Latest TCS Aptitude Question SOLUTION: sum of the cp's of two cars is Rs.150,000.ist car is sold at a profit of 20% and the second car at the loss of 20%.however,their S.p's are same. what is the cost price of the 1st c

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Profit and Loss

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40153709) TCS Numerical Ability question Keep an EYE

sum of the cp's of two cars is Rs.150,000.ist car is sold at a profit of 20% and the second car at the loss of 20%.however,their S.p's are same. what is the cost price of the 1st car?

Asked In TCS karthika (8 years ago)

Unsolved Read Solution (10)

Is this Puzzle helpful?   (23)   (18) Submit Your Solution Profit and Loss

Read Solution (10) :

ROHIT YADAV  (8 years ago)

cp1+ cp2 =150000

write a equation in cp

100sp/120 +100sp/80=150000
solve equation
sp is same


sp=72000

so now put the equation in sp format and find cp

cp 1= 60000
cp 2= 90000

Is this solution Helpfull? Yes (15) | No (1)  

Sayandeep Jana  (8 years ago)

we get,
CP of the 1st car is=x & 2nd car is=y
x+y=150,000
20% profit means SP of 1st car is=120x/100
20% loss means SPof 2nd car is=80y/100
according to the question,
120x/100=80y/100
or,y=3x/2
then x+3x/2=150,000
x=60000

Is this solution Helpfull? Yes (10) | No   

SK INTAJ MOHAMMAD  (8 years ago)

let x and y is the cost price of 1st and 2nd car respectively.
so, x+y=150,000..................(1)
now, s.p. of the 1st car is (x*120)/100=(6*x)/5
and s.p. of the 2nd car is (y*80)/100=(4*y)/5
but their s.p. are same. So, (6*x)/5=(4*y)/5 or, y=3*x/2
put these value in the equation (1),we will get x+(3*x)/2=150,000 or,x=60000
and y=90000 is the cost price of the 2nd car.
therefore cost price of the 1st car is 60000.....(ans)

Is this solution Helpfull? Yes (6) | No   

Arpan Roy  (6 years ago)

let,cost price=100;
1st car selling price=120
2nd car selling price=80
1st:2nd=80:120=2:3
cost price of 1st car=(2/5)*150000=60000

Is this solution Helpfull? Yes (5) | No   

Nemai Layek  (8 years ago)

let cp1=x so cp2=150000-x
now x+x/5=(150000-x)-(150000-x)/5
solving the above eqsn we have cp1 i.e. x=60000
you can cross check the ans i.e. 60000+20%=72000==90000-20% (proved)

Is this solution Helpfull? Yes (2) | No   

shagufta  (8 years ago)

let cp of ist car be X and cp of 2nd car be Y
accordng to qstn,
X+Y=150000
also, sp of 1st car =12X/10
and sp of 2nd car=8Y/10
as given sp's are same
so,12X/10=8Y/10
X=2/3
on solving,
we get, X=60000, Y=90000

Is this solution Helpfull? Yes (2) | No   

DHEERAJ KUMAR SINGH  (8 years ago)

let cp's be c1& c2 res.
so,c1+c2=150000
c1+c1/5=c2-c2/5
c2-c1=30000
c1=90000

Is this solution Helpfull? Yes (1) | No (6)  

ZAKIR HUSSAIN  (8 years ago)

either 90000 or 60000

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

B.sudarsana kapoor  (6 years ago)

20 -20
100
3 2
gives sp are same
so ration is 2 :3
2/5*150000
60000

Is this solution Helpfull? Yes (1) | No   

Aritra Chakraborty  (8 years ago)

car1+car2 = 150000 ... 1.2*car1 = 0.8car2 => car2 = 3/2*car1.. 5/2*car1 = 150000 => car1= 60000

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.