Quantitative Aptitude - Arithmetic Ability Questions

S.I. for 3 years = Rs. (12005 - 9800) = Rs. 2205.

S.I. for 5 years = Rs.22053×5= Rs.3675

Principle = Rs.(9800-3675) = Rs.6125

Hence, Rate = 100×36756125×5 =12%

We will show that the number of odd days between last day of February and last day of October is zero. .

March April May June July Aug. Sept. Oct.

31 + 30 + 31 + 30 + 31 + 31 + 30 + 31

= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = 0.

Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows

Concentration of petrol in    A        B        C

                                                1/2      3/5     4/5

Quantity of petrol taken from A = 1 litre out of 2 litre

Quantity of petrol taken from B = 1.8litre out of 3 litre

Quantity of petrol taken from C = 0.8 litre out of 1 litre

Therefore, total petrol taken out from A, B and C = 1+1.8+0.8 =3.6 litres

So, the quantity of kerosen =(2+3+1) - 3.6 =2.4 litre

Thus, the ratio of petrol to kerosene = 3.6/2.4 = 3/2

Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = ( 22 + x ), which must be divisible by 3.

When x=2. [(22 + 2) = 24 is divisible by 3]

So, the answer is 2

The digits on either side of the letter in each of the groups 1P6, 2T0, and 1R8 indicate the place value of the letter in the alphabet.

This pattern is not followed by 2Y4.

required percentage = πr22πr×100 =250%

Let the number of candidates appeared from each state be x.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000