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2122

Question 1:Â The product of ages of Harris and Sharma is 240 .If twice the age of Sharma is more than Harris age by 4 years ,what is Sharma age in years?

a)Â 12

b)Â 20

c)Â 10

d)Â 14

Question 2:Â If the ages of P and R added to twice the age of Q ,the total becomes 59 .If the age of Q and R are added to thrice age of P,the total becomes 68 and if the age of the P is added to thrice age of Q and thrice the age age of R the total becomes 108 .What is the age of P?

a)Â 15 years

b)Â 19 years

c)Â 17 years

d)Â 12 years

e)Â None of these

Question 3:Â Jayesh is twice as old as Vijay and half as old as Suresh .If sum of Sureshâ€™s and Vijayâ€™s age is 85 years what is the Jayeshâ€™s age in years?

a)Â 34

b)Â 36

c)Â 68

d)Â Cannot determined

e)Â None of these

Question 4:Â 4 years ago, the respective ratio between ${1 \over 2}$ of Aâ€™s age at that time and four times of Bâ€™s age at that time was 5: 12. Eight years hence ${1 \over 2}$ of Aâ€™s age at that time will be less than Bâ€™s age at that time by 2 years. What is Bâ€™s present age ?

a)Â 10 years

b)Â 14 years

c)Â 12 years

d)Â 5 years

e)Â 8 years

Question 5:Â The sum of the present ages of P and Q is 25 years more than the age of R. The present age of Q is 5 years more than that of R. Find the present ageÂ  of P

a)Â 20 years

b)Â 25 years

c)Â 21 years

d)Â 22 years

e)Â None of these

Question 6:Â 7.69% of 130 + 6.66% of 150 = ?

a)Â 20

b)Â 21

c)Â 22

d)Â 23

e)Â 24

Question 7:Â $\frac{1}{7}$ of $2\frac{5}{8} \div \frac{3}{4} + \frac{1}{3} + \frac{1}{6}$

a)Â 1

b)Â 0

c)Â $\frac{1}{7}$

d)Â 2

e)Â None of the above

Question 8:Â $125^{4} \div 25^{4} * 5 ^{5} = 5 ^{?}$

a)Â 7

b)Â 9

c)Â 11

d)Â 13

e)Â none of the above

Question 9:Â What is the LCM of 1/3, Â¼ and 1/5?

a)Â 1

b)Â 2

c)Â Â½

d)Â 1/6

e)Â None of the above

Question 10:Â 134% of 3894 + 38.94 of 134 = ?

a)Â 11452

b)Â 10000

c)Â 10452

d)Â 1100

e)Â None of these

Question 11:Â In order to pass in an exam a student is required to get 780 marks out of the aggregate marks. Sonu got 728 marks and was declared failed by 5 per cent. What are the maximum aggregate marks a student can get in the examination.

a)Â 1040

b)Â 1100

c)Â 1000

d)Â Cannot be determined

e)Â None of these

Question 12:Â The average of four consecutive numbers A, B, C and D respectively is 49.5. What is the product of B and D?

a)Â 2499

b)Â 2352

c)Â 2450

d)Â 2550

e)Â None of these

Question 13:Â Nandita scored 80 per cent marks in five subjects together viz, Hindi, Science, Maths, English and Sanskrit together wherein the maximum marks of each subject were 105. How many marks did Nandita score in Science if she scored 89 marks in Hindi, 92 marks in Sanskrit, 98 marks in Maths and 81 marks in English?

a)Â 60

b)Â 75

c)Â 65

d)Â 70

e)Â None of these

Question 14:Â Find the average of the following sets of sources.
253, 124, 255, 534, 836, 375, 101, 443, 760

a)Â 427

b)Â 413

c)Â 441

d)Â 490

e)Â None of these

Question 15:Â Ramolaâ€™s monthly income is three times Ravinaâ€™s monthly income, Ravinaâ€™s monthly income is fifteen percent more that Ruchikaâ€™s monthly income. Ruchikaâ€™s monthly income is Rs. 32,000. What is Ramolaâ€™s annual income?

a)Â Rs. 1, 10, 400

b)Â Rs. 13, 24, 800

c)Â Rs. 36, 800

d)Â Rs. 52, 200

e)Â None of these

Question 16:Â Seven men, five women and eight children were given an assignment of distributing 2000 books to students in a school over a period of three days. All of them distributed books on the first day. On the second day two women and three children remained absent and on the third day three men and five children remained absent. If the ratio of the number of books distributed in a day by a man, a woman and a child was 5: 4: 2 respectively, a total of â€˜â€™approximatelyâ€™â€™ how many books were distributed on the second day ?

