M4maths Previous Puzzles

16 Jan 2014 Solve

The breadth of a plate increases from one corner to other in a continuous manner. Breadth of the plate changes from 10 cm. to 20 cm. and length of the plate is 50 cm., then area of the plate is

OPtion

1) 250
2) 860
3) 1455
4) 4800
5) 475
6) 500
7) 625
8) 750
9) 125
10)None of these

Solution

We can assume that the plate is formed of three geometries
First is a right angled triangle of two perpendicular sides 50 & 5 cm.
Second is a rectangle of dimensions 50 & 10 cm.
Third is a right angled triangle of two perpendicular sides 50 & 5 cm.
Total area of the plate is 1/2(50+5) + 50×10 + 1/2(50+5) = 750

Correct Option: 8 Best Solution (13)

Dr.Dipin Singh   11 years AGO

Area of plate = 10*50 + 50*(20-10)/2= 500+250=750 sq cms.

Like? Yes (8) | No (1)  

madanpal vishwakarma    11 years AGO

initaialy the the length is=50 and breadth = 10 cm
so area= l*b = 50*10=500 cm^2

now the breadth increased to 20 cm so the additional area increased is a right angled triangle

so increased area= 1/2* 50* 10= 250 cm^2
so total area = 500 + 250 =750 cm^2

Like? Yes (9) | No (5)  

KRISHNAN   11 years AGO

The Breadth changed from 10 to 20 cm and the height is 50cm

So the Area of the Trapezium is found by 1/2(a+b) x h
where a and b are the two breadths and h is the height
Therefore 1/2(10+20)x 50 = 750 cm is the area hence option 8 is the correct answer

Like? Yes (3) | No (1)  

Udhaya kumar   11 years AGO

(1/2)bh + lb
(1/2) * 50 * 10 = 250
50*10=500
500+250 = 750..

Like? Yes (2) | No (1)  

Sudipta Mukherjee   11 years AGO

width of one corner=10, another one is =20,
length=50,
it is a trapezium
so,area of the plate=1/2*(sum of parallel sides)*hight;
1/2*(10+20)*50=750 cm^2

Like? Yes (2) | No (1)  

Gurmeet Singh   11 years AGO

Required area=
50*10 + (1/2*10*50)= 750

Like? Yes (1) | No   

Srinivasa Vamsi Krishna Dodda   11 years AGO

Trepizoid area .5*[10+20]*50

Like? Yes (2) | No (1)  

Vijay Chauhan   11 years AGO

suppose first breadth is 10cm and length is 50cm...so area is 500cm...
now breadth of the plate is change only one side 10 to 20 cm...so become new triangle for that..10 X 50...
so its area is 250...so add 500 + 250 = 750

Like? Yes (1) | No (1)  

Naveen Pandey   11 years AGO

its a trapezium

Like? Yes (1) | No (1)  

Sukanya   11 years AGO

Split the plate into two parts.
The upper part is a right triangle with base 50 and height 10 so its area is 1/2*50*10=250
The lower part is a rectangle with length 50 & breadth 10 so its area is 50*10=500.
Area of the plate is 750.

Like? Yes (1) | No (1)  

ARJYA HALDER   11 years AGO

The above area can be divided into two parts, i.e,
1.A right angled triangle with height=10 cm & base=50 cm
2.A rectangle with length=50cm & breadth=10 cm.
Area of required figure is=
Area of triangle+Area of rectangle=
[(1/2)*10*50]+[50*10]=
750 cm^2
So, option 8 is correct.

Like? Yes (1) | No (1)  

VIKASH RAY   11 years AGO

The above area can be divided into two parts, i.e,
1.A right angled triangle with height=10 cm & base=50 cm
2.A rectangle with length=50cm & breadth=10 cm.
Area of required figure is=
Area of triangle+Area of rectangle=
[(1/2)*10*50]+[50*10]=
750 cm^2
So, option 8 is correct

Like? Yes (1) | No (1)  

Naveen Kumar   11 years AGO

area of trapezium=(1/2)*(sum of length of parallel sides)*height between these parallel sides

Like? Yes (1) | No (1)  

15 Jan 2014 Solve

A man purchases three objects of consecutive prices. The number of purchased objects are 1, 2 & 3. He pays Rs 115 to shopkeeper. If all prices are taken in integers, then out of 10 choices, the price of that object, whose 3 pieces are purchased, is

OPtion

1) 18
2) 17
3) 21
4) 22
5) 24
6) 20
7) 16
8) 26
9) 12
10)28

Solution

6 cases arise
1x + 2(x + 1) + 3(x + 2) = 115 => x not equal integers
1x + 3(x + 1) + 2(x + 2) = 115 => x = 18
2x + 1(x + 1) + 3(x + 2) = 115 => x = 18
2x + 3(x + 1) + 1(x + 2) = 115 => x not equal integer
3x + 1(x + 1) + 2(x + 2) = 115 => x not equal integer
3x + 2(x + 1) + 1(x + 2) = 115 => x not equal integer
Since x = 18, hence solutions are 19 & 20 according to 2nd and 3rd conditions given above,
Out of 10 choices, 20 is correct

Correct Option: 6 Best Solution (28)

Dr.Dipin Singh   11 years AGO

18*2+19*1+20*3= 36+19+60=115
so the price of that object, whose 3 pieces are purchased, is Rs 20.

Like? Yes (8) | No (5)  

somashekhar   11 years AGO

18*2->36
19*1->19
20*3->60
-->115
So, option # 6 is the right answer

Like? Yes (1) | No   

SHIKHA SUMAN   11 years AGO

let prices of three objects be x, x+1, x+2
now x+2(x+1)+3(x+2)=115
=> 6x + 8= 115
=> x = 17.833
=18
x+2=18+2=20

Like? Yes (1) | No   

ramakrishna   11 years AGO

Object prices are consecutive so,
Let we take ,the prices as
X,X+1,X+2
The number of purchased objects are 1,2,3
so,
X+2(X+1)+3(X+2)=115
X+2X+2+3X+6=115
6X+8=115
6X=115-8
X=107/6
X=17.8(appx 18)
All are taken in Integers so ,prices are
18,19,20
he asked 3 Pieces object price so answer is 20
Option 6 is answer

Like? Yes (1) | No   

Dayashankar Rajhans   11 years AGO

2x+(x+1)+3(x+2)=115
so,x=18
and,x+2=20

Like? Yes (1) | No   

Satya   11 years AGO

let the price are like x,x+1 and x+2
so he will purchase like 3 objects from price of x+2
2 objects from price of x
and 1 object from price of x+1

so the eqn becomes 3(x+2)+2(x)+(x+1)=115
=> 3x+6+2x+x+1=115
=>6x=108
=>x=18

as the man purchased 3 objects from price of x+2 .so the ans is 20

Like? Yes (4) | No (6)  

Udhaya kumar   11 years AGO

x+2(x+1)+3(x+2) = 115
x=17.8
ans x+2 = 19.8 = 20.......

Like? Yes (3) | No (6)  

DHIRENDRA    11 years AGO

let three consicutive price are
x,x+1,x+2
given
The number of purchased objects are 1, 2 & 3. He pays Rs 115
x+2(x+1)+3(x+2)=115
x+2x+2+3x+6=115
6x+8=115
6x=115-8
x=107/6
x=17.833333
x=18(approximate)
then

1 objects price =18
2 objects price =19
3 objects price =20

Like? Yes (1) | No (4)  

KRISHNAN   11 years AGO

Let x be the price of First object
then 2 object will be x +1 and 3 rd object x + 2

So x + 2(x+1) + 3(x+2) = 115
or 6x = 107
0r x = 17.83 or 18
so price of 3 object bought is 18+2 or 20

Like? Yes (1) | No (5)  

Vinay Sipani    11 years AGO

Let the prices of objects be (x-1),x,(x+1)

Let no. of pieces of each objecg that was purchased be a,b,c
where a,b,c = 1,2,3 (not in ordered form)

=> a(x-1) + bx + c (x+1) = 115
=> x (a+b+c) + (c-a) = 115

a+b+c=6 => 6x = 115 - (c - a)

The RHS must be even to be divisible by 6
=> (c-a) must be odd

114 is divisible by 6 => (c - a) = 1
=> c,a = (3,2) or (2,1)

1) If (a,b,c) = (2,1,3) => x = 19
=> The price of object with 3 pieces = 20

2) If (a,b,c) = (1,3,2) => x= 19
=> The price of object with 3 pieces = 19

So possible answers are Rs.19 and Rs.20
=> Option 6

Like? Yes (2) | No (6)  

Naveen Pandey   11 years AGO

ITS EASY :)

Like? Yes (1) | No (5)  

Vijay Chauhan   11 years AGO

he purchase,
1 for 19
2 for 18 and
3 for 20.....

Like? Yes (2) | No (6)  

Krishna Chaitanya   11 years AGO

Given a man purchases three objects of consecutive prices. The number of purchased objects are 1, 2 & 3. He pays Rs 115 to shopkeeper.
From the given option let the 3rd object price be Rs 20 then 2nd object price is Rs 19 and 1st object price is Rs 18.
Therefore total price of three objects=18+2(19)+3(20)=116 which is nearly equal to 115.

Like? Yes (2) | No (6)  

satyen nabar   11 years AGO

Let n be price of the cheapest item
So (n+1) and (n+2) are the prices of the other 2 items.
The only equation that gives n as a whole integer is
2n + (n+1) + 3(n+2)= 115.
6n+7 =115.
n= 18.
Price of item whose 3 pieces are purchased is 20.

Like? Yes (2) | No (6)  

Chandni Maurya   11 years AGO

ans=20
20*3+18*2+19=115

when we assume consecutive prices of purchased object 1,2,3 resp. then it must be even
as (x-2)+2(x-1)+3x or (x+2)+2(x+1)+3x is always even whether x is even or odd .

so,
again lets assume x is the price of 3rd object
and 2,1,3 resp are in consecutive prices
(x-1)+2(x-2)+3x =115
6x-5=115
x=120/6=20

Like? Yes (2) | No (6)  

madanpal vishwakarma    11 years AGO

let the first object price is x
then second should be x+1
and next x+3
so
according to the problem
x+(x+1)*2+(x+2)*3=115
so after solving
x=18
so
3rd object price is
x+2=18+2=20

Like? Yes (3) | No (7)  

ASHISH MITTAL   11 years AGO

x+2(x+1)+3(x+2)=115
x+2x+2+3x+6=115
6x=115-8
x=107/6
18(approx)
so with 3 piece object price would be 18+2=20

Like? Yes (2) | No (6)  

pankaj chandra joshi   11 years AGO

let first price =x
2nd is x+1
3rd is x+2
1st purchased
+1peace 2nd 2,3rd 3
totel paid =115
x+2(x+1)+3(x+2)=115
x=17.8
18
so x+2=19.8
20 approx
option 6

Like? Yes (1) | No (5)  

swastik   11 years AGO

let the price are like x,x+1 and x+2
so he will purchase like 3 objects from price of x+2
2 objects from price of x
and 1 object from price of x+1

so the eqn becomes 3(x+2)+2(x)+(x+1)=115
=> 3x+6+2x+x+1=115
=>6x=108
=>x=18

as the man purchased 3 objects from price of x+2 .so the ans is 20

Like? Yes (4) | No (8)  

mahima srivastava   11 years AGO

the prices will be x, x+1, x+2
quantity is 1, 2 and 3

x+2(x+1)+3(x+2)=115
x=17.333
but it is not an integer value

now next equation can be,
2x+x+1+(x+2)3=115
x=18
the object whose three pieces are purchased is of price x+2
so its price is 20

Like? Yes (1) | No (5)  

upendra singh chahar   11 years AGO

3(x)+2(x-1)+(x-2)=115
here x will not an integer.
so other choice is 3(x)+(x-2)+2(x-1)=115
after solving
x=20

Like? Yes (1) | No (5)  

VIKASH RAY   11 years AGO

Let, the prices of the three integer is-(x-2),(x-1),x.
then,
price of first object is-(x-2)
price of 2nd object is-2(x-1)
price of 3rd object is-3(x)
according to question,
(x-2)+2(x-1)+3(x)= 115
=> x=19.833333
=20 (appx.)
so, option 6 is correct.

Like? Yes (1) | No (5)  

Devendra Marghade   11 years AGO

If prices of three objects are x,(x+1),(x+2) then
x+2(x+1)+3(x+2)=115
6x+8=115, x=107/6
the price of that object, whose 3 pieces are purchased=(107/6)+2=119/6=19.833=20 approx.

Like? Yes (1) | No (5)  

sudipta sadhukhan   11 years AGO

18*2+19*1+20*3=115

Like? Yes (3) | No (7)  

Ramneek    11 years AGO

let the prices of 3 obejcts be
x,x+1,x+2 and the number of objects purchased in order 2,1,3 resp

2x+(x+1)+3(x+2)=115
x=18
price of that object, whose 3 pieces are purchased, is
x+2=20

Like? Yes (3) | No (7)  

B.Senthil Kumar   11 years AGO

x+2x+2+3x+3=115
6x+5=115
6x=110
x=18.3 so approximately 18 and answer is 20 i.e. x+2

Like? Yes (1) | No (5)  

susmit   11 years AGO

1*x+2*(x+1)+3*(x+2)=115
solving x=18 (near integer)
x+2=20
option no.6

Like? Yes (1) | No (6)  

Akanksha Mishra   11 years AGO

Consider purchases three objects of consecutive prices are x,(x+1),(x+2)

According to the question
The number of purchased objects are 1, 2 & 3

so prices of that object x , 2(x+1) , 3(x+2)

Total cost = 115 Rs

x + 2(x+1) + 3(x+2) = 115
x + 2x +2 + 3x + 6 = 115
6x + 8 = 115
6x = 107
x = 107/6
x = 17.833
he price of that object, whose 3 pieces are
= (x+2) = 17.833 + 2 = 19.833
prices are taken in integer no.
so ans will 20(option 6)

Like? Yes (15) | No (20)  

14 Jan 2014 Solve

A fractional number is such that its numerator is 2 less then denominator. If both numerator & denominator increase by 1 the fraction becomes 4/5. What will be the numerator of the number

OPtion

1) 1
2) 2
3) 3
4) 4
5) 5
6) 6
7) 7
8) 8
9) 9
10)None of these

Solution

Let the number is x/x+2
(x+1)5 = (X+3)4
5x+5 = 4X+12
x=7

option 7)

Correct Option: 7 Best Solution (59)

Akanksha Mishra   11 years AGO

Consider denominator is x
numerator is 2 less then denominator
so;
Nominator will be x -2

According to the question
If both numerator & denominator increase by 1 the fraction becomes 4/5

(x-2)+1/x+1 = 4/5
x-1/x+1 = 4/5
5(x-1) = 4(x+1)
5x - 5 = 4x + 4
x = 9

so denominator (x) = 9
nominator (x-2) = 9 - 2 = 7

so answer will be option 7

Like? Yes (16) | No (11)  

MP    11 years AGO

4/5=8/10 when diff in numerator and denominator is 2.
so initial fraction = (8-1)/(10-1)=7/9
so numerator =7

Like? Yes (4) | No (1)  

Krishna Chaitanya   11 years AGO

Given numerator is 2 less than denominator
Let the required fraction be (x/x+2)
Given numerator & denominator increase by 1 the fraction becomes
Therefore (x+1/x+2+1)=4/5
x+1/x+3=4/5 =)x=7

Like? Yes (2) | No   

julee verma   11 years AGO

x-2+1/x+1=4/5
on solving we get numerator=7
so answer=7

Like? Yes (2) | No   

AMIT JAISWAL   11 years AGO

let the required fraction be x/y
and x = y-2

so,
y-2/y=4/5
y-2+1/y+1=4/5
5y-5=4y+4
y=9
then, x = y-2 =7

Like? Yes (2) | No   

navneet shankhwar   11 years AGO

acc. to qustion eqution will be
x-2+1/x+1=4/5
on solving x will be 9
therefor, fraction will be 7/9
and 7 will be the numerator of the number.

