M4maths Previous Puzzles

03 Jun 2015 combination

There are 13 members in a committee having 7 men and 6 women. In how many ways a team of 4 persons can be selected in such a way that at least 1 men would be there in the team.

OPtion

1) 355
2) 435
3) 515
4) 585
5) 645
6) 695
7) 700
8) 725
9) 785
10) 835

Solution

Number of ways with 1 men = 7/1 * 6*5*4/3*2*1 = 140
Number of ways with 2 men = 7*6/2*1 * 6*5/2*1 = 315
Number of ways with 3 men = 7*6*5/3*2*1 * 6/1 = 210
Number of ways with 4 men = 7*6*5*4/4*3*2*1 = 35
Total number of ways = 700

Correct Option: 7 Best Solution (41)

Supriya Mondal   9 years AGO

There are 4 ways by which we can form a team
1 man and 3 women=7C1*6C3=140
2 men and 2 women=7C2*6C2=315
3 men and 1 woman=7C3*6C1=210
4 men=7C4=35
So, total=140+315+210+35=700
So, option 7)700 is the right answer

Like? Yes (2) | No   

viswambher   9 years AGO

since there must be atleast 1 men in the team
so the possibilities are

1 men and 3 women
2 men and 2 women
3 men and 1 women
4 men and o women

so this can be done using combinations such as ncr
ncr=(n!/[(n-r)!r!])

the no of ways of selecting 1 man and 3 women or 2men and 2 women or 3men and 1 woman or 4 men and 0 women is

7c1*6c3 + 7c2*6c2 + 7c3*6c1 + 7c4*6c0

Like? Yes (2) | No   

Ramdas Singathiya   9 years AGO

total men=7
total women=6
select 1 man and 3 women= (7c1) *(6c3) = 140
select 2 man and 2 women= (7c2) *(6c2) = 315
select 3 man and 1 women= (7c3) *(6c1) = 210
select 4 man and 0 women= (7c4) *(6c0) = 35

so total ways=700

Like? Yes (1) | No   

ABHINAV PUSHPAM   9 years AGO

=> 7c1*6c3 + 7c2*6c2 + 7c3*6c1 + 7c4
=> 140+315+210+35 = 700
So, answer is 700
i.e. option 7)

Like? Yes (1) | No   

Surit Datta   9 years AGO

7C4 + (7C3*6C1) + (7C2*6C2) + (7C1*6C3) = 700
so option 7
that's it.

Like? Yes (1) | No   

Deepthi   9 years AGO

(7C1*6C3)+(7C2*6C2)+(7C3*6C1)+7C4 =700

Like? Yes (1) | No   

Aashish gohel   9 years AGO

there is at least one men in committee so the no of ways will b
(7C1)(6C3)+(7C2)(6C2)+(7C3)(6C1)+(7C4)(6C0)=700

Like? Yes (1) | No   

aman aditya   9 years AGO

140+315+210+35=700

Like? Yes (1) | No   

W J Krishna Teja   9 years AGO

total no of ways=7c1*6c3+7c2*6c2+7c3*6c1+7c4

Like? Yes (1) | No   

Venkatesh kadali   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4*6c0=700

Like? Yes (1) | No   

Karthick Raja   9 years AGO

TOTAL POSSIBILITIES
13!/(4!*9!)=715 (4 MEMBERS IN A GROUP)
WE CAN CALCULATE how many ways a team of 4 persons can be selected in such a way that at least 1 men would be there in the team SO WE SUBTRACT ALL 4 MEMBERS ARE WOMEN FROM TOTAL POSSIBILITIES


POSSIBILITIES OF ALL 4 ARE WOMEN
6!/(4!*2!)=15

POSSIBILITIES OF a team of 4 persons can be selected in such a way that at least 1 men would be there in the team =715-15=700

Like? Yes (1) | No   

PRASHANT KUMAR   9 years AGO

ANS=7c4*6c0+7c3*6c1+7c2*6c2+7c1*6c3=700

Like? Yes (1) | No   

SHAIK RAFFEE   9 years AGO

the combination of 4 members out o 7 men and 6 women such that atleast one man should present is
7c1*6c3 + 7c2*6c2 +7c3*6c1 +7c4*6c0
=140+315+210+35
=700

Like? Yes (1) | No   

Yathiraj   9 years AGO

13C4 => 715
Total 715 combinations of 4 people committee can be done from group of 13 people.

Combinations of only women
6C4 => 15

So, the committe with at least one men is 715 - 15 = 700.

Option 7.

Like? Yes (1) | No   

MURUGESAN.P   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4*6c0=700

Like? Yes (1) | No   

nikhil kumar   9 years AGO

7C1.6C3 + 7C2.6C2 + 7C3.6C1 + 7C4 = 700

Like? Yes (1) | No   

naina shukla    9 years AGO

we ve to select 4 person out of 13.
there will be atleast 1 men i.e. at most 4 men can be there
There are 7 men and 6 women. so
(7c1*6c3)+(7c2*6c2)+(7c3*6c1)+(7c4*6c0)
=(7*20)+(21*5)+(35*6)+(35)
=700
so option 7 is correct.

Like? Yes (1) | No   

Ayush Gupta   9 years AGO

7C1*6C3+7C2*6C2+7C3*6C1+7C4*6C0=700

Like? Yes (1) | No   

sompon temtheerakij   9 years AGO

1man+3women = 140 ways
2men+2women = 315 ways
3 men+1women = 210 ways
4men+0women = 35 ways
total ways = 140+315+210+35 = 700 ways

Like? Yes (1) | No   

Meenakshi   9 years AGO

=>7C1*6C3 + 7C2*6C2 + 7C3*6C1+7C4*6C0
solving this answer is 700

Like? Yes (1) | No   

Manoj   9 years AGO

4 members in group but at least one man combination is
7c1*6c3 + 7c2*6c2 +7c3*6c1 + 7c4*6c0
=140+315+210+35
=700

Like? Yes (1) | No   

muthu   9 years AGO

7c1*6c3 +7c2*6c2 +7c3*6c3 + 7c4 =700

Like? Yes (1) | No   

B.venkata suresh   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4=700

Like? Yes (1) | No   

Mehul D Parmar   9 years AGO

here in a question ...atleast 1 man would be there in the team then we take 1 man or more than one like 2,3,and 4 also ...than

7 6 7 6 7 6 7
= c * c + c * c + c * c + c
1 3 2 2 3 1 4

=140+210+315+35

=700

Like? Yes (1) | No   

khaleelulla   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4

Like? Yes (1) | No   

Gaurav Malik   9 years AGO

it is pretty simple !
as it is given in the question atleast one man should be there among 4 persons so you can calculate the ans as:
7C1*6C3+7C2*6C2+7C3*6C1+7C4*6C0=700

Like? Yes (1) | No   

Devendra Marghade   9 years AGO

From 7 men and 6 women, atleast 1 men among 4 can be selected as (1M,3W) or (2M,2W) or (3M,1W) or (4M,0W)
So (7C1*6C3) + (7C2*6C2) + (7C3*6C1) + (7C4*6C0)
=140+315+210+35
=700
So answer is option (7)

Like? Yes (1) | No   

akas   9 years AGO

The No Of all the combinations are =n!/(n-r)!r!
therefore solution:
=7!/6! 1! *6!/3! 3! +7!/5! 2! *6!/4! 2! +7!/4! 3!*6!/5! 1! +7!/3! 4!*6!/6! 0!
= 7*20+21*15+35*6+35
=700

Like? Yes (1) | No   

keerthi   9 years AGO

7c1+6c3+7c2*6c2+7c3*6c1+7c4+6c1
700

Like? Yes (1) | No   

venkatesh   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4*6c0

Like? Yes (1) | No   

Krishan Kumar   9 years AGO

atleast 1 men
fr 1 man 3 women=7C1*6C3
fr 2 men 2 women= 7C2*6C2
fr 3 men 1 women= 7C3*6C1
fr 4 men 0 women= 7C4*6C0

Like? Yes (1) | No   

Gowtham   9 years AGO

7c1*6c3 + 7c2*6c2 + 7c3*6c1 + 7c4*6c0 = 700.

Like? Yes (1) | No   

saranya.s   9 years AGO

7 men and 6 women.
A team 4 persona can be selected, such that
NC base R combination
700 ways are there

Like? Yes (1) | No   

NAVEEN   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4=700

Like? Yes (1) | No   

p.supriya   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4
140+315+210+35
=700

Like? Yes (1) | No   

Pavithra    9 years AGO

atleast one men must be in the team.So 7c4+(7c3*6c1)+(7c2*6c2)+(7c1*6c3)

Like? Yes (1) | No   

kousik tankala   9 years AGO

total(715)- 0 boys case(15)=700

Like? Yes (1) | No   

Deepak Joshi   9 years AGO

140 + 315 + 210 + 35

Like? Yes (1) | No   

anuradha   9 years AGO

7C1*6C3+7C2*6C2+7C3*6C1+7C4

Like? Yes (1) | No   

Rajat Jain   9 years AGO

(7c1*6c3)+ (7c2*6c2) + (7c3*6c1) + (7c4*6c0)
140+315+210+35
700

Like? Yes (1) | No   

ADERU CHENCHU NITHISH   9 years AGO

no.of ways = 7c4+(7c3*6c1)+(7c2*6c2)+(7c1*6c3)
=35+210+315+140=700

Like? Yes (1) | No   

02 Jun 2015 Number of boys

A sum of Rs.41 was divided among 50 boys and girls. Each boy gets 90 paise and a girl 65 paisa. The number of boys.

