## Miscellaneous Problems : Set 2

**1. Angela deposited 15000 rupees in a bank at a rate of 9 p.c.p.a. She got simple interest amounting to 5400 rupees. For how many years had she deposited the amount?**

**Given:**

P = ₹ 15000

R = 9 p.c.p.a

Simple Interest = ₹ 17400

**To find: **

No of years

**Solution:**

Total interest = \(\large \frac{P\, ×\, R\, ×\, T}{100} \)

∴ 5400 = \( \large \frac{15000\, ×\, 9\, ×\, T}{100} \)

∴ 5400 = 150 × 9 × T

∴ 5400 = 1350 × T

∴ \( \large \frac{5400}{1350} \) = T

∴ T = 4 years

**Ans: **Angela had deposited the amount for 4 years.

**2. Ten men take 4 days to complete the task of tarring a road. How many days would 8 men take?**

**Solution:**

10 men take 4 days to tar the road

∴ 8 men would take ? hours to tar the road

Let 8 men take x days to tar the road

**The number of men and the number of days required to tar the road are in inverse proportion**

∴ 8 × x = 10 x 4

∴ 8x = 40

∴ x = \(\large \frac {40}{8}\)

∴ x = 5

**Ans:** 8 men will require 5 days to tar the road.

**3. Nasruddin and Mahesh invested ₹ 40,000 and ₹ 60,000 respectively to start a business. They made a profit of 30%. How much profit did each of them make?**

**Given: **

Nasruddin invested for business = ₹ 40,000

Mahesh invested for business = ₹ 60,000

Profit made by them = 30%

**Solution:**

Total amount invested by Nasruddin and Mahesh = ₹ 40,000 + ₹ 60,000

∴ Total amount invested by Nasruddin and Mahesh = Rs 1,00,000

Profit earned by them = 30%

∴ Total profit = 30% of 100000

∴ Total profit = \(\large \frac {30}{100}\) × 100000

∴ Total profit = ₹ 30000

Proportion of investment = 40000 : 60000

∴ Proportion of investment = 2 : 3 …[Dividing by 20000]

Let Nasruddin’s profit be Rs 2x and Mahesh’s profit be Rs 3x.

∴ 2x + 3x = 30000

∴ 5x = 30000

∴ x = \(\large \frac {30000}{5}\)

∴ x = 6000

So,

Nasruddin’s profit = 2x

∴ Nasruddin’s profit = 2 × 6000

∴ Nasruddin’s profit = ₹ 12000

And, Mahesh’s profit = 3x = 3 × 6000 = Rs 18000

∴ Mahesh’s profit = 3 × 6000

∴ Mahesh’s profit = ₹ 18000

**Ans:** The profit of Nasruddin is ₹ 12000 and that of Mahesh is ₹ 18000.

**4. The diameter of a circle is 5.6 cm. Find its circumference.**

**Given:**

Diameter of the circle = 5.6 cm

**To find:**

Circumference of the circle

**Solution:**

Circumference of the circle = πd

∴ Circumference of the circle = \(\large \frac {22}{7}\) × 5.6

∴ Circumference of the circle = 22 × 0.8

∴ Circumference of the circle = 17.6 cm

**Ans:** The circumference of the circle is 17.6 cm.

**5. Expand.**

**(i) (2a – 3b)² **

**Solution: **

Here,

a = 2a and b = 3b

We know that,

(a – b)² = a² – 2ab + b²

∴ (2a – 3b)² = (2a)² – 2 × 2a × 3b + (3b)²

∴ (2a – 3b)² = 4a² – 12ab + 9b²

**Ans:** (2a – 3b)² = 4a² – 12ab + 9b²

**(ii) (10 + y)²**

**Solution:**

Here,

a = 10 and b = y

We know that,

(a + b)² = a² + 2ab + b²

∴ (10 – y)² = (10)² + 2 × 10 × y + (y)²

∴ (10 – y)² = 100 + 20y + y²

**Ans:** (10 – y)² = 100 + 20y + y²

**(iii) \((\large \frac {p}{3} + \large \frac {q}{4})\)²**

**Solution:**

Here,

a = \(\large \frac {p}{3}\) and b = \(\large \frac {q}{4}\)

We know that,

(a + b)² = a² + 2ab + b²

∴ \((\large \frac {p}{3} + \large \frac {q}{4})\)² = (\(\large \frac {p}{3}\))² + 2 × \(\large \frac {p}{3}\) × \(\large \frac {q}{4}\) + (\(\large \frac {q}{4}\))²