Like? Yes (2) | No   

Ashutosh Nilesh Ujwala Patkar   11 years AGO

Let numerator be 'x'
denominator 'x+2'
when 1 is added to both numbers then fraction becomes 4/5
therefore x+1/x+3=4/5
x=7,x+2=9
numerator is 7

Like? Yes (1) | No   

madanpal vishwakarma    11 years AGO

(x-2+1)/(x+1)=(4/5)
after solving this u get x=9;
so numerator is =7

Like? Yes (1) | No   

KRISHNAN   11 years AGO

Let the denomenator be x then the numerator is x-2

now if we increase both the number by one we get
denomenator as x + 1 and numerator as x - 1

so x - 1 / x + 1 = 4 / 5
or 4x + 4 = 5 x - 5
or x = 9
so the numerator x - 2 = or 7

Like? Yes (1) | No   

NAGENDRA P   11 years AGO

given..N=D-2...(1)
(N+1)/(D+1)=4/5....(2)
cross multiplying ...(2)..
we get 5N-4D=-1....(3)
sub..(1) in (3)..we get..D=9..N=9-2=7..
SO..NUMERATOR=7..option 7..

Like? Yes (1) | No   

venkatesh.p   11 years AGO

let the denominator of the fraction be x,then numerator is x - 2
the fraction is x - 2 / x
if both the numerator & denominator increase by 1 the fraction becomes 4/5
therefore x - 2 + 1/ x + 1 = 4/5
x - 1 / x+ 1 = 4/5

4x + 4 = 5x - 5
there fore x = 9 therefore fraction is 9 -2 /9 = 7 / 9
therefore numerator of fraction is 7

Like? Yes (1) | No   

Sumit Roy   11 years AGO

let numerator=n, denominator=d
given, n+2=d
and, (n+1)/(d+1)=4/5
solving for n and d, we get n=7, d=9

Like? Yes (1) | No   

Devi Nallamilli   11 years AGO

denominator is x
then numerator is x-2
by increasing 1 for both numerator and denominator it will be
x-1/x+1=4/5
5x-5 = 4x+4
x=9
numerator is x-2 = 7

Like? Yes (1) | No   

Chandni Maurya   11 years AGO

ans=7
let x be the denominator,
according to ques
(x-2+1)/(x+1)=4/5
5x-s=4x+4
x=9
=>x-2=9-2=7

Like? Yes (1) | No   

DHIRENDRA    11 years AGO

let denominator =x
then numerator = x-2
given
If both numerator & denominator increase by 1 the fraction becomes 4/5
(x-1)/(x+1)=4/5
5x-5=4x+4
x=9
denominator =9
numerator =9-2
7

Like? Yes (1) | No   

Ankit Goyal   11 years AGO

let the fraction be (x-2)/x

when 1 is added to both the fraction becomes
(x-1)/(x+1) = 4/5
x=9

fraction = 7/9

Numerator = 7

Like? Yes (2) | No (1)  

rakesh   11 years AGO

let the fraction be x/y [ were x =y-2 ]

Given (x+1) / ( y+1) = 4/5

5x+5 = 4y + 4 [ since y= x+2 ]

5x+5 = (4x+8)+4

5x + 1 = 4x +8

5x-7 =4x

x = 7

so numerator is 7

Like? Yes (1) | No   

Surya Narayana K   11 years AGO

Let's take denominator = x, then numerator =x-2

If both numerator & denominator increase by 1 the fraction becomes 4/5.So
(x-2+1)/(x+1)=4/5
==>(x-1)/(x+1)=4/5
==>5(x-1)=4(x+1)
==>5x-5=4x+4
==> x=9

So, the numerator will be (9-2)=7 (Option 7).

Like? Yes (1) | No   

satyen nabar   11 years AGO

7/9
increase by 1 -- 8/10 = 4/5

Like? Yes (1) | No   

Naveen Pandey   11 years AGO

x-1/x+1=4/5..solving we get 7

Like? Yes (1) | No   

AVIJIT   11 years AGO

let nem=x
den=y
x=y-2;
again x+1/y+1=4/5;
put y=x+2
get x=7;

Like? Yes (1) | No   

pankaj chandra joshi   11 years AGO

fractn=[a/a+2]
botb num and deno. incrsed than frctn=4/5
(a+1)/(a+3)=4/5
a=7
optn 7

Like? Yes (1) | No   

mounika   11 years AGO

1st condition: (x-2)/x [consider (x-2) is the numerator and x is the denominator ]
2nd condition: (x-2+1)/(x+1)=(x-1)/(x+1)=4/5
5x-5=4x+4
x=9(denominator)
x-2=9-2=7
so 7 is the numerator.

Like? Yes (1) | No   

OM PRAKASH JHA   11 years AGO

Let the denominator be x then numerator will be x-2.
Now according to question,
(numerator+1)/(denominator+1)=4/5
i.e. (x-2+1)/(x+1)=4/5
solving we get =9
hence numerator will be x-2 i.e. 7

Like? Yes (1) | No   

LOGAN WOLVERINE   11 years AGO

fraction x/y;
x=y-2;
so y-2/y;
increase by 1 is 4/5;
so y-2+1/y+1=4/5;
y-1/y+1=4/5;
5y-5=4y+4;
y=9;
x=7;
fraction x/y=7/9;

Like? Yes (1) | No   

OM PRAKASH JHA   11 years AGO

Let the denominator be x then numerator will be x-2.
Now according to question,
(numerator+1)/(denominator+1)=4/5
i.e. (x-2+1)/(x+1)=4/5
solving we get =9
hence numerator will be x-2 i.e. 7

Like? Yes (1) | No   

santhosh   11 years AGO

7/9-numerator is 2 less than denom. both increases by 1 it becomes 8/10=4/5

Like? Yes (1) | No   

Ganesh Hembram   11 years AGO

let x be the denominator,
then numerator will be x-2
So,
x-2/x = 4/5
5(x-2)= 4x
5x-10 = 4x
5x-4x = 10
x = 10

Numerator is (x-2)
10-2 = 8
and both numerator & denominator are increases by 1, then the value of numerator will be
8-1 = 7

therefore,the value of numerator is 7

Like? Yes (1) | No   

Sukanya   11 years AGO

The fraction is 7/9.
If 1 is added on numerator & denominator it becomes 8/10 which is equal to 4/5.

Like? Yes (1) | No   

Satyendra Singh   11 years AGO

if y is denominator than
y-2+1/y+1 = 4/5
5y-10+5= 4y+4
y = 9
numerator is y-2 = 7

Like? Yes (1) | No   

Priya Dharshini   11 years AGO

8/10=4/5;
if both numerator and denominator is reduced by 1 den 7/9;so ans is 7

Like? Yes (1) | No   

swethakottakota   11 years AGO

let the given fraction = (x-2)/x
if both the numerator and denominator increases by 1, (x-1)/(x+1) = 4/5
solving this, we can get , x=9.
so numerator of the number = 7

Like? Yes (1) | No   

gopal   11 years AGO

assume the number is x then
numerator is lessthan denamirator so x-2
x-2/x now we add +1 in numerator n denominator then it eqals to 4/5 so equation is x-1/x+1=4/5 solve this equation we get x=9,bt in numeratot it its less than 2 so 9-2=7

Like? Yes (1) | No   

Vijay Chauhan   11 years AGO

4/5 is final answer...multiply by 2
8/10...
diffence is 2...then reduce both by 1...
means 7/9

Like? Yes (1) | No   

lovely verma   11 years AGO

let we take the denominator number is a and numerator is 2 less than denominator so numerator is a-2
so the fraction number is (a-2)/a
after increasing 1 number in both numerator and denominator the fraction number is
= [(a-2)+1]/[a+1]=4/5
= [a-1]/[a+1]=4/5
after solving to the equation :
=(5*a)-5=(4*a)+4
= 5a-4a=4+5
= a=9 which is denominator
and numerator is a-2= 9-2 =7
so numerator is 7 ans.

Like? Yes (2) | No (1)  

Dhanashri   11 years AGO

if we take numerator as x,then
(x+1)/x+3=4/5
x=7
ans is 7

Like? Yes (1) | No   

priyanka jogin   11 years AGO

they have given numerator is 2 less than denominator so we can find it easily by taking num as d-2 and equate it to new fraction by adding 1 to both num and deno and get the answer

Like? Yes (1) | No   

P.SWETHA   11 years AGO

let numerator=p,denominator=q
given (p+1)/(q+1)=4/5,
p=q-2
q=p+2
substituting in equation we get
p+1/p+3=4/5
p=7

Like? Yes (1) | No   

Devendra Marghade   11 years AGO

If Numerator=x, Denominator=y
Given, x=y-2 ---(i) &
(x+1)/(y+1)=4/5 ---(ii)
Solving (i) & (ii),x=7,y=9

Like? Yes (1) | No   

Megha Aggarwal   11 years AGO

X/X+2 is the original fraction
X+1/X+3 is equal to 4/5
after calculating X=7

Like? Yes (1) | No   

Krutika Neelak   11 years AGO

(d-2+1)/(d+1)=4/5
d=9
n=7

Like? Yes (1) | No   

Ranjana Sinha   11 years AGO

I found the answer by the 'process of elimination' from the given options. First I tried 1/3 which on increasing the numerator and denominator by 1 became 2/4, similarly I tried all the options, until 7. It gave out 8/10 when added 1 to both the numerator and denominator, which is in turn 4/5.

Like? Yes (1) | No   

amit chandel    11 years AGO

let den is \\\'x\\\',den (x-2)/x,now adding 1 in both num and den it will (x-1)/(x+1)=4/5....den can be solved...x=9....den x-2=7

Like? Yes (1) | No   

DHAKAD NEETENDRA SINGH   11 years AGO

let D=x
so N= x-2
according to question,
(x-2)+1 / x+1 = 4/5
after cross multiplication we get x=9
but N= x-2
= 9-2
= 7

Like? Yes (1) | No   

ekta rani   11 years AGO

Let numerator=x
denominator=y
x=y-2.............1)
x+1/y+1=4/5
y-1/y+1=4/5{from 1}
5y-5=4y+4
y=9
x=7{from 1}

Like? Yes (1) | No   

amit bhagat   11 years AGO

let the required fraction be x/y
and x = y-2

so,
y-2/y=4/5
y-2+1/y+1=4/5
5y-5=4y+4
y=9
then, x = y-2 =7

Like? Yes (1) | No   

aasha   11 years AGO

7/9 where 9= 7+2
and (7+1)/(9+1) = 8/10 in turn 4/5

Like? Yes (1) | No   

Sneha   11 years AGO

let numerator be x then denominator be x+2.
x+1/x+2+1=4/5
=>x+1/x+3=4/5
=>5x+5=4x+12
=>x=7

Like? Yes (1) | No   

P Subrahmanya Sarma   11 years AGO

7/9 (before adding 1) after adding 1 to both numerator and denominator it becomes 8/10 which is equal to 4/5

Like? Yes (1) | No   

Prabhat Yadav   11 years AGO

(X-2+1/x+1)=4/5
by solveing above equation we get x=9 since numerator is 2 less then denomenator,hence ans is 7

Like? Yes (1) | No   

Arati Sethi   11 years AGO

let x be the denominator so x-2 is numerator
given x-2/x
after increase by 1,x-2+1/x+1=4/5
after solving x=9
so numerator=x-2=9-2=7

Like? Yes (1) | No   

Siba Kumar Sahu   11 years AGO

let denominator be x
then numerator=x-2
i.e x/x-2
a/t qustn: x-2+1/x+1=x-1/x+1
or x-1/x+1=4/5
or x=9
and numerator= x-2=9-2=7
so option 7 is correct

Like? Yes (1) | No   

TAREK HAYDER SK   11 years AGO

4*2/5*2=8/10;
so, (8-1)/(10-1)=7/9

Like? Yes (1) | No   

raghav kumar   11 years AGO

let numerator be x and denominator be y
x=y-2
and x+1/y+1=4/5
solving we get x=7 and y=9

Like? Yes (1) | No   

vishal singh thapa   11 years AGO

according to info in ques.;
a=b-2;
a+1/b+1=4/5;
b-1/b+1=4/5;
hence
b=9;
and
a=7;

Like? Yes (1) | No   

harika narreddy   11 years AGO

(x-2/x)=4/5; by solving this we can get numerator as
7

Like? Yes (1) | No   

Vinay Sipani    11 years AGO

Let numerator be x => denominator is (x + 2)

If both increased by 1 then ratio = 4/5

=> (x + 1) / (x + 3) = 4 / 5
=> 5x + 5 = 4x + 12
=> x = 7

So numerator is 7.

Like? Yes (2) | No (2)  

Harsh Poddar   11 years AGO

denominator=x
numerator=x-2
x-2/x is the fraction
numerator and denominator both increase by 1
numerator=x-2+1=x-1
denominator=x+1
new fraction=4/5
(x-1)/(x+1)=4/5
solve x=9
numerator=9-2=7
option 7

Like? Yes (1) | No (1)  

VIKASH RAY   11 years AGO

After adding 1 in both numerator and denominator the fraction is-4/5.and the numerator is 2 less than denominator.
s0,option 7 is correct.

Like? Yes (1) | No (1)  

13 Jan 2014 Sequence

What is the next number of the following sequence
1,1,2,6,24,....