OPtion

1) 16
2) 36
3) 24
4) 18
5) 22
6) 32
7) 30
8) 40
9) 14
10)34

Solution

Average money received by each = Rs.(41/50) = 82 pais.
Ration of boys and girls = 17:8
Therefore
Number of boys = 50*(17/25) = 34 paisa

Option 10)

Correct Option: 10 Best Solution (71)

Ramdas Singathiya   9 years AGO

assume No. of boys=X
No. of Girls=Y
so

X+Y=50....................(1)
90X+65Y=4100...........(2)

solve these two we get X=34

Like? Yes (34) | No (9)  

viswambher   9 years AGO

let the no of boys =x
no. of girls =y
total no. of boys and girls = x+y =50
each boys gets 90 paise
each girl 65paise
all the boys get 90x paise
girls get 65y paise
total amount 4100 paise

solving
x+y=50
90x+65y=4100
we get x=34
so total no. of boys =34

Like? Yes (11) | No (1)  

ANANTHI.P   9 years AGO

NUMBER OF BOYS=X
NUMBER OF GIRLS=50-X
Xx90+(50-X)x65==4100 PAISE
X=34

Like? Yes (9) | No   

devvrata   9 years AGO

X + Y=50
90X +65Y=4100
By solving this equation we will get X=34

Like? Yes (6) | No   

Radhika   9 years AGO

Let the number of boys be 'x'
let the number of girls be 'y'
then x+y=50 -------------------------- (eq 1)
and 90x+65y=4100 (in paise) --------------(eq 2)
By solving (eq 1) and (eq 2) we get the number of boys is 34 and girls is 16.
Therefore, ans is 34

Like? Yes (6) | No (1)  

prabhat kumar   9 years AGO

let
x no. of boys
y no. of girls
x+y=50
X×90/100+Y×65/100=41
X=34

Like? Yes (5) | No (1)  

padam   9 years AGO

use alligation techniques convert 41rs in paise and divie by 50 to get average
now on soving aligation ratio of boys and girls is 17:8 and boys+girls =50 so boys are 34 and girls are 16

Like? Yes (8) | No (4)  

vishnuvardhan   9 years AGO

0.9b+0.65g=41
b+g=50
b=34
g=16

Like? Yes (4) | No   

poojasaisri   9 years AGO

b+g=50 eqn 1
90b+65g=4100 eqn 2
solving eqn 1 and eqn 2
we get
90b+65g=4100
65b+65g=3250
25b=850
b=34

Like? Yes (3) | No   

ABHINAV PUSHPAM   9 years AGO

Let X be number of boys and (50-X) be number of girls.
=>.90x+.65(50-x)=41
=>.25x=8.5
=>x=34
option 10)

Like? Yes (4) | No (2)  

Yathiraj   9 years AGO

Let B be no. of Boys and G be no. of Girls
B+G = 50
Each boy gets 90 paise and each girl gets 65 paisa and total is 41 rupees
Hence,
0.9B + 0.65G = 41
G=50-B
0.9B + 0.65(50-B)=41
0.9B + 32.5 - 0.65B = 41
0.25B = 8.5
B=34

Option 10.

Like? Yes (2) | No   

SAKTHEESVARI S   9 years AGO

90*boys+65*girls=4100paise:
hence,90*34+65*(50-34)=4100 paise;
so 34 boys

Like? Yes (2) | No   

vivek   9 years AGO

x+y = 50
90x+65y=4100
so, x= 34

Like? Yes (2) | No   

mahak saha   9 years AGO

let the no girls & boys be x & y resptvly
then, x+y=50
and .9y+.65x=41
solve for x & y
then y=34 is the ans

Like? Yes (2) | No   

Supriya Mondal   9 years AGO

Let, number of boys x
Then we can write,
.9*x+.65*(x-50)=41
Solving this equation we get x=34
So, option 10)34 is the right answer

Like? Yes (2) | No   

aman aditya   9 years AGO

Solving equation
X+y=50
And
.90x+.65y=41
Where x= boys
Y=girls
X=34
Y=16

Like? Yes (1) | No   

Karthick Raja   9 years AGO

ASSUME BOYS X AND ASSUME GIRLS Y
X + Y = 50 ------>EQN1
BOYS GET 0.90
GIRLS GET 0.65
0.90X + 0.65 Y = 41 ----> EQN2

(EQN 1 * 0.65) - EQN 2
0.65X + 0.65Y = 32.5 (-)
0.90X + 0.65Y = 41
-0.25X = -8.5
X = (8.5/0.25)
X = 34
TOTAL NUMBER OF BOYS = 34

Like? Yes (1) | No   

MRITUNJAY   9 years AGO

x*0.9+(50-x)*0.65=41
solve this we get
0.25*x=8.5
x= 34
so number of boys is 34
option 10

Like? Yes (1) | No   

Akhil Malhotra   9 years AGO

Let number of boys=x nd number of girls =y
90x+65y=4100
x+y=50
Put value of y frm dis in first
90x+65(50-x)=4100
Solving this we get x=34

Like? Yes (1) | No   

sravani   9 years AGO

let the boys be x.
x(90)+(50-x)65=41*100Paise
x=34.

Like? Yes (1) | No   

N.Dayanithi   9 years AGO

No of Boys and Girls = 50
If Boys = 34, 34*0.9 = 30.6
Then girls will be 16, 16*0.65 = 10.4
30.6 + 10.4 = 41
So no.of Boys = 34

Like? Yes (1) | No   

NESAB JAFAR   9 years AGO

x+y=50
90X+65Y=4100
x=34 num of boys

Like? Yes (1) | No   

Ankita Anand   9 years AGO

90x+65(50-x)=4100
90x-65x=4100-3250
25x=850
x=34

Like? Yes (1) | No   

sompon temtheerakij   9 years AGO

let number of boys = A , number of girl = B
A+B = 50.....1
90A+65B = 4100.....2
A = 34

Like? Yes (1) | No   

Shubham Khandelwal   9 years AGO

X = Number of boys and Y = Number of girls.
X + Y = 50
also 90X + 65Y = 4100
Equating these equations will give X = 34 and Y = 16

Like? Yes (1) | No   

Aishwariya    9 years AGO

Average amount is 82 p(41*2) by allgation and mixture ratio of amounts are
for grls 90-82= 8
for boys it is 82-65 = 17
the ratio of boys to girls are 17:8
the no of boys are: (17/25)*50 = 34

Like? Yes (1) | No   

Girjesh kumar   9 years AGO

Let number of boys are X and number of girls are Y.
then according to problem ....
0.9X + 0.65Y = 41
so now....
Y = (4100 - 90x)/65 --------
and the total number of boys and girls are 50..so
X + Y = 50 ---------
after solving these 2 equation...we get
X = 34
so the number of boys are 34

Like? Yes (1) | No   

shivangi   9 years AGO

b=90paise
g=65paise
90b+65g=41*100paise=4100
n total money=45*100paise
hence b+g=4500
on solving 2 equations ,we get
g=16 and 50-16=34 boys

Like? Yes (1) | No   

naina shukla    9 years AGO

we can check the ans from solution
we take 34 then 34*90=3060
i.e. rs30.60
then no og girls =16, i.e. 16*65=1040. i.e rs 10.40
10.40+30.60=41
so , option (10) is correct answer.

Like? Yes (1) | No   

Deepthi   9 years AGO

let no of boys -X
let no of girls-Y
Total money is 41
i.e 90/100*x+65/100*y=41
we get 18X+13Y=820--------------1
Total members -50
so X+Y=50--------------------2
solving 2 eqns we get X=34 and Y=16

Like? Yes (1) | No   

Pradyumna K S   9 years AGO

Let x be the number of boys. Then 50-x is the number of girls.
given that each boy gets 90 paise and a girl 65 paisa.
This can be written as
90x+(50-x)65=4100
solving.....
x=34
therefore the answer is option 10

Like? Yes (1) | No   

ravi singh   9 years AGO

answer is 34

since, no. of girls and boys are integer.
by putting all value only option 10 will satisfied.
boys=34 and girls=16

Like? Yes (1) | No   

SUSHANT KUMAR   9 years AGO

X+Y=50
90X+65Y=4100
SOLVING FOR X WE GET 34.

Like? Yes (1) | No   

vivek kumaR   9 years AGO

let x is no. of boys
then no. of girl is: 50-x
0.9*x+0.65*(50-x)=41
when we sove this equation we got x=34

Like? Yes (1) | No   

Aashish gohel   9 years AGO

let ,
x be no of boys &
y be no of girls
so as per question
x+y=50 (total no of boys&girls)
90x+65y=4100(90 paisa to boy & 65 paisa to girl & total 4100 paisa that is Rs 41 )
solving both equations we get x=34
that is the no of boys

Like? Yes (1) | No   

nikhil kumar   9 years AGO

let x=no of boys and y=no of girls

x+y =50
and
0.90x + 0.65y = 41 (in rs)
solving the two equations we get
x=34

Like? Yes (1) | No   

Aarti Galar   9 years AGO

Average money received by each of them Rs. 41 = 82paisa
Sum received by each boy 90 paise
Sum received by each girl 65 paise
Average 82 paisa in the ratio 17 : 8
By solving we getno. of boys = 34
therefore the ans. is 34

Like? Yes (1) | No   

vini   9 years AGO

Total strength 50, let 'x' be no.of boys and 50-x be no.of girls

(0.90*x) + (0.65)*(50-x) = 41
0.90x + 32.5 - 0.65x = 41
0.25x= 8.5
x= 34
so, no.of boys = 34

Like? Yes (1) | No   

Meenakshi   9 years AGO

Average money received by each =Rs.41=82 p.
Sum received by each boy=90 p
Sum received by each girl=65 p
Average=82 p
By alligation rule,
ratio of boys=82-65=17
ratio of girls=90-82=8
Ratio of boys and girls=17:8
So,number of boys=17*2=34
option 10)

Like? Yes (1) | No   

Bashitwa Roy   9 years AGO

Let number of boys be x. then 41=.9*x+.65*(50-x). Then solve for x.

