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IBPS Clerk Prelims Quantitative Aptitud...

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IBPS Clerk Prelims Quantitative Aptitude Practice Sets

shape 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.

shape Aptitude

Simplification:

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

1. (13056 ÷ 32)(513 ÷ 19) = ? + 95 \( \sqrt{1104 + 40 × 30} \)

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


Answer: Option A

Solution: (13056 ÷ 32)(513 ÷ 19) = ? + 95 \( \sqrt{1104 + 40 × 30} \)

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

11016 = ? + 95 \( \sqrt{2304} \)

? = 11016 – 95 x 48

? = 11016 – 4560

? = 6456

2. \( \sqrt{29 × 1240 ÷ 31 + 173 × 5} \) × 36 =\( ?^2 \) + 29 × 16

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


Answer: Option E

Solution: \( \sqrt{29 × 1240 ÷ 31 + 173 × 5} \) × 36 =\( ?^2 \) + 29 × 16

\( \sqrt{29 × 40 + 865} \) × 36 = \( ?^2 \) + 464

\( \sqrt{1160 + 865} \) × 36 = \( ?^2 \) + 464

\( ?^2 \) = 45 x 36 – 464

\( ?^2 \) = 1620 – 464

\( ?^2 \) = 1156

\( ?^2 \) = \( \sqrt {1156}\)

\( ?^2 \) = 34

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

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


Answer: Option A

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

(16980 – 14558) ÷ 14 ? + 150

2422 ÷ 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


Answer: Option A

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


Answer: Option B

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

\( \frac{16 − 7} {26}\) \( ?^2 \) = 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


Answer: Option C

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 \( \frac{8} {15}\)


Answer: Option A

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


Answer: Option B

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 ÷ 33)) – 240

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


Answer: Option D

Solution: 3\( \sqrt{?}\) = (1849 – (2112 x 25 ÷ 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 \( (32)^2 \) ÷ \( (8)^5 \) = \( (2)^{P^3+3} \)

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


Answer: Option E

Solution: \( (16)^4 \) x \( (32)^2 \) ÷ \( (8)^5 \) = \( (2)^{P^3+3} \)

\( 2^{4 × 4} \)

\( (2)^{4 × 4} \) x \( (2)^{2 × 5} \) ÷ \( (2)^{5 × 3} \) = \( (2)^{P^3+3} \)

\( (2)^{16} \) × \( (2)^{10} \) ÷ \( (2)^{15} \) = \( (2)^{P^3+3} \)

\( (2)^{16 – 5}\) = \( (2)^{P^3+3} \)

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

\( (?){^3} \) = 11 – 3 = 8

? = 3\( \sqrt{8} \) = 2

Simplification:

Directions for Questions (1 – 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


Answer: Option C

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

\( \frac{32} {100} \) x \( \frac{72} {100} \)x 225 ≈ ? + \( \frac{19} {100}\) x 1101

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

? + 209 ≈ 544 + 162

? = 706 – 209 = 497

2. 27.04 x 116.95 ÷ 12.98 + \( (87.01)^2 \) = ? + 7312

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


Answer: Option B

Solution: 27.04 x 116.95 ÷ 12.98 + \( (87.01)^2 \) = ? + 7312

27 x 117 ÷ 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 ÷ 24.02)

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


Answer: Option D

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

\( \frac{1} {43} \) 4817 + (82 x 21) ≈ ? + (37272 ÷ 24)

? + 150

? + 1553 = 112. 02 + 1722

? ≈ 169 + 112

? = 281

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

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


Answer: Option A

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

\( \frac{48} {100}\) x 6410 + \( \frac{24} {100}\) x 8108 ≈ ? \( – (45)^2 \)

– 2025 = 48 x 64.10 + 71 x 81.08

– 2025 ≈ 3072 + 5751

? = 8823 + 2025

? = 10848

5. 27.01 of \( \frac{5} {81}\) of 8700 + \( \sqrt{27.02 × 19 + 9 × 24} \) = \( ?^2 \) – 24

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


Answer: Option C

Solution: 27.01 of \( \frac{5} {81}\) of 8700 + \( \sqrt{27.02 × 19 + 9 × 24} \) = \( ?^2 \) – 24

\( \frac{27} {100}\) \(\frac{5} {81} \) x 8700 + \( \sqrt{27 × 19 + 9 × 24} \) = \( ?^2 \) – 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 – 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


Answer: Option B

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


Answer: Option A

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


Answer: Option B

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


Answer: Option E

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


Answer: Option A

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


Answer: Option A

Solution: According to the question,

Perimeter of a square = 4 × side

Perimeter of a square = 4 × 33 = 132m

Circumference of a circle 2πr

132 = 2 \( \frac{22} {7}\) × r

r = 21m

Diameter of a circle = 2r

Diameter of a circle = 2 × 21 = 42m

Breadth of a rectangle = 42m

Area of a rectangle = 756 \( m^2 \)

