Maths

Let the interest rate be r percent interest earned from 1.2 Lacs at the end of year

= 1,200,000 * r /100

= 1200r

From 1.8 lakh, = 8/12 * 1, 80,000 * 2r/100

= 2400r

Total interest = 3600r = 54000

r = 15 percent

Let the quotient obtained when the numbers is divided by 3, 5 and 6 be x₁, x₂ and x₃

Respectively

N = 3 x₁ + 1

x₁ = 5 x₂ + 3

x₂ = 6 x₃ + 2

N = 3 [5 (6 x₃ + 2) + 3] + 1

=90 x₃ + 40 < 1000

x₃ can take 11 values

Assuming that

b – c = p b = c + p

c – a = q c = a + q

Hence b = c + p

= a + q + p

$?$ $y+\frac{1}{(b-c)a(c-a)}$ can be written as a + q + p + $\frac{1}{a.p.q}$

Apply AM GM in equally concept in above equation

$\frac{a+q+p\frac{1}{a.p.q}}{4}\ge 4\sqrt[]{a*p*q*\frac{1}{a.p.q}}$

$\frac{a+q+p\frac{1}{a.p.q}}{4}\ge 1$

$a+q+p\frac{1}{a.p.q}\ge 4$

Hence, minimum value is 4

Let the distance between Delhi (D) and Jaipur (J) be 8x and Akbar's initial speed be 4a and final speed be 7a.

Let the point where he increased his speed at P and let Q be the point on DJ such that $DP=\frac{5}{8}(DJ)$

$\frac{PQ}{PJ}=\frac{4}{7}$

Let PQ = 4y,

PJ = 7y

QJ = 7y – 4y = 3x

3y = 3x

y= x

PQ = 4x and PJ =7x

Required ratio = $\frac{DP}{DJ}$

= $\frac{x}{8x}$ = 1/8

f (a) = a² – 2ax + y

f (x) = x² – 2x² + y = 0

– x² + y =0

y = x²                           ………. (1)

f (y) = y² – 2xy + y = 0

y² – 2xy + x² = 0

  (y – x)² = 0

y – x = 0

y = x

Put value of y in equation 1

We get

x = y = 1

So, f (4) = 4² – 2 (4*1) + 1

= 16 – 8 +1

= 9

Let the airline has a free luggage allowance →'f' kg

Luggage by M →m kg

Luggage by S →s kg 

Rate charger beyond f kg →'e'/kg

Extra luggage of m = (m – f) kg

Extra luggage of s = (s – f) kg

ATQ

e (2m – f) = 2400 ______equation 1

e (2s – f) = 900 ______equation 2

add 1 and 2

e (m + s –f ) = 1650

Also, e (m – f) + e (s – f) = 1050

e (m + s –f ) – ef = 1050

1650 – ef= 1050

ef = 600

Extra luggage from m = e (m – f)

= $\frac{e(2m-2f)}{2}=\frac{e(2m-f)-ef}{2}$

$=\frac{2400-600}{2}$

= Rs. 900

$\frac{1}{2lo{g}_{a}10}+\frac{1}{2lo{g}_{b}10}+\frac{1}{2lo{g}_{c}10}+\frac{1}{2lo{g}_{d}10}$

$=\frac{lo{g}_{10}a}{2}+\frac{lo{g}_{10}b}{2}+\frac{lo{g}_{10}c}{2}+\frac{lo{g}_{10}d}{2}$

$=\frac{lo{g}_{10}a+lo{g}_{10}b+lo{g}_{10}c+lo{g}_{10}d}{2}$

$=\frac{lo{g}_{10}abcd}{2}$

$=\frac{lo{g}_{10}10^8}{2}$

=8 * ½ = 4

x (x – 6) > 2x – 12

x² – 6x > 2x – 12

x² – 8x > 12x > 0

(x – 2) (x – 6) > 0

x < 2 or x >6

Given equation is

x² – 25 <5

–5 < x² – 25 < 5

20 < x² < 30

$\sqrt[]{20}$ < x < $\sqrt[]{30}$

Hence x = 5

x = 5