Quant Boosters by VP  Set 1

EXAMINATION has 11 letters in total
And out of which
E > 1
X > 1
M >1
T > 1
O > 1
(A) > 2
I > 2
N > 28 distinct letters.
 4 letters selected, which are all distinct: 8C4 = 70
 2 letters alike, and 2 distinct
(eg: AAEM)
= 3c1 x 7c2 = 63  2 letters alike, and 2 letters alike
(eg: NNII)
= 3c2 = 3
So answer is, 70 + 63 + 3 = 136

Q10) Find the 1000th term of the sequence : 1,3,4,7,8,9,10,11,13,14,... in which there is no number which contain digit 2,5 or 6.

Method:1 (Not recommended)
_ : 6 numbers ; _ _ : 6 * 7 = 42 numbers ; _ _ _ : 6 * 7 * 7 = 294 numbers
342 numbers added till 999
1 _ _ _ : 7 * 7 * 7 = 343 added = 685 numbers till now. Remaining 1000  685 = 315
3 0 _ _ : 49 added
3 1 _ _ : 49 added
3 3 _ _ : 49 added
3 4 _ _ , 3 7 _ _ , 3 8 _ _ = 49*3 = 147
Total 294 more. Remaining 21
3 9 0 _ : 7
3 9 1 _ : 7
3 9 3 _ : 7
3939Method:2
Base method approach
0 > 0
1 > 1
2 > 3
3 > 4
4 > 7
5 > 8
6 > 9
1000th term : 1000 in base 7 = 2626 <  > 3939

Q11) One day Vikram was out bicycling. After entering a oneway tunnel and after having ridden onefourth of the distance through it, he looked back over his shoulder and saw a bus approaching the tunnel entrance at a speed of 80 miles/hr. Doing a quick mental exercise, Vikram realized that if he accelerated immediately to his top speed, he could just escape with his life, whichever direction he rode. What is Vikram's top biking speed in miles/hr?

Bus .......... A....... Vikram ...B
AB is a tunnel
And suppose person is at point C
AC : CB = 1:3
Now the time at which bus reaches A
Vikram would reach from point C to point A
And the time at which bus reaches to B
Vikram reaches from C to B
Let assume
Length of tunnel is 4km
So if Vikram moves 1 km forward towards B
Bus would reach at A
So remaining distance covered by Vikram is 2km
And bus covered the whole distance 4km(length of tunnel) in the same time with a speed of 80km/hr
So speed of Vikram = 80/2 => 40km/hr

Q12) Find the ten's place digit of 625^246  441^128

(a5)^even ends with 25
Last two digits
(41)^28
=> (81)^14
=> (61)^7
=> 21
Or (41)^28
=> (4 * 8)1
=> 21So 2521 = 04
digit at tenth place = 0

Q13) 25 males and 12 females can complete a piece of work in 12 days. They worked together for 8 days and the women left the work.the remaining men completed the work in 6 days. Find out in how many days 15women can complete the entire work?

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.