OPtion

1) 36
2) 48
3) 54
4) 64
5) 72
6) 78
7) 81
8) 120
9) 124
10)None of these

Solution

1!/1 = 1
2!/2 = (1*2)/2 = 1
3!/3 = (1*2*3)/3 = 2
4!/4 = (1*2*3*4)/4 = 6
5!/5 = (1*2*3*4*5)/5 = 24
Similarly
6!/6 = (1*2*3*4*5*6)/6 = 120

Option 8)

Correct Option: 8 Best Solution (79)

Akanksha Mishra   11 years AGO

look upon the series carefully

1,1,2,6,24
1 * 1 = 1
1 * 2 = 2
2 * 3 = 6
6 * 4 = 24
24* 5 = 120
so answer will be 120

Like? Yes (20) | No (15)  

Chandni Maurya   11 years AGO

ans=120
0!=1
1!=1
2!=1*2=2
3!=1*2*3=6
4!=1*2*3*4=24
similarly,
5!=1*2*3*4*5=120

Like? Yes (4) | No (1)  

ROHIT UPADHYAY   11 years AGO

Factorial
fifth no. is -----> 5! = 5*4*3*2*1 =120

Like? Yes (2) | No (1)  

julee verma   11 years AGO

1*1=1,
1*2=2
2*3=6
6*4=24
24*5=120
so answer =120

Like? Yes (1) | No   

YUVAN SATHAIAH   11 years AGO

1
1*1=1
2*1=2
3*2=6
4*6=24
5*24=120
option is 8)120

Like? Yes (1) | No   

lovely verma   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No   

LOGAN WOLVERINE   11 years AGO

720,5040 because you multiply continuously to the previous term i.e 1*2,2*3,6*4,24*5,120*6.....
so 120*6=720 an 720*7=5040

Like? Yes (1) | No   

R.JOTHILAKSHMI   11 years AGO

1 2 6 24 120...
1 4 18 96
3 14 78
where each term is the previous term multiplied by n.

Like? Yes (1) | No   

Devendra Marghade   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No   

Ganesh Hembram   11 years AGO

1 * 1 = 1
1 * 2 = 2
2 * 3 = 6
6 * 4 = 24
24* 5 = 120

Like? Yes (1) | No   

anand madhav   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No   

RAJESH KUMAR SINGH   11 years AGO

1
1*1=1;
2*1=2;
3*2=6;
4*6=24;
5*24=120;

Like? Yes (1) | No   

Dhanashri   11 years AGO

(1+1)*1=2;
(1+2)*2=6;
(2+6)*3=24;
(6+24)*4=120;

Like? Yes (1) | No   

jayakumar   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No   

satyen nabar   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120.

Like? Yes (1) | No   

Vijay Chauhan   11 years AGO

0! = 1,
1! = 1,
2! = 2,
3! = 6,
4! = 24,
5! = ?(120)...
factorial of 5 = 120 so answer is 120

Like? Yes (1) | No   

chandan   11 years AGO

check them first

count 24=5th position

24*5=120

Like? Yes (1) | No   

madanpal vishwakarma    11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
then
24*5=120

Like? Yes (1) | No   

g.soujanya   11 years AGO

1 1 2 6 24 120
multiplied by 1,3,5

Like? Yes (1) | No   

bhavana wali   11 years AGO

1x1=1
1x2=2
2x3=6
6x4=24
24x5=120

Like? Yes (1) | No   

NITESH KUMAR   11 years AGO

since first number i.e 1 is multiplied with 1 we get second number i.e 1 ,then the second number is multiplied with 2 we get third number i.e 2 and soon.
i.e:1
1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No   

mahima srivastava   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120
so option 8

Like? Yes (1) | No   

Hemalatha   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120 so ans is D

Like? Yes (1) | No   

Somya chaturvedi   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No   

Sudipta Mukherjee   11 years AGO

1*(1+0)=1
1*(1+1)=2
2*(1+2)=6
6*(1+3)=24
24*(1+4)=120

Like? Yes (1) | No   

Arunabh Saxena   11 years AGO

If we divide 2nd no. from 1st no. and do it throughout the series.
the value of quotient is increasing by 1.
like
1\\1=1
2\\1=2
6\\2=3
24\\6=4
then it turn to have 5 in quotient
so we multiply 24 to 5
and the answer is 120.
thanks you all for reading my answer.

Like? Yes (1) | No   

nandish   11 years AGO

1,1=(1+1)*1=2,
1,2=(1+2)*2=6,
2,6=(2+6)*3=24,
6,24=(6+24)*4=120,
24,120=(24+120)*5=720,
so series goes as 1,1,2,6,24,120,720......
hence answer is 120.

Like? Yes (1) | No   

Chithra   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No   

arun prakash singh   11 years AGO

multiplay with increasing no. so 120

Like? Yes (1) | No   

somashekhar   11 years AGO

1*1->1
1*2->2
2*3->6
6*4->24
24*5->120
So option # 8 is the right answer

Like? Yes (1) | No   

ganachari nandini   11 years AGO

1x1,1x2,2x3,6x4,24x5

Like? Yes (1) | No   

vishal singh thapa   11 years AGO

the next term as
1/1=1;
2/1=2;
6/2=3;
24/6=4;
120/24=5.
hence 120.

Like? Yes (1) | No   

TEJASHWI PARIMAL   11 years AGO

1*1=1 (2nd no)
1*2=2 (3rd no)
2*3=6 (4th no)
6*4=24(5th no)
&
now;
24*5=120 (answer)

Like? Yes (1) | No   

DHIRENDRA    11 years AGO

1
1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No   

Gopal_Singh   11 years AGO

1 divide by 1=1
2 divide by 1=2
6 divide by 2=3
24 divide by6=4
120 divide by24=5

Like? Yes (1) | No   

P.SWETHA   11 years AGO

(1+1)*1=2
(2+1)*2=6
(6+2)*3=24
(24+6)*4=120

Like? Yes (1) | No   

shaik akram   11 years AGO

1+1=2x1=2
2+1=3x2=6
6+2=8x3=24
6+24=30x4=120

Like? Yes (1) | No   

Ankit Goyal   11 years AGO

1 * 1 = 1
1 * 2 = 2
2 * 3 = 6
6 * 4 = 24
24 * 5 = 120

Like? Yes (1) | No   

Anand Sanmani   11 years AGO

Incrementing with multiple of respective number

Like? Yes (1) | No   

DIVYA MURUGESH   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No   

Mrinal Kishor   11 years AGO

1 * 1 = 1
1 * 2 = 2
2 * 3 = 6
6 * 4 = 24
24 * 5 = 120

Like? Yes (1) | No   

Krishna Chaitanya   11 years AGO

Given 1,1,2,6,24
the series is 1*1=1
and 1*2=2
2*3=6
5*4=24
24*5=120

Like? Yes (1) | No   

RAKESH ROSHAN   11 years AGO

1 * 1 = 1
1 * 2 = 2
2 * 3 = 6
6 * 4 = 24
so next no will be
multiply of 24 by 5,
then we get 120
24 * 5 = 120

Like? Yes (3) | No (2)  

Dipak Shaw   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No   

Deepak Sharma   11 years AGO

1+1=2 > 2*1=2
1+2=3 > 3*2=6
2+6=8 > 8*3=24
24+6=30> 30*4=120

Like? Yes (1) | No   

surendhar   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No   

ASHISH MITTAL   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No   

Shreya   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120=ans

Like? Yes (1) | No   

pratibha Tiwari    11 years AGO

1
1*1=1
1*1*2=2
1*1*2*3=6
1*1*2*3*4=24
1*1*2*3*4*5=120
see diagonally..it is in sequence 1,2,3,4

Like? Yes (1) | No   

NAGENDRA P   11 years AGO

1*1=1.1*2=2..2*3=6...6*4=24...
in the same way..24*5=120...
so option 8..

Like? Yes (1) | No   

ponprakash   11 years AGO

1 x 2 = 2
2 x 3 = 6
6 x 4 = 24
24 x 5 = 120 ..

Like? Yes (1) | No   

Abhishek Agarwal   11 years AGO

24*5=120

Like? Yes (1) | No   

shiva   11 years AGO

it is a series of factorials
1! , 2! , 3! , 4! , 5!

Like? Yes (1) | No   

lavanya   11 years AGO

1*1=1, 1*2=2, 2*3=6, 6*4=24, 24*5=120

Like? Yes (1) | No   

amit bhagat   11 years AGO

here ,

0!=1
1!=1
2!=2
3!=6
4!=24
5!=120

Like? Yes (1) | No   

saima amin   11 years AGO

it is because
(1+1)*1=2
(2+1)*2=6
(6+2)*3=24
(6+24)*4=120

Like? Yes (1) | No   

swastik   11 years AGO

1x1=1
2x3=6
4x6=24
5x24=120
6x120=720
and so on

Like? Yes (2) | No (1)  

Satya   11 years AGO

1X1=1
1X2=2
2X3=6
6X4=24
24X5=120
120X6=720

Like? Yes (2) | No (1)  

K.Shyam Prasad Rao   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (2) | No (1)  

mohan lal garg   11 years AGO

series in based on factorial
0!=1
1!=1
2!=1*2=2
3!=1*2*3=6
4!=1*2*3*4=24
same as next term is
5!=1*2*3*4*5=120

so ans ia 120
option 8

Like? Yes (6) | No (6)  

Vinay Sipani    11 years AGO

1 , 1 , 2 , 6 , 24 , __

0! , 1! , 2! , 3! , 4!
=> Next term = 5! = 120

Like? Yes (1) | No (1)  

FAISAL RIZWAN   11 years AGO

Given series 1,1,2,6,24,..
1*1=1
1*2=2
2*3=6
6*4=24
24*5=120
So option 8 is correct

Like? Yes (2) | No (2)  

Dr.Dipin Singh   11 years AGO

1,1,2,6,24,..
0!=1
1!=1
2!=2
3!=6
4!=24
next 5!=120..... option 8

Like? Yes (2) | No (2)  

RAKESH   11 years AGO

1,1,2,6,24,..
0!,1!,2!,3!,4!,..
next is 5!=120
(8) 120

Like? Yes (1) | No (1)  

KRISHNAN   11 years AGO

Multiply the number you got in the previous answer by the next number in the 1,2,3... progression.

1*1 = 1
1*2 = 2
2*3 = 6
6*4= 24
24*5 = 120

Like? Yes (3) | No (3)  

Udhaya kumar   11 years AGO

factorial of 0,1,2,3,4,5 respectively...........
0!=1
1!=1
2!=2
3!=6
4!=24
5!=120.........

Like? Yes (2) | No (2)  

Naveen Pandey   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No (1)  

Venkatesh KK   11 years AGO

It is a factorial series..
0!=1, 1!=1, 2!=2*1=2, 3!=3*2*1=6, 4!=4*3*2*1=24,

Therefore,
5!=5*4*3*2*1=120(option 8)

Like? Yes (1) | No (1)  

Avishek Pattanaik   11 years AGO

0!=1
1!=1
2!=2
3!=6
4!=24
5!=120

Like? Yes (1) | No (1)  

swethakottakota   11 years AGO

1, 1*1, 1*2 , 2*3, 6*4, 24*5, 120*6,.....this is the series.......

Like? Yes (1) | No (1)  

AMIT JAISWAL   11 years AGO

1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (1) | No (1)  

sudipta sadhukhan   11 years AGO

Multiply by 1, then multiply by 2,
then multiply by 3, and so on.

Like? Yes (1) | No (1)  

pankaj chandra joshi   11 years AGO

1
,1*1=1
,1*2=2
,2*3=6
,6*4=24
,24*5=120

Like? Yes (1) | No (1)  

rakesh   11 years AGO

1*1 =1 ,1*2=2,2*3=6,6*4=24,24*5=120

Like? Yes (1) | No (1)  

Darsana K   11 years AGO

sequence is like ds
0! 1! 2! 3! 4! 5!
0!=1
1!=1
2!=2
3!=6
4!=24
5!=120

Like? Yes (1) | No (1)  

Sourish Barman   11 years AGO

series is 1,1,2,6,24
so 1*1=1
1*2=2
2*3=6
6*4=24
24*5=120

Like? Yes (3) | No (3)  

susmit   11 years AGO

1*1 = 1
1*2 = 2
2*3 = 6
6*4 = 24
Hence
24*5=120
Hence option no.(8)

Like? Yes (2) | No (3)  

salma   11 years AGO

1*1 = 1
1*2 = 2
2*3 = 6
6*4= 24
24*5 = 120

Like? Yes (1) | No (2)  

Devi Nallamilli   11 years AGO

24*5=120

Like? Yes (1) | No (2)  

10 Jan 2014 Coded Language

If in coded language
MOHAN = 518
SURESH = 656
SHIVANI = 657
Then
GAURAV = ?

OPtion

1) 1056
2) 957
3) 1236
4) 756
5) 676
6) 1144
7) 625
8) 1256
9) 476
10)None of these

Solution

A(1), B(2), C(3), D(4), E(5), F(6), G(7), H(8), I(9), J(10), K(11), L(12), M(13), N(14), O(15), P(16), Q(17), R(18), S(19), T(20), U(21), V(22), W(23), X(24), Y(25), Z(26)
MOHAN = (M+O+H+A)*N = (13+15+8+1)*14 = 37*14 = 518
SURESH = (S+U+R+E+S)*H = (19+21+18+5+19)*8 = 82*8 = 656
SHIVANI = (S+H+I+V+A+N)*I = (19+8+9+22+1+14)*9 = 73*9 = 657
Then
GAURAV = (G+A+U+R+A)*V = (7+1+21+18+1)*22 = 48*22 = 1056
Option 1)

Correct Option: 1 Best Solution (7)

Dr.Dipin Singh   11 years AGO

MOHAN = 518... (13+15+8+1)*14 = 37*14=518
SURESH = 656... ( 19+21+18+5+19)*8 =82*8=656
SHIVANI = 657
Similarly
GAURAV.... (7+1+21+18+1)*22=48*22=1096

Like? Yes (13) | No (4)  

Devendra Marghade   11 years AGO

Multiplying the letters position sum upto last but one letter with last letter
MOHAN =(13+15+8+1)*14=518
SURESH =(19+21+18+5+19)*8=656
SHIVANI =(19+8+9+22+1+14)*9=657
GAURAV =(7+1+21+18+1)*22=1056

Like? Yes (7) | No (6)  

amit bhagat   11 years AGO

MOHAN = 518 = (13+15+8+1)*14
SURESH = 656 = (19+21+18+5+19)*8
SHIVANI = 657 = (19+8+9+22+1+14)*9

SO,
GAURAV = (7+1+21+18+1)*22 = 48*22 =1056

Like? Yes (2) | No (3)  

Chandni Maurya   11 years AGO

ans=1056
considering their alphabetic order no.

