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.

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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Find 0.42-0.35

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

A. \(0.7\)

B. 0.7

C. \(\frac{7}{11}\)

D. \(\frac{4}{13}\)

E. \(\frac{7}{99}\)

Find 0.42-0.35


.42 nonterminating repeating will be equivalent to 42/99

Similarly .35 (NTR)= 35/99

42/99-35/99
=7/99

E

Posted from my mobile device

Difference would be 0.0707070.....

Obviously (A) and (B) are wrong and we can easily see that (C) and (D) are greater than 0.1

So Answer must be (E)
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given n is an +ve integer and check whether n(n^2+2) is divisible by 12
#1
n is prime ; for n = 2 we get yes and for any other value no ;
insufficient
#2
n>10 again yes for n=12 and no for n=11 insufficient
from 1 &2
n is prime and >10 , n(n^2+2) is not divisible by 12
IMO C

X+Y+Z=36
X,Y,Z>0

X=A+1
Y=B+1
Z=C+1
A,B,C are non-negative integers

then
A+B+C=33

A|B|C
Total number of ways to arrange 33 objects and 2 lines=\(\frac{(33+2)!}{33!2!}\)=35*34/2 = 595

The numbers you have taken do not give a possible scenario:

0.3P + 0.6E = 4
3P + 6E = 40
3*(P + 2E) = 40
P + 2E = 40/3

If P and E are integers, their sum cannot be a fraction.

Instead of $4, take $4.2 and you will get your solution.
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RESERVED................................
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RESERVED................................
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Hi math experts, try this question and provide feedback...


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Hi Janvisahu,

Should be D.

From Stmnt 1: Angle BCA = 60. Hence Triangle, GDC is equilateral Triangle. So, Sufficient.

From Stmnt 2: Triangle ABC is Similar to Triangle GHC. Using this you can find the length of GC. So, Sufficient.

Hope this Helps.

How could you come to the conclusion that angle BCA is 60 degrees from statement 1??

Posted from my mobile device

Sin C = \(8/16/\sqrt{3}\)
C = 60.

Hi
It is easy to prove that angle BCA is 60.
triangle ABC is 30:60:90 triangle as we know by the ration of sides.
Sides are in ratio-1:root 3:2




[/quote]

Hi Janvisahu,

Should be D.

From Stmnt 1: Angle BCA = 60. Hence Triangle, GDC is equilateral Triangle. So, Sufficient.

From Stmnt 2: Triangle ABC is Similar to Triangle GHC. Using this you can find the length of GC. So, Sufficient.

Hope this Helps.[/quote]


How could you come to the conclusion that angle BCA is 60 degrees from statement 1??

Posted from my mobile device[/quote]
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New Question:

If from a well-shuffled deck of 52 cards, five cards are drawn at random one by one with replacement. What is the probability of fifth card being spade?
A) 1/3
B) 1/4
C) 2/5
D) 3/13
E) 3/26

source:self


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Challenging Question:- Earn kudos for correct solution

If from a well-shuffled deck of 52 cards, five cards are drawn at random one by one WITHOUT replacement. What is the probability of fifth card being spade?
A) 1/3
B) 1/4
C) 2/5
D) 3/13
E) 3/26


source:self
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genxer123
You might like this problem...


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Total way to select (9 good & 1 bad pen) with 4 pen defective
G*G*G*D*G*G*G*G*G*G => 9*8*7*1*6*5*4*3*2*1 => 9!
Total way with no pre condition 10!

Probability=> 9!/10! => 1/10