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We have introduced "the 17 days study plan to crack IPPB Scale-1 Prelims 2017". This plan is specifically dedicated to Quantitative Aptitude and cover all types of questions asked in Bank PO exams. This will help you to prepare in a more efficient and systematic way. So, follow this post to keep updated with the plan.
A. 14
B. 21
C. 28
D. 35
E. 7
Sol. Let length and breadth are 7a, 3a
Perimeter = 20a
Circumference of circle =20a+8
Radius of circle =(1/2)×7a=3.5a
2a+8=(44/7)×(7a/2) , a=4
Thus, Length = 28cm.
2. A hemispherical bowl of internal diameter 54 cm contains a liquid. The liquid is to be filled in cylindrical bottles of radius 3 cm and height 9 cm. How many bottles are required to empty the bowl ?
A. 221
B. 343
C. 81
D. 243
E. None of these
Sol. Sol. Volume of hemi-sphere = (2/3)×π×(27)³
Volume of cylinder bottle = πr²h = π×(3)²×9
Required No. of Bottles = 2×3×27 = 162 bottles.
3. The perimeter of the rectangular field is 240 m. The ratio of length and breadth is 8:7. Find the area of the rectangle ?
A. 3854
B. 3584
C. 3485
D. 3845
E. None of these
Sol. Perimeter of the rectangle = 2×(l + b) = 240
Also given, l/b = 8/7, using this in 1st eqn,
8b/7 + b = 120
b = 56
l = 64
Area of the rectangle = 56*64 = 3584 sq.m.
4. The height of a right circular cone is decreased by 20%. By what percent should the diameter of the base be decreased so that the volume of the right circular cone is decreased by 35.2%?
A. 6.66%
B. 9%
C. 10%
D. 11.11%
E. None of these
Sol. Suppose the original radius and height are 10 each.
Since (1/3)p is a constant, we can simple work with r^2h and say that the original volume is 1000.
The new height is 8 and the new volume is 648.
If the new radius is R, then R^2 = 648/8 = 81 ? R = 9.
Thus the radius and therefore the diameter is reduced by 10%.
5. A cube of side 8 feet is to be painted from outside. The cost of paint is Rs 25 per kg. If 48 square feet are covered by 1 kg, then what is the cost of painting the cube ?
A. Rs 200
B. Rs 250
C. Rs 280
D. RS 230
E. None of these.
Sol. Surface area of cube = 6a² = 6×8² = 384 sq. feet
So quantity of paint required = (384/48) = 8 kg
So cost of painting = 25×8 = Rs 200.
6. If the length of a rectangular field is doubled and its breadth is halved(i.e. reduced by 50%). What is percentage change in its area ?
A. 0%
B. 10%
C. 25%
D. 33.33%
E. None of these
Sol. Let the length is 'l' and breadth is 'b'
Area = l×b
New area = 2l×0.5b = lb
Change in area = 0%.
7. If the wheel of a bicycle makes 560 revolutions in travelling 1.1 km, what is its radius?
(a) 31.25 cm
(b) 37.75 cm
(c) 35.15 cm
(d) 11.25 cm
(e) none of these
Sol. Perimeter = 1.1×1000/560
2×22/7×r = 1.1×100/56
r = 110×7/(56×22×2) = 5/16
= 31.25 cm.
8. A cistern 6 m long and 4 m wide contains water up to a breadth of 1 m 25 cm. Find the total area of the wet surface.
(a)42 m sqaure
(b)49 m sqaure
(c)52 m sqaure
(d)64 m square
(e)none of these
Sol.Total Surface Area of wet surface
= 2 (l + B) × h + lb
= 2 (6 + 4) 1.25 + 6 × 4
= 20 × 1.25 + 24
= 25 + 24
= 49 m square
9. The smallest side of a right-angled triangle is 8 cm less than the side of a square of perimeter 56 cm, The second largest side of the right-angled triangle is 4 cm less than the length of the rectangle of area 96 sq cm and breadth 8 cm. What is the largest side of the right-angled triangle ?
A. 20 cm
B. 15 cm
C. 10 cm
D. 25 cm
E. None of these
Sol. Smallest side of a right-angled triangle = 1/4 (Perimeter of Square) – 8 = 1/4 (56) – 8 = 14 – 8 = 6
second largest side of the right-angled triangle = 96/8 – 4 = 8
largest side of the right-angled triangle = sq.root of (6 * 6 + 8* 8) = 10 cm
10. The ratio of the adjacent angles of a parallelogram is 7 : 8. Also, the ratio of the angles of quadrilateral is 5 : 6 : 7 : 12. What is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral ?
A. 168
B. 188
C. 224
D. 216
E. None of these
Sol. Sum of the adjacent angles of Parallelogram = 180°
Smaller angle of Parallelogram = 7/15 * 180 = 84°
Second largest angle of the quadrilateral = 7/30 * 360 = 84°
Sum = 84° + 84° = 168°
11. The largest and the second largest angles of a triangle are in the ratio of 7 : 3 respectively. The smallest angle is 20% of the sum of the largest and the second largest and the second largest angles. What is the sum of the smallest and the second largest angles ?
A. 80
B. 75
C. 90
D. 85
E. None of these
Sol. Sum of 3 angles of triangle is 180
= (7x + 3x)*20/100
= 2x
Thus, 7x + 3x + 2x = 180
x =15
Sum = 5x = 5*15 = 75.
12. The angles of a quadrilateral are in the ratio of 3 : 5 : 11 : 1. The smallest angle of the quadrilateral is equal to the smallest angle of the triangle. The ratio of smallest and largest angle of triangle is 1:5. What is the second largest angle of the triangle?
A. 92
B. 72
C. 62
D. 82
E. None of these
Sol. 3x + 5x + 11x + x = 360
x = 18
Smallest angle of triangle = 18
x = 18 ; 5x = 90
Second largest angle of a triangle = 180 - (18+90) = 72.
13. The base and the other side of an isosceles triangle is 8cm and 12cm. Find its triangle ?
A)45.2 cm2
B)54.8 cm2
C)40.5 cm2
D)38.7 cm²
E)None of these
Sol. Area = b/4√ (4a² –b²)
= 8/4√ (4×144 – 64)
= 2(22.6) = 45.2cm²
14. If the sides of a equilateral triangle is increased by 10% , 30%and 60%then a new triangle is formed. By what % perimeter of the triangle is increased ?
A)40.50%
B)32.45%
C)33.33%
D)35.67%
E)None of these
Sol. Let a = side of the triangle
Perimeter = 3a
New perimeter = a×110/100 + a×130/100 + a×140/100
= a(11+13+16)/10 = 4a
% increased in perimeter = (4a – 3a/3a)×100
= 100/3 = 33.33%
15. A circle is drawn circumscribing a rectangle of sides 8cm and 6cm, find the area of the circle
A)78.5cm2
B)82.7cm2
C)65.4cm2
D)54.9cm2
E)None of these
Sol. Diagonal of rectangle = Diameter of circle
Diameter = √ (64+36) = 10
Radius = 5cm
Area of circle = 3.14×5×5 = 78.5cm².
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