M O H A N=> 13 15 8 1 14
=>(13+15+8+1)*14 = 37*14 =518

S U R E S H => 19 21 18 5 19 8
=>(19+21+18+5+19)*8 = 82*8 =656

S H I V A N I=>19 8 9 22 1 14 9
=>(19+8+9+22+1+14)*9 = 73*9 = 657

G A U R A V => 7 1 21 18 1 22
=> (7+1+21+18+1)*22 = 48*22 = 1056

Like? Yes (2) | No (3)  

Naveen Pandey   11 years AGO

sum of all alphabet except last * last alphabet

Like? Yes (1) | No (3)  

AMIT JAISWAL   11 years AGO

MOHAN = 518 = (13+15+8+1)*14
SURESH = 656 = (19+21+18+5+19)*8
SHIVANI = 657 = (19+8+9+22+1+14)*9

SO,
GAURAV = (7+1+21+18+1)*22 = 48*22 =1056 MOHAN = 518 = (13+15+8+1)*14
SURESH = 656 = (19+21+18+5+19)*8
SHIVANI = 657 = (19+8+9+22+1+14)*9

SO,
GAURAV = (7+1+21+18+1)*22 = 48*22 =1056

Like? Yes (2) | No (4)  

sudipta sadhukhan   11 years AGO

G-------7
A-------1
U-------21
R-------18
A-------1
V-------22


so 7+1+21+18+1=48
and last alphabet 22
so 48*22=1056

Like? Yes (2) | No (4)  

09 Jan 2014 A.P.

If the sum of m terms of and A.P. is the same as the sum of its n terms, Then the sum its(m+n) terms is?

OPtion

1) m/2
2) n/2
3) (m+n)/2
4) m+(n/2)
5) mn
6) mn/2
7) 0
8) m
9) n
10)None of these

Solution

By Using AP
Solve using Sn = n/2(2a+(n-1)d)

S(m+n) = 0

option 7)

Correct Option: 7 Best Solution (33)

Hardik Patel   11 years AGO

according to question


m/2 * (2a + (m-1)d) = n/2 * (2a + (n-1)d)

cutting 2

we get



m(2a + (m-1)d) - n(2a+ (n-1)d) = 0

2am + m^2d - md -2an -n^2d +nd =0

2a(m-n) + (m^2 - n^2)d -(m - n)d =0

2a(m-n) + ( (m + n) (m- n) ) d - (m - n )d = 0

taking (m-n) common


2a + ( m + n -1) d = 0------------ 1


S m+n = m+n/2( 2a + (m+n -1)d

we know that 2a + (m+n)d is 0 from eqn. 1


therefore S m+n = 0

Like? Yes (107) | No (25)  

mahima srivastava   11 years AGO

Sn=(n/2)(2a+(n-1)d) ....eq(i)
Sm=(m/2)(2a+(m-1)d) ....eq(ii)
Sm+n=((m+n)/2)(2a+(m+n-1)d) .....eq(iii)
given, Sn=Sm
therefore,
2am+dm^2-md = 2an+dn^2-nd
2a(m-n)+d(m^2-n^2-(m-n))=0
m-n(2a+d(m+n-1)=0
2a+d(m+n-1)=0 .... eq(iv)
from eq(iii) and eq(iv) we get
Sm+n=0
so, option 7

Like? Yes (54) | No (17)  

Chandni Maurya   11 years AGO

ans=0
given,
sum of m terms= sum of n terms
Sm=Sn
(m/2)[2a+(m-1)d]=(n/2)[2a+(n-1)d]
m[2a+(m-1)d] = n[2a+(n-1)d]
2a(m-n)= d[n(n-1)-m(m-1)]
2a(m-n)= d[(n^2-m^2)-(n-m)
2a(m-n)= d[(n-m)(n+m)-(n-m)]
2a(m-n)= -d[(m-n)(m+n-1)]
2a= -d(m+n-1)
2a+d(m+n-1)=0---(1)
now,
sum of (m+n) terms
S(m+n)= {(m+n)/2}[2a+(m+n-1)d]
from(1)
S(m+n) ={(m+n)/2}[0] = 0

Like? Yes (28) | No (16)  

ROHIT UPADHYAY   11 years AGO

m terms=m/2[2a+(m-1)d]
n terms=n/2[2a+(n-1)d]
because m terms = n terms
m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m^2d-md=2an+n^2d-nd
2a(m-n)+(m-n)(m+n)d - md + nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0
(2a-d)(m-n)=2a(m-n)- md + nd
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a ---------eq(1)
m+n terms=n/2*[2a+(m+n-1)d]
=n/2*[2a-2a]
=0

Like? Yes (22) | No (12)  

Bharat Bhushan Jha   11 years AGO

let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a (equation 1)
sum of m+n terms = m=n/2*[2a+(m+n-1)d] (equation 2)
put the vaue of equation 1 to equation 2 you will get the zero

Like? Yes (12) | No (5)  

Abhimanyu singh   11 years AGO

Sm=Sn

->m[2a+(m-1)d]/2=n[2a+(n-1)d]/2
->2a(m-n)+{m(m-1)-n(n-1)}d=0
->2a(m-n)+{(m^2-n^2)-(m-n)}d=0
->(m-n)[2a+(m+n-1)d]=0
->2a+(m+n-1)d=0

.'. Sm+n=(m+n)[2a+(m+n-1)d]/2
=(m+n)[0]/2
=0

Like? Yes (11) | No (5)  

Puja Jha   11 years AGO

let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a (equation 1)
sum of m+n terms = m=n/2*[2a+(m+n-1)d] (equation 2)
put the value of equation 1 to equation 2 you will get the zero.

Like? Yes (14) | No (10)  

Pargat Pal Singh   11 years AGO

m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
solving it, we get
(m+n-1)d= -2a
sum of m+n terms = [(m+n)/2]*[2a+(m+n-1)d] =0

Like? Yes (12) | No (8)  

LOGAN WOLVERINE   11 years AGO

let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a (equation 1)
sum of m+n terms = m=n/2*[2a+(m+n-1)d] (equation 2)
put the vaue of equation 1 to equation 2 you will get the zero

Like? Yes (5) | No (3)  

Krishna Chaitanya   11 years AGO

Given sum of m terms of and A.P. is the same as the sum of its n terms
We know sum of n terms of AP= n/2(2a+(n-1)d)
Therefore sum of m terms =m/2(2a+(m-1)d)----(1)
sum of n terms=n/2(2a+(n-1)d)-----(2)
According to the problem (1)=(2)
m/2(2a+(m-1)d)=n/2(2a+(n-1)d)
2ma+(m2-m)d=2na+(n2-n)d
(2m-2n)a=(n2-n)d-(m2-m)d
2(m-n)a=n2d-nd-m2d+md
2a(m-n)=-d(m-n)(m+n-1)
a=-d/2(m+n-1)----(3)
From (3) we got a,sum of m+n terms=m+n/2(2(-d/2(m+n-1)+(m+n-1)d)
which is 0
Hence sum of its(m+n) terms is Zero

Like? Yes (10) | No (8)  

Subham Mahanta   11 years AGO

The sum of m terms of the A.P. is = m/2(m+1)
the sum of n terms of the A.P. is = n/2(n+1)
from the question, we get,
m/2(m+1) = n/2(n+1)
=> m2+m = n2+n
=>(m2-n2)+(m-n)=0
=>(m+n)(m-n)+(m-n)=0
=>(m-n)(m+n+1)=0
=>(m+n+1)=0
=>{(m+n)/2}*(m+n+1)=0*{(m+n)/2}
=>{(m+n)/2}*(m+n+1)=0
=> the sum of (m+n) terms of the A.P.=0

Like? Yes (7) | No (5)  

shivam tiwari   11 years AGO

Let AP be first term = A and common difference = D, then Question gives
SM = SN gives M(2A + MD - D) = N(2A + ND - D) whence [2A + (M+ N-1)D] = 0
hence SM+N =(M+N)[2A + (M+N-1)D]/2 =0

Like? Yes (5) | No (3)  

AMIT JAISWAL   11 years AGO

here , sn=sm

m/2[2a+(m-1)d]=n/2[2a+(n-1)d]
am+(m^2-m)d/2=an+(n^2-n)d/2
2am+(m^2-m)d=2an+(n^2-n)/2
2a(m-n)=(n^2-n)d-(m^2-m)d
now adding mn both sides we get-
2a(m-n)+mn=(mn+n^2-n-m^2+m)d
2a(m-n)= (n^2+mn-n-m^2-mn+m)
2a(m-n)= n(m+n-1)d - m(m+n-1)d
2a(m-n)= (n-m)(m+n-1)d
2a = (1-m-n)d

so. s(m+n) = (m+n)/2[2a+(m+n-1)d]
= (m+n)/2[(1-m-n)d+ (m+n-1)d]
= 0

Like? Yes (7) | No (5)  

RagunathKC   11 years AGO

let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d];
sum of n terms=n/2[2a+(n-1)d];
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a ....(eq1)
sum of m+n terms = (m+n)/2*[2a+(m+n-1)d].... (eq2)
put the vaue of eq1 to eq2 you will get the zero

Like? Yes (6) | No (5)  

Ankit Goyal   11 years AGO

This is property of an AP

Like? Yes (13) | No (12)  

Prateek Kumar   11 years AGO

S(n)=n/2[2a+(n-1)d]
S(m)=m/2[2a+(m-1)d]
S(n)=S(m)
so;on equating ,we get
[2a+(m+n-1)d]=0 ;m not equals n
S(m+n)=(m+n)/2[2a+(m+n-1)d]=0

Like? Yes (5) | No (4)  

OM PRAKASH JHA   11 years AGO

Let AP be first term = a and common difference = d, then Question gives
Sm = Sn gives m(2a + md -- d) = n(2a + nd -- d) whence [2a + (m+ n -- 1)d] = 0
hence Sm+n =(m+n)[2a + (m+n--1)d]/2 =0

Like? Yes (6) | No (5)  

julee verma   11 years AGO

sn=sm

m/2[2a+(m-1)d]=n/2[2a+(n-1)d]
am+(m^2-m)d/2=an+(n^2-n)d/2
2am+(m^2-m)d=2an+(n^2-n)/2
2a(m-n)=(n^2-n)d-(m^2-m)d
now adding mn both sides we get-
2a(m-n)+mn=(mn+n^2-n-m^2+m)d
2a(m-n)= (n^2+mn-n-m^2-mn+m)
2a(m-n)= n(m+n-1)d - m(m+n-1)d
2a(m-n)= (n-m)(m+n-1)d
2a = (1-m-n)d

so. s(m+n) = (m+n)/2[2a+(m+n-1)d]
= (m+n)/2[(1-m-n)d+ (m+n-1)d]
= 0

Like? Yes (2) | No (1)  

Rahul gulati   11 years AGO

according to question


m/2 * (2a + (m-1)d) = n/2 * (2a + (n-1)d)

cutting 2

we get



m(2a + (m-1)d) - n(2a+ (n-1)d) = 0

2am + m^2d - md -2an -n^2d +nd =0

2a(m-n) + (m^2 - n^2)d -(m - n)d =0

2a(m-n) + ( (m + n) (m- n) ) d - (m - n )d = 0

taking (m-n) common


2a + ( m + n -1) d = 0------------ 1


S m+n = m+n/2( 2a + (m+n -1)d

we know that 2a + (m+n)d is 0 from eqn. 1


therefore S m+n = 0

Like? Yes (3) | No (2)  

KRISHNAN   11 years AGO

The sum of p terms of an arithmetic series is
p2(2a+(p−1)d) where a is the first term and d is the difference between each term.
We can express what m and n equal to by putting p equal to n and m respectively.

m/2 * (2a + (m-1)d) = n/2 * (2a + (n-1)d)

cutting 2

we get



m(2a + (m-1)d) - n(2a+ (n-1)d) = 0

2am + m^2d - md -2an -n^2d +nd =0

2a(m-n) + (m^2 - n^2)d -(m - n)d =0

2a(m-n) + ( (m + n) (m- n) ) d - (m - n )d = 0

taking (m-n) common


2a + ( m + n -1) d = 0------------ 1


S m+n = m+n/2( 2a + (m+n -1)d

we know that 2a + (m+n)d is 0 from eqn. 1


therefore S m+n = 0

Like? Yes (3) | No (3)  

Naveen Pandey   11 years AGO

let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a (equation 1)
sum of m+n terms = m=n/2*[2a+(m+n-1)d] (equation 2)
put the vaue of equation 1 to equation 2 you will get the zero

Like? Yes (4) | No (4)  

Surya Narayana K   11 years AGO

Given that , Sum of m terms is equal to sum of n terms in A.P. So, m(m+1)/2 = n(n+1)/2.

=>m^2+m=n^2+n
=>m^2-n^2=n-m
=>(m+n)(m-n)=-(m-n)
=>m+n=-1

Sum of its (m+n) terms is (m+n)(m+n+1)/2.

so,Sum of its (m+n) terms is Zero.(Since m+n=-1==>m+n+1=0).

Therefore, Sum of its (m+n) terms is zero(Option 7).