Area of a rectangle = Length of a rectangle × 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


Answer: Option C

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


Answer: Option D

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) × 5 ≡ (14 girls + 6 boys) × 6

(70 girls + 50 boys) × 5 ≡ (84 girls + 36 boys) × 6

14 boys ≡ 14 girls

1 boy ≡ 1 girl

(19 girls + 50 boys) ≡ (19 girls + 16 boys) = 35 boys

M1 D1 = M2 D2

24 x 5 = 35 D2

D2 = \( \frac{5 × 24} {35}\)

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. \( \frac{32} {55}\)


Answer: Option A

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

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

(\( \frac{x} {y}\)) = \( \frac{4 × 8} {11 × 5 }\)

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

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


Answer: Option D

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 × 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\( \frac{3} {11}hours \)


Answer: Option E

Solution: According to the question

Part of cistern filled by Pipe J in first 2 hours = \( \frac{1} {14} × 2 \) = \( \frac{1} {7}\)

357x – 323x – 204

34x = 204

x = 6

Number of students in Hindi class = 19x = 19 × 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 \( \frac{23/28} {11/84} \)

= \( \frac{23 × 84} {28 × 11} \)

= \( \frac{69} {11} \)

= 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%


Answer: Option D

Solution: According to the question,

C.P. of 1st article \( \frac{720 × 100} {120} \) = Rs600

C.P. of 2nd article \( \frac{640 × 100} {80} \) = Rs800

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

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

Loss = 1400 – 1360 = Rs 40

Loss% = \( \frac{40 × 1400} {100} \) = \( \frac{40} {14} \)

= 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


Answer: Option E

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. \( \frac{784} {4845} \)


Answer: Option E

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

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

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

= !\( \frac{!8} {!2 × !8 – 2} \) = \( \frac{8 × 7 } {2} \) X \( \frac{8 × 7 } {2} \)

= 28 X 28 = 784

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

= 5 × 3 × 19 × 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


Answer: Option E

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-

= \( \frac{!6 × !5} {!2} \)

= \( \frac{6 × 5 × 4 × 3 × 2 × 1} {2 × 1} \)

= 720 × 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%


Answer: Option D

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%


Answer: Option B

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% ≈ 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


Answer: Option E

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%


Answer: Option D

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


Answer: Option A

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

Correct Answer: 3

2. Answer of following (a³b)4is ________.

  1. a14b5
  2. a8b4
  3. a12b4
  4. a10

Correct Answer: 3

3. Tan(α + β) = ___________.

  1. tanα-tanβ/1 + tanαtanβ
  2. tanα+tanβ/1 – tanαtanβ
  3. cotα+cotβ/1-cotαcotβ
  4. cotα-cotβ/1 + cotαcotβ

Correct Answer: 2

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

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

Correct Answer: 4

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

Correct Answer: 1

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

Correct Answer: 3

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

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

Correct Answer: 3

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

Correct Answer: 1

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 = 72° , angle B = 48° and c = 9 cm then Ĉ is

  1. 69°
  2. 66°
  3. 60°
  4. 63°

Correct Answer: 3

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

Correct Answer: 2

Solution:

  1. matches of first player with other 5 players
  2. matches of second player with 4 players other than the first player

III. matches of third player with 3 players other than the first player and second player.

  1. matches of fourth player with 2 players other than the first player, second player and third player.
  2. matches of fifth player with 1 player other than the first player, second player, third player and fourth player.

So total matches will be 5+4+3+2+1 = 15


1. What is the area of an equilateral triangle of side 16 cm?

  1. 48√3 cm2
  2. 128√3 cm2
  3. 9.6√3 cm2
  4. 64√3 cm2

Correct Answer: 4
Solution:
Area of an equilateral triangle = √3/4 S2
If S = 16, Area of triangle = √3/4 * 16 * 16 = 64√3 cm2


2. The mean of first n natural numbers is ______.

  1. (n+1)/2
  2. n(n+1)/2
  3. (n-1)/2

Correct Answer: 1


3.The means of the frequency distribution is ______.

x 1 2 3 4 5 6
f 45 25 19 8 2 1
  1. 1
  2. 2
  3. 3
  4. 4

Correct Answer: 2


4. The modal class of data given below is 10-15 then ________.

x 0-5 5-10 10-15 15-20 20-25
Frequency 7 6 f 4 3
  1. f < 8
  2.  ≥ 8
  3. f > 8 only
  4. f < 7

Correct Answer: 2


5. The number of family members of 30 families of a village is according to the following table. Find the mode.

  1. 2
  2. 4
  3. 3
  4. 6

Correct Answer: 4


6. The wickets taken over by a bowler in 10 cricket matches are as follows 2 6 4 5 0 1 2 1 3 2 3 . Find the mode of this data.

  1. 0
  2. 1
  3. 2
  4. 3

Correct Answer: 3


7. The cumulative frequency of the class 55-58 is how much greater than the frequency of the class 58-61 in the following distribution.

