Quantitative Aptitude - Arithmetic Ability Questions

All powers of 6 has 6 in units place so answer will definitely be 1.

(0.75)31-0.75+0.75+(0.75)2+1=(0.75)3+(1-0.75)(1)2+(0.75)2+1×0.751-0.75 

= (0.75)3+(1)3-(0.75)31-0.75 =10.25=4=2

For a triangle, we need 3 non-collinear points. So with 12 points (when all the 12 are such that any three non-collinear is12C3. But among them 8 points are collinear.

If all these 8 points are different we get 8C3 triangles as they are collinear.

In 12C3 triangles, we do not get 8C3 triangles

Therefore, The number of triangles we get = 12C3-8C3 = 164

Let the required number of bottles be x.

More weavers, More mats (Direct Proportion)

More days, More mats (Direct Proportion)

$$\begin{gathered} Wavers 4:8 \\ Days 4:8 \end{gathered}$$  :: 4:x

 4 x 4 x x = 8 x 8 x 4

=> x= 8*8*44*4

=>  x = 16.

Let speed of train B be x m/s And length of train B be y m
Then length of train A is 2y m
Speed of train A = 84 × 5/18 = 210/9 m/s = 70/3 m/s
A.T.Q, (2y+y)/10 = 70/3 − x .............(i)
and (2y+y)/ 22.5 = 70/3 − 2x ................(ii)
solving (i) and (ii), y = 50 m

PW = Amount1+R100T=28001+121002   = Rs.2500 

TD = Amount - PW = 2800 - 2500 = Rs.300

Let 's' be the number of small packets and 'b' the number of large packets sold on that day.

Therefore, s + b = 5000 ... eqn (1)

Each small packet was sold for Rs.150.
Therefore, 's' small packets would have fetched Rs.150s.

Each large packets was sold for Rs.250.
Therefore, 'b' large packets would have fetched Rs.250b.

Total value of sale = 150s + 250b = Rs. 10.5 Lakhs (Given)

Or 150s + 250b = 10,50,000 ... eqn (2)

Multiplying equation (1) by 150, we get 150s + 150b = 7,50,000 ... eqn (3)

Subtracting eqn (3) from eqn (2), we get 100b = 3,00,000
Or b = 3000

We know that s + b = 5000
So, s = 5000 - b = 5000 - 3000 = 2000.

2000 small packets were sold.

Here,n = 4(children)

P(girl)= 0.5

P(of atleast one girl)= 1 - P(no girls)

                             = 1 - 0.0625 = 0.9375