Quant Boosters by Hemant Malhotra  Set 16

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
If x + y + z = 13, then what is the maximum value of (x3)(y2)(z+1)?
a) 26
b) 27
c) 32
d) 24x+y+z=13
(x3)+(y2)+(z+1)=9
now (x3) * (y2) * (z+1) will be max when all are equal so 3 * 3 * 3 = 27A number X is when divided by 11, 4 and 3 leaves remainders 9, 2 and 1 respectively. The least such X when divided successively by 3, 4 and 11 leaves remainders
(1) 6,0,0
(2) 0,1,2
(3) 0,0,6
(4) 1,3,10difference of divisor and remainder is same 119=2
so number will be lcm(11,4,3)2=130
now number is 130
130 mod 3=1
3* 43 + 1=130
now 43 mod 4=3
so 4 * 10+3=43
now 10 mod 11=10
so 1,3,10The sum of the squares of the first 15 positive integers is equal to 1240. What is the sum of the squares of the second 15 positive integers?
a) 2480
b) 3490
c) 6785
d) 82155 * 31 * 61  1240 = 8215
sum till 30  sum till 15
so 1^2+2^2.....30^2(1^2+2^2+.................15^2)
= 5 * 31 * 611240=8215Find the smallest number which when divided by 11, 12 and 15 gives remainders 1, 2 and 3 respectively.
11a+1 = 12b+2 = 15c+3
now 15c+3 is divisible by 3
now 12b+2 mod 3=2
so this is not possible in any case OA=no such number existIf a = 13 then find the value of a^5 – 14 a^4 + 14 a^3 – 14 a^2 + 14a – 1
Method1 (a1) * (a^413a^3+a^213a+1)
Put a=13
12(13^413^4+13^213^2+1)
Ans: 12Method 2
a^5  13a^4a^4+13a^3+a^313a^2a^2+13a+131
a^4 (a13) a^3(a13) +a^2 (a13) a(a13) +12
since a13 = 0 so 12Method3 Every question of this type has some pattern and in cat u will never find any question which will take more than 12 minutes .. If that is the case then u are not following the basic approach or options to solve that question
here a^514a^4+14a^314a^2+14a1
here i m concentrating on 14 .. and sign +++ so it will follow same pattern
if i replace 14 to 2 then also it will follow same pattern but to apply these things you have to give time .
then a^52a^4+2a^32a^2+2a1
so 13= 1
so 12+22+21=0
so our ans is 11=0
so in every case our ans will follow same pattern
so OA will be 131=12
if here 14=3 and value at y=2 then value will be 1If A is less then or equal to 6 and B is less then or equal to 8, what is the difference between the maximum and minimum possible value of AB
A < = 6 so 6 < = A < = 6
B < = 8 so 8 < = B < = 8
when A=B then AB will be min=0
when A is positive and B is negative then it will be max or B is positive and A is negative
A=6 and B=8 so AB=14 so AB=14
so maxmin=140=14In a room 2/5 of all the people are wearing gloves and ¾ of the people are wearing hats. What is the minimum number of people in the room wearing both hat and gloves
Atleast 15% of the total population wears both hat and gloves.
20 is the least total population where 2/5 and 3/4 can be a whole number.
so 15% of 20 is 3.If a is one of the roots of a^5 – 1 = 0 and a# 1, then what is the value of a^15 + a^16 + a^17 + …. + a^50
a^51=0
(a1)(1+a+a^2+a^3+a^4)=0
so 1+a+a^2+a^3+a^4=0
now a^15(1+a+a^2+a^3+a^4)+a^20(1+a+a^2+a^3+a^4)+
a^25(1+a+a^2+a^3+a^4+a^5).................a^45((1+a+a^2+a^3+a^4)+a^50
so every value except a^50 will be zero
now a^5=1
so a^50=1
so OA=1Ram, a milkman buys pure milk and adds some water in it before selling. He sells the mixture at a price 10% less than the price of pure milk and makes a profit of 10%. Find the quantity (in liters) of water than he adds to every two liters of pure milk.
Method1= C.P of 1 litre pure milk = 100
S.P. of 1 litre milk + water = 90 ( 10 % profit)
C.P of 1 litre milk + water = 90/ 1.1
100  1 litre pure milk
90/1.1 9/11 litres
water/litre = 2/11
To 9/11 liter pure milk, water added = 2/11
so to 2 liters pure milk, water added = 4/9Method2 Ram spends 100% for milk and sells the mixture for 90 %
so he loses 10 % of milk, which he compensates by 90 % of water.
10% > 90 %
He compensates for 10% loss and he also makes 10% more so on the whole he will make 20 % (of the milk price)
So,90% of money of water(or milk) is equivalent to 20% of the profit
hence ratio 2/9 per litre . So for 2 litres it would be 4/9Six jars of cookies contain 18 , 19 , 21 , 23 , 25 and 34 cookies respectively. One jar contains almond cookies only. The other 5 jars contain no almond cookies. A takes three of the jars and B takes two of the others. Only the jars of almond cookies remains. If A gets twice as many cookies as B , how many almond cookies are there ?
method1 A gets twice So he has even number of cookies
even =even+even+even
even=odd+odd+even
three even is not here so he will choose two odd and one even
so 34+(((19+21) or ((34+19+23) or ((34+19+25) or ((34+21+23))
or ((34+23+25))
so 74 or 76 or 78 or 78 or 82
now Amya sir has 37 or 38 or 39 or 41
37 =18+19 ((( 19 is already used so not possible)
38 is not possible
39=18+21 (( this is possible combination))
41=23+18
so 23 almond cookies are theremethod2 Sum of all the jars is 140
So it is of the form 3k+2
So to make it multiple of 3 we should subtract a number of the form 3k+2. The only possible number is 23