Like? Yes (2) | No (2)  

OM PRAKASH JHA   11 years AGO

Let AP be first term = a and common difference = d, then Question gives
Sm = Sn gives m(2a + md -- d) = n(2a + nd -- d) whence [2a + (m+ n -- 1)d] = 0
hence Sm+n =(m+n)[2a + (m+n--1)d]/2 =0

Like? Yes (4) | No (4)  

vignesh   11 years AGO

let a and c be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a (equation 1)
sum of m+n terms = m=n/2*[2a+(m+n-1)d] (equation 2)
put the vaue of equation 1 to equation 2 you will get the zero

Like? Yes (2) | No (2)  

vishal singh thapa   11 years AGO

SM={2A + (M-1)D}M/2
SN= {2B+(N-1)D}N/2
SINCE
SM=SN
{2A+D(M-1)}M/2={2B+(N-1)D}N/2
2MA+M(M-1)D=2NA+N(N-1)D
2A(M-N)=(N^2-N-M^2+M)D
-2A=(M+N-1)D ....EQ(1)
HENCE
S(M+N)={2A+(M+N-1)D}(M+N)/2
S(M+N)=0 {FROM EQ(1)}

Like? Yes (2) | No (2)  

Sana   11 years AGO

let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+(m*m*d)-md=2an+(n*n*d)-nd
2am+(m*m*d)-md-2an-(n*n*d)+nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
2a(m-n)+(m-n)(m+n)d-d(m-n)=0
2a(m-n)+(m-n)(m+n)d-d(m-n)=0
2a+(m+n)d-d=0;
(m+n-1)d=-2a;[equation 1]
sum of m+n terms = m=(m+n)/2*[2a+(m+n-1)d] (equation 2)
put the vaue of equation 1 to equation 2 you will get the zero

Like? Yes (2) | No (2)  

VIKASH RAY   11 years AGO

let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a (equation 1)
sum of m+n terms = m=n/2*[2a+(m+n-1)d] (equation 2)
put the vaue of equation 1 to equation 2 you will get the zero

Like? Yes (4) | No (5)  

RAKESH   11 years AGO

Sm=Sn
m/2[2a+(m-1)d]=n/2[2a+(n-1)d]
or 2a(m-n)+d(m^2-m-n^2+n)=0
or 2a(m-n)+d(m-n)(m+n-1)=0
or (m-n)[2a+(m+n-1)d]=0
or [2a+(m+n-1)d]=0 ......(1)
so S(m+n)=(m+n)/2 *[2a+(m+n-1)]=0 from eqn(1)

Like? Yes (2) | No (3)  

chirag jain   11 years AGO

Sp=Sq
p/2[2a +(p-1)d]=q/2 [2a+(q-1)d]
p[2a +pd - d]=q[2a+qd - d ]
2ap + p2d - pd =2aq + q2d -qd
2ap-2aq +p2d -q2d -pd +qd =0
2a (p-q) +(p+q)(p-q)d -d(p-q)=0
(p-q)[2a + (p+q)d - d ]=0
2a + (p+q)d - d=0
2a + [(p+q)-1]d=0
hence prove that the sum of its first (p+q) terms is 0.

Like? Yes (3) | No (4)  

Dinesh thatti   11 years AGO

let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a (equation 1)
sum of m+n terms = m=n/2*[2a+(m+n-1)d] (equation 2)
put the vaue of equation 1 to equation 2 you will get the zero

Like? Yes (2) | No (3)  

navneet shankhwar   11 years AGO

sum of m term is equal to sum of n term.
m/2[2a+(m-1)d]=n/2[2a+(n-1)d]
on solving above eq., then we get
2a+(m+n-1)d=0
sum of m+n terms is
m+n/2[2a+(m+n-1)d]=m+n/2[0]=0

Like? Yes (5) | No (6)  

Rituraj Singh   11 years AGO

sum of m terms = sum of n terms
=> m/2(2a+(m-1)d) = n/2(2a+(n-1)d)

on solving this equation we get;

2a + (m+n-1)d = 0..............................(1)

therefore,
sum of m+n terms is

(m+n)/2(2a + (m+n-1)d)
using (1) we get it equal to 0(zero)

Like? Yes (2) | No (3)  

NILOTPAL MRINAL   11 years AGO

n/2[2a+(n-1)d]=m/2[2a+(m-1)d]
after solving it
we get the value of m+n..
which will come 0

Like? Yes (3) | No (5)  

08 Jan 2014 Solve

After dividing an amount into two parts, Mr. Suresh invests these parts in two different business. In first, he gains 50% profit whereas in second he has a loss of 25%. Total profit of Mr. Suresh in first business is 1888 rupee. What is the total investment of Mr. Suresh if he gains 23% profit over all.

OPtion

1) 5775
2) 6465
3) 5328
4) 6732
5) 6963
6) 5421
7) 5900
8) 5855
9) 5515
10)None of these

Solution

Let total investment = P & part investments are x & (P – x)
According to given condition
x/2 = 1888 implies x = 3776
Similarly according to other condition
3x/2 +(3/4)(P – x) = P 123/100
=>3/2 * 3776 +(3/4)(P – 3776) = P 123/ 100
=>3 * 1888 + (3 /4) P – 3 * 944 = 123 P/100
=>5664 – 2832 = 123 P /100 – (3/4) P
=>2832 = (123 P – 75 P)/100
=>2832 = 48 P/100
=>P = 2832 * 100/ 48 = 5900.

Correct Option: 7 Best Solution (29)

ROHIT UPADHYAY   11 years AGO

in first he gain 50%= 100+50=150
in second he loss 25%= 100-25=75
total profit is 23%= 100+23=123
(150*1888)+(75*x)=123(1888+x)
2832200 + 75x =232224+123x
48x =50976
x=1062
Total investment=2(1888+1062)=5900

Like? Yes (9) | No (4)  

Jitendra   11 years AGO

Assume - A is the total amount.
X = investment in 1st business
Y = investment in 2nd business
A = X + Y
Profit in 1st Business = 1888
0.5X = 1888 = 3776 rupees
Overall profit is 23%
A + 0.23A = X + 0.5X + Y - 0.25Y
substituting A = X + Y and also the value of X with 3776
Y = 2124
So, the total Investment A = 3776 + 2124 = 5900

Like? Yes (2) | No (1)  

SHIKHA SUMAN   11 years AGO

let total investment=x
investment in first business=y
then investment in second business=x-y
now for first business:
CP=y
profit=1888
SP=profit+CP
=1888+y
profit%= (profit/CP)*100
50= (1888/y)*100
so y=3776
now for second investment we have:
CP=x-y
loss%=25
loss%= (loss/CP)*100
25=(loss/(x-y))*100
25=(loss/(x-3776)*100
loss=(x-3776)/4
for overall investment:
profit%=23
profit=(23*x)/100
so overall profit= profit in 1st business-loss
in 2nd business
23x/100=1888-(x-3776)/4
x=5900
hence total investment=5900 rupees

Like? Yes (1) | No   

Gurmeet Singh   11 years AGO

Let the two parts be x and y. So the total investment is x+y.
As per statement
x/2=1888
So x=3776
and x+x/2 + y-y/4= x+y + 23/100(x+y)
Solving this we get 9x=16y
So y=2124
x+y = 2124+3776= 5900 is the total investment

Like? Yes (1) | No   

Arati Sethi   11 years AGO

let two parts be x & y
overall profit=23%(x+y)=0.23(x+y)=0.23x+0.23y
profit=50% of x=0.5x
loss=25% of y=0.25y
total=0.5x-0.25y
we know that 0.5x=1888 so x=3776
&0.23x+0.23y=0.5x-0.25y or 0.27x=0.48y or 0.27*3776=0.48y or 1019.52=0.48y or y=2124
so total investment =3776+2124=5900
option 7 is correct.

Like? Yes (3) | No (3)  

ashu baweja   11 years AGO

let the 2 investment be y and x
50% profit on y

50/100(y)=1888

y=3776

25 % loss on x

total profit is 23 %

so equation is

50/100(y) - 25/100(x)=23/100(y+x)

50/100(3776) - 25/100(x)=23/100(3776+x)

188800 -25x = 86848 + 23x

x=2124

total investment = y+x
=3776 + 2124

= 5900

Like? Yes (5) | No (5)  

OM PRAKASH JHA   11 years AGO

Total investment = Rs x
Investment in the 1st business = Rs y
Investment in the 2nd business = Rs (x - y) Profit in the 1st business = 50% of Rs y = Rs y/2 = Rs 1888 => y = 1888*2 = 3776 ------- (1)
Loss in the 2nd business = 25% of Rs (x - y) = Rs (x - y)/4
Net profit in the business = Rs {y/2 - (x - y)/4} => y/2 - (x - y)/4 = (23x)/100 (as per the question)
=> 2y - x + y = (23x * 4)/100 = (23x)/25 => 50y - 25x + 25y = 23x = 75y = 48x ------ (2)
Using the relation (1), we have 48x = 75*3776 => x = 75*3776/48 = 5900 =>
Total investment = Rs 5900

Like? Yes (1) | No (1)  

OM PRAKASH JHA   11 years AGO

Total investment = Rs x
Let Investment in the 1st business = Rs y Investment in the 2nd business = Rs (x - y) Profit in the 1st business = 50% of Rs y = Rs y/2 = Rs 1888 => y = 1888*2 = 3776 ------- (1)
Loss in the 2nd business = 25% of Rs (x - y) = Rs (x - y)/4 Net profit in the business = Rs {y/2 - (x - y)/4} => y/2 - (x - y)/4 = (23x)/100 (as per the question) => 2y - x + y = (23x * 4)/100 = (23x)/25 => 50y - 25x + 25y = 23x = 75y = 48x ------ (2) Using the relation (1), we have 48x = 75*3776 => x = 75*3776/48 = 5900 => Total investment = Rs 5900

Like? Yes (1) | No (1)  

RAKESH   11 years AGO

x,y be parts
then 50x/100=1888
or x=3776
also 150x/100 + 75y/100 =(x+y)123/100
solving
x=3776, y=2124
x+y=5900

Like? Yes (1) | No (2)  

Surya Narayana K   11 years AGO

Let x is investment in first business and y is investment in second business.
It is given Mr.Suresh gains profit 50% in first business and it is given as 1888 rupee.That is (50/100)*x=1888 ==> x=3776.
And also given that 25% loss in second business and 23% profit in total investment. So
(150/100)*x +((100-25)/100)*y = (123/100)*(x+y)
==> 27*x = 48*y
==> y= (27*3776)/48
==> y= 2124

So total investment is x+y= 3776 + 2124 =5900

so answer is 5900(option 6)

Like? Yes (1) | No (2)  

Leelanand Kumar Yalla   11 years AGO

Mr. Suresh In First Investment He Gains 50%. I.e., Equal to 1888. 100% is 3776.
We got First Investment i.e., 3776..
Second Investment, Let us assume 'X'.
In which He got 25% loss..
3776*150/100+X*75/100= (3776+X)*123/100;
By Solving This we will get X as 2124.. So this is Second Investment..
Total Investment is 3776+2124= 5900;
So, This is the answer....

Like? Yes (3) | No (4)  

Suresh   11 years AGO

Consider, Investment in Business 'A' = x , Investment in Business 'B' = y

Total Investment = x+y;

Profit in business 'A' = 50% of x = 0.5x=1888, so x=3776

So now,

1.5x + 0.75y = 1.23(x+y)

since x=3776, y=2124

So, Total Investment , x+y = 5900 (Option 7)

Like? Yes (1) | No (2)  

Harshit Daga   11 years AGO

Assume,
total investment of Suresh in both business= x

Investment in Ist business = y
Investment in IInd business= z

i.e. x = y + z ---------eq (1)

Gain in Ist business= 50% i.e. y/2
Loss in IInd Business = 25% i.e. z/4

Total profit in first business = 1888
it means,
y/2=1888
y= 3776 -------- eq (2)

suresh gains 23% profit over all
it means

y/2 - z/4 = (23/100)x ------ eq(3)

from eq 1,2 and 3, we get

x= 5900
and,
z = 2124

so total investment of suresh in both the business = 5900

Like? Yes (1) | No (2)  

Krishna Chaitanya   11 years AGO

Given Mr.Suresh divides amount and invests in different business
In first he gains 50% profit =)SP1=(150/100)*CP1
In second he loses 25% =) SP2=(75/100)*CP2
Given total profit of Mr. Suresh in first business is 1888 rupee =)SP1-CP1=1888
3/2CP1-CP1=1888=)CP1=3776
Given total overall profit=23%
SP1+SP2=(123/100)*(CP1+CP2)
5664+3/4CP2=(123/100)*(3776+CP2)
On solving we get CP2=2124
Total investment of Mr.Suresh=3776+2124=5900

Like? Yes (2) | No (3)  

ANU   11 years AGO

let T = total investment, f = investment in first business, then in 2nd investment is T-f
profit = 50% in first and is 1888 so you have 0.5 f = 1888 and f = 3776
you also have total profit 23% = 0.23 T = 1888 - 0.25(T-f) = 1888 - .25T +944
which simplifies to (.23 + .25)T = 2832 and T = 2832 / .48 = 5900

Like? Yes (1) | No (2)  

sudipta sadhukhan   11 years AGO

Let X is the total AMT

and also one part is Y.so another part is X-Y

now A/Q X*50/100=1888 ==>X=3776

SO now
1888-((3776-Y)*25/100)=X*23/100

==>X=5900

Like? Yes (1) | No (2)  

sandy   11 years AGO

150x/100+75y/100=123z/100
5666+75y/100=(123(x+y))
5666+75y/100=(123(3776+y))
on solving y=2128
x=3776
x+y=5900

Like? Yes (1) | No (2)  

Harsh Poddar   11 years AGO

First investment=1888+1888=3776
let total investment=x
the according to question
x+23x/100=5664+x-3776-((x-3776)4)
by solving x=5900
option 7

Like? Yes (1) | No (2)  

Muhammad Nadeem Haider Naseem   11 years AGO

Suppose
T = total investment, f = investment in first business,
Similarly in 2nd investment is T-f
profit = 50% in first and is 1888 so you have 0.5 f = 1888 and f = 3776
you also have total profit 23% = 0.23 T = 1888 - 0.25(T-f) = 1888 - .25T +944
which simplifies to (.23 + .25)T = 2832 and T = 2832 / .48 = 5900

Like? Yes (3) | No (4)  

Praveen kumar   11 years AGO

Total investment = Rs x
Investment in the 1st business = Rs y
=> Investment in the 2nd business = Rs (x - y)
Profit in the 1st business = 50% of Rs y = Rs y/2 = Rs 1888
=> y = 1888*2 = 3776 ------- (1)
Loss in the 2nd business = 25% of Rs (x - y) = Rs (x - y)/4
Net profit in the business = Rs {y/2 - (x - y)/4}
=> y/2 - (x - y)/4 = (23x)/100 (as per the question)
=> 2y - x + y = (23x * 4)/100 = (23x)/25
=> 50y - 25x + 25y = 23x
= 75y = 48x ------ (2)
Using the relation (1), we have
48x = 75*3776
=> x = 75*3776/48 = 5900
=> Total investment = Rs 5900

Like? Yes (1) | No (2)  

amit bhagat   11 years AGO

here let the total investment be p.
then, investment in first buisness = q

and investment in second business = (p-q)
now, in first buisness the profit is 50% of q = 1888
therefore value of p = Rs 3776
and in second business net loss is 25% of (p-q)

so, the net gain in overall investment is -
50% of q - 25% of (p-q) = 23% of p
q/2 - (p-q)/4 = 23p/100
2q-p+q = 23p/25
3q = 23p/25+p
3*3776 = 48p/25

p = (3*3776*25)/48
p = 236*25
p = RS 5900

Like? Yes (1) | No (2)  

AMIT JAISWAL   11 years AGO

here let the total investment be p.
then, investment in first buisness = q

and investment in second business = (p-q)
now, in first buisness the profit is 50% of q = 1888
therefore value of p = Rs 3776
and in second business net loss is 25% of (p-q)

so, the net gain in overall investment is -
50% of q - 25% of (p-q) = 23% of p
q/2 - (p-q)/4 = 23p/100
2q-p+q = 23p/25
3q = 23p/25+p
3*3776 = 48p/25

p = (3*3776*25)/48
p = 236*25
p = RS 5900

Like? Yes (1) | No (2)  

julee verma   11 years AGO

let the total investment=x
and let the first part of investment=y, then second part of investment will be x-y
y*50%=1888
according to the question-
y*50%-(x-y)*25%=x*23%
1888-(x-3776)*1/4=x*23/100
on solving we get total investment x=5900

Like? Yes (1) | No (2)  

vishal singh thapa   11 years AGO

LET D BE THE TOTAL AMOUNT.
SP1=X+(50/100)*X
SP1=(150/100)*X .....EQ(1)
SP2=X2+(25/100)*X2
SP2=(75/100)*X2
SINCE
(50/100)*X=1888
SO
X=3776 RS. ....EQ(2)
NOW
(150X+75X2)/100={D+(23/100)*D}
75*(X+D)=123D
48D=75X
D=(75/48)*3776
HENCE
D=5900
which is required amount.

