Quantitative Aptitude


1. Three containers A, B and C are having mixtures of milk and water in the ratio 1 : 5, 3 : 5 and 5 : 7, respectively. If the capacities of the containers are in the ratio 5 : 4 : 5, then find the ratio of the milk to the water if the mixtures of all the three containers are mixed together.
(a) 51 : 115
(b) 52 : 115
(c) 53 : 115
(d) 54 : 115
(e) None of these
Ans.c

Ratio of milk in the containers are,

5 × 16 : 4 × 38 : 5 × 512 = 56 : 32 : 2512

and the ratio of water in the containers are,

5 × 56 : 4 × 58 : 5 × 712 = 256 : 52 : 3512

Ratio of mixture of milk and water in the containers

= (16 × 5 + 38 × 4 + 512 × 5) : (56 × 5 + 58 × 4 + 712 × 5)

= 106 : 230 = 53 : 115

2. Groundnut oil is now being sold at 27 per kg. During last month its cost was 24 per kg. Find by how much % a family should reduce its consumption, so as to keep the expenditure same.
(a) 1119%
(b) 11111%
(c) 11910%
(d) 9110%
(e) None of these
Ans.a

% change in rate = \frac{27 - 24}{24} \times 100 = \frac{100}{8}\%

For fixed expenditure, % change in consumption

= \frac{\% change \; in \; rate}{100 + \% \; change \; in \; rate}

= \frac{100/8}{100\left [ 1 + \frac{1}{8} \right ]} \times 100 = \frac{100}{9}\% = 11\tfrac{1}{9}\%

3. An ice-cream company makes a popular brand of ice-cream in rectangular shaped bar 6 cm long, 5 cm wide and 2 cm thick. To cut the cost, the company has decided to reduce the volume of the bar by 20%, the thickness remaining the same, but the length and width will be decreased by the same percentage amount. The new length L will satisfy :
(a) 5.5 < L < 6
(b) 5 < L < 5.5
(c) 4.5 < L < 5
(d) 4 < L < 4.5
(e) None of these
Ans.b

L × B × 2 = 48

⇒ L × B = 24

Now, 6 – 6 × 10% = 5.4,

5 – 5 × 10% = 4.5 and

Therefore, 5.4 × 4.5 = 24.3

Clearly, 5 < L < 5.5

4. A sum of ₹ 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of ₹ 362.50 more is lent but at the rate twice the former. At the end of the year, ₹ 33.50 is earned as interest from both the loans. What was the original rate of interest?
(a) 3.6%
(b) 4.5%
(c) 5%
(d) 3.46%
(e) None of these
Ans.d

Let the original rate be R%. Then,new rate= (2R)%.

\left ( \frac{725 \times R \times 1}{100} \right ) + \left ( \frac{362.50 \times 2R \times 1}{100 \times 3} \right ) = 33.50

⇒ (2175 + 725)R = 33.50 × 100 × 3 = 10050

⇒ R = \frac{10050}{2900} = 3.46%

5. The difference between compound interest and simple interest on a sum for 2 years at 10% per annum, when the interest is compounded annually is ₹ 16. If the interest were compounded half-yearly, the difference in two interests would be:
(a) ₹ 24.81
(b) ₹ 26.90
(c) ₹ 31.61
(d) ₹ 32.40
(e) None of these
Ans.a

For first year,S.I. = C.I.

Now, ₹ 10 is S.I.on ₹ 100.

∴ 16 is S.I. on ₹ (10010 × 16) = ₹ 160.

So, S.I. on principal for 1 year at 10% is ₹ 160

∴ Principal = ₹\left (\frac{100 \times 160}{10 \times 1} \right ) = ₹ 1600.

Amount for 2 years compounded half yearly

= ₹\left [ 1600 \times \left ( 1 + \frac{5}{100} \right )^{4} \right ] = ₹ 1944.81

∴ C.I. = ₹(1944.81 – 1600) = ₹ 24.81.

S.I. = ₹ \left ( \frac{1600 \times 10 \times 2}{100} \right ) = ₹ 320.

∴ (C.I.) – (S.I.) = ₹ (344.81 – 320) = ₹ 24.81.

