Total sum of money = ₹3000 Let the number of winners of ₹100 be x. And those who are not winners = 63-x According to the question, 100Xx+25X(63-x)=3000 ⇒ 100x+1575-25x=3000 ⇒ 75x+1575=3000 ⇒ 7x+1575-1575=3000-1575 [Subtracting 1575 from both sides] ⇒ 7x-1425 ⇒ 7x/7=1425/7 [Dividing both sides by 7]Read more
Total sum of money = ₹3000
Let the number of winners of ₹100 be x.
And those who are not winners = 63-x
According to the question, 100Xx+25X(63-x)=3000
⇒ 100x+1575-25x=3000
⇒ 75x+1575=3000
⇒ 7x+1575-1575=3000-1575 [Subtracting 1575 from both sides]
⇒ 7x-1425
⇒ 7x/7=1425/7 [Dividing both sides by 7]
⇒ x=19
Hence the number of winner is 19.
Class 8 Maths Chapter 2 Exercise 2.2 Solution in Video
Total sum of money = ₹300 Let the number of ₹5 coins be x, number of ₹2 coins be 3x and number of ₹1 coins be 160-(x+3x)=160-4x According to question, 5Xx+2X(3x)+1X(160-4x)=300 ⇒ 5x+6x+160-4x=300 ⇒ 7x+160=300 ⇒ 7x+160-160=300-160 [Subtracting 160 from both sides] ⇒ 7x-140 ⇒ 7x/7=140/7 [Dividing bothRead more
Total sum of money = ₹300
Let the number of ₹5 coins be x, number of ₹2 coins be 3x and number of ₹1 coins
be 160-(x+3x)=160-4x
According to question, 5Xx+2X(3x)+1X(160-4x)=300
⇒ 5x+6x+160-4x=300
⇒ 7x+160=300
⇒ 7x+160-160=300-160 [Subtracting 160 from both sides]
⇒ 7x-140
⇒ 7x/7=140/7 [Dividing both sides by 7]
⇒ x=20
Hence, the number of coins of ₹5 denomination = 20
Number of coins of ₹2 denomination = 3 x 20 = 60
Number of coins of ₹1 denomination = 160 – 4 x 20 = 160 – 80 = 80
Class 8 Maths Chapter 2 Exercise 2.2 Solution in Video
Let number of notes be 2x,3x and 5x. According to question, 100x2x+50x3x+10x5x=4,00,000 ⇒ 200x+150x+50x=4,00,000 ⇒ 400x-4,00,000 ⇒ 400x/400=4,00,000/400 [Dividing both sides by 400] ⇒ x=1000 Hence, number of denominations of ₹100 notes = 2 x 1000 = 2000 Number of denominations of ₹50 notes = 3 x 100Read more
Let number of notes be 2x,3x and 5x.
According to question, 100x2x+50x3x+10x5x=4,00,000
⇒ 200x+150x+50x=4,00,000
⇒ 400x-4,00,000
⇒ 400x/400=4,00,000/400 [Dividing both sides by 400]
⇒ x=1000
Hence, number of denominations of ₹100 notes = 2 x 1000 = 2000
Number of denominations of ₹50 notes = 3 x 1000 = 3000
Number of denominations of ₹10 notes = 5 x 1000 = 5000
Therefore, required denominations of notes of ₹100, ₹50 and ₹10 are 2000, 3000 and
5000 respectively.
Class 8 Maths Chapter 2 Exercise 2.2 Solution in Video
Let the rational number be x. According to the question, 5/2x+2/3=-7/12 ⇒ 5/2 x+2/3-2/3=-7/12-2/3 [Subtracting 2/3 from both sides] ⇒ 5x/2=-7-8/12 ⇒ 5x/2=-15/12 ⇒ 5xX12=-15x2 ⇒ 60x=-30 ⇒ 60x/60=-30/60 [Dividing both sides by 60] ⇒ x=-1/2 Hence, the rational number is -1/2 for more answers vist to: hRead more
Let the rational number be x.
According to the question, 5/2x+2/3=-7/12
⇒ 5/2 x+2/3-2/3=-7/12-2/3 [Subtracting 2/3 from both sides]
⇒ 5x/2=-7-8/12 ⇒ 5x/2=-15/12
⇒ 5xX12=-15×2 ⇒ 60x=-30
⇒ 60x/60=-30/60 [Dividing both sides by 60]
⇒ x=-1/2
Hence, the rational number is -1/2
Let Ravi’s present age be x years. After fifteen years, Ravi’s age = 4x years. Fifteen years from now, Ravi’s age = (x+15) years. According to question, 4x=x+15 ⇒ 4x-x=15 [Transposing x to L.H.S.] ⇒ 3x=15 ⇒ 3x/3=15/3 [Dividing both sides by 3] ⇒ x=5 years Hence, Ravi’s present age be 5 years. for moRead more
Let Ravi’s present age be x years. After fifteen years, Ravi’s age = 4x years.
Fifteen years from now, Ravi’s age = (x+15) years.
According to question, 4x=x+15
⇒ 4x-x=15 [Transposing x to L.H.S.]
⇒ 3x=15
⇒ 3x/3=15/3 [Dividing both sides by 3]
⇒ x=5 years
The organizers of an essay competition decide that a winner in the competition gets a prize of ₹100 and a participant who does not win, gets a prize of ₹25. The total prize money distributed is ₹3,000. Find the number of participants is 63.