Like? Yes (1) | No   

Preethi   9 years AGO

34*90=30.60
16*65=10.40

Like? Yes (1) | No   

kiran   9 years AGO

let no. of boys be x
=> no. of girls = 50-x

according to the problem,
x*90/100 + (50-x)65/100 = 41

on solving x = 34 = no. of boys

Like? Yes (1) | No   

satthi reddyzz   9 years AGO

34*0.90+16*0.65=41

Like? Yes (1) | No   

m.lokeswari   9 years AGO

0.90x+0.65(50-x)=41
Finally x=34

Like? Yes (1) | No   

SUNNY KAMA   9 years AGO

x+y=50 & 90x+65y=4100 on solving x=34

Like? Yes (1) | No   

Rishika Vishwakarma   9 years AGO

let d nos. of boys=x
nos of grls= 50-x

eqation becomes-
(0.91)x+(50-x)(0.65)=41
x=34

Like? Yes (1) | No   

arun sharma   9 years AGO

34*.90=30.6
16*.65=10.4
30.4+10.6=41
so 34 boys

Like? Yes (1) | No   

Rohit   9 years AGO

90b+65g=4500
90b+65g=4100

Like? Yes (1) | No   

Mani Sachdev   9 years AGO

.90x+.65y=41------(i)
x+y=50--------------(ii)
Solve eq.
x(boys)=34 Ans

Like? Yes (1) | No   

BALAJI   9 years AGO

90×34=3060
65×16=1040
so ans:34

Like? Yes (1) | No   

B.venkata suresh   9 years AGO

34*0.90+0.65*65=41

Like? Yes (1) | No   

Dhananjay Gupta   9 years AGO

x+y=50 and 90x+65y=4100.after solving these two equation we will get x=34 which is the number of boys

Like? Yes (1) | No   

PRASHANT KUMAR   9 years AGO

x+y=50
.9x+.65y=41

Like? Yes (1) | No   

anmol jain   9 years AGO

let boys be x n girls be 50-x
then
x *.9 + (50-x)*.65=41

Like? Yes (1) | No   

Y M ANIEF   9 years AGO

90 65
82
17 : 8

25--50
17--34

Like? Yes (1) | No   

Deepak kushwaha   9 years AGO

here
b+g=50........(1)
Rs41==4100 paisa
90b+65g=4100.........(2)
from (1) & (2)
we get b=34

Like? Yes (1) | No   

a.k.gayathridevi   9 years AGO

34*90+16*65=41

Like? Yes (1) | No   

Ayush Gupta   9 years AGO

x+y=50
90x+65y=4100

solve both
x=34(boys)
y=16(girls)

Like? Yes (1) | No   

sahil arora   9 years AGO

let no.of boys be x
no.of girls be y
x+y=50
.90x+.65y=41
on solving both equation y=16
therefore x=34

Like? Yes (1) | No   

Soumen Dofadar   9 years AGO

equation1: x+y=50
equation2: 90x+65y=4100

Like? Yes (1) | No   

neha gupta   9 years AGO

let x=no of boys
y=no of girls

x+y=50
0.90*x + 0.65*y =41

on solving above two equations
we get

0.25x=8.5
x=8.5/0.25
x=850/25

x=34

Like? Yes (1) | No   

Ashiya Bhuyan   9 years AGO

4100=90.b+65(50-b)

Like? Yes (1) | No   

krishna chandrika   9 years AGO

let x boys nd y girls..
x+y=50
90x+65y=4100
solving above 2 eqs x=34,y=16
no.of boys=34

Like? Yes (1) | No   

Aditi Khandagale   9 years AGO

total money=Rs.41
total boys & girls=50

so 34*90=30.60
16*65=10.40

30.60+10.40=41

so no. of boys=34
& no. of girls=16

Like? Yes (1) | No   

Archie   9 years AGO

let x be no. of boys and y be no. of gals
Then,
x+y=50 ---(1)
&
0.90x+0.65y =41 -----(2) [Since amt. distributed among boys+amt. distributed among gals=actual sum]
Solving (1)&(2) we get,
x=34 and y=16
Hence no. of boys=34 and correct option is 10 .

Like? Yes (1) | No   

hari prasath   9 years AGO

assume b-boys,g-girls

From the statement the equation forms

b+g=50
90b+65g=4100

if we solve above two equations than the boys are 34

Like? Yes (1) | No   

anuradha   9 years AGO

0.90*x+0.65*(50-x)=41

Like? Yes (1) | No   

Ajayakumar   9 years AGO

there is total 50 peoples (boys&girls), so take "x"number of boys and remaining is (50-x)are girls. If a boy get 90p than 90X will be the total boys got paise, and (50-X)65 of girls having. Than sum of all paise =4100
i,e: 90x+65(50-X) =4100 => X=34.

Like? Yes (1) | No   

Surit Datta   9 years AGO

The number of boys are 34
so option 10
that's it.

Like? Yes (1) | No (1)  

Divyabala R   9 years AGO

Given 50 boys and girls so consider X(boys) + Y(girls) = 50 ;
And 90 paise to each boy and 65 paise to each girls equals, 41 rupees converting this into paise we get 4100 paise =>
90X+65Y =4100 on solving both equation we get Girls is 16 and boys are 34 .

Like? Yes (1) | No (1)  

akas   9 years AGO

solution:
34(90)+16(65)=4100paise
therefore 34 boys and 16 girls

Like? Yes (1) | No (1)  

01 Jun 2015 Sequence

What is the next number of the following sequence
37, 67, 145, 155, 166, 179, ......

OPtion

1) 169
2) 166
3) 192
4) 186
5) 214
6) 236
7) 196
8) 189
9) 268
10)201

Solution

37
37+(3+7)*3 = 37+30 = 67
67+(6+7)*6 = 67+78 = 145
145+(1+4+5)*1 = 145+10 = 155
155+(1+5+5)*1 = 155+11 = 166
166+(1+6+6)*1 = 166+13 = 179
Similarly
179+(1+7+9)*1 = 179+17 = 196
Option 7)

Correct Option: 7 Best Solution (17)

Supriya Mondal   9 years AGO

3+7=10
10*3=30
So, 37+30=67
6+7=13
13*6=78
so,67+78=145
1+4+5=10
10*1=10
so,145+10=155
1+5+5=11
11*1=11
so, 155+11=166
1+6+6=13
13*1=13
so, 166+13=179
1+7+9=17
17*1=17
so,179+17=196
so, option 7)196 will be right answer

Like? Yes (2) | No   

Devendra Marghade   9 years AGO

Multiply by the first digit to the sum of the digits to get the next term:
37+3*(3+7)=67
67+6*(6+7)=145
145+1*(1+4+5)=155
155+1*(1+5+5)=166
166+1*(1+6+6)=179
179+1*(1+7+9)=196
So answer is option (7)

Like? Yes (1) | No   

RAKESH   9 years AGO

179 + 1*(1+7+9) = 196

Like? Yes (1) | No   

Pradyumna K S   9 years AGO

37, 67, 145, 155, 166, 179, ......
The above sequence can be written as.......
37 = 37
37 + 3(3+7) = 67
67 + 6(6+7) = 145
145 + 1(1+4+5) = 155
155 + 1(1+5+5) = 166
166 + 1(1+6+6) = 179
179 + 1(1+7+9) = 196
Therefore the required number is 196 i.e., option 7

Like? Yes (1) | No   

Ashwin Dinesh   9 years AGO

179 + [1*(1+7+9)]
= 179 + 17 = 196

Like? Yes (1) | No   

Josepheena Joseph   9 years AGO

37 + (3*(3+7)) = 67
67 + (6*(6+7)) = 145
145 + (1*(1+4+5)) = 155
155 + (1*(1+5+5)) = 166
166 + + (1*(1+6+6)) = 179
179 + (1*(1+7+9)) = 196

Like? Yes (1) | No   

Ayush Gupta   9 years AGO

To get the next digit add all the digits of the number and multiply this with the first digit of the number and add this value to the number

37=3+7=10*first digit(3)=30 add this to 37=67
67=6+7=13*6=78 add this to 67=145
similarly..
179=1+7+9=17*1=17 add this to 179=196

196 is the ans

Like? Yes (1) | No   

hari prasath   9 years AGO

add all digits in number then multiply it with left most digit.Make addition to that number.

3+7=10 10*3=30 37+30=67
6+7=13 13*6=78 67+78=145
1+4+5=10 10*1=10 145+10=155
.........
1+7+9=17 17*1=17 179+17=196

The ans is OPTION(7) 196

Like? Yes (1) | No   

Jotshnesh Patel   9 years AGO

Ans is 196
Sol:
145 => 1+4+5 = 10,
145+10 = 155.

155 => 1+5+5 = 11,
155+11= 166.

