of 120 students is 120*35=4200
let x who passed and 120-x who failed
then 39x+(120-x)15=4200
24x+1800=4200
24x=2400
x=100

Like? Yes (1) | No (1)  

17 Aug 2015 Percentage

The price of rice has increased by 60%. In order to restore to the original price, the new price must be reduced by

OPtion

1) 67/2 %
2) 40 %
3) 65/2 %
4) 45 %
5) 79/2 %
6) 65 %
7) 75/2 %
8) 55 %
9) 45/2 %
10)None of these

Solution

Let original price of a unit = Rs. 100
Increased price = Rs. 160
Reduction on Rs. 160 = Rs. 60
Reduction on Rs. 100 = 60/160 * 100 = 37.5 %
75/2 %

Option 7)

Correct Option: 7 Best Solution (84)

Tarun Sai   9 years AGO

Let actual price of rice be 100
Now it is Increased by 60% i.e.. 100+60=160(New Price)
In order to Obtain old Price
(60/160) *100=75/2% is to reduced.

Like? Yes (9) | No (2)  

Abhishek Anand   9 years AGO

let the initial price of wheat is Rs. 100 and it gets increased by 60%
Now price becomes 100 + 60 = 160
In order to reduce the present cost we have to do 60/160 X 100 = 75/2 % Ans.

Like? Yes (8) | No (2)  

aman aditya   9 years AGO

let number be 100.
after 60% increment=160.
now (60/160)*100=37.5%

Like? Yes (5) | No   

Devendra Marghade   9 years AGO

If original price=x, With 60% increase price becomes x*160/100=8x/5
To restore original price must be reduced by %=
100*[(8x/5)-x]/(8x/5)=75/2
Option 7)

Like? Yes (8) | No (4)  

pooja methi   9 years AGO

i assume the no is 1 when it increased by 12%it become 8/5 now the total is this if i decrease 1 from this it willl be 3/5 that is 75/2 % of 8/5

Like? Yes (3) | No   

Ganesh kumar   9 years AGO

If orginal is 100 let it be. So it becomes 160 now. To make it 100 we should reduce (75/2)%. soooo...................

Like? Yes (2) | No   

Vivek   9 years AGO

Considering the price of rice to be Rs 100 increase in price will be 60% of Rs100 =Rs 60
So total amount = Rs 160
So to reduce to Rs 100 the percentage decrease in price
= 60/160 * 100 = 75/2 %
Therefore the option is 7)

Like? Yes (2) | No   

Sushmita Ghanty   9 years AGO

let previous price was 100
now price is 160
let price reduced by x%, so
160-(160*x/100)=100
or, 16x/100=60
or, x=60*100/16
or,x=75/2

Like? Yes (2) | No   

rahul shaw   9 years AGO

let C.P at first be 100
so new C.P= 160
diff=60
reducing %=(60*100)/160=75/2
option 7

Like? Yes (2) | No   

sanjeev kumar   9 years AGO

suppose original price=x
x+x*(60/100)=(8*x)/5
suppose decreased price=y
(8*x)/5 - ((8*x)/5)*y/100=x
y=75/2%

Like? Yes (2) | No   

sagi sahithi   9 years AGO

Suppose the price of the rice is 100 then, after 60% increase price becomes 160 .To reduce the price to 100 again the price should be reduced by 75/2%.

Like? Yes (2) | No   

Golluru Lalitha   9 years AGO

let original price=100,
if A greater than B by R% then B lesser than a by R/(R+100) %
lly R=increase in price= 60
reduced in new price=R/(R+100)=75/2
decrease in %=75/2

Like? Yes (1) | No   

Md Ehsan   9 years AGO

Initial Rs 100.now it is 160..to restore Rs 100..60 of 160 should be removed i.e60/160*100=75/2%

Like? Yes (1) | No   

Math Lovers   9 years AGO

Let Current price = 100
after 60% rise price = 160
reduction required = 160 - 100 = 60
%age reduction required = 60*100/160 = 75/2 %

Like? Yes (1) | No   

ramya   9 years AGO

Let original price be 100Rs
So new price must be 100+60% of 100=160 Rs
To get the old price, new price must be reduced to (60/160)*100=37.5 i.e 75/2%

Like? Yes (1) | No   

NITESH KUMAR SINGH   9 years AGO

let price be 100 then the new price will be 160.
then to restore to original price we must->60*100/160=75/2%

Like? Yes (1) | No   

Chandrajeet Yadav   9 years AGO

let assume that the new price must be reduced by x% then
160-160*(x/100) = 100
by solving this equation we will get x=75/2
So Answer = 75/2 %

Like? Yes (1) | No   

Rishikesh Ranjan   9 years AGO

let initial price is rs.100;
so now the price of rice will be =100+60= rs.160;
now let the price is reduced by x%;
so, 160-(160*x/100)=100
or, x=75/2

