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

### Set A

1.When 15 is included in list of natural numbers, their mean is increased by 2. When 1 is included in this new list, the mean of the numbers in the new list is decreased by 1. How many numbers were there in the original list?

(1) 4

(2) 5

(3) 6

(4) 8

2.A shopkeeper marks his goods at such a price that after allowing a discount of 12.5% on the marked price, he still earns a profit of 10% The marked price of an article which costs him Rs. 4,900 is

(1) Rs. 5,390

(2) Rs. 6,160

(3) Rs. 5,490

(4) Rs. 5,390

3.Ravi can walk a certain distance in 40 days when he rests 9 hours a day. How long will he take to walk twice the distance, twice as fast and rest twice as long each days?

(1) 40 days

(2) 80 days

(3) 50 days

(4) 100 days

4.A solid is in the form of a right circular cylinder with hermisphereical ends. The total length of the solid is 35 cm. The diameter of the cylinder is 1/4 of its height. The surface area of the solid is (Take π=22/7)

(1) 462 cm^2

(2) 693cm^2

(3) 750 cm^2

(4) 770 cm^2

5.A sum of money lent out at compound interest increases in value by 50% in 5 years. A person wants to lend three different sums x, y and z for 10, 15 and 20 years respectively at the above rate in such a way that he gets back equal sums at the end of their respective periods. The ratio x : y : z is

(1) 6 : 9 : 4

(2) 9 : 4 : 6

(3) 9 : 6 : 4

(4) 6 : 4 : 9

6.The number (2^48 – 1) is exactly divisible by two numbers between 60 and 70. The numbers are :

(1) 63 and 65

(2) 63 and 67

(3) 61 and 65

(4) 65 and 67

7.The six -digit number 5ABB7A is a multiple of 33 for digits A and B. Which of the following could be possible value of A + B?

(1) 8

(2) 9

(3) 10

(4) 14

8.The product of two 2 -digit numbers is 2028 and their HCF is 13. The sum of the numbers is:

(1) 65

(2) 78

(3) 91

(4) 169

9.Cost of a diamond various directly as the square of its weight. A diamond broke into four places with their weight in the ratio 1 : 2 : 3 : 4. If the loss in the total value of the diamond was Rs. 70, 000, the price of the original diamond was

(1) Rs. 1, 00, 000

(2) Rs. 1, 40, 000

(3) Rs. 1, 50, 000

(4) Rs. 1, 75, 000

10.Two candles having the same lengths are such that one burns out completely in 3 hours at a uniform rate and the other in 4 hours. At what time (P.M.)., the lenghts of one is twice the length of the other?

(1) 1 : 24

(2) 1 : 30

(3) 1 : 36

(4) 1 : 32

Answers and Solution

1.(1) 15 = 2 * 5 + 5

SO Original mean = 5

New mean = 7

SO Sum of five numbers

= 7 * 5 = 35

After adding 1, new mean

= 36/6 = 6

2.(2) SP of the article

= 4900 * 110/100 = Rs. 5390

SO Marked Price = 100/87.5 * 5390

= Rs. 6160

3.(4) Working hours per day

= 24 – 9 = 15 hours

Total time

= 15 * 40 = 600 hours

Decrease in walking hours per day after increase in hours of rest = 24 – (2*9) = 6 hours

Required time = 600/6 = 100 days

4.(4)

5.(3) According to the question.

According to the question.

(3/2)^2 x = (3/2)^3 y = (3/2)^4 z =k let

=>x = (2/3)^2 k,y= (2/3)^3 k

and z = (2/3)^4 k

SO x:y:z = (2/3)^2 k∶ (2/3)^3 k∶ (2/3)^4 k

= 1 : 2/3 : (2/3)^2=9∶6∶4

6.(1) 2^48 – 1 = (2^24 + 1) (2^24 – 1)

= (2^24 + 1) (2^12 + 1) (2^12 – 1)

= (2^24 + 1) (2^12 + 1) (2^6 +1) (2^6 -1)

SO Required numbers

= 2^6 + 1 and 2^6 – 1

= 65 and 63

7.(2) A number is divisible by 33 if it is divisible by 3 and 11 both For divisiblity by 11,

so 5 + B + 7 = A + B + A

so 2A = 12

=>A = 6

so For divisibility of 5 6 B B 7 6 by 3, B = 3

so Number = 563376

so A + B = 6 + 3 = 9

8.(3) Let the number be 13x and 13y where x and y are prime to each other.

so 13x * 13y = 2028

=>xy = 2028/13 * 13 = 12 = 3 * 4

so Numbers = 13 * 13 = 39 and 13 * 4 = 52

so Sum of numbers

= 39 + 52 = 91

9.(1) Let the weights of the four prices of diamond be x, 2x, 3x and 4x units respectively.