Height 52-55 55-58 58-61 61-64
No of students 10 20 25 10
  1. 2
  2. 3
  3. 4
  4. 5

Correct Answer: 4

Solution: 

Cumulative frequency of the class 55-58 = 10+20 = 30

Frequency of the class 58-61 = 25


8. The mean and median of a data are respectively 20 and 22. The value of mode is ________.

  1. 20
  2. 26
  3. 22
  4. 21

Correct Answer: 2

Solution: 

3 median =  mode + 2 mean

3 * 22 = Mode + 2*20

Mode = 26


9. If mean = (3medain – mode)/k, then K = ______.

  1. 1
  2. 1/2
  3. 2
  4. None of these

Correct Answer: 3


10. Find the mode of the following distribution.

Size of the shows 4.5 5.0 5.5 <6.0/td> 6.5 7.0 7.5 8.0 8.5 9.0
No of shoes 1 2 3 4 5 15 30 60 95 82 75
  1. 6
  2. 7
  3. 8
  4. 9

Correct Answer: 3

1. Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.

  1. 225 cm2
  2. 275 cm2
  3. 285 cm2
  4. 315 cm2

Correct Answer: 3
Solution:
Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm2


2. Find the area of a parallelogram with base 24 cm and height 16 cm.

  1. 262 cm2
  2. 384 cm2
  3. 192 cm2
  4. 131 cm2

Correct Answer: 2
Solution:
Area of a parallelogram = base * height = 24 * 16 = 384 cm2


3. The ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is 6912 sq cm. Find the ratio of the breadth and the area of the rectangle?

  1. 1 : 96
  2. 1 : 48
  3. 1 : 84
  4. 1 : 68

Correct Answer: 1
Solution:

Let the length and the breadth of the rectangle be 4x cm and 3x respectively.
(4x)(3x) = 6912
12x2 = 6912
x2 = 576 = 4 * 144 = 22 * 122 (x > 0)
=> x = 2 * 12 = 24
Ratio of the breadth and the areas = 3x : 12x2 = 1 : 4x = 1: 96.


4. The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle?

  1. 8
  2. 12
  3. 6
  4. 2

Correct Answer: 4
Solution:

Let the sides of the rectangle be l and b respectively.
From the given data,
(√l2 + b2) = (1 + 108 1/3 %)lb
=> l+ b2 = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l2 + b2)/lb = 25/12
12(l2 + b2) = 25lb
Adding 24lb on both sides
12l2 + 12b2 + 24lb = 49lb
12(l2 + b2 + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)2 = 49lb
=> 12(14)2 = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l – b = 8 – 6 = 2m.


5. An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of carpet is Rs. 45 per sq m?

  1. Rs. 3642.40
  2. Rs. 3868.80
  3. Rs. 4216.20
  4. Rs. 4082.40

Correct Answer: 4
Solution:

Length of the first carpet = (1.44)(6) = 8.64 cm
Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)
= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m
Cost of the second carpet = (45)(12.96 * 7) = 315 (13 – 0.04) = 4095 – 12.6 = Rs. 4082.40


6. The circumferences of two circles are 264 meters and 352 meters. Find the difference between the areas of the larger and the smaller circles.

  1. 4192 sq m
  2. 4304 sq m
  3. 4312 sq m
  4. 4360 sq m

Correct Answer: 3
Solution:

Let the radii of the smaller and the larger circles be s m and l m respectively.
2∏s = 264 and 2∏l = 352
s = 264/2∏ and l = 352/2∏
Difference between the areas = ∏l2 – ∏s2
= ∏{1762/∏2 – 1322/∏2}
= 1762/∏ – 1322/∏
= (176 – 132)(176 + 132)/∏
= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq m


7. A cube of side one meter length is cut into small cubes of side 10 cm each. How many such small cubes can be obtained?

  1. 10
  2. 100
  3. 1000
  4. 10000

Correct Answer: 3
Solution:

Along one edge, the number of small cubes that can be cut
= 100/10 = 10
Along each edge 10 cubes can be cut. (Along length, breadth and height). Total number of small cubes that can be cut = 10 * 10 * 10 = 1000


8. The length of a rectangle is two – fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 1225 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units?

  1. 140
  2. 156
  3. 175
  4. 214

Correct Answer: 1
Solution:

Given that the area of the square = 1225 sq.units

=> Side of square = √1225 = 35 units
The radius of the circle = side of the square = 35 units Length of the rectangle = 2/5 * 35 = 14 units
Given that breadth = 10 units
Area of the rectangle = lb = 14 * 10 = 140 sq.units


9. The ratio of the volumes of two cubes is 729 : 1331. What is the ratio of their total surface areas?
  1. 81 : 121
  2. 9 : 11
  3. 729 : 1331
  4. 27 : 121

Correct Answer: 1
Solution:

Ratio of the sides = ³√729 : ³√1331 = 9 : 11

Ratio of surface areas = 92 : 112 = 81 : 121