Like? Yes (1) | No (2)  

Pargat Pal Singh   11 years AGO

If suresh invested amt x and y in 2 parts respectively,
0.5x-0.25y = 0.23(x+y)>> y=x*9/16
0.5x =1888; x= Rs 3776
y= Rs 2124
total investment = x+y= Rs 5900

Like? Yes (2) | No (3)  

shivam tiwari   11 years AGO

Total investment = Rs x
Investment in the 1st business = Rs y
=> Investment in the 2nd business = Rs (x - y)
Profit in the 1st business = 50% of Rs y = Rs y/2 = Rs 1888
=> y = 1888*2 = 3776 ------- (1)
Loss in the 2nd business = 25% of Rs (x - y) = Rs (x - y)/4
Net profit in the business = Rs {y/2 - (x - y)/4}
=> y/2 - (x - y)/4 = (23x)/100 (as per the question)
=> 2y - x + y = (23x * 4)/100 = (23x)/25
=> 50y - 25x + 25y = 23x
= 75y = 48x ------ (2)
Using the relation (1), we have
48x = 75*3776
=> x = 75*3776/48 = 5900
=> Total investment = Rs 5900

Like? Yes (2) | No (3)  

Devendra Marghade   11 years AGO

If investment in Ist & 2nd business=x & y respectively
Given 1.5x-x=1888,x=3776
Also 1.5x+0.75y=1.23(x+y)
1.5*3776+0.75y=1.23*3776+1.23y, y=2124
Total investment=x+y=3776+2124=5900

Like? Yes (1) | No (3)  

KRISHNAN   11 years AGO

Let the one part be x and the other part be y

So Suresh got 50% profit from x or 0.5x
50% profit is 1888 so x is 1888 x 2 or 3776
and 25% loss from y or .25y

and over all profit 23% or .23(x+y)

0.5x - 0.25y = 0.23(x+y)
0r 0.27x = 0.48y
or 0.27 x 3776 = 0.48 y or y = 2124

so x + y = 3776 +2124 or 5900
so replacing
Now putting it in equation

0.5x - .25y = .23(x

Like? Yes (1) | No (3)  

Akanksha Mishra   11 years AGO

Let,total investment is Rs x

Investment in the 1st business = Rs y

=> Investment in the 2nd business = Rs (x - y)

Profit in the 1st business = 50% of Rs y = Rs y/2 = Rs 1888
=> y = 1888*2 = 3776 ------- (1)

Loss in the 2nd business = 25% of Rs (x - y) = Rs (x - y)/4

Net profit in the business = Rs {y/2 - (x - y)/4}
=> y/2 - (x - y)/4 = (23x)/100 (as per the question)

=> 2y - x + y = (23x * 4)/100 = (23x)/25
=> 50y - 25x + 25y = 23x
= 75y = 48x ------ (2)

Using the relation (1), we have
48x = 75*3776
=> x = 75*3776/48 = 5900

=> Total investment = Rs 5900

Like? Yes (13) | No (15)  

07 Jan 2014 Solve

A rectangle of perimeter 34 units is inscribed in a circle of diameter 13 units, find its sides.

OPtion

1) 13,17 units
2) 12,15 units
3) 13,5 units
4) 17,12 units
5) 5,12 units
6) 17,5 units
7) 13,12 units
8) 13,15 units
9) 15,5 units
10)None of these

Solution

Let x be one of side
then adjacent side is (34/2)-x = 17-x
Applying pythagoras theorem
solve
x^2-17x+60 = 0
solve
hence sides are 5, 12
option 5)

Correct Option: 5 Best Solution (52)

Akanksha Mishra   11 years AGO

the diameter of the circle is also the diagonal of the rectangle, which means it's the hypotenuse of two congruent right triangles.

the two sides of the triangle (or rectangle) sum to 17 since they are half of the perimeter.

let x = one side
17 - x = the other side

use the Pythagorean theorem:

x^2 + (17 - x)^2 = 13^2

x^2 + 289 - 34x + x^2 = 169

2x^2 - 34x + 120 = 0

x^2 - 17x + 60 = 0

(x - 5)(x - 12) = 0

x = 5 or x = 12

the sides of the rectangle are 5 and 12 units.

that's it! ;)

Like? Yes (15) | No (6)  

ROHIT UPADHYAY   11 years AGO

the diameter of the circle is also the diagonal of the rectangle
the two sides of the rectangle sum to 17 since they are half of the perimeter.

let x = one side
17 - x = the other side

use the Pythagorean theorem:

base^2+height^2=hypotenuse^2

x^2 + (17 - x)^2 = 13^2

x^2 + 289 - 34x + x^2 = 169

2x^2 - 34x + 120 = 0

x^2 - 17x + 60 = 0

(x - 5)(x - 12) = 0

x = 5 or x = 12

Like? Yes (9) | No (6)  

pratibha Tiwari    11 years AGO

The diameter of the circle is the
diagonal of the rectangle.
The width and length of the
rectangle and the diagonal form a
right triangle.
From the pythagorean theorem
w^2 + l^2 = 13^2
w^2 + l^2 = 169
The perimeter of the rectangle
2w + 2l = 34
w + l = 17
l = 17 - w
Substituting
w^2 + (17 - w)^2 = 169
w^2 + 289 - 34w + w^2 = 169
2w^2 - 34w + 289 = 169
2w^2 - 34w + 120 = 0
w^2 - 17 + 60 = 0
Factor
(w - 5)(w - 12) = 0
w = {5, 12}
Plugging into
2w + 2l = 34
2(5) + 2l = 34
10 + 2l = 34
2l = 12
l =12
The rectangle is 5 units by 12
units

Like? Yes (3) | No (1)  

Jigyasa Kumar   11 years AGO

Answer : 5 by 12 units
The diameter of the circle is the diagonal of the rectangle.
The width and length of the rectangle and the diagonal form a right triangle.

From the pythagorus theorem
w^2 + l^2 = 13^2
w^2 + l^2 = 169

The perimeter of the rectangle
2w + 2l = 34
w + l = 17
l = 17 - w

Substituting
w^2 + (17 - w)^2 = 169
w^2 + 289 - 34w + w^2 = 169
2w^2 - 34w + 289 = 169
2w^2 - 34w + 120 = 0
w^2 - 17 + 60 = 0

Factor
(w - 5)(w - 12) = 0

w = {5, 12
keeping into
2w + 2l = 34
2(5) + 2l = 34
10 + 2l = 34
2l = 12
l =12

The rectangle is 5 units by 12 units

Like? Yes (4) | No (3)  

Aditya Verma   11 years AGO

simple perimeter of a rectangle is = 2*(l+h)
now cross check then we get the single answer to satisfy above equation that is 5 and 12
2*(5+12)=34
answer is option 5

Like? Yes (5) | No (4)  

Leelanand Kumar Yalla   11 years AGO

The diameter of the circle is the diagonal of the rectangle.
So, according to Hypotenus Theorem;
Hyp^2 = (Length)^2+(Breadth)^2;
169= (Length)^2+(Breadth)^2;------->(1)
Perimeter of a Rectangle is 2(L+B)= 34;------->(2)
L+B= 17;
17-L= B;
Substitute in Above Equation (1)
169= (L)^2+(17-L)^2;
L^2+289+L^2-34L = 169;
2L^2-34L+120=0;
L=5,12;
SUBSTITUTE IN (2);
2(5+B)= 34;
Therefore B=12;
Therefore The sides of a rectangle are 5,12......

Like? Yes (1) | No   

KRISHNAN   11 years AGO

The diameter of the circle is the diagonal of the rectangle.
The width and length of the rectangle and the diagonal form a right triangle.

From the pythagorean theorem
w^2 + l^2 = 13^2
w^2 + l^2 = 169

The perimeter of the rectangle
2w + 2l = 34
w + l = 17
l = 17 - w

Substituting
w^2 + (17 - w)^2 = 169
w^2 + 289 - 34w + w^2 = 169
2w^2 - 34w + 289 = 169
2w^2 - 34w + 120 = 0
w^2 - 17 + 60 = 0

Factor
(w - 5)(w - 12) = 0

w = {5, 12}

Plugging into
2w + 2l = 34
2(5) + 2l = 34
10 + 2l = 34
2l = 12
l =12

The rectangle is 5 units by 12 units

Like? Yes (2) | No (1)  

Rahul gulati   11 years AGO

Firstly, Only 5,12 is the combination of sides which gives perimeter as 34.

On further checking, if its in a circle then:
(5/2)^2 + (12/2)^2 = (13/2)^2

This is satisfied, hence it is a correct answer.

Like? Yes (1) | No   

NILOTPAL MRINAL   11 years AGO

the two sides of the triangle (or rectangle) sum to 17 since they are half of the perimeter.

let x = one side
17 - x = the other side

use the Pythagorean theorem:

x^2 + (17 - x)^2 = 13^2

x^2 + 289 - 34x + x^2 = 169

2x^2 - 34x + 120 = 0

x^2 - 17x + 60 = 0

(x - 5)(x - 12) = 0

x = 5 or x = 12

the sides of the rectangle are 5 and 12 units.

Like? Yes (2) | No (1)  

vigneshwari   11 years AGO

given that, diameter of the circle = 13
and, rectangle is inscribed in the circle;
so that, diagonal of the rectangle = 13
that is, sqrt(l^2 + b^2) = 13
(l^2 + b^2) = 169....(1)
Given that, perimeter = 2(l+b) = 34
(l+b) = 17
(l+b)^2 = 289
l^2 + b^2 + 2lb = 289
from (1), we have,
169 + 2lb = 289
2lb = 120 => lb = 60 ...(2)
Solving (1) and (2), we get,
(l,b) = (12,5).

Like? Yes (1) | No   

NEERAV KHANDELWAL   11 years AGO

The diameter of the circle is the diagonal of the rectangle.
The width and length of the rectangle and the diagonal form a right triangle.

From the pythagorean theorem
w^2 + l^2 = 13^2
w^2 + l^2 = 169

The perimeter of the rectangle
2w + 2l = 34
w + l = 17
l = 17 - w

Substituting
w^2 + (17 - w)^2 = 169
w^2 + 289 - 34w + w^2 = 169
2w^2 - 34w + 289 = 169
2w^2 - 34w + 120 = 0
w^2 - 17 + 60 = 0

Factor
(w - 5)(w - 12) = 0

w = {5, 12}

Plugging into
2w + 2l = 34
2(5) + 2l = 34
10 + 2l = 34
2l = 12
l =12

The rectangle is 5 units by 12 units

Like? Yes (2) | No (1)  

ANU   11 years AGO

The diameter of the circle is the diagonal of the rectangle.
The width and length of the rectangle and the diagonal form a right triangle.

From the pythagorean theorem
w^2 + l^2 = 13^2
w^2 + l^2 = 169

The perimeter of the rectangle
2w + 2l = 34
w + l = 17
l = 17 - w

Substituting
w^2 + (17 - w)^2 = 169
w^2 + 289 - 34w + w^2 = 169
2w^2 - 34w + 289 = 169
2w^2 - 34w + 120 = 0
w^2 - 17 + 60 = 0

Factor
(w - 5)(w - 12) = 0

w = {5, 12}

Plugging into
2w + 2l = 34
2(5) + 2l = 34
10 + 2l = 34
2l = 12
l =12

The rectangle is 5 units by 12 units

Like? Yes (3) | No (2)  

navneet shankhwar   11 years AGO

perimeter of rectangle is 34 then side will be 12 and 5
therfore, 12+5+12+5=34

Like? Yes (2) | No (1)  

Praveen kumar   11 years AGO

draw the diagram!

the diameter of the circle is also the diagonal of the rectangle, which means it's the hypotenuse of two congruent right triangles.

the two sides of the triangle (or rectangle) sum to 17 since they are half of the perimeter.

let x = one side
17 - x = the other side

use the Pythagorean theorem:

x^2 + (17 - x)^2 = 13^2

x^2 + 289 - 34x + x^2 = 169

2x^2 - 34x + 120 = 0

x^2 - 17x + 60 = 0

(x - 5)(x - 12) = 0

x = 5 or x = 12

the sides of the rectangle are 5 and 12 units.

that's it! ;)

Like? Yes (2) | No (1)  

sravani   11 years AGO

12+5=17
2*17=34.
more over sides must be < 13.

Like? Yes (1) | No   

VISHNUVARDHANA   11 years AGO

If rectangle inscribed in a circle if and only if both the sides of rectangle will less than the diameter of circle
here 12,5 satisfy the perimeter 2*(5+12) = 34

so correct option is 5

Like? Yes (1) | No   

pankaj chandra joshi   11 years AGO

diaa= 13=diagonal of rectngl
pythagoras thorm
13^2=a^2+b^2
also (a+b)^2=a^2+b^2+2*ab
givn 2*(a+b)=34
a+b=17
PUT VALUE ABOVE
AB=60
A+B=17
a=12 b=5

Like? Yes (1) | No   

Chandni Maurya   11 years AGO

ans=>w=5,L=12
let w,L are width and length resp.
since it lies inside the circle of 13 unit dia.
=> the diagonalof rectangle will be 13 unit
=>as given 2(W+L)=34
=> W+L= 17---(1)
also, W^2 + L^2= 13^2---(2) (for the triangle made by diagonal)
by trial and error method we can get easily
5+12=17 will satisfy the condition
ans=>w=5,L=12

Like? Yes (1) | No   

palak diwan   11 years AGO

diameter= x*x+y*y = 13*13
2(x+y)= 34
y=12 or 5
x=5 or 12

Like? Yes (1) | No   

RAKESH   11 years AGO

as 5^2+12^2=13^2

Like? Yes (1) | No   

kishore   11 years AGO

2(12+5)=34

Like? Yes (1) | No   

IFFAT IRAM   11 years AGO

a+b=17
b=17-a
d^2=a^2+(17-a)^2
169=a^2+289+a^2-34a
2a^2-34a+120=0
a^2-17a+60=0
(a-5)(a-12)=0
5,12

Like? Yes (1) | No   

P.SWETHA   11 years AGO

sum of l+b=17 we can check from options,also diameter =13 ==> diagonal of a rectangle = 13
so by using Pythagoras theorem l^2+b^2=(13)^2
==>5^2+12^2=25+144=169.

