Quant Boosters by VP  Set 2

(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.?

Let the 5th term of the A.P. be ‘a’ and the common
difference be ‘d’.
The 6th term will be (a + d) and the
9th term will be (a + 4d).So
a×(a + d) = 300 (i)
And
5a + 4 = (a + 4d)
=> d = a + 1  (ii)
Solving (i) and (ii),
we get a = 12 or 25/2If a = 25/2
=> then the value of ‘d’ will also be
negative, which is not possible in an increasing A.P.Therefore,
a = 12 and d = 13.
And
The first term will be
= (a – 4d)
= 12 – 52
= –40.

Q27) Of the applicants who passed a certain test, 15 applied to both colleges X and Y. If 20% of the applicants who applied to college X and 25% of the applicants who applied to college Y applied to both college X and Y, how many applicants applied only to college X or college Y?

X intersection Y = 15
20% of the applicants who applied for X applied for both X and Y
I.e 20% of Total X = 15
Total X = 75
Only X = 60
Similarly Total Y = 15 * 4 = 60
Only Y = 45
So 60 +45 = 105

Q28) The price of an article reduces to 576 after two successive discounts. The markup is 80% above the cost price of Rs. 500. What is the new profit percentage if instead of two successive discounts the markup price was further increased successively two times by the same percentage?

CP = 500
MP = 1.8 * 500 = 900
900 * (1 R/100) ^ 2 = 576
(1 R/100) ^2 = .64
.8^2
I.e 20% discount
Now
MP = 900 * (1.2)^2 = 1296
CP = 500
I.e profit % = 796/500 * 100 = 159.2 %

Q29) Sameer and Sumer started running from the same point in opposite directions on a circular track of length 120 m. Their speeds are 20 m/s and 40 m/s respectively. After every second, Sameer increases his speed by 2 m/s whereas Sumer decreases his speed by 2 m/s. How many times would they have met on the track by the time Sumer comes to rest?

Sumer speed is 40m/s
His speed decreased by 2 m/s every second
And he travels till he comes to rest
40/2 = 20 seconds
So he travels only 20 seconds
Now , Their relative velocity is not changing so in every seconds they will cover a distance of 60 meter and Sumeer stops after 20 seconds
So in 20 secs the distance covered is 20 * 60 = 1200 and track is of 120m
So they met = 1200/120 = 10times

Q30) In how many ways can 1000 be written as a sum of ‘n’ consecutive natural numbers, where ‘n’ is greater than 1?

Concept : numbers of consecutive natural numbers sum to n = (number of odd factors of n ) 1
1000 = 5^3 * 2^3
here number of odd factors = 4 and required answer = 4  1 = 3