IBPS Clerk Prelims Quantitative Aptitud...

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IBPS Clerk Prelims Quantitative Aptitude Practice Sets

Introduction

Quantitative Aptitude is an important section in the employment related competitive exams in India. In particular, exams like IBPS, SBI and other bank related employment exams have Quantitative Aptitude questions along with Reasoning. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.This article presents IBPS Clerk Prelims Quantitative Aptitude Practice Sets for acing the IBPS Clerk Prelims Examination.

Aptitude

Simplification:

Directions for Questions (1 â€“ 10): What value should come in place of the question marks (?) in the following questions?

1. (13056 Ã· 32)(513 Ã· 19) = ? + 95 $$\sqrt{1104 + 40 Ã— 30}$$

A. 6456
B. 6455
C. 6457
D. 6454
E. 6458

Solution: (13056 Ã· 32)(513 Ã· 19) = ? + 95 $$\sqrt{1104 + 40 Ã— 30}$$

408 x 27 = ? + 95 $$\sqrt{1104 + 200}$$

11016 = ? + 95 $$\sqrt{2304}$$

? = 11016 – 95 x 48

? = 11016 – 4560

? = 6456

2. $$\sqrt{29 Ã— 1240 Ã· 31 + 173 Ã— 5}$$ Ã— 36 =$$?^2$$ + 29 Ã— 16

A. 32
B. 33
C. 35
D. 31
E. 34

Solution: $$\sqrt{29 Ã— 1240 Ã· 31 + 173 Ã— 5}$$ Ã— 36 =$$?^2$$ + 29 Ã— 16

$$\sqrt{29 Ã— 40 + 865}$$ Ã— 36 = $$?^2$$ + 464

$$\sqrt{1160 + 865}$$ Ã— 36 = $$?^2$$ + 464

$$?^2$$ = 45 x 36 – 464

$$?^2$$ = 1620 – 464

$$?^2$$ = 1156

$$?^2$$ = $$\sqrt {1156}$$

$$?^2$$ = 34

3. (16980 – 9910 – 4648) Ã· 14 = ? + (16650 Ã· 111)

A. 23
B. 21
C. 22
D. 24
E. 25

Solution: (16980 – 9910 – 4648) Ã· 14 = ? + (16650 Ã· 111)

(16980 – 14558) Ã· 14 ? + 150

2422 Ã· 14 ? + 150

? + 150 = 173

? = 173 – 150 = 23

4. 73% of 4936 + 24% of 5612 – 55% of 3848 = ?

A. 2833.76
B. 2831.56
C. 2832.12
D. 2834.54
E. 2835.45

Solution: 73% of 4936 + 24% of 5612 – 55% of 3848 = ?

$$\frac{73} {100}$$ x 4936 + $$\frac{24} {100}$$ x 5612 – $$\frac{55} {100}$$ x 3848

? = 73 x 49.36 + 24 x 56.12 – 55 x 38.48

? = 3603.28 + 1346.88 – 2116.4

? = 4950.16 – 2116.4 = 2833.76

5. $$\frac{8} {13}$$ – $$\frac{7} {26}$$ x $$?^2$$ = 234

A. 25
B. 26
C. 27
D. 24
E. 23

Solution: $$\frac{8} {13}$$ – $$\frac{7} {26}$$ x $$?^2$$ = 234

$$\frac{16 âˆ’ 7} {26}$$ $$?^2$$ = 234

$$\frac{9} {26}$$$$?^2$$ = 234

$$?^2$$ = 676

? = $$\sqrt{676}$$= 26

6. $$\frac{5} {17}$$ of $$\frac{51} {81}$$ of 54% of 180 = $$\sqrt{?}$$ – $$9^2$$

A. 9804
B. 9803
C. 9801
D. 9802
E. 9805

Solution: $$\frac{5} {17}$$ of $$\frac{51} {81}$$ of 54% of 180 = $$\sqrt{?}$$ – $$9^2$$

$$\frac{5} {17}$$ x $$\frac{51} {81}$$ x 54 x 180 = $$\sqrt{?}$$ – $$9^2$$

$$\frac{5} {27}$$ 54 x 1.8 = $$\sqrt{?}$$ – 81

18 = $$\sqrt{?}$$ – 81

? = $$\sqrt{99}$$ = 9801

7. 87 $$\frac{3} {8}$$ + 54 $$\frac{5} {24}$$ – 110 $$\frac{7} {60}$$ = ?

