Latest TCS Aptitude Question SOLUTION: 20 people going to one temple. (some time waste data) They want Rs. 20. For men-Rs.3, women-Rs.2, children-50 paisa. How much money spends for men, children and women?

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20 people going to one temple. (some time waste data) They want Rs. 20. For men-Rs.3, women-Rs.2, children-50 paisa. How much money spends for men, children and women?

Asked In TCS Sri Reddy (10 years ago)

Solved Ann Theressa Read Solution (9)

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Read Solution (9) :

Akanksha Yadav  (10 years ago)

we cannot solve for three variables using two equations

Is this solution Helpfull? Yes (18) | No (1)  

Ann Theressa  (10 years ago)

Rs.3,Rs.7,Rs.10

Let number of men be x, women y and children z

x+y+z = 20 ---(1)

men gives 3 rupees each --->total money spend by men = 3x
women gives 2 rupees each--->total money spend by women =2y
children gives 50 paisa each ---> total money spend by children = 0.5z
Total 20 rupees is given by all

3x+2y+0.5z = 20 ---(2)

Solving (1) and (2)
x=1 , y=5 , z = 14

total money spend by men = 3x = Rs3
total money spend by women =2y = 2*5 = Rs10
total money spend by children = 0.5z = 0.5*14 = Rs7

Is this solution Helpfull? Yes (17) | No (20)  

RAKESH  (10 years ago)

M=1 => 1*3= 3
W=5 => 5*2= 10
C=14 => 14*.5= 7

1+5+14= 20 & 3+10+7=20

Is this solution Helpfull? Yes (11) | No (1)  

GAUTAM SHAHI  (9 years ago)

x+y+z=20
3x+2y+0.5z=20
now verify it from option

Is this solution Helpfull? Yes (3) | No   

Shruti  (9 years ago)

Ann Theressa , plz explain how did u solve the value of 3 unknowns using 2 equations?????

Is this solution Helpfull? Yes (1) | No (1)  

Ann Theressa  (9 years ago)

@JYOTSHANA
by trail and error method.

the 2 equations are

x+y+z = 20 ---(1)
and
3x+2y+0.5z = 20 ---(2)

multiply eqn (1) by 2

2x+2y+2z = 40 ---(3)

eqn (3) - (2) gives

-x + 3/2*z = 20
3/2 *z = 20 + x

Since x,y and z are positive numbers,from this equation 3/2*Z must be greater than 20.(z > 13)

From equation (1) z is less than or equal to 18.(min value of x and y can be 1)
Also from equation (2) , z must be an even number.

So z can be 14,16 or 18
Lets take z = 14

3/2*z = 20+x
x = 1
x+y+z = 20
so y = 5

Substituting these values in eqn (2)
3x + 2y + 0.5z = 3+10+7 = 20
which satisfies both the equations.

Is this solution Helpfull? Yes (1) | No (5)  

sarathkumar  (9 years ago)

its a random methodology.
since fr women rs 2 it is even .so rs fr men and child to be odd so tat total 20rs is even.
choose appropriately u vl get answer within 3 possible inputs

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vinay kumar  (7 years ago)

0.5*14=7 3*3=9 2*3=6 men 3 women 3 children 14 total 20 rs 20

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Prerna bisen  (5 years ago)

a) Rs.6-men, Rs.4-children, Rs.10- women

b)Rs.3-men, Rs.7-children, Rs.10- women

c)Rs.3-men, Rs.8-children, Rs.9- women

d)Rs.9-men, Rs.3-children, Rs.8- women

Sol: By using options we calculate it easily.
From option b
Men - 1men*Rs.3= Rs.3
Children - 14children* 50paisa= Rs.7
Women - 5women*Rs.2= Rs.10

Is this solution Helpfull? Yes | No   

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