Total sum of money = ₹3000 Let the number of winners of ₹100 be x. And those who are not winners = 63-x According to the question, 100Xx+25X(63-x)=3000 ⇒ 100x+1575-25x=3000 ⇒ 75x+1575=3000 ⇒ 7x+1575-1575=3000-1575 [Subtracting 1575 from both sides] ⇒ 7x-1425 ⇒ 7x/7=1425/7 [Dividing both sides by 7]Read more
Total sum of money = ₹3000
Let the number of winners of ₹100 be x.
And those who are not winners = 63-x
According to the question, 100Xx+25X(63-x)=3000
⇒ 100x+1575-25x=3000
⇒ 75x+1575=3000
⇒ 7x+1575-1575=3000-1575 [Subtracting 1575 from both sides]
⇒ 7x-1425
⇒ 7x/7=1425/7 [Dividing both sides by 7]
⇒ x=19
Hence the number of winner is 19.
Class 8 Maths Chapter 2 Exercise 2.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
I have a total of ₹300 in coins of denomination ₹1, ₹2 and ₹5. The number of ₹2 coins is 3 times the number of ₹5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Total sum of money = ₹300 Let the number of ₹5 coins be x, number of ₹2 coins be 3x and number of ₹1 coins be 160-(x+3x)=160-4x According to question, 5Xx+2X(3x)+1X(160-4x)=300 ⇒ 5x+6x+160-4x=300 ⇒ 7x+160=300 ⇒ 7x+160-160=300-160 [Subtracting 160 from both sides] ⇒ 7x-140 ⇒ 7x/7=140/7 [Dividing bothRead more
Total sum of money = ₹300
Let the number of ₹5 coins be x, number of ₹2 coins be 3x and number of ₹1 coins
be 160-(x+3x)=160-4x
According to question, 5Xx+2X(3x)+1X(160-4x)=300
⇒ 5x+6x+160-4x=300
⇒ 7x+160=300
⇒ 7x+160-160=300-160 [Subtracting 160 from both sides]
⇒ 7x-140
⇒ 7x/7=140/7 [Dividing both sides by 7]
⇒ x=20
Hence, the number of coins of ₹5 denomination = 20
Number of coins of ₹2 denomination = 3 x 20 = 60
Number of coins of ₹1 denomination = 160 – 4 x 20 = 160 – 80 = 80
Class 8 Maths Chapter 2 Exercise 2.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
Lakshmi is a cashier in a bank. She has currency notes of denominations ₹100, ₹50 and ₹10 respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹4,00,000. How many notes of each denomination does she have?
Let number of notes be 2x,3x and 5x. According to question, 100x2x+50x3x+10x5x=4,00,000 ⇒ 200x+150x+50x=4,00,000 ⇒ 400x-4,00,000 ⇒ 400x/400=4,00,000/400 [Dividing both sides by 400] ⇒ x=1000 Hence, number of denominations of ₹100 notes = 2 x 1000 = 2000 Number of denominations of ₹50 notes = 3 x 100Read more
Let number of notes be 2x,3x and 5x.
According to question, 100x2x+50x3x+10x5x=4,00,000
⇒ 200x+150x+50x=4,00,000
⇒ 400x-4,00,000
⇒ 400x/400=4,00,000/400 [Dividing both sides by 400]
⇒ x=1000
Hence, number of denominations of ₹100 notes = 2 x 1000 = 2000
Number of denominations of ₹50 notes = 3 x 1000 = 3000
Number of denominations of ₹10 notes = 5 x 1000 = 5000
Therefore, required denominations of notes of ₹100, ₹50 and ₹10 are 2000, 3000 and
5000 respectively.
Class 8 Maths Chapter 2 Exercise 2.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get -7/12. What is the number?
Let the rational number be x. According to the question, 5/2x+2/3=-7/12 ⇒ 5/2 x+2/3-2/3=-7/12-2/3 [Subtracting 2/3 from both sides] ⇒ 5x/2=-7-8/12 ⇒ 5x/2=-15/12 ⇒ 5xX12=-15x2 ⇒ 60x=-30 ⇒ 60x/60=-30/60 [Dividing both sides by 60] ⇒ x=-1/2 Hence, the rational number is -1/2 for more answers vist to: hRead more
Let the rational number be x.
According to the question, 5/2x+2/3=-7/12
⇒ 5/2 x+2/3-2/3=-7/12-2/3 [Subtracting 2/3 from both sides]
⇒ 5x/2=-7-8/12 ⇒ 5x/2=-15/12
⇒ 5xX12=-15×2 ⇒ 60x=-30
⇒ 60x/60=-30/60 [Dividing both sides by 60]
⇒ x=-1/2
Hence, the rational number is -1/2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Let Ravi’s present age be x years. After fifteen years, Ravi’s age = 4x years. Fifteen years from now, Ravi’s age = (x+15) years. According to question, 4x=x+15 ⇒ 4x-x=15 [Transposing x to L.H.S.] ⇒ 3x=15 ⇒ 3x/3=15/3 [Dividing both sides by 3] ⇒ x=5 years Hence, Ravi’s present age be 5 years. for moRead more
Let Ravi’s present age be x years. After fifteen years, Ravi’s age = 4x years.
Fifteen years from now, Ravi’s age = (x+15) years.
According to question, 4x=x+15
⇒ 4x-x=15 [Transposing x to L.H.S.]
⇒ 3x=15
⇒ 3x/3=15/3 [Dividing both sides by 3]
⇒ x=5 years
Hence, Ravi’s present age be 5 years.
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/