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__SBI Clerk Prelims 2016- Practice Aptitude Questions (Mensuration) Set-41__
Dear Readers, Important Practice Aptitude Questions with solution for Upcoming SBI Clerk Exam, candidates those who are preparing for those exams can use this practice questions.

**1).**The area of a square is 196 sq cm whose side is half of the radius of a circle. The circumference of the circle is equal to breadth of a rectangle, if perimeter of rectangle is 712 cm. What is the length of rectangle?

a)
196 cm

b)
186 cm

c)
180 cm

d)
190 cm

e)
None of these

**2).**What is the area of a circle whose circumference 132 cm?

a)
1562 sq cm

b)
1386 sq cm

c)
1434 sq cm

d)
1836 sq cm

e)
None of these

**3).**If the radius of a circle is increased by 50%, then its circumference will increase by

a)
25%

b)
50%

c)
75%

d)
100%

e)
None of these

**4).**The area of a square is 1024 sq cm. What is respective ratio between the length and the breadth of a rectangle whose length is twice the side of square and breadth is 12 cm less than the side of square?

a)
5 : 18

b)
16 : 7

c)
14 : 5

d)
32 : 5

e)
None of these

**5).**The angles of a quadrilateral are in the ratios 2 : 4 : 7 : 5. The smallest angle of the quadrilateral equal to the smallest angle of a triangle. One of the angles of the triangle is twice the smallest angle of triangle. What is the second largest angle of triangle?

a)
80

^{°}
b)
60

^{°}
c)
120

^{°}
d)
Cannot be determined

e)
None of these

**6).**The smallest angle of a triangle is equal to two- third the smallest angle of a quadrilateral. The ratio between the angles of the quadrilateral is 3 : 4 : 5 : 6. The largest angle of the triangle is twice it’s the smallest angle. What is the sum of second largest angle of the triangle and largest of the quardrilateral?

a)
160

^{°}
b)
180

^{°}
c)
190

^{°}
d)
170

^{°}
e)
None of these

**7).**The ratio between the adjacent angles of a parallelogram are 2 : 3. The smallest angle of the quadrilateral is equal to the half of the smallest angle of a parallelogram. The highest angle of a quadrilateral times greater than its smallest angle. What is the sum of the highest angle of quadrilateral and the smallest angles of a parallelogram?

a)
252

^{°}
b)
226

^{°}
c)
144

^{°}
d)
180

^{°}
e)
None of these

**8).**One of the angles of a triangle is two – third angle of sum of adjacent angles of parallelogram. Remaining angles of the triangle are in ratio 5 : 7, respectively. What is the value of the second largest angle of the triangle?

a)
25

^{°}
b)
40

^{°}
c)
35

^{°}
d)
Cannot be determined

e)
None of these

**9).**The ratio between the angles of a quadrilateral is 3 : 4 : 6 : 5. Two – third the largest angle of the quadrilateral is equal to the smaller angle of a parallelogram. What is the value of adjacent angle of the parallelogram?

a)
120

^{°}
b)
110

^{°}
c)
100

^{°}
d)
130

^{°}
e)
None of these

**10).**The total area of a circle and a rectangle is equal to 1166 sq cm. The diameter of the circle is 28 cm. What is the sum of the circumference of the circle and the perimeter of the rectangle, if the length of the rectangle is 25 cm?

a)
186cm

b)
182 cm

c)
184 cm

d)
Cannot be determined

e)
None of these

__Solution:__**1)**. Area of square (a)

^{2}= 196

A=√(196)
= 14cm

Radius
of a circle = 14 ᵡ 2 = 28cm

Circumference

=(22/7)
× 2 × 28 = 176cm

Now,
according to the question,

b=176cm

Also,
2(l + b) =712

2(l
+176) =712

l
+ 176 = 336

I
= 336 – 176 à/=180cm

**Answer: c)**

**2)**. Circumference of a circle = 132cm

2πr
= 132 × (22/7) × r =132

r=(132×7)/
(2 ×22) àr = 21cm

Hence,
area of a circle = πr

^{2}
=(
22/7) × 21 × 21 = 1386 sq cm

**Answer: b)**

**3).**Let the original radius be r.

New
radius =150% of r = (3r/2)

