**Question 1 :**

**If xy>0. Which of the following must be greater than 0?**

**i.x+y ii. x**^{2}+y iii (x+y)^{2}+1 iv (x-y)^{2}

**Solution: **let’s eliminate answer options by substituting values

Identify the conditions…… xy>0 and each answer option must be greater than 0. For any value of x and y the answer option >0

since xy>0. You can substitute

- x as +ve or y as + ve
- x as -ve or y as – ve
- x =y

x = -1 y = -6… xy>0 but x+y is not greater than 0. So rule out option i and ii

x = -1 y = -6… xy>0 but **x**^{2}+y is not greater than 0. So rule out option ii

x =1 y=1 xy>0 but (x-y)^{2} is not greater than 0. So rule out option iv

**option 3 is the answer.**

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**Question 2:**

**For which value of x is the following inequality true: x**^{2} < 2x

**1.x<0 2. 0<x<2 3.-2<x<2 4. x<2 5. x>2**

Solution:

**Tip: When ever you get ranges.. substitute the lowest possible value and the highest possible value. **You can eliminate options faster.

Let x =-1. Hence x^{2} < 2x .. (-1)^{2} < 2(-1)… not true..

Hence eliminate options 1/4/3

Lets take x = 100 Hence x^{2} < 2x .. (100)^{2} < 2(100)… not true..

Hence eliminate options 5

**Option 2 is the answer.**

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**Question 3:**

**The numbers a,b and c are all positive integers. a is a factor of c and b is also factor of c.if a<b , then which of the following must also be a positive integer factor of c?**

**1. a+b 2. a*b 3. c/a 4.c/(b*a)**

**Solution:**

a is a factor of c and b is also factor of c

2 is a factor of 6 and 3 is also factor of 6

2+3 is not a factor of 6. Hence eliminate option 1

**Note- take big numbers too.. don’t stick to single digit numbers**

Lets take c as 100. a is 20 and b is 50

Then 100/( 20*50) = 0.1 is not a integer factor of 100. Hence eliminate option 4

Lets take c as 24. a is 8 and b is 12

8*12 =96 is not a factor of 24. Hence eliminate option 2

**Option 3 is the answer.**

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**Question 4:**

**The difference of two number is 2. If the sum of their reciprocals is 7/24. What is the sum of the two numbers?**

**1.20 2.14 3.15**

**Solution:**

Let the two numbers be x and y. x-y=2 (1/x) + (1/y) = (7/24)

We can solve by quadratic equation. But that’s cumbersome. Lets solve by substituting numbers

(1/x) + (1/y) = (7/24)

Simplifying the fractions: (y+x)/(xy) = 7/24

Lets break the fraction. y+x =7 yx=24 and x-y=2

Now lets substitute numbers .

if you start with x-y=2.. there are so many possibilities.

If you start with yx=24. Then possibilities are (24,1) (6,4),(8,3)…

None of the pairs satisfy all the 3 equations

Lets try again

(y+x)/(xy) = 7/24

Since this is a ratio

Lets try the next iteration of the ratio.. Multiply the numerator and denominator with 2

(y+x)/(xy) = 14/48

Lets break the fraction

y+x =14 yx=48 and x-y=2

yx=48. Then possibilities are (24,2) (8,6),(12,4)…

8,6 satisfies all the conditions.

**Answer is 14**

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**Question 5:**

**Which of the following number is the product of two integers whose sum is 11?**

**1. -28 2. -42 3. 12 4.26 5. 32**

** Solution:**

Let us assume the numbers are a and b. a+b =11.

a and b can be positive/negative or 0.

Lets take option 1

Break 28 into prime factors….. -28 = (-1)*7*2*2

Combination of prime factors should give two numbers whose sum is 11

Possibilities are

(-14, 2) (-7,4), (14,-2)…………

None of the pairs give a difference of 11.

Lets take option 2

-42 = a*b

-42 =(-1)*2*7*3

Combination of prime factors should give two numbers whose sum is 11

(-14, 3) (-7,6), …………

-14 and 3 gives 11

**Hence option 2 is the answer**

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**Question 6:**

**If the quotient a/b is positive, which of the following must be true?**

**1. a>0 2. b>0 3. ab>0 4. a-b>0 5. a+b>0**

**Option 3 is the answer**

Solution ..refer to the solution for question 3

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**Question 7:**

**If x and y are prime numbers. Which of the following cannot be the sum of x and y?**

**1.5 2.9 3.13 4.16 5.23**

Option 5 is the answer

Solution ..refer to the solution for question 5

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**Question 8:**

**If 0 <= x <=4 and y<12**

**Which of the following cannot be a value of xy?**

**1.-2 2.0 3.6 4.24 5.48**

Option 5 is the answer

Solution .refer the solution for question 3

**Hope you understood both the methods. Even if you are out of touch with math ..**

**With logic you can ace the math section…**

**If you need any help math help in your prep… Contact me…**

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