# Aptitude Question with Solved Answers for Competitive Exam

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If bank, ssc, upsc, railway and other competitive exams students face difficulties in solving Aptitude questions related paper. So in this blog, here are the important aptitude question with solved answers, so that you can solve aptitude section and its difficulty level.

These aptitude questions with solutions are given here will be useful for exam preparations. You should try to solve fist your self and improve your mental ability with the help of these questions.

## Important Solved Aptitude Questions with Answers:

**1. 106 ****× 106 – 94 ****× 94 =?**

Solution

106 × 106 – 94 × 94

=(106)^{2} – (94)^{2} = (106 + 94) (106 – 94) [(a+b)^{2}+(a-b)^{2}=2(a^{2}+b^{2})]

= (200 × 12) = 2400

**Q.2. H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is 2 ^{4} **

**× 3**

^{5}**× 5**

^{2}**× 7**

^{2}. The third number is3240 = 2_{}^{3}×3^{4}×5; 3600 = 2^{4}×3^{2}×5^{2}; H.C.F = 36 = 2^{2}×3^{2}.

Since H.C.F is the product of lowest powers of common factors, so the third number must have (2^{2}×3^{2}) as its factor.

Since L.C.M. is the product of highest powers of common prime factors. So the third number must have 3^{5} and 7^{2} as its factors.

⸫ Third number = 2^{2}×3^{5}×7^{2}.

**Q.3. Evaluate:**

(1) 35 ÷ .07

(2) 2.5 ÷ 0.0005

(3) 136.09 ÷ 43.9

Solution :

(1) 35/.07 = 35*100 / .07*100 = 3500/7 = 500.

(2) 2.5/ 0.0005 = 2.5*10000 / 0.0005*10000 = 25000/5 = 5000.

(3) 136.09/43.9 = 136.09*10/43.9*10 = 1360.9/439 = 3.1.

**Q.4. Simplify : b-[b-(a+b)-{b-(b-a-b)}+2a].**

Given expression = b-[b-(a+b)-{b-(b-a+b)}+2a]

= b-[b-a-b-{b-2b+a}+2a]

= b-[-a-{b-2b+a+2a}]

= b-[-a-{-b+3a}]

= b-[-a+b-3a]

= b-[-4a+b]= b+4a-b=4a.

**Q.5. Simplify : **

$$\sqrt { [(12.1)^2-(8.1)^2]÷[(0.25)^2+(0.25)(19.95)]} \ $$

$$ =\sqrt {{(12.1+8.1)(12.1-8.1)\over(0.25)(0.25+19.95)} } \ $$

$$ =\sqrt { {20.2*4\over0.25*20.2}} \ $$

$$=\sqrt {{4\over0.25} } \ $$

$$ =\sqrt {400\over25} \ $$

$$ =\sqrt {16} \ $$

$$=4$$

**Q.6. Of the tree number, second is twice the first and is also thrice the third. If the average of the three numbers is 44, find the largest number.**

Solution :

Let the third number be x. Then, second number = 3x. First number = 3x/2

⸫ x+3x+ 3x/2 = (44*3) or 11x/2 = 44*3 or x = 24.

So, largest number = 2^{nd} number = 3x = 72.

**Q.7. The sum of a rational number and its reciprocal is 13/6. Find the number.**

Solution

Let the number be x.

Then x+ 1/x = 13/6 ↔ x^{2}+1/x = 13/6 ↔ 6x^{2}-13x+6=0

↔ 6x^{2}-9x-4x+6=0 ↔ (3x-2)(2x-3)=0

↔ x = 2/3 or x = 3/2.

Hence, the required number is 2/3 or 3/2

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