Question

i want to know the solution and the answer​

Answer 1

Answer:

Hey mate here is your answer

Step-by-step explanation:

(a)

Given:

Let zinc be zn,

and copper be cu;

According to question

$\frac{zn}{cu } = \frac{5}{2}$

5x + 2x = 17×12kg=>7x=352kg=>x=52kg

5x + 2x = 17×12kg=>7x=352kg=>x=52kg∴copper=> 5x=52×5=252kg

5x + 2x = 17×12kg=>7x=352kg=>x=52kg∴copper=> 5x=52×5=252kgzinc=>2x=52×2=5kg

5x + 2x = 17×12kg=>7x=352kg=>x=52kg∴copper=> 5x=52×5=252kgzinc=>2x=52×2=5kgNow after adding 1.250 kg of zinc

5x + 2x = 17×12kg=>7x=352kg=>x=52kg∴copper=> 5x=52×5=252kgzinc=>2x=52×2=5kgNow after adding 1.250 kg of zinc => zinc becomes

5x + 2x = 17×12kg=>7x=352kg=>x=52kg∴copper=> 5x=52×5=252kgzinc=>2x=52×2=5kgNow after adding 1.250 kg of zinc => zinc becomes=> 5 + 1.250 = 6.250 kg

5x + 2x = 17×12kg=>7x=352kg=>x=52kg∴copper=> 5x=52×5=252kgzinc=>2x=52×2=5kgNow after adding 1.250 kg of zinc => zinc becomes=> 5 + 1.250 = 6.250 kg= 614=254∴ New ratio becomes

5x + 2x = 17×12kg=>7x=352kg=>x=52kg∴copper=> 5x=52×5=252kgzinc=>2x=52×2=5kgNow after adding 1.250 kg of zinc => zinc becomes=> 5 + 1.250 = 6.250 kg= 614=254∴ New ratio becomescopper : zinc = 252:254

614=254∴ New ratio becomescopper : zinc = 252:254= 2 : 1

b)Sum of 15 numbers = 15×7=105

Sum of 15 numbers = 15×7=105Sum of 8 numbers = 8×6.5=52

Sum of 15 numbers = 15×7=105Sum of 8 numbers = 8×6.5=52Sum of Last 8 numbers = 8×9.5=76

Sum of 15 numbers = 15×7=105Sum of 8 numbers = 8×6.5=52Sum of Last 8 numbers = 8×9.5=76middle numbers is = 76 + 52 - 105 = 23

c)

LCM of (3, 5, 8, 12)

LCM of (3, 5, 8, 12)⇒ 3 × 5 × 4 × 2 = 120

LCM of (3, 5, 8, 12)⇒ 3 × 5 × 4 × 2 = 120⇒ Now greatest five digits number is 99999

LCM of (3, 5, 8, 12)⇒ 3 × 5 × 4 × 2 = 120⇒ Now greatest five digits number is 99999On dividing 99999 by = 120 (LCM)

LCM of (3, 5, 8, 12)⇒ 3 × 5 × 4 × 2 = 120⇒ Now greatest five digits number is 99999On dividing 99999 by = 120 (LCM)We get remainder = 99999120,

LCM of (3, 5, 8, 12)⇒ 3 × 5 × 4 × 2 = 120⇒ Now greatest five digits number is 99999On dividing 99999 by = 120 (LCM)We get remainder = 99999120, remainder = 39

LCM of (3, 5, 8, 12)⇒ 3 × 5 × 4 × 2 = 120⇒ Now greatest five digits number is 99999On dividing 99999 by = 120 (LCM)We get remainder = 99999120, remainder = 39⇒ By subtracting remainder from 99999 we get the greatest five digits number which is completely divisible by given numbers (3, 5, 8, 12)

LCM of (3, 5, 8, 12)⇒ 3 × 5 × 4 × 2 = 120⇒ Now greatest five digits number is 99999On dividing 99999 by = 120 (LCM)We get remainder = 99999120, remainder = 39⇒ By subtracting remainder from 99999 we get the greatest five digits number which is completely divisible by given numbers (3, 5, 8, 12)∴ 99999 - 39 = 99960

LCM of (3, 5, 8, 12)⇒ 3 × 5 × 4 × 2 = 120⇒ Now greatest five digits number is 99999On dividing 99999 by = 120 (LCM)We get remainder = 99999120, remainder = 39⇒ By subtracting remainder from 99999 we get the greatest five digits number which is completely divisible by given numbers (3, 5, 8, 12)∴ 99999 - 39 = 99960⇒ Now, we required greatest five digit number which when divided by (3, 5, 8, 12) leaves remainder 2 in each case.

LCM of (3, 5, 8, 12)⇒ 3 × 5 × 4 × 2 = 120⇒ Now greatest five digits number is 99999On dividing 99999 by = 120 (LCM)We get remainder = 99999120, remainder = 39⇒ By subtracting remainder from 99999 we get the greatest five digits number which is completely divisible by given numbers (3, 5, 8, 12)∴ 99999 - 39 = 99960⇒ Now, we required greatest five digit number which when divided by (3, 5, 8, 12) leaves remainder 2 in each case.⇒ add 2 in the 99960

LCM of (3, 5, 8, 12)⇒ 3 × 5 × 4 × 2 = 120⇒ Now greatest five digits number is 99999On dividing 99999 by = 120 (LCM)We get remainder = 99999120, remainder = 39⇒ By subtracting remainder from 99999 we get the greatest five digits number which is completely divisible by given numbers (3, 5, 8, 12)∴ 99999 - 39 = 99960⇒ Now, we required greatest five digit number which when divided by (3, 5, 8, 12) leaves remainder 2 in each case.⇒ add 2 in the 99960⇒ 99960 + 2

LCM of (3, 5, 8, 12)⇒ 3 × 5 × 4 × 2 = 120⇒ Now greatest five digits number is 99999On dividing 99999 by = 120 (LCM)We get remainder = 99999120, remainder = 39⇒ By subtracting remainder from 99999 we get the greatest five digits number which is completely divisible by given numbers (3, 5, 8, 12)∴ 99999 - 39 = 99960⇒ Now, we required greatest five digit number which when divided by (3, 5, 8, 12) leaves remainder 2 in each case.⇒ add 2 in the 99960⇒ 99960 + 2⇒ 99962

Answer 2

Answer:

sorry muje nhi ata agar 8 ka quation ho to batna bye