a)Â 1000

b)Â 800

c)Â 650

d)Â 900

e)Â Cannot be determined

Question 17:Â Are all the four friends viz. A, B, C and D who are sitting around a circular table, facing the centre?
I) B sits second to right of D. D faces the centre. C sits to immediate right of both B and D.
II) A sits to immediate left to B. C is not an immediate neighbour of A. C sits to immediate right of D.
III) D is an immediate neighbour of both A and C. B sits to the immediate left of A. C sits to the immediate right of B.

a)Â If the data in Statement I and II are sufficient to answer the question, while the data in Statement III are not required to answer the question.

b)Â If the data in Statement I and III are sufficient to answer the question, while the data in Statement II are not required to answer the question.

c)Â If the data in Statement II and III are sufficient to answer the question, while the data in Statement I are not required to answer the question.

d)Â If the data in either Statement I alone or Statement II alone or Statement III alone are sufficient to answer the question.

e)Â If the data in all the Statements I, II and III together are necessary to answer the question.

Question 18:Â How is ‘one’ coded in the code language?
I) ‘one of its kind’ is coded as ‘zo pi ko fe’ and ‘in kind and cash’ is coded as ‘ga to ru ko’
II) ‘Its point for origin’ is coded as ‘ba le fe mi’ and ‘make a point clear’ is coded as ‘yu si mi de’
III) ‘make money and cash’ is coded as ‘to mi ru hy’ and ‘money of various kind’ is coded as ‘qu ko zo hy’.

a)Â If the data in Statement I and II are sufficient to answer the question, while the data in Statement III are not required to answer the question.

b)Â If the data in Statement I and III are sufficient to answer the question, while the data in Statement II are not required to answer the question.

c)Â If the data in Statement II and III are sufficient to answer the question, while the data in Statement I are not required to answer the question.

d)Â If the data in either Statement I alone or Statement II alone or Statement III alone are sufficient to answer the question.

e)Â If the data in all the Statements I, II and III together are necessary to answer the question.

Question 19:Â The height of a triangle is equal to the perimeter of a square whose diagonal is 14.14m and the base of the same triangle is equal to the side of the square whose area is 784 m2. What is the area of the triangle? (in m2)

a)Â 504

b)Â 560

c)Â 478

d)Â 522

e)Â 496

Question 20:Â The diameter of a wheel is 49 m. How many revolutions will it make to cover a distance of 3200 m?

a)Â 17

b)Â 27

c)Â 24

d)Â 21

e)Â 18

Question 21:Â The area of a rectangle is equal to the area of a circle with circumference equal to 220 m What is the length of the rectangle of its breadth is 50 m ?

a)Â 56 m

b)Â 83 m

c)Â 77 m

d)Â 69 m

e)Â None of these

Question 22:Â The perimeter of a square is one-fourth the perimeter of a rectangle If the perimeter of the square is 44 cm and the length of the rectangle is 51 cm what is the difference between the breadth of the rectangle and the side of the square ?

a)Â 30 cm

b)Â 18 cm

c)Â 26 cm

d)Â 32 cm

e)Â None of these

Question 23:Â What would be the area of a square whose diagonal measures 28 cm?

a)Â 288 $cm^{2}$

b)Â 514 $cm^{2}$

c)Â 428 $cm^{2}$

d)Â 392 $cm^{2}$

e)Â None of these

Question 24:Â 0ne-eighth of a number is 17.25. What will 73% of the number be ?

a)Â 100.74

b)Â 138.00

c)Â 96.42

d)Â 82.66

e)Â None of these

Question 25:Â â€˜ Aâ€™ ,â€™Bâ€™ and â€˜Câ€™ are three consecutive even integers such that four times â€˜Aâ€™ is equal to three times â€˜Câ€™. What is the value of Bâ€™?

a)Â 12

b)Â 10

c)Â 16

d)Â 14

e)Â None of these

Question 26:Â What is the LCM of the following fractions? 3/11, 2/5, 1/9

a)Â 0

b)Â 1

c)Â 2

d)Â 3

e)Â 6

Question 27:Â What will be the the compound interest acquired on sum of Rs 12,000/- for 3 years at the rate of 10 % per annum ?

