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Math Expert
Joined: 02 Sep 2009
Posts: 60779

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11 Jul 2018, 22:57
1
lfc wrote:
1) Clearly not sufficient.

2) Gives me the following equations $$A=3B, B=A-1$$. On solving I get the number of cakes for A as $$3/2$$

Doubt Since the number of cakes has to be an integer, and 2) gives me a non-integer answer, should the answer be E or B?

PS: This is a self made question, made it to clarify this doubt, can such questions show up on the GMAT?

No, you won't get such question on the tests. If a variable must be an integer then it will be an integer. So, the answer to your question is none of the options, it's flawed by GMAT standards.

P.S. Please post self-made questions here only: https://gmatclub.com/forum/users-self-m ... 54835.html Thank you.
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Joined: 08 Jul 2018
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11 Jul 2018, 23:55
Bunuel wrote:
lfc wrote:
1) Clearly not sufficient.

2) Gives me the following equations $$A=3B, B=A-1$$. On solving I get the number of cakes for A as $$3/2$$

Doubt Since the number of cakes has to be an integer, and 2) gives me a non-integer answer, should the answer be E or B?

PS: This is a self made question, made it to clarify this doubt, can such questions show up on the GMAT?

No, you won't get such question on the tests. If a variable must be an integer then it will be an integer. So, the answer to your question is none of the options, it's flawed by GMAT standards.

Okay! Thanks
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Joined: 13 Feb 2018
Posts: 494
GMAT 1: 640 Q48 V28

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16 Jul 2018, 04:42
Napoleon sent two corps from the french town Lille to capture the city of Brussels. First corps’ (consisted of cavalry) average speed was 40 km/h and had to rest every 30 minutes to feed horses. After 2 hours and 40 minutes when the first corps resumed the march, they were 20 km far from Brussels. What was the average speed of the second corps, if they marched restless and reached Brussels after 4 hours?

A) 25 km/h
B) 30 km/h
C) 35 km/h
D) 40 km/h
E) 45 km/h

VP
Joined: 19 Oct 2018
Posts: 1297
Location: India

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10 May 2019, 18:15
2
1
Mr. White is an approximately forty years old father with 4 sons of distinct ages. Writing his age 3 times in succession, we get a 6-digit number that is equal to the product of his age, his wife's age and his 4 sons' ages. What is the sum of his wife's age and all 4 sons' ages?

A. 25
B. 30
C. 41
D. 51
E. 61
Intern
Joined: 26 Jan 2019
Posts: 3

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10 May 2019, 21:18
404040= 40*37*13*7*3*1

Wife's age = 37
Sons' age = 13,7,3,1
Sum = 37+13+7+3+1
= 61

Posted from my mobile device
Retired Moderator
Joined: 27 Oct 2017
Posts: 1410
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)

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23 Jun 2019, 09:01
2
The shape of a window is a combination of semicircle and rectangle with semicircle QR on top and rectangle PQRS at the bottom. If arc QR is a semicircle and PQRS is a rectangle. What is the area of the window?

(1) The perimeter of rectangle PQRS is 28 feet
(2) Each diagonal of rectangle PQRS is 10 feet long

Attachment:

Window.PNG [ 8.14 KiB | Viewed 384 times ]

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24 Jun 2019, 11:37
Suppose side QR=Length (L), and side RS = Breath (B)
From I,
2(L+B)=28
L+B=14 ..................... (i) , NOT SUFFICIENT

from II,
L^2+B^2=H^2
L^2+B^2=10^2

L^2+2LB+B^2=100+2LB
(L+B)^2=100+2LB ........... (ii) NOT SUFFICIENT

putting valu of L+B in eq ii
14^2=100 +2LB
LB=48
L=48/B .............................(iii)

Putting value of L in eq (i)
48/B+B=14
B^2-14B+48 = 0
B=6 and 8,
Hence Either (PS, SR) = (6,8) or (PS, SR) = (8,6)

Hence the radius of the semicircle cannot be determined.

