Dear Readers, Important Practice Aptitude Questions for NICL AO & RBI Grade B 2017 Exams was given here with Solutions. Aspirants those who are preparing for Bank exams & all other Competitive examination can use this material.
1.A rectangular shaped pit of length 10 ft and breath 8 ft is made in a rectangular field of length 32 ft and breath 20 ft. The earth dug out of the pit is uniformly spread on the remaining area of the field. If the increase in level of the remaining area is 1.25 ft then what will be the depth of the pit?
1.A rectangular shaped pit of length 10 ft and breath 8 ft is made in a rectangular field of length 32 ft and breath 20 ft. The earth dug out of the pit is uniformly spread on the remaining area of the field. If the increase in level of the remaining area is 1.25 ft then what will be the depth of the pit?
1). Area of the rectangular field = 32 × 20 = 640 sq. ft
Area of the rectangular pit = 8 × 10 = 80 sq. ft
∴ The area of remaining field = 640 – 80 = 560 sq. ft
Hence, Volume of the pit = Area of remaining field × Increased level
l × b × h = 560 × 1.25
10 × 8 × h = 700
∴ h = 700/80 = 8.75 ft
Hence, the depth of the pit is 8.75 ft
Answer: C
2.Rectangles of length 4 cm and breath 2 cm each are cut from each of the corners of a rectangular metal sheet of length 22 cm and breadth 16 cm. What will be the perimeter of the remaining portion of the metal sheet?Area of the rectangular pit = 8 × 10 = 80 sq. ft
∴ The area of remaining field = 640 – 80 = 560 sq. ft
Hence, Volume of the pit = Area of remaining field × Increased level
l × b × h = 560 × 1.25
10 × 8 × h = 700
∴ h = 700/80 = 8.75 ft
Hence, the depth of the pit is 8.75 ft
Answer: C
2).
The perimeter of the remaining portion of the metal sheet is,
= 4(4) + 4 (2)+ 2(14)+ 2(12) = 16 +8 + 28 + 24 = 76 cm
Answer: A
= 4(4) + 4 (2)+ 2(14)+ 2(12) = 16 +8 + 28 + 24 = 76 cm
Answer: A
3.The diameter of the circular ground is one- fourth of the area of the rectangle.
A fence is to be drawn around a circular ground at a cost of 130 per meter. What will be the total cost of fencing the ground, if the area of the rectangle is 56 m?
3).The diameter of the circular ground = (1/4) × 56 = 14 m
∴ the radius of the circular ground = 7 m
The circumference of the circular ground = 2πr
= 2 × 22/7 × 7 = 44 m
∴ the cost of fencing the circular ground = 130× 44 = 5720
Answer: C
∴ the radius of the circular ground = 7 m
The circumference of the circular ground = 2πr
= 2 × 22/7 × 7 = 44 m
∴ the cost of fencing the circular ground = 130× 44 = 5720
Answer: C
4.Find the area of the shaded area in the given diagram. If the wide of the shaded area is 2 m.
4).The Area of the outer rectangle = 28 × 22 = 616 sq. m
The Area of the inner rectangle= (28-(2+2)) × (22-(2+2))
= (28-4) × (22-4) = 24 × 18 = 432 sq. m
Hence,
The Area of the shaded portion= Area of outer rectangle – Area of inner rectangle
= 616 – 432 = 184 sq. m
Answer: D
5.All the four lateral walls and the ceiling of a room of length 16 ft, breadth 12 ft and height 10 ft are to be painted. The total cost of painting is Rs.12,636. If the cost of painting is Rs. 18 per sq ft then what is the area occupied by the doors and windows?The Area of the inner rectangle= (28-(2+2)) × (22-(2+2))
= (28-4) × (22-4) = 24 × 18 = 432 sq. m
Hence,
The Area of the shaded portion= Area of outer rectangle – Area of inner rectangle
= 616 – 432 = 184 sq. m
Answer: D
5).The Lateral Surface Area (LSA) = 2(l +b) ×h = 2 (16+12) × 10 = 560
Area of the ceiling = (l × b) = 16 × 12 = 192
Total cost = LSA + Area of ceiling – (Windows and door occupied area= x )
12636 = (560 +192) × 18 – x × 18
Hence, x= (13536 – 12636)/ 18 = 50 sq ft
∴ the area occupied by the doors and windows = 50 sq. ft
Answer: A
Area of the ceiling = (l × b) = 16 × 12 = 192
Total cost = LSA + Area of ceiling – (Windows and door occupied area= x )
12636 = (560 +192) × 18 – x × 18
Hence, x= (13536 – 12636)/ 18 = 50 sq ft
∴ the area occupied by the doors and windows = 50 sq. ft
Answer: A
10 cm
6).The radius of the larger sphere is 20/2 = 10 cm
Let n be the number of small balls
Volume of larger sphere = n × Volume of small balls
4/3 × π × R3 = n × 4/3 × π × r3
4/3 × π × 103 = 8 × 4/3 × π × r3
r3 = 1000/8 = 125
∴ r = 5 cm
Answer: C
Let n be the number of small balls
Volume of larger sphere = n × Volume of small balls
4/3 × π × R3 = n × 4/3 × π × r3
4/3 × π × 103 = 8 × 4/3 × π × r3
r3 = 1000/8 = 125
∴ r = 5 cm
Answer: C
7.If the length of a rectangle is increased by 10% and the breadth of the rectangle is decreased by 6%, then what is the change in the area?
7).Percentage change in area = [10 – 6 – (10× 6 /100)]
= 4 – 0.6 = 3.4%
∴ the area is increases 3.4%
Answer: A
8.The inner circumference of a circular race track 7 m wide is 176 m. what is the circumference of the outer circle?= 4 – 0.6 = 3.4%
∴ the area is increases 3.4%
Answer: A
8).Let, r and R be the inner and outer radius of the circle respectively,
Inner circumference = 2πr = 176
r = 176 × 7/22 × 1/2
r = 28 m
∴ The outer radius = 28 + 7 = 35 m
Hence, the circumference of the outer circle = 2πr = 2 × 22/7 × 35 = 220 m
Answer: C
9. Ratio of length and breadth of a rectangle is 4: 3 and the perimeter of the rectangle is 84 cm. What is the area of the rectangle?Inner circumference = 2πr = 176
r = 176 × 7/22 × 1/2
r = 28 m
∴ The outer radius = 28 + 7 = 35 m
Hence, the circumference of the outer circle = 2πr = 2 × 22/7 × 35 = 220 m
Answer: C
9).l : b = 4 : 3
2 (l+b) =84
l + b = 42
Length of the rectangle = 4/7 × 42 = 24
∴ Breadth of the rectangle = 42 – 24 = 18
∴ The area of the rectangle = 24 × 18 = 432 sq cm
Answer: A
2 (l+b) =84
l + b = 42
Length of the rectangle = 4/7 × 42 = 24
∴ Breadth of the rectangle = 42 – 24 = 18
∴ The area of the rectangle = 24 × 18 = 432 sq cm
Answer: A
10.The diagonal of a square is √648 cm. What is the perimeter of the square?
10).The diagonal (a√2) of a square is √648 =√(324 × 2) = 18 √2
a √2 = 18 √2
a =18
∴ The perimeter of the square is 4a = 4 × 18 = 72 cm
Answer: B
a √2 = 18 √2
a =18
∴ The perimeter of the square is 4a = 4 × 18 = 72 cm
Answer: B
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