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| Author: | u2lover [ 08 Aug 2006, 12:49 ] |
| Post subject: | Each digit in the two-digit number G is halved to form a new |
Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H? A. 153 B. 150 C. 137 D. 129 E. 89 |
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| Author: | Bunuel [ 13 Aug 2013, 23:50 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
u2lover wrote: Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H? A. 153 B. 150 C. 137 D. 129 E. 89 Two-step solution: G + G/2 = 3G/2 --> the sum is a multiple of 3. G is a two-digit number --> G < 100 --> 3G/2 < 150. Among the answer choices the only multiple of 3 which is less than 150 is 129. Answer: D. |
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| Author: | ps_dahiya [ 08 Aug 2006, 13:22 ] |
| Post subject: | |
D That means H is half of G. So G+ H = 3H. This means sum of G and H must be divisible by 3. C and E are out. Lets see other choices A - 153 then H = 51, G = 102  G is a two gidit number
B - 150 then H = 50, G = 100 G is a two gidit number
D - 129 then H = 43, G = 86  ANSWER
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| Author: | paddyboy [ 08 Aug 2006, 13:02 ] |
| Post subject: | |
D it is... The way to do it is to try and break the units digit into numbers that satisfy the criterion that it should be the sum of a number and its double... i.e. x + 2x... A satisfies, as 3 = 1 + 2, but 15 cannot be broken into such a form. The next choice that satisfies is D; 9 = 3 + 6. Also, 12 can be written as (8+4)... Hence D |
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| Author: | haas_mba07 [ 08 Aug 2006, 13:12 ] |
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D 129 Given G =10x+y H= 10 (x/2) + y/2 G+H = 3/2(10x+y) = 3/2G G < 100 => G+H < 150 A & B are ruled out (G+H)x2/3 = G an integer. Only D is an integer. Answer: D |
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| Author: | u2lover [ 08 Aug 2006, 13:14 ] |
| Post subject: | |
Can you enlighten me why not E?  what am I missing?
46 and 23 is 89 |
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| Author: | IanStewart [ 05 May 2011, 13:45 ] |
| Post subject: | Re: odds and evens! |
AnkitK wrote: Each digit in the 2 digit number G is halved to form a new 2 digit number H.Which of the following could be the sum of G and H? A.153 B.150 c.137 D.129 E.89 For the question to make sense, G must be even. So we are adding an even number G to G/2, and the answer will be 3*(G/2), and thus must be a multiple of 3. Further, G < 100, so 3G/2 is less than 150. The only possible answer is thus D. |
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| Author: | Nayimoni [ 05 May 2011, 15:23 ] |
| Post subject: | Re: odds and evens! |
G = 10x + y H = 5x + y/2 so G + H = 15x + 3/2y. Multiply by 2 and you get 2 ( G+H )= 3 (10 x + y) so G+H must be a multiple of 3 and G a multiple of 2 (obviously otherwise we cant divide G) and we know G less than or equal to 88 (highest 2 digits even) and so H less than or equal to 44 (half G), so G+H less than 132. ABC out and E out because not a multiple of 3 Answer is D Hope this is helpful |
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| Author: | ygdrasil24 [ 05 Sep 2013, 20:24 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
Bunuel wrote: u2lover wrote: Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H? A. 153 B. 150 C. 137 D. 129 E. 89 Two-step solution: G + G/2 = 3G/2 --> the sum is a multiple of 3. G is a two-digit number --> G < 100 --> 3G/2 < 150. Among the answer choices the only multiple of 3 which is less than 150 is 129. Answer: D. What could be the minimum number ? |
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| Author: | Bunuel [ 05 Sep 2013, 22:51 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
ygdrasil24 wrote: Bunuel wrote: u2lover wrote: Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H? A. 153 B. 150 C. 137 D. 129 E. 89 Two-step solution: G + G/2 = 3G/2 --> the sum is a multiple of 3. G is a two-digit number --> G < 100 --> 3G/2 < 150. Among the answer choices the only multiple of 3 which is less than 150 is 129. Answer: D. What could be the minimum number ? Assuming G is a positive number, the least value of G+G/2 will be 20+10=30. G must be even and cannot be less that 20. If it's an even number less than 20, then G/2 will not be a two-digit number. Hope it's clear. |
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| Author: | ygdrasil24 [ 05 Sep 2013, 23:08 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
What could be the minimum number ?[/quote] Assuming G is a positive number, the least value of G+G/2 will be 20+10=30. G must be even and cannot be less that 20. If it's an even number less than 20, then G/2 will not be a two-digit number. Hope it's clear.[/quote] Yes it is thanks  So basically G ranges from 20 to 198 for all G >0 |
