Quantitative Aptitude - Arithmetic Ability Questions

Let the two numbers be x and y

Given x : y = 3 : 7 .....(1)

Now, x+6 : y+6 = 5 : 9 .....(2)

From (1), x = 3y/7

From (2), 5y - 9x = 24

=> 5y - 9(3y/7) = 24

=> y = 21

=> From(1), x = 9

Hence, the two numbers be 9 and 21

Given that Rs. 1860 will become Rs. 2641.20 at 12%

=> Simple Interest = 2641.20 - 1860 = Rs. 781.20

We know I = PTR/100

=> 781.20 x 100 = 1860 x T x 12

=> T = 78120/1860x12

=> T = 78120/22320

=> T = 3.5 years.

Let the rate be R% p.a.

I  gives, P = Rs. 8000 and T = 4 years.

II gives, S.I = Rs. (8800 - 8000) = Rs. 800.

R = [100 x S.I] / [p x t ]= (100 x 800)/(8000 x 4) = 2 ½ % p.a

Thus, I and II both are needed to get the answer.

Let C = x.
Then B = x/2
and A = x/4

A:B:C = 1:2:4.
C's share Rs.[(4/7)*700) = 400

Let the speed of the boat upstream be p kmph and that of downstream be q kmph

Time for upstream = 8 hrs 48 min = 845hrs

Time for downstream = 4 hrs

Distance in both the cases is same.

=> p x  845= q x 4

=> 44p/5 = 4q

=> q = 11p/5

 Now, the required ratio of Speed of boat : Speed of water current

= q+p2:q-p2

=> (11p/5 + p)/2  : (11p/5 - p)/2

=> 8 : 3

1 million distinct 3 digit initials are needed.

Let the number of required alphabets in the language be ‘n’.

Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials.

Note distinct initials is different from initials where the digits are different.

For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.

This n3 different initials = 1 million 

i.e. n3=106  (1 million = 106)

  => n = 102 = 100

Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.

Let Karan’s present age be x years and Shiva’s present age be y years.
Then, according to the first condition,
x - 10 = 3(y - 10) or x - 3y = - 20 ..(1)
Now, Karan’s age after 10 years = (x + 10) years
Shiva’s age after 10 years = (y + 10)
(x + 10) = 2 (y + 10) or x - 2y = 10 ..(2)
Solving (1) and (2), we get x = 70 and y = 30
Karan’s age = 70 years and Shiva’s age = 30 years.

C.P. of 50 kg rice = Rs. (36 x 20 + 14 x 36) = Rs. (720 + 504) = Rs.1224.  

S.P. of 56 kg rice = Rs. (50 x 30) = Rs.1500.

Profit = 1500 - 1224 = Rs.276 

Profit % = ProfitC.P×100  = 2761224×100 = 22.54%.