Aptitude Question with Solved Answers for Competitive Exam

Important Solved Aptitude Questions with Answers:

Q.8. The product of the age of Ankit and Nikita is 240. If twice the age of Nikita is more than Ankit’s age by 4 years, what is Nikita’s age ?

Solution

Let Ankit’s age be x years. Then, Nikita’s age = 240/x years.

⸫ 2*240/x = 4 ↔ 480 – x²+4x-480=0

↔ (x+24)(x-20)=0 ↔ x=20

Hence, Nikita’s age = (240/40) years = 12 years.

Q.9. (1000)⁷ ÷ 10¹⁸ =?

Solution

(1000)⁷ ÷ 10¹⁸ = (1000)⁷/10¹⁸ = 10^(3*7)/10¹⁸ = 10^(21-18) = 10³= 1000.

Q.10. Evaluate :

Q.10. (1) 2 is what percent of 50?

(2) 1/2 is what percent of 1/3 ?

           (3) What percent of 7 is 84 ?

       (4) What percent of 2 metric tonnes is 40 quintals ?

        (5) What percent of 6.5 litres is 130 ml ?

Solution

(1) Required percentage = [(2/50) * 100)]% = 4%

(2) Required percentage = [(1/2) * (3/1) * 100)% = 150%

(3) Required percentage = [(84/7)*100)]% = 1200%

(4) 1 metric tonne = 10 quintals

⸫ Required percentage = [(40/2*10) * 100)]% = 200%

(5) Required percentage = [(130/ 6.5*1000)* 100)]% = 2%

Q.11. A person incurs 5% loss by selling a watch for Rs. 1140. At what price should the watch be sold to earn 5% profit?

Solution

Let the new S.P be Rs. x. Then,

(100 – loss%) : (1^st S.P) = (100+gain%) : (2^nd S.P)

→ [(100-5)/1140] = [(100+5)/x] → x = [(105*1140)/95]=1260.

⸫ New S.P = Rs. 1260.

Q.12. A mixture contains alcohol and water in the ratio 4 : 3. If 5 litres of water is added to the mixture, the ratio becomes 4 : 5. Find the quantity of alcohol in the given mixture.

Solution

Let the quantity of alcohol and water be 4x litres and 3x litres respectively. Then,

[4x/ (3x+5) = 4/5 ↔ 20x = 4(3x+5) ↔ 8x = 20 ↔ x = 2.5.

⸫ Quantity of alcohol = (4*2.5) litres = 10 litres.

Q.13. A,B and C enter into partnership. A invests 3 times as much as B invests and B invests two third of what C invests. At the end of the year, the profit earned is Rs. 6600. What is the share of B ?

Solution

Let C’s capital = Rs. x. Then, B’s capital = Rs. 2/3 x.

$$ A's \ capital = Rs.( 3×{2\over3}x)=Rs. 2x.$$

 $$ ⸫ \ Ratio \ of \ their \ capitals \ = 2x: {2\over3}x:x=6:2:3.$$

 $$ Hence, B's \ share = Rs.( 6600×{2\over11})= Rs.1200.$$

Q.14. if the wages of 6 men for 15 days-be Rs. 2100, then find the wages of 9 men for 12 days.

Solution

Let the required wages be Rs. x.

More men, More wages         (Direct Proportion)

Less days, Less wages             (Direct Proportion)

(Men 6:9) (Days 15:12) : :   2100 : x

⸫ (6*15*x)=9*12*2100) ↔ x = [(9*12*2100)/(6*15)]=2520.

Hence, the required wages are Rs. 2520.

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