**Find solutions for all the questions at the end of each set:**

### Set A

1.There are some boys and girls in a room. The square of the number of the girls is less than the square of the number of boys by 28. If there were two more girls, the number of boys would have been the same as that of the girls. The total number of the boys and girls in the room are

(a) 56

(b) 14

(c) 10

(d) 7

2.The H.C.F of two numbers, each having three digits, is 17 and their L.C.M. is 714. The sum of the numbers will be:

(a) 289

(b) 391

(c) 221

(d) 731

3.The first period of a class starts at 10 : 30 hours and fourth ends at 13 : 45 hours. If periods are of equal duration and after each period a break of 5 minutes is given to the students, the exact duration of each period is :

(a) 35 minutes

(b) 42 minutes

(c) 45 minutes

(d) 40 minutes

4.The smallest positive integer n, for which 864n is a perfect cube, is :

(a) 1

(b) 2

(c) 3

(d) 4

5.A, B and C can complete a work in 10, 12 and 15 days respectively. The started the work to gether. But A left the work before 5 days of its completion. B also left the work 2 days after A left. In how many days was the work completed?

(a) 4

(b) 5

(c) 7

(d) 8

6.A can complete a piece of work in 10 days. B in 15 days and C in 20 days. A and C worked together for two days and then A was replaced by B. In how many days, altogether, was the work completed?

(a) 12

(b) 10

(c) 6

(d) 8

7.Two pipes A and B can fill a water tank in 20 and 24 minutes respectively and a third pipe C can empty at the rate of 3 drums per minute. If A, B and C opened together fill the tank in 15 minutes, the capacity (in drums) of the tank is :

(a) 180

(b) 150

(c) 120

(d) 60

8.If 10 men or 20 boys can make 260 mats in 20 days, then how many mats will be made by 8 men and 4 boys in 20 days?

(a) 260

(b) 240

(c) 280

(d) 520

9.A fan is listed at Rs. 1,400 and the discount offered is 10%. What additional discount must be given to bring the net selling price to Rs. 1,200?

(a) 16 2/3 %

(b) 5%

(c) 4 16/21 %

(d) 6%

10.A man goes from A to B at a uniform speed of 12 kmph and returns with a uniform speed of 4 kmph His average speed (in kmph) for the whole journey is :

(a) 8

(b) 7.5

(c) 6

(d) 4.5

ANSWERS AND SOLUTION :

1.(b) Let the number of boys and girls in the room be x and y respectively.

According to the question,

x^2 = y^2 + 28

=> x^2 – y^2 = 28 …………………..(i)

and

x = y + 2

=> x – y = 2 ……………………….(ii)

On dividing equation (i) by equation (ii), we have

(x^2- y^2)/(x-y)=28/2

=> ((x+y)(x-y))/(x-y)=14

=> x + y = 14

so Total number of boys and girls = 14

2.(c) Let the numbers be 17x and 17y where x and y are co-prime.

LCM of 17x and 17y = 17 xy

According to the question.

17xy = 714

=> xy = 714/17 = 42 = 6 * 7

=> x = 6 and y = 7

or x = 7 and y = 6

so first number = 17x

= 17 * 6 = 102

Second number = 17y

= 17 * 7 = 119

so Sum of the numbers

= 102 + 119 = 221

3.(c) Time interval between the first period and last (fourth) period = 1:45 pm. – 10 : 30 am.

= 3 hours 15 minutes

= Break after each period

= 5 minutes

so Total break = 15 minutes

so Time period of classes = 3 hours 15 minutes – 15 minutes.

= 3 hours

so Exact duration of each period = 3/4 hour = 45 minutes.

4.(b)

5.(c) Let the work be completed in x days.

According to the question,

(x-5 )/10+(x - 3)/12 + x/15 = 1

=>(6x - 30 + 5x – 15 + 4x)/60 = 1

=> 15x – 45 = 60

=> 15x = 105 => x = 105/15 = 7

Hence, the work will be completed in 7 days.

6.(d) Work done by (A +C) in 2 days = 2 (1/10+1/20)

= 2 ((2 + 1)/20)=6/20=3/10

Remaining work = 1 – 3/10 = 7/10

(B + C)’s 1 day’s work

= 1/15 + 1/20 = (4 + 3)/60 = 7/60

so Time taken by (B + C) to finish 7/10 parts of the work

= 60/7 * 7/10 = 6 days

so Total time = 2 + 6 = 8 days

7.(c) Let the capacity of the tank be x drum.

Quantity of water filled in the tank in 1 minute when all the pipes A, B and C are opened simulataneously

= x/20 + x/24 -3

According to the question.

x/20 + x/24 – 3 = x/15.

=> x/20 + x/24 -x/15 = 3

=>(6x+5x-8x)/120 = 3

=> 3x = 3 * 120

=> x = (3 * 120)/3 = 120 drums

8.(a) 10 men = 20 days

so 1 man = 2 boys

so 8 men + 4 boys

= (16 + 4) boys = 20 days

Hence 8 men and 5 boys will make 260 mates in 20 days.

