Quantitative Aptitude - Arithmetic Ability Questions

But at x=-1/3, log x is not defined.

The only admissible value of x is 1.

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.

B runs around the track in 10 min.

 i.e ,    Speed of B = 10 min per round

 Therefore,  A beats B by 1 round

  Time taken  by A to complete 4 rounds 

  = Time taken by B to complete 3 rounds  

 = 30 min 

Therefore, A's speed = 30/4 min per round = 7.5 min per round

 Hence, if the race is only of one round A's time over the course = 7 min 30 sec

 L.C.M of 5, 6, 7, 8 = 840

Therefore, Required Number is of the form 840k+3.

Least value of k for which (840k+3) is divisible by 9 is k = 2 

Therefore, Required  Number = (840 x 2+3)=1683

Given,

P can fill in 12 hrs

Q can fill in 15 hrs

R can fill in 20 hrs

=> Volume of tank = LCM of 12, 15, 20 = 60 lit

=> P alone can fill the tank in 60/12 = 5 hrs

=> Q alone can fill the tank in 60/15 = 4 hrs

=> R alone can fill the tank in 60/20 = 3 hrs

Tank can be filled in the way that

(P+Q) + (P+R) + (P+Q) + (P+R) + ....

=> Tank filled in 2 hrs = (5+4) + (5+3) = 9 + 8 = 17 lit

=> In 6 hrs = 17 x 6/2 = 51 lit

=> In 7th hr = 51 + (5+4) = 51 + 9 = 60 lit

=> So, total tank will be filled in 7 hrs.

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.

Given Five boxes of bananas sell for Rs. 30.

=> 1 Box of Bananas for = 30/5 = Rs. 6

Then, for Rs. 9

=> 9/6 = 3/2 = 1.5

Hence, for Rs. 9, 1.5 box of bananas can buy.

3/5 of  350 - 4/7 of 210 = 210 - 120 = 90.