**>>CLICK HERE TO DOWNLOAD PDF FILE of NUMBERS..**

In my last post, we have discussed simple problems based on Numbers but when difficulty factor goes on increasing, it becomes difficult to crack the puzzle.

Let see how to solve difficult part in Numbers

Let see how to solve difficult part in Numbers

**Problem1: The ratio of two numbers is 3:2. If 10 and the sum of two numbers are added to the product. Square of sixteen is obtained. What could be the smaller number?**Solution:

**Step1:**Let the two numbers are 3x and 2x

**Step2:**If 10 and the sum of two numbers are added to the product.

**10 + (3x + 2x) + (3x * 2x) = (16)**

^{2}**Step3:**Solve the equation

**6x**

6x

(6x + 41) (x-6) = 0

X = 6 or -41/6

^{2}+ 5x- 246 = 06x

^{2}+ 41x - 36x - 246 = 0(6x + 41) (x-6) = 0

X = 6 or -41/6

**Step4:**Negative number cannot be accepted. So, take x=6

**Step5:**Smaller Number is 2x = 2 x 6 =12

**Problem2: The positive number when decreased by 4 is equal to 21 times the reciprocal of the number. The number is ?**Solution:

**Step1:**Let the number be X

**Step2:**The positive number when decreased by 4 is equal to 21 times the reciprocal of the number.

**X-4 = 21/X**

X

X

X (X-7) + 3 (X-7) = 0

(X-7) (X+3) = 0

X = 7

X

^{2}- 4x -21 = 0X

^{2}- 7x + 3x -21 = 0X (X-7) + 3 (X-7) = 0

(X-7) (X+3) = 0

X = 7

**Step3:**Neglecting -3, The number is X=7

**Problem3: A certain number of two digits is three times the sum of the its digits and if 45 be added to it, the digits are reversed. The number is?**Solution:

**Step1:**Let the unit digit of the number = X and ten’s digit = Y

**Step2:**A number of two digit is three times the sum of its digit

**3 (X + Y) = 10y + X**

2X – 7Y = 0 [Equation 1]

2X – 7Y = 0 [Equation 1]

**Step3:**If 45 be added to it, the digits are reversed

**10Y + X + 45 = 10X + Y**

9X – 9Y = 45

X – Y = 5 [Equation 2]

9X – 9Y = 45

X – Y = 5 [Equation 2]

**Step4:**Solving both the equations X =7, Y=2

**Step5:**The number is XY =27

**Problem4: If a number is subtracted from the square of its one half, the result is 48. The square root of the number is?**Solution:

**Step1:**Let the number is X

**Step2:**If a number is subtracted from the square of its one half, the result is 48

**(X/2)**

X

(X-16) (X+12) = 0

X= 16

^{2}–X = 48X

^{2}/4 – X =48(X-16) (X+12) = 0

X= 16

**Step3:**The square root of the number 16 is 4

**Problem5: A two digit number is seven times the sum of its digits. If each digit is increased by 2, the number, thus obtained is 4 more than six times the sum of its digits. Find the number?**Solution:

**Step1:**Let the two digit number is 10X + Y

**Step2:**A two digit number is seven times the sum of its digits.

**10X + Y = 7 (X+ Y)**

X= 2Y [Equation 1]

X= 2Y [Equation 1]

**Step3:**If each digit is increased by 2, the number, thus obtained is 4 more than six times the sum of its digits.

**10 (X+2) + (Y+2) = 6 (X+Y+4) + 4**

4X – 5Y =6 [Equation 2]

4X – 5Y =6 [Equation 2]

**Step4:**Solving equation 1 and 2

We get

**X=4 and Y=2**

**Step5:**Number is XY = 42

**Problem6: 1/5 of a number is equal to 5/8 of the second number. If 35 is added to the first number, then it becomes 4 times of the second number. What is the value of the second number?**Solution:

**Step1:**Let the two digit number X and Y

**Step2:**1/5 of a number is equal to 5/8 of the second number

**1/5X = 1/5Y**

X/Y = 25/8 [Equation 1]

X/Y = 25/8 [Equation 1]

**Step3:**If 35 is added to the first number, then it becomes 4 times of the second number.

**X + 35 = 4Y**

**25/8Y + 35 = 4Y**

Y = 40

Y = 40

**Step4:**Value of second number is 40

**Problem7: If a fraction numerator is increased by 1 and the denominator is increased by 2 then the fraction becomes 2/3. But when the numerator increased by 5 and the denominator is increased by 1, then the fraction becomes 5/4. What is the value of the original fraction?**Solution:

**Step1:**Let the fraction be X/Y

**Step2:**If a fraction numerator is increased by 1 and the denominator is increased by 2 then the fraction becomes 2/3.

**X + 1/ Y+2 = 2/3**

3X = 2Y + 1 [Equation 1]

3X = 2Y + 1 [Equation 1]

**Step3:**When the numerator increased by 5 and the denominator is increased by 1, then the fraction becomes 5/4.

**(X + 5) /(Y + 1) = 5/4**

4X = 5Y – 15 [Equation 2]

4X = 5Y – 15 [Equation 2]

**Step4:**Solving both the equations

**(2Y + 1) / 3 = (5Y – 15) / 4**

8Y + 4 =15Y – 45

Y = 7

8Y + 4 =15Y – 45

Y = 7

**X = (2Y +1) / 3 =[(2*7) + 1] / 3**

X = 5

X = 5

**Step5:**Required original fraction is X/Y =5/7

**>>CLICK HERE TO DOWNLOAD PDF FILE of NUMBERS..**

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