Topic - Quant Mixed Bag

Solved ? - Yes

Source - Elite's Grid Prep Forum ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - Elite's Grid Prep Forum ]]>

a) 26

b) 27

c) 32

d) 24 ]]>

(x-3) + (y-2) + (z+1) = 9

now (x-3) * (y-2) * (z+1) will be max when all are equal so 3 * 3 * 3 = 27 ]]>

(1) 6,0,0

(2) 0,1,2

(3) 0,0,6

(4) 1,3,10 ]]>

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 1,3,10 ]]>

a) 2480

b) 3490

c) 6785

d) 8215 ]]>

sum till 30 - sum till 15

so 1^2+2^2.....30^2-(1^2+2^2+.................15^2)

= 5 * 31 * 61-1240=8215 ]]>

now 15c + 3 is divisible by 3

now 12b + 2 mod 3 = 2

so this is not possible in any case ]]>

Put a=13

12(13^4-13^4+13^2-13^2+1)

Ans: 12

Method 2

a^5 - 13a^4-a^4+13a^3+a^3-13a^2-a^2+13a+13-1

a^4 (a-13) -a^3(a-13) +a^2 (a-13) -a(a-13) +12

since a-13 = 0 so 12

Method3- Every question of this type has some pattern and in cat u will never find any question which will take more than 1-2 minutes .. If that is the case then u are not following the basic approach or options to solve that question

here a^5-14a^4+14a^3-14a^2+14a-1

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^5-2a^4+2a^3-2a^2+2a-1

so 13= 1

so 1-2+2-2+2-1=0

so our ans is 1-1=0

so in every case our ans will follow same pattern

so OA will be 13-1=12

if here 14=3 and value at y=2 then value will be 1

|B| < = 8 so -8 < = B < = 8

when A=B then |A-B| 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 A-B=14 so |A-B|=14

so max-min=14-0=14 ]]>

20 is the least total population where 2/5 and 3/4 can be a whole number.

so 15% of 20 is 3. ]]>

(a-1)(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=1 ]]>

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 = 4/9

Method2- 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 4/9

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 there

method2- 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

a) 125.50

b) 110

c) 115.50

d) NOT ]]>

so amount =210 rs

he returned 100 rs so 110 rs remaining

now SI=110 * 5/100=5.5

so amount =110+5.5=115.5 ]]>

x+3 = increasing

2x-1 = increasing

2-x decreasing

so compare

x+3=2-x so x=-1/2

2x-1=2-x

so x=1

now check for these two values

at x=1 y=min(4,1,1)=1

and x=-1/2 , y=min((5/3,-2,5/2)=-2

so max of min will be 1 ]]>