6. A person lent out a certain sum on simple interest and the same sum on compound interest at certain rate of interest per annum. He noticed that the ratio between the difference of compound interest and simple interest of 3 years and that of 2 years is 25 : 8. The rate of interest per annum is:
(a) 10%
(b) 11%
(c) 12%
(d) 1212%
(e) None of these
Ans.d

Let the principal be ₹ P and rate of interest be R% per annum.

Difference of C.I. and S.I. for 3 years

= \left [ p \times \left ( 1 + \frac{R}{100} \right )^{3} - p \right ] - \left ( \frac{P \times R \times 3}{100} \right )

= \frac{PR^{2}}{10^{4}}\left ( \frac{300 + R}{100} \right )

Difference of C.I. and S.I.for 2 years = P(R100)2

\frac{\frac{PR^{2}}{10^{4}}\left ( \frac{300 + R}{100} \right )}{\frac{PR^{2}}{10^{4}}} = \frac{25}{8}

\frac{300 + R}{100} = \frac{25}{8}

⇒ R = 1008 = 1212%

7. A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days,4/7 of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day ?
(a) 80
(b) 81
(c) 82
(d) 83
(e) None of these
Ans.b

Let x additional men employed.

117 men were supposed to finish the whole work in 46 × 8 = 368 hours.

But 117 men completed 47 of the work in 33 × 8 = 264 hours

∴ 117 men could complete the work in 462 hours.

Now (117 + x) men are supposed to do 37 of the work, working 9 hours a day, in 13 × 9 = 117 hours, so as to finish the work in time.

i.e. (117 + x) men are supposed to complete the whole work in 117 × 73 = 273 hours.

∴ (117 + x) × 273 = 117 × 462

⇒ (117 + x) × 7 = 3 × 462

⇒ x + 117 = 3 × 66 = 198 ⇒ x = 81

∴ Required number of additional men to finish the work in time = 81.

8. Two pipes A and B can fill a cistern in 30 minutes and 40 minutes respectively. Both the pipes are opened. Find when the second pipe B must be turned off so the cistern may just be full in 10 minutes.
(a) 2623 min.
(b) 25 min.
(c) 4023 min.
(d) 4223 min.
(e) None of these
Ans.a

Let A and B together work for x minutes than amount of water filled in the period = x(130 + 140) = 7120x

Remaining part = 1 – 7120x = \left ( \frac{120 - 7x}{120} \right )

Work done by A in (10 – x) minutes = \left ( \frac{120 - 7x}{120} \right )

= 1 - 7120x

7120x + \frac{10 - x}{30} = 1 or 7x + 40 – 4x = 120

3x = 120 – 40 = 80

x = 2623 min

9. A train leaves station X at 5 a.m. and reaches station Y at 9 a.m. Another train leaves station Y at 7 a.m. and reaches station X at 10 : 30 a.m. At what time do the two trains cross each other ?
(a) 7 : 36 am
(b) 7 : 56 am
(c) 8 : 36 am
(d) 8 : 56 am
(e) None of these
Ans.b

Let the distance between X and Y be x km. Then, the speed of A is x4 km/h and that of B is 2x7 km/h.

Relative speeds of the trains

= (x4 + 27x) = 1528x km/h

Therefore the distance between the trains at 7 a.m. = x - x2 = x2 km

Hence, time taken to cross each other

= \frac{\frac{x}{2}}{\frac{15x}{28}} = \frac{x}{2} \times \frac{28}{15x} = \frac{14}{15} \times 60 = 56 min

Thus, both of them meet at 7 : 56 a.m.

10. A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of water current respectively?
(a) 2 : 1
(b) 3 : 2
(c) 8 : 3
(d) Cannot be determined
(e) None of these
Ans.c

Let the man's upstream speed be Su kmph and downstream speed be Sd kmph. Then, Distance covered upstream in 8 hrs 48 min.

d = Distance covered downstream in 4 hrs.

⇒ (Su × 845) = (Sd × 4) ⇒ 445Su = 4Sd ⇒ Sd = 115Su.