a)Â 2,652

b)Â 3,972

c)Â 3,960

d)Â 3852

e)Â None of these

Question 28:Â Ms suchi deposits an amount of 24000 to obtain in a simple interest at the rate of S.I 14 p.c.p.a for 8 years .what total amont will ms suchi gets at the end of 8 years

a)Â 52080

b)Â 28000

c)Â 50880

d)Â 26880

e)Â none of these

Question 29:Â What would be the simple interest obtained on a amount 5670 at the rate of 6 pcpa after 3 years?

a)Â 1020.60

b)Â 1666.80

c)Â 1336 .80

d)Â 1063.80

e)Â none of these

Question 30:Â A, B and C started a business by investing Rs. 20,000 Rs. 28,000 and Rs. 36,000 respectively. After 6 months, A and B withdrew an amount of Rs. 8,000 each and C invested an additional amount of Rs. 8,000. All of them invested for equal period of time. If at the end of the year. C got Rs. 12,550 as his share of profit, what was the total profit earned ?

a)Â Rs. 25,100

b)Â Rs. 26,600

c)Â Rs. 24,300

d)Â Rs. 22,960

e)Â Rs. 21,440

Question 31:Â Sarita started a boutique investing an amount of Rs. 50,000. Six months later Neeta joined her with an amount of Rs. 80,000. At the end of one year they earned a profit of Rs. 18,000. What is Saritaâ€™s share in the profit?

a)Â Rs. 9000

b)Â Rs. 8000

c)Â Rs. 12000

d)Â Rs. 10000

e)Â None of these

Question 32:Â In 1 kg of a mixture of sand and iron, 20% is iron .How such sand should be added so that the proportion of iron becomes 10%

a)Â 1 kg

b)Â 200gm

c)Â 800 gm

d)Â 1.8 kg

e)Â None of these

Question 33:Â P. Q and R have a certain amount of money with themselves. Q has 25% more than what P has, and R has ${1 \over 5}$th of what Q has. If P. Q and R together have Rs. 150, then how much money does P alone have? (in Rs.)

a)Â 40

b)Â 70

c)Â 80

d)Â 60

e)Â 50

Question 34:Â Among five people – A, B, C, D and E â€” each scoring different marks, only one person scored less marks than B. D scored more marks than B but less than A. A did not score the highest marks. Who scored the second highest marks?

a)Â E

b)Â Cannot be determined

c)Â A

d)Â C

e)Â D

Question 35:Â A, B and C have a certain amount of money with themselves. C has ${3 \over 4}$ of what A has and B has Rs. 50 less than C. If A, B and C together have Rs. 250, then how much does A alone have? (in Rs.)

a)Â 75

b)Â 160

c)Â 80

d)Â 120

e)Â 140

Question 36:Â If an amount of Rs. 97836 is distributed equally amongst 31 children, how much amount would each child get ?

a)Â Rs. 3756

b)Â Rs. 3556

c)Â Rs. 3356

d)Â Rs. 3156

e)Â None of these

Question 37:Â 12 men can finish a project in 20 days. 18 women can finish the same project in 16 days and 24 children can finish it in 18 days. 8 women and 16 children worked for 9 days and then left. In how many days will 10 men complete the remaining project ?

a)Â $10\frac{1}{2}$

b)Â 10

c)Â 9

d)Â $11\frac{1}{2}$

e)Â $9\frac{1}{2}$

Question 38:Â Sixteen men and twelve women can complete a work in 8 days, if 20 men can complete the same work in 16 days, in how many days 16 women can complete the same piece of work ?

a)Â 12

b)Â 8

c)Â 10

d)Â 15

e)Â 20

Question 39:Â What will come in the place of question mark (?) in the following series ?
2 Â 9 Â 28 Â 65 ?

a)Â 96

b)Â 106

c)Â 126

d)Â 130

e)Â None of these

Question 40:Â What will come in place of the question mark (?) in the following number series?
9Â  10Â  39Â  220Â  ?Â  14382

a)Â 1589

b)Â 1598

c)Â 1958

d)Â 1985

e)Â 1835

let age of sharma = x

let age of harris = y

xy = 240

2x = y+4

on solving these two equations

y = 20 or -24

age must be positive, so y = 20

x = 12

Let ages of P,Q and R be $p,q,r$ respectively

According to ques, => $(p+r)+2q=59$ ————(i)

and $(q+r)+3p=68$ ————-(ii)

and $p+3q+3r=108$ ————(iii)

