Practice Quantitative Aptitude Questions For IBPS 2017 Exams (Mensuration):

Dear Readers, Important Practice Aptitude Questions for IBPS Exams 2017 was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this material.

2).Let the length and breadth be x and y respectively.
2 × 4 [(x + y) + (2 × 4)] = 1104
x + y + 8 = 1104 / 8 = 138
x + y = 130
Answer: C

3).Here the edge of an ice cube is 14 cm.
Radius of the cylinder = 14/2 = 7 cm
Height of the cylinder = 14 cm
Volume of the largest cylinder = πr^2h
= 22/7 × 7 × 7 ×14
= 2156 cube cm.
Answer: C

4.If the length of a rectangle is increased by 230%, its area becomes 828 sq cm and perimeter 162 cm. What is the perimeter of the original rectangle?

4).Length of the new rectangle
= √((162)^2)/16 ×828 + 162/4 = (114/4+ 162/4) = 69
Breadth = 162/9 – 69=12 cm
Length of the original rectangle
= 69 × 100/230=30 cm
Breadth is same, ie 12 cm
Perimeter = 2(30 + 12) = 84 cm.
Answer: B

5.What will be the cost of fencing a circular field whose area is 1386 sq m? The cost of fencing is 5 per square metre.

5).Area of the circle = πr^2
Now, πr^2 = 1386
or, r^2 = 1387 ×7/22=441
r = √441 = 21 cm
Circumference of the circle = 2πr
= 2 × 22/7 × 21 = 132 cm
Hence, the cost of fencing = 132 × 5 = 660.
Answer: A

6.A horse is tethered to a peg with a 14 metre long rope at the corner of a 40 metre long and 24 metre wide rectangular grass-field. What area of the field will the horse graze?

7. The cost of fencing a circular plot at the rate of 15 per m is 3300. What will be the cost of flooring the plot at the rate of 100 per sq m?

7).Circumference of circular plot = 3300/15 = 220
2πr = 220
r = 220 / (2 × 22) × 7
= 55 × 7/11 = 35 m
Total cost of flooring the plot = πr^2 × 100
= 22/7 × 35 × 35 × 100
= 385000.
Answer: A

8. Area of a rectangle is equal to the area of the circle whose radius is 21 cms. If the length and the breadth of the rectangle are in the ratio of 14 : 11 respectively, what is its perimeter ?

8).Area of rectangle = Area of circle
= 22/7 × 21 × 21 = 1386 sq. cm.
Let the length and breadth of rectangle be 14x and 11x cm respectively. Then 14x × 11x = 1386
x^2 = 1386/14 × 11 = 9
x = √9 = 3
Perimeter of rectangle = 2 (14x + 11x) = 50x
= 50 × 3 = 150 cm.
Answer: D

9.Area of a rectangular field is 3584 m^2 and the length and the breadth are in the ratio 7 : 2 respectively. What is the perimeter of the rectangle?

9). Let the common ratio be x
Length of the rectangular field = 7x m
Breadth of the rectangular field = 2x m
According to the question,
7x × 2x = 3584
x = 16
Required perimeter = 2 (7 × 16 + 2 × 16)
= 2 (112 + 32) = 2 × 144 = 288 m.
Answer: C

10.The width of a rectangular park is 10/21 of its length. If the area of the park is 3360 sq m, then what is the difference between the length and the width of the park?

10).Let the length of the park be x metres.
Width = x × 10 / 21 = 10x / 21
Area = x × 10x / 21 = 3360
or, x^2 = 3360 × 21 / 10 = 7056
or, x = 84
Hence, length = x = 84m and width = 84 × 10 / 21 = 40 metres
Difference = 84 – 40 = 44 metres.
Answer: B

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Practice Questions on Quantitative Aptitude section For IBPS 2017 Exams (Mensuration)

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