∴ Required Ratio :

= \left ( \frac{S_{d} + S_{u}}{2} \right ) : \left ( \frac{S_{d} - S_{u}}{2} \right ) = \left ( \frac{16S_{u}}{5} \times \frac{1}{2} \right ) : \left ( \frac{6S_{u}}{5} \times \frac{1}{2} \right ) = \frac{8}{5} : \frac{3}{5}

= 8 : 3

DIRECTIONS (Qs. 11 - 15) : Study the pie charts given below and answer the following questions.

11. What is the cost of Gear Box?
(a) ₹ 9000
(b) ₹ 6000
(c) ₹ 3000
(d) ₹ 15,000
(e) None of these
Ans.c

Cost of gear box = 20 × \frac{1,00,000}{100} \times \frac{15}{100} = 3000

12. What percentage of total cost is contributed by the brake?
(a) 5.5%
(b) 6.6%
(c) 6%
(d) 5.4%
(e) None of these
Ans.c

Cost of brake = \frac{20 \times 1,00,000}{100} \times \frac{30}{100} = 6000

∴ Required percentage = \frac{6000}{1,00,000} \times 100 = 6.0\%

13. If the price of tyres goes up by 25%, by what amount should be the sale price be increased to maintain the amount of profit?
(a) ₹ 750
(b) ₹ 2250
(c) ₹ 3750
(d) ₹ 375
(e) None of these
Ans.a

Price of tyres = \frac{20 \times 1,00,000}{100} \times \frac{15}{100} = 3000

Increased price of tyres = 3000 × 125100 = 3750

∴ Price should be increased = 3750 – 3000 = ₹ 750

14. If transmission cost increases by 20%, by what amount is the profit reduced (total price of car remains same)?
(a) ₹ 3000
(b) ₹ 4000
(c) ₹ 6000
(d) Cannot be determined
(e) None of these
Ans.b

Increased transmission cost = 20,000 × 120100 = 24000

∴ increase in transmission cost = 24000 – 20000 = ₹ 4000

Here, this increase will reduce the profit by 4000.

15. What % of sale price is contributed by clutch?
(a) 6%
(b) 2%
(c) 3%
(d) Cannot be determined
(e) None of these
Ans.a

Price of clutch = 30% of 20,000 = 6,000

∴ Required percentage = \frac{6000}{1,00,000} \times 100 = 6\%

DIRECTIONS (Qs. 16 - 20) : Read the following information and answer the questions that follow.

In a huge Jewellery shop, the electric gadgets being used are 17 tubelights of 40 W each, 14 fans of 80 W each, 16 bulbs of 60 W each, 11 bulbs of 100 W each, 11 AC’s of 2100 W each, 9 laptops of 200 W each and 10 TV’s of 120 W each. In a day, tube lights and TV’s are used for 8 h but 60 W bulbs are used for 7 h and 100 W bulbs are used for 9 h whereas laptops and AC's are used for 5 h and 9 h respectively. However, fans are used for 11 h.(Note : 1000 W = 1 unit, 1 month = 30 days).

16. What is the total electric energy consumed (in units) by 60 W bulbs in the whole month?
(a) 432
(b) 576
(c) 67.2
(d) 201.6
(e) None of these
Ans.d

Total electric energy consume by 60 W bulb in whole month = 16 × 60 × 7 × 30 W = 201.6 unit

17. Electricity consumed by all fans is what per cent of energy consumed by all the laptops?
(a) 132.2%
(b) 136.88%
(c) 122.68
(d) 169.62%
(e) None of these
Ans.b

Electricity consumed by all fans = 14 × 80 × 11 × 30 W

Electricity consumed by all laptops = 9 × 200 × 5 × 30

Required% = \frac{14 \times 80 \times 11 \times 30}{9 \times 200 \times 5 \times 30} \times 100 = 136.88\%

18. Out of the following, which type of gadgets consume more electricity in the shop?
(a) Fans
(b) Tubelights
(c) Laptops
(d) TV’s
(e) None of these
Ans.a

Electricity consumed by all fans

= 14 × 80 × 11 × 30 = 369600 W

Electricity consumed by all laptops

= 9 × 200 × 5 × 30 = 270000 W

Electricity consumed by all tubelights

= 17 × 40 × 8 × 30 = 163200 W

Electricity consumed by all TV’s

= 17 × 120 × 8 × 30 = 288000 W

Electricity consumed by 100 W bulb

= 11 × 100 × 9 × 30 = 297000 W

hence fans consumed more electricity.

