Quant Boosters by VP  Set 1

They worked for 8 days
I.e completed 2/3 work
So Remaining (1/3) work is completed by men in 6 days
I.e males complete this work in 18 days
So 18 * X /(18+X) = 12
=> 3x /(18+X) = 2
=> 3x = 36 + 2x
=> X = 36
I.e 12 women complete this work in 36 days
So 15 women in 12 * 36/15
I.e 144/5 days

Q14) X can do a piece of work in 20 days working 7 hours a day. The work is started by X and on the second day one man whose capacity to do the work is twice that of X, joined. On the third day another man whose capacity is thrice that of X, joined and the process continues till the work is completed. In how many days will the work be completed, if everyone works for four hours a day?

Total work = 20 * 7 = 140 units
Efficiency of X => 1unit /hourFirst day
1 * 4Second day
Another man joined
Whose efficiency is 2
So total work on second day
(1+2) * 4So basically
4 [ 1 + 1+2 + 1+2+3 + 1+2+3+4 + 1+2+3+4+5 ] = 140
So 5 days

Q15) How many multiples of 18 are there which are less than 3500 and also 2 more than the square of a natural number

18k = t^2 + 2 (1)
Now t^2 mod 18
Min Possibility 0
and max possibility  17
so for t^2 +2 to be divisible by 18
t^2 must give remainder as 16 .
so t must be of type 18k+ 4 or 18k  4 ..
√3600 = 60
√3500 = 59 (approx)
So maximum value of t can be 59
so t = 4 (18 * 0 +4)
14 (18 * 1  4)
22 ,
32,
40,
50 ,
58(18 * 3 + 4 )
So 7

Q16)

Area of triangle using medians
M = (16+20+24)/2 => 30
Area of triangle
=> 4/3 √ [30 * (3016)(3020)(3024) ]
=> 4/3 √ [ 30 * 14 * 10 * 6]
=> 80 √7Now we know medians divide
Triangle into 6 equal parts
So area of one smaller triangle who height is BG
Is 1/6 (80√7)
And base = 1/3 of (16)
So
(1/2) * (1/3) * 16 * h = 1/6 (80√7)
=> H = 5√7

Q17) If a > 0, b > 0 then the number of real roots of the equation 2x^7  ax^5  3x^4  bx^2 + 7 = 0 can be
a) 2
b) 3
c) 4
d) 0

Use Descartes rule
And see the sign changes in F(X)
And F(x)
If Number of sign changes in F(X) =k
number of Positive roots can be k , k2 ,k4 and so on
Similarly
Number of sign Changes in F(x) = k
number of negative roots = k , k2 , k4 and so on

Q18) Find the largest 5 digit number which when divided by 8 leaves a remainder of 5 and when divided by 7 leaves a remainder of 2.

7k +2 => 8p +5
=> 7k = 8p + 3
For p = 4
It satisfiesI.e least number which satisfies the condition is
8 * 4 +5 = 37
And number is of the form
56k + 37
Now to find largest 5 digit number
99999/56
=> 1785.xx
So k = 1785
56 * 1785 + 37
=> 99997

Q19) From the first 25 natural numbers, how many arithmetic progressions of 6 terms can be formed such that common difference of the AP is a factor of the 6th term.

Let the first term of the AP be a and the common difference be d.
The sixth term of the series will be a+5d
Given that d should be a factor of a+5d
=> a+5d is divisible by d
=> a should be divisible by d
So the required cases are
d = 1, a = 1,2,3.......20
d= 2 , a = 2,4,6.......14
d = 3, a = 3, 6,9
d= 4, a = 4
So the required number of AP’s are 20 + 7 + 3 + 1 = 31

Q20) 101010101010..... (94 digits) / 375. Find the remainder.

375 => 125 * 3
Use CRT
With 3
Sum of digits
47 mod 3 = 2
Now ,
With 125
You have to check only last three digits
As 10^3 mod 125 = 0
So last three digits would be 010
So 125k+10
3b +2 = 125k +10
=> For k = 2
It satisfies
So 260

Q21) Four Prime Numbers are written in ascending order. The product at first three is 7429 and the product at last three is 12673. Find the sum of first and last Number?

7429 = 17 x 19 x 23
12673 = 19 x 23 x 29
∴ four prime numbers are 17 x 19 x 23 x 29.
∴ The sum is = 17 + 29 = 46.

Q22) Hypoteneus of a Right angle triangle is 26. What could be its perimeter?
(a) 51
(b) 56
(c) 45
(d) 60

We know 5  12  13 is a Pythagorean triplet
∴ 10 – 24 – 26.
Will also be a Pythagorean triplet. ∴ Perimeter is 60. Hence (d).

Q23) P, 2P, q + P, 2P – 2q – 6 is an A.P. then what is its 99th term?
(a)–99
(b) –98
(c) 99
(d) 100