Let the three consecutive integers be x,x+1 and x+2. According to the question, x+x+1+x+2=51 ⇒ 3x+3=51 ⇒ 3x+3-3=51-3 [Subtracting 3 from both sides] ⇒ 3x=48 ⇒ 3x/3=48/3 Dividing both sides by 3] ⇒ x=16 Hence, first integer = 16, second integer = 16 + 1 = 17 and third integer = 16 + 2 = 18. for moreRead more
Let the three consecutive integers be x,x+1 and x+2.
According to the question, x+x+1+x+2=51
⇒ 3x+3=51
⇒ 3x+3-3=51-3 [Subtracting 3 from both sides]
⇒ 3x=48
⇒ 3x/3=48/3 Dividing both sides by 3]
⇒ x=16
Hence, first integer = 16, second integer = 16 + 1 = 17 and third integer = 16 + 2 = 18.
Let the two numbers be 5x and 3x According to question, 5x-3x=18 ⇒ 2x=18 ⇒ 2x/2=18/2 [Dividing both sides by 2] ⇒ x=9 Hence, first number = 5 x 9 = 45 and second number = 3 x 9 = 27. for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
Let the two numbers be 5x and 3x
According to question, 5x-3x=18
⇒ 2x=18
⇒ 2x/2=18/2 [Dividing both sides by 2]
⇒ x=9
Hence, first number = 5 x 9 = 45 and second number = 3 x 9 = 27.
Sum of two number = 95 Let the first number be x, then another number be x+15. According to the question, x+x+15=95 ⇒ 2x+15=95 ⇒ 2x+15-15=95-15 [Subtracting 15 from both sides] ⇒ 2x=80 ⇒ 2x/2=80/2 [Dividing both sides by 2] ⇒ x=40 So, the first number = 40 and another number = 40 + 15 = 55 Hence, thRead more
Sum of two number = 95
Let the first number be x, then another number be x+15.
According to the question, x+x+15=95
⇒ 2x+15=95
⇒ 2x+15-15=95-15 [Subtracting 15 from both sides]
⇒ 2x=80
⇒ 2x/2=80/2 [Dividing both sides by 2]
⇒ x=40
So, the first number = 40 and another number = 40 + 15 = 55
Hence, the two numbers are 40 and 55.
⇒ 62/15=4/3+x+x ⇒ 62/15=4/3+2x ⇒ 62/15-4/3=4/3-4/3+2x [Subtracting 4/3 from both the sides] ⇒ 62-20/15=2x ⇒ 42/15=2x ⇒ 42/15x2=2x/2 [Dividing both sides by 2] ⇒ 7/5=x ⇒ x=7/5cm Hence, each equal side of an isosceles triangle is x=7/5cm. for more answers vist to: https://www.tiwariacademy.com/ncert-sRead more
⇒ 62/15=4/3+x+x
⇒ 62/15=4/3+2x
⇒ 62/15-4/3=4/3-4/3+2x [Subtracting 4/3 from both the sides]
⇒ 62-20/15=2x ⇒ 42/15=2x
⇒ 42/15×2=2x/2 [Dividing both sides by 2]
⇒ 7/5=x ⇒ x=7/5cm
Hence, each equal side of an isosceles triangle is x=7/5cm.
Let the breadth of the pool be x m. Then, the length of the pool = (2x+2)m Perimeter = 2(l+b) ⇒ 154 = 2(2x+2+x) ⇒ 154/2 = 2(2x+2+x)/2 [Dividing both sides by 2] ⇒ 77=3x+2 ⇒ 77-2=3x+2-2 [Subtracting 2 from both sides] ⇒ 75=3x ⇒ 75/3=3x/3 [Dividing both sides by 3] ⇒ 25=x ⇒ x=25 Length of the pool= 2xRead more
Let the breadth of the pool be x m.
Then, the length of the pool = (2x+2)m
Perimeter = 2(l+b)
⇒ 154 = 2(2x+2+x)
⇒ 154/2 = 2(2x+2+x)/2 [Dividing both sides by 2]
⇒ 77=3x+2
⇒ 77-2=3x+2-2 [Subtracting 2 from both sides]
⇒ 75=3x
⇒ 75/3=3x/3 [Dividing both sides by 3]
⇒ 25=x ⇒ x=25
Length of the pool= 2x+2=2×25+2=50+2=52 m, Breadth of the pool=25m Hence, the length of the pool is 52 m and breadth is 25 m.
Three consecutive integers add up to 51. What are these integers?
