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# Arithmetic Questions For IBPS PO Set-2 PDF

Download important Arithmetic Questions PDF based on previously asked questions in IBPS PO and other MBA Exams. Practice Arithmetic Question and Answers for IBPS PO Exam.

InstructionsInstructions: What will come in place of question mark (?) in the following questions?

Question 1:Â 3463 Ã— 295 – 18611 = ? + 5883

a)Â 997091

b)Â 997071

c)Â 997090

d)Â 999070

e)Â None of these

Question 2:Â (23.1)2 + (48.6)2 – (39.8)2 = ? + 1147.69

a)Â (13.6)2

b)Â $\sqrt{12.8}$

c)Â 163.84

d)Â 12.8

e)Â None of these

Question 3:Â $\frac{28}{65}$ x $\frac{195}{308}$ Ã· $\frac{39}{44}$ + $\frac{5}{26}$ = ?

a)Â $\frac{1}{3}$

b)Â $0.75$

c)Â 1$\frac{1}{2}$

d)Â $\frac{1}{2}$

e)Â None of these

Question 4:Â [(3$\sqrt{8}$+ $\sqrt{8}$ ) (8$\sqrt{8}$ + 7$\sqrt{8}$ )] – 98 = ?

a)Â 2$\sqrt{8}$

b)Â 8$\sqrt{8}$

c)Â $382$

d)Â $386$

e)Â None of these

Question 5:Â $\sqrt{11449}$x$\sqrt{6241}$ – (54)2 = $\sqrt{?}$ + (74)2

a)Â 3844

b)Â 3721

c)Â 3481

d)Â 3638

e)Â None of these

InstructionsInstructions: What approximate value should come in place of question mark (?) in the following questions? (Note: You are not expected to calculate the exact value.)

Question 6:Â 39.897% of 4331 + 58.779% of 5003 =?

a)Â 4300

b)Â 4500

c)Â 4700

d)Â 4900

e)Â 5100

Question 7:Â 43931.03 Ã· 2111.02 Ã— 401.04 =?

a)Â 8800

b)Â 7600

c)Â 7400

d)Â 9000

e)Â 8300

Question 8:Â $\sqrt{6354}$ x 34.999 = ?

a)Â 3000

b)Â 2800

c)Â 2500

d)Â 3300

e)Â 2600

Question 9:Â $\sqrt[3]{4663}$ + 349 = ? Ã· 21.003

a)Â 7600

b)Â 7650

c)Â 7860

d)Â 7560

e)Â 7680

Question 10:Â 59.88 Ã· 12.21 Ã— 6.35 =?

a)Â 10

b)Â 50

c)Â 30

d)Â 70

e)Â 90

InstructionsInstructions: What will come in place of the question mark (?) in the following questions?

Question 11:Â 4003Ã—77 – 21015 = ?Ã—116

a)Â 2477

b)Â 2478

c)Â 2467

d)Â 2476

e)Â None of these

Question 12:Â [(5$\sqrt{7}$ + $\sqrt{7}$ ) Ã— (4$\sqrt{7}$ + 8$\sqrt{7}$ )] – (19)2 = ?

a)Â 143

b)Â 72$\sqrt{7}$

c)Â 134

d)Â 70$\sqrt{7}$

e)Â None of these

Question 13:Â (4444Ã·40) + (645Ã·25) + (3991Ã·26) = ?

a)Â 280.4

b)Â 290.4

c)Â 295.4

d)Â 285.4

e)Â None of these

Question 14:Â $\sqrt{33124}$Ã—$\sqrt{2601}$ – (83)2 = (?)2 + (37)2

a)Â 37

b)Â 33

c)Â 34

d)Â 28

e)Â None of these

Question 15:Â 5$\frac{17}{37}$ Ã— 4$\frac{51}{52}$ Ã— 11$\frac{1}{7}$ + 2$\frac{3}{4}$ = ?

a)Â 303.75

b)Â 305.75

c)Â 303$\frac{3}{4}$

d)Â 305$\frac{1}{4}$

e)Â None of these

InstructionsInstructions: What approximate value should come in place of the question mark (?) in the following questions. (Note: You are not expected to calculate the exact value)

Question 16:Â 8787 Ã· 343 Ã— $\sqrt{50}$ = ?