166 => 1+6+6 = 13,
166+13 = 179.

and
179 => 1+7+9 = 17,
179+17 = 196.

Answer is 196.

Like? Yes (1) | No   

Sagar Mate   9 years AGO

Ans = number + (addition of all digits * first digit)

Like? Yes (1) | No   

Sambit S. Mondal   9 years AGO

Add the adition of digits multipled by the first digit to get the next number.

Like? Yes (1) | No   

sompon temtheerakij   9 years AGO

37
67 = 37+3x7+3^2
145 = 67+6x7+6^2
155 = 145+1x5+5
166 = 155+1x5+6
179 = 166+1x6+7
196 = 179+1x9+8

Like? Yes (1) | No   

naina shukla    9 years AGO

37+10*3=67
67*2+11=145
1+4+5=10 , 145+10=155
1+5+5=11 ,155+11=166
1+6+6=13, 166+13=179
1+7+9=179, 179+17=196

Like? Yes (1) | No   

ravi singh   9 years AGO

Answer is 196

(Sum of digit)*(1st digit) + no.

(3+7)*3+37=67

(6+7)*6+67=145
.
.
.
(1+7+9)*1+179=196

Like? Yes (1) | No   

Bashitwa Roy   9 years AGO

The logic behind is add up the digits of each numbers and multiply it with the first digit. Add the result to it and get the next number. 37= 3+7=10. 10*3=30. 37+30=67.
Similarly, 179= 1+7+9= 17. 17*1=17. 179+17=196.

Like? Yes (1) | No   

Soumen Dofadar   9 years AGO

to get nth number we have to sum up all digits of the (n-1)th number then multiply it with first digit and then add it with (n-1)th number.

Like? Yes (1) | No   

Karthick Raja   9 years AGO

37=3+7=10. 10*3=30. 37+30=67
67=6+7=13. 13*6=78. 67+78=145
145=1+4+5=10. 10*1=10. 145+10=155
155=1+5+5=11. 11*1=11. 155+11=166
166=1+6+6=13. 13*1=13. 166+13=179
179=1+7+9=17. 17*1=17. 179+17=196

Like? Yes (1) | No   

29 May 2015 Probability

If x+y+z = 9; then probability of having x = y = z = 3 will be, x, y, z are positive integers

OPtion

1) 1/28
2) 1/7
3) 1/14
4) 1/6
5) 1/3
6) 1/2
7) 1/16
8) 1/10
9) 1/8
10)1/56

Solution

For x = 1, y and z can be arranged in 7 ways
for x = 2, y and z can be arranged in 6 ways
for x = 3, y and z can be arranged in 5 ways
Similarly
------------
------------
for x = 7, y and z can be arranged in 1 ways
Total ways = 7+6+5+4+3+2+1 = 28
So probability 1/28
Option 1)

Correct Option: 1 Best Solution (14)

Radhika   9 years AGO

the total of x+y+z=9. therefore, the combinations of values of x,y,z will be
1,3,5 can be arranged in 3! ways=6
1,2,6 can be arranged in 3! ways=6
2,3,4 can be arranged in 3! ways=6
1,1,7 can be arranged in 3 ways=3
1,4,4 can be arranged in 3 ways=3
2,2,5 can be arranged in 3 ways=3
Now, 3,3,3 can be arranged in 1 way=1
The total combinations are = 6+6+6+3+3+3+1=28
therefore, the probability of getting (x=y=z=3) = 1/28

Like? Yes (10) | No   

W J Krishna Teja   9 years AGO

required probability=3/9c3

Like? Yes (1) | No   

Karthick Raja   9 years AGO

Total possibilities is 28
so probability is 1/28
117,126,135,145,154153,162,171
216,225,234,243,252,261
315,324,333,343,351
414,423,432,441
513,522,531
612,621
711

Like? Yes (1) | No   

vivek   9 years AGO

total no of outcomes = 28
required probability is 1/28

Like? Yes (1) | No   

Vaibhav Srivastava   9 years AGO

Favorable case = 1
Sample space = 28

Like? Yes (1) | No   

ravi singh   9 years AGO

answer is 1/28
total possible conditions are 28 because zero is neither positive nor negative,
hence probability = 1/28

Like? Yes (1) | No   

Mehul D Parmar   9 years AGO

because x=y=z=3 is possible only one time from 28 cases.here favourable case is 1 and total case is 28 than 1/28 is right ans.


here , give positive integers if non negative integers than find dircectly

n+r-1 9+3-1
c = c
r-1 3-1


bt here positive integer than find manually.

Like? Yes (1) | No   

anubhav verma   9 years AGO

number of combination with 7
with 2. 6 and so on
therefore number of combination with 7 at hundereds place are 1
total number of combination made is 28 and all three be equal to 3 is one combination of it hence 1:28

Like? Yes (1) | No   

keerthi   9 years AGO

they gave eq x+y+z=9 i.e., either x=9 or y=9 or z=9 so that in denominator we can write it as 9*9*9

Like? Yes (1) | No   

Abdullah khan   9 years AGO

Total number of cases will be ::
x + y = z = 6 { x>=0 ,y>=0 ,z>=0 }
(6+3-1)
C (3-1) =====
8
C =================28
2
hence the probaility of x=y=z=3 is 1/28

Like? Yes (1) | No   

PRASHANT KUMAR   9 years AGO

117 216 315................... 711
126 225 324.......
135 234 333.....
. . .
. . .
. . 351
. 261
171
total terms =7+6+5+4+3+2+1=28
no. of favourable terms =1 (333 only)
so P=1/28

Like? Yes (1) | No   

vibhu raj gupta   9 years AGO

total favorable cases 28
probability=1/28

Like? Yes (1) | No   

Paresh   9 years AGO

117 126 135 144 162 153 171
216 225 234 243 252 261
315 324 333 342 351
414 423 432 441
513 522 531
612 621
711

option 1

Like? Yes (1) | No (2)  

vivek raj   9 years AGO

total outcomes
7/1/1-3
6/2/1-6
5/2/2-3
5/3/1-6
4/3/2-6
4/1/4-3
3/3/3-1

=28
favorable outcome 3/3/3=1

1/28

Like? Yes (1) | No (3)  

28 May 2015 Value of k

If x = 29k+7 and x-k is divisible by 5, then k may be

OPtion

1) 2
2) 3
3) 4
4) 5
5) 6
6) 7
7) 8
8) 9
9) 0
10)12

Solution

x = 29k + 7
x-k = 28k + 7 = 30k+(7-2k)
Since (x-k) is divisible by 5 so k may be 6

Option 5)

Correct Option: 5 Best Solution (41)

RAKESH   9 years AGO

x-k = 7(4k+1)

x-k is divisible by 5, so k should be 6.

Like? Yes (4) | No (1)  

Aashish gohel   9 years AGO

let x-k=5m
where m=1,2,3,......
x-29k=7
(x-k)-28k=7
5m-28k=7
k=5m-7/28
k=6

Like? Yes (3) | No (1)  

sompon temtheerakij   9 years AGO

X = 29k+7........1
X-k = 5A...........2
from 1,2 ;- 28k+7 = 5A
The product can divided by 5 , k = 1 , 6
ans. k= 6

Like? Yes (1) | No   

Venkatesh kadali   9 years AGO

x=29k+7 we substitute k value than we x value.wecan subtract k value from x ,nd the result is divisible by 5 than we get k=6.

Like? Yes (1) | No   

Radhika   9 years AGO

Put the values of k from the options in the given equation
x=29k+7 by putting the value of k in this equation we get the value of x.
Then calucate x-k if its unit place is 0 or 5 ten it is divisible by 5 and the value of k which we have taken is the required answer.
i.e. by taking k=6 x=29*6+7=181 that means x=181 and we have k=6 now x-k=181-6=175 which is divisible by 5.
therefore, this is our required answer

Like? Yes (2) | No (1)  

Jotshnesh Patel   9 years AGO

put the value of x into x-k,
We'll get one equation that is 28k+7.
put the value of k (k=6) and answer 175 which is divisible by 5.

Like? Yes (1) | No   

atmakuri jagadish   9 years AGO

x=29k-7
so , x-k = 28k+7
As x-k is divisible by 5
So, 28K + 7 should also be divisible by 5
Hit and trial of the options given results in only one correct option
So answer is 6 , option 5

Like? Yes (1) | No   

sai   9 years AGO

taking -k on both sides
x-k= 28k+7 this is divisible by 5 go through options 6 is answer so k=6

Like? Yes (1) | No   

KUMAR HARSH   9 years AGO

X=29*k+7
X-k=28*k+7
28*6+7=175
Which is divisible by 5

Like? Yes (1) | No   

Aditi Khandagale   9 years AGO

x=29k+7 and x-k divisible by 5
x=29(6)+7
=181

and x-k=181- 6
so x-k=175
as last digit of the answer is 5 the number is divisible by 5

Like? Yes (1) | No   

Kavitha Lakshmi A N V   9 years AGO

if a number is divisible by 5 then tha last number must be 0 or 5....

Like? Yes (1) | No   

sivaramakrishna   9 years AGO

just check the unit digit for every option.
if K=6,
29*6+7=----1
----1-6=---5
number divisible by 5.

Like? Yes (1) | No   

Debayan   9 years AGO

x-k = 28k+7
Now, 28k+7 can be divisible by 5 only when it either ends with 5 or 0 i.e 28k must end either with 6 or 3.
Since 28k will always be even, hence 28k must end with 6.
Now, K must take values which when multiplied by 8 result in 6 at unit digit's place.
Therefore, k = 6,16,26...