Let the three consecutive integers be x,x+1 and x+2. According to the question, x+x+1+x+2=51 ⇒ 3x+3=51 ⇒ 3x+3-3=51-3 [Subtracting 3 from both sides] ⇒ 3x=48 ⇒ 3x/3=48/3 Dividing both sides by 3] ⇒ x=16 Hence, first integer = 16, second integer = 16 + 1 = 17 and third integer = 16 + 2 = 18. for moreRead more
Let the three consecutive integers be x,x+1 and x+2.
According to the question, x+x+1+x+2=51
⇒ 3x+3=51
⇒ 3x+3-3=51-3 [Subtracting 3 from both sides]
⇒ 3x=48
⇒ 3x/3=48/3 Dividing both sides by 3]
⇒ x=16
Hence, first integer = 16, second integer = 16 + 1 = 17 and third integer = 16 + 2 = 18.
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Let the two numbers be 5x and 3x According to question, 5x-3x=18 ⇒ 2x=18 ⇒ 2x/2=18/2 [Dividing both sides by 2] ⇒ x=9 Hence, first number = 5 x 9 = 45 and second number = 3 x 9 = 27. for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
Let the two numbers be 5x and 3x
According to question, 5x-3x=18
⇒ 2x=18
⇒ 2x/2=18/2 [Dividing both sides by 2]
⇒ x=9
Hence, first number = 5 x 9 = 45 and second number = 3 x 9 = 27.
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Sum of two number = 95 Let the first number be x, then another number be x+15. According to the question, x+x+15=95 ⇒ 2x+15=95 ⇒ 2x+15-15=95-15 [Subtracting 15 from both sides] ⇒ 2x=80 ⇒ 2x/2=80/2 [Dividing both sides by 2] ⇒ x=40 So, the first number = 40 and another number = 40 + 15 = 55 Hence, thRead more
Sum of two number = 95
Let the first number be x, then another number be x+15.
According to the question, x+x+15=95
⇒ 2x+15=95
⇒ 2x+15-15=95-15 [Subtracting 15 from both sides]
⇒ 2x=80
⇒ 2x/2=80/2 [Dividing both sides by 2]
⇒ x=40
So, the first number = 40 and another number = 40 + 15 = 55
Hence, the two numbers are 40 and 55.
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 62/15 cm. What is the length of either of the remaining equal sides?
⇒ 62/15=4/3+x+x ⇒ 62/15=4/3+2x ⇒ 62/15-4/3=4/3-4/3+2x [Subtracting 4/3 from both the sides] ⇒ 62-20/15=2x ⇒ 42/15=2x ⇒ 42/15x2=2x/2 [Dividing both sides by 2] ⇒ 7/5=x ⇒ x=7/5cm Hence, each equal side of an isosceles triangle is x=7/5cm. for more answers vist to: https://www.tiwariacademy.com/ncert-sRead more
⇒ 62/15=4/3+x+x
⇒ 62/15=4/3+2x
⇒ 62/15-4/3=4/3-4/3+2x [Subtracting 4/3 from both the sides]
⇒ 62-20/15=2x ⇒ 42/15=2x
⇒ 42/15×2=2x/2 [Dividing both sides by 2]
⇒ 7/5=x ⇒ x=7/5cm
Hence, each equal side of an isosceles triangle is x=7/5cm.
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/
The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and breadth?
Let the breadth of the pool be x m. Then, the length of the pool = (2x+2)m Perimeter = 2(l+b) ⇒ 154 = 2(2x+2+x) ⇒ 154/2 = 2(2x+2+x)/2 [Dividing both sides by 2] ⇒ 77=3x+2 ⇒ 77-2=3x+2-2 [Subtracting 2 from both sides] ⇒ 75=3x ⇒ 75/3=3x/3 [Dividing both sides by 3] ⇒ 25=x ⇒ x=25 Length of the pool= 2xRead more
Let the breadth of the pool be x m.
Then, the length of the pool = (2x+2)m
Perimeter = 2(l+b)
⇒ 154 = 2(2x+2+x)
⇒ 154/2 = 2(2x+2+x)/2 [Dividing both sides by 2]
⇒ 77=3x+2
⇒ 77-2=3x+2-2 [Subtracting 2 from both sides]
⇒ 75=3x
⇒ 75/3=3x/3 [Dividing both sides by 3]
⇒ 25=x ⇒ x=25
Length of the pool= 2x+2=2×25+2=50+2=52 m, Breadth of the pool=25m Hence, the length of the pool is 52 m and breadth is 25 m.
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/