Like? Yes (1) | No   

prerna   9 years AGO

initial:100
now:160
new price must be reduced by :60/160 *100
=75/2%

Like? Yes (1) | No   

UDIT   9 years AGO

100-->(60% increase)160-->100(? % decrease)
?=(60/160)*100=(75/2)%

Like? Yes (1) | No   

saiharsha   9 years AGO

let price=100
increased price=160
x/100 *160=60
x=37.5=75/2%

Like? Yes (1) | No   

Mohit Shrivastava   9 years AGO

After 60% increase the price became 160% of original hence to get back we have to reduce the price by (160-100)*100/160

Like? Yes (1) | No   

Santosh Kumar Kasaudhan   9 years AGO

price increased by 60%
eqn will be=
1.6 *x=1
so
x=1/1.6
now reduced by
1-x
=75/2

Like? Yes (1) | No   

Sanjeeb Chatterjee   9 years AGO

let price per kg be Rs 100
hence price becomes rs 160
ratio is 100/160=5/8
now we reverse this ratio as new price has to be reduced hence 8/5
new price has to be decreased by (8-5)/8 *100=75/2%

Like? Yes (1) | No   

pooja gupta   9 years AGO

let original price=rs100
increase in price=60%
new price=160rs
now,
to restore t0 rs 100, new price must be decreased by rs 60 i.e= 60/160*100%=75/2%

Like? Yes (1) | No   

apoorv gupta   9 years AGO

60% increase in 10 = 10 + 10*(60/100) = 16
To get the original price, we have to decrease new price by 6.
So what percentage of 16 is 6 ?
6 = 16*(x/100)
x = 75/2 %

Like? Yes (1) | No   

Ashwin Dinesh   9 years AGO

Let us assume original price of rice is Rs 100/-.
60% increase means increased price is Rs. 160/-.
Now to reduce it to the original price, we need to subtract Rs 60 from 160.
(60/160)*100 = 75/2
hence option 7) 75/2%

Like? Yes (1) | No   

IYTHA VENKATA RAMAIAH   9 years AGO

Let the price of rice be Rs.100
then rce of 60% be 160.
now bring back to original price we reduce 60 and base value will be 160
=((160-100)/(160))*100=(3/8)*100=37.5 which is option 75/2

Like? Yes (1) | No   

vivek   9 years AGO

Let the price of rice be 100. Thn increased price will be 160. Let x% is reduced from New price to get original price.
160*x/100=60
x=75/2%

Like? Yes (1) | No   

Balaselvaraj M   9 years AGO

let us consider price of rice is Rs.100.
it is increased by 60%,now the price of rice is Rs.160.
to get actual price,Rs.60 must be reduced.
percentage of reduced cost is (60/160)*100=75/2%

Like? Yes (1) | No   

Nsr King   9 years AGO

Let the initial value be 100
after increasing 60% it becomes 160.
To get it back to 100 we need to reduce 60. So 60 in terms of % will be:-
60*100/160=75/2%

Like? Yes (1) | No   

sainath somisetty   9 years AGO

old price: 100
new price: 160
change in price: 60
percentage with respect to 160 is (60/160)*100=75/2%

Like? Yes (1) | No   

ritu yadav   9 years AGO

let initial price be Rs x
then new price that is 60% increased value of initial will be Rs. 1.6x
to restore original price new price should be reduced by
= (1.6x-x)/1.6*100
=75/2%

Like? Yes (1) | No   

THUMMALA VINEESHA   9 years AGO

100/160=0.625
1-0.625=0.375=37.5%=75/2%

Like? Yes (1) | No   

ADITYA KUMAR   9 years AGO

suppose first100 was the price after increaseing by 60% it will be 160. now
let x% has to be reduced in order to restore the original price..
160-(x*160)/100=100
hence x=75/2%

Like? Yes (1) | No   

Deepanshu Chaturvedi   9 years AGO

let initial price be 100
new price is 160
so, reduction in new price to make it back to 100 is 60
percentage is (60/160)*100=37.5% or 75/2%

Like? Yes (1) | No   

shivank garg   9 years AGO

let price of rice=100
then after 60% increment price=160
price increased=160-100=60
price increased percent=60/160*100=75/2

Like? Yes (1) | No   

TARUN CHAUHAN   9 years AGO

we have direct formula for it
if 'A' value increase by r% then 'B' value will decrease by (r/(100+r)*100) % in order to keep expenditure same.

