### CAT Pipes and Cisterns - Do-able but atypical

**Question**

**Correct Answer: (A)**

**Explanatory Answer:**

**Answer Choice (A)**

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A drain pipe can drain a tank in 12 hours, and a fill pipe can fill the same tank in 6 hours. A total of n pipes – which include a few fill pipes and the remaining drain pipes – can fill the entire tank in 2 hours. How many of the following values could ‘n’ take?

a) 24

b) 16

c) 33

d) 13

e) 9

f) 8

A. 3

B. 4

C. 2

D. 1

Two drain pipes can drain the same volume that one fill pipe fills. This means that a D-D-F combination has to have a net volume effect of 0.

In spite of this, the tank still gets filled. Only the fill pipes can manage to fill the tank. In addition to all the net zero effect pipes, we need three more fill pipes in order to fill the tank in 2 hours.

So, we can have as many D-D-Fs as we want, but we need one F-F-F at the end to ensure that the tank gets filled in 2 hours.

So the number of pipes will be → (D – D - F).......(D – D - F) + (F – F - F)

The number of pipes has to be a multiple of 3. Only options A, C and E fit the description.

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Traders A and B buy two goods for Rs. 1000 and Rs. 2000 respectively. Trader A marks his goods up by x%, while trader B marks his goods up by 2x% and offers a discount of x%. If both make the same profit, find x

A. 25%

B. 12.5%

C. 37.5%

D. 40%

SP of trader A = 1000 (1 + x)

Profit of trader A = 1000 (1 + x) - 1000

MP of trader B = 2000 (1 + 2x)

SP of trader B = 2000 (1 + 2x) (1 - x)

Profit of trader B = 2000(1 + 2x) (1 - x)- 2000

Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 - x)- 2000

1000x = 2000 – 4000x^{2} + 4000x – 2000x - 2000

4000x^{2} -1000x = 0

1000x (4x - 1) = 0

ð x = 25%

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(|x|
-3) (|y| + 4) = 12. How many pairs of integers (x, y) satisfy this equation?

- 4
- 10
- 6
- 8

Product
of two integers is 12. Further |y| + 4 __>__ 4 . So, we know that one of
the numbers has to be greater than 4.

So,
we can have

1
x 12 => x = __+__4, y = __+__8

2
x 6 => x = __+__5, y = __+__2

3
x 4 => x = __+__6, y = 0

4
+ 4 + 2 = 10 possible solutions

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How many of the following
statements have to be true?

- No year can have 5 Sundays in the month of May and 5 Thursdays in the month of June
- If Feb 14
^{th}of a certain year is a Friday, May 14^{th}of the same year cannot be a Thursday - If a year has 53 Sundays, it can have 5 Mondays in the month of May

A. 0

B. 1

C. 2

D. 3

I.
No year can have 5 Sundays in the month of May
and 5 Thursdays in the month of June

A year has 5 Sundays in the month
of May => it can have 5 each of Sundays, Mondays and Tuesdays, or 5 each of
Saturdays, Sundays and Mondays, or 5 each of Fridays, Saturdays and Sundays.
Or, the last day of the Month can be Sunday, Monday or Tuesday.

Or, the 1^{st} of June
could be Monday, Tuesday or Wednesday. If the first of June were a Wednesday,
June would have 5 Wednesdays and 5 Thursdays. So, statement I need not be true.

II.
If Feb 14^{th} of a certain year is a
Friday, May 14^{th} of the same year cannot be a Thursday

From Feb 14 to Mar 14, there are
28 or 29 days, 0 or 1 odd day

Mar 14 to Apr 14, there are 31
days, or 3 odd datys

Apr 14 to May 14 there are 30 days
or 2 odd days

So, Feb 14 to May 14, there are
either 5 or 6 odd days

So, if Feb 14 is Friday, May 14
can be either Thursday or Wednesday. So, statement 2 need not be true.

III.
If a year has 53 Sundays, it can have 5 Mondays
in the month of May

Year has 53 Sundays => It is
either a non-leap year that starts on Sunday, or leap year that starts on
Sunday or Saturday.

Non-leap year starting on Sunday:
Jan 1^{st} = Sunday, jan 29^{th} = Sunday. Feb 5^{th}
is Sunday. Mar 5^{th} is Sunday, Mar 26 is Sunday. Apr 2^{nd}
is Sunday. Apr 30^{th} is Sunday, May 1^{st} is Monday. May
will have 5 Mondays.

So, statement C can be true.

Only one of the
three statements needs to be true. **Answer
Choice (B)**

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A.
2

B.
3

C.
1

D.
More than one possible value of z exists

xy = 4

xy could be 2 x 2 or 4 x 1

221

222 These are the possible triangles

223

441

22x will be a triangle if x
is 1, 2 or 3 (trial and error)

44x is a triangle only if x
is 1.

Ø
221 is acute. 1^{2} + 2^{2}>
2^{2}

Ø
222 is equilateral. So acute.

Ø
223 is obtuse. 2^{2} + 2^{2}<
3^{2}

Ø
144 is acute. 1^{2} + 4^{2}>
4^{2}

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Sides of a triangle are 6,
10 and x for what value of x is the area of the D the maximum?

A. 8
cms

B. 9
cms

C. 12
cms

D. None
of these

Side
of a triangle are. 6, 10, x.

Area
= 1/2 x 6 x 10 sin ÐBAC.