Like? Yes (2) | No (1)  

Piyush Kumar   9 years AGO

Put 6 in 1st eqn, x = 181
181-6 = 175 which is divisible by 5.

Like? Yes (1) | No   

PRASHANT KUMAR   9 years AGO

x-k=5n;
so
(28k+7)/5=n;.............................so for k=6....n is an integer

Like? Yes (1) | No   

akas   9 years AGO

If k=6
X=29(6)+7
X=181 X-K=181-6=175
IT IS DIVISIBLE BY 5.
THEREFORE OPTION 5)6

Like? Yes (1) | No   

prashanth   9 years AGO

29k+7-k=28k+7
put k=6 then we get 175/5=35 so that
k=6

Like? Yes (1) | No   

L KOUSHIK KUMAR   9 years AGO

since (x-k) is divisible by 5, the last digit of (x-k) should be either 0 or 5.
multiply last digit 9 with different values and add it with 7. from result obtained, subtract the given value. if it is divisible by 5, then that is the solution.

Like? Yes (1) | No   

Karthick Raja   9 years AGO

x=29k+7
if k=6
x=(29*6)+7
x=234+7
x=241
x-k=241-6
x-k=235
235 is divisible by 5
so k is 6

Like? Yes (1) | No   

T.Aravinth kumar   9 years AGO

sub the k value (choices) in
x=27k+7
if sub k=6
then get x=181
now check x-k is divisible by 5..
x-k=181-6=175
175/5=35...
so
and 5)6..

Like? Yes (1) | No   

nasrul   9 years AGO

as if the k is 6, hence the x will be 181. thus, it is applicable for x-k will be divisible with 5 as it will be 175.

Like? Yes (1) | No   

Amartya Bhattacharyya   9 years AGO

x-k=7*(4k+1)
now, check options to find out for which value of k, (4k+1) will be multiple of 5

Like? Yes (1) | No   

ganesh babu   9 years AGO

x-k=28k+7 divisible by 5
may be k=1 and 6

Like? Yes (1) | No   

Pranati Kundu   9 years AGO

put the values of k from the options...

Like? Yes (1) | No   

Chitra.S   9 years AGO

sub x = 29k+7 in (x-k)/5,
then...from the options check..,.
if k=6,then condition satisfies.

Like? Yes (1) | No   

vibhu raj gupta   9 years AGO

write eqation as this x-k=28k+7
multiply 28 by 6=168+7=175
which is divided by 5

Like? Yes (1) | No   

aman aditya   9 years AGO

x=29k+7
x=(29*6)+7
x=174+7=181
now
x-k=181-6=175 i.e divisible by 5

Like? Yes (1) | No (1)  

devvrata   9 years AGO

x-k=28k+7 If x-k is divisible by 5 then 28k+7 will also be divisible by....put k =6 we will get 175 which is divisible by 5

Like? Yes (1) | No (1)  

Shiva Teja   9 years AGO

x=29 k+7
x-k=29 k +7 -k=28 k+7
By substituting the given values
the expression (x-k) i.e, (28 k+7) is divided by 5...When k=6

Therefore from the given option k=6.

Like? Yes (1) | No (1)  

padam   9 years AGO

29*6+7=181
181-5=175 divisible by 5

Like? Yes (1) | No (1)  

ABHINAV PUSHPAM   9 years AGO

X=29*6=174+6=181
181-6=175 which is divisible by 5
So K = 6

Like? Yes (1) | No (1)  

MRITUNJAY   9 years AGO

putting the value of k from 1 to 9
it only satisfy at 6
so k will be 6.

Like? Yes (1) | No (1)  

ANANTHI.P   9 years AGO

29*6=174+7=181
x=181;k=6
DIFF=175 WHICH IS DIVISIBLE BY 5

Like? Yes (1) | No (1)  

PAWAN KUMAR   9 years AGO

When k=6, x=29*6+7 = 181
x-k = 181-6 = 175, which is divisible by 5.

Like? Yes (1) | No (1)  

Ankita Anand   9 years AGO

x=29k+7
and x-k%5=0
by solving we get k=6

Like? Yes (1) | No (1)  

chepuru sai priyanka   9 years AGO

given that x-k is divisible by 5
let (x-k)/5=a
x-k=5a => x=5a+k
given, x=29k+7
5a+k=29K+7
28k=5a-7
let a=7 then k=1 not in option try for next possible value
a=35 then k=6

Like? Yes (1) | No (1)  

poojasaisri   9 years AGO

x=(29*6)+7
x=181
x-k=175
175 is divisible by 5

Like? Yes (1) | No (1)  

ravi singh   9 years AGO

answer is 6

as given,
X=29*K + 7
then, X - K = 28*K +7

put the value of k only 6 is verified.
hence answer is 6

Like? Yes (1) | No (1)  

Ayush Gupta   9 years AGO

29*6=174 174+7=181
181-6=175 divisible by 5

Like? Yes (1) | No (2)  

syed kashif ullah   9 years AGO

28*6+7 =175
it is divisible by 5

Like? Yes (1) | No (2)  

MURALI KUDUPUDI   9 years AGO

28*6+7=175

Like? Yes (1) | No (2)  

27 May 2015 Cost of 1 book

Total cost of 45 books and 65 copies is 1290 and that of 65 books and 45 copies is 1500. Cost of 1 book will be

OPtion

1) 15.84
2) 16.52
3) 16.98
4) 17.93
5) 18.25
6) 18.76
7) 19.63
8) 19.99
9) 20.05
10)20.23

Solution

According to the given condition
45x + 65y = 1290
65x + 45y = 1500
Solve this
x = 17.93

Option 4)

Correct Option: 4 Best Solution (50)

poojasaisri   9 years AGO

45b+65c=1290 eqn 1
65b+45c=1500 eqn 2
eqn 1+eqn 2 =>
110b+110c=2790 =>b+c=25.35 eqn3
eqn 1-eqn 2 =>
20b-20c=210 =>b-c=10.5 eqn4
eqn3+eqn4
2b=35.85 =>b=17.93

Like? Yes (2) | No   

Shiva Teja   9 years AGO

cost of 45 books and 65 copies is 1290
(45*B)+(65*C)=1290
cost of 65 books and 45 copies is 1500
(65*C)+(45*B)=1500

Buy solving the two equations we get,
Price of one book=17.931

Like? Yes (1) | No   

ravi singh   9 years AGO

Answer is 17.93

Let, represent Book=B
Copy=C
Then,
45*B + 65*C = 1290______eq(1)

And
65*B + 45*C = 1500_____eq(2)
From eq(1) & eq(2)

B=17.93

Like? Yes (1) | No   

chepuru sai priyanka   9 years AGO

let book=X and copies=Y then,
45X+65Y=1290
65X+45Y=1500
if u solve X=17.93
y=7.43
so 1 book will be 17.93

Like? Yes (1) | No   

MRITUNJAY   9 years AGO

45B+65C=1290
65B+45C=1500
after solving this equation b=17.93 and c=7.43
so answer will be 17.93
option 4.

Like? Yes (1) | No   

Surit Datta   9 years AGO

Solving as linear equation we get
PRICE OF BOOK IS 17.93 Rupee.

Like? Yes (1) | No   

Pranati Kundu   9 years AGO

let cost of 1 book be x and cost of 1 copy be y
45x+65y=1290
65x+45y=1500
solving this we get x=17.93

Like? Yes (1) | No   

Ayush Gupta   9 years AGO

45x+65y=1290
65x+45y=1500
Solve these equations and get
x=17.93(price of 1 book)
y=7.43 (price of 1 copy)

Like? Yes (1) | No   

RAKESH   9 years AGO

1/2 * [(1290+1500)/ (45+65) + (1500-1290)/(65-45)] = 17.93

Like? Yes (1) | No   

Vijendra singh   9 years AGO

let x as a book and y as a copy
45x +65y=1290
65x +45y=1500
then x=17.93 ans

Like? Yes (1) | No   

THUMMALA VINEESHA   9 years AGO

45x+65y=1290
65x+45y=1500
by solving these two equations we will get x=17.931 that is the value of 1 book

Like? Yes (1) | No   

umesh   9 years AGO

45b+65c=1290......(1)*(9/13)
65b+45c=1500......(2)

solving the eqns we get 17.93

Like? Yes (1) | No   

Karthick Raja   9 years AGO

45B+65C=1290 -----> EQN1
65B+45C=1500 ----->EQN2

(EQN1 x45) - (EQN2 x65)

2200B=39450
B=39450/2200
B=17.93

Like? Yes (1) | No   

Venkatesh kadali   9 years AGO

45b+65c=1290
65b+45c=1500
solve this two equation now we get one book is=17.93

Like? Yes (1) | No   

Prayani Singh   9 years AGO

45b+65c=1290
65b+45c=1500
solving above 2 equations:
20b-20c=210
b-c=10.5
replace c=B-10.5 in equation 2:
110b-472.5=1500
b=17.926

Like? Yes (1) | No   

sompon temtheerakij   9 years AGO

let books = A , copies = B
45A+65B = 1290
65A+45B = 1500
by solve these two equations
A = 17.93

Like? Yes (1) | No   

ANANTHI.P   9 years AGO

45B+65C=1290
65B+45C=1500
ON SOLVING C=7.43
SUBSTITUTE IN ANY OF THE 2 EQUATIONS;
B=17.93

Like? Yes (1) | No   

sravani   9 years AGO

45b+65c=1290
65b+45c=1500

Like? Yes (1) | No   

Meenakshi   9 years AGO

45B+65C=1290-->(1)
65B+45C=1500--(2)
C=1290-45B/65
substitute C value in equation (2)
65B+45(1290-45B/65)=1500
845B+11610-405B=19500
440B=19500-11610
440B=7890
B=7890/440
B=17.9318
cost of one book=17.93

Like? Yes (1) | No   

Radhika   9 years AGO

Hit and Trial Method
We'll choose from options to calculate the cost of copies first i.e. when we take the cost of book as 17.93 than we get the cost of copies as 7.43
1290-45*17.93=65*cost of copies
from this we get the cost of copies than put it in the 2nd equation if it comes 1500 than our ans is correct.

