a multiplication game

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Fri Apr 10, 2009 5:17 am

In a certain deck of cards, each card has a positive integer written on it. In a multiplication game, a child draws a card and multiplies the integer on the card by the next larger integer. If each possible product is between 15 and 200, then the least and greatest integers on the cards could be
A. 3 & 15
B. 3 & 20
C. 4 & 13
D. 4 & 14
E. 5 & 12

OA C

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Source: Beat The GMAT — Problem Solving | Fri Apr 10, 2009 5:17 am

kapsii: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Thu Feb 05, 2009 8:43 am; Thanked: 5 times

by kapsii » Fri Apr 10, 2009 5:41 am

Distilling the question for objective, we find it stating that for every number X in the given set of numbers multiply it by x+1 and the results should be between 15 and 200. Find the minimum and maximum number in the set.

I would start by examining the perfect squares between the ranges, they would give an indication of the numbers contained in the deck.

The nearest perfect square to 15 is 16 (4^2).
Nearest perfect square to 200 is 14^2.

examining the number 3, 3*4 is < 15, so, the minimum number has to be 4.
and 14^2 (196) is closest to 200, so we max value of product obtainable has to be 13*14 or 14*15 (which ever does not break the limit imposed by the range), so, go for 13*14, thus maximum number contained in the deck is 13.

Cheers,
Dubes

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pallasy: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Tue Nov 15, 2011 8:07 pm

by pallasy » Sun Mar 18, 2012 10:14 am

kapsii wrote:Distilling the question for objective, we find it stating that for every number X in the given set of numbers multiply it by x+1 and the results should be between 15 and 200. Find the minimum and maximum number in the set.

I would start by examining the perfect squares between the ranges, they would give an indication of the numbers contained in the deck.

The nearest perfect square to 15 is 16 (4^2).
Nearest perfect square to 200 is 14^2.

examining the number 3, 3*4 is < 15, so, the minimum number has to be 4.
and 14^2 (196) is closest to 200, so we max value of product obtainable has to be 13*14 or 14*15 (which ever does not break the limit imposed by the range), so, go for 13*14, thus maximum number contained in the deck is 13.

Why did you automatically assume 14 must be multiplied by 15? I understand the instructions say to multiply by the next highest integer, but since it's a "deck of cards," wouldn't 14 be the highest integer (J=11, Q=12, K=13, A=14)? If so, wouldn't 14 just need to be multiplied by 14 since there is no larger integer? 14x14 = 196, which means the answer would be 4 & 14? Please explain why this is not the case.

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killer1387: Legendary Member; Posts: 581; Joined: Sun Apr 03, 2011 7:53 amFollowed by: 5 members; Thanked: 52 times

by killer1387 » Mon Mar 19, 2012 4:13 am

sanju09 wrote:In a certain deck of cards, each card has a positive integer written on it. In a multiplication game, a child draws a card and multiplies the integer on the card by the next larger integer. If each possible product is between 15 and 200, then the least and greatest integers on the cards could be
A. 3 & 15
B. 3 & 20
C. 4 & 13
D. 4 & 14
E. 5 & 12

OA C

by options itself,
3*4=12<15
hence A/B out
4*5=20>15 ok
for 14*15=210>200 hence D is out
question asks of least E is out

hence C

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PhDmessi: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Mon Oct 03, 2011 1:33 pm; Thanked: 1 times

by PhDmessi » Mon Mar 19, 2012 5:26 am

If 15 is the smallest product, start by picking numbers:
- If the card is a 3, then eventually 3*4 = 12 --> too small
- Try 4: 4*5 = 20

Thus, least integer is 4, eliminate answer choice A,B and E.

Anwer choice C and D have 13 and 14 left, try both:
- 13*14 = 182
- 14*15 = 210 --> too large

Thus, 13 is the greatest integer on the cards.

Correct answer is C.

Never give up!

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