a)Â 250

b)Â 140

c)Â 180

d)Â 100

e)Â 280

Question 17:Â $\sqrt[3]{54821}$Ã—(303Ã·8) = (?)2

a)Â 48

b)Â 38

c)Â 28

d)Â 18

e)Â 58

Question 18:Â $\frac{5}{8}$ of 4011.33 + $\frac{7}{10}$ of 3411.22 = ?

a)Â 4810

b)Â 4980

c)Â 4890

d)Â 4930

e)Â 4850

Question 19:Â 23% of 6783 + 57% of 8431 = ?

a)Â 6460

b)Â 6420

c)Â 6320

d)Â 6630

e)Â 6365.7

Question 20:Â 335.01Ã—244.99 Ã· 55 = ?

a)Â 1490

b)Â 1550

c)Â 1420

d)Â 1590

e)Â 1400

3463 Ã— 295 = 1021585
1021585 – 18611 = 1002974
1002974 – 5883 = 997091

(23.1)2 = 533.61
(48.6 )2 = 2361.96
(39.8)2 = 1584.04
533.61 + 2361.96 – 1584.04 = 1311.53
1311.53 – 1147.69 = 163.84

We solve the problem as per BODMAS rule
$\frac{195}{308}$ Ã· $\frac{39}{44}$ = $\frac{5}{7}$
$\frac{28}{65}$ x $\frac{5}{7}$ = $\frac{4}{13}$
$\frac{4}{13}$ + $\frac{5}{26}$ = $\frac{13}{26}$ = $\frac{1}{2}$

[ $\sqrt{8}$ (3 +1) x $\sqrt{8}$(8 + 7)] – 98
= [4$\sqrt{8}$ x 15 x $\sqrt{8}$ ] – 98
= [60 x $\sqrt{8}$] – 98
= 480 – 98 = 382

$\sqrt{11449}$ x $\sqrt{6241}$ – (54)2 – (74)2 = $\sqrt{?}$
$\sqrt{?}$ = [107 x 79] – 2916 – 5476
= 8453 – 2916 – 5476 = 61
$\sqrt{?}$= (61)2 = 3721

40 x $\frac{4330}{100}$ + 59 + $\frac{5000}{100}$
= 1732 + 2950
= 4682

$\frac{43931}{2111}$ x 401 = 8345.01 â‰ˆ 8300

After taking approximation equation will be

$\sqrt{6354}$ x 34.993 = 80 x 35 = 2800

$\sqrt[3]{4663}$ â‰ˆ 17
17 + 349 = ? Ã· 21
366 x 21 = ?
? = 7686 â‰ˆ 7680

After taking approximation equation will be 60 Ã· 12 x 6 = 30

4003 x 77 = 308231
308231 – 21015 = 287216
$\frac{287216}{116}$ = 2476

[(5$\sqrt{7}$ + $\sqrt{7}$ ) Ã— (4$\sqrt{7}$ + 8$\sqrt{7}$ )] – (19)2
= [6$\sqrt{7}$ x 12$\sqrt{7}$] – 192
= [72 x 7] – 192
= 143

$\frac{4444}{40}$ = 111.1

$\frac{645}{25}$ = 25.8

$\frac{3991}{26}$ = 153.5

Total = 111.1 + 25.8 + 153.5 = 290.4

$\sqrt{33124}$Ã—$\sqrt{2601}$ – (83)2
= 2393
2393 – 372
= 1024
$\sqrt{1024}$ = 32

5$\frac{17}{37}$ = $\frac{202}{37}$

4$\frac{51}{52}$ = $\frac{259}{52}$

11$\frac{1}{7}$ = $\frac{78}{7}$

2$\frac{3}{4}$ = $\frac{11}{4}$

$\frac{202}{37}$ x $\frac{259}{52}$ x $\frac{78}{7}$ = 303

303 + $\frac{11}{4}$ = 305.75

8787 Ã· 343 = 25.61
$\sqrt{50}$ = 7.07
25.61 x 7.07 = 181.06 â‰ˆ 180

303 Ã· 8 =37.875 â‰ˆ 38
$\sqrt[3]{54821}$ = 37.98 â‰ˆ 38
382

$\frac{5}{8}$ x 4011.33 = 2507.08
$\frac{7}{10}$ of 3411.22 = 2387.854
2507.08 + 2387.854 = 4894.934 â‰ˆ 4890