Simplification Practice Set 1

5 Steps - 3 Clicks

Simplification Practice Set 1

Introduction

Simplification Practice Set 1 is useful for the aspirants of several bank exams and competitive exams. The Simplification Practice Set 1 contains important questions from Simplification which match the pattern of Banking exams. Simplification Quiz Questions will consume more time. BODMAS technique is very useful to score maximum marks in Simplification concept related questions.

Quiz

1. $$\frac{4}{9}$$Ã— 1701 + $$\frac{2}{11}$$Ã— 1386

A. 180
B. 1080
C. 1008
D. 1800

Explanation: $$\frac{4}{9}$$Ã— 1701 + $$\frac{2}{11}$$Ã— 1386 = ?
? = (4 Ã— 189) + (2 Ã— 126)
? = 756 + 252 = 1008.
Hence, option C is correct.

2. 9 $$\frac{3}{7}$$ + 6 $$\frac{4}{7}$$ – ? = 14 $$\frac{4}{7}$$Ã— 1701

A. $$\frac{3}{7}$$
B. $$\frac{4}{7}$$
C. 2 $$\frac{4}{7}$$
D. 1$$\frac{3}{7}$$

Explanation: ? = 9 $$\frac{3}{7}$$ + 6 $$\frac{4}{7}$$ – ? = 14 $$\frac{4}{7}$$Ã— 1701
â‡’ ? = 9 + $$\frac{3}{7}$$ + 6 + $$\frac{4}{7}$$ – 14 – $$\frac{4}{7}$$
On solving the whole number part separately, we get

â‡’ ? = 1 + $$\frac{3}{7}$$ + $$\frac{4}{7}$$ – $$\frac{4}{7}$$
â‡’ ? = 1 + $$\frac{3}{7}$$ = 1$$\frac{3}{7}$$

3. $$\frac{4}{5}$$ x ? x $$\frac{3}{7}$$ = $$\frac{16}{105}$$

A. $$\frac{8}{9}$$
B. $$\frac{5}{7}$$
C. 2 $$\frac{4}{9}$$
D. 1$$\frac{3}{7}$$

Explanation: Let’s the number (?) be x, then

From the given equation,
$$\frac{4}{5}$$ x X x $$\frac{3}{7}$$ = $$\frac{16}{105}$$
$$\frac{12x }{35}$$ = $$\frac{16}{105}$$ â‡’ x = $$\frac{16 Ã— 35}{105 Ã— 12}$$
â‡’ x = $$\frac{4}{9}$$

4. $${5}^{2}$$ + $${13}^{2}$$ – $${11}^{2}$$ = $${?}^{3}$$ â€“ 52

A. 6
B. 3
C. 4
D. 5

Explanation: From the given equation

â‡’ 25 + 169 â€“ 121 = $${?}^{3}$$ â€“ 52

â‡’ 25 + 48 + 52 = $${?}^{3}$$

â‡’$${?}^{3}$$ = 125

â‡’ $${?}^{3}$$ = $${5}^{3}$$ â‡’ ? = 5.

5. 98643 â€“ 21748 = 51212 + ?

A. 24383
B. 24713
C. 25683
D. 25973

Explanation: Given eqn,

98643 â€“ 21748 = 51212 + ?

â‡’ ? = 98643 â€“ 21748 â€“ 51212

â‡’ ? = 25683

1. (764 Ã— ?) Ã· 250 = 382

A. 115
B. 145
C. 135
D. 125

Explanation: $$\frac{760 Ã—}{250}$$ = 382
? = $$\frac{382 Ã— 250}{764}$$ = 125

2. $$\frac{1}{4}$$ Ã— (4856 Ã— 0.5) Ã— 12 = ?

A. 7284
B. 7462
C. 7262
D. 7414

3. 853 + ? Ã· 17 = 1000

A. 2516
B. 2482
C. 2499
D. 16147

Explanation:
853 + $$\frac{7}{?}$$ = 1000
$$\frac{7}{?}$$ = = 1000 â€“ 853 = 147
? = 17 Ã— 147 = 2499

4. 6156 Ã· âˆš? Ã— 53 = 4028

A. 6889
B. 6241
C. 5929
D. 6561

Explanation:
853 + $$\frac{6153 Ã— 53}{âˆš?}$$ = 4028
â‡’ {âˆš?} = 80.96 â‰ˆ 81
â‡’ ? = 6561

5. 9643 â€“ 7750 + ? = 4990

A. 3079
B. 3097
C. 3090
D. 4010

Explanation:
853 + $$\frac{6153 Ã— 53}{âˆš?}$$ = ? = 4990 â€“ 9643 + 7750 = 3097

Hence, option B is correct.

1. (608.40 Ã— ?) Ã· 225 + 37 = 375

A. 115
B. 135
C. 130
D. 125

Explanation: $$\frac{6040 Ã— ?}{225}$$+ 37 = 375
â‡’ 608.40 Ã— ? = (375 â€“ 37) Ã— 225 = 76050 â‡’ ? = 125

2. 12.25 Ã— 7.2 + 84.33 = ?

A. 182.51
B. 177.44
C. 174.33
D. 172.53

Explanation: 12.25 Ã— 7.2 + 84.33 = 88.2 + 84.33 = 172.53

3. âˆš? + 416 = (60% of 920) â€“ 110

A. 576
B. 676
C. 784
D. 1024

Explanation: âˆš? + 416 = (60% of 920) â€“ 110
â‡’ âˆš? = (50% of 920 + 10% of 920) â€“ 110 â€“ 416
â‡’ âˆš? = (460 + 92) â€“526
â‡’ âˆš? = 552 â€“ 526 = 26

â‡’ ? = $${26}^{2}$$ = 676

4. (14896 Ã· 19) Ã· 16 = ?

A. 49
B. 54
C. 58
D. 62

Explanation: (14896 Ã· 19) Ã· 16 = ?
â‡’ ? = $$\frac{14896}{19}$$Ã· 16

$$\frac{784}{16}$$ = 49

5. 12% of 555 + 15% of 666 = ?

A. 166.5
B. 167.5
C. 168.5
D. 169.5

Explanation: ? = 12% of 555 + 15% of 666

= (10 + 2)% of 555 + (10 + 5)% of 666

= 10% of 555 + 2% of 555 + 10% of 666 + 5% of 666

= 55.5 + 11.1 + 66.6 + 33.3 = 166.5

Exams

Competitive Exams – College Entrance Exams
Diploma NITC New Delhi Goa Diploma Admissions 2019

Competitive Exams – Recent Job Notifications
Category
Banking SSC Railway
Defence Police Insurance