Original
circumference = 2πr

New
circumference

=
2π × (3r/2) = 3πr

Increase
percentage

{[(3πr
- 2πr) /2πr] × 100} = 50%

or
increase in circumference will be same as increase in radius

r
= 50%

**Answer: b)**

**4)**. Area of square =1024cm

Side
= √(1024) = 32cm

Length
of rectangle = 2 × 32 = 64cm

Breadth
=32 – 12 =20cm

Ratio
= 64 : 20 =16:5

**Answer: e)**

**5).**Sum of angles of quadrilateral =360

^{°}

2x
+ 4x +7x + 5x =360

^{°}
18x
= 360

^{° }; x=20^{°}
Smallest
angles of quadrilateral = 2 × 20

^{° }= 40^{°}
Smallest
angle of triangle = 40

^{°}
Second
angle of triangle =2 × 40

^{°}= 80^{°}
Third
angle of triangle =180

^{°}– (40^{°}+80^{°}) =60^{°}
Hence,
the second largest angle of triangle is 60

^{°.}**Answer: b)**

**6)**. Let the angles of the quadrilateral be 3x, 4x, 5x and 6x, respectively.

Then,
3x +4x +5x +6x =360

^{°}
18x=360

^{°}à x=20^{°}
Smallest
angle of the triangle = (3 × 20

^{°}) × (2/3) = 40^{°}
Largest
angle of the triangle =40

^{0}× 2 =80^{°}
Second
largest angle of the triangle =180

^{°}– (40^{°}+80^{°}) = 60^{°}
and
the largest angle of the quadrilateral =6x = 6 × 20

^{°}=120^{°}
Hence,
required sum =60

^{°}+120^{°}=180^{°}**Answer: b)**

**7).**Suppose adjacent angle of parallelogram be 2x

^{°}and 3x

^{°}.

Then,
according to the theorem,

2x

^{°}+ 3x^{° }=180^{°}
=5x

^{°}= 180^{°}àx^{° }=180^{°}/5 =36^{°}
Smaller
angle of parallelogram = 2 × 36

^{°}= 72^{°}
Smaller
angle of quadrilateral =36

^{°}
Highest
angle = 4 × 36

^{°}= 144^{°}
Hence,
required sum of angles = 144

^{°}+36^{°}=180^{°}**Answer: d)**

**8).**An angle of a triangle = 2/3 ᵡ 180

^{°}=120

^{°}

Remaining
=180

^{°}– 120^{°}=60^{°}
Which
is the ratio of 5:7

So,
5x + 7x = 60

^{°}; 12x =60^{°}
x
= 5

^{°}
So,
angles are 5 × 5 =25

^{°}
And
7 × 5 = 35

^{°}and 120^{°}
So,
value of the second largest angle of triangle is 35

^{°}.**Answer: c)**

**9).**3x + 4x +6x +5x = 360

^{0 },x = 20

^{°}

Largest
angle of quadrilateral = 6x = 6 × 20

^{°}=120^{°}
Smaller
angle of parallelogram =120

^{°}× (2/3) =80^{°}
So,
the adjacent angle =100

^{°}**Answer: c)**

**10).**The area of the circle, πr

^{2}= (22/7) × 14 ×14 = 616sq cm

Let
the length and breadth of the rectangle be x cm and y cm, respectively.

Then,
x × y =1166 -616

=
25 × y =550

=
y=550/25 =22cm

Perimeter
of the rectangle =2(x +y)

=
2(25 +22) =94cm

And
circumference of the circle =2 πr

2
× 22/7 ᵡ 14 =88cm

Hence,
required sum =94 + 88 =182cm

**Answer: b)**

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