1. If you subtract ½ from a number and multiply the result by ½, you get 1/8 what is the number?
Solution:
Let us assume the number to be x.
According to the given 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
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 the breadth of the pool?
Solution:
Given,
Perimeter of rectangular swimming pool = 154 m
Now, Let the breadth of rectangle be = x
According to the given question,
Length of the rectangle = 2x + 2
By the formula 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
Hence,
Breadth = x = 25 cm
Now,
Length = 2x + 2
= (2 × 25) + 2
= 50 + 2
= 52 m
Length = 52 m
3. The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 4 2/15 cm. What is the length of either of the remaining equal sides?
Solution:
Given, Base of isosceles triangle = 4/3 cm
Perimeter of triangle = 4 2/15
4 2/15 = 62/15
Let us assume the length of equal sides of triangle be x.
According to the given 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
Hence, the length of the remaining equal sides are 7/5 cm.
4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution:
Let us assume the one of the number be= x.
According to the question
The second number becomes x + 15
Now,
x + x + 15 = 95
⇒ 2x + 15 = 95
⇒ 2x = 95 – 15
⇒ 2x = 80
⇒ x = 80/2
⇒ x = 40
First number = x = 40
Second number = x + 15 = 40 + 15 = 55
5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Solution:
Let the both numbers be 5x and 3x.
According to the given question,
5x – 3x = 18
⇒ 2x = 18
⇒ x = 18/2
⇒ x = 9
So,
The numbers are:
5x = 5 × 9 = 45
And 3x = 3 × 9 = 27.
Hence, required numbers are 45 and 27.
6. Three consecutive integers add up to 51. What are these integers?
Solution:
Let us assume 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
Hence, the required integers are 16, 17 and 18.
7. The sum of three consecutive multiples of 8 is 888. Find the multiples.
Solution:
Let us assume 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
Hence, the required numbers are 288, 296 and 304.
8. 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.
Solution:
Let us assume the three consecutive integers are 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
Hence, the required numbers are 7, 8 and 9.
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?
Solution:
Let us assume the ages of Rahul and Haroon be 5x and 7x.
After four years, 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
Hence,
Present age of Rahul = 5x = 5×4 = 20
Present age of Haroon = 7x = 7×4 = 28
10. 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?
Solution:
Let us assume 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
Hence,
Number of boys = 7×4 = 28
Number of girls = 5×4 = 20
Total number of students = 20+28 = 48
11. 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?
Solution:
Let us assume the age of Baichung’s father be x.
Then,
The age of Baichung’s grandfather = (x+26)
Age of Baichung = (x-29)
According to the question,
x + (x+26) + (x-29) = 135
⇒ 3x + 26 – 29 = 135
⇒ 3x – 3 = 135
⇒ 3x = 135 + 3
⇒ 3x = 138
⇒ x = 138/3
⇒ x = 46
Hence,
Age of Baichung’s father = x = 46
Age of Baichung’s grandfather = (x+26) = 46 + 26 = 72
Age of Baichung = (x-29) = 46 – 29 = 17
12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Solution:
Let us assume the present age of Ravi be x.
After fifteen years, age of Ravi will be x+15 years.
According to the question,
x + 15 = 4x
⇒ 4x – x = 15
⇒ 3x = 15
⇒ x = 15/3
⇒ x = 5
Hence, Present age of Ravi = 5 years.
13. 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?
Solution:
Let us assume the rational be x.
According to the question,
x × (5/2) + 2/3 = -7/12
⇒ 5x/2 + 2/3 = -7/12
⇒ 5x/2 = -7/12 – 2/3
⇒ 5x/2 = (-7- 8)/12
⇒ 5x/2 = -15/12
⇒ 5x/2 = -5/4
⇒ x = (-5/4) × (2/5)
⇒ x = – 10/20
⇒ x = -½
Hence, the rational number is -½.
14. 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?
Solution:
Let us assume the numbers of notes of ₹100, ₹50 and ₹10 be 2x, 3x and 5x respectively.
Value of ₹100 = 2x × 100 = 200x
Value of ₹50 = 3x × 50 = 150x
Value of ₹10 = 5x × 10 = 50x
According to the question,
200x + 150x + 50x = 4,00,000
⇒ 400x = 4,00,000
⇒ x = 400000/400
⇒ x = 1000
Hence,
Numbers of ₹100 notes = 2x = 2000
Numbers of ₹50 notes = 3x = 3000
Numbers of ₹10 notes = 5x = 5000
15. 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?
Solution:
Let us assume the number of ₹5 coins be x.
Then,
number ₹2 coins = 3x
number of ₹1 coins = (160 – 4x)
Now,
Value of ₹5 coins = x × 5 = 5x
Value of ₹2 coins = 3x × 2 = 6x
Value of ₹1 coins = (160 – 4x) × 1 = (160 – 4x)
According to the question,
5x + 6x + (160 – 4x) = 300
⇒ 11x + 160 – 4x = 300
⇒ 7x = 140
⇒ x = 140/7
⇒ x = 20
Hence,
Number of ₹5 coins = x = 20
Number of ₹2 coins = 3x = 60
Number of ₹1 coins = (160 – 4x) = 160 – 80 = 80
16. The organisers 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 winners, if the total number of participants is 63.
Solution:
Let us assume the numbers of winner be x.
Then,
The total number of participants who didn’t win = 63 – x
Now,
Total money given to the winner = x × 100 = 100x
Total money given to participant who didn’t win = 25×(63-x)
According to the question,
100x + 25×(63-x) = 3,000
⇒ 100x + 1575 – 25x = 3,000
⇒ 75x = 3,000 – 1575
⇒ 75x = 1425
⇒ x = 1425/75
⇒ x = 19
Hence, the numbers of winners are 19.
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