Question

a cooler costs ₹4156.60 and another cooler costs ₹3084.75. what is the difference between their cost prices.

Answer 1

SOLUTION

GIVEN

A cooler costs ₹4156.60 and another cooler costs ₹3084.75.

TO DETERMINE

The difference between their cost prices.

EVALUATION

Here it is given that a cooler costs ₹ 4156.60 and another cooler costs ₹ 3084.75.

Cost of first cooler = ₹ 4156.60

Cost of second cooler = ₹ 3084.75

Hence the difference between their cost prices

= ₹ ( 4156.60 - 3084.75)

= ₹ 1071.85

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Answer 2

Question and answer -

Let's understand the question 1st -

★ This question says that the cost of a cooler is Rupees 4156.60 and the another cooler cost Rupees 3084.75 We have to find the difference between their cost price.

★ And we know that to find the difference between any two terms we have to just subtract the digits.

Given that -

✯ CP (cost price) of first cooler = Rupees 4156.60

✯ Cost price (CP) of another cooler = Rupees 3084.75

To find -

✯ Difference between both cooler cost price.

Solution -

✯ Difference between both cooler cost price = Rupees 1071.85

Full solution -

➝ Rupees ( 4156.60 - 3084.75 )

➝ Rupees 1071.85

Knowledge booster -

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi b^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Circumference \: of \: circle \: = \: 2 \pi r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: cube \: = \: 6(side)^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto LSA \: of \: cube \:= \: 4(side)^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cube \: = \: (side)^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diagonal \: of \: cube \: = \: \sqrt(l^{2} + b^{2} + h^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: cube \: = \: 4(l+b+h)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: cuboid \: = \: 2(l \times b + b \times h + l \times h}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto LSA \: of \: cuboid \: = \: 2h(l+b)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cuboid \: = \: L \times B \times H}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diagonal \: of \: cuboid \: = \: \sqrt 3l}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: cuboid \: = \: 12 \times Sides}}}$