Question

1)question
the rate of piece of gift paper is Rs. 0.40 per paper. how many pieces of paper can we purchase for Rs. 78.40
2)question
A shopkeeper sold a toy for Rs. 96 at a loss of 20% find the loss
3)question
if a car needs 9 litres of petrol for a journey of 162km. Find how many litres of petrol is required for 306km?
4)question
solve the equation 3x-1.5 by 0.9-1.5=0 and verify the solution
5)question
A room is triangular in shape. Its base is 9.4m and height is 8.6m.Find at the rate in rs. 250 per square metre.
6)question
subtract
${3}x ^{5} - {4}x^{4} + {8}x^{3} - 6 from {8}x^{5} + {5}x^{4} - {3}x^{3} + 4x + 3.$
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Answer 1

Answer:

Question 1

Given

• The rate of piece of gift paper is Rs. 0.40 per paper.

So,

78.40 ÷ 0.40 = 196

Thus, We can purchase 196 pieces of paper.

Question 2

Given

• A shopkeeper sold a toy for Rs. 96 at a loss of 20%.

Here,

Loss of 20% means Rs 96 represents 80% of cost price.

Therefore,

CP = 96/80 × 100 = 120

So,

₹120 - ₹96 = ₹24

Thus, The loss is ₹24

Question 3

Given

• A car needs 9 litres of petrol for a journey of 162km.

Here, We will find that how much km the car drive in 1 litre of petrol,

So,

162 ÷ 9 = 18 km

Now, we need to find for 306 km.

306 ÷ 18 = 17

Thus, 17 litres of petrol required for 306 km.

Question 4

(3x - 1.5) ÷ (0.9 - 1.5) = 0

(3x - 1.5) ÷ (-0.6) = 0

3x - 1.5 = 0

3x = 1.5

x = 1.5/3 = 0.5

VERIFICATION

(3x - 1.5) ÷ (0.9 - 1.5) = 0

[3(0.5) - 1.5] ÷ (0.9 - 1.5) = 0

(1.5 - 1.5) ÷ (-0.6) = 0

0 ÷ (-0.6) = 0

0 = 0

Hence Verified!!

Question 5

Given

• A room is triangular in shape.
• Base (b) = 9.4m
• Height (h) = 8.6m

Formula

• Area = ½ × Base × Height

So,

A = 1/2 × 9.4 × 8.6 = 40.42 m²

We are given that the rate of per square metre is ₹250.

So,

40.42 × 250 = ₹10,105

Thus, the rate of 40.42 m² is ₹10,105.

Question 6

(8x⁵ + 5x⁴ - 3x³ + 4x + 3)

(3x⁵ - 4x⁴ + 8x³ - 0x - 6)

- + - + +

5x⁵ + 9x⁴ - 11x³ + 4x + 9

Thus, The answer is (5x⁵ + 9x⁴ - 11x³ + 4x + 9).