Quant Boosters  Kamal Lohia  Set 5

Number of Questions  30
Topic  Quant Mixed Bag
Solved ? : Yes
Source

Q1) All the water in a big tank A is emptied into two smaller empty tanks B and C. The volume of water in C is 33 1/3% of that in A. If 300 liters of water, which had gone into B had instead gone into C, B would have 50% more water than C. What was the volume of water in A?

Let total volume of water in A = 900x liters.
So as per given data, we know that
600x  300 = 1.5(300x + 300)
=> 150x = 750
=> x = 5
=> volume of water in A = 900x = 900 × 5 = 4500 liters.

Q2) Find out the number of digits in the expression 2^150

2^10~10^3
So 2^150 ~ 10^45 i.e. 45 + 1 = 46 digits.

Q3) In a locality, there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are next to each other?

H H H H S H S H S H
Any two of the three selected houses are not to be adjacent. That means there will be atleast one house between any two of these selected house. See the scenario from reverse now.
H  H  H  H  H  H H 
Among the remaining seven houses, we need to locate for the places from where these three houses have been selected. Also between any two of those remaining seven houses, at most one of the selected houses may be placed.
So out of total 8 available places, between the remaining seven houses as shown above, we need to find/select the 3 places where the three selected houses actually belonged to.
And this can be accomplished in C(8, 3) ways i.e. = 56 ways.

Q4) Considering an obtuse angle triangle having sides with integer side as 8, 15 and 'X'. how many such different triangles can be formed?

This is a CAT2008 question.
Just apply the triangle inequality on X to get that 15  8 < X < 15 + 8
=> 7 < X < 23
=> 8 ≤ X ≤ 22.
Now we have one more condition that triangle should be obtuse. So if c is the longest side in a triangle with a, b, c as three sides such that c² > a² + b² then triangle is obtuse.
So two cases are possible now that X is the longest side OR X is not the longest side. Just apply the above condition and find the possible range for X.

Q5) F(a + b) = F(a) + F(b) and F(5)=12. Find F(12)

Given is: f(a + b) = f(a) + f(b)
For a = b, we have: f(2a) = 2f(a)
Similarly if we put b = 2a in given equation, we get: f(3a) = 3f(a) using second equation also.
So in general, f(na) = nf(a).
Now given is: f(5) = 12.
So f(12) = f((12/5)×5) = (12/5)×f(5) = (12/5)×12 = 144/5

Q6) What are the number of integer coordinates that lies with in the area bounded by curves for example X^2 + Y^2 = 9

You can solve it using equation like:
x² + y² ≤ 9
=> x² ≤ 9  y²
=> √(9  y²) ≤ x ≤ √(9  y²).
We just need to find integer pairs (x, y) which satisfy this.
See √(9  y²) cannot be greater than 3 so x is bounded between [3, 3]. So we just need to find all pairs (x, y) which satisfy this with above mentioned bound on x.
Also √(9  y²) can be (√9 = 3), (√8 = 2.something), (√5 = 2.something) or 0 as y becomes 0, 1, 2 or 3 respectively.
So when y = 0, x can take 7 values from [3, 3]
when y = ±1, x can take 5 values from [2, 2]
when y = ±2, x can take 5 values from [2, 2]
and when y = ±3, x will be 0 only.
Hence total number of pairs are ≡ 7 + 10 + 10 + 2 = 29.

Q7) P and Q are two opposite ends of a straight running track. Ria and Tia start running towards each other simultaneously from P and Q respectively. As soon as any of them reaches an end, she turns back and starts running towards the other end. Their first meeting happens at a point 210 m away from P and the second meeting happens at a point 150 m away from Q. If the speeds (in m/s) of Ria and Tia are x and y respectively, such that 2x > y > x, then what is the distance between P and Q?

Till first meeting point, sum of the distance travelled of the two runners is one complete track length, say d (which is the distance between PQ, i.e. to be determined).
And when they are meeting for the second time, sum of the distances travelled (from starting position) of the two runners is 3d (just check (how??) by drawing the diagram of movement of the runners).
That means the runners has travelled for thrice the time taken earlier or each runner has travelled thrice the distance as earlier.
Now till first meeting Ria travels a distance of 210m (measured from starting position) and till second meeting she travels a distance of (d + 150)m (measured from starting point).
From earlier observation we know that ratio of distance travelled is 1 : 3
that means d + 150 = 3 × 210
or d = 630  150 = 480m.

Q8) Three people A, B and C start moving around a circular track of 100 m simultaneously with speeds of 2 m/s, x m/s and y m/s respectively in clock wise direction. They meet for the first time after t seconds and they meet for the first time at their common starting point after 3t seconds. Which of the following can never be the value of x?
a) 2
b) 3
c) 5
d) 8

First understand one thing that when two runners, running in same direction on a circular track and also starting at same point and same time, meets first time then difference in their distance travelled is one complete round say 1R. Vice versa, if difference of distance travelled becomes one complete round between the two runners, moving in same direction on a circular track, then they have met once.
Now we know that one of the runner has speed 2 m/s and other runner has speed x m/s.
Let's examine the options,
if x = 2 m/s, that means these two runners are always together, so it's possible that third runner is meeting them thrice only till the time they are meeting at starting point again. So x = 2 m/s is possible.
If x = 3 m/s, that means in the time when first runner had travelled 2R, this second runner must have travelled 3R and both are at starting point. Also because difference in distance travelled by two runner, by this point of time, is 1R so they have met for the first time after start. But according to question, we want that the three runners must met in between some where (after t seconds) before meeting at starting point (after 3t seconds). So x cannot be 3 m/s.
If x = 5 m/s, then in the time first runner has travelled a distance of 2R, second runner would have travelled 5R i.e. creating a difference of 3R that means they would have met three times till the time they are meeting at starting point which is in accordance the question statement. So x = 5 m/s is possible.
If x = 8 m/s, then again for 1R distance of first runner, second runner would have travelled a distance of 4R creating a difference of 3R as earlier. So x = 8 m/s is also possible.

Q9) A watch, which gains time uniformly, was 5 minutes behind the correct time when it showed 11:55 AM on Monday. It was 10 minutes ahead of the correct time when it showed 06:10 PM on the next day. When did the watch show the correct time?
a) 6 AM, Tuesday
b) 6 PM, Monday
c) 2 PM, Tuesday
d) 10 PM, Monday

Just see that nonuniform watch is gaining 15 minutes in 30 hrs (from actual time of 12:00 noon on Monday to 6:00 pm on Tuesday it shows 11:55am  6:10pm).
So we just need to find that in how many hours it'll gain 5 minutes (to compensate previous gap). It should be very clear that if it is gaining 15 minutes in 30 hrs, it should gain 5 minutes in 10 hrs.
So it will shoe correct time on 10:00 pm Monday. (d)

Q10) In how many different ways can a cube be painted if each face has to be painted either red or blue?