Area
is maximum, when ÐBAC
= 90^{o}

x = sqrt(100 + 36) = sqrt(136)

There
is a more algebraic method using hero’s formula. Try that also.

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Consider a class of 40 students whose
average weight is 40 kgs. m new students join this class whose average weight
is n kgs. If it is known that m + n = 50, what is the maximum possible average
weight of the class now?

- 40.18 kgs
- 40.56 kgs
- 40.67 kgs
- 40.49 kgs

If the overall average
weight has to increase after the new people are added, the average weight of
the new entrants has to be higher than 40.

So, n > 40

Consequently, m has to be
<10 (as n + m = 50)

Working
with the “differences” approach, we know that the total additional weight added
by “m” students would be (n - 40) each, above the already existing average of
40. m(n - 40) is the total extra additional weight added, which is shared
amongst 40+m students. So, m (n – 40)/ (m + 40) has to be maximum for the
overall average to be maximum.

At
this point, use the trial and error approach (or else, go with the answer
options) to arrive at the answer.

The
maximum average occurs when m = 5,and n = 45

And
the average is 40 + (45 – 5) * 5/45 = 40 + 5/9 = 40.56

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51 -60 - 3 students

61 -70 - 8 students

71 -80 - 7 students

81 -90 - 4 students

91 -100 - 3 students

Furthermore, we know that at least as many
students scored 76 or more as those who scored below 75. What is the minimum
possible average overall of this class?

- 72
- 71.2
- 70.6
- 69.2

Let's employ the idea of a total of 25 students (all of the same
weight) sitting on a see-saw, which has numbers from 51 to 100 marked on it. At
least as many students are sitting on 76 (or to its right), as there are
sitting to the left of 75. Now this means that you can have only one person
sitting to the left of 75 and all the rest sitting beyond 76. But you can't do
that, as you have other constraints as well.

First of all, you have to seat 3 students from 51 to 60, and 8
students from 61 to 70. Secondly, you also have to make sure that the average
is the least. This means that the see-saw should be tilting as much to the left
as possible, which in turn means that the number of people sitting to the left
of 75 should be the highest possible.

This makes it 12 students to the left of 75, and the remaining 13
students on 76 or to its right.

Next, how do you ensure that the average is least, i.e. how do you
ensure that the balance tilts as much as possible to the left? Make each
student score as little as possible given the constraints.

So, the first 3 students only score 51 each. The next 8 students
score only 61 each. 11 Students are now fixed. The 12^{th} student has
to be below 75, so seat him on 71. The remaining 6 students (who are in the 71
to 80 range) have to score 76. The next 4 score 81 and the next 3 score 91.

This would give you the least average.

The lowest possible average would be:

= [(3 * 51) + (8 * 61) + 71 + (6 * 76) + (4* 81) + (3 * 91)]/25

= 70.6

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How many onto functions can be defined from the set A = {1, 2, 3, 4} to
{a, b, c}?

- 81
- 79
- 36
- 45

First let us think of the number of potential functions possible. Each
element in A has three options in the co-domain. So, the number of possible
functions = 3^{4} = 81.

Now, within these, let us think about functions that are not onto. These
can be under two scenarios

Ø Let us assume that elements are mapped into A and B. Number of ways in
which this can be done = 2^{4} – 2 = 14

o
2^{4} because the number
of options for each element is 2. Each can be mapped on to either A or B

o
-2 because these 2^{4}
selections would include the possibility that all elements are mapped on to A
or all elements being mapped on to B. These two need to be deducted

Ø The elements could be mapped on B & C only or C & A only. So,
total number of possible outcomes = 14 * 3 = 42.

Total number of onto functions = Total number of functions – Number of
functions where one element from the co-doamin remains without a pre-image -
Number of functions where 2 elements from the co-doamin remain without a
pre-image

ð Total number of onto functions = 81 – 42 – 3 = 81 – 45 = 36

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A. 2

B. 4

C. 0

D. 1

We
start with the knowledge that the modulus of a number can never be negative,
though the number itself may be negative.

The
first equation is a pair of lines defined by the equations

y
= 8 – x ------- (i) {when y is
positive}

y
= x – 8 ------- (ii) {when y is
negative}

With
the condition that x ≤ 8 (because if x becomes more
than 8, |y| will be forced to be negative, which is not allowed)

The
second equation is a pair of lines defined by the equations:

y
= 6 – x ------- (iii) {when x is
positive}

y
= 6 + x ------- (iv) {when x is
negative}

with
the condition that y cannot be greater than 6, because if y > 6, |x| will
have to be negative.

On
checking for the slopes, you will see that lines (i) and (iii) are parallel.
Also (ii) and (iv) are parallel (same slope).

Lines
(i) and (iv) will intersect, but only for x = 1; which is not possible as
equation (iv) holds good only when x is negative

Lines
(ii) and (iii) do intersect within the given constraints. We get x = 7, y = -1.
This satisfies both equations.

Only
one solution is possible for this system of equations.

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Qn. f(x + y) = f(x)f(y)
for all x, y, f(4) = +3 what is f(-8)?

f(x + 0) = f(x) f(0)

f(0) = 1

f(4 + -4) = f(0)

f(4 + -4) = f(4) f(-4)

1 = +3 x f(-4)

f(-4) = 1/3

f(-8) = f(-4 + (-4)) = f(-4) f(-4)

f(-8) = 1/3 x 1/3 = 1/9

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