Question - 1 : - 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?

Answer - 1 : -

Base of isosceles triangle = 4/3 cm

Perimeter of triangle =

image cm = 62/15

Let the length of equal sides of 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 are 7/5 cm.

Question - 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?

Answer - 2 : -

Let the number of ₹5 coins be x.

Then,

number ₹2 coins = 3x

and, 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

Number of ₹5 coins = x = 20

Number of ₹2 coins = 3x = 60

Number of ₹1 coins = (160 – 4x) = 160 – 80 = 80

Question - 3 : - 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.

Answer - 3 : -

Let the numbers of winner be x.

Then, the number of participants who didn’t win = 63 – x

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

Therefore, the numbers of winners are 19.

Question - 4 : - If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8 what is the number?

Answer - 4 : -

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

Question - 5 : - 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?

Answer - 5 : -

Given that,

Perimeter of rectangular swimming pool = 154 m Let the breadth of 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

Question - 6 : - Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Answer - 6 : -

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

Question - 7 : - Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Answer - 7 : -

Let the two numbers be 5x and 3x. According to the question,

5x – 3x = 18

⇒ 2x = 18

⇒ x = 18/2

⇒ x = 19

Thus,

The numbers are 5x = 5 × 9 = 45

And 3x = 3 × 9 = 27.

Question - 8 : - Three consecutive integers add up to 51. What are these integers?

Answer - 8 : -

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

Question - 9 : - The sum of three consecutive multiples of 8 is 888. Find the multiples.

Answer - 9 : -

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

Question - 10 : - 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.

Answer - 10 : -

Let 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

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