19. If one electric unit costs ₹ 2.70 and power (used by AC's) unit costs ₹ 3.70, then what money is paid to the electricity department for one month?
(a) ₹ 27368
(b) ₹ 28683
(c) ₹ 78600
(d) ₹ 2900
(e) None of these
Ans.a

One unit cost = ₹ 2.70.

Power (used by AC’s) Unit cost = ₹ 3.70

Electricity consumed by all equipment (except AC’s)

= (201.6 + 369.6 + 270 + 163.2 + 288 + 297) unit

= (1292.4 + 297.00) unit = 1589.4 unit

Cost for these unit = 1589.4 × 2.7 = 4291.38

Electricity consumed by AC’s

= 11 × 2100 × 9 × 30 W = 623700 W = 6237 Unit

Cost for it = 6237 × 3.7 = 23076.9

Total cost = 23076.9 + 4291.38 = ₹ 27368

20. What is the ratio of consumption of electricity in units by 60 W and 100 W bulbs in a month?
(a) 5 : 6
(b) 4 : 5
(c) 3 : 4
(d) 2 : 3
(e) None of these
Ans.d

Required ratio = 201.6297 = 23

DIRECTIONS (Qs.21 – 25): In each of the following questions two equations are given. You have to solve them and give answer accordingly.

(a) If x > y
(b) If x < y
(c) If x = y
(d) If x ≥ y
(e) If x ≤ y

21.
I. 2x2 + 5x + 1 = x2 + 2x – 1
II. 2y2 – 8y + 1 = – 1
Ans.b

I. 2x2 + 5x + 1 = x2 + 2x – 1

x2 + 3x + 2 = 0

x2 + 2x + x + 2 = 0

x(x + 2) + 1 (x – 2) = 0

(x + 2)(x + 1) = 0

x = –2 , –1

II. 2y2 – 8y + 1 = –1

2y2 – 8y + 2 = 0

y2 – 4y + 1 = 0

\frac{+ 4 \pm \sqrt{16 - 4 \times 1 \times 1}}{2 \times 1}

= 2 ± √12 = 2 ± 2√3

Hence, y > x

22.
I. x22 + x - 12 = 1
II. 3y2 – 10y + 8 = y2 + 2y – 10
Ans.b

I. x2 + 2x – 1 = 2

x2 + 2x – 3 = 0

x + 3x – x – 3 = 0

x (x + 3) – 1 (x + 3) = 0

(x + 3)(x – 1) = 0

x = –3 , 1

II. 2y2 – 12y + 18 = 0

y2 – 6y + 9 = 0

(y – 3)2 = 0

y = 3, 3

Hence, y > x

23.
I. 4x2 – 20x + 19 = 4x – 1
II. 2y2 = 26y + 84
Ans.b

I. 4x2 – 24x + 20 = 0

x2 – 6x + 5 = 0

x2 – 5x – x + 5 = 0

x(x – 5) – 1(x – 5) = 0

(x – 5) (x – 1) = 0

x = 5, 1

II. y2 – 13y + 42 = 0

y2 – 7y – 6y + 42 = 0

y (y – 7) – 6 (y – 7) = 0

(y – 7) (y – 6) = 0

y = 7, 6

Hence, y > x.