∴ \((\large \frac {p}{3} + \large \frac {q}{4})\)² = \(\large \frac {p²}{9}\) + \(\large \frac {pq}{6}\)+ \(\large \frac {q²}{16}\)

**Ans:** \((\large \frac {p}{3} + \large \frac {q}{4})\)² = \(\large \frac {p²}{9}\) + \(\large \frac {pq}{6}\)+ \(\large \frac {q²}{16}\)

**(iv) \((y \, – \large \frac {3}{y})\)²**

**Solution:**

Here,

a = y and b = \(\large \frac {3}{y}\)

We know that,

(a – b)² = a² – 2ab + b²

∴ \((y \, – \large \frac {3}{y})\)² = (y)² – 2 × y × \(\large \frac {3}{y}\) + \((\large \frac {3}{y})\)²

∴ \((y \, – \large \frac {3}{y})\)² = y² – 6 + \(\large \frac {9}{y²}\)

**Ans:** \((y \, – \large \frac {3}{y})\)² = y² – 6 + \(\large \frac {9}{y²}\)

**6. Use a formula to multiply.**

**(i) (x – 5) (x + 5) **

**Solution:**

Here, a = x and b = 5

We know that,

(a + b)(a – b) = a² – b²

∴ (x – 5) (x + 5) = x² – 5²

∴ (x – 5) (x + 5) = x² – 25

**Ans:** (x – 5) (x + 5) = x² – 25

**(ii) (2a – 13) (2a + 13)**

**Solution:**

Here, a = 2a and b = 13

We know that,

(a + b)(a – b) = a² – b²

∴ (2a – 13) (2a + 13) = (2a)² – 13²

∴ (2a – 13) (2a + 13) = 4a² – 169

**Ans:** (2a – 13) (2a + 13) = 4a² – 169

**(iii) (4z – 5y) (4z + 5y) **

**Solution:**

Here, a = 4z and b = 5y

We know that,

(a + b)(a – b) = a² – b²

∴ (4z – 5y) (4z + 5y) = (4z)² – (5y)²

∴ (4z – 5y) (4z + 5y) = 16z² – 25y²

**Ans:** (4z – 5y) (4z + 5y) = 16z² – 25y²

**(iv) (2t – 5) (2t + 5)**

**Solution:**

Here, a = 2t and b = 5

We know that,

(a + b)(a – b) = a² – b²

∴ (2t – 5) (2t + 5) = (2t)² – 5²

∴ (2t – 5) (2t + 5) = 4t² – 25

**Ans:** (2t – 5) (2t + 5) = 4t² – 25

**7. The diameter of the wheel of a cart is 1.05 m. How much distance will the cart cover in 1000 rotations of the wheel?**

**Given:**

Diameter of the wheel = 1.05 m

No of rotations = 1000

**To find:**

Distance covered by cart in 1000 rotations

**Solution:**

Distance covered by cart in 1 rotation = Circumference of the wheel

∴ Distance covered by cart in 1 rotation = πd

∴ Distance covered by cart in 1 rotation = \(\large \frac {22}{7}\) × 1.05

∴ Distance covered by cart in 1 rotation = 3.3 m

Now,

Distance covered by cart in 1000 rotations = 1000 × 3.3 m

∴ Distance covered by cart in 1000 rotations = 3300 m

∴ Distance covered by cart in 1000 rotations = \(\large \frac {3300}{1000}\) km …[∵ 1 km = 1000 m]

∴ Distance covered by cart in 1000 rotations = 3.3 km

**Ans:** The distance covered by the cart in 1000 rotations of the wheel is 3.3 km.

**8. The area of a rectangular garden of length 40 m, is 1000 sq m. Find the breadth of the garden and its perimeter. The garden is to be enclosed by 3 rounds of fencing, leaving an entrance of 4 m. Find the cost of fencing the garden at a rate of 250 rupees per metre.**

**Given:**

Length of the rectangular garden = 40 m

Area of the rectangular garden = 1000 sq. m.

Rate of fencing the garden = ₹ 250 per m.

