AP ICET 2019 Question Paper Shift-1 (26th April)

The given number in the question is $$=2035$$

Now, after subtracting x from the given number, new number will be $$2035-x$$

When we are dividing the $$2035-x$$ by $$9,10,15$$ it is leaving the remainder 5.

Now, the LCM of the number $$90$$

Now, when we are dividing the 2035 by 90, then it is leaving the remainder 55.

But as per the condition, the remainder should be 5. hence the required number $$=55-5=50$$

Let the present age of Rahim $$=x$$ years

So, the present age of David$$=x-6$$

and the present age of Amar$$=x-6-8=x-14$$years

After the 8 years, age of Rahim will be $$x+8$$

After the 8 years, age of Amar will be $$x-14+8=x-6$$

As per the condition given in the question, $$x+8=2(x-6)$$

Hence, $$2x-x=12+8$$

$$x=20$$years

Hence, the present age of David $$=20-6=14$$years

4 years ago, the age of David $$=14-4=10$$years.

10 consecutive integers are:-

$$ x,(x+1),(x+2), (x+3), (x+4),--------(x+9)$$

sum of the 5 least number of the set is 265

   $$\Rightarrow x+(x+1)+(x+2)+(x+3)+(x+4) = 265$$

   $$\Rightarrow 5x+10 = 265$$

  $$\Rightarrow 5x = 255$$

  $$\Rightarrow x = \dfrac{255}{5}$$

  $$\Rightarrow x = 51$$

Therefore, the sum of 4 greatest number of the set will be

  $$\Rightarrow (x+6)+(x+7)+(x+8)+(x+9)$$

  $$\Rightarrow 4x+30$$

 $$\Rightarrow 4\times 51+30$$ 

  $$\Rightarrow 204+30$$

   Hence the sun of the greatest number will be  234

Let the terms of the GP are $$\dfrac{a}{r},a, ar$$

Where a is the first term and r is the common ratio.

As per the condition, $$\dfrac{a}{r}+a+ar=26------------(i)$$

$$\dfrac{a}{r}(1+r+r^2)=26$$

Now, squaring both side,

$$\dfrac{a^2}{r^2}(1+r+r^2)^2=676$$

As per the the second condition,

$$(\dfrac{a}{r})^2+a^2+(ar)^2=364$$

$$\dfrac{a^2}{r^2}(1-r+r^2)(1+a^2+a^2r^2)=364-------------(ii)$$

From equestion (i) and (ii)

$$\Rightarrow \dfrac{1-r+r^2}{1+r+r^2}=\dfrac{364}{676}$$

$$\Rightarrow \dfrac{1-r+r^2}{1+r+r^2}=\dfrac{7}{13}$$

$$\Rightarrow 13 - 13r + 13r^2 = 7 + 7r + 7r^2$$

$$\Rightarrow (3r-1)(r-3) =0$$

So,$$r=3$$ and $$r=\dfrac{1}{3}$$

Now, substituting the value of r=3,

So, $$a(1+r+r^2)=26$$

$$a(1+3+9)=26$$

$$a(13)=26$$

$$a=\dfrac{26}{13}=2$$

Hence, the product of these numbers $$=a\times \dfrac{a}{r}\times ar =a^3=6^3=216$$

The natural number between 100 and 1000 are $$105, 110, 115......995$$

It is making an arithmetic series.

Sum of the terms of arithmetic series $$S_n=\dfrac{n(2a+(n-1)d)}{2}$$

$$t_n=a+(n-1)d$$

$$995=105+(n-1)5$$

$$n-1=\dfrac{995-105}{5}=178$$

$$n=178+1=179$$

Hence, the required sum $$S_n=\dfrac{179(2\times 105+(179-1)\times 5)}{2}$$

$$S_n=\dfrac{179\times(210+178\times 5)}{2}=550\times 179=98450$$

Hence, the required sum $$S_n=98450$$