Like? Yes (1) | No   

lovely verma   9 years AGO

45x+65y=1290...............(1)
65x+45y=1500.................(2)
now we multiply by 9 in !st Equation And 13 in second equation
+405x+585y=11610.........(3)
+845x+585y=19500.........(4)
(-) (-) (-)
-------------------------------------------
After Subtraction we get:-
-440x=-7890
x=789/44
x=17.925
x=17.93 Ans

Like? Yes (1) | No   

pragya sahu   9 years AGO

let b=book
c=copy
45b+65c=1290.........1st
65b+45c=1500.........2nd
so,
c=7.4318182
when put in 1st eq
we get the value of b=17.93

Like? Yes (1) | No   

Ankita Anand   9 years AGO

let teh cost of a book is Rs x and of a copy is Rs y.
now 45x+65y=1290--(i)
and 65x+45y=1500--(ii)
now multiplying i by 45 and multiplying ii by 65 and subtracting i from ii we get;
2200x=39450
so x=17.93

Like? Yes (1) | No   

Divyabala R   9 years AGO

Simple Equation type question consider Books as X,Copies as Y which gives 45X + 65Y = 1290,
65X + 45Y = 1500

on solving this u l get X = 17.93

which is the answer. :)

Like? Yes (1) | No   

FOLSHIYA SATHYARAJ   9 years AGO

Divide the numbers b

Like? Yes (1) | No   

SANTOSH   9 years AGO

45b+65c=1290
65b+45c=1500
-------------------------
b=17.93

Like? Yes (1) | No   

SAKTHEESVARI S   9 years AGO

we get 2 eqn,45x+65y=1290;
65x+45y=1500;
solving these two, we get value of x (i.e) cost of 1 book=17.93 Rs

Like? Yes (1) | No   

Dharmveer Prajapat   9 years AGO

45x + 65 y =1290
65 x + 45 y = 1500
solve this you will get x=17.93

Like? Yes (1) | No   

Preethi   9 years AGO

books 17.93*45=806.85 copies 17.93 *65=483.15

books 17.93*65=1165.45 copies 17.93 *45=334.55.

each book= 17.93
copies = 7.43

Like? Yes (1) | No   

Penchal kumar   9 years AGO

45*x+65*y=1290
65*x+45*y=1500

Like? Yes (1) | No   

vivek   9 years AGO

45b +65c =1290..........(i)
65b +45c =1500..........(ii)
eq (i)+(ii),
b+ c= 25.36
eq (ii)- (i),
b- c=10.5
so, b = 17.93

Like? Yes (1) | No   

karthikeya varma   9 years AGO

45B+65C=1290-------(1)
65B+45C=1500-------(2)
-------------------------------------
SOLVE THE GIVEN EQUATIONS YO WILL GET
B=17.93
C=7.43
SO HERE AFTER THE COST OF SINGLE BOOK IS B=17.93

Like? Yes (1) | No   

g.soujanya   9 years AGO

By solving the two equations we can get the answer as 17.93

Like? Yes (1) | No   

Kavitha Lakshmi A N V   9 years AGO

boks are more than copies then price is more.

Like? Yes (1) | No   

T LAVANYA   9 years AGO

45b+65c=1290
65b+45c=1500
c=(1290-45b)/65
then
65b+45(1290-45b)/65=1500
b=17.93

Like? Yes (1) | No   

Rohit Gurjar   9 years AGO

45b+65c=1290
65b=45c=1500

solve this Equation Book
b=17.93

Like? Yes (1) | No   

DHANIKONDA RATNA BHAVANI   9 years AGO

Let X be Books
Y be Copies
(45X+65Y=1290 ) * 65
(65X+45Y=1500 ) * 45
2925x+4225y-2925x-2025y=83850-67500
2200y=16350
y(copies)=7.431818
45x=1290-65(7.431818)
45x=1290-806.93183

x(books)=17.93181844

Like? Yes (1) | No   

deep   9 years AGO

45x+65y=1290.....(1)
65x+45y=1500...(2)
solve these equation
n we get x=17.9

Like? Yes (1) | No   

padam   9 years AGO

make the equation and solve for x and y book price is 17.93

Like? Yes (1) | No   

Udhaya kumar   9 years AGO

let cost of book and copy. as x and y
then by given conditions
45x+65y = 1290
65x+45y = 1500
by solving x = 17.93...

Like? Yes (1) | No   

Payal Garg   9 years AGO

45b+65c =1290
65b +45c=1500

Solving above two , we get c=7.4 and b= 17.9

Like? Yes (1) | No   

sai   9 years AGO

45B + 65C=1290
65B+45C=1500
subtract we get 20B-20C=210
b-c = 10.5
now solve any two with above
45b+65(b-10.5)= 1290
1 book cost = 17.93

Like? Yes (1) | No   

prashanth   9 years AGO

117x+81(7.43)=2700
x=2700-81(7.43)/117=17.93

Like? Yes (1) | No   

Rahul Lone   9 years AGO

45books + 65 copies=1290
65 books+ 45 copies=1500
by solving
cost of 1 books =17.93

Like? Yes (1) | No   

Aashish gohel   9 years AGO

make equation such that
x= cost of books
y= cost of copies
45x +65y= 1290
65x+45y=1500
solve this both
you will get x=17.93
that is cost of book

Like? Yes (1) | No   

Rajesh Kumaresan   9 years AGO

45X+65Y=1290
65X=45y=1500
solving these 2 equations we get X=17.93

Like? Yes (1) | No   

T.Aravinth kumar   9 years AGO

45x+65y=1290
65x+45y=1500
solve
x=17.93

Like? Yes (1) | No   

Ram   9 years AGO

45x+65y=1290; 65x+45y=1500
By simplying we get x=17.93

Like? Yes (1) | No   

P R KANAKA MAHA LAXMI   9 years AGO

let us consider x is the cost of one book and y is the cost of one copy
45x+65y=1290
65x+45y=13500
by solving we get x=17.93

Like? Yes (1) | No   

Piyush Kumar   9 years AGO

45b + 65c = 1290 ---- (1)
.'. 9b + 13c = 258 ---- (2)
65b + 45c = 1500 ---- (3)
Adding 1 and 3 we get,
11b + 11c = 279 ---- (4)

Now, (4) - (2)
2b - 2c = 21 --- (5)

From (4) we get:
2b + 2c = 558/11 ----(6)

from (5) and (6) we get
b = 789/44 = 17.93 ----(Ans)

Like? Yes (1) | No   

26 May 2015 Simple Interest

The rate of interest on a sum of money is 4% per annum for the first 2 years, 6% per annum for the next 4 years and 8% per annum for the period beyond 6 years. If the simple interest accrued by the sum for a total period of 9 years is Rs. 1120, what is the sum?

OPtion

1) Rs.1500
2) Rs.1800
3) Rs.2000
4) Rs.2400
5) Rs.2500
6) Rs.3000
7) Rs.3500
8) Rs.4000
9) Rs.4500
10)Rs.4800

Solution

Let the sum be x. Then,
{(x*4*2)/100}+{(x*6*4)/100}+{(x*8*3)/100} = 1120
x = 1120*100/56 = 2000

Option 3)

Correct Option: 3 Best Solution (21)

RAKESH   9 years AGO

x/100 *[4*2+6*4+8*3] = 1120 => x=2000

Like? Yes (5) | No (2)  

sompon temtheerakij   9 years AGO

total interest = 4+4+6+6+6+6+8+8+8 = 56%
sum = 1120x100/56 = 2000

Like? Yes (2) | No   

MRITUNJAY   9 years AGO

1120 (x*4*2+x*6*4+x*8*3)/100
1120=56*x/100
x=2000

Like? Yes (3) | No (2)  

VIVEK KUMAR   9 years AGO

1120=(8p+24p+24p)/100
112000=56p
p=2000

Like? Yes (1) | No   

kranthi kanchanapally   9 years AGO

Let sum=x
x*4*2/100+x*6*4/100+x*8*3/100=1120
56x=112000
x=2000

Like? Yes (1) | No   

Ayush Gupta   9 years AGO

Let us assume the sum as x
S.I. for first two years will be (x*4*2)/100 .... (1)
similarly for next 4 years (x*6*4)/100 ......(2)
for next 3 years at rate of 8% S.I. will be (x*8*3)/100 ...(3)

So total interest for 9 years is 1120..
Adding (1),(2) and (3) gives
56x=112000
x=2000
:)

Like? Yes (4) | No (3)  

Pradeep   9 years AGO

question should be 8% per annum for the period beyond 9years instead of 6
=> 4*2*x/100 +6*4*x/100 +8*3*x/100 =1120
=>56x/100 =1120
=>x= 20000

Like? Yes (1) | No   

Dr.Dipin Singh   9 years AGO

If sum is Rs 100x, then total interest = 8x+24x+24x
=56x= Rs. 1120
x= rs 20.
so sum = Rs 2000.