Total weight of the original diamond

= x + 2x + 3x + 4x = 10x units

Price of the original diamond

k (100x^2) where k is a constant. Cost of four pieces of diamond

= k(x^2 + 4x^2 + 9x^2 + 16x^2)

= k.30x^2

Loss in the cost of diamond

=k(100x^2 = 30x^2) = k.70x^2

so k.70x^2 = 70000

=>kx^2 = Rs. 1000

so Cost of the original diamond

= k(100x^2)

= 1000 * 100 = Rs. 100000

10.(3) If the length of each candle be l then rate of burning of the first

= l/4 and that of the second = l/3

According to the question.

l – l*t/4 = 2(l – lt/3)

=>l – lt/4 = 2l – 2lt/3

=> 2/3t – t/4 = 2 – 1

=>((8 -3)/12)t = 1

=>5t/12 = 1

so t = 12/5 hours

= 2 hours 24 minutes

So Required time = 4 hours – 2 hours 24 minutes = 1 hour 36 minutes

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

### Set B

1.The ratio of investment of two partners is 11:12 and the ratio of their profits 2:3 . if A invested the money for 8 months. Find for how much time B invested his money?

A. 11 months

B. 10 months

C. 9 months

D. 8 months

2. A number lies between the cubes of 15 and 16.If the number is divisible by the square of 12 as well as by 7,what is number?

A. 3469

B. 4032

C. 4045

D. 5249

3. A man deposite a certain sum in a bank.He gets 4% per annum intrest for first 3 years, 5% for next 2 years and 6% for the period beyond that. If he gets Rs.2000 as simple intrest for 8 years, how much money did he deposite in the bank.?

A.Rs.8000

B.Rs.6000

C.Rs.4000

D.Rs.5000

4.The age of shiva is five times that of his son.6 years ago his age was nine times that of his son.what is shiva’s present age?

A. 60 years

B. 55 years

C. 50 years

D. 45 years

5. Trees are to be planted at equal intervals of 10 ft.around a rectangular lawn whose length is twice its breadth. If the total number of trees planted is 36.find the area of the lawn.

A. 7200 sq.ft

B. 6000 sq.ft

C. 5000 sq.ft

D. 4800 sq.ft

6. A coin box contains 50 paise and 1 rupee coins.The total number of coins is 36. The total value of 50 paise coins equal the total value of repess coins.What is the total amount in the box.

A. Rs.18

B. Rs.20

C. Rs.22

D. Rs.24

7. A hallow sphere of internal and external diameter of 4 c.m.and 8.cm.respectively is melted in to a cone of base diameter 8cm.The height of the cone is –

A. 14 cm.

B. 12 cm.

C. 15 cm.

D. 18 cm.

8.A man starts walking. He walked 2 km in the first hour. then, he walked two-third of the previous hour in each next hour.If he walked continuously then how long maximum could he walk?

A. 60 km.

B. 6 km.

C. 12 km.

D. 8 km.

9. The ratio of between the school ages of Neeta and sameer is 5:6 respectively.If the ratio between the one-third age of Neeta and half of the sameer’s age is 5:9, then what is the school age of sameer?

A. 30 years

B. 25 years

C. 36 years.

D. Data indequate

10. A boat covers a distance of 30 km downstream in 2 hours while it takes 6 hours to cover the same distance upstream.If the speed of current is half of the speed of the boat then, what is the speed of boat in km per hour?

A. 15 kmph

B. 5 kmph

C. 10 kmph

4. Data inadequate.

Answers and solution:

1(A) Suppose A invested Rs.11 for 8 months and B invested Rs.12 for y months.

Then ratio of investment of A and B = (11×8) : (12xy)

= 88 : 12y

88/12y = 2/3

24y = 88×3

y = 264/24 =11 months.

2(B)

(15)^3= 3375

(16)^3= 4096

(12)^2= 144

The number lies between 3375 and 4096.

The number is divisible by 144 as well as 7.

L.C.M. of 144 and 7.= 1008

so the number must be a multiple of 1008.

1008×3 = 3024

1008×4 = 4032

4032 lies between 3375 and 4096.

3(D)

I = I1 + I2 + I3

I = p/100(r1t1+r2t2+r3t3)

I = p/100(4×3+5×2+6×3)

P = (100xI)/40

P = (100×200)/4

P == Rs. 5000

4(A)

Let the present age of Shva’s son be x years.

so shiva’s present age = 5x

6 year ago:

son ‘s age = (x-6) yrs.

