Quantitative Aptitude - Arithmetic Ability Questions

Given year, 2013 when divided by 4 leaves a remainder of 1.

NOTE: When the remainder is 1, 11 is substracted from the given year to get the result.

So, 2013 - 11 =2002

It won't requires that much solving.

Here given the increase in weekly collection by Rs. 1,05,000

That is this increase denotes for all 7 days.

=> Gross Collection increase per day = Rs. 1,05,0007 = Rs. 15,000

Hence, the gross collection increase per day = Rs. 15,000.

43 x 616.6/100 + 37 x 217/100

=> 265.138 + 80.29

=>  345.428

Here n(S) = 6 x 6 = 36

E={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3) ,(6,6),(1,3),(2,2),(2,6),(3,1),(3,5), (4,4),(5,3),(6,2)}

=> n(E)=20

Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.

Let the Oranges be 'O', Bananas be 'B' and Apples be 'A'

From the given data,

2O + 3B + 4A = 15 ...... (I)

3O + 2B + A = 10 ..... (II)

On adding (I) and (II),

5O + 5B + 5A = 25

O + B + A = 5

4O + 4B + 4A = 20.

$$\begin{gathered} 6.46=6+0.46 \\ =6+4699 \\ 594+4699 \\ =>64099 \end{gathered}$$

Speed of lorry = 36012 = 30 kmph

Speed of van =250100 x 30 = 75 kmph

Speed of bike = 35 x 75 = 45 kmph

Therefore, now required average speed of bike and van  = 75 + 452= 60 kmph.

We know that,

$$\begin{gathered} i2 = -1 \\ i3 = i2 x i = -1 x i = -i \\ i4 = i2 x i2 = -1 x -1 = 1 \\ Hence, i233 =i4 x 58 + 1 = i232 x i1 = 1 x i = i \end{gathered}$$