Topic - Quant Mixed Bag

Solved ? : Yes

Source : Learningroots forum ]]>

Topic - Quant Mixed Bag

Solved ? : Yes

Source : Learningroots forum ]]>

i) x^2 + 0.2x = 4.83

ii) y^2 + 4.62 = 4.3y

a) x > y

b) x ≥ y

c) x < y

d) x < y

e) x = y ]]>

(x+2.3)(x-2.1)=0

462=2×231=2×3×77=2×3×7×11=22×21

(y-2.2)(y-2.1)=0

So x < = y ]]>

original SP ->y , final 1.1y

y-x / x = 0.15 ; 1.1y-x-100 / x+100 = .1

Solve for x = 2000/3 ]]>

Hence, to make the gap of 90m, B must cover 900m ]]>

a. 80640

b. 13440

c. 18480

d. 1680 ]]>

(1) 36°

(2) 35°

(3) 34°

(4) All of these ]]>

If y=0, x^2-5x=0, x=0 or 5

(0,0)(0,2)(0,3)(5,0) are the only points (0,0) was common to both the cases. ]]>

2^(3 - 1) = 4

1 * 270

2 * 135

27 * 10

5 * 54

[There's a direct formula - 2^(no of prime factors - 1)]

]]>2^7 * 3^4 * 5 * 7

5 and 7 cannot be increased further so they have to be present as it is. We can reduce the numbers by joining various powers of 2 and 3. The highest power of 2 that is a single digit is 8 and the highest power of 3 that is a single digit is 9. So, we are left with 2 * 8 * 8 * 9 * 9 * 5 * 7. We cannot tweak these numbers any further and so, it will be 2578899. ]]>

Understand that i have not considered the first digit to be non 0 in the second case because there won't be a number starting with a 0 that will satisfy p>q and the fact that it reduces our work.

]]>Any even number that is in the form of 2^a * x^b * y^c... (in short, an even number not in the form of 2^n) can be expressed as a sum of three or more consecutive integers

The only type that remains is numbers in the form of 2^n. These cannot be expressed as a sum of at least 2 consecutive integers

So power of 2 lying between 200-1500 => 256+512+1024 =>1792 ]]>

Total 7 * 6 * 5 * 3 = 630 ways. ]]>