Like? Yes (1) | No   

sai pavan   11 years AGO

ans is 5
Given that
Perimeter of rectangle = 34 =2(Length+Width)
Diameter of circle=13

so Length+width=34/2=17
from the option 5 5+12=17

Checking with diameter of circle
sqrt(5^2+12^2)=>sqrt(169)=13
So No doubt ans is option 5

Like? Yes (1) | No   

vishal singh thapa   11 years AGO

perimeter=2*(l+b)=34
l+b=17 ......eq(1)
and taking diameter of the circle eill be the diagonal of rectangle.
hence in right angle triangle thus formed;we get by pithagorian theorem
l^2+b^2=diagonal^2
l^2+(17-l)^2=13^2 [from eq(1)]
l^2+289+l^2-34l=169
2l^2-34l+120=0
l^2-17l+60=0
l^2-12l-5l+60=0
(l-5)*(l-12)=0
l=5,12
hence
b=12,5. [from eq(1)]
hence option 5 is correct.

Like? Yes (1) | No   

Siba Kumar Sahu   11 years AGO

perimeter of rectangle= 2(a+b) where a & b are sides of rectangle
so p=2(a+b)=34
(a+b)=17
since rectangle is inscribed in a circle of dia 13 unit so by pythagoras theorem, a^2+b^=13^2 only 12 & 5 of sides
of rectangle possible.
so option 5 is correct.

Like? Yes (1) | No   

G.veeranjaneya vara prasad   11 years AGO

12+5=17 by substitute this in 2(a+b) we will get 34.it is also satisfy circle equation also

Like? Yes (2) | No (1)  

saranya sankar   11 years AGO

The diameter of the circle is the diagonal of the rectangle.
The width and length of the rectangle and the diagonal form a right triangle.

From the pythagorean theorem
w^2 + l^2 = 13^2
w^2 + l^2 = 169

The perimeter of the rectangle
2w + 2l = 34
w + l = 17
l = 17 - w

Substituting
w^2 + (17 - w)^2 = 169
w^2 + 289 - 34w + w^2 = 169
2w^2 - 34w + 289 = 169
2w^2 - 34w + 120 = 0
w^2 - 17 + 60 = 0

Factor
(w - 5)(w - 12) = 0

w = {5, 12}

Plugging into
2w + 2l = 34
2(5) + 2l = 34
10 + 2l = 34
2l = 12
l =12

The rectangle is 5 units by 12 units

Like? Yes (1) | No   

saluwaji shushma   11 years AGO

given rectangle perimeter is 34.
we know dat rectangle perimeter =2(l+b)
so 2(l+b)=34
now we substitute the options in the above formula to satisfy the above condition. here 5th option is enough.

Like? Yes (1) | No   

saurabh anand   11 years AGO

here rectangle is inscribed in circle.
Let length be a and breadth be b.
as, perimeter of rectangle = 34 units
so, a + b = 17 units....... equ(1)

as, length of diameter = 13 units
so, by following pythagoras theoram; where diameter as hypotenus of the right triangle and length and breadth be right angling sides.
a^2 + b^2 = 13^2.......... equ(2)

so, only possible combination for equ (1) & equ (2) is 5 and 12 units.
so, option 5 is correct.

Like? Yes (2) | No (2)  

abhilasha   11 years AGO

The diameter of the circle is the diagonal of the rectangle.
The width and length of the rectangle and the diagonal form a right triangle.

From the pythagorus theorem
w^2 + l^2 = 13^2
w^2 + l^2 = 169

The perimeter of the rectangle
2w + 2l = 34
w + l = 17
l = 17 - w

Substituting
w^2 + (17 - w)^2 = 169
w^2 + 289 - 34w + w^2 = 169
2w^2 - 34w + 289 = 169
2w^2 - 34w + 120 = 0
w^2 - 17 + 60 = 0

Factor
(w - 5)(w - 12) = 0

w = {5, 12
keeping into
2w + 2l = 34
2(5) + 2l = 34
10 + 2l = 34
2l = 12
l =12

The rectangle is 5 units by 12 units

Like? Yes (1) | No (1)  

Aarush Singh   11 years AGO

The diameter of the circle is the diagonal of the rectangle.
The width and length of the rectangle and the diagonal form a right triangle.

From the pythagorus theorem
w^2 + l^2 = 13^2
w^2 + l^2 = 169

The perimeter of the rectangle
2w + 2l = 34
w + l = 17
l = 17 - w

Substituting
w^2 + (17 - w)^2 = 169
w^2 + 289 - 34w + w^2 = 169
2w^2 - 34w + 289 = 169
2w^2 - 34w + 120 = 0
w^2 - 17 + 60 = 0

Factor
(w - 5)(w - 12) = 0

w = {5, 12
keeping into
2w + 2l = 34
2(5) + 2l = 34
10 + 2l = 34
2l = 12
l =12

The rectangle is 5 units by 12 units

Like? Yes (3) | No (3)  

Gurmeet Singh   11 years AGO

Options are silly just by looking at the options you have 2(l+b)=34 for 12 and 5 :D
Moreover l should be less than 13.

Like? Yes (1) | No (1)  

sudipta sadhukhan   11 years AGO

as diametre 13 unts

so 13^2=5^2+12^2

Like? Yes (1) | No (1)  

Ashutosh Nilesh Ujwala Patkar   11 years AGO

perimeter of a rectangle is 2[length + breath]
therefore here length +breadth = 17 units
as 12+5 =17 therefore sides are 12 and 5
as to verify it
6^2+ 2.5^2= radius^2[by pythagoras theorem]
36+6.25=6.5^2
42.25=42.25
ans = 12,5

Like? Yes (1) | No (1)  

Gopal   11 years AGO

perimeter = 34
so add of sides = 17
here (5)^2+(12)^2=(13)^2

Like? Yes (3) | No (3)  

Surya Narayana K   11 years AGO

Diameter of circle = Diagonal of Rectangle
Let L be the length and B be the Breadth of rectangle then
L^2 + B^2 =169 and 2(L+B)=34

The possible solution above two equations is L=12 and B=5.
so it sides are 5,12 units (Option 5)

Like? Yes (1) | No (1)  

Naveen Pandey   11 years AGO

a^2+b^2=169
a+b=17..
solving we get a,b=12,5

Like? Yes (3) | No (3)  

mahima srivastava   11 years AGO

diameter is 13 so none of the side can be more than 13 units
all the other options have value more than except option 5 ie. 5,12 units
moreover 2(5+17)=34
and 34 is the perimeter of rectangle
so option 5

Like? Yes (1) | No (1)  

stutisahoo   11 years AGO

sum of sides is half of perimrter

Like? Yes (2) | No (2)  

TAREK HAYDER SK   11 years AGO

perimeter=34; so x+y=17;x^2+y^2=13^2
so, x=12,y=5(as 169=144+25)

Like? Yes (1) | No (1)  

Doddi Venkata Rao   11 years AGO

2(l+b)=34


so,l+b=17

Like? Yes (1) | No (1)  

Sukanya   11 years AGO

The sides will be less than the diameter as it is inscribed inside the circle.
So from the options 5,12
The perimeter =2(l+b)=2(12+5)=34 is the correct option.

Like? Yes (1) | No (1)  

Parveen Soni   11 years AGO

a+b=34/2=17......... (1)
a^2+b^2=13^2..........(2)
solving 1 & 2 we have
a-b= -7 or +7 and ab=60
hence option 5

Like? Yes (2) | No (2)  

ashu baweja   11 years AGO

as perimeter =34
2(l+b)=34
l+b= 17
apply pythagoras theorem in rectangle where hypatoneus is the diagonal
diadonal^2 = l^2 + b^2
13^ = l^2 + b^2

as l+b =17
therefore on solving equation l=12 ,b=5

Like? Yes (3) | No (3)  

Sourish Barman   11 years AGO

the diagonal of rectangle is = the diameter of circle is 13 units so applying pythagoras theorem
the two sides of rectangle are 5 and 12 as 5^2+12^2=13^2=13 so 5 length=12 breadth =5.

Like? Yes (1) | No (1)  

MD JARJIS HOSSAIN   11 years AGO

2(a+b)=34
=>a+b=17
now trial and error method a=5 & b=12

Like? Yes (1) | No (1)  

kumari julie   11 years AGO

as we know diagonal of rectangle is the diameter of the circle so from pythagores theorem,a^2+b^2=169....(1)
and 2(a+b)=34.....(2)
after solving these equ we get the ans a=12 and b=5
where a nd b are the sides of the rectangle.

Like? Yes (1) | No (1)  

sonali jha   11 years AGO

perimeter=34 that means 2(l+b)=34 => l+b=17
so 5+12=17

Like? Yes (1) | No (1)  

Sudipta Mukherjee   11 years AGO

5^2+12^2=169 ,

Like? Yes (1) | No (1)  

samrat nayak   11 years AGO

perimeter 34 meter
so length+width of rectangular= 17 meter
as diameter of circle is 13 meter = diagonal of rectangle
so 12 and 5 is the correct
12^2 +5^2 = 13^2

Like? Yes (1) | No (1)  

Sneha   11 years AGO

The diameter of the circle is the diagonal of the rectangle.
The width and length of the rectangle and the diagonal form a right triangle.
From the pythagorean theorem
w^2 + l^2 = 13^2
w^2 + l^2 = 169
The perimeter of the rectangle
2w + 2l = 34
w + l = 17
l = 17 - w
Substituting
w^2 + (17 - w)^2 = 169
w^2 + 289 - 34w + w^2 = 169
2w^2 - 34w + 289 = 169
2w^2 - 34w + 120 = 0
w^2 - 17 + 60 = 0
Factor
(w - 5)(w - 12) = 0
w = {5, 12}
The rectangle is 5 units by 12 units

Like? Yes (1) | No (1)  

06 Jan 2014 Sequence

What is the next number of the following sequence
3, 2.5, 2.333, 2.25, 2.2,......

OPtion

1) 2
2) 1.86
3) 3.166
4) 3.5
5) 3.8
6) 2.166
7) 2.35
8) 4
9) 3
10)None of these

Solution

(1+2)/1 = 3/1 = 3
(2+3)/2 = 5/2 = 2.5
(3+4)/3 = 7/3 = 2.333
(4+5)/4 = 9/4 = 2.25
(5+6)/5 = 11/5 = 2.2
Similarly
(6+7)/6 = 2.166

Option 6)

Correct Option: 6 Best Solution (28)

Akanksha Mishra   11 years AGO

Look upon the following sequence carefully
3, 2.5, 2.333, 2.25, 2.2,......

3 = 3/1
2.5 = 5/2
2.333 = 7/3
2.25 = 9/4
2.2 = 11/5

3/1, 5/2 , 7/3 , 9/4 , 11/5 ,...
SO next number will be 13/6
13/6 = 2.166

Like? Yes (27) | No (9)  

saurabh anand   11 years AGO

By carefully observing the series; We get the series as :-
3 = 2 + 1
2.5 = 2 + 1/2
2.333 = 2 + 1/3
2.25 = 2 + 1/4
2.2 = 2 + 1/5
so, actual series is 1, 1/2, 1/3, 1/4, 1/5...

thus, next term of this series is 1/6; which is equal to 0.166

thus succedding term of original series will be 2 + 0.166 = 2.166 ans.

so, correct option is 6.

Like? Yes (7) | No (3)  

Naveen Pandey   11 years AGO

decreasing sequence
2+1=3
2+(1/2)=2.5
2+(1/3)=2.33
2+(1/4)=2.25
2+(1/5)=2.2
2+(1/6)=2.1666

Like? Yes (1) | No   

sairam polaki   11 years AGO

3,3-(1/2),3-(2/3),3-(3/4)........
3-(5/6)=2.166

Like? Yes (1) | No   

prakash kumar   11 years AGO

in generalized form = 2+1/n
n=1, 2+1=3,
n=2, 2+1/2=2.5
similarly n=6,2+1/6=2.166

Like? Yes (1) | No   

Dr.Dipin Singh   11 years AGO

3, 2.5, 2.333, 2.25, 2.2,......
3=2+1/1
2.5=2+1/2
2.333= 2+1/3
2.25=2+1/4
2.2=2+1/5
next number will be 2+1/6=2.1666 --- option 6

Like? Yes (2) | No (1)  

Sneghapriya Ravindran   11 years AGO

Ans: 2.1666

series:
3/1=3
5/2=2.5
7/3=2.333
9/4=2.25
11/5=2.2
13/6=2.166

Like? Yes (1) | No   

ROHIT UPADHYAY   11 years AGO

for first no. ---->1000/1=1000 so---->3.000
for second no.---->1000/2=500 so---->2.500
for third no. ---->1000/3=333 so---->2.333
for fourth no.---->1000/4=250 so---->2.250
for fifth no. ---->1000/5=200 so---->2.200
for six no. ---->1000/6=166 so---->2.166

Like? Yes (5) | No (4)  

vaibhav singh   11 years AGO

3-1/2=2.5
3-2/3=2.333
3-3/4=2.25
3-4/5=2.2
3-5/6=2.166

Like? Yes (1) | No (1)  

Nitesh Joshi   11 years AGO

its sequence of 1/2,1/3,1/4,1/5... so next will be 1/6 = 0.16666 so ans will be 2.166

Like? Yes (1) | No (1)  

YUVAN SATHAIAH   11 years AGO

1/1+2=1+2=3
1/2+2=0.5+2=2.5
1/3+2=0.333+2=2.333
1/4+2=0.25+2=2.25
1/5+2=0.2+2=2.2
then
1/6+2=0.166+2=2.166

Like? Yes (1) | No (1)  

RAKESH   11 years AGO

3, 2.5, 2.333, 2.25, 2.2,......
3/1,5/2,7/3,9,4,11/5,13/6

Like? Yes (1) | No (1)  

ponprakash   11 years AGO

ans =2.166..

