Question Detail

A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is.

200 m
150 m
148 m
140 m

Answer: Option A

Explanation:

Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then
\begin{aligned}
\text{Area of triangle =}\frac{1}{2}*b*h1\\
\text{Area of rectangle =}b*h2\\
\text{As per question }\\
\frac{1}{2}*b*h1 = b*h2 \\

\frac{1}{2}*b*h1 = b*100 \\

h1 = 100*2 = 200 m \\

\end{aligned}

1. A courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is :

16000
18000
20000
22000

Answer: Option C

Explanation:

\begin{aligned}
\text{Number of bricks =}\frac{\text{Courtyard area}}{\text{1 brick area}} \\
= \left( \frac{2500 \times 1600}{20 \times 10} \right) \\
= 20000
\end{aligned}

2. What will be the cost of gardening 1 meter boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs. 10 per square meter ?

Rs. 3430
Rs. 3440
Rs. 3450
Rs. 3460

Answer: Option B

Explanation:

In this question, we are having perimeter.
We know Perimeter = 2(l+b), right
So,
2(l+b) = 340
As we have to make 1 meter boundary around this, so
Area of boundary = ((l+2)+(b+2)-lb)
= 2(l+b)+4 = 340+4 = 344

So required cost will be = 344 * 10 = 3440

3. What are the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?

Answer: Option B

Explanation:

In this type of questions, first we need to calculate the area of tiles. With we can get by obtaining the length of largest tile.
Length of largest tile can be obtained from HCF of length and breadth.
So lets solve this,

Length of largest tile = HCF of (1517 cm and 902 cm)
= 41 cm

Required number of tiles =
\begin{aligned}
\frac{\text{Area of floor}}{\text{Area of tile}} \\
= \left(\frac{1517\times902}{41 \times 41}\right)\\
= 814

\end{aligned}

4. The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares .

22 cm
24 cm
26 cm
28 cm

Answer: Option B

Explanation:

We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm

Area of third Square = 10*10 - 8*8
= 36 cm

So side of third square = 6 [because area of square = side*side]
Perimeter = 4*Side = 4*6 = 24 cm

5. What will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of the rectangle as base and a vertex on the opposite side of the rectangle ?

Answer: Option D

Explanation:

As far as questions of Area or Volume and Surface area are concerned, it is all about formulas and very little logic. So its a sincere advice to get all formulas remembered before solving these questions.

Lets solve this,

\begin{aligned}
\text{Area of rectangle =}l*b\\
\text{Area of triangle =}\frac{1}{2}l*b\\
\text{Ratio =}l*b:\frac{1}{2}l*b \\
= 1:\frac{1}{2} \\
= 2:1

\end{aligned}

One little thing which should be taken care in this type of question is, be sure you are calculating ration in the given order of the question.
If it is ratio of triangle and rectangle then we have to write triangle formula first. cheers :)