[SOLVED] RBI Assistant Previous Year Question Paper 2017 - Quantitative Aptitude

Q. (1) Find the missing number in the given series.

7, 21, 5, 23, 3, (?)

- 25
- 28
- 27
- 33
- 32

Answer: 1

Solution: Given series is based on the following pattern:

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• 21-7=14=(10+4)

• 21-5=16=(10+6)

• 23-5=18=(10+8)

• 23-3=20=(10+10) Therefore, 3 + (10 + 12) = 25 Hence, the required number = 25

Q. (2) Find the missing number in the given series.

15, 22, 32, 46, 65, (?) - 80
- 82
- 85
- 90
- 94

Answer: 4

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3

Solution: Given series is based on the following pattern:

• 22-15=7

• 32-22=10=(7+3)

• 46-32=14=(10+4)

• 65-46=19=(14+5) Therefore, 65 + (19 + 6) = 90 Hence, the required number = 90

Q. (3) Find the missing number in the given series.

9, 10, 18, 27, 91, (?) - 110
- 112
- 116

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4 - 121
- 124

Answer: 3

Solution: Given series is based on the following pattern:

• 10-9=1=12

• 18-10=8=23

• 27-18=9=32

• 91-27=64=43 Therefore, 91 + 52 = 91 + 25 = 116

So, the required number in the series is 116.

Q. (4) Find the missing number in the given series.

17, 23, 35, 59, (?), 203

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5 - 117
- 107
- 127
- 97
- None of these

Answer: 2

Solution: Given series is based on the following pattern:

• 23-17=6

• 35 - 23 = 12 = (6 x 2)

• 59 - 35 = 24 = (12 x 2) Therefore, 59 + (24 x 2) = 107 Hence, the required number = 107

Q. (5) Find the missing number in the given series.

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6, 7, 16, 51, 208, (?) - 1010
- 1045
- 1035
- 1038
- None of these

Answer: 2

Solution: Given series is based on the following pattern:

• (6×1)+1=7

• (7×2)+2=16

• (16×3)+3=51

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• (51×4)+4=208

• (208 × 5) + 5 = 1045

So, the required number in the series is 1045.

Direction Q. (6 - 10): What value should come in place of the question mark (?) given in each of the

following questions?

Q. (6) (4)4 ÷ (16)3 × 256 = 4(? – 6) - 3
- 6
- 9
- 5

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8 - None of these

Answer: 5

Solution: (4)4 ÷ (16)3 × 256 = 4(? – 6)

⇒ (4)4 ÷ (42)3 x (4)4 = 4(? - 6)

⇒ (4)-2 x (4)4 = 4(? - 6)

⇒ (4)2÷ 4=?-6

⇒ ?=4+6=10

Q. (7)? % of 650 + 844 = 1000 - 54
- 24
- 34

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9 - 14
- None of these

Answer: 2

Solution:? % of 650 + 844 = 1000

⇒? % of 650 = 156

⇒ (? /100) x 650 = 156

⇒ ? = 1560/65

⇒ ?=24

Q. (8) (?)9/4/ 324 = (?)1/4 / 9 - 27
- (36)2
- 6

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10 - 36
- 6

Answer: 3

Solution: (?)9/4/ 324 = (?)1/4 / 9

⇒ (?)9/4 ÷ (?)1/4 = 324/9

⇒ (?)9/4 - 1/4 = 36

⇒ (?)2 = 62

⇒ ? = 6

Q.(9)2¼+1⅓-4½=(?) - -1 ½
- 11/12

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11 - -11/12
- 1 ½
- None of these

Answer: 5

Solution: 2 ¼ + 1 ⅓ - 4 ½ = (?)

⇒ (2)1/2 + 1/2 + ∛1 - (22)1/2 = (?)

⇒ √2 x √2 = ? + 1

⇒ (√2)2=?+1

⇒ ?=2-1=1

Q. (10) 126 ÷ 14 × (9)2 – 53 = (?)2 - 26
- 26

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12 - -729
- 27
- 27

Answer: 1

Solution: 126 ÷ 14 × (9)2 – 53 = (?)2

⇒ 9 x (9)2.- 53 = (?)2

⇒ (9)3- 53 = (?)2

⇒ 729 - 53 = (?)2

⇒ (?) = √676

⇒ ?=26

Q. (11) 724 - 336 + 499 = (?) + 112

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13 - 765
- 745
- 551
- 641
- None of these

Answer: 5

Solution: 724 - 336 + 499 = (?) + 112

⇒ 388 + 499 = (?) + 112

⇒ 887 - 112 = (?)

