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Ram ordered for 6 black toys And some additional brown toys. The price of black toy is 2.5 times that of a brown toy. While preparing the bill, the clerk interchanged the number of black toys and brown toys which increased the bill by 45%. Find the number of brown toys.
Read Solution (Total 8)
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- let cost of brown toy = x and black toy = 2.5x
let y be no of brown toys
(2.5*x*y + 6*x)-(6*2.5x+xy) = 45/100*(6*2.5x+xy)
on solving y=15 - 11 years agoHelpfull: Yes(28) No(1)
- cost of black= bl, brown= br.
bl=1.5br
no. of black=6, brown= x
6bl+x.br=cost=br(x+15)
interchached= 6.br+x.bl=br(2.5x+6)
profit=1.5x-9=45%
x=15=n0. of brown toys - 11 years agoHelpfull: Yes(8) No(0)
- no of brown toys is 15....
15 is the right answer
- 11 years agoHelpfull: Yes(4) No(0)
- ans=15
let x= no. of brown toys
let p= price of one brown toy
=> actual price=6*2.5p+xp=A(let)...... eqn(1)
bill prepared by clerk=x*2.5p+6p=A+0.45A ..... eqn(2)
dividing above eqn 91) by (2)
=> x=15
- 11 years agoHelpfull: Yes(4) No(0)
- answer is 62
- 11 years agoHelpfull: Yes(1) No(12)
- i think ans is 8
- 11 years agoHelpfull: Yes(1) No(12)
- cost of black toy=2.5*brown
so
interchange blak, brown
6*2.5=15
let cost of brown toy=1 rs thn
total no. of brown toys=15
- 10 years agoHelpfull: Yes(1) No(0)
- Let x and y be the cost of black and brown toys respectively.
6x+by+(6x+by)*45/100=6y+bx
20(6x+by)+9(6x+by)=20(6x+by)
174x+29by=120y+20bx
substitute x=2.5y
435y+29by=120y+21by
315y=21by
b= 315/21 =15 - 10 years agoHelpfull: Yes(1) No(0)
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