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# Reena and Nitu had a pre-determined number of baskets, at a ratio of 5

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Math Expert
Joined: 02 Sep 2009
Posts: 62496

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27 May 2016, 08:02
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Difficulty:

35% (medium)

Question Stats:

72% (01:39) correct 28% (02:13) wrong based on 63 sessions

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Reena and Nitu had a pre-determined number of baskets, at a ratio of 5:3. Reena offered 10 of her baskets to Nitu so that the ratio of number of baskets became 7:5. How many more baskets does Reenu now have than Nitu?

(A) 30
(B) 40
(C) 60
(D) 80
(E) 90

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27 May 2016, 09:44
1
Bunuel wrote:
Reena and Nitu had a pre-determined number of baskets, at a ratio of 5:3. Reena offered 10 of her baskets to Nitu so that the ratio of number of baskets became 7:5. How many more baskets does Reenu now have than Nitu?

(A) 30
(B) 40
(C) 60
(D) 80
(E) 90

$$\frac{5x-10}{3x+10}$$ = $$\frac{7}{5}$$

5 *(5x-10) = 7 * (3x+10)

25 x -50 = 21 x + 70

4 x = 120

x = 30

Reena now has: 5x-10 = 140 baskets
Neetu now has: 3x+10 = 100 baskets

Reena has 40 more baskets than Neetu.

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27 May 2016, 09:45
Let number of baskets that Reena has = R
and number of baskets that Nitu has = N

R/N = 5/3
=>3R -5N = 0 --- equation 1

(R-10)/(N+10) = 7/5
=>5R - 50 = 7N+70
=>5R-7N=120 --- equation 2
Multiply equation 1 by 5 and equation 2 by 3 , we get
15R - 25N = 0
and 15R - 21N = 360
Now , 4N = 360
=> N = 90
R= 150
Number of baskets Reenu now has more than Nitu = (150-10) - (90+10)
=140-100
=40
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Re: Reena and Nitu had a pre-determined number of baskets, at a ratio of 5   [#permalink] 27 May 2016, 09:45
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