AP ICET 26th April 2019 Shift-2

Two digit number $$=12,16, 20, 24........96$$

Hence, it is making the AP series $$12+16+20+........96$$

$$n=24-2=22$$

$$d=4$$

$$a=12$$

$$S_n=\dfrac{n(2a+(n-1)d)}{2}$$

Hence, the required sum $$S_n=\dfrac{22(2\times 12+(22-1)\times 4)}{2}$$

$$\Rightarrow S_n=11\times (24+84)$$

$$\Rightarrow S_n=11\times 108$$

$$\Rightarrow S_n=1188$$

As per the given question,

$$ 8625 =3\times 5\times 5\times 5\times 23$$

$$8625=3^1\times 5^3\times 23^1$$

Hence the number of divisors $$=(3+1)(1+1)(1+1)=4\times 2 \times 2=16$$

After excluding 1, the number of divisors $$16-2=14$$

The LCM of 3, 4, 10, 15 and 24

LCM= 120

Hence ,the smallest positive integer is 120.

Hence the required integer is $$=120-9=111$$

As per the given data in the question,

When these number are divided by m, then 1541, 2444, 2659  it is leaving the same remainder.

Now, (2444-1541=903) and (2659-2444=215) and (2659-1541=1118)

HCF, of these numbers, $$903=43\times 7\times 3$$

$$215=5\times 43$$

And $$1118=43\times 13\times 2$$

Hence, from the above it is clear that 43 is the largest value of m.

Let his salary was Rs x.

Amount paid for the house rent $$=0.1 x$$Rs.

Amount spent on the provisions $$\dfrac{90x}{100}\times \dfrac{15}{100}=0.9\times 0.15x$$

Amount spent on the children's education $$=\dfrac{0.90\times 0.85 x\times 10}{100}=0.9\times 0.85\times 0.1 x$$

The left amount $$=24786Rs.$$

Hence, his salary $$0.1x +0.9\times 0.15x+0.9\times 0.85\times 0.1 x+24786=x$$

$$\Rightarrow 0.3115x+24786=x$$

$$\Rightarrow x-0.3115x=24786$$

$$\Rightarrow 0.6885 x=24786$$

$$\Rightarrow x=\dfrac{24786}{0.6885}=36000 Rs.$$