Multiplying equation (ii) by 3, => $3q+3r+9p = 204$ ———-(iv)

Subtracting equation (iii) from (iv), we getÂ :

=> $(9p-p)=(204-108)$

=> $8p=96$

=> $p=\frac{96}{8}=12$ years

=> Ans – (D)

Let Vijay’s age = $x$ years

=> Jayesh’s age = $2x$ years and Suresh’s age = $4x$ years

Sum of Suresh’s and Vijay’s ages = $(4x+x)=85$

=> $x=\frac{85}{5}=17$

$\therefore$ Jayesh’s age = $2x=2 \times 17=34$ years

=> Ans – (A)

Let B’s present age = $x$ years

and A’s present age = $y$ years

Acc. to ques, => $\frac{\frac{1}{2} (y – 4)}{4 (x – 4)} = \frac{5}{12}$

=> $6 (y – 4) = 20 (x – 4)$

=> $6y – 24 = 20x – 80$

=> $10x – 3y = 28$ ————(i)

Also, $(x + 8) – \frac{1}{2} (y + 8) = 2$

=> $2x + 16 – y – 8 = 4$

=> $2x – y = -4$ ————-(ii)

Applying the operation, (i) – 3*(ii), we getÂ :

=> $(10x – 6x) + (-3y + 3y) = 28 + 12$

=> $x = \frac{40}{4} = 10$ years

Let present ages of P,Q andÂ R be $p,q,r$ years respectively.

=> $q = r + 5$ ————(i)

and $p + q = r + 25$ ————(ii)

Subtracting eqn(i) from (ii), we getÂ :

=> $p = 25 – 5 = 20$

$\therefore$ Present age of P = 20 years

10+10 = 20 (7.69% = 1/13 and 6.66%=1/15)

$\frac{1}{7}$ of $2\frac{5}{8} \div \frac{3}{4}$ = 0.5

$125^{4} is equal to 5^{12} and 25^{4} is equal to 5^{8}$

LCM of fractions = LCM of numerators/HCF of denominators
HCF of (3, 4, 5) = 1
So, LCM of the three fractions = 1/1 = 1

[(3900 x 134)Â Ã· 100] + [134 x 39]
= 10452

Let the maximum aggregate marks be x
780/x – 728/x = 5/100
52/x = 1/20
x = 52*20 = 1040

Since the ages of A, B, C, D are consecutive

Let the ages of A, B, C, D be n, n+1,n+2,n+3

$\frac{n+n+1+n+2+n+3}{4} = 49.5$

4n+6 = 49.5*4 = 198

4n = 192

n = 48

Ages of A, B, C, D = 48, 49, 50 ,51

Product of ages of B and D = 49*51 = (50-1)(50+1) =$50^2-1$ = 2500-1 = 2449.

The maximum marks of each subjec is 105

80% of 105 = 84

Since there are 5 subjects

Average marks =$\frac{Sum\ of\ all\ the\ marks}{5} = 84$

Sum of all the marks = 84*5 = 420

89+92+98+81+Science = 420

360+Science = 420

Science = 420-360 = 60

There are 9 numbers in the series:

Average =$\frac{253+124+255+534+836+375+101+443+760}{9}$ = 3681/9 = 409

Ruchika’s monthly income = Rs 32000
Ravina’s monthly incomeÂ = 32000 x (1 + $\frac{15}{100}$) = 32000 x $\frac{115}{100}$ = Rs. 36800
Ramola’s monthly income = 3 x 36800Â = 110400
Ramola’s annual income = 12 x 110400 =Â 1324800

Let the number of book distributed in a day by a man = 5x

woman = 4x & child = 2x

Day 1 : There were 7 men, 5 women, 8 children

=> No. of books sold = (7 * 5x) + (5 * 4x) + (8 * 2x)

= 35x + 20x + 16x = 71x

Day 2 : There were 7 men, 3 women, 5 children [As, 2 women & 3 children were absent]

=> No. of books sold = (7 * 5x) + (3 * 4x) + (5 * 2x)

= 35x + 12x + 10x = 57x

Day 3 : There were 4 men, 5 women, 3 children [As, 3 men & 5 children were absent]

=> No. of books sold = (4 * 5x) + (5 * 4x) + (3 * 2x)

= 20x + 20x + 6x = 46x

Now, total books distributed on the course of three days = 71x + 57x + 46x = 2000

=> x = 2000/174

No. of books distributed on the second day = 57x = 57 * $\frac{2000}{174}$ = 655.17 = ~650

Each of the 3 statements individually statements that all four of them aren’t facing the centre. Hence choice (d) is the correct option.