⇒ (?) = 775

Q. (12) 869.4 + 604.8 = [489.5 - 398.5] x (?) - 18
- 14.1

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14 - 14.5
- 16.8
- 16.2

Answer: 5

Solution: 869.4 + 604.8 = [489.5 - 398.5] x (?)

⇒ 1474.2 = 91 x (?)

⇒ (?) = 1474.2/91

⇒ (?) = 16.2

Q. (13) √ (24)4 + 224 = (?) x (20)2 - 20
- 4

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15 - 2
- 16
- None of these

Answer: 3

Solution: √ (24)4 + 224 = (?) x (20)2

⇒ (24)2 + 224 = (?) x 400

⇒ 576 + 224 = (?) x 400

⇒ (?) = 800/400

⇒ (?)=2

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Q. (14) A boat running downstream covers a distance of 20 km in 2 hours. While coming back, the boat

takes 4 hours to cover the same distance. What is the speed of the boat in still water (in kmph)? - 6.5
- 7.5
- 8.5
- 9
- None of these

Answer: 2

Solution: Let, the speed of the boat in still water be x kmph and that of the stream be y kmph.

∴ 20/(x + y) = 2

⇒ x + y = 10(i)*_*

20/(x - y) = 4

⇒ x - y = 5(ii) Adding (i) & (ii) we get,*___*

2x = 15

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⇒ x = 7.5

Therefore, the speed of the boat in still water = 7.5 kmph

Q. (15) Prachi deposits an amount of ₹ 78,000 to obtain a simple interest at the rate of 13 p.c.p.a. for 3

years. What total amount will Prachi get at the end of 3 years? - ₹ 30,420
- ₹ 1,05,420
- ₹ 1,12,420
- ₹ 1,08,420
- None of these

Answer: 1

Solution: Principal (here the amount) = Simple Interest x [100/ (rate x time)]

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⇒ 78,000 = Simple Interest x [100/ (13 x 3)]

⇒ Simple Interest = (78000 x 13 x 3)/ 100

⇒ Simple Interest = Rs. 30,420

Q. (16) The simple interest accrued on an amount of ₹ 84000 at the end of 3 yr is ₹ 30240. What would

be the compound interest accrued on the same amount at the same rate in the same period? - ₹ 30013.95
- ₹ 31013.95
- ₹ 32013.95
- ₹ 33013.95
- ₹ 34013.95

Answer: 5

Solution: Let, the rate of interest be y.

∴ Principal (here, the amount) = Simple Interest × 100/ Rate × Time

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⇒ 84000 = 30240 x 100/y × 3

⇒ y = 30240/(840 x 3)

⇒ y = 12

Hence, compound interest = P (1 + r/100 )n - P

= [84000 (1 + 12/100)3] - 84000

= [84000 x (112/100) x (112/100) x (112/100)] - 84000

= 118013.95 - 84000

= Rs. 34013.95

Q. (17) 6 women and 10 children together take six days to complete a piece of work. How many days will

10 children take to complete that piece of work if six women together can complete the same piece of

work in 10 days? - 15

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20 - 8
- 12
- 10
- None of these

Answer: 1

Solution: Work done by 6 women and 10 children in 1 day = 1/6

Work done by 6 women in 1 day = 1/10

Therefore, work done by 10 children in 1 day = 1/6 − 1/10 = 1/15

Hence, 10 children can complete the work in 15 days.

Directions Q. (18 - 22): Study the following table carefully to answer the following questions.

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Number of units manufactured (M) and sold (S) in hundreds by five different countries over the years

Q. (18) What is the total number of units manufactured by Company C over all the years together? - 1420

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22 - 1030
- 1230
- 1320
- None of these

Answer: 3

Solution: The total number of units manufactured by Company C over all the years together

= 260 + 220 + 210 + 280 + 260

= 1230

Q. (19) What is the approximate percent increase in the number of units sold by Company E in the year

2007 from the previous year? - 17
- 36

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23 - 27
- 32
- 21

Answer: 1

Solution: Percentage increase = [(170 - 140)/ 170] x 100 = 300/17 = 17% (approx.)

Q. (20) The number of units sold by Company D in the year 2006 is what percent of the number of units

manufactured by it in that year? (rounded off to two digits after decimal) - 52.63
- 61.57
- 85.15
- 73.33
- None of these

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Answer: 4

Solution: Company D in 2006, M = 300 and S = 220

Let the no. of units sold be x % of no. of units manufactured.

