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vennila 1 year ago

Raj divided 50 into two parts such that the sum of their reciprocal is 1/12 ,we get the parts as

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KICK 1 year ago

IF it is a exam point of view question i suggest you always go by options bro
if u want explanation here it follows
let one of the parts = x implies another part is 50-x
so given that reciprocals sum is 1/12 so 1/x + 1/50-x == 1/12
solve it and u get value of x ! do the math
so x is 20 another part is 30
verification : 1/20+1/30 = 50/600 = 1/12 ( proved )

if u face difficulty in solving x heres the thing :
1/x+ 1/ (50-x) = 1/12
taking cross multiplication (50-x+x)/ (50x-x^2) = 1/12
==} 50x-x^2 = 12*50
==} 50x-x^2 = 600
==} x^2 -50x + 600 = 0
===} x^2 -20x -30x +600= 0
===} x(x-20) - 30 ( x-20) =0
=====} ( x-30) ( x-20) = 0
=====} x= 30, 20 ! gotcha

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vennila 1 year ago

If (P/Q) – (Q/P) =21/10 ;
(4P/Q) + (4Q/P) =?

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Karan 1 year ago

If (P/Q) – (Q/P) =21/10 ;

(P/Q) + (Q/P) = 29/10
(4P/Q) + (4Q/P) = 4*29/10= 58/5

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vennila 1 year ago

We are given the following sequence
PROBLEMSOLVINGPROBLEMSOLVINGPRO……. If the pattern continues, what letter will be in the 2015thposition

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vennila 1 year ago

We are given the following sequence
PROBLEMSOLVINGPROBLEMSOLVINGPRO……. If the pattern continues, what letter will be in the 2015thposition

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vennila 1 year ago

If (P/Q) – (Q/P) =21/10 ;
(4P/Q) + (4Q/P) =?

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vennila 1 year ago

In the sixth, seventh, eighth and ninth basketball games of the session a player scored 23, 14, 11, 20 points respectively. Hereafter nine games than it was after the first five games. If her average after ten games was points per game average was higher greater than eighteen what is the least number of points that she could have scored in the tenth game?

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vennila 1 year ago

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siva 11 Months ago

The sum of the scores for games 6 through 9 is 68. The average in these four games is 68/4 = 17.

The total points in all ten games is greater than 10*18 = 180. Thus, it must be at least 181.

There are at least 181 - 68 = 113 points in the other six games: games 1-5 and game 10.

Games 1-5 must have an average of less than 17. Thus we cannot put more than 16 + 17 + 17 + 17 + 17 = 84 points in those five games.

Thus, the tenth game must have at least 113 - 84 = 29 points

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vennila 1 year ago

2/3rd of the balls in a bag are blue, the rest are pink. If 5/9th of the blue balls and 7/8th of the pink balls are defective. Find the number of balls in the bag, if the number of non defective balls is 146?

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Karan 1 year ago

Total balls in bag are 432.
288 blue and 144 pink balls.
160 blue balls are defective. 126 pink balls are defective.
out of 146 non defective balls, 128 blue balls and 18 pink balls are non defective.

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Devendra Marghade 1 year ago

Ans: 432

Let total number of balls be x, then
Blue balls = 2x/3, Pink balls=x/3
Total defective balls =(5/9) of blue + (7/8) of pink = (5/9)*(2x/3) + (7/8)*(x/3)=143x/216
So non-defective balls=x - 143x/216= 73x/216
A's 73x/216=146, so x=432

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