A. 31 $$\frac{7} {15}$$
B. 31 $$\frac{7} {13}$$
C. 31 $$\frac{8} {13}$$
D. 31 $$\frac{6} {13}$$
E. 31 $$\frac{8} {15}$$

Solution: 87 $$\frac{3} {8}$$ + 54 $$\frac{5} {24}$$ – 110 $$\frac{7} {60}$$ = ?

(87 + 54)$$\frac{3} {8}$$ + $$\frac{5} {24}$$– 110 $$\frac{7} {60}$$

(141)$$\frac{7} {12}$$ – 110 $$\frac{7} {60}$$

? = (141 – 110) $$\frac{7} {12}$$ – $$\frac{7} {60}$$

? = 31$$\frac{28} {60}$$

? = 31$$\frac{7} {15}$$

8. 12450 – 8421 + 5641 – 4378 = ? + 24 x (36% of 275)

A. 2917
B. 2916
C. 2915
D. 2918
E. 2919

Solution: 12450 – 8421 + 5641 – 4378 = ? + 24 x (36% of 275)

18091 – 12799 = ? + 24 x (36% of 275)

? + 24 x (36% of 275) = 5292

? + 24 x $$\frac{36} {100}$$ x 275 = 5292

? + 24 x 99 = 5292

? = 5292 – 2376

? = 2916

9. 3$$\sqrt{?}$$ = (1849 – (2112 x 25 Ã· 33)) – 240

A. 728
B. 731
C. 730
D. 729
E. 727

Solution: 3$$\sqrt{?}$$ = (1849 – (2112 x 25 Ã· 33)) – 240

3$$\sqrt{?}$$= (1849 – 3$$\sqrt{?}$$ – (64 x 25)) – 240

3$$\sqrt{?}$$= (1849 – 3$$\sqrt{?}$$ – (1600) – 240

3$$\sqrt{?}$$= 249 – 240

3$$\sqrt{?}$$= 9

? = $$9^3$$ = 729

10. $$(16)^4$$ x $$(32)^2$$ Ã· $$(8)^5$$ = $$(2)^{P^3+3}$$

A. 1
B. 3
C. 4
D. 5
E. 2

Solution: $$(16)^4$$ x $$(32)^2$$ Ã· $$(8)^5$$ = $$(2)^{P^3+3}$$

$$2^{4 Ã— 4}$$

$$(2)^{4 Ã— 4}$$ x $$(2)^{2 Ã— 5}$$ Ã· $$(2)^{5 Ã— 3}$$ = $$(2)^{P^3+3}$$

$$(2)^{16}$$ Ã— $$(2)^{10}$$ Ã· $$(2)^{15}$$ = $$(2)^{P^3+3}$$

$$(2)^{16 – 5}$$ = $$(2)^{P^3+3}$$

$$(?){^3}$$ + 3 = 11

$$(?){^3}$$ = 11 – 3 = 8

? = 3$$\sqrt{8}$$ = 2

Simplification:

Directions for Questions (1 â€“ 5): 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)

1. 32% of 1699.98 + 71.93% of 225.10 = ? + 19% of 1101

A. 498
B. 499
C. 497
D. 500
E. 496

Solution: (32% of 1699.98 + 71.93% of 225.10 = ? + 19% of 1101

$$\frac{32} {100}$$ x $$\frac{72} {100}$$x 225 â‰ˆ ? + $$\frac{19} {100}$$ x 1101

? + 19 x 11.01 = 32 x 17 + 9 x 18

? + 209 â‰ˆ 544 + 162

? = 706 – 209 = 497

2. 27.04 x 116.95 Ã· 12.98 + $$(87.01)^2$$ = ? + 7312

A. 501
B. 500
C. 502
D. 499
E. 503

Solution: 27.04 x 116.95 Ã· 12.98 + $$(87.01)^2$$ = ? + 7312

27 x 117 Ã· 13 + $$(87.01)^2$$ = ? + 7312

? + 7312 = 27 x 9 + 7569

? + 7312 = 27 x 9 + 7569

? = 257 + 243

? = 500

3. $$\frac{1} {43}$$ of 4817 + (82.01 x 21) = ? + (37272 Ã· 24.02)

A. 285
B. 286
C. 284
D. 281
E. 278

Solution: $$\frac{1} {43}$$ of 4817 + (82.01 x 21) = ? + (37272 Ã· 24.02)

$$\frac{1} {43}$$ 4817 + (82 x 21) â‰ˆ ? + (37272 Ã· 24)