220 = x% of 300

220 = (x/100) x 300

220 = 3x

x = 220/3

x = 73.33

Q. (21) What is the respective ratio of the total number of units manufactured by Company A and B

together in the year 2009 to those sold by them in the same year? - 5:2
- 3:1

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25 - 7:5
- 3:2
- None of these

Answer: 5

Solution: For Company A in 2009, M = 100, S = 40

For Company B in 2009, M = 240, S = 130

Hence, required number = (100 + 240) / (40 + 130) = 2:1

Q. (22) What is the average number of units sold by Company D over all the years together? - 166
- 158
- 136

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26 - 147
- None of these

Answer: 2

Solution: The average number of units sold by Company D over all the years together

= (220 + 190 + 150 + 120 + 110)/5

= 790/5

= 158

Q. (23) The cost of 14 kgs of rice is 672, the cost of 12 kgs of wheat is 432 and the cost of 18 kgs of sugar

is 504. What is the total cost of 20 kgs of rice, 15 kgs of wheat and 16 kgs of sugar? - 1,898

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27 - 1,948
- 2,020
- 1,964
- None of these

Answer: 2

Solution: Cost of 1 kg rice = 672/14 = 48

Cost of 1 kg wheat = 432/12 = 36

Cost of 1 kg sugar = 504/18 = 28

Hence, cost of 20 kg rice = 48 × 20 = 960

Cost of 15 kg wheat = 36 × 15 = 540

Cost of 16 kg Sugar = 28 × 16 = 448

Therefore, total cost = 960 + 540 + 448 = 1948

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Q. (24) The strength of a university is 4500. The number of boys and girls is increased by 12% and 17%

respectively. As a result, the strength of the university becomes 5125. Find the number of girls in the

university. - 2800
- 2400
- 1700
- 2100
- 1500

Answer: 3

Solution: Let the no. of girls be y.

So, the no. of boys = 4500 – y

The no. of boys and girls is increased by 12% and 17% respectively. As a result, the strength of the

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university becomes 5125.

So, we can write now,

[(4500 – y) × 12%] + [y x 17/100] = 5125 – 4500

⇒ 54000 – 12y + 17x = 62500

⇒ 5y = 62500 – 54000

⇒ y = 8500/5

⇒ x = 1700

∴ The no. of girls = 1700

Q. (25) In a survey, the satisfaction level of the customers with a product was analysed. It was found that

32% of the customers were satisfied with the product, 26% were dissatisfied with the product and the

rest 189 customers were neither satisfied nor dissatisfied with the product. What was the total number

of customers who were either satisfied or dissatisfied with the product? - 260
- 270

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30 - 254
- 285
- 261

Answer: 5

Solution: Total no. of customers = (189 x 100) / [100 – (32 26)]

= 18900/42 = 450

Number of customers who are neither satisfied nor dissatisfied = 189

Hence, the number of customers who are either satisfied or dissatisfied = 450 – 189 = 261

Q. (26) At present, Aruna's age is 1.5 times the age of Pujan and twice the age of Somy. After six years,

Aruna will be 1.4 times the age of Pujan and Somy will be 0.8 times the age of Pujan then, what is the

present age of Somy?

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31 - 30 years
- 36 years
- 24 years
- 18 years
- None of these

Answer: 4

Solution: Let, Punjan’s present age be x years.

Therefore, Aruna’s present age = 1.5x years

Somy’s present age = 1.5x/2 years

After 6 years,

Punjan’s age = (x + 6) years

Aruna’s age = (1.5x + 6) years

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Now, 1.4(x + 6) = (1.5x + 6)

⇒ 1.4x + 8.4 = 1.5x + 6

⇒ 0.1x = 2.4

⇒ x = 24 years

Therefore, Somy’s age = (1.5 x 24)/2 = 18 years.

Q. (27) A train moves with a speed of 30 km/hr for 12 minutes and for the next 8 minutes at a speed of

45 km/hr. the average speed of the train is: - 37.5 km/hr
- 36 km/hr
- 48 km/hr
- 30 km/hr

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33 - None of these

Answer: 2

Solution: Distance = Speed × Time

Distance covered by the train with the speed of 30 kmph in 12 minutes = 30 × 1260 = 6km

Distance covered by the train with the speed of 45 kmph in 8 minutes = 45 × 860 = 6km

Average speed = (Total distance)/(Total time)

= (6+6)/(12+8) km/min

= (12/20) × 60 kmph

= 36 kmph

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Q. (28) A job can be completed by a person A alone in 24 days and by another person B alone in 40 days.

A and B work on the job together and C works with them for 6 days. The three complete the job in 12

days. In how many days can C alone complete the job? - 33 days
- 36 days
- 24 days
- 30 days
- None of these

Answer: 4

Solution: A can complete 1 work in 24 days.

