Quantitative Aptitude - Arithmetic Ability Questions

Since first second varieties are mixed in equal proportions, so their average price = Rs.(126+135)/2= Rs.130.50

So, Now the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say Rs. 'x' per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find 'x'.

Cost of 1 kg tea of 1st kind         Cost of 1 kg tea of 2nd kind 

x-153/22.50 = 1  => x - 153 = 22.50  => x=175.50.

 Hence, price of the third variety = Rs.175.50 per kg.

Given that the student got 125 marks and still he failed by 40 marks

=> The minimum pass mark = 125 + 40 = 165

Given that minimum pass mark = 33% of the total mark

=> total mark =33/100 =165

=> total mark = 16500/33 = 500

Let the son's present age be x years .Then, (38 - x) = x => x= 19. 

Son's age 5 years back = (19 - 5) = 14 years

One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.

C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.

Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.

Remaining Work = 7/10, which was done by A,B and C in the initial number of days. 

Number of days required for this initial work = 7 days. 

Thus, the total numbers of days required = 4 + 7 = 11 days.

Let the original price be Rs. 100.

New final price  = 120 %  of (75 % of Rs. 100) = Rs. [(120/100) * (75/100) * 100] = Rs. 90.

Decrease = 10%

Let the edge = a cm   

So increase by 50 % = a + a/2 = 3a/2  

Total surface Area of original cube = 6a2  

TSA of new cube = 63a22 =69a24=  13.5a2   

Increase in area = 13.5a2-6a2 =7.5a2  

 7.5a2 Increase % =7.5a26a2×100 = 125%

Let the total income be x. 

Then, income left = (100 - 80)% of [100 -  (35 + 25)] % of x = 20% of 40% of x = [(20/100) * (40/100) * 100] % of x = 8 % of x.

Let principal = P, Then, S.I.= P and Time = 8 years

We know that S.I. = PTR/100

Rate = [(100 x P)/ (P x 8)]% = 12.5% per annum.