? + 150

? + 1553 = 112. 02 + 1722

? â‰ˆ 169 + 112

? = 281

4. 48% of 6410 + 71% of 8108 = ? $$– (45.02)^2$$

A. 10848
B. 10849
C. 10847
D. 10846
E. 10845

Solution: 48% of 6410 + 71% of 8108 = ? $$– (45.02)^2$$

$$\frac{48} {100}$$ x 6410 + $$\frac{24} {100}$$ x 8108 â‰ˆ ? $$– (45)^2$$

– 2025 = 48 x 64.10 + 71 x 81.08

– 2025 â‰ˆ 3072 + 5751

? = 8823 + 2025

? = 10848

5. 27.01 of $$\frac{5} {81}$$ of 8700 + $$\sqrt{27.02 Ã— 19 + 9 Ã— 24}$$ = $$?^2$$ – 24

A. 13
B. 12
C. 14
D. 15
E. 11

Solution: 27.01 of $$\frac{5} {81}$$ of 8700 + $$\sqrt{27.02 Ã— 19 + 9 Ã— 24}$$ = $$?^2$$ – 24

$$\frac{27} {100}$$ $$\frac{5} {81}$$ x 8700 + $$\sqrt{27 Ã— 19 + 9 Ã— 24}$$ = $$?^2$$ – 24

$$?^2$$ – 24 = 5 x 29 + $$\sqrt{513 + 216}$$

$$?^2$$ – 24 = 145 + $$\sqrt{729}$$

$$?^2$$ = 145 + 27 + 24 = 196

$$?^2$$ = $$\sqrt{196}$$

= 14

Number Series:

Directions for Questions (6 â€“ 10): What will come at the place of question mark (?) in the following number series?

6. 652, 211, ?, 166, 742, 117

A. 694
B. 695
C. 696
D. 697
E. 698

Solution: The pattern is

652 – $$21^2$$ = 211

211 + $$22^2$$ = 695

695 – $$23^2$$ = 166

166 + $$24^2$$ = 742

742 – $$25^2$$ = 117

7. 225, 255, 327, 383, ?, 545

A. 455
B. 454
C. 453
D. 456
E. 457

Solution: The pattern is

225 + 5 x 6 = 255

255 + 6 x 7 = 327

327 + 7 x 8 = 383

383 + 8 x 9 = 455

455 + 9 x 10 = 545

8. 652, 211, ?, 166, 742, 117

A. 694
B. 695
C. 696
D. 697
E. 698

Solution: The pattern is

652 – $$21^2$$ = 211

211 + $$22^2$$ = 695

695 – $$23^2$$ = 166

166 + $$24^2$$ = 742

742 – $$25^2$$ = 117

9. 1, 17, 49, ?, 161, 24

A. 95
B. 96
C. 98
D. 99
E. 97

Solution: 1 + $$4^2$$ x 1 = 17

17 + $$4^2$$ x 1 = 49

49 + $$4^2$$ x 1 = 97

97 + $$4^2$$ x 1 = 161

161 + $$4^2$$ x 1 = 241

10. 19, 10, ?, 18, 38, 97.5

A. 11
B. 10
C. 12
D. 14
E. 13

Solution: 19 x .5 +.5 = 10

10 x 1 + 1 = 11

11 x 1.5 + 1.5 = 18

18 x 2 +2 = 38

38 x 2.5 + 2.5 = 97.5

Mensuration:

1. The breadth of a rectangle is equal to the diameter of a circle. The circumference of a circle is equal to the perimeter of a square of side 33m. What is the length of a rectangle. If the area of a rectangle is 756$$m^2$$?

A. 18m
B. 19m
C. 20m
D. 17m
E. 16m

Solution: According to the question,

Perimeter of a square = 4 Ã— side

Perimeter of a square = 4 Ã— 33 = 132m

Circumference of a circle 2Ï€r

132 = 2 $$\frac{22} {7}$$ Ã— r

r = 21m

Diameter of a circle = 2r

Diameter of a circle = 2 Ã— 21 = 42m

Breadth of a rectangle = 42m

Area of a rectangle = 756 $$m^2$$

Area of a rectangle = Length of a rectangle Ã— Breadth of a rectangle

Length of a rectangle = $$\frac{756} {42}$$ = 18m

Compound Interest:

2. What will be the compound interest on a sum of Rs. 44,375 at compound interest compounded annually at 12% per annum in two years?

