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If you subtract ½ from a number and multiply the result by ½, you get 1/8. What is the number?
Let the number be x.
According to the question,
(x – 1/2) × ½ = 1/8
x/2 – ¼ = 1/8
x/2 = 1/8 + ¼
x/2 = 1/8 + 2/8
x/2 = (1+ 2)/8
x/2 = 3/8
x = (3/8) × 2
x = ¾
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 of the pool?
Given that,
The perimeter of the rectangular swimming pool = 154 m. Let the breadth of the rectangle be = x
According to the question,
Length of the rectangle = 2x + 2 We know that,
Perimeter = 2(length + breadth)
⇒ 2(2x + 2 + x) = 154 m
⇒ 2(3x + 2) = 154
⇒ 3x +2 = 154/2
⇒ 3x = 77 – 2
⇒ 3x = 75
⇒ x = 75/3
⇒ x = 25 m
Therefore, Breadth = x = 25 cm
Length = 2x + 2
= (2 × 25) + 2
= 50 + 2
= 52 m
The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 
cm. What is the length of either of the remaining equal sides?
Base of isosceles triangle = 4/3 cm
Perimeter of triangle =
image cm = 62/15
Let the length of equal sides of the triangle be x.
According to the question,
4/3 + x + x = 62/15 cm
⇒ 2x = (62/15 – 4/3) cm
⇒ 2x = (62 – 20)/15 cm
⇒ 2x = 42/15 cm
⇒ x = (42/30) × (½)
⇒ x = 42/30 cm
⇒ x = 7/5 cm
The length of either of the remaining equal sides is 7/5 cm.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Let one of the numbers be = x.
Then, the other number becomes x + 15. According to the question,
x + x + 15 = 95
⇒ 2x + 15 = 95
⇒ 2x = 95 – 15
⇒ 2x = 80
⇒ x = 80/2
⇒ x = 40
First number = x = 40
And, other number = x + 15 = 40 + 15 = 55
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 the question,
5x – 3x = 18
⇒ 2x = 18
⇒ x = 18/2
⇒ x = 9
Thus,
The numbers are 5x = 5 × 9 = 45
And 3x = 3 × 9 = 27.
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 = 51 – 3
⇒ 3x = 48
⇒ x = 48/3
⇒ x = 16
Thus, the integers are
x = 16
x + 1 = 17
x + 2 = 18
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Let the three consecutive multiples of 8 be 8x, 8(x+1) and 8(x+2). According to the question,
8x + 8(x+1) + 8(x+2) = 888
⇒ 8 (x + x+1 + x+2) = 888 (Taking 8 as common)
⇒ 8 (3x + 3) = 888
⇒ 3x + 3 = 888/8
⇒ 3x + 3 = 111
⇒ 3x = 111 – 3
⇒ 3x = 108
⇒ x = 108/3
⇒ x = 36
Thus, the three consecutive multiples of 8 are:
8x = 8 × 36 = 288
8(x + 1) = 8 × (36 + 1) = 8 × 37 = 296
8(x + 2) = 8 × (36 + 2) = 8 × 38 = 304
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 = 74 – 11
⇒ 9x = 63
⇒ x = 63/9
⇒ x = 7
Thus, the numbers are:
x = 7
x + 1 = 8
x + 2 = 9
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 ages of Rahul and Haroon be 5x and 7x. Four years later,
The ages of Rahul and Haroon will be (5x + 4) and (7x + 4), respectively. According to the question,
(5x + 4) + (7x + 4) = 56
⇒ 5x + 4 + 7x + 4 = 56
⇒ 12x + 8 = 56
⇒ 12x = 56 – 8
⇒ 12x = 48
⇒ x = 48/12
⇒ x = 4
Therefore, Present age of Rahul = 5x = 5×4 = 20
And, present age of Haroon = 7x = 7×4 = 28
The number of boys and girls in a class is in the ratio of 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?
Let the number of boys be 7x, and girls be 5x.
According to the question,
7x = 5x + 8
⇒ 7x – 5x = 8
⇒ 2x = 8
⇒ x = 8/2
⇒ x = 4
Therefore, number of boys = 7×4 = 28
And, number of girls = 5×4 = 20
Total number of students = 20+28 = 48