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Beng, Chandra and Danial have some stickers. Chandra and Danial have 7/10 of the stickers. Beng and Danial have 6/7 of the stickers. Beng and Chandra have a total of 620 stickers. How many more stickers did Danial collect than Beng?

 Feb 11, 2015

Best Answer 

 #7
avatar+128408 
+5

I'm just doing this one to see if I can solve it without "cheating" and looking at other's answers.....!!!

Let N be the total number of stamps....so we have

C + D   = 49/70N   (1)

B + D  = 60/70N    (2)

B + C = 620

Adding (1) and(2), we have

620 + 2D  = (109/70)N  → D = (109/140)N - 310

So

B + C + D  = N

620 + (109/140)N - 310  = N

310 = (31/140)N →   N = 1400

So D has  (109/140)(1400) - 310 = 780

And B has

780 +  = (6/7)(1400) → D = 1200 - 780 = 420

So D - B = 780 - 420 = 360 more

 

 Feb 11, 2015
 #1
avatar+33615 
+5

Using an obvious notation we have:

Sticker numbers

.

 Feb 11, 2015
 #2
avatar
+5

Beng and Danial have 6/7 of the stickers ,so chandra have 1/7 of the total stickers

so Danial have 7/10-1/7=39/70 of the total stickers , so beng have 6/7-39/70=21/70=3/10

Beng and chandra have 1/7+3/10=31/70,so the Danial ,Beng and chandra have 620/31*70=1400

Danial collect  more 1400*(39/70-3/10)=1400*(9/70)=180 stickers than Beng

 Feb 11, 2015
 #3
avatar+118608 
0

I think we need an adjudicator on this one - I don't have time at present   

Thank you   :))

 Feb 11, 2015
 #4
avatar
+5

i made a mistook.

39/70-3/10=39/70-21/70=18/70

(18/70)*1400=360 ,so Danial collect more 360 than Beng.

i am so sorry guys.

 Feb 11, 2015
 #5
avatar+26367 
+5

Beng, Chandra and Danial have some stickers. Chandra and Danial have 7/10 of the stickers. Beng and Danial have 6/7 of the stickers. Beng and Chandra have a total of 620 stickers. How many more stickers did Danial collect than Beng

$$\small{\text{
All Stickers $=x$
}}\\\\
\begin{array}{lcccc}
(1) & b+c+d=x & \quad c+d=x-b &\quad b+d=x-c &\quad b+c=x-d\\\\
\hline \\
(2) & c+d = \frac{7}{10}x &\frac{7}{10}x = x-b \\\\
(3) & b+d = \frac{6}{7}x & &\frac{6}{7}x = x-c \\\\
(4) & b+c = 620 & && 620 = x-d \\\\
\hline \\
& b+c+d=x & b=\frac{3}{10}x & c=\frac{1}{7}x & d=x-620\\\\
\end{array}\\
\begin{array}{rrcl}
& \frac{3}{10}x + \frac{1}{7}x + x-620 &=& x\\ \\
& x &=& 1400
\end{array}$$

$$\\b=\frac{3}{10}*1400 = 420\\\\
c=\frac{1}{7}*1400 = 200\\\\
d=1400-620 = 780\\\\
d-b=780-420 = 360$$

How many more stickers did Danial collect than Beng?  360

 Feb 11, 2015
 #6
avatar+118608 
0

Thank you Heureka and Alan 

 

We all make mistakes anon.  Don't sweat it 

 Feb 11, 2015
 #7
avatar+128408 
+5
Best Answer

I'm just doing this one to see if I can solve it without "cheating" and looking at other's answers.....!!!

Let N be the total number of stamps....so we have

C + D   = 49/70N   (1)

B + D  = 60/70N    (2)

B + C = 620

Adding (1) and(2), we have

620 + 2D  = (109/70)N  → D = (109/140)N - 310

So

B + C + D  = N

620 + (109/140)N - 310  = N

310 = (31/140)N →   N = 1400

So D has  (109/140)(1400) - 310 = 780

And B has

780 +  = (6/7)(1400) → D = 1200 - 780 = 420

So D - B = 780 - 420 = 360 more

 

CPhill Feb 11, 2015
 #8
avatar+128408 
0

Yep..that looks correct.....I love these "Brain Teasers".....!!!

 

 Feb 11, 2015

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