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| Author: | Bunuel [ 05 Sep 2013, 23:12 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
ygdrasil24 wrote: Yes it is thanks So basically G ranges from 20 to 198 for all G >0 No. G must also be a two digit number, so it ranges from 20 to 88. |
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| Author: | ygdrasil24 [ 05 Sep 2013, 23:21 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
Bunuel wrote: ygdrasil24 wrote: Yes it is thanks So basically G ranges from 20 to 198 for all G >0 No. G must also be a two digit number, so it ranges from 20 to 88. Hmm... blunder as always  By the way why cant G(max) be 98 , H(max) be 49 in that case |
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| Author: | Bunuel [ 05 Sep 2013, 23:23 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
ygdrasil24 wrote: Bunuel wrote: ygdrasil24 wrote: Yes it is thanks So basically G ranges from 20 to 198 for all G >0 No. G must also be a two digit number, so it ranges from 20 to 88. Hmm... blunder as always By the way why cant G(max) be 98 , H(max) be 49 in that case We are told that EACH digit in the two-digit number G is halved, thus both digits of G must be even. |
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| Author: | JeffTargetTestPrep [ 24 Jan 2018, 09:39 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
u2lover wrote: Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H? A. 153 B. 150 C. 137 D. 129 E. 89 We can let a = the tens digit of H and b = units digit of H; thus, H = 10a + b and G = 20a + 2b and the sum of H and G is: H + G = (10a + b) + (20a + 2b) = 30a + 3b = 3(10a + b) = 3H Since the sum G + H is a multiple of 3, we can eliminate choices C and E. Now let’s analyze the remaining three choices: A) 153 3H = 153 H = 51 and G = 102 However, G is a two-digit number, so A couldn’t be the answer. B) 150 3H = 150 H = 50 and G = 100 However, G is a two-digit number, so B couldn’t be the answer. Therefore, the answer must be D. Let’s verify it anyway. D) 129 3H = 129 H = 43 and G = 86 Answer: D |
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| Author: | asrithareddyk [ 25 Jul 2021, 10:50 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
u2lover wrote: Can you enlighten me why not E?  what am I missing?46 and 23 is 89 Hey 46+23 is 69 So 89 is not answer Also 89 not divisible by 3 Posted from my mobile device |
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| Author: | Archit3110 [ 27 Sep 2021, 09:47 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
u2lover wrote: Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H? A. 153 B. 150 C. 137 D. 129 E. 89 two digit number 10+b and half ; 5a+b/2 sum ; 10+b+5a+b/2 ; 30a+3b/2 or say ; 3*(10a+b)/2 use plugin we see at 129 3*(10a+b) = 258 10a+b =86 which is two digit number sufficient option D is correct |
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| Author: | Fdambro294 [ 06 Oct 2021, 19:24 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
To take a two digit number G and half each digit to get a new two digit integer H, then the digits of G must be EVEN Max we can make G and still get an H integer is: G = 88 In which case: H = 44 MAX sum of (G + H) = 88 + 44 = 132 Eliminate A, B, C Answer (D) 129 is pretty close to 132 (it is -3 less) If we drop the units digit of G from 8 to 6 (gives us -2 less towards the sum) ——-> H’s unit digit would drop from 4 to 3 (gives us -1 less towards the sum) Thus, G = 86 should work G = 86 ——-> H = 43 (86 + 43) = 129 D is the answer Posted from my mobile device |
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| Author: | vaibhav1221 [ 06 Jan 2022, 00:08 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
My approach was as a bit question specific and allowed me to avoid algebra. The approach is as follows: Because we need to half each of the digits of the two digit number, both the numbers must be even. The largest single digit even number is 8. So, the largest two digit number eligible would be 88. So, Largest value of G = 88, then H = 44 and G+H = 132. This leaves on option D and E because the largest value G+H can take is 132. 129 would be much easier to check and if it doesn't fit, 89 would be the answer without checking. Because 129 is very close to 132, we must decrease the value of G from 88 to 86 and value of H to 43. This gives us the value G+H = 129. D |
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| Author: | iamcabbage [ 31 Oct 2022, 11:10 ] |
| Post subject: | Re: Each digit in the two-digit number G is halved to form a new |
Hi, why is 3G/2 less than 150? |
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