AP ICET 26th April 2019 Shift-1

Any number is divisible by 11 if,

The sum of odd placed number - sum of even placed number will be divisible by 11, then the number will be divisible by 11.

$$3+x+2-(0+9)=x-4$$

It will be divisible by 11, if $$x-4=0$$

Or $$x=4$$

The numbers which are divisible by 7, in between 200 to 400 are given below,

$$203,210, 217..........399$$

We know for the AP series,

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

$$\Rightarrow 399=203+7(n-1)$$

$$\Rightarrow 7(n-1)=399-203=196$$

$$\Rightarrow n-1=\dfrac{196}{7}$$

$$n=28+1=29$$

As per the given question,

The given numbers $$6,9,12,15$$

It is giving remainder 5, if we are dividing a  by these given number.

So, the required number is=LCM of the given number +5

Hence the required number =$$180+5=185$$

As per the given question,

The first type of oil have 462 liters

2nd type of oil have 286 liters

3rd type of oil have 187 litrs of oil

Hence $$462=2\time 3\times 7 \times 11$$

$$286=2\times 11\times 13$$

$$187=11\times 17$$

Hence the required capacity= $$11 liter$$

The number of tin required $$=\dfrac{462}{11}+\dfrac{286}{11}+\dfrac{187}{11}$$

Hence, the number of tin of 11 liter capacity required $$=42+26+17=85$$

Let he have the total amount of x Rs.

Percentage of money gave to charity $$=\dfrac{12x}{100}=0.12x$$

The remaining amount $$=x-0.12x=0.88x$$

Now money gave to charity=80%

Hence the remaining amount $$\dfrac{0.88x\times 20}{100}=264$$

$$=0.176x=264$$

Hence $$x=\dfrac{264}{0.176}=1500$$

Hence, The amount of money he have in the begining $$=1500Rs.$$