To figure out how “one” is coded in that encryption, we need to identify the codes for every other word in “one of its kind”. That information is available in each of the 3 statements provided. Hence the correct option is (e)

We know that, diagonal = 1.414* side
Here, diagoal = 14.14, side = 10 m
Height = Perimeter of square = 10*4 = 40m
Now, base = side of square with 784 $m^2$ area.
base = 28
Area of triangle = .5*28*40 = 560 $m^2$

Option B is correct option.

Circumference of th wheel = pi * Diameter = (22/7) * 49 = 154
Number of revolutions of the wheel = 3200/154 = 20.77~21
The wheel will make around 21 revolutions.

Ciircumference = 2*(22/7)*r = 220
Hence, r = 35
Now, area of circle = (22/7)*35*35 = 3850
Area of rectangle is same as that of circle
Area of rectangle = 3850
length = 3850/50 = 77m
Therefore, option C is correct answer.

Perimeter of rectangle = 4(44) = 176 cm
Difference between side of square and breadth of rectangle = 37 – 11 = 26 cm
Therefore, the correct option is option C.

We know that,
Diagonal = side * 1.414
Therefore, side = 28/1.414 = 19.8
Area of square = 19.8 *19.8 = 392.04 ~ 392
Therefore, the correct option is option D.

Let the number be $8x$

Acc to ques,

=> $\frac{1}{8} * 8x = 17.25$

=> $x = 17.25$

$\therefore$ 73 % of the number = $\frac{73}{100} * 8x$

= 0.73 * 8 * 17.25 = 100.74

let A be 2x ,B be 2x +2 and C be 2x +4

Given that 4A = 3C

4(2x) = 3 (2x +4)

8x = 6x +12

2x = 12

x = 6

B = 2*6 + 2 = 14

The LCM of the fractions = LCM of the numerators/ HCF of the denominators
LCM of the numerators = LCM of 3, 2 and 1 = 6
HCF of the denominators = HCF of 11, 5 and 9 = 1
So, the required LCM = 6/1 = 6

Principal amount = Rs. 12,000

Time period = 3 years and rate of interest = 10% under compound interest.

=>Â $C.I. = P [(1 + \frac{R}{100})^T – 1]$

= $12,000 [(1 + \frac{10}{100})^3 – 1]$

= $12,000 [(\frac{11}{10})^3 – 1] = 12,000 (\frac{1331 – 1000}{1000})$

= $12 \times 331 = Rs. 3,972$

Amount deposited = Rs. 24,000

Rate = 14 % and time = 8 years under simple interest

=> $S.I. = \frac{P \times R \times T}{100}$

= $\frac{24000 \times 14 \times 8}{100}$

= $240 \times 112 = Rs. 26,880$

$\therefore$Total amount will ms Suchi gets at the end of 8 years

= $24000 + 26880 = Rs. 50,880$

Principal amount = Rs. 5,670

Time period = 3 years and rate = 6% under simple interest.

=> $S.I. = \frac{P \times R \times T}{100}$

= $\frac{5670 \times 6 \times 3}{100}$

= $56.7 \times 18 = Rs. 1,020.6$

Amount invested by A,B and C =Â Rs. 20,000 Rs. 28,000 and Rs. 36,000 respectively.

Ratio of profits shared by A : B : C

= $[(20,000 \times 6) + (12,000 \times 6)]$ : $[(28,000 \times 6) + (20,000 \times 6)]$ : $[(36,000 \times 6) + (44,000 \times 6)]$

= $32 : 48 : 80 = 2 : 3 : 5$

Let total profit earned = $Rs. x$

Profit received by C = $\frac{5}{(2 + 3 + 5)} \times x = 12550$

=> $\frac{x}{2} = 12550$

=> $x = 12550 \times 2 = Rs. 25,100$

Amount invested by Sarita = Rs. 50,000 and amount invested by Neeta = Rs. 80,000

Ratio of amount invested by Sarita : Neeta = 5Â : 8

Time period in which Sarita invested = 12 months and Neeta = 6 months

Ratio of time periods of Sarita :Â Neeta = 2Â : 1

=> Ratio of profits earned by Sarita : Neeta = $(5 \times 2)$:$(8 \times 1)$

= 5Â : 4

Total profit earned = Rs. 18,000

$\therefore$Â Saritaâ€™s share in the profit = $\frac{5}{(5+4)} \times 18,000$

= $5 \times 2,000 = Rs. 10,000$

=> Ans – (D)