Shiva’s age = (5x-6)yrs.

So 5x-6 = 9(x-6)

5x-6 = 9x-54

4x = 48

x = 12

so shiva’s present age = 5x = 60 yrs.

5(A)

Perimeter of the lawn =2(l+b)= 36×10 ft.

So l+b = 180

But l = 2b so 3b = 180

b = 60 ft

so l= 120 ft

Area of the lawn = lb

= 120 x 60 sq.ft = 7200 sq.ft

6(D)

7(A)

Volume of cone = volume of the spherical shell

1/3πr^2h = 4/3π(R1^3 – R2^3)

h = 4(R1^3-R2^3)/r^2

= 4(4^3-2^3)/4^2 = 14cm.

8(B)

Required distance

=2((1+(2/3)+(2/3)^2+(2/3)^3+……….)

= 2 x 1/(1-2/3) = 2 x 3 = 6 km.

9(D)

10(C)

Downstream speed = 30/2 =15 km/h

Upstreams speed = 30/6 = 5 km/h

so speed of the boat = 1/2(15+5)

=1/2 x 20 = 10 km/h

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

### Set C

1.The HCF of two numbers 12906 and 14818 is 478. Their LCM is

(1)400086

(2)200043

(3)600129

(4)800172

2.The H.C.F and L.C.M of two numbers are 8 and 48 respectively. If one of the numbers is 24, then the other number is

(1)48

(2)36

(3)24

(4)16

3.Find the smallest multiple of 13 such that on being divided by 4, 6, 7 and 10. It gives remainder 2 in each case.

(1)420

(2)840

(3)2470

(4)2522

4.The least number which when divided by 4, 6, 8 and 9 leave zero remainder in each case and when divided by 13 leaves a remainder of 7 is :

(1)144

(2)72

(3)36

(4)85

5. 4 bells ring at intervals of 30 minutes, 1 hour, 1 ½ hour and 1 hours 45 minutes respectively.All the bells ring simultaneously at 12 noon. They will again ring simultaneously at:

(1)12 mid night

(2)3 a.m

(3)6 a.m

(4)9 a.m

6.Three bells ring simultaneously at 11 a.m. They rign at regular intervals of 20 minutes. 30 minutes. 40 minutes respectively. The time when all the three ring together next is

(1)2 p.m.

(2)1 p.m.

(3)1.15 p.m.

(4)1.30 p.m.

7.The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets same number of pens and same number of pencils, is :

(1)91

(2)910

(3)1001

(4)1911

8.A milkman has 75 litres milk in one cane and 45 litres in another. The maximum capacity of container which can measure milk of either container exact number of times is :

(1)1 litre

(2)5 litres

(3)15 litres

(4)25 litres

9.The LCM of two positive integers is twice the larger number. The difference of the smaller number and the GCD of the two numbers is 4. The smaller number is :

(1)12

(2)6

(3)8

(4)10

10.The HCF (GCD) of a, b is 12, a. b are positive integers and a > b > 12. The smallest values of (a,b) are respectively .

(1)12, 24

(2)24, 12

(3)24, 36

(4)36, 24

ANSWERS AND SOLUTION

1.(1)

Product of two numbers = HCF * LCM

= 12906 * 14818 = LCM * 478

LCM = (12906 * 148)/478

= 400086

2.(4) first number x second number

= HCF * LCM

= 24 * second number = 8 * 48

so Second Number = (8 * 48)/24 = 16

3.(4)

4 = 2^2

6 = 2 * 3

7 = 7

10 = 2 * 5

LCM = 4 * 3 * 7 * 5 = 420

Number = 420 k + 2

= 416 k + (4k + 2)

When K = 6 then 4k + 2

= 4 * 6 + 2 = 26 which is divisible by 13.

so Required number

= 420 k + 2

= 420 * 6 + 2 = 2522

4.(2)

5.(4)

6.(2) LCM of 20, 30 and 40 minutes = 120 minutes

Hence, the bells will toll together again after 2 hours i.e. at 1 p.m.

7.(1) Maximum number of students

= The greatest common divisor

= HCF of 1001 and 910 = 91

8.(3) Required maximum capacity of container

= HCF of 75 l and 45 l

Now, 75 = 5 * 5 * 3

45 = 5 * 3 * 3

so HCF = 15 litres

9.(3) Let the numbers be xH and yH where H is the HCF and yH > xH.

so LCM = xyH

xyH = 2yH

x = 2

Again, xH = H = 4

2H – H = 4

H = 4

Smaller number = xH = 8

10.(4) HCF of a and b = 12

so Numbers = 12x and 12y

Where x and y are prime to each other.

A > b > 12