Like? Yes (1) | No   

sommya   9 years AGO

since the total increase is 60% so to reduce it original price

formula used - (R/R+100)
60/(60+100) = 75/2 % ANS

Like? Yes (1) | No   

Murali Krishna.P   9 years AGO

let the price be 100.
60% increase=160
and to restore same price, 100-60/160*100=75/2%

Like? Yes (1) | No   

Durga Prasad Mahato   9 years AGO

if initial price of rice =100
after increment 60% it will be 160
then,
to restore the original price we have to reduce 160 into 100
so we have to reduce 60
and 60 is the 75/2% of 160
hence we have to reduce 75/2% to restore price from increased price.

Like? Yes (1) | No   

pravin kumar   9 years AGO

old price =160/100*new price
so old price = 0.625new price
75/2% will be the reduction

Like? Yes (1) | No   

siddhi datri   9 years AGO

let the initial price be Rs100.
increased price=Rs100+60=160
Therefore reduced %=160-100/160*100=75/2

Like? Yes (1) | No   

kumaravel   9 years AGO

Increase in price(d) = 60%
d/(d+100)*100=60/(160)*100=75/2%

Ans:75/2%

Like? Yes (1) | No   

Prabhat Burnwal   9 years AGO

100*(1+60/100)*(1-x/100)=100
calculate the value of x

Like? Yes (1) | No   

santosh kumar patel   9 years AGO

Let original price = Rs 100
increase price = rs 160
Decrease on rs 160 = rs 60
Decrease on rs 100 = (60/160*100)%
=75/2%

Like? Yes (1) | No   

Shiva Sharma   9 years AGO

let 100 rs is starting price
increase by 60% so 160
now decrease by x% to regain 100 rs
so x=37.5

Like? Yes (1) | No   

lavanya   9 years AGO

Reduction in price=(( R/(R+100))*100 )%
= (60/160)*100 %
=75/2 %

Like? Yes (1) | No   

Pratyay   9 years AGO

If the old price is Rs. 100, new price is Rs. 160. So, 60*100/160 % = 37.5% must be reduced to get the previous price.

Like? Yes (1) | No   

SAURABH SINGH   9 years AGO

initial price : new price = 5:8
in order to restore itmust be 8:5 = 3*100/8%
=75/2%

Like? Yes (1) | No   

raj   9 years AGO

let price is 100 rs now,so increased price is 160.now, for restoring the price the reduced price is 100*100/160=125/2%.
so total reduced %age is 100-125/2=75/2%

Like? Yes (1) | No   

Tarang gupta   9 years AGO

let the price be X rs.
after 60 % increase new price will be X+.6X=1.6X

price deducted to get again price X will be rs .6X

so will be = (.6X/1.6X)*100=600/16=75/2%

Like? Yes (1) | No   

banupriya   9 years AGO

if price is increased by 1/n then it will be decreased by
(1/n+1) to return back to original.
so 60% means 3/8,it will be decreased by 3/8+1,so 37.5

Like? Yes (1) | No   

Kshitiz Khandelwal   9 years AGO

let rate be x
after increase in price
x + 60% of x = 1.6 x

to get old price again
60% of x / 1.6 x * 100= 75/2%

Like? Yes (1) | No   

archana   9 years AGO

if price increased by r% then (r/100+r.100)%
(60/100+60 .100)% = 75/2%

Like? Yes (1) | No   

akanksha prakash   9 years AGO

let the original price= Rs 100
then ,60% of 100= Rs 60
therefore new price = Rs 160
to restore the original price : Rs(160-100)=Rs 60 must be reduced .
therefore % reduction = (60/160)*100
=75/2
=37.5%

Like? Yes (1) | No   

srikanth   9 years AGO

let original price is 100 it increased by 60% then new value must be 160 but we have to reduce it to maintain the original price then we have to find out percentage of increase
ie 160-100/160*100=75/2
ie 37 1/2

Like? Yes (1) | No   

keerthi   9 years AGO

let price is 100
it is increased by 60% then it is 160...
to get the original price we have to reduce 160 to 100
(60/160)*100 = 75/2 %

Like? Yes (1) | No   

Manikanta   9 years AGO

let price of rice is 100 then
the increased price by 60% then the new price is 160
1% of new price is 1.60 so the percent increased to compensate the old price is 100-(100/1.6)=37.5=(75/2)

Like? Yes (1) | No   

vikas kulshrestha   9 years AGO

old price=160/100*new price
new price =100/160=0.625*100
reduction=37.5%

Like? Yes (1) | No   

Surabhi Yadav   9 years AGO

Let original price be 100
% increased by 60%
Therefore 160
160-100=60
60/160*100=
75/2%

Like? Yes (1) | No   

shravan kumar   9 years AGO

let 100/- be the original price
when it was increased by 60% i.e., 160/-
now 160 should be reduced to 100
so 37.5%
option 7

Like? Yes (1) | No   

tadimarri jyothi   9 years AGO

cost of rice =Rs.100 say
so,cost of rice after increment=Rs.160
now.. price of Rs.160 rice should be decreased to original amount==>Rs.100
% in decrement in cost=(1-(100/160))*100=37.5%=75/2%

Like? Yes (1) | No   

VARUN DUTT   9 years AGO

let original price is 100
new price 160
to bring back the increased price to 100 it shud be decreased by 75/2 %
160-x/100*160=100
solve x

Like? Yes (1) | No   

Nikhil Gajjala   9 years AGO

ex: let original price=100
60% of 100=60
new price=100+60=160
np-op=160-100=60
reduce %=60*100/160=75/2

Like? Yes (1) | No   

Tru Jan   9 years AGO

160:60
100:?

