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Q:

The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.

A) B=900;H=300 B) B=300;H=900
C) B=600;H=700 D) B=500;H=900
 
Answer & Explanation Answer: A) B=900;H=300

Explanation:

Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
13.5*10000m2=135000m2 
Let altitude = x metres  and  base = 3x metres.
Then, 12*3x*x=135000x2=90000x=300 

Base = 900 m and Altitude = 300 m.

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Q:

Find the area of a right-angled triangle whose base is 12 cm and hypotenuse is 13cm.

A) 30 B) 40
C) 50 D) 60
 
Answer & Explanation Answer: A) 30

Explanation:

Height of the triangle = 132-122=25 = 5 cm.

 

Its area =12*base*height12*12*530cm2.

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Q:

Compute the sum of 4 digit numbers which can be formed with the four digits 1,3,5,7, if each digit is used only once in each arrangement.

A) 105555 B) 106665
C) 106656 D) 108333
 
Answer & Explanation Answer: C) 106656

Explanation:

The number of arrangements of 4 different digits taken 4 at a time is given by 4P4 = 4! = 24.All the four digits will occur equal number of times at each of the position,namely ones,tens,hundreds,thousands.

 

 

 

Thus,each digit will occur 24/4 = 6 times in each of the position.The sum of digits in one's position will be 6 x (1+3+5+7) = 96.Similar is the case in ten's,hundred's and thousand's places.

 

 

 

Therefore,the sum will be 96 + 96 x 10 + 96 x 100 + 96 x 100 = 106656

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Q:

Find the area of a triangle whose sides measure 13 cm, 14 cm and 15 cm.

A) 84 B) 64
C) 44 D) 22
 
Answer & Explanation Answer: A) 84

Explanation:

Let a = 13, b = 14 and c = 15. Then, s=12a+b+c=21

 (s- a) = 8, (s - b) = 7 and (s - c) = 6.


Area =ss-as-bs-c =21×8×7×6 = 84 sq.cm   

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Q:

A room is half as long again as it is broad. The cost of carpeting the at Rs. 5 per sq.m is Rs. 270 and the cost of papering the four walls at Rs. 10 per sq.m is Rs. 1720. If a door and 2 windows occupy 8 sq. m, find the dimensions of the room.

A) b=6; l=18; H=6 B) b=5; l=6; H=18
C) l=6; b=18; H=15 D) l=5; b=18; H=18
 
Answer & Explanation Answer: A) b=6; l=18; H=6

Explanation:

Let breadth = x metres, length = 3x metres, height = H metres. 

Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m 

x×3x2=54x2=54×2x=6  

So, breadth = 6 m and length =362 = 9 m.

Now, papered area = (1720/10) =  172 sq.m 

Area of 1 door and 2 windows = 8 sq.m 

Total area of 4 walls = (172 + 8) sq.m = 180 sq.m 

2×9+6H=180H=6

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Q:

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

A) 20 B) 30
C) 40 D) 50
 
Answer & Explanation Answer: D) 50

Explanation:

Let x and y be the length and breadth of the rectangle respectively.
Then, x - 4 = y + 3 or x - y = 7 ----(i)
Area of the rectangle =xy; Area of the square = (x - 4) (y + 3)
(x - 4) (y + 3) =xy <=> 3x - 4y = 12 ----(ii)
Solving (i) and (ii), we get x = 16 and y = 9.
Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

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Q:

If each side of a square is increased by 25%, find the percentage change in its area.

A) 56.25% B) 36.25%
C) 16.25% D) 12.25%
 
Answer & Explanation Answer: A) 56.25%

Explanation:

Let each side of the square be a. Then, area = .a2
New side =125a100=5a4. New area = 5a42 = 25a216

 

Increase in area = 25a216-a2=9a216
Increase% = 9a216*1a2*100 % = 56.25%.

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Q:

In how many different ways  can 3 students be associated with 4 chartered accountants,assuming that each chartered accountant can take at most one student?

A) 12 B) 36
C) 24 D) 16
 
Answer & Explanation Answer: C) 24

Explanation:

Number of permutations = 4P3 = 24

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