Collection of 8 DS questions

Stand Corrected.

My assumption Two adjecent angles can not be 90 otherwise the other two will also be 90 was wrong.
Answer should be E.

The questions in this thread are pretty simple compared to the more difficult that Bunuel has posted.
Don't be complacent if you can solve all the above "700" questions correctly - just look in other sets of questions and you will find room for improvements.

This is good one!

Could you please explain the answers for the 1st one..I chose C as the answer

because for statement1

24
26
28
36
39
48

these show that the tens digit is factor of units digit..but >40

55
66
77

Tens digit is not a factor but multiple of units digit...am I right? I often get confused with these terminologies of factors and multiples:( could you take time to explain...thanks alot

Integer \(a\) is a multiple of integer \(b\) is the same as integer \(b\) is a factor of integer \(a\) and can be expressed as \(a=b*integer\). For example 6 is a multiple of 1, 2, 3, and 6 itself and 1, 2, 3, and 6 are factors (divisors) of 6.

In case when \(a=b\neq{0}\) then integer \(a\) is both a factor and a multiple of integer \(b\) (and vise-versa). For example 5 is the greatest factor (divisor) of 5 and 5 is also the least positive multiple of 5.

As for the question: algebraic solution is given in my post on the previous page.

Alternately you can notice than the question basically asks whether n is a prime number more than 20. Now, all primes (except 2 and 5) end with 1, 3, 7, or 9, as (1) says that the tens digit of n is a factor of the units digit then n can not possibly be a prime as the only 2-digit integers more than 20 satisfying this and having either of these digits as units digit are: 33, 77, 39, and 99 and neither is a prime number.

So the answer is A.

For more on this issues check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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The answers below may be incorrect. I haven't matched them with OA. If you see any problem with the logic or answer; please comment.


Sol:
20<n<100
(1)
Possible values of n:
22,24,26,28 ; Between 21 and 29, inclusive; 2 is a factor of 2,4,6,8
30,33,36,39 ; Between 30 and 39, inclusive; 3 is a factor of 0,3,6,9
40,44,48; Between 40 and 49, inclusive; 4 is a factor of 0,4,8
50,55
60,66
70,77
80,88
90,99

Every possible value is composite. Sufficient.

(2)
Unit's digit can be 3 or 6 giving at least 2 possibilities; 23 or 26, one of which is prime and other composite.
Not sufficient.

Ans: "A"

Sol:
(1) If the opposite angles are \(90^{\circ}\), then the quadrilateral is a rectangle or square. None of the interior angles is \(60^{\circ}\)
If the adjacent angles are \(90^{\circ}\), then the quadrilateral may have the rest two angles are any set of supplementary angles which also includes {60,120}. Maybe one of the interior angles is \(60^{\circ}\). Not sure though.

Not sufficient.

(2) The adjacent angles are \(90^{\circ}\), then the quadrilateral may have the rest two angles be any set of supplementary angles which also includes {60,120}. Maybe one of the interior angles is \(60^{\circ}\). Not sure though.

Not sufficient.

Combing both;
Two interior angles are \(90^{\circ}\). They can't be opposite angles as there are also two angles that are in ratio 1:2. If two angles are 90. Then their sum is 180. Sum of all interior angles of a quadrilateral is 360. The sum of the other two angles is 180.
Also they are in ratio;
1/2

So; x+2x=180
x=60
2x=120.

For sure there is one angle which is \(60^{\circ}\)

Sufficient.

Ans: "C"

Sol:
(1) x<8/9
y can be 1000 or 1/1000.
Not sufficient.

(2)y<1/8
x can be 1000 or 1/1000.
Not sufficient.

Combing both;
\(x+y < \frac{8}{9} + \frac{1}{8}\)
\(x+y < \frac{73}{72}\)

x+y can be 722/720 or 1/16.
Not sufficient.

Ans: "E"

Sol:

(1)
\(x^2+y^2 > z^2\)
Squaring both sides
\(x^4+y^4+2x^2y^2 > z^4\)
Now,
\(2x^2y^2\) will always be equal to or greater than 0 because square is always +ve. "+ve*+ve*2" will always be positive or greater than equal to 0.

Hence;
\(x^4+y^4>z^4\)
Sufficient.

(2)
x+y > z
I don't know the formula for double squaring this one. So; just substituting some values;

-1-1>-10
-2>-10
16<10000

2+2>1
16+16>1
32>1
Not Sufficient.

Ans: "A"

Sol:

Adult's ticket sold: a
Children's ticket sold: c
Total Price: 5a+2c
Average Cost:

(1)
\(\frac{c}{a}=\frac{3}{2}\)
\(2c=3a\)
\(c=(3/2)a\)
\(\frac{(5a+2c)}{(a+c)}=\frac{5a+3a}{a+(3/2)a}=\frac{16}{3}\)
Sufficient.