Like? Yes (1) | No (1)  

Vijay Chauhan   11 years AGO

3,
only consider after decimal digit
2.5(1/2)
2.333(1/3)
2.25(1/4)
2.2(1/5)
so
2.166(1/6)

Like? Yes (1) | No (1)  

Aditya Verma   11 years AGO

because each number can be written as
3 = 2+1/1
2.5 = 2+1/2
2.33 = 2+1/3
2.25 = 2+1/4
2.20 = 2+1/5
x = 2+1/6
ten x is 2.166
answer is option 6

Like? Yes (1) | No (1)  

Dipak Shaw   11 years AGO

3 (1/1),2 (1/2), 2 (1/3),2 (1/4), 2 (1/5),
so next will be 2(1/6)

Like? Yes (1) | No (1)  

OM PRAKASH JHA   11 years AGO

Here from the series we see that
3-1/2=2.5
2.5-1/(2X3)=2.333
2.333-1/(2X)=2.25
2.25-1/(2X)=2.2
2.2-1/(2X)=2.16
So answer is 2.16

Like? Yes (1) | No (1)  

OM PRAKASH JHA   11 years AGO

Here from the series we see that
3-1/2=2.5
2.5-1/(2X3)=2.333
2.333-1/(2X)=2.25
2.25-1/(2X)=2.2
2.2-1/(2X)=2.16
So answer is 2.16

Like? Yes (1) | No (1)  

Bala Jagadish Peddi   11 years AGO

2+(1/1)=3
2+(1/2)=2.5
2+(1/3)=2.333
2+(1/4)=2.25
2+(1/5)=2.2
2+(1/6)=2.1666

Like? Yes (1) | No (1)  

susmit   11 years AGO

Given sequence is 3/1, 5/2, 7/3, 9/4 , 11/5
Hence the next number is 13/6(Numerator is increasing by 2 and denominator is consecutive)
13/6 = 2.166
Hence the next number will be 2.166(option no.6)

Like? Yes (1) | No (2)  

Surya Narayana K   11 years AGO

Coded language is as follows

2+(1/1)=3
2+(1/2)=2.5
2+(1/3)=2.33
2+(1/4)=2.25
2+(1/5)=2.2
Similarly next number is
2+(1/6) =2.166
So answer is 2.166 (Option 6).

Like? Yes (1) | No (2)  

Prabhat Yadav   11 years AGO

=2+=3
similarly
=2+=2.5
hence
=2+=2.166
ans is 2.166

Like? Yes (1) | No (2)  

sudipta sadhukhan   11 years AGO

3/1=3
5/2=2.5
7/3=2.333
9/4=2.25
11/52.2
13/6=1.1666

Like? Yes (1) | No (2)  

Jitendra   11 years AGO

series is:
(3-1)+1/2, (3-1)+1/3, (3-1)+1/4, (3-1)+1/5, (3-1)+1/6
Answer = 2.166 (option 6)

Like? Yes (1) | No (2)  

Devendra Marghade   11 years AGO

Sequence is 3/1, 5/2, 7/3, 9/4, 11/5, 13/6,...
So 13/6=2.166

Like? Yes (1) | No (2)  

swethakottakota   11 years AGO

3-(1/2 * 1) = 2.5
3-(2/3) = 2.333
3-(3/4) = 2.25
3-(4/5) = 2.2
3-(5/6)=2.1667

Like? Yes (1) | No (2)  

NEERAV KHANDELWAL   11 years AGO

5.5/2....

Like? Yes (1) | No (2)  

S Prakash   11 years AGO

Solution:
1/2=0.5

3-0.5=2.5

(1/2)-(1/3)=0.167

2.5-0.167=2.333

(1/3)-(1/4)=0.083

2.333-0.083=2.25

(1/4)-(1/5)=0.05

2.25-0.05=2.2

(1/5)-(1/6)=0.0333

2.2-0.0333=2.1667
so
Answer is 2.1667 (option is 6)

Like? Yes (1) | No (4)  

03 Jan 2014 Coded language

If in coded language
732 = 2
824 = 4
642 = 8
Then
616 = ?

OPtion

1) 9
2) 8
3) 7
4) 6
5) 5
6) 4
7) 3
8) 2
9) 1
10)None of these

Solution

732 = 7*3*2 = 42 = 2 (digits multiply and use once digit)
824 = 8*2*4 = 64 = 4
642 = 6*4*2 = 48 = 8
then similarly
616 = 6*1*6 = 36 = 6

option 4)

Correct Option: 4 Best Solution (46)

Akanksha Mishra   11 years AGO

Look upon coded language carefully

732 = 7*3*2 = 42
Last digit of 42 is 2 so
732 = 2

824 = 8*2*4 = 64
Last digit of 64 is 4 so
824 = 4

642 = 6*4*2 = 48
Last digit of 48 is 8 so
642 = 8

as well as
616 = 6*1*6 = 36
Last digit of 36 is 6

so 616 = 6
answer will be option 4

Like? Yes (32) | No (7)  

RAKESH ROSHAN   11 years AGO

7*3*2 = 42
8*2*4 = 64
6*4*2 = 48
so
every last digit no is coded
thats why
6*1*6 = 36
so Ans will be 6

Like? Yes (6) | No (2)  

P Niranjan   11 years AGO

in the number multiply with one another and the units digit is the right hand side number

7*3*2 = 42 ==> 2

8*2*4 = 64 ==> 4

6*4*2 = 48 ==> 8

6*1*6 = 36 ==> 6

Like? Yes (4) | No (1)  

kongar Ajaykumar   11 years AGO

7*3*2=42 unit digit 2
8*2*4=64 unit digit 4
6*4*2=48 unit digit 8
then
6*1*6=36 unit digit 6

Like? Yes (3) | No   

ROHIT UPADHYAY   11 years AGO

7*3*2=42----->2
8*2*4=64----->4
6*4*2=48----->8

6*1*6=36----->6

Like? Yes (3) | No   

OM PRAKASH JHA   11 years AGO

7x3x2=42
8x2x4=64
6x4x2=48
Taking the last digit of the answer i.e. unit digit...it is 2,4 and 8. Then 6x1x6=36 so answer is 6

Like? Yes (4) | No (2)  

Sourangsu Bandyopadhyay   11 years AGO

The solution can be derived as:
732=7*3*2=42 (Take Last Digit of Answer)=2
824=8*2*4=64 (Take Last Digit of Answer)=4
642=6*4*2=48 (Take Last Digit of Answer)=8

Then,
616=6*1*6=36 (Answer)=6

Like? Yes (2) | No   

pranitha karnati   11 years AGO

7*3*2= 42 , last digit is 2
8*2*4= 64 ---> 4
6*4*2= 48 ---> 8
hence, 6*1*6= 36 ---> 6 , last digit is 6
answer: 6

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Dr.Dipin Singh   11 years AGO

732 = 7*3*2 =42 ... last digit 2
824 = 8*2*4=64.. last digit 4
642 = 6*4*2= 48.. last digit 8
Then
616 = 6*1*6=36.. last digit 6

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Udhaya kumar   11 years AGO

It is the last digit of product of individual digits.....
6*1*6 = 36 .. last digit 6..........

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Surya Narayana K   11 years AGO

7*3*2 =42 unit digit is 2
8*2*4 = 64 unit digit is 4
6*4*2 = 48 unit digit is 8
so
6*1*6 =36 unit digit is 6.

So answer is 6 (Option 4).

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RAGHAVENDRA RAO   11 years AGO

7*3=21*2=42 last digit is 2
8*2=16*4=64 last digit 4 and so on....

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Ashutosh Nilesh Ujwala Patkar   11 years AGO

Multiply all the digits of the numbers
the digit in the unit's place of the answer is the coded number
eg: 732=2
732=7*3*2=42 digit in unit place 2
616=6
616=6*1*6=36 digit in unit's place 6

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Dinesh thatti   11 years AGO

7*3*2=42 take last digit => 2
8*2*4=64 take last digit => 4
6*4*2=48 take last digit => 8
6*1*6=36 take last digit => 6

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OM PRAKASH JHA   11 years AGO

7x3x2=42
8x2x4=64
6x4x2=48
Taking the last digit of the answer i.e. unit digit...it is 2,4 and 8. Then 6x1x6=36 so answer is 6

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Krishna Chaitanya   11 years AGO

Given 732=2
824=4
642=8
to find 616=?
Here we need to multiply the numbers to get result and unit's digit is the required result
7*3*2=42 so 2 is the answer
8*2*4=64 hence 4
6*4*2=48 hence 8
Similarly 6*1*6=36 so 6 is the answer

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vigneshwari   11 years AGO

consider a number abc.
According to that coded language,
abc = unit digit of (a * b * c)
732 = unit digit of 7x3x2 = unit digit of 42 = 2
824 = unit digit of 8x2x4 = 4
642 = unit digit of 6x4x2 = 8
And, 616 = unit digit of 6x1x6 = 6.

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pankaj chandra joshi   11 years AGO

7*3*2=42----2
8*2*4=64-----4
6*4*2=48------8
so
6*1*6=36.-------6

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g.soujanya   11 years AGO

7*3*2=42 unit digit is 2
8*2*4=64 unit digit is 4
6*4*2=48 unit digit is 8
6*1*6=36 unit digit is 6

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eeranna   11 years AGO

7*3*2=42
8*2*4=64
6*4*2=48
6*1*6=36
take first digit of all the numbers we get 2,4,8,6.. so 6 is the answer.

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Devendra Marghade   11 years AGO

Answer=Unit digit after multiplying the digits of given number.
732=7*3*2=42,Unit digit=2
824=8*2*4=64,Unit digit=4
642=6*4*2=48,Unit digit=8
616=6*1*6=36,Unit digit=6

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Sneghapriya Ravindran   11 years AGO

Ans: 6
7*3*2=42 => unit\'s place =2
8*2*4=64 => unit\'s place =4
6*4*2=48 => unit\'s place =8
so,
6*1*6=36 => unit\'s place =6

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anil   11 years AGO

7*3*2=42 in coded language it is last digit =2;
8*2*4=64 in coded language it is last digit=4;
6*4*2=48 in coded language it is last digit=8;
in the same way

6*1*6=36 in coded language it is last digit=6;

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Rajasekhar M   11 years AGO

7*3*2=42 unit digit is 2
8*2*4 = 64 ====> 4
642 = 48 =====> 8
Then
616 = 36 ===> 6

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jainithish   11 years AGO

732=2 means 7*3*2=42
824=4 so 8*2*4=64
642=8 6*4*2=48
616=6 6*1*6=36

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Keshav Kishlay   11 years AGO

7*3*2=42 so, 2
8*2*4=64 so, 4
6*4*2=48 so, 8
hence 6*1*6=36 so, 6

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S Prakash   11 years AGO

multiply three number and u get 2 digit numbers.
Answer taken has last digit.
Answer is = 6
732=7*3*2=42=2
824=8*2*4=64=4
642=6*4*2=48=8
616=6*1*6=36=6

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AMIT JAISWAL   11 years AGO

7*3*2 = 42
8*2*4 = 64
6*4*2 = 48
6*1*6 = 36

so all the unit digit are being written in R.H.S =
2, 4,8,6

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amit bhagat   11 years AGO

7*3*2 = 42
8*2*4 = 64
6*4*2 = 48
6*1*6 = 36

so all the unit digit are being written in R.H.S =
2, 4,8,6

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Naveen Pandey   11 years AGO

multiply all the three digits and take the unit's digit of each..e.g 7*3*2=42->2
8*2*4=64->4
6*4*2=48->8
hence, 6*1*6=36->6

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KRISHNAN   11 years AGO

7*3*2=42
8*2*4=64
6*4*2=48
6*1*6=36
taken last digits as the answer

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Aditya Verma   11 years AGO

7*3*2=42 last digit of 42 is =2
8*2*4=64 last digit of 64 is =4
6*4*2=48 last digit of 48 is =8
then
6*1*6=36 last digit of 36 is =6
Answer is 6
option 4

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siri   11 years AGO

7*3*2=42
8*2*4=64
6*4*2=48
6*1*6=36
here numbers in one digit are considered.
hence the answer is 6.

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pratibha Tiwari    11 years AGO

7*3*2=42
8*2*4=64
6*4*2=48
6*1*6=36
taken last digits as the ans
so 6 is the ans
Other trick is 1+1=2,2+2=4,4+4=8,8+8=16
So according options unit digit of 16 is 6 so 6will be ans or none of these according to this trick

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Ranjeet   11 years AGO

7*3*2=42(=>last digit is 2)
8*2*4=64(=>last digit is 4)
6*4*2=48(=>last digit is 8)
6*1*6=36(=>last digit is 6)

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Dayashankar Rajhans   11 years AGO

7*(3*2)=42=2
8*(2*4)=64=4
6*(4*2)=48=8
so,6*(1*6)=36=6

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Vikash jain   11 years AGO

Accroding to me in this question
(1)732=7*3*2=>42=take last digit than 2
(2)824=8*2*4=>64=take last digit than 4
(3)642=6*4*2=>48=take last digit than 8
same as:616=6*1*6=>36=take last digit than 6

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sai pavan   11 years AGO

product of 3 digits of each number
And consider the unit digit as answer
7*3*2=42 =>2
8*2*4=64 =>4
6*4*2=48 =>8

6*1*6=36 =>6

so ans is 6

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saurabh anand   11 years AGO

7*3*2=42
8*2*4=64
6*4*2=48
6*1*6=36
taken last digits as the ans
so 6 is the ans

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Sneha   11 years AGO

7*3*2=42
8*2*4=64
6*4*2=48
6*1*6=36
taken last digit as the ans
so 6 is the ans

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navneet shankhwar   11 years AGO

732=7*3*2=42=2
824=8*2*4=64=4
642=6*4*2=48=8
taking last digit as answer
same as 616=6*1*6=36=6
option 4 is correct

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shivam tiwari   11 years AGO

7*3*2=42 take 2
8*2*4=64 take 4
6*4*2=48 take 8
now 6*1*6=36 take 6

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YUVAN SATHAIAH   11 years AGO

answer is last digit number
6*1*6=36

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pavithra   11 years AGO

next series in the form of 2*2=4,4*2=8,8*2=16
i think so last digit is considered...

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Sivanjali   11 years AGO

7*3*2 = 42 unit digit 2
8*2*4 = 64 unit digit 4
6*4*2 =48 unit digit 8
so unit digit of 616 is 6(6*1*6=36)
ans 6.

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Satyendra Singh   11 years AGO

7*3*2= 42 last number is 2
8*2*4 = 64 last number is 4
6*4*2 = 48 last number is 8
6*1*6 = 36 last number is 6

Like? Yes (1) | No (1)