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nick1816
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The cost of 40 pens, 44 pencils and 50 erasers is $392. The cost of 4 Updated on: 21 Jun 2019, 07:47
Originally posted by nick1816 on 20 Jun 2019, 20:30.
Last edited by nick1816 on 21 Jun 2019, 07:47, edited 1 time in total.
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The cost of 40 pens, 44 pencils and 50 Erasers is $392. The cost of 46 pens, 54 pencils and 60 erasers is $466. If the cost of 109 pens, 125 pencils and 140 Erasers is $N, find N.

A. 545.5
B. 1091
C. 2182
D. 2500
E. Data Insufficient
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B
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The cost of 40 pens, 44 pencils and 50 erasers is $392. The cost of 4 29 Jul 2021, 05:54
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Aastha1908
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nick1816
The cost of 40 pens, 44 pencils and 50 Erasers is $392. The cost of 46 pens, 54 pencils and 60 erasers is $466. If the cost of 109 pens, 125 pencils and 140 Erasers is $N, find N.

A. 545.5
B. 1091
C. 2182
D. 2500
E. Data Insufficient

Solution -

Let cost one pen, one pencil, one eraser be dollars x, y, z respectively.

Therefore, \(40x+44y+50z = 392\) …………. (1)

\(46x+54y+60z = 466\) ………………… (2)

Multiply equation (2) by \(\frac{3}{2}\);

\(69x + 81y + 90z = 699 \)…………. (3)

Adding equations (1) and (3); we get –

\(109x + 125y + 140z = 1091\).

Answer Choice-B


--------------------
How do you know you need to multiply the second equation by "3/2". Was it trial and error? If so, wouldn't that take up a lot of time?
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There's a small catch in the Question.

Difference between options is high enough to make a calculated guess.

Assume 40 pens , 40 pencils and 50 erasers to cost 400

If you multiple by 3.

120 pens , 120 pencils and 150 erasers would cost 1200.

So 109 pens, 125 pencils and 140 erasers, so... answer should be lesser than 1200 since we've decreased the the number.

525 is too small. Option B fits.

Posted from my mobile device
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The cost of 40 pens, 44 pencils and 50 erasers is $392. The cost of 4 29 Jul 2021, 05:44
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Aastha1908 rohitagr999

I did it with hit and trial and didn't take more than 2 min to solve.
Though we can solve it algebraically too.

40x+44y+50z=392

or 20x+22y+25z=196.....(1)

46x+54y+60z=466

or 23x+27y+30z=233.....(2)


suppose we have 2 constants a and b such that (1)*a +(2)*b= 109x+125y+140z

(20x+22y+25z)*a + (23x+27y+30z)*b = 109x+125y+140z

(20a+23b)x + (22a+27b)y + (25a+30b)z = 109x+125y+140z

20a+23b=109.....(3)

25a+30b=140 or 20a+24b=112.....(4)

From (3) and (4)

b=3 and a=2


109x+125y+140z= 196*2+233*3
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The cost of 40 pens, 44 pencils and 50 erasers is $392. The cost of 4 21 Jun 2019, 07:45
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nick1816
The cost of 40 pens, 44 pencils and 50 Erasers is $392. The cost of 46 pens, 54 pencils and 60 erasers is $466. If the cost of 109 pens, 125 pencils and 140 Erasers is $N, find N.

A. 545.5
B. 1091
C. 2182
D. 2500
E. Data Insufficient
Show more

40a+44b+50c=392.....(1)

46a+54b+60c=466.......(2)

Multiply equation (2) by 3/2, and add to equation (1), we get

109a+125b+140c=1091
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The cost of 40 pens, 44 pencils and 50 erasers is $392. The cost of 4 08 Jul 2021, 05:11
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nick1816
The cost of 40 pens, 44 pencils and 50 Erasers is $392. The cost of 46 pens, 54 pencils and 60 erasers is $466. If the cost of 109 pens, 125 pencils and 140 Erasers is $N, find N.