24.
I. y2 + y - 1 = 4 - 2y - y2
II. x22 - 32x = x - 3
Ans.a

I. 2y2 + 3y – 5 = 0

2y2 + 5y – 2y – 5 = 0

y (2y + 5) – 1 (2y + 5) = 0

(2y + 5) (y – 1) = 0

y = -52, 1

II. x2 – 3x = 2x – 6

x2 – 5x + 6 = 0

x2 – 3x – 2x + 6 = 0

x (x – 3) – 2 (x – 3) = 0

(x – 3) (x – 2) = 0

x = 3, 2

Hence, x > y

25.
I. 6x2 + 13x = 12 – x
II. 1 + 2y2 = 2y + 5y6
Ans.e

I. 6x2 + 14x = 12

3x2 + 7x – 6 = 0

(x + 3) (3x – 2) = 0

x = –3 , 23

II. 1 + 2y2 = 176y

12y2 – 17y + 6 = 0

12y2 – 8y – 9y + 6 = 0

4y (3y – 2) – 3 (3y – 2) = 0

(3y – 2) (4y – 3) = 0

y = 23, 34

Hence, x ≤ y

DIRECTIONS (Qs. 26 - 30): Study the following graph carefully to answer these questions.

26. Shri Giridhar invested total amount of ₹ 25000 in 2009 for one year in the two companies together and got a total dividend of ₹ 3340. What was the amount invested in Company A?
(a) ₹ 12000
(b) ₹ 9000
(c) ₹ 16000
(d) Cannot be determined
(e) None of these
Ans.b

Suppose Giridhar invested ₹ x in company A.

\frac{x \times 14}{100} + \frac{\left ( 25000 - x \right ) \times 13}{100} = 3340

or, 14100x + 3250 - 13100x = 3340

or, x100 = 90 or, x = ₹ 9000

27. Anuja invested ₹ 35000 in Company B in 2011. After one year she transferred the entire amount with dividend to Company A in 2012 for one year. What amount will be received back by Anuja including dividend?
(a) ₹ 49980
(b) ₹ 49000
(c) ₹ 48300
(d) ₹ 49563.50
(e) None of these
Ans.a

Amount of dividend received by Anuja in 2011 from company B

= \frac{35000 \times 19}{100} = ₹ 6650

Total amount invested by Anuja in 2012 in Company A

= 35000 + 6650 = ₹ 41650

Reqd amount = 41650 × 120100 = ₹ 49980

28. An amount of ₹ 18000 was invested in Company A in 2012. After one year the same amount was re-invested for one more year. What was the total dividend received at the end of two years?
(a) ₹ 5805
(b) ₹ 6300
(c) ₹ 6480
(d) ₹ 6840
(e) None of these
Ans.b

Total dividend = 18000 × (20100 + 15100) = ₹ 6300

29. Bhushan invested different amounts in Companies A and B in 2015 in the ratio of 5 : 8. What will be the ratio between the amounts of dividends received from Companies A and B respectively?
(a) 2 : 3
(b) 5 : 6
(c) 3 : 4
(d) Cannot be determined
(e) None of these
Ans.c

Reqd ratio = \frac{5 \times 12}{8 \times 10} = 3 : 4

30. In the year 2014, Suraj invested ₹ 56000 in Company B. How much more or less dividend would he have received had the amount been invested in Company A?
(a) ₹ 1640 more
(b) ₹ 1640 less
(c) ₹ 1860 less
(d) ₹ 1680
(e) None of these
Ans.d

From the graph it is obvious that Suraj will get less dividend in 2014 from company A than from B.

Reqd less amount = 3% of 56000 = ₹ 1680.

DIRECTIONS (Qs. 31 - 35) : Study the following table to answer the given questions.

31. The production of Company AVC in 2000 is approximately what per cent of its average production over the given years?
(a) 300
(b) 110
(c) 136
(d) 118.25
(e) None of these
Ans.b

Production of company AVC in 2012 = 360 crore units

Average production of AVC over the given years = 19706

Hence, required percent = \frac{360 \times 6}{1970} \times 100 = 109.64\% \; \approx \; 110\%

32. For SIO, which year was the per cent increase or decrease in production from the previous year the highest?
(a) 2013
(b) 2010
(c) 2014
(d) 2012
(e) None of these
Ans.c

Approximate per cent increase or decrease introduction from the previous year for SIO are as follows :

2010 = 285 × 100 = 2.35%

2011 = \frac{2 \times 100}{87} = 2.29%

2012 = \frac{2 \times 100}{89} = 2.24%

2013 = \frac{1 \times 100}{91} = 1.09%

2014 = \frac{4 \times 100}{92} = 4.35%

Quicker method : See the difference of produced units between two consecutive years. The difference is maximum for 2013 to 2014, and production during all these years is almost same.Hence, in the year 2014, SIO registered maximum increase in production over the previous year.