**To find:**

Breadth of the garden

Perimeter of the garden

Cost of fencing the garden

**Solution:**

Area of the rectangular garden = length of the garden × breadth of the garden

∴ 1000 = 40 × breadth of the garden

∴ \(\large \frac {1000}{40}\) = breadth of the garden

∴ breadth of the garden = 25 m

Perimeter of the garden = 2 × length of the garden + 2 × breadth of the garden

∴ Perimeter of the garden = 2 × 40 + 2 × 25

∴ Perimeter of the garden = 80 + 50

∴ Perimeter of the garden = 130 m

Now,

Length of one round of fence = Perimeter of the garden – width of the entrance

∴ Length of one round of fence = 130 – 4

∴ Length of one round of fence = 126 m

Total length of fencing = Length of one round of fencing × number of rounds

∴ Total length of fencing = 126 × 3

∴ Total length of fencing = 378 m

And,

Total cost of fencing = Total length of fencing × rate of fencing the garden

Total cost of fencing = 378 × 250

Total cost of fencing = ₹ 94500

**Ans:** The cost of fencing the garden is ₹ 94500.

**9. From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC.**

**Given:**

AB = 20 units

BC = 21 units

∠ABC = 90°

**To find:**

The value of x

**Solution:**

In ∆ABC, ∠ABC = 90°

∴ Side AC is the hypotenuse

According to Pythagoras’ theorem,

l(AC)² = l(AB)² + l(BC)²

∴ l(AC)² = 20² + 21²

∴ l(AC)² = 400 + 441

∴ l(AC)² = 841

∴ l(AC)² = 29²

∴ l(AC) = 29 units

Perimeter of ∆ABC = l(AB) + l(BC) + l(AC)

∴ Perimeter of ∆ABC = 20 + 21 + 29

∴ Perimeter of ∆ABC = 70 units

**Ans:** Perimeter of ∆ABC is 70 units.

**10. If the edge of a cube is 8 cm long, find its total surface area.**

**Given:**

Side = 8 cm

**To find:**

Total surface area of cube

**Solution:**

Total surface area of the cube = 6l²

∴ Total surface area of the cube = 6 × (8)²

∴ Total surface area of the cube = 6 × 64

∴ Total surface area of the cube = 384 sq. m.

**Ans:** Total surface area of the cube is 385 sq. m.

**11. Factorise : **

**365y⁴z³ – 146y²z⁴**

**Solution:**

365y⁴z³ – 146y²z⁴ = 73 (5y⁴z³ – 2y²z⁴)

∴ 365y⁴z³ – 146y²z⁴ = 73y² (5y²z³ – 2z⁴)

∴ 365y⁴z³ – 146y²z⁴ = 73y²z³ (5y² – 2z)

**Ans:** 365y⁴z³ – 146y²z⁴ = 73y²z³ (5y² – 2z)

**Multiple Choice Questions **

**Choose the right answers from the options given for each of the following questions.**

**1. If the average of the numbers 33, 34, 35, x, 37, 38, 39 is 36, what is the value of x ?**

**(i) 40 **

**(ii) 32 **

**(iii) 42 **

**(iv) 36**

**Solution:**

We know that,

Average = \( \large \frac{\text { Sum of all scores in the given data }}{\text { Total number of scores }}\)

∴ 36 = \(\large \frac{33\, +\, 34\, +\, 35\, +\, x\, +\, 37\, +\, 38\, +\, 39}{7}\)

∴ 36 × 7 = 216 + x

∴ 252 = 216 + x

∴ 252 – 216 = x

∴ x = 36

**OPTION (iv) :** 36

**2. The difference of the squares, (61² – 51²) is equal to ……………… .**

**(i) 1120 **

**(ii) 1230 **

**(iii) 1240 **

**(iv) 1250**

**Solution:**

In (61² – 51²),

a = 61 and b = 51

We know that,

a² – b² = (a + b)(a – b)

∴ 61² – 51² = (61 + 51)(61 – 51)

∴ 61² – 51² = (112)(10)

∴ 61² – 51² = 1120

**OPTION (i) :** 1120

**3. If 2600 rupees are divided between Sameer and Smita in the proportion 8 : 5, the share of each is …………… and …………… respectively. **

**(i) ₹ 1500, ₹ 1100 **

**(ii) ₹ 1300, ₹ 900**

**(iii) ₹ 800, ₹ 500 **

**(iv) ₹ 1600, ₹ 1000**

**Solution:**

₹ 2600 is to be divided among Sameer and Smita in the ratio = 8 : 5

Total = 8 + 5 = 13

∴ Sameer’s share = \(\large \frac {8}{13}\) × 2600

∴ Sameer’s share = ₹ 1600

Ans,

Smita’s share = \(\large \frac {5}{13}\) × 2600

∴ Smita’s share = ₹ 1000

**OPTION (iv) :** ₹ 1600, ₹ 1000