Question

Please answer........

Solve the following ​

Answer 1

Answers:

Question 1:

Let the unknown value be x.

Given,

12% of x = 1080

= $\dfrac{12}{100}\timesx=1080$

= $\dfrac{3}{25}\timesx=1080$

= $x=1080\times\dfrac{25}{3}$

= x = 360 x 25

= Rs 9000

Question 2:

Let the total salary be x.

Given,

10% of x = 400

=$\dfrac{10}{100}\timesx=400$

=$\dfrac{1}{10}\timesx=400$

= x = 400 x 10

= Rs 4000

Question 3:

No. of games won the previous year = 4

No. of games won this year = 6

Increase in games won = 6 - 4 = 2

Increase % = $\mathtt{\dfrac{Increase\:in\:games\:won}{No.\:of\:games\:won\:the\:previous\:year}\times100}$

= $\dfrac{2}{4}\times100$

= 50%

Question 4:

P = Rs 1200

R = 12% p.a

T = 3 years

Simple Interest = $\mathtt{\dfrac{PRT}{100}}$

= $\dfrac{1200\times\12\3}{100}$

= 12 x 12 x 3

= 144 x 3

Rs 432

Amount [A] = S.I + P

= 432 + 1200

= Rs 1632

Question 5:

1) LCM = 99

$\dfrac{27}{99}+\dfrac{55}{99}$

$\dffrac{27+55}{99}$

$\dfrac{82}{99}$

2) LCM = 88

$\dfrac{-33}{88}+\dfrac{-56}{88}$

$\dfrac{-89}{88}$

$-1\dfrac{1}{88}$

3) Here, we need to simply multiply the numerators and denominators.

$\dfrac{2\times-5}{3\times9}$

$\dfrac{-10}{27}$

4) In order to divide two fractions, we need to multiply the first fraction with the reciprocal of the second.

$\dfrac{-7}{12}\times\dfrac{-13}{2}$

$\dfrac{91}{24}$

Answer 2

Questions:-

1. Find the whole quantity if 12% of it is Rs. 1080.
2. Meeta saves Rs. 400 from her salary. If this is 10% of her salary , what is her salary ?
3. A school team won 6 games this years against 4 games won last year. What is the percent increase ?
4. Find the amount to be paid at the end of 3 years if principal = Rs 1200 at 12 % pa.
5. Find :

• 3/11 + 5/9

• - 3/8 - 7/11

• 2/3 × ( - 5/9)

• ( - 7/12) ÷ ( - 2/13)

Answers:-

1) Let the whole quantity be x.

So, 12% of x = 1080

⟹ 100% of x = x

According to the question,

12/100 = 1080/x

⟹ 12 × x = 1080 × 100

⟹ x = 1080 × 100/12

⟹ x = Rs. 9000

∴ The total amount is Rs. 9000.

_________________________________

2) Given:

10% of Meeta's salary = Rs. 400.

100% of Meeta's salary = ?

Let Meeta's salary be Rs. x.

So,

⟹ 10/100 = 400/x

⟹ 10 × x = 400 × 100

⟹ x = 400 × 100/10

⟹ x = Rs. 4000

∴ Meeta's salary is Rs. 4000.

________________________________

3) Given:

Number of matches won by the school team last year = 4

Number of matches won by the school team this year = 6.

Hence,

% increase = { (6 - 4) / 4 } × 100 = 2/4 (100) = 50%

___________________________________

4) Given:

Principal (P) = Rs. 1200

Rate of interest (R) = 12% p.a.

Time (T) = 3 years.

We know that,

Simple interest (SI) = PTR/100

So,

⟹ SI = (1200)(3)(12)/100

⟹ SI = Rs. 432

Now,

Amount = Principal + Interest.

⟹ Amount = 1200 + 432

⟹ Amount = Rs. 1632

∴ Amount to be paid after 3 years is Rs. 1632.

___________________________________

5)

(i) 3/11 + 5/9

LCM of 11,9 = 99

= (3 × 9 + 5 × 11) / 99

= (27 + 55) / 99

= 82/99

(ii) - 3/8 - 7/11

LCM of 8 , 11 = 88

= ( - 3 × 11 - 7 × 8 ) / 88

= ( - 33 - 56 ) / 88

= - 89/88

(iii) 2/3 × ( - 5/9)

= (2 × ( - 5)) / 3 × 9

= - 10/27

(iv) ( - 7/12) ÷ ( - 2/13)

= ( - 7/12) × ( - 13/2)

= { ( - 7) × ( - 13) } / (12 × 2)

= 91/ 24