(2)
80 adult tickets were sold doesn't tell us anything about the number of children's tickets sold.
Not Sufficient.

Ans: "A"

Sol:
(1) Not sufficient. Doesn't tell us anything about goats.
(2) Not sufficient. Doesn't tell us anything about cows.

Combining both;
We know that all cows are goats as well as lions. But; there are also few lions that are not cows but goat. Thus, some goats are not cows; they are lions.

Sufficient.

Ans: "C"

Sol:
(1) Not Sufficient. Putting away the dishes may take less than 5 minutes or more than 5 minutes.
(2) Sufficient. Irrespective of how long it takes for vacuuming or laundry, putting away the dishes alone will cross the 1 hour deadline. 55 minutes will be spent washing and say Pat is already done done with his laundry and vacuuming; he can't complete all his tasks before at least 1hour 2minutes > 1hour.

Ans: "B"

Sol:
(1)
There can be fifteen 4's making the sum=60 and all of the elements equal.
There can be fourteen 1's and one 46 making the sum=60 and not all elements equal.
Not Sufficient.

(2)
Don't know by what principle of mathematics it is so. But; looks like it is possible only when all of them are 4.
Sufficient.

Ans: "B"

Thanks for the post Bunuel.

Squaring both sides

Now,
will always be equal to or greater than 0 because square is always +ve. "+ve*+ve*2" will always be positive or greater than equal to 0.

Hence;

Sufficient.

I used same method as Fluke. Can anyone clarify why E and not A for this one?

Refer to my first post on this page: set-of-8-ds-questions-85290-20.html#p768034
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@Bunnel, pls explain statement 1 in #2

Is is possible to draw a quad with two 90 degree on the same side or opp side and still satisfy the condition angles on the same side sum upto 180?

I think it is not possible to get 60 with statement 1.

Is the measure of one of the interior angles of quadrilateral ABCD equal to 60?

Sum of inner angels of quadrilateral is 360 degrees. (Sum of inner angles of polygon=180*(n-2), where n is # of sides)

(1) Two of the interior angles of ABCD are right angles --> angles can be 90+90 + any combination of two angels totaling 180. Not sufficient.

(2) The degree measure of angle ABC is twice the degree measure of angle BCD --> <ABC=2<BCD. Not sufficient

(1)+(2) Angles can be 90+90+45+135 Or 90+90+60+120. Not sufficient.

Answer: E.
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@Bunuel, would u plz explain problem no # 3??

Sure.

Is x+y<1

(1) \(x<\frac{8}{9}=\frac{64}{72}\). Not sufficient by itself.

(2) \(y<\frac{1}{8}=\frac{9}{72}\). Not sufficient by itself.

(1)+(2) \(x+y<\frac{64}{72}+\frac{9}{72}=\frac{73}{72}\), so \(x+y<\frac{73}{72}\) --> \(x+y\) may or may not be less than 1 (\(x+y\) could be less than 1 as well as more than 1, in the range \((1,\frac{73}{72})\)). Not sufficient.

Answer: E.
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You can draw numerous trapezoids wherein two adjacent angles are 90 and the other two in various combinations

Refer BUNUEL's solution.

When the question says two of the angles are right angles, can I then assume that the other two angles are NOT right angles?

Sure, since if 3 or 4 of the angles of ABCD are right, then saying that "2 of the angles of ABCD are right" would be clearly wrong.
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6. Are some goats not cows?
(1) All cows are lions
(2) All lions are goats.


Hi Bunuel - Is the below sol OK?

10 cows then 10 lions
20 lions then 20 goats
Q Are some goats not cows? this becomes may be because we don't about the 10 G and also we can mark strait E as we don't know how many goats are there if there are 100 G then no info abt 80 Gs....

This reasoning correct?

I don't understand what you mean there. Anyway solution is given here: collection-of-8-ds-questions-85290.html#p639290 and here: collection-of-8-ds-questions-85290.html#p666509

Hope it helps.
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Hi Bunuel,

My answer for this question is A. Please tell me where I am going wrong. This is how I solved the question:

From A:
x^2 + y^2 > z^2
Squaring both the sides, we get
x^4 + y^4 + 2x^2 * y^2 > z^4
=> x^4 + y^4 < z^4

How did you get that x^4 + y^4 < z^4 from x^4 + y^4 + 2x^2 * y^2 > z^4?

Correct answer is E, check here: collection-of-8-ds-questions-85290-20.html#p768034
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Oops! Realized my mistake. Thanks for the questions though. They are quite helpful.