Quantitative Aptitude - Arithmetic Ability Questions

Number of non - negative integral solutions =n+r-1 Cr-1 = C223 = 253 

The length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high is

=>  d2 = 122 + 42 + 32 = 13mts.

p=d×1002R2 

Where d is difference, r is rate of interest. 

=> (48 x 100 x 100) / 8 x 8 = Rs.7500

Let the actual distance travelled be x km.
Then x/8=(x+20)/12
=> 12x = 8x + 160
=> 4x = 160
=> x = 40 km.

   Sum = [(B.D.xT.D.)/ (B.D.-T.D.)]

              = [(B.D.xT.D.)/B.G.]

T.D./B.G.  = Sum/ B.D.

   =1650/165

               =10

Thus, if B.G. is Re 1, T.D. = Rs. 10.

If B.D.is Rs. ll, T.D.=Rs. 10.

If B.D. is Rs. 165, T.D. = Rs. [(10/11)xl65]

            =Rs.150

 And, B.G. = Rs. (165 - 150) = Rs, 15.

The cow can graze an area of 2826 sq.m i.e it is able to cover area of circle of 2826 sq.m

Therefore, πr2 =2826
r2 =(2826×7/22) = 899.18 = 900
r = 30 m.

Therefore, the radius of the circle implies the length of the rope = 30 mts.

Probability of occurrence of an event, 

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)

⇒ Probability of getting head in one coin = ½, 

⇒ Probability of not getting head in one coin = 1- ½ = ½, 

Hence, 

All the 11 tosses are independent of each other.

∴ Required probability of getting only 2 times heads =122×129=1211

Let, Hema attempted 'k' sum correctly, then

k x 3 -2 x(35-k) = 60
5k = 130
k = 26

so 26 correct sums.