A. Rs.11,288
B. Rs.11,290
C. Rs.11,289
D. Rs.11,287
E. Rs.11,291

Solution: According to the question,

P = Rs.44,375

r = 12%

t = 2 years

C.I = P({1 + $$\frac{r} {100})^t$$ – 1}

C.I = 44375{(1 + $$\frac{12} {100})^2$$– 1}

C.I = 44375{( $$\frac{25 + 3} {25})^2$$– 1}

C.I = 44375{$$\frac{784 } {625}$$ – 1}

C.I = 44375{$$\frac{784 – 625} {625}$$}

C.I = 44375 x $$\frac{159} {625}$$

C.I = 71 x 159 = Rs.11,289

Time and work:

3. 14 girls and 10 boys can do a piece of work in 5 days while 14 girls and 6 boys can do the same piece of work in 6 days. In how many days can 19 girls and 16 boys do the same piece of work?

A. 3$$\frac{6} {7}$$ days
B. 3$$\frac{5} {7}$$ days
C. 3$$\frac{2} {7}$$ days
D. 3$$\frac{3} {7}$$ days
E. 3$$\frac{4} {7}$$ days

Solution: 14 girls and 10 boys 1 day work

14 girls and 6 boys 1 day work

According to the question,

(14 girls + 10 boys) Ã— 5 â‰¡ (14 girls + 6 boys) Ã— 6

(70 girls + 50 boys) Ã— 5 â‰¡ (84 girls + 36 boys) Ã— 6

14 boys â‰¡ 14 girls

1 boy â‰¡ 1 girl

(19 girls + 50 boys) â‰¡ (19 girls + 16 boys) = 35 boys

M1 D1 = M2 D2

24 x 5 = 35 D2

D2 = $$\frac{5 Ã— 24} {35}$$

D2 = 3$$\frac{3} {7}$$

Percentage:

4. If the numerator of a fraction is increased by and denominator is also increased by then the value of fraction is $$\frac{4} {7}$$. The original fraction is:

A. $$\frac{29} {50}$$
B. $$\frac{30} {55}$$
C. $$\frac{34} {55}$$
D. $$\frac{31} {55}$$
E. $$\frac{32} {55}$$

Solution: According to the question,

Let the original fraction be $$\frac{100 x} {100 y}$$ then,

($$\frac{100 + 450} {100 + 300}$$)$$\frac{x} {y}$$ = $$\frac{4} {5}$$

The original fraction

($$\frac{550} {400}$$)$$\frac{x} {y}$$ = $$\frac{4} {5}$$

($$\frac{x} {y}$$) = $$\frac{4 Ã— 8} {11 Ã— 5 }$$

($$\frac{x} {y}$$) = $$\frac{32} {55}$$

The original fraction = $$\frac{32} {55}$$

Ratio and Proportion:

5. In a college the students in Math and Hindi classes are in the ratio 17:19. When 12 more students join Hindi class the ratio becomes 17:21. How many students are there in the Hindi class?

A. 115
B. 110
C. 112
D. 114
E. 113

Solution: Let the number of students in Math and Hindi be , then According
to the question,

($$\frac{17 x} {19 x + 12}$$) = $$\frac{17} {21}$$

357x = 323x + 204

357x – 323x – 204

34x = 204

x = 6

Number of students in Hindi class = 19x = 19 Ã— 6 =114

Pipes and Cistern:

6. Pipe J can fill a tank in 14hours while pipe K alone can fill it in 28hours and pipe L can fill a tank in 42hours. Pipe J was opened and after one hour Pipe K was opened and after two hours from start of Pipe J, Pipe L was also opened. Find the time in which the cistern is just full?

A. 3$$\frac{3} {11}hours$$
B. 8$$\frac{3} {11}hours$$
C. 7$$\frac{3} {11}hours$$
D. 5$$\frac{3} {11}hours$$
E. 6$$\frac{3} {11}hours$$

Solution: According to the question

Part of cistern filled by Pipe J in first 2 hours = $$\frac{1} {14} Ã— 2$$ = $$\frac{1} {7}$$

