Quant Boosters by Hemant Malhotra  Set 20

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Thirty two men can complete a work in 16 days and 48 women can complete the same work in 12 days. Sixteen men and 36 women started working together and worked for 8 days. If the remaining work has to be completed in 2 days, how many additional men would be required?
let total work=512 unit
32 men per day work=512/16=32 unit
so 1 men per day=1 unit
in same way per day work of women=8/9 units
now 16 M+ 36 W = 8 day work=16 * 1 * 8 + 36 * 8 * 8/9 = 384 units
so remaining work=128 units
let x more men needed
so (16+x) * 2 + (36 * 8/9 * 2) = 128
so x = 16The roots of x^3  ax^2 + bx  c = 0 are p, q and r while the roots of x^3 + dx^2 + ex  90 = 0 are p+3, q+3 and r+3. what is the value of 9a + 3b + c ?
x^3  ax^2 + bx  c = 0
Sum of roots p + q + r = a      (i)
pq + qr + pr = b        (ii)
pqr = c      (iii)x^3 + dx^2 + ex  90 = 0
(p+3)(q+3)(r+3) = 90
pqr + 3(pq + pr + qr) + 9(p + q + r) + 27 = 90Put (i), (ii) and (iii) in the above equation, It will be
9a + 3b + c = 63A test has ten questions. Points are awarded as follows:
• Each correct answer is worth 3 points.
• Each unanswered question is worth 1 point.
• Each incorrect answer is worth 0 points.
How many different scores are possible ?
a) 31
b) 30
c) 29
d) 28
e) 27a= correct
b= incorrect
c= unanswered
a+b+c=10
now 3a+c= k where k is different score
when a=10 then max score = 30
when a=9 then you can give rest 1 to b or c when b=1 then score =27
when c=1 then score 27+1=28
now a=8 then rest 2 can be distributed as b=2, c=0 or b=0 , c=2 or b=1,c=1
so score 26,25,24
so now you are getting consecutive values so stop here
so missed out values = 29
so total 0 to 30 except 29 so 30 valuesFor how many positive integer values of ‘x’ is x – 1 – 2 – 3 – 4 < 5
Approach1
Manual counting will work hereApproach  2
We can clearly see that x = 1 is satisfying, so minimum value of x will be 1.Now for the highest possible value of integer ‘x’ which satisfies the given inequality, say x  maximum , all the modulus brackets will open with a positive sign
x – 1 > 0, x – 1 – 2 > 0, x – 1 – 2 – 3 > 0
When all the brackets open with a positive sign, x – 1 – 2 – 3 – 4 < 5 or x < 15 or x's maximum value will be 14
Total values will be 114 so 14 valuesIt is given that a straight line L intersects the curve y = 3x^3  15x^2 + 7x  8 at three distinct points (x1, y1), (x2, y2), (x3, y3). Find the value of x1 + x2 + x3
Straight line is y = mx + c
so 3x^3  15x^2 + 7x  8 = mx + c
so 3x^3  15x^2 + (7  m)x  (8 + c) = 0
so sum of roots = 15/3=5Find all real numbers x such that x[x[x[x]]]= 88 ([.]greatest integer less than equal to x)
(x1)^4 < x[x[x[x]]] =< x^4
so (x1)^4 < 88 < = x^4
so (x1) < 88^(1/4) < = x
so x1 > 88^(1/4) so x>1+88^(1/4)
so 88^(1/4)< = x < (88)^1/4 +1
88/x= [x[x[x]]
so 88/x should be integral value
and x will be in 88^1/4 < = x < 88^(1/4)+1
approx value of 88^1/4=3.066 and approx value of 88^1/4 +1 =4.066
and 88/x should be integer so look for those values of x which will give 88/x as integers
because we are looking for x in form of m/n where m is factor of 88
x could be 4 or 44/13 or 44/12 or 44/11 or 88/23 or 88/25 or 88/27 , 88/28 ..
now out of these only x=88/28 is satisfying our condition so OA=88/28Three are n – 1 red balls, n green balls, and n + 1 blue balls in a bag. The number of ways of choosing two balls from the bag that have different colours is 299. What is the value of n?
(n1)n+n(n+1)+(n+1)(n1)=299
so 3n^21 = 299
so 3n^2 = 300
so n = 10If x^2 + x = 19, Find (x+5)^2+1/(x+5)^2
x^2 + x = 19
(x+5)^2=x^2+10x+25
x^2=19x
so x^2+10x+25=19x+10x+25=9x+44
(9x+44)+1/(9x+44)
(9x+44)^2 +1)/(9x+44)
(81x^2+44^2+88 * 9 * x+1)/(9x+44)
((81((19x)+44^2+88 * 9 * x+1))/9x+44
= 79If x and y are real numbers such that x^2  10x + y^2 + 16 = 0. Determine the maximum value of the ratio y/x
Let y/x =k so y=kx
So x^2 10x+(k^2x^2)+16=0
x^(1+k^2)10x+16=0
x is real so D>=0
100  4 * 16 * (1+k^2)>=0
So 100/64 >=1+k^2
So 36/64 >=k^2
So 6/8 < = k < = 6/8
So max k = 3/4For positive integers x and y, 3x + 8y = 68. Find the maximum value of x^2 y^3
3x+8y=68
we want to maximize x^2 * y^3
3x/2 =8y/3 =68/5
so x=68 * 2/(15)
y= (68 * 3)/40
so x integral will be 12 and y integral will be 4
max (x^2 y^3) would be 9216