Quant Boosters by VP  Set 2

While Badal travelled 20m, Carol travelled 3421=13m
so speed ratio B:C=20:13
so B/C = 20/13 = (L20)/(L34)
L=60m

Q17) I wanted to buy 2 dozen bananas but I am 30 Rupees short. So I bought 20 bananas and 2 rupees is left with me. Price of 1 banana?

24 bananas cost= > 30 more than what I have
20 bananas cost = > 2 less than what I have
So 4 bananas cost 30+2=32rupees
1 banana cost 8 rupees

Q18) 100 kg grapes contains 98% water. After few days, due to evaporation some water evaporates and it contains 94% of water. Find weight of grapes.

98% water means 2% mass=2kg mass
When water evaporates, water=94% so mass =6%
but mass is constant so 2 kg = 6%
Numerator(mass) is same but fraction increased 3 times. So denominator decreased 3 times
100/3 = 33.33 kg new weight of grapes.

Q19) If 8 coins are tossed, what is the probability that no two heads appear consecutively.

Use Fibonacci function
F(0) = 1
F(1) = 2
F(2) = 1 + 2 = 3
F(3) = 2 + 3 = 5
F(4) = 3 + 5 = 8
F(5) = 5 + 8 = 13
F(6) = 8 + 13 = 21
F(7) = 13 + 21 = 34
F(8) = 21 + 34 = 55So 55/2^8 Ans
It's like 3 coins ho tab find f(3)
4 coins ho tab f(4)
N coins k liye f(n) that's generalisation
Example try for 3 coins
hhh
hht
hth
htt
thh
tht
tth
ttt
5/8 HaiOr directly f(3)/2^8 = 5/8

Q20) There were x pigeons and y mynahs in a cage. One fine morning p of them escaped to freedom. If the bird keeper, knowing only that p = 7, was able to figure out without looking into the cage that at least one pigeon had escaped, then which of the following does not represent a possible (x,y) pair?
a) (10, 8)
b) (7, 2)
c) (25, 6)
d) (12, 4)

For the bird keeper to figure out that at least 1 pigeon had escaped, the number of mynahs has to be less than 7. In other words, y < 7. Hence, the pair (10,8) is not a valid one.

Q21) Anuj likes to jog in a park. While jogging he also calculates his speed and the number of steps taken by him. He observes that the steps taken by him per minute are 5 times his speed in km/hr. What is the distance he covers per step if he jogs at a uniform speed ?

let x km per hr be the speed
so 5x steps per min = 5x * 60 = 300x steps per hour
so 300x steps = x km
300x steps = 1000x m
1 step = 10/3 m ?

Q22) The probability of a car passing a certain intersection in a 20 minute windows is 0.9. What is the probability of a car passing the intersection in a 5 minute window? (Assuming a constant probability throughout)

Probability of a car passing in a 20 minute window = 1 – (probability of no car passing in a 20 minute window)
Probability of a car passing in a 20 minute window = 1 – (1 – probability of a car passing in a 5 minute window)^4
0.9 = 1 – (1 – x)^4
(1 – x)^4 = 0.1
1  x = .1^.25
x = 1  .1^.25

Q23) When three brands of juices are mixed in the ratio of 3 : 4 : 5 and 4 : 5 : 6, the cost price of the mixture of these two juices comes out to be Rs. 20 and Rs. 25 per litre respectively. Find the cost price of a litre of mixture of juices in which the three brands are mixed in the ratio of 6 : 7 : 8.

(3x + 4y + 5z)/(3 + 4 + 5) = 20
=> 3x + 4y + 5z = 20 * 12  (1)
(4x + 5y + 6z)/(4 + 5 + 6) = 25
4x + 5y + 6z = 25 * 15.....(2)
6x + 7y + 8z=?
3 * (2)  2 * (1)
=> 6x + 7y + 8z = 3 * 25 * 15  2 * 20 * 12
=645
per litre cost
= 645/(6+7+8)
=645/21
= 215/7

Q24) A Contractor employed a certain number of workers to finish constructing a road in a certain scheduled time. Sometime later, when a part of work had been completed, he realised that the work would get delayed by threefourth of the scheduled time, so he at once doubled the no of workers and thus he managed to finish the road on the scheduled time. How much work he had been completed, before increasing the number of workers?

Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100D) days.
D * x +(100 D) * 2x= 175x
=> D= 25 days
Now , the work done in 25 days = 25x
Total work = 175x
therefore, workdone before increasing the no of workers = 25x/175x = 1/7

Q25) The sequence 1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17,…. has one odd number followed by the next two even numbers, then the next three odd numbers followed by the next four even numbers and so on. What is the 2003rd term of the sequence?

We have the following alternate sequences of odd
and even terms:
Number of Terms Terms
1 1
2 2,4
3 5,7,9
4 10,12,14,16
5 17,19,21,23,25
If you observe the sequences carefully the last term in
any sequence is the square of the number of terms,
i.e. when n = 3, last term = 9;
when n = 4, last term
= 16; and so on
Also, the total number of terms in the sequence is the
sum of the number of terms in the alternate sequences
of even and odd terms.
Since 62 × 63 /2 = 1953
we can say that the 2003rd
term will lie in a sequence of odd terms and will be the
50th term in that sequence. The last term in the
sequence of even terms with n = 62 will be
622 = 3844. Hence, the next odd sequence begins at
3845. The 50th term in this sequence will be
3845 + 49 × 2 = 3943

Q26) In an increasing Arithmetic Progression, the product of the 5th term and the 6th term is 300. When the 9th term of this A.P. is divided by the 5th term, the quotient is 5 and the remainder is 4. What isthe first term of the A.P.?