357x – 323x – 204

34x = 204

x = 6

Number of students in Hindi class = 19x = 19 Ã— 6 = 114

Part of cistern filled by Pipe K in one hour = $$\frac{1} {28}$$

Part of cistern filled by Pipe K in one hour = $$\frac{1} {28}$$

Remaining to the filled = 1 – $$\frac{1} {7}$$ – $$\frac{1} {28}$$

Pipe J, K and L together 1hour filled = $$\frac{1} {14}$$ + $$\frac{1} {28}$$ + $$\frac{1} {42}$$ = $$\frac{6 + 3 + 2} {84}$$

$$\frac{23} {28}$$ part of cistern is filled by pipe J, K and L $$\frac{23/28} {11/84}$$

= $$\frac{23 Ã— 84} {28 Ã— 11}$$

= $$\frac{69} {11}$$

= 6$$\frac{3} {11}$$hours

Profit and Loss:

7. A shopkeeper sells one article for Rs.720 at a gain of 20% and another for Rs.640 at a loss of 20%. His total gain or loss percent is:

A. 7.12%
B. 9.22%
C. 5.12%
D. 2.86%
E. 4.28%

Solution: According to the question,

C.P. of 1st article $$\frac{720 Ã— 100} {120}$$ = Rs600

C.P. of 2nd article $$\frac{640 Ã— 100} {80}$$ = Rs800

Total C.P. = 600 + 800 = Rs 1400

Total S.P. = 720 + 640 = Rs 1360

Loss = 1400 – 1360 = Rs 40

Loss% = $$\frac{40 Ã— 1400} {100}$$ = $$\frac{40} {14}$$

= 2.86%

Average:

8. The sum of three numbers is 195. If the ratio of the first and the second is 4:5 and that of the second and the third is 5:4, then what is the sum of first and third numbers?:

A. 114
B. 118
C. 110
D. 115
E. 120

Solution: According to the question,

Let the three numbers be a, b, and c, then

a : b = 4 : 5

b : c = 5 : 4

a : b : c = 20 : 25 : 20

a = $$\frac{195 } {20 + 25 + 20}$$ x 20

a = $$\frac{195 } {65}$$ x 20 = 60

c = $$\frac{195 } {20 + 25 + 20}$$ x 20

c = $$\frac{195 } {65}$$ x 20 = 60

sum of first and third number = 60 + 60 = 120

Probability:

9. A box contains 4 blue, 8 grey and 8 white balls. If 4 balls are picked at random. What is the probability that two are grey and two are white?:

A. $$\frac{784} {4845}$$
B. $$\frac{788} {4845}$$
C. $$\frac{781} {4845}$$
D. $$\frac{787} {4845}$$
E. $$\frac{784} {4845}$$

Solution: Total number of balls in the box = 4 + 8 + 8 = 20

P = $$\frac{n(E)} {n(S)}$$

n(E) = $$8_{{C}_{2}}$$ x $$8_{{C}_{2}}$$

= !$$\frac{!8} {!2 Ã— !8 – 2}$$ = $$\frac{8 Ã— 7 } {2}$$ X $$\frac{8 Ã— 7 } {2}$$

= 28 X 28 = 784

n(S)= $$20_{{C}_{4}}$$ = !$$\frac{!120} {!4 Ã— !120 – 4}$$ = $$\frac{20 Ã— 19 Ã— 18 Ã— 17} {4 Ã— 3 Ã— 2 Ã— 1}$$

= 5 Ã— 3 Ã— 19 Ã— 17 = 4845

P = $$\frac{784} {4845}$$

Permutation and Combination:

10. In how many different ways can the letters of the word “HELICOPTER” be arranged so that the consonant always come together?:

A. 720
B. 120
C. 5040
D. 40320
E. 43200

Solution: In word “HELICOPTER” there are 10 letters in which consonants are H, L, C, P, T and R and vowels are E, I, O, and E. E occurs twice So it can be arranged in following ways-

= $$\frac{!6 Ã— !5} {!2}$$

= $$\frac{6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1} {2 Ã— 1}$$

= 720 Ã— 60 = 43200

Data Interpretation:

Directions for Questions (1 – 5) Study the following line-chart carefully to answer the given questions.

1. What is the percentage increase number of male workers working in factory G in 2011 as compared to the previous year?

A. 52%
B. 48%
C. 55%
D. 50%
E. 45%

Solution: Number of male workers working in factory G in 2011

= 180 x 100 = 18000

Number of male workers working in factory G in 2010

120 x 100 = 12000

Percentage increase = $$\frac{18000 – 12000} {12000}$$ x 100

= $$\frac{6000} {12000}$$ x 100

= 50%

2. Number of male workers working in factory H in 2014 and 2015 together are approximately what percent of number of male workers working in factory I in 2010 and 2015 together?