37.5 = 75/2

Like? Yes (1) | No   

bhavini   9 years AGO

new cost price becomes 1.6 of original and hence profit is calculated

Like? Yes (1) | No   

punnapu swetha   9 years AGO

if price of rice=100
it increased 60%=160
to get old price=60*100/160=75/2

Like? Yes (1) | No   

vishal varma   9 years AGO

let us take rice cost to be 100:
given that it rised to 60% r8 60% of 100=60
so total will b 160
v asked that ? % of new price must be reduced to make it original r8 160-100%
100-?% by cross multiplying v get x=125/2
now v asked how much it must be removed so 100-125/2 =75/2

Like? Yes (1) | No   

shiva   9 years AGO

the cost of rice is 100 and after increacsed by 60% its cost is 160
then to restore the original price the rice cost must be reduced by
(100-(100/160*100))
=37.5
i.e., 75/2

Like? Yes (1) | No   

rejju   9 years AGO

new price =160/-
to reduce it to original price the % is 37.5%

Like? Yes (1) | No   

Anubrata Bhowmick   9 years AGO

let us consider the original price to be 100..
after increase.. price 160
need to decrease..60

new price must be reduced by.. (60/160)*100 = 75/2

Like? Yes (1) | No   

kartik dolas   9 years AGO

let initial price be 100 rs.
then it is increased by 60%
finaly so final cost is 160
for bringing it to original price we have to decrease by 60/160*100
=75/2

Like? Yes (1) | No   

Swapnil Gupta   9 years AGO

suppose the initial price of rice is 100/- and then after 60% increment it becomes 160 now to restore the original price we must decrease 60 from 160 which is equal to 75/2% of 160.hence, 7th option is the answer.

Like? Yes (1) | No   

Puja Dhar   9 years AGO

let, the previous price was 100. then after 60% increment of price it would be a60.. let x% should be reduced. then,
160-160*x=100
or, x=(75/2)%

Like? Yes (1) | No   

Sagar Wadjikar   9 years AGO

Let initial be 100 so 60% increment means 160 now we have to make it again 100 so (60/160)*100 is the ans. i.e 75/2%

Like? Yes (1) | No   

Archie   9 years AGO

New Price is 60% more than hence to reach original price must reduce by (60/(100+60))*100 = 75/2%
i.e. option 7

Like? Yes (1) | No   

Aditya Grover   9 years AGO

suppose initial price is Rs. 100, increased price is Rs. 160. To get it back to 100, we have to reduce it by (60/160)*100 = 75/2 %

Like? Yes (1) | No   

Chitra.S   9 years AGO

To reduce price Use formula n/d=n/d+n
60/100=60/160
Hence,75/2%
option(7).

Like? Yes (1) | No   

rahul bhartiyam   9 years AGO

let the price of the rice=100,it's rate is increased by 60% that means it's new price is 160,160*x%=60.by which we found the answer is 75/2%

Like? Yes (1) | No   

Prakhar Sandhu   9 years AGO

Let the original price be x
Then after increment, price =8x/5
Let the reduction in new price be y
Then 8x/5-y=x
so y=3x/5

so now (3x/5) /(8x/5)*100=75/2%

Like? Yes (1) | No   

priya saxena   9 years AGO

suppose initial price is 100 rs.then after increasing it by 60% new price will be 160 rs.to restore original price i.e. 100rs the new price 160rs will be reduced to 75/2% see the solution
160-160*x/100=100
16x/10=60
x=600/16=75/2%

Like? Yes (1) | No   

Vishal   9 years AGO

using formula for effective % change: where eff. change =0;
so 0 = x+y+xy/100
x=60 we get y = -75/2 %
so decrease by 75/2%

Like? Yes (1) | No (1)  

G Mounika   9 years AGO

Let the initial price of rice= Rs X.
It is increased by 60%
That means New price of rice = X+(60% of X) = 16X/10.
In order to restore the original price , the new price must be reduced by Y%.
so
(16X/10) - Y% of 16X/10 = X
6X/10 = Y % of 16X/10
6X/10 = (Y/100) * (16X/10)
by solving this Y= 75/2 %.

Like? Yes (1) | No (1)