Like? Yes (3) | No (2)  

s kumar   9 years AGO

let sum be P.
SI=PRT/100

1120=[(P*4*2) +(P*6*4) +(P*3*8)]/100

P=20,00

Like? Yes (2) | No (2)  

Divyabala R   9 years AGO

Sum = (Simple Interest * 100)/( Rate of interest *No of years)
So => Sum = (1120 *100)/((4*2)+(6*4)+(8*3))

=>Sum = 112000/56 => 2000

Hence,Sum is 2000

Like? Yes (2) | No (2)  

Rahul Mahadik   9 years AGO

pnr100=si
4*2*x100+x*4*6100+x*3*8100=1120
by solving x=2000

Like? Yes (2) | No (2)  

Karthick Raja   9 years AGO

ASSUM SUM = X
FIRST 2 YEARS = (X*4*2/100)
NEXT 4 YEARS =(X*6*4/100)
NEXT 3 YEARS =(X*8*3/100) (TOTAL PERIOD FOR 9 YEARS SO WE CAN TAKE 3 YEARS)
(X*4*2)/100+(X*6*4/100)+(X*8*3/100)=1120
8X+24X+24X=1120*100
56X=112000
X=112000/56
X=2000

Like? Yes (1) | No (2)  

Shiva Teja   9 years AGO

Let the sum is equal to x
Simple interest on the sum for first two years=2*.04*x
Simple interest on the sum for next four years=4*.06*x
Simple interest on the sum for 7,8 and 9 years=3*.08*x

Total interest for 9 years equals to (2*.04+4*.06+3*.08)*X
=0.56*X
simple interest accrued by the sum for a total period of 9 years is Rs. 1120
So, 0.56*X=1120
Therefore, sum=X=1120/.56=Rs.2000/-

Like? Yes (1) | No (2)  

ANANTHI.P   9 years AGO

1120=(p*4*2/100)+(p*4*6/100)+(p*8*3/100)
56p=1120*100
p=2000

Like? Yes (1) | No (2)  

aman aditya   9 years AGO

Let principle be 100
Interest for first two year=4*2=8
. next 4 yr=6*4=24
Next 3yr =8*3=24
Total=56
So (1120/56)*100=2000

Like? Yes (1) | No (2)  

ROHIT KUMAR   9 years AGO

Let Sum(or principal amount )=p
then,
p*4*2/100+p*6*4/100+p*8*3/100=1120
8p+24p+24p=1120*100
56p=1120*100
p=2000/-

Like? Yes (1) | No (2)  

Devendra Marghade   9 years AGO

If sum=x, then
(x*2*4/100) +(x*4*6/100) +(x*8*3/100)= 1120
8x+24x+24x=112000
x=2000
So answer is option (3)

Like? Yes (1) | No (2)  

sravani   9 years AGO

sigma(PTR/100)=1120
substitue respective R's and T's in the above equation.
then will get P as 2000

Like? Yes (1) | No (2)  

Md Irfan   9 years AGO

let sum be x
x*4*2/100+x*6*4/100+x*3*8/100=1120
x=20000

Like? Yes (1) | No (2)  

syed kashif ullah   9 years AGO

8x/100+24x/100+24x/100=1120
on solving
x=2000

Like? Yes (1) | No (2)  

Aditi Khandagale   9 years AGO

ans:3

Let sum be Rs x.
Given: ( x*4*2/100)+(x*6*4/100)+(x*8*3/100)=1120
56x =112000
x=112000/56=Rs.2000

Like? Yes (1) | No (2)  

25 May 2015 Sequence

What is the next number of the following sequence
4, 65, 5, 57, 6, 69, 7,

OPtion

1) 56
2) 74
3) 89
4) 99
5) 117
6) 107
7) 104
8) 101
9) 81
10)127

Solution

4
4^2 = 16 = reverse = 61+number = 61+4 = 65
5
5^2 = 25 = reverse = 52+number = 52+5 = 57
6
6^2 = 36 = reverse = 63+number 63+6 = 69
7
7^2 = 49 = reverse = 94+number = 94+7 = 101
Option 8)

Correct Option: 8 Best Solution (3)

RAKESH   9 years AGO

7^2 ------> 49 rev = 94+7 = 101

Like? Yes (7) | No   

Venkatesh kadali   9 years AGO

4 seqre=16 reverse 61 61+4=65 thats way 7 seqre =49 reverse 94 94+7=101

Like? Yes (1) | No   

Karthick Raja   9 years AGO

4 65 5 57 6 69 7 x 8

4*5=20 5*6 = 30 6*7=42 7*8=56
20+45=65 30+27=67 42+27=69 56+45=101

x=101

Like? Yes (3) | No (4)  

22 May 2015 Time and work

9 men finish one-third work in 10 days. The number of additional men required for finishing the remaining work in 2 more days will be

OPtion

1) 81
2) 80
3) 78
4) 64
5) 55
6) 47
7) 30
8) 28
9) 25
10)21

Solution

Time taken by 9 men to finish the whole work = 30 days
Remaining work = (1 - 1/3) = 2/3
More work, more men (direct)
Less days, more men (indirect)
Work 1/3 : 2/3
Days 2 : 10
:: } 9:x
Therefore x = (2/3 * 10 * 9 * 3 * 1/2) = 90
Hence, additional number of men = 90-9 = 81
Option 1)

Correct Option: 1 Best Solution (25)

sivaramakrishna   9 years AGO

m1*d1/w1=m2*d2/w2

(9*10)/(1/3)=(?)*2/(2/3)
?=90 men required.
already 9 members are there.
extra men needed=90-9=81

Like? Yes (21) | No   

aman aditya   9 years AGO

let ,
9 men do work in 10 days=10*9=90 unit
work left=180 unit
work to be done in one day=180/2=90 unit
additional men=90-9=81

Like? Yes (19) | No (2)  

MRITUNJAY   9 years AGO

M*D*H/W*C=M*D*H/W*C

9*10*3/1=X*2*3/2
X=90
SO NUMBER OF MORE WORKER =90-9=81

Like? Yes (6) | No   

tanu goel   9 years AGO

1st case:
No. of men's engaged for work:9
No. of hours:10hrs
work done:1/3

2nd case:
No. of men require for work:(9+x)
Additional Days :2
work to be done: 2/3

Acc. to formula:

(M1*D1)/W1=(M2*D2)/W2

(9*10)/(1/3)=((9+X)*2*3)/2

Upon solving ,u ll get x=81 workers

Like? Yes (6) | No   

Muhammad Nadeem Haider Naseem   9 years AGO

9 men finish 1/3 work in 10 days, so 9 men finish all the work in 10 * 3 = 30 days
9 men finish all work in 30 days, so 1 man finish all work in 30 * 9 = 270 days
This implies 1 man will finish 2/3 work in (2/3) * 270 = 180 days
180 / 2 = 90 men will complete 2/3 work in 2 days

Since 90 men can complete the remaining 2/3 work, and there are already 9 men here, the required additional men will be = 90 - 9 = 81

Like? Yes (4) | No   

ABHINAV PUSHPAM   9 years AGO

[9*10/(1/3)]=[x*2/(2/3)]
x=90
so, additional men required = 90-9= 81 men
option 1)

Like? Yes (4) | No (1)  

Sundar Kumar   9 years AGO

Time taken by 9 men to do the whole work = 30 days
Remaining Work = (1-(1/3)) = 2/3
since , more work, more men ( direct)
and less days, more men (Indirect)
Work =(1/3):(2/3)::9:x
days 2:10::9:x
x=(2/3)*10*9*3*(1/2) = 90
Additional number of men = (90-9) =81
Answer : Option 1) 81

Like? Yes (3) | No   

chepuru sai priyanka   9 years AGO

same 9 mens required another 20 days to complete remaining two-third work...to complete it in 2 days 90 mens are required....already 9 mens are there we need 81 mens in additional to them

Like? Yes (2) | No   

Devendra Marghade   9 years AGO

9 Men's 1 day work in 1/(3*10)=1/30 days.
So 1 Man's 1 day work=1/(9*30)=1/270
As remaining work=2/3, which is to be completed in 2 days with existing 9 men & 'x' additional men, then
2*(1/30) + 2*(x/270) =2/3
(9+x)/270=1/3, x=81

Like? Yes (1) | No   

Shiva Teja   9 years AGO

9 men finishes 1/3 of work in 10 days
One man in one day completes (1/3)*(1/9)*(1/10)=1/270 of work

To complete 2/3 of work in 2 days
We need ........(2/3) / ((1/270)*2) no of workers
no of workers needed =90
No of additional workers needed=90-9=81

Like? Yes (1) | No   

sompon temtheerakij   9 years AGO

2/3 work use 2x9x10 = 180 man/2days
= 90 men / day
The number of additional men = 90-9 = 81 men

Like? Yes (1) | No   

Aditi Tenguria   9 years AGO

time taken by 9 men to finish whole work=30 days
remaining work=(1-1/3) =>2/3
..
work 1/3:2/3::9:x
days 2:10::9:x
=>x=2/3*10*9*3*1/2=90
additional no of men=90-9=81

Like? Yes (1) | No   

vivek   9 years AGO

using chain rule,
3*2*x= 9*10*6
x= 90
SO required no. of man = 90-9=81

Like? Yes (2) | No (1)  

poojasaisri   9 years AGO

(m1*d1*h1)/w1=(m2*d2*h2)/w2
(9*10)/(1/3)=(m2*2)/(2/3)
m2=90
already 9 men are there
additional men are 81

Like? Yes (1) | No   

Deepak Puri   9 years AGO

9 men finish 1/3 work in 10 days,
so 9 men finish all the work in 10 *3 = 30 days
9 men finish all work in 30 days,
so1 man finish all work in 30 * 9 = 270 days
This implies 1 man will finish 2/3 work in (2/3) * 270 = 180 days
180 / 2 = 90 men will complete 2/3work in 2 daysSince 90 men can complete the remaining 2/3 work,
and there are already 9 men here, the required additional men will be = 90 - 9 = 81

Like? Yes (1) | No   

KRISHAN MURARI   9 years AGO

1-1/3=2/3
so 2/3 work will be done by 90(9*10) men in 2 days
additional men=90-9=81

Like? Yes (1) | No   

Divyabala R   9 years AGO

When each men works in equal rate then 9 men * 10 days / work done(1/3) = x men * 2 days / work to be done (2/3)
gives , total of 90 men to complete the remaining work in 2 days .