So, the Area cannot be determined. NOT SUFFICIENT

gmatbusters wrote:
The shape of a window is a combination of semicircle and rectangle with semicircle QR on top and rectangle PQRS at the bottom. If arc QR is a semicircle and PQRS is a rectangle. What is the area of the window?

(1) The perimeter of rectangle PQRS is 28 feet
(2) Each diagonal of rectangle PQRS is 10 feet long

Attachment:
Window.PNG

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Joined: 14 Aug 2019
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21 Aug 2019, 05:42
A car runs on four tyres and has one extra tyre. If each tyre lasts for 10,000 km then what is the
maximum distance (in km) that the car can travel using the five tyres?
(a) 10,000 (b) 11,000 (c) 12,500 (d) 15,000

Posted from my mobile device
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Posts: 1297
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12 Sep 2019, 07:14
1
Find the number of odd (positive) divisors of 9!

A. 12
B. 14
C. 16
D. 18
E. 20
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Posts: 382
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12 Sep 2019, 07:34
1
9! = 9x8x7x6x5x4x3x2x1
= $$2^7$$ x $$3^4$$x 5 x 7

We know that the number of factors of a number will be expressed by the formula (p+1)(q+1)(r+1).....

To find odd factors/divisors, get rid of powers of 2 as they give even factors.

Odd factors/divisors of 9! = (4+1) x (1+1) x (1+1) = 20..........Option E
Math Expert
Joined: 02 Aug 2009
Posts: 8322

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12 Sep 2019, 07:46
2
nick1816 wrote:
Find the number of odd (positive) divisors of 9!

A. 12
B. 14
C. 16
D. 18
E. 20

For this we have to get 9! in prime factorization..

$$9!=1*2*3*4*5*6*7*8*9=2*3*2*2*5*2*3*7*2*2*2*3*3=2^7*3^4*5*7$$
Total factors = (7+1)(4+1)(1+1)(1+1)=8*5*2*2=160

Odd positive numbers will depend upon $$3^4*5*7$$....(4+1)(1+1)(1+1)=5*2*2=20

E
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12 Sep 2019, 07:53
1
nick1816 wrote:
Find the number of odd (positive) divisors of 9!

A. 12
B. 14
C. 16
D. 18
E. 20

To find the odd factors we need to remove the factors of 2 present in 9!

All the prime factors below 9 are 2,3,5,7

And we can find the powers of each prime factor in 9!

we get that 9! = 2^7 * 3^4 * 5^1 * 7 ^1.
when we remove all the factors of 2.
we are left with 3^4 * 5^1 * 7 ^1.

the total number of ODD factors will be - (4+1)(1+1)(1+1) = 5*2*2 = 20 factors.

IMO option E is the answer.

If you like my solution, do give kudos!

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Joined: 03 Jun 2019
Posts: 1950
Location: India
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12 Sep 2019, 09:45
1
nick1816 wrote:
Find the number of odd (positive) divisors of 9!

A. 12
B. 14
C. 16
D. 18
E. 20

Asked: Find the number of odd (positive) divisors of 9!

9! = 9*8*7*6*5*4*3*2*1 = 3^4 * 5*7 * even number
We have 5 choices for 3, 2 choices for 5 and 2 choices for 7
Number of odd divisors of 9! = 5*2*2 = 20

IMO E
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Posts: 1060
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14 Sep 2019, 09:22
nick1816 wrote:
Mr. White is an approximately forty years old father with 4 sons of distinct ages. Writing his age 3 times in succession, we get a 6-digit number that is equal to the product of his age, his wife's age and his 4 sons' ages. What is the sum of his wife's age and all 4 sons' ages?