B can complete 1 work in 40 days.

Also, A can complete 1/24 work in 1 day.

∴ B can complete 1/40 work in 1 day.

Therefore, work done by A+B in 1 day = 1/24 + 1/40 = (5 + 3)/120 = 8/120 = 1/15

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Work done by A+B in 12 days = (1/15) x 12 = 12/15 = 4/5 Therefore, work left = 1- (4/5) = 1/5

Now, work done by C in 6 days = 1/5

Work done by C in 1 day = (1/5)/6 = 1/30

Therefore, C can complete 1/30 work in 1 day.

Hence, C can complete 1 work in 30 days.

Q. (29) The present population of state A is 5/4 times the present population of village B. If a year ago

the population of village B was 24,400 and is increased at a rate of 15% pa, what is the present

population of state A? - 30,355
- 39,360
- 34,456
- 35,075

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36 - 38,545

Answer: 4

Solution: Present population of State B

= 24400 x 115/100

= 244 × 115

= 28060

Present population of State A = 28060 × 5/4

= 35075

Q. (30) The perimeter of a square is 144m. The area of a rectangle is 339 sq m less than the area of the

given square. If the breadth of the rectangle is 7m less than the side of the square, what is its perimeter?

(in m) - 120

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37 - 132
- 128
- 124
- 136

Answer: 4

Solution: Perimeter of square = 4a

Then, 4a = 144

⇒ a = 144 4 = 36m

∴ Area of the square = 36 × 36 = 1296 sq m Area of the rectangle = 1296 – 339 = 957 sq m Hence,

breadth of the rectangle = 36 – 7 = 29 m Length of the rectangle = 957/29 = 33

∴ Perimeter of the rectangle = 2(33 + 29) = 124 m

Q. (31) A man has 972 in the denominations of one-rupee, two-rupee, five-rupee and ten-rupee. If

there are equal no. of coins of each denomination, what is the total number of coins that he has?

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38 - 216
- 232
- 224
- 236
- 212

Answer: 1

Solution: There are equal no. of coins of each denomination.

Let the no.of coins of each denomination be x.

∴ x + 2x + 5x + 10x = 972 18x = 972

x = 972/18 = 54

Total no. of coins = 54 × 4 = 216 Value of one-rupee coins = 54

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Value of two-rupee coins = 54 × 2 = 108 Values of five-rupee coins = 5 × 54 = 270 Value of ten-rupee

coins = 10 × 54 = 540

Q. (32) If a man can row 13.3 km downstream in 19 minutes and his rowing speed in still water is

38kmph, how much distance can he cover upstream in 15 minutes? (in km) - 7.75
- 9.5
- 8.5
- 8.25
- 9.25

Answer: 3

Solution: Downstream speed = (13.3 x 60)/19 = 42 kmph

Speed of man in still water = 38 kmph

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∴ Current’s speed = 42 – 38 = 4 kmph Upstream speed = 38 – 4 = 34 kmph Therefore, in 60

minutes, he can row 34 km

Hence, in 15 minutes he can row = (34/60) x 15 = 8.5 km

Q. (33) Pinky while selling an article incurred a loss of 15%. Had she sold it for `96 more, she would have

earned a profit of 17%. What is the cost price of the article? - 350
- 300
- 346
- 385
- 320

Answer: 2

Solution: Let, the cost price be Rs. x

∴ (117x/100) - (85x/100) = 96

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⇒ 32x/100 = 96

⇒ x = (96 x 100)/32 = 300

Q. (34) The simple interest obtained when a certain sum of money is invested in scheme A for 3 years is

38.8 less than the simple interest obtained when the same sum of money is invested in scheme B for 4

years. If the rates of interest offered by scheme A and scheme B are 14% pa and 12.5% pa respectively,

how much money was invested in these schemes individually? - 440
- 450
- 490
- 475
- 485

Answer: 5

Solution: Let the sum be P.

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Then, [(P x 12.5 x 4)/100] - [(P x 14 x 3)/100] = 38.8

∴ (50P – 42P)/100 = 38.8

⇒ 8P = 38.8 × 100 = 3880

⇒ P = 3880/8 = 485

Q. (35) A pipe can fill a tank in 8 hrs, but due to leakage it took 29 ⅓ hours to fill the tank. If the tank is

full, in how much time will the tank become empty due to the leakage? - 11 hrs
- 12 hrs
- 9 ½ hrs
- 9 hrs
- 11 ½ hrs

Answer: 1

Solution: The leakage can empty the full tank in (⅛ - 3/88)

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= (11 – 3)/88

= 8/88

= 1/11 = 11 hours