33. Which company has less average production in the last three years compared to that of first three years?
(a) No company
(b) CTU
(c) ZIR
(d) TP
(e) None of these
Ans.d

Sum of the productions of companies in the first three years and the last three years is as follows :

Company First three Years Last three Years
TP 358 349
ZIR 238 267
AVC 900 1070
CTU 836 852
PEN 90 127
SIO 261 279
34. The total production of the six companies in the first two given years is what per cent of that of last two given years?(round off up to two decimal places)
(a) 87.08
(b) 104.55
(c) 90.40
(d) 10.62
(e) None of these
Ans.c

Total production of the six companies in first two given years = 863 + 927 = 1790

Again, total production of the six companies in last two given years = 989 + 991 = 1980

Therefore, required per cent = \frac{1790 \times 100}{1980} = 90.40\%

35. For ZIR, which of the following is the difference between production in 2013 and that in 2014?
(a) 10,00,00,000
(b) 1,00,00,000
(c) 10,00,000
(d) 40,00,000
(e) None of these
Ans.b

The required difference = (91 – 92) crore units = 1 × 10000000 = 10000000 units

36. When the price of a radio was reduced by 20%, its sale increased by 80%. What was the net effect on the sale?
(a) 44% increase
(b) 44% decrease
(c) 66% increase
(d) 75% increase
(e) None of these
Ans.a

Net effect = x + y + xy100 = -20 + 80 + \left (\frac{-20 \times 80}{100} \right )

= 60 – 16 = 44% increased

37. Two sides of a plot measure 32 metres and 24 metres and the angle between them is a perfect right angle. The other two sides measure 25 metres each and the other three angles.
What is the area of the plot?
(a) 768
(b) 534
(c) 696.5
(d) 684
(e) None of these
Ans.d

(32 - y)2 + (24 - x)2 = 625 .....(1)

x2 + y2 = 625 .....(2)

⇒ (24)2 + (32)2 - 64y - 48x = 0 (From (1) & (2))

⇒ 64y + 48x = 576 + 1024

⇒ 4y + 3x = 36 + 64 = 100

or y = \left (\frac{100 - 3x}{4} \right )

∴ x2 + \left (\frac{100 - 3x}{16} \right )^{2} = 625 (From(2))

⇒ -600x + 16x2 + 10000 + 9x2 = 625 × 16

⇒ 25x2 - 600x + 10000 - 625 × 16 = 0

⇒ x = 24 and y = 7

∴ Area = (24 × 25) + 12 × 24 × 7 = 684

38. In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
(a) 360
(b) 480
(c) 720
(d) 5040
(e) None of these
Ans.c

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

∴ Required number of ways = (120 × 6) = 720.

39. In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
(a) 2146
(b) 25117
(c) 150
(d) 325
(e) None of these
Ans.a

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

= 25C3

= \left ( \frac{25 \times 24 \times 23}{3 \times 2 \times 1} \right ) = 2300

n(E) = (10C1 × 15C2)

= \left [ 10 \times \frac{\left ( 15 \times 14 \right )}{\left ( 2 \times 1 \right )} \right ]

= 1050.

P\left ( E \right ) = \frac{n\left ( E \right )}{n\left ( S \right )} = \frac{1050}{2300} = \frac{21}{46}

40. Gauri went to the stationery and bought things worth ₹ 25,out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?
(a) ₹ 15
(b) ₹ 15.70
(b) ₹ 19.70
(d) ₹ 20
(e) None of these
Ans.c

Let the amount taxable purchases be Rs. x.

Then, 6% of x = 30100

⇒ x = (30100 × 1006) = 5

∴ Cost of tax free items = Rs. [25 - (5 + 0.30)] = ₹ 19.70

DIRECTIONS (Qs 41 - 45): In this type of questions, one term in the number series is wrong. Find out the wrong term.