A. 25
B. 30
C. 41
D. 51
E. 61

Well, with all the due respect, the question says approximately 40 years old. That no way signals we need to consider the age as 40 years. So I guess I will just pass and move on.
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Updated on: 14 Sep 2019, 11:58
1
Assume a 2-digit number=xy
xyxyxy=xy*a*b*c*d*e
xy(10101)=xy*a*b*c*d*e
10101=a*b*c*d*e
a*b*c*d*e=37*13*7*3*1

You actually don't need the actual age of Mr. White to solve this question.

TheNightKing wrote:
nick1816 wrote:
Mr. White is an approximately forty years old father with 4 sons of distinct ages. Writing his age 3 times in succession, we get a 6-digit number that is equal to the product of his age, his wife's age and his 4 sons' ages. What is the sum of his wife's age and all 4 sons' ages?

A. 25
B. 30
C. 41
D. 51
E. 61

Well, with all the due respect, the question says approximately 40 years old. That no way signals we need to consider the age as 40 years. So I guess I will just pass and move on.

Originally posted by nick1816 on 14 Sep 2019, 09:31.
Last edited by nick1816 on 14 Sep 2019, 11:58, edited 2 times in total.
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14 Sep 2019, 09:39
nick1816 wrote:
Assume a 2-digit number=xy
xyxyxy=xy*a*b*c*d*e
xy(100001)=xy*a*b*c*d*e
100001=a*b*c*d*e
a*b*c*d*e=37*13*7*3*1

You actually don't need the actual age of Mr. White to solve this question.

You meant 10101*xy? I guess so.

Thank you for providing the explanation.!
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14 Sep 2019, 09:46
Yup. That was a typo.
Edited.

TheNightKing wrote:
nick1816 wrote:
Assume a 2-digit number=xy
xyxyxy=xy*a*b*c*d*e
xy(100001)=xy*a*b*c*d*e
100001=a*b*c*d*e
a*b*c*d*e=37*13*7*3*1

You actually don't need the actual age of Mr. White to solve this question.

You meant 10101*xy? I guess so.

Thank you for providing the explanation.!
SVP
Joined: 03 Jun 2019
Posts: 1950
Location: India
GMAT 1: 690 Q50 V34

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14 Sep 2019, 11:56
1
nick1816 wrote:
Mr. White is an approximately forty years old father with 4 sons of distinct ages. Writing his age 3 times in succession, we get a 6-digit number that is equal to the product of his age, his wife's age and his 4 sons' ages. What is the sum of his wife's age and all 4 sons' ages?

A. 25
B. 30
C. 41
D. 51
E. 61

Given:
1. Mr. White is an approximately forty years old father with 4 sons of distinct ages.
2. Writing his age 3 times in succession, we get a 6-digit number that is equal to the product of his age, his wife's age and his 4 sons' ages.

Asked: What is the sum of his wife's age and all 4 sons' ages?

Let the age of Mr White be m, his wife be w, his sons ages be s1, s2, s3 & s4.
s1, s2, s3 & s4 are distinct positive integers

Let m be written as xy, where x is 3 or 4
xyxyxy = xy * w * s1*s2*s3*s4
10101(xy) = xy * w * s1*s2*s3*s4
10101 = w*s1*s2*s3*s4
10101 = 37*13*7*3*1

Sum of his wife's age and all 4 sons' ages = 37 + 13 + 7 + 3 + 1 = 61

IMO E
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Updated on: 16 Sep 2019, 22:23
1
Find 0.42-0.35

0.ab=0.ababababa......

A. $$0.7$$

B. 0.7

C. $$\frac{7}{11}$$

D. $$\frac{4}{13}$$

E. $$\frac{7}{99}$$

Originally posted by nick1816 on 16 Sep 2019, 22:13.
Last edited by nick1816 on 16 Sep 2019, 22:23, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 60779

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16 Sep 2019, 22:16
nick1816 wrote:
Find 0.42-0.35

0.ab=0.ababababa......

A. $$0.7$$

B. 0.7

C. $$\frac{7}{9}$$

D. $$\frac{4}{13}$$

E. $$\frac{7}{99}$$

Options B and C are the same. What is the source of this question?
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