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
Let Baichung’s age be x years, then Baichung’s father’s age = (x+29) years and Baichung’s granddaughter’s age = (x+29+26) = (x+55) years. According to condition, x+x+29+x+55=135 ⇒ 3x+84=135 ⇒ 3x+84-84=135-84 [Subtracting 84 from both sides] ⇒ 3x=51 ⇒ 3x/3=51/3 [Dividing both sides by 3] ⇒ x=17 yearsRead more
Let Baichung’s age be x years, then Baichung’s father’s age = (x+29) years and
Baichung’s granddaughter’s age = (x+29+26) = (x+55) years.
According to condition, x+x+29+x+55=135
⇒ 3x+84=135
⇒ 3x+84-84=135-84 [Subtracting 84 from both sides]
⇒ 3x=51
⇒ 3x/3=51/3 [Dividing both sides by 3]
⇒ x=17 years
Hence, Baichung’s age = 17 years, Baichung’s father’s age = 17 + 29 = 46 years and
Baichung’s granddaughter’s age = 17 + 29 + 26 = 72 years.
Let the number of girls be x. Then, the number of boys = x+8. According to the question, x+8/x=7/5 ⇒ 5(x+8)=7x ⇒ 5x+40=7x ⇒ 5x-7x = -40 [Transposing 7x to L.H.S. and 40 to R.H.S.] ⇒ -2x=-40 ⇒ -2x/-2=-40/-2 [Dividing both sides by 2] ⇒ x=20 Hence the number of girls = 20 and number of boys = 20 + 8Read more
Let the number of girls be x. Then, the number of boys = x+8.
According to the question, x+8/x=7/5
⇒ 5(x+8)=7x ⇒ 5x+40=7x
⇒ 5x-7x = -40 [Transposing 7x to L.H.S. and 40 to R.H.S.]
⇒ -2x=-40
⇒ -2x/-2=-40/-2 [Dividing both sides by 2]
⇒ x=20
Hence the number of girls = 20 and number of boys = 20 + 8 = 28.
Let the present ages of Rahul and Haroon be 5x years and 7x years respectively. According to condition, (5x+4)+(7x+4)=56 ⇒ 12x+8=56 ⇒ 12x+8-8=56-8 [Subtracting 8 from both sides] ⇒ 12x/8 ⇒ 12x/12=48/12 [Dividing both sides by 12] ⇒ x=4 Hence, present age of Rahul = 5 x 4 = 20 years and present age oRead more
Let the present ages of Rahul and Haroon be 5x years and 7x years respectively.
According to condition, (5x+4)+(7x+4)=56
⇒ 12x+8=56
⇒ 12x+8-8=56-8 [Subtracting 8 from both sides]
⇒ 12x/8
⇒ 12x/12=48/12 [Dividing both sides by 12]
⇒ x=4
Hence, present age of Rahul = 5 x 4 = 20 years and present age of Haroon = 7 x 4 = 28 years.
Let the three consecutive integers be x,x+1 and x+2 According to the question, 2x+3(x+1)+4(x+2)=74 ⇒ 2x+3x+3+4x+8=74 ⇒ 9x+11=74 ⇒ 9x+11-11=74-11 [Subtracting 11 from both sides] ⇒ 9x+63 ⇒ 9x/9=63/9 [Dividing both sides by 9] ⇒ x=7 Hence first integer = 7, second integer = 7 + 1 = 8 and third integerRead more
Let the three consecutive integers be x,x+1 and x+2
According to the question, 2x+3(x+1)+4(x+2)=74
⇒ 2x+3x+3+4x+8=74
⇒ 9x+11=74
⇒ 9x+11-11=74-11 [Subtracting 11 from both sides]
⇒ 9x+63
⇒ 9x/9=63/9 [Dividing both sides by 9]
⇒ x=7
Hence first integer = 7, second integer = 7 + 1 = 8 and third integer = 7 + 2 = 9.
Let the three consecutive multiples of 8 be x,x+8 and x+16. According to question, x+x+8+x+16=888 ⇒ 3x+24=888 ⇒ 3x+24-24=888-24 [Subtracting 24 from both sides] ⇒ 3x=864 ⇒ 3x/3 = 864/3 ⇒ x=288 Hence, first multiple of 8 = 288, second multiple of 8 = 288 + 8 = 296 and third multiple of 8 = 288 + 16 =Read more
Let the three consecutive multiples of 8 be x,x+8 and x+16.
According to question, x+x+8+x+16=888
⇒ 3x+24=888
⇒ 3x+24-24=888-24 [Subtracting 24 from both sides]
⇒ 3x=864
⇒ 3x/3 = 864/3
⇒ x=288
Hence, first multiple of 8 = 288, second multiple of 8 = 288 + 8 = 296 and third multiple of 8 = 288 + 16 = 304.