41.  93,    309,    434,    498,    521,    533
(a) 309
(b) 434
(c) 498
(d) 521
(e) None of these
Ans.d

The correct pattern is +63, +53, +43, +33, .....

So, 521 is wrong and must be replaced by (498 + 33) i.e.525.

42.  46080,    3840,    384,    48,    24,    2,    1
(a) 384
(b) 48
(c) 24
(d) 2
(e) None of these
Ans.c

The correct pattern is ÷ 12, ÷ 10, ÷ 8,÷ 6,.....

So,24 is wrong and must be replaced by (48 + 6) i.e. 8.

43.  5,    27,    61,    122,    213,    340,    509
(a) 27
(b) 61
(c) 122
(d) 509
(e) None of these
Ans.a

The terms of the series are (23 – 3), (33 – 3),(43 – 3),(53 – 3), (63 – 3), (73 – 3), (83 – 3).

So, 27 is wrong and must be replaced (33 – 3) i.e. 24.

44. 11,    5,    20,    12,    40,    26,    74,    54
(a) 5
(b) 20
(c) 40
(d) 26
(e) None of these
Ans.c

The given sequence is a combination of two series :

I. 11,  20,  40,  74 and

II. 5,  12,  26,  54

The correct pattern in I is + 9, + 18,+ 36,.....

So, 40 is wrong and must be replaced by (20 + 18) i.e. 38.

45.  1,    3,    10,    21,    64,    129,    356,    777
(a) 21
(b) 129
(c) 10
(d) 356
(e) None of these
Ans.d

The correct pattern is × 2 + 1, × 3 + 1, × 4 + 1, × 3 + 1 + 1,.....

So, 356 is wrong and must be replaced by (129 × 3 + 1)i.e. 388.

DIRECTIONS (46 - 50): What approximate value will come in place of question mark (?) in the following questions (You are not expected to calculate the exact value).

46. 9228.789 – 5021.832 + 1496.989 = ?
(a) 6500
(b) 6000
(c) 6300
(d) 5700
(e) 5100
Ans.d

Having a glance at the given options one can find out that the two nearest values have a difference of 300.

So round off the numbers to the nearest ten's values.

9228.789 ≈ 9230; 5021.832 ≈ 5020 and 1496.989 ≈ 1500

Now the equation will become

9230 – 5020 + 1500 = ?

? = 5710

But the nearest value is 5700.

47. 1002 ÷ 49 × 99 – 1299 = ?
(a) 700
(b) 600
(c) 900
(d) 250
(e) 400
Ans.a

It can be rounded off to the nearest ten's places.

1002 ≈ 1000; 49 ≈ 50; 99 ≈ 100 and 1299 ≈ 1300

Now the equation will become

1000 ÷ 50 × 100 – 1300 = ?

20 × 100 – 1300 = ?

2000 – 1300 = ?

? = 700

48. 29.8% of 260 + 60.01% of 510 – 103.57 = ?
(a) 450
(b) 320
(c) 210
(d) 280
(e) 350
Ans.d

The difference between two nearest values is 70 (210 and 280). So round off the numbers to the nearest integers. 29.8% of 260 ≈ 30% of 260; 60.01% of 510 ≈ 60% of 510 and 103.57 ≈ 104

Now the equation will become

30% of 260 + 60% of 510 – 104 = ?

30/100 × 260 + 60/100 × 510 – 104 = ?

78 + 306 – 104 = ?

? = 384 – 104 = 280

49. (21.98)2 – (25.02)2 + (13.03)2 = ?
(a) 25
(b) 120
(c) 10
(d) 65
(e) 140
Ans.a

(21.98)2 ≈ (22)2

(25.02)2 ≈ (25)2

and (13.03)2 ≈ (13)2

The equation will becomes

222 – 252 + 132 = ?

484 – 625 + 169 = ?

653 – 625 = ?

? = 28 so the nearest value is 25

50. \frac{\sqrt{2498} \times \sqrt{625}}{\sqrt{99}} = ?
(a) 110
(b) 90
(c) 200
(d) 160
(e) 125
Ans.e

\frac{50 \times 25}{10} = \frac{1250}{10} = 125