Question

Solved the question 9, 10, 11, 12, 13, 14

Class 7th ​

Answer 1

Questions

(i) The sum of three numbers is $\sf 2\frac{1}{36}$ . If two of them are $\sf \frac{7}{12}$ and $\sf \frac{11}{18}$, find the third number.

(ii) The sum of two consecutive integers is 39. Find the two integers.

(iii) One-third of a number when added to 1 gives 15. Find the number.

(iv) What is the number which gives -24 when (-46) is subtracted from it?

(v) What are prime numbers? Write all prime numbers between 60 and 80.

(vi) Evaluate the given equation 2.789×6.328 and round off to the hundredth place.

$\rule{300}{0.5}$

Answers

(i) Step 1: Add the two numbers given.

$\sf \frac{7}{12}+\frac{11}{18}$

Find the LCM of the denominators i.e. 12 and 18.

$\begin{array}{r | l} 2&12,18 \\ \cline{2-2} 3 & 6,9 \\ \cline{2-2} & 2,3 \\ \end{array}$

LCM of 12 and 18 = 2×2×3×3 = 4×9 = 36

Make the denominators equal.

$\sf \frac{7\times 3 }{12\times 3 }+\frac{11\times 2 }{18\times 2 }$

$\sf \frac{21 }{36 }+\frac{22 }{36 }$

Add the numerators and let the denominator be the same.

$\sf \frac{21 }{36 }+\frac{22 }{36 }$

$\sf \frac{43}{36 }$

Step 2: Subtract $\sf \frac{43}{36 }$ from $\sf 2\frac{1}{36} =\frac{73}{63}$.

$\sf \frac{73}{63} - \frac{43}{36}$

Find the LCM of the denominators i.e. 63 and 36.

$\begin{array}{r | l} 3 & 63,36 \\ \cline{2-2} 3 & 21, 12\\ \cline{2-2} & 7,4 \\ \end{array}$

LCM of 63 and 36 = 3×3×4×7 = 9×28 = 252

Make the denominators equal.

$\sf \frac{73\times 4 }{63\times 4 } - \frac{43\times 7 }{36\times 7 }$

$\sf \frac{292}{252} - \frac{301}{252}$

Add the numerators and let the denominator be the same.

$\sf \frac{-9}{252}$

Simplify the fraction.

$\sf \frac{-1}{28}$

∴ The third number is $\bf \frac{-1}{28}$.

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(ii) Let one of the numbers be 'y'.

Our equation now would be ⇒ $\sf y+y+1 = 39$

Step 1: Simplify the equation.

$\sf y+y+1 = 39$

$\sf 2y+1 = 39$

Step 2: Subtract 1 from both sides of the equation.

$\sf 2y+1 -1 = 39-1$

$\sf 2y =38$

Step 3: Divide 2 from both sides of the equation.

$\sf \frac{2y}{2}=\frac{38}{2}$

$\sf y = 19$

∴ One of the numbers is ⇒ y = 19

∴ The other number is ⇒ y+1 = 19+1 = 20

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(iii) We have to solve this equation for this question ⇒ $\sf \frac{1}{3}x+1=15$

Step 1: Subtract 1 from both sides of the equation.

$\sf \frac{1}{3}x+1-1=15-1$

$\sf \frac{1}{3}x=14$

Step 2: Multiply 3 from both sides of the equation.

$\sf 3 \times \frac{1}{3}x=3 \times 14$

$\sf x = 42$

∴ The number is 42.

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(iv) To find the number we need to solve this equation ⇒ $\sf p-(-64)=-24$

Step 1: Simplify both sides of the equation.

$\sf p +64=-24$

Step 2: Subtract 64 from both sides.

$\sf p +64-64=-24-64$

$\sf p= -88$

∴ The number is (-88).

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(v) Numbers that are divisible to itself ae known as prime numbers. There are only 2 factors for a prime number. They are one and the number itself.

Prime numbers between 60 and 80 are ⇒ 61, 67, 71, 73, and 79.

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(vi) 2.789×6.328 = 17.648792

When we round this number we get ⇒ 17.65

$\rule{300}{0.5}$

Answer 2

Answer:

hii.................