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24 Oct 2022 Arithmetic

On this Diwali, Akash a tech guy, decorated his home by a series of LED lights of three different colours: blue, red and green where he uses total 100 LED lights and also spends exactly Rs. 100 on making this series. If cost of blue LEDs is Rs. 4 for 3, red LEDs Rs. 3 for 9 and green LEDs Rs. 7 for 10, then which of the following can be the number of blue LEDs in the series ?

OPtion

1) 10
2) 62
3) 20
4) 40
5) 63
6) 54
7) 27
8) 14
9) 8
10) None of these

Solution

B + R + G = 100 --- (i) and
B*(4/3) + R*(3/9) + G*(7/10) = 100 ⇒ 40B + 10R + 21G = 3000 ---(ii)
Multiplying eqn. (i) by 10 and subtracting it from (ii), we get 30B + 11G = 2000
B = (2000 - 11G)/30
We get the integer value of B for G=10 and 40, so that count of total lights is 100.
Therefore combinations of Blue, Red & Green are B=63, R=27, G=10 OR B=52, R=8, G=40
Correct Option 5)

Correct Option: 5 Best Solution (1)

G.S.Kantharao   2 years AGO

Let Number of Blue,Red,Green are =B,,R,G
B+R+G=100
Cost of each Blue,Red,Green=
₹ 4/3,₹1/3,₹7/10
4B/3 + R/3 +7G/10=100
40B+10R+21G=3000
30B+11G+10(B+R+G)=3000
30B+11G+10(100)=3000
30B+11G=2000
For B=63
30*63+11G=2000
11G=2000-1890=110=>G=10
B=63,R=27,G=10
Cost
B=63*4/3+27*1/3+10*7/10
=84+9+7=100
So my option is 5 👈

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