A. 545.5
B. 1091
C. 2182
D. 2500
E. Data Insufficient
Show more

Solution -

Let cost one pen, one pencil, one eraser be dollars x, y, z respectively.

Therefore, \(40x+44y+50z = 392\) …………. (1)

\(46x+54y+60z = 466\) ………………… (2)

Multiply equation (2) by \(\frac{3}{2}\);

\(69x + 81y + 90z = 699 \)…………. (3)

Adding equations (1) and (3); we get –

\(109x + 125y + 140z = 1091\).

Answer Choice-B
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The cost of 40 pens, 44 pencils and 50 erasers is $392. The cost of 4 28 Jul 2021, 08:19
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Implementing the requisite equations leads to

40x+44y+50z=39240x+44y+50z=392 …………. (1)

46x+54y+60z=46646x+54y+60z=466 ………………… (2)

modifying (2) by 3/2 multiplication ;

69x+81y+90z=69x+81y+90z=699…………. (3)

Adding of (1) and (3) leads us to

109x+125y+140z=1091
Therefore IMO B
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The cost of 40 pens, 44 pencils and 50 erasers is $392. The cost of 4 29 Jul 2021, 00:48
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The cost of 40 pens, 44 pencils and 50 Erasers is $392. The cost of 46 pens, 54 pencils and 60 erasers is $466. If the cost of 109 pens, 125 pencils and 140 Erasers is $N, find N.

A. 545.5
B. 1091
C. 2182
D. 2500
E. Data Insufficient

Solution -

Let cost one pen, one pencil, one eraser be dollars x, y, z respectively.

Therefore, \(40x+44y+50z = 392\) …………. (1)

\(46x+54y+60z = 466\) ………………… (2)

Multiply equation (2) by \(\frac{3}{2}\);

\(69x + 81y + 90z = 699 \)…………. (3)

Adding equations (1) and (3); we get –

\(109x + 125y + 140z = 1091\).

Answer Choice-B
Show more


--------------------
How do you know you need to multiply the second equation by "3/2". Was it trial and error? If so, wouldn't that take up a lot of time?
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The cost of 40 pens, 44 pencils and 50 erasers is $392. The cost of 4 09 Jun 2025, 06:32
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How do we know to multiply by 3/2? Thanks Bunuel
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The cost of 40 pens, 44 pencils and 50 erasers is $392. The cost of 4 09 Jun 2025, 06:45
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How do we know to multiply by 3/2? Thanks Bunuel
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We multiply by 3/2 because it helps align the coefficients in equation (2) so that, when added to equation (1), the result matches the target expression: 109a + 125b + 140c.
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The cost of 40 pens, 44 pencils and 50 erasers is $392. The cost of 4 09 Jun 2025, 07:09
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(1) 40a + 44b + 50c = 392

=> 20a + 22b + 25c = 196

(2) 46a + 54b + 60c = 466

=> 23a + 27b + 30c = 233

(3) What we need -> 109a + 125b + 140c

Hit and trial approach: I observed the 20a and 23a. How to get 109a from this?

To get unit digit 9, 23 x 3, perhaps? (23 x 3) + (20 x 2) works.

So, what happens with this to the whole equation?

[(20a + 22b + 25c) x 2] + [(23a + 27b + 30c) x 3] = 109a + 125b + 140c (woah!). I got lucky here!

The answer then -> (196 x 2) + (233 x 3) = 392 + 699 = 1091. Choice B.


I also see that an awesome algebraic approach has been shared already.

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The cost of 40 pens, 44 pencils and 50 erasers is $392. The cost of 4 01 Aug 2026, 06:57
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Did some approximation...

By subtracting the sum and difference of 40a + 44b + 50c = 392 & 46a + 54b + 60c = 466 you get:

40a + 44b + 50c = 392
Multiply by 3 as required equation is close to 3 times of derived equation - 3*(40a + 44b + 50c = 392) =>> 120a + 132b + 150c = 1161

Only option B matches

Of course, this would not work if options were close together.
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