**>>CLICK HERE TO DOWNLOAD PDF FILE of 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

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