Quantitative Aptitude Questions (Mensuration) for SBI PO 2018 Day-131

Dear Readers, SBI is conducting Online Examination for the recruitment of probationary officer. To enrich your preparation here we have providing new series of Mensuration – Quantitative Aptitude Questions. Candidates those who are appearing in SBI PO Exam can practice these Quantitative Aptitude average questions daily and make your preparation effective.

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1. The total area of a Square and rectangle is 496 Sq cm. The side of the square is 14 cm. What is the sum of the perimeter of square and rectangle, if length of the rectangle is 25 cm?

1. 120 cm
2. 140 cm
3. 130 cm
4. 160 cm
5. None of these

2. The perimeter of the rectangular field is 2 times the perimeter of square field. If the side of the square field is 18 m and the breadth of the rectangular field is 45 m, then find the area of the rectangular field?

1. 1215 Sq m
2. 1336 Sq m
3. 1050 Sq m
4. 1152 Sq m
5. None of these

3. The cost of fencing a square plot at the rate of Rs. 15 per metre is Rs. 3780. What will be the cost of flooring the plot at the rate of Rs. 3 per square meter?

1. 9856
2. 10548
3. 12563
4. 11907
5. None of these

4. The circumference of a circle is equal to the perimeter of a square whose area is 4356 sq cm. What is the area of the circle?

1. 4678 Sq cm
2. 5544 Sq cm
3. 5126 Sq cm
4. 4282 Sq cm
5. None of these

5. Find the volume of a cylinder whose radius is one – third of the radius of a circle having area 1386 cm². Height of the cylinder is double its radius?

1. 2584 cm³
2. 3562 cm³
3. 2156 cm³
4. 3245 cm³
5. None of these

6. The side of the equilateral triangle is equal to the diameter of the circle. The area of the equilateral triangle is 196√3 Sq cm. Find the circumference of the circle?

1. 92 cm
2. 88 cm
3. 74 cm
4. 66 cm
5. None of these

7. If the breadth of a parallelogram is increased by 30% while the height of the parallelogram is decreased by 20% then find percentage change in area of the parallelogram?

1. 8 % increased
2. 5 % decreased
3. 9 % decreased
4. 4 % increased
5. None of these

8. The circumference of a circle is half of the perimeter of a rectangle. The area of the circle is 2464 sq. m. What is the area of the rectangle if the breadth of the rectangle is 80 m?

1. 7680 Sq m
2. 5572 Sq m
3. 6128 Sq m
4. 7054 Sq m
5. None of these

9. The radius of the circle is equal to two-fifth of the side of the square. The area of the circle is 616 sq cm. Find the perimeter of a square?

1. 152 cm
2. 166 cm
3. 140 cm
4. 178 cm
5. None of these

10. A Circular path is surrounding the circular plot is being graveled at a total cost of Rs. 3080 at Rs. 4 per square meter. Find the width of the path, if the radius of the circle is 14 m?

1. 12 meter
2. 7 meter
3. 15 meter
4. 20 meter
5. None of these

 

Answers:

1.Answer c

a² + lb = 496

14² + 25b = 496

196 + 25b = 496

25b = 496 – 196

25b = 300

B = 12 cm

Perimeter of square = 4a = 4*14 = 56 cm

Perimeter of rectangle = 2*(l + b) = 2*(25 + 12)

= > 2 * 37 = 74 cm

Required sum = 56 + 74 = 130 cm

2.Answer a

Side of square (a) = 18 m

Perimeter of rectangle = 2*4a = 2*4*18 = 144

= > 2(l + 45) = 144

= > l + 45 = 72

= > l = 72 – 45 = 27 m

Area of rectangle = lb = 45*27 = 1215 Sq m.

3.Answer d

Perimeter of square = 3780/15 = 252

4a = 252

Side (a) = 252/4 = 63

Area of square = a² = 63² = 3969

Cost of flooring the plot = 3979*3 = Rs. 11907

4.Answer b

The circumference of a circle = Perimeter of a Square

Area of Square = a² = 4356

Side (a) = 66 cm

Perimeter of square = 4a = 66*4 = 264 cm

According to the question,

2πr = 264

2*(22/7)*r = 264

r = 42

Area of the circle= πr² = (22/7)*42*42 = 5544 Sq cm

5.Answer c

πr² = 1386

=> (22/7)* r² = 1386

=> r² = 1386 x (7/22)

=> r² = 441

=> r = 21 cm

Radius of the cylinder = 21*(1/3) = 7 cm

Height of the cylinder = 7*2 = 14 cm

Volume of the cylinder = πr²h = (22/7)*7*7*14 = 2156 cm³

6.Answer b

The area of the equilateral triangle = 196√3 Sq cm

The area of the equilateral triangle = (√3/4)*a²

(√3/4)*a² = 196√3

a² = 196*4

Side (a) = 14*2 = 28 cm

The diameter of the circle = 28 cm

Radius (r) = 14 cm

Circumference of the circle = 2πr = 2*(22/7)*14 = 88 cm

7.Answer d

Let the breadth and height of the parallelogram is 10 cm and 10cm,

Normal area= 10*10= 100

New length= 10*130/100= 13

New height= 10*80/100= 8

New area= 13*8= 104

Required percentage = [(104 – 100)/100]*100 = 4 % increased

8.Answer a

Area of a circle= πr²

2464 = 22r²/7

2464*(7/22) = r²

r² = 784

r = 28 m

Circumference = 2*22/7 * 28 = 176 sq. m

Perimeter of the rectangle = 2*176 = 352 sq. m

352 = 2(l + 80)

176 = l + 80

l = 176 – 80

l = 96

Area of the rectangle = 96*80 = 7680 sq. m

9.Answer c

Radius = (2/5)*side

Area of circle = πr ² = 616

(22/7)*r² = 616

r² = 616*(7/22) = 196

Radius (r) = 14 cm

Side = Radius*(5/2) = 14*(5/2) = 35 cm

Perimeter of the square = 4a

= > 35*4

= > 140 cm

10.Answer b

Radius of the circular plot = 14 m

Area of the circular path = 3080/4 = 770

Area of the path = π(r + x)² – πr² (Here x is the width of the path)

= > (22/7) [(14+x)² – 14²]

= > (22/7) [196 + 28x + x² – 196]

= > (22/7) [28x + x²]

(22/7) [28x + x²] = 770

28x + x² = 770*(7/22)

28x + x² = 245

X² + 28x – 245 = 0

(x+35) (x – 7) =0

X = 7 meter

 

Daily Practice Test Schedule | Good Luck

Topic	Daily Publishing Time
Daily News Papers & Editorials	8.00 AM
Current Affairs Quiz	9.00 AM
Quantitative Aptitude “20-20”	11.00 AM
Vocabulary (Based on The Hindu)	12.00 PM
General Awareness “20-20”	1.00 PM
English Language “20-20”	2.00 PM
Reasoning Puzzles & Seating	4.00 PM
Daily Current Affairs Updates	5.00 PM
Data Interpretation / Application Sums (Topic Wise)	6.00 PM
Reasoning Ability “20-20”	7.00 PM
English Language (New Pattern Questions)	8.00 PM

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