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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 an

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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?

Guest Feb 11, 2015

#7
+93025
+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

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
#1
+27246
+5

Using an obvious notation we have:

.

Alan  Feb 11, 2015
#2
+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

Guest Feb 11, 2015
#3
+94202
0

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

Thank you   :))

Melody  Feb 11, 2015
#4
+5

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.

Guest Feb 11, 2015
#5
+20711
+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

heureka  Feb 11, 2015
#6
+94202
0

Thank you Heureka and Alan

We all make mistakes anon.  Don't sweat it

Melody  Feb 11, 2015
#7
+93025
+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

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
+93025
0

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

CPhill  Feb 11, 2015