We already have 9 men so additionally we need 81 and thats it.

Like? Yes (1) | No   

Dr.Dipin Singh   9 years AGO

Total mandays required to complete job = 9*10*3=270
mandays utilised in 10 days = 9*10=90 days
Balance mandays for 2 days = (270-90)=180
No of men required = 180/2=90
Additional men reqd. = 90-9=81

Like? Yes (1) | No   

kambalanaresh   9 years AGO

90 members finish the remaining wok in 2 days

Like? Yes (1) | No   

Sanjiv   9 years AGO

2/3 work remaining which can be done by 180 men in a day

so we need to complete it into 2 days .So totally 90 men required and 9 are already there so we need 81 more men to complete the task in text two days.

Like? Yes (1) | No   

sravani   9 years AGO

45 for 2 says
for 1/3 work
for 2/3 its 90 men
90-9 =81men more

Like? Yes (1) | No   

Aditi Khandagale   9 years AGO

9 men finish 1/3 work in 10 days, so 9 men finish all the work in 10 * 3 = 30 days
9 men finish all work in 30 days, so 1 man finish all work in 30 * 9 = 270 days
This implies 1 man will finish 2/3 work in (2/3) * 270 = 180 days
180 / 2 = 90 men will complete 2/3 work in 2 days

Since 90 men can complete the remaining 2/3 work, and there are already 9 men here, the required additional men will be = 90 - 9 = 81

Like? Yes (1) | No   

Balakumaran N.A   9 years AGO

9 men finish 1/3 work in 10 days, so 9 men finish all the work in 10 * 3 = 30 days
9 men finish all work in 30 days, so 1 man finish all work in 30 * 9 = 270 days
This implies 1 man will finish 2/3 work in (2/3) * 270 = 180 days
180 / 2 = 90 men will complete 2/3 work in 2 days

Since 90 men can complete the remaining 2/3 work, and there are already 9 men here, the required additional men will be = 90 - 9 = 81

Like? Yes (1) | No   

THUMMALA VINEESHA   9 years AGO

9 men can do all the work in 30 days . 1 men can do all the work in 270 days. This shows that 1 men can do the remaining work 2/3 in 180 days. 90 men can complete the 2/3 work in 2 days. total =90. needed=90-9=81

Like? Yes (3) | No (2)  

kranthi kanchanapally   9 years AGO

Time taken to 9 men to complete the work is 30 days
remining wrk is =1-1/3=2/3
work 1/3:2/3::9:x
days 2:10::9:x
then x=2/3*10*9*3*1/2
x=90
then 90-9=81 is teh answer

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21 May 2015 How much coal will be needed

If 5 engines consume 6 metric tonnes of coal when each is running 9 hours a day, how much coal (in metric tonnes) will be needed for 8 engines, each running 10 hours a day, it being given that 3 engines of the former type consume as much as 4 engines of latter type?

OPtion

1) 26.68
2) 25/8
3) 6.48
4) 77/9
5) 25/3
6) 77/3
7) 11.58
8) 27
9) 18
10)8

Solution

More engines, more coal (Direct)
More hours a day, more coal (Direct)
More rate of consumption, more coal (Direct)
Engines ------------ 5:8
Hrs/day ------------ 9:10
Rae of consumption - (1/3):(1/4)
6:x
Therefore
x = (8*10*(1/4)*6*(1/5)*(1/9)*3) = 8 metric tonnes.
Option 10)

Correct Option: 10 Best Solution (18)

ABHINAV PUSHPAM   9 years AGO

[(5*9/6)*4]=[(8*10/X)*3]
on solving x=8
answer is 8

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Aditi Tenguria   9 years AGO

LET THE REQUIRED CONSUMPTION OF COAL BE "X"UNIT
NO.OF ENGINES=5:8
WORKING HOURS=9:10::6:X
RATE OF CONSUMPTION=1/3:1/4
....
5*9*1/3*X =8*10*6*1/4
=>X=8 METRIC TONNES

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Shiva Teja   9 years AGO

Coal consumed by one engine of type 1
=6/(5*9)
=2/15 metric tonnes per hour
3 engines of type 1 consume as much as 4 engines of type 2
Therefore,
Coal consumed by one engine of type 2=(3/4)*(2/15)
=1/10 MT/hour

Coal consumed by 8 engines (of type 2),running 10 hours a day
=(1/10)*8*10
=8 metric tonnes.

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Sundar Kumar   9 years AGO

Engines 5 : 8
hours / day 9 : 10
Rate of consumption 1/3:1/4

Therefore x= 8*10*(1/4)*6*(1/5)*(1/9)*3
x=8 metric tonnes

Answer: Option: 10) 8

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Karthick Raja   9 years AGO

RATIO

ENGINE 5:8
HOURS 9:10 (PER DAYS)
3 engines of the former type consume as much as 4 engines of latter type (1/3):(1/4)

THESE THREE RATIOS ARE EQUAL TO 6:x

x=(8*10*(1/4)*6*(1/5)*(1/9)*3)
X=8 METRIC TONNES

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Dileepkumar Palli   9 years AGO

5 engines consumes 6 metric tonnes of coal in 9 hours a day.
so 1 engine consumes of coal in 1 hour of a day is 6/(5*9).
and 3 former engines is equal to 4 latter engines.
so 8 latter engines= 6 former engines.
so 8 latter engines used coal in 10 hours a day= 6*(1 former engine consumes coal in 1 hour a day)*10 hours.
=6*(6/(5*9))*10.
=8.
so ans is 8

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SUSHANT KUMAR   9 years AGO

engine=5:8
hours/day=9:10
rate of consumption=1/3:1/4
coal=6:x
so x=8

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Mohit   9 years AGO

after solving the answer is found to be this 8 metric tonne

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vivek   9 years AGO

let 3 engines of former type consumes 1 unit in 1 hr.
then, 4 engines of former type consumes 1 unit in 1 hr.
let the required consumption of coal be x unit.
so according to chain rule,
5*9*4*x= 8*10*3*6
or, x= 2*2*2=8 metric tones

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padam   9 years AGO

engine 5:8
hours/day 9:10
rate of consumption 1/3:1/4
tonnes 6:x
x=8*10*1/4*6*1/5*1/9*3=8 answer option 10

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Ashwin Dinesh   9 years AGO

6*(8/5)*(10/9)*(3/4) = 8

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sompon temtheerakij   9 years AGO

8 engines , 10 hr/day consume = 8x10x6/5x9 = 10 2/3 tons
=10 2/3 x 3/4 = 8 tons

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karan   9 years AGO

3 engines of former type consume : 3.6 metric tonne (6/5=1.2*3=3.6)
4engines of latter type=3engines of former type=3.6 metric tonne
7engine consume=3.6+3.6=7.2 metric tonne
1engine consume=1.3 metric tonne
8engine consume=7.2+1.38.55=77/9 metric tonne

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siddharth prakash singh   9 years AGO

using chain rule.
more engine consume more coal (direct relation)
more hour of running consume more coal (direct relation)
more rate more consuming of coal (direct relation)
so, 5:8,9:10,1/3:1/4::6:x
giving x=8

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Divyabala R   9 years AGO

Its given that 3 engines of former equals the 4 engines of latter in coal intake so we need to find the solution using former i.e,
from the given 5E[9H] = 6MT of coal;

so we should find 6E[10H] = ? which gives the answer for 8E[10h] of latter .

Consider 5E[9H] = 6 MT of coal then,

1E[1H] = 6 / (5* 9) = 2/15

6E[1h] = [2/15] * 6 =12/15

6E[10H] = [12/15] * 10 = 8

Thats it then we can pass this to 8E[10H] of the latter .

So the answer is 8.

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sivaramakrishna   9 years AGO

5f*9=6==>f=6/45, 3f=4s==>s=3f/4=3*6/45*4
substitute s in the equation
8s*10=?==>8

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YUVRAJ BALINGE   9 years AGO

Consumtion is directly proportional to no. of engine and time
of hour it works.
Consumption of former type engine=2/15 metric tonne/hour/engine
Consumption of later=8

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aman aditya   9 years AGO

First type of machine work 9*5=45 hr
energy consumed per hr=6/45
second type of machine work 10*8=80hr
as 3 of first machine take as much as energy of 4 machine of second type.
so,
(2*3*20)/(45*4)=8

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