Quantitative Aptitude - Arithmetic Ability Questions

Let the average expenditure per head be Rs. p
Now, the expenditure of the mess for old students is Rs. 44p
After joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3
Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33
Therefore, 59(p-3) = 44p + 33
59p - 177 = 44p + 33
15p = 210
=> p = 14
Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.

Initial wheat per kg cost be Rs. 100

Then, monthly expenditure of the family = 30 x 100 = 3000

After increase in cost,

Let P be new monthly consumption of family.

P x 132 = 110/100 x 3000

=> P = 25 kg.

Let the required number of bottles be x.

More machines, More bottles (Direct Proportion)

More minutes, More bottles (Direct Proportion)

 machines6:10time(in min)1:4 ⋮⋮ 270:x 

=> 6 * 1 * x = 10 * 4 * 270

=> x= 1800

S.P of Rs. 5000 stock = Rs.156100*5000= Rs. 7800. 

Income from this stock = Rs.12100*5000 = Rs. 600.  

Let investment in 8 % stock be x and that in 9 % stock = (7800 - x).

Therefore,  

x*890+7800-x*9108=600+70  

4x45+7800-x12=670⇔x=3600 

Therefore,  Money invested in 8 % stock at 90 = Rs. 3600.

Money invested in 9 % at 108 = Rs. (7800-3600) = Rs. 4200.

When we divide 1234 by 8, remainder is 2

When we divide 1235 by 8, remainder is 3

When we divide 1237 by 8, remainder is 5

---> 2 x 3 x 5 = 30 

As 30 will not be the remainder because it is greater than 8,

when 30 divided by 8, remainder = 6.

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.

Hence, number of ways in which we can select the black balls

= 4C1 + 4C2 + 4C3 + 4C4
= 24-1 ........(A)

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red balls

Hence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
=23-1........(B)

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.

Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + … + 5C5
= 25..............(C)

From (A), (B) and (C), required number of ways
=  2524-123-1

We have M1 D1 H1 / W1 = M2 D2 H2 / W2 (Variation rule)
(60 x 30 x 8)/ 150 = (35 x D2 x 6) / 200
D2 = (60 x 30 x 8 x 200) / (150 x 35 x 6) => D2 = 91.42 days =~ 91.5 days.