A. 71%
B. 67%
C. 74%
D. 64%
E. 62%

Solution: Number of male workers working in factory H in 2014 and 2015 together

(180 + 260) x 100 = 44000

Number of male workers working in factory I in 2010 and 2015 together

(320 + 340) x 100 = 44000

Percentage = $$\frac{44000} {66000}$$ x 100

= $$\frac{4400} {6600}$$

= 66.7% â‰ˆ 67%

3. What is the difference between number of male workers working in factory I in 2010, 2011, 2012, 2013 and 2015 together and number of male workers working in factory G, H, I and J together in 2013?

A. 22120
B. 22080
C. 22050
D. 22100
E. 22000

Solution: Number of male workers working in factory I in 2010, 2011, 2012, 2013 and 2015 together

= (320 + 240 + 180 + 140 + 340) x 100

= 12200

= Number of male workers working in factory G, H, I and J together in 2013

= (240 + 180 + 140 + 340) x 100

= 10000

Difference = 12200 – 10000

= 22000

4. What is the percentage decrease in number of male workers working in factory H in 2013 as compared to the previous year?

A. 17.6%
B. 15.5%
C. 10.3%
D. 12.5%
E. 14.2%

Solution: Number of male workers working in factory H in 2013

280 X 100 = 28000

Number of male workers working in factory H in 2012

320 X 100 = 32000

Percentage decrease = $$\frac{32000 – 28000} {32000}$$ x 100

= $$\frac{4000} {32000}$$ x 100

= $$\frac{100} {8}$$

= 12.5%

5. What is the ratio between number of male workers working in factory G and J together in 2014 to number of male workers working in factory H and I together in 2015?

A. 29 : 30
B. 19 : 20
C. 29 : 20
D. 30 : 19
E. 30 : 29

Solution: Number of male workers working in factory G and J together in 2014

Number of male workers working in factory G and J together in 2014

= (300 + 280) x 100

= 58000

Number of male workers working in factory H and I together in 2015

= (260 + 340 x 100

= 60000

Ratio = $$\frac{58000} {60000}$$

= $$\frac{58} {60}$$

= 29 : 30

1. Prime factors of 40 are _______.

1. 2 x 2 x 3 x 5 x 5
2. 2 x 3 x 5 x 5
3. 2 x 2 x 2 x 5
4. 2 x 3 x 5

2. Answer of following (aÂ³b)4is ________.

1. a14b5
2. a8b4
3. a12b4
4. a10bÂ³

3. Tan(Î± + Î²) = ___________.

1. tanÎ±-tanÎ²/1 + tanÎ±tanÎ²
2. tanÎ±+tanÎ²/1 – tanÎ±tanÎ²
3. cotÎ±+cotÎ²/1-cotÎ±cotÎ²
4. cotÎ±-cotÎ²/1 + cotÎ±cotÎ²

4. Path described by any moving point is classified as ___________.

1. ordinate ray rays
2. line segment
3. rays
4. line

5. The sector of a circle has radius of 21 cm and central angle 135o. Find its perimeter?

1. 91.5 cm
2. 93.5 cm
3. 94.5 cm
4. 92.5 cm

Solution:

Perimeter of the sector = length of the arc + 2(radius)

= (135/360 * 2 * 22/7 * 21) + 2(21)

= 49.5 + 42 = 91.5 cm

6. The middle value of an ordered array of numbers is the

1. Mode
2. Mean
3. Median
4. MidPoint

7. Which of the following is not a measure of central tendency?

1. Percentile
2. Quartile
3. Standard deviation
4. Mode

Solution:

Standard Deviation is measure of Dispersion or variability its not measure of central tendency.

8. What will be the cost of building a fence around a square plot with area equal to 289 sq ft, if the price per foot of building the fence is Rs. 58?

1. Rs. 3944
2. Rs. 3828
3. Rs. 4176
4. Cannot be determined

Solution:

Let the side of the square plot be a ft.

a2 = 289 => a = 17

Length of the fence = Perimeter of the plot = 4a = 68 ft.

Cost of building the fence = 68 * 58 = Rs. 3944.

9. In a triangle ABC, if angle A = 72Â° , angle B = 48Â° and c = 9 cm then Äˆ is

1. 69Â°
2. 66Â°
3. 60Â°
4. 63Â°

10. In a chess tournament each of six players will play every other player exactly once. How many matches will be played during the tournament?

1. 12
2. 15
3. 30
4. 36