9.(c) Marked price of the fan = Rs. 1400

SP after allowing a discount of 10% = 90% of 1400

= (1400 * 90)/100 = Rs. 1260

Second discount

= Rs. (1260 – 1200) = Rs. 60

Let the second discount be x%

so X% of 1260 = 60

=> x = (60 * 100)/1260 = 100/21 = (4) 16/21%

10.(c) If two equal distances are covered at two unequal speeds of x kmph and y kmph, then average speed

= (2xy/(x+y)) kmph

= ((2 ×12 ×4)/(12+4 )) =96/16 = 6 kmph

**Find solutions for all the questions at the end of each set:**

### Set B

1.The sum of first 20 odd natural numbers is equal to :

(1) 210

(2) 300

(3) 400

(4) 420

2.What number should be subtracted from both term of the ratio 15 : 19 in order to make it 3 : 4?

(1) 9

(2) 6

(3) 5

(4) 3

3.A starts business with Rs. 3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. What is B's contribution in the capital?

(1) Rs. 8000

(2) Rs. 8500

(3) Rs. 9000

[4] Rs. 7500

4.A, B and C rent a pasture. A puts in 10 oxen for 7 months, B 12 oxen for 5 months and C 15 oxen for 3 months for grazing. If the rent of the pasture is Rs. 175/per, how much must C pay as his share of rent?

(1) Rs. 45

(2) Rs. 50

(3) Rs. 55

(4) Rs. 60

5.A reduction of 20% in the price of oranges enables a man to buy 5 oranges more for Rs. 10/ -. The price of an orange before reduction was:

(1) 20 paise

(2) 40 paise

(3) 50 paise

(4) 60 paise

6.In an examination, a student who gets 20% of the maximum marks fails by 5 marks. Another student who scores 30% of the maximum marks gets 20 marks more than the pass marks. The necessary percentage required for passing is :

(1) 32%

(2) 23%

(3) 22%

(4) 20%

7.If 3 men or 6 women can do a piece of work in 16 days. in how many days can 12 men and 8 women do the same piece of work?

(1) 4 days

(2) 5 days

(3) 3 days

(4) 2 days

8.On what sum does the difference between the compound in interest and the simple interest for 3 years at 10% is Rs. 31 ?

(1) Rs. 1500

(2) Rs. 1200

(3) Rs. 1100

(4) Rs. 1000

9.A builder borrows Rs. 2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 years in two equal yearly instalments. How much will each instalment be ?

(1)Rs. 1352

(2)Rs. 1377

(3)Rs. 1275

(4)Rs. 1283

10.At what percent per annum will Rs. 3000/ - amount to Rs. 3993 in 3 years if the interest is compounded annually?

(1) 9%

(2) 10%

(3) 11%

(4) 13%

ANSWERS AND SOLUTION

1.(3) Series of first 20 odd natural numbers is an arithmetic progression with 1 as the first term and the common difference 2.

Sum of the n terms in arithmetic progression is given by.

Sn = 1/2 n[2a + (n-1)d]

Where a : First term

d : common difference

so (S)20 = 1/2 * 20 {2 * 1) + (20 – 1) * 2}

= 10[2 + 38] = 10 * 40 = 400

2.(4) Let x be subtracted from each term of 15/19

so (15 – x)/(19 – x) = 3/4

=> 57 – 3x = 60 – 4x

=> x = 3

3.(3) A’s investment of Rs. 3500 is for 12 months

B’s investment (let it be Rs. x) is for 7 months only.

At the end of the year the profit is divided in the ratio 2 : 3 and it must be equal to the ratio of the product. (Amount * time)

(12 * 3500)/7x = 2/3

Or x = (12 * 3500)/7 x 3/2

Or x = 9000

so B’s investment is Rs. 9000.

4.(1) Share of rent = (number of oxen x time)

A : B : C = (10 * 7) : (12 * 5) : (15 * 30)

A : B : C = 70 : 60 : 45

A : B : C = 14 : 12 : 9

C’s share of rent

= 9/ (14 + 12 + 9) * 175 = 9/35 * 175 = 45

C’s share of rent is Rs. = 45

5.(3) 20% of Rs. 10

= 20/100 * Rs. 10 = Rs. 2

Reduced price of 5 oranges

= Rs. 2

so Reduced price of 1 orange

= Rs. 2/5 = 200/5 paise = 40 paise

Original price of 1 orange

= 40/(1 – 0.20) = 40/0.8 = 400/8 = 50 paise

6.(3) Let the maximum marks be x.

According to question.

20% of (x + 5) = 30% of (x – 20)

(30 – 20)% of x = 25

x = (25 * 100)/10 = 250

so Pass marks

= 20% of 250 + 5 = 55

so % Pass marks = 55/250 * 100 = 22%

7.(3) 3m = 6w

so m = 2w

12m + 8w = (12 * 2w) + 8w = 32w

so 6 women can do the work in 16 days.

so 32 women can do the work (16 * 6)/32 = 3 days

8.(4)

9.(1)

10.(2)

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