Aptitude

Showing 1 - 10 of 13 questions

1

If each edge of a cube is increased by 50%, find the percentage increase in its surface area.

a

125%

b

150%

c

175%

d

110%

Let the edge = a cm

So increase by 50% = a+a/2 = 3a/2

Total surface Area of original cube = 6a^2

TSA of new cube = 6(3a/2)^2

6 xx 9a^2/4 = 13.5a^2

Increase in area = 13.5a^2 - 6a^2 = 7.5a^2

7.5a^2 Increase % = (7.5a^2)/(6a^2) xx 100 = 125%

2

The diagonal of a rectangle is $\inline \sqrt{41}$ cm and its area is 20 sq. cm. The

perimeter of the rectangle must be:

a

18

b

28

c

38

d

48

sqrt(l^2+b^2) = sqrt(41) (or) l^2+b^2 = 41

Also, lb = 20

(l+b)^2 = l^2 + b^2 + 2lb

= 41 + 40 = 81

(l + b) = 9

Perimeter = 2(l + b) = 18 cm

3

An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square?

a

4.04 %

b

2.02 %

c

4 %

d

2 %

Error = 2% while measuring the side of a square.

Let the correct value of the side of the square = 100
Then the measured value = 100 + (2% of 100) = 100 +2 = 102 (∵ error 2% in excess)

Correct Value of the area of the square = 100 × 100 = 10000

Calculated Value of the area of the square = 102 × 102 = 10404

Error = 10404 - 10000 = 404

Percentage Error=4.04%
ErrorActual Value×100=40410000×100=4.04%

4

A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area?

a

30 %

b

28 %

c

32 %

d

26 %

Solution 1:
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Lost 20% of length
=> New length = Original length - (20% of Original Length)
100- (20% of 100) = 100-20 = 80

= 100 - (10% of 100) = 100-10 = 90

New area = 80 × 90 = 7200

Decrease in area
= Original Area - New Area
= 10000 - 7200 = 2800

Percentage of decrease in area = (2800/10000) * 100 = 28%

5

The length of a room is 5.5 m and width is 3.75 m. What is the cost of paying the floor by slabs at the rate of Rs. 800 per sq. metre.

a

Rs.12000

b

Rs.16000

c

Rs.18000

d

Rs.16500

Area = 5.5 × 3.75 sq. metre.
Cost for 1 sq. metre. = Rs. 800
Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500

6

The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle?

a

18 cm

b

16 cm

c

40 cm

d

20 cm

Then length = 2x cm
Area = lb = x × 2x = 2x2

New length = (2x - 5)
New breadth = (x + 5)
New Area = lb = (2x - 5)(x + 5)

But given that new area = initial area + 75 sq.cm.
=> (2x - 5)(x + 5) = 2x2 + 75
=> 2x2 + 10x - 5x - 25 = 2x2 + 75
=> 5x - 25 = 75
=> 5x = 75 + 25 = 100
=> x = 1005 = 20 cm

Length = 2x = 2 × 20 = 40cm

7

If the difference between the length and breadth of a rectangle is 23 m and its perimeter is 206 m, what is its area?

a

2800 m2

b

2740 m2

c

2520 m2

d

2200 m2

l - b = 23 ....(Equation 1)

perimeter = 2(l + b) = 206
=> l + b = 103 ....(Equation 2)

(Equation 1) + (Equation 2) => 2l = 23 + 103 = 126
=> l = 1262 = 63 metre

Substituting this value of l in (Equation 1), we get
63 - b = 23
=> b = 63 - 23 = 40 metre

Area = lb = 63 × 40 = 2520 m2

8

What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

a

814

b

802

c

836

d

900

l = 15 m 17 cm = 1517 cm
b = 9 m 2 cm = 902 cm
Area = 1517 × 902 cm2

Now we need to find out HCF(Highest Common Factor) of 1517 and 902.

Hence, HCF of 1517 and 902 = 41

Hence, side length of largest square tile we can take = 41 cm
Area of each square tile = 41 × 41 "cm"^2

Number of tiles required = (1517 * 902)/(41 * 41) = 37 * 22 = 814

9

The length of a room is 5.5 m and width is 3.75 m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq. metre.

a

Rs.15,600

b

Rs.15,550

c

Rs.15,000

d

Rs.16,500

Area of the floor = (5.5 x 3.75) m2 = 20.625 m2.

Cost of paving = Rs. (800 x 20.625) = Rs.16500.

10

A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, then what is the area of the parking space in square feet?

a

81

b

126

c

46

d

252

Clearly, we have : length(l) = 9 ft.

and l+2b = 37 ft.

i.e., 9+2b = 37

2b = 37-9

2b = 28

So, the breadth(b) = 14 ft.

Area = (l x b) = (9 x 14) sq.ft. =126 sq.ft.