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/
Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Let Baichung’s age be x years, then Baichung’s father’s age = (x+29) years and Baichung’s granddaughter’s age = (x+29+26) = (x+55) years. According to condition, x+x+29+x+55=135 ⇒ 3x+84=135 ⇒ 3x+84-84=135-84 [Subtracting 84 from both sides] ⇒ 3x=51 ⇒ 3x/3=51/3 [Dividing both sides by 3] ⇒ x=17 yearsRead more
Let Baichung’s age be x years, then Baichung’s father’s age = (x+29) years and
Baichung’s granddaughter’s age = (x+29+26) = (x+55) years.
According to condition, x+x+29+x+55=135
⇒ 3x+84=135
⇒ 3x+84-84=135-84 [Subtracting 84 from both sides]
⇒ 3x=51
⇒ 3x/3=51/3 [Dividing both sides by 3]
⇒ x=17 years
Hence, Baichung’s age = 17 years, Baichung’s father’s age = 17 + 29 = 46 years and
Baichung’s granddaughter’s age = 17 + 29 + 26 = 72 years.
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
The number of boys and girls in a class are in the ratio 7: 5. The number of boys is 8 more than the number of girls. What is the total class strength?
Let the number of girls be x. Then, the number of boys = x+8. According to the question, x+8/x=7/5 ⇒ 5(x+8)=7x ⇒ 5x+40=7x ⇒ 5x-7x = -40 [Transposing 7x to L.H.S. and 40 to R.H.S.] ⇒ -2x=-40 ⇒ -2x/-2=-40/-2 [Dividing both sides by 2] ⇒ x=20 Hence the number of girls = 20 and number of boys = 20 + 8Read more
Let the number of girls be x. Then, the number of boys = x+8.
According to the question, x+8/x=7/5
⇒ 5(x+8)=7x ⇒ 5x+40=7x
⇒ 5x-7x = -40 [Transposing 7x to L.H.S. and 40 to R.H.S.]
⇒ -2x=-40
⇒ -2x/-2=-40/-2 [Dividing both sides by 2]
⇒ x=20
Hence the number of girls = 20 and number of boys = 20 + 8 = 28.
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?
Let the present ages of Rahul and Haroon be 5x years and 7x years respectively. According to condition, (5x+4)+(7x+4)=56 ⇒ 12x+8=56 ⇒ 12x+8-8=56-8 [Subtracting 8 from both sides] ⇒ 12x/8 ⇒ 12x/12=48/12 [Dividing both sides by 12] ⇒ x=4 Hence, present age of Rahul = 5 x 4 = 20 years and present age oRead more
Let the present ages of Rahul and Haroon be 5x years and 7x years respectively.
According to condition, (5x+4)+(7x+4)=56
⇒ 12x+8=56
⇒ 12x+8-8=56-8 [Subtracting 8 from both sides]
⇒ 12x/8
⇒ 12x/12=48/12 [Dividing both sides by 12]
⇒ x=4
Hence, present age of Rahul = 5 x 4 = 20 years and present age of Haroon = 7 x 4 = 28 years.
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Let the three consecutive integers be x,x+1 and x+2 According to the question, 2x+3(x+1)+4(x+2)=74 ⇒ 2x+3x+3+4x+8=74 ⇒ 9x+11=74 ⇒ 9x+11-11=74-11 [Subtracting 11 from both sides] ⇒ 9x+63 ⇒ 9x/9=63/9 [Dividing both sides by 9] ⇒ x=7 Hence first integer = 7, second integer = 7 + 1 = 8 and third integerRead more
Let the three consecutive integers be x,x+1 and x+2
According to the question, 2x+3(x+1)+4(x+2)=74
⇒ 2x+3x+3+4x+8=74
⇒ 9x+11=74
⇒ 9x+11-11=74-11 [Subtracting 11 from both sides]
⇒ 9x+63
⇒ 9x/9=63/9 [Dividing both sides by 9]
⇒ x=7
Hence first integer = 7, second integer = 7 + 1 = 8 and third integer = 7 + 2 = 9.
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Let the three consecutive multiples of 8 be x,x+8 and x+16. According to question, x+x+8+x+16=888 ⇒ 3x+24=888 ⇒ 3x+24-24=888-24 [Subtracting 24 from both sides] ⇒ 3x=864 ⇒ 3x/3 = 864/3 ⇒ x=288 Hence, first multiple of 8 = 288, second multiple of 8 = 288 + 8 = 296 and third multiple of 8 = 288 + 16 =Read more
Let the three consecutive multiples of 8 be x,x+8 and x+16.
According to question, x+x+8+x+16=888
⇒ 3x+24=888
⇒ 3x+24-24=888-24 [Subtracting 24 from both sides]
⇒ 3x=864
⇒ 3x/3 = 864/3
⇒ x=288
Hence, first multiple of 8 = 288, second multiple of 8 = 288 + 8 = 296 and third multiple of 8 = 288 + 16 = 304.
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/