Total mixture of sand and iron = 1 kg

Quantity of iron = $\frac{20}{100} \times 1 = 0.2$ kg

Let $x$ kg of sand should be added, thus total iron in the mixture

=> $0.2=\frac{10}{100} \times (x+1)$

=> $2=x+1$

=> $x=2-1=1$ kg

=> Ans – (A)

Let P has = $Rs. 100x$

=> Amount with Q = $100x + \frac{25}{100} \times 100x = Rs. 125x$

=> Amount with R = $\frac{1}{5} \times 125x = Rs. 25x$

Total amount together = $100x + 125x + 25x = 150$

=> $x = \frac{150}{250} = \frac{3}{5}$

=> $x = 0.6$

$\therefore$ Amount with P alone = $100 \times 0.6 = Rs. 60$

Let us rank the person according to the marks scored by them, where 1 -> highest marks and 5 -> lowest marks.

Only one person scored less marks than B, => B = 4

Also, AÂ > DÂ >Â B and $A \neq 1$

=> A = 2 and D = 3

Thus, ranking from 1-5 = C/E , A , D , B , E/C

$\therefore$ A scored the second highest marks.

Amount with A = $Rs. 4x$

=> Amount with C = $\frac{3}{4} \times 4x = Rs. 3x$

=> Amount withÂ B = $Rs. (3x – 50)$

Total amount with A,B & C = $4x + 3x + (3x – 50) = 250$

=> $10x = 250 + 50 = 300$

=> $x = \frac{300}{10} = 30$

$\therefore$ Amount with A = $4 \times 30 = Rs. 120$

Total amount = 97836

No. of children = 31

Since amount is distributed equally

=> Amount each child will get = $\frac{97836}{31}$ = Rs. 3,156

12 men can finish the project in 20 days.

=> 1 day work of 1 man = $\frac{1}{12 \times 20} = \frac{1}{240}$

Similarly,Â => 1 day work of 1 woman = $\frac{1}{18 \times 16} = \frac{1}{288}$

=> 1 day work of 1 children = $\frac{1}{24 \times 18} = \frac{1}{432}$

8 women and 16 children worked for 9 days

=> Work done in 9 days = $9 \times (8 \times \frac{1}{288}) + (16 \times \frac{1}{432})$

= $9 \times (\frac{1}{36} + \frac{1}{27}) = 9 \times \frac{7}{108}$

= $\frac{7}{12}$

=> Work left = $1 – \frac{7}{12} = \frac{5}{12}$

$\therefore$ Number of days taken by 10 men to complete the remaining work

= $\frac{\frac{10}{240}}{\frac{5}{12}} = \frac{1}{24} \times \frac{12}{5} = \frac{1}{10}$

Thus, 10 men will complete the remaining the work in 10 days.

Let work done by 1 man be $x$ and 1 woman be $y$

Now, 16 men and 12 women complete work in 8 days.

=> $16x + 12y = \frac{1}{8}$ ———Eqn(i)

Also, $20x = \frac{1}{16}$

=> $16x = \frac{1}{20}$

Putting it in eqn(i), we getÂ :

=> $\frac{1}{20} + 12y = \frac{1}{8}$

=> $12y = \frac{1}{8} – \frac{1}{20} = \frac{3}{40}$

=> $y = \frac{3}{40 \times 12} = \frac{1}{160}$

Thus, 16 women can complete the work in = $16 \times \frac{1}{160} = \frac{1}{10}$

$\therefore$ 16 women can complete the work in 10 days.

Each number is of the form $(n^3+1)$ where $n$ is a natural number

$1^3+1$ = 2

$2^3+1$ =Â 9

$3^3+1$ =Â 28

$4^3+1$ =Â 65

$5^3+1$ =Â 126

=> Ans – (C)

9 $\times 1 + 1^2$ = 10
10Â $\times 3 + 3^2$ =Â 39
39Â $\times 5 + 5^2$ =Â 220
220Â $\times 7 + 7^2$ =Â 1589
1589Â $\times 9 + 9^2$ =Â 14382