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

### Set A

1. n! has 23 zeroes . what is the maximum possible value of n ?

A. 99

B. 100

C. 98

D. none of these

2. n! has 13 zeroes. The highest and least value of n are ?

A. 59 and 55

B. 59 and 56

C. 59 and 57

D. none of these

3. Find the maximum value of n such that 50! is perfectly divisible by 2520n.

A. 6

B. 8

C. 7

D. none of these

4.The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number ?

A. 240

B. 270

C. 295

D. 360

5. If the number 517*324 is completely divisible by 3, then the smallest whole number in the place of * will be:

A. 0

B. 1

C. 2

D. None of these

6.The sum of first 45 natural numbers is:

A. 1035

B. 1280

C. 2070

D. 2140

7. The difference between the local value and the face value of 7 in the numeral 32675149 is

A. 75142

B. 64851

C. 5149

D. 69993

8. On dividing a number by 56, we get 29 as remainder. On dividing the same number by 8, what will be the remainder ?

A. 4

B. 5

C. 6

D. 7

9.What will be remainder when (67^67 + 67) is divided by 68 ?

A. 1

B. 63

C. 66

D. 67

10. How many natural numbers are there between 23 and 100 which are exactly divisible by 6 ?

A. 8

B. 11

C. 12

D. 13

ANSWERS KEY AND SLUTION:

1(d) This can never happen at 99! number of zeroes is 22 and at 100! the number of zeroes is 24.

2(a)

3(B)Explanation:

2520 = 7x3^2x2^3x5

The value of n would be given by the value of the number of 7s in 50!

This value is equal to (50/7) + 50/49)= 7+1 = 8

4(B)Explanation:

Let the smaller number be x. Then larger number = (x + 1365).

so x + 1365 = 6x + 15

=>5x = 1350

=> x = 270

Smaller number = 270.

5(C)Explanation:

Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = (22 + x), which must be divisible by 3.

x = 2.

6(A)Explanation:

Let Sn =(1 + 2 + 3 + ... + 45). This is an A.P. in which a =1, d =1, n = 45.

Sn = n/2 [2a + (n - 1)d] = 45/2 x [2 x 1 + (45 - 1) x 1]

=> (45/2 x 45)= (45 x 23)

= 45 x (20 + 3)

= 45 x 20 + 45 x 3

= 900 + 135

= 1035.

Shorcut Method:

Sn = n(n + 1)/2 = 45(45 + 1)/2 = 1035.

7(D)Explanation:

(Local value of 7) - (Face value of 7) = (70000 - 7) = 69993

8(B)Explanation:

29 as remainder is divided by 8 than remainder is 5

9(C)Explanation:

(x^n + 1) will be divisible by (x + 1) only when n is odd.

so (67^67 + 1) will be divisible by (67 + 1)

(67^67 + 1) + 66, when divided by 68 will give 66 as remainder.

10(D)Explanation:

Required numbers are 24, 30, 36, 42, ..., 96

This is an A.P. in which a = 24, d = 6 and l = 96

Let the number of terms in it be n.

Then tn = 96 a + (n - 1)d = 96

=>24 + (n - 1) x 6 = 96

=> (n - 1) x 6 = 72

=> (n - 1) = 12

=> n = 13

Required number of numbers = 13.

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

### Set B

1. If p - q = 6 and p^2 + q^2 = 116, what is the value of pq?

A. 30

B. 40

C. 20

D. 50

2.The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?

A. Rs. 1200

B. Rs. 2400

C. Rs. 4800

D. Cannot be determined

3. If a^2+b^2+c^2= 2(a-b-c)-3, then find the value of 2a-3b+4c.

A. 3

B. 1

C. 2

D. 4

4. To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to three-fifth of its present?

A. 63

B. 64.5

C. 62.5

D. 60.5

5.A sum of Rs. 1360 has been divided among A, B and C such that A gets 2/3 of what B gets and B gets 1/4 of what C gets. B's share is:

A. Rs. 120

B. Rs. 160

C. Rs. 240

D. Rs. 300

6.A man has Rs. 312 in the denominations of one-rupee notes, five-rupee notes and twenty-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?

A. 36

B. 24

C. 28

D. 32

7.Simplfy b - [b -(a+b) - {b - (b - a+b)} + 2a]

A. a

B. 2a

C. 4a

D. 0

8.Free notebooks were distributed equally among children of a class. The number of notebooks each child got was one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distributed?

A. 602

B. 528

C. 423

D. 512

9 .a * b = 2a - 3b + ab, then 3*5 + 5*3 = ?

A. 20

B. 21

C. 22

D. 23

10.One-third of Rahul's savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,80,000 as total savings, how much has he saved in Public Provident Fund?

A. Rs. 72000

B. Rs. 44000

C. Rs. 58000

D. Rs. 92000

Note : Solution will be provided within an hour. you can give your answer in comment section.

ANSWER AND SOLUTION:

1(B)Explanation :

(a-b)^2=a^2-2ab+b^2

so (p - q)^2 = p^2 - 2pq + q^2

(p - q)^2 = (p^2 + q^2)- 2pq

6^2 = 116 - 2pq

36 = 116 - 2pq

2pq = 80

pq = 40

2(B)Explanation:

Let the price of a saree and a shirt be Rs. x and Rs. y respectively.

Then, 2x + 4y = 1600 .... (i)

and x + 6y = 1600 .... (ii)

Divide equation (i) by 2, we get the below equation.

=> x + 2y = 800. --- (iii)

Now subtract (iii) from (ii)

x + 6y = 1600 (-)

x + 2y = 800

----------------

4y = 800

----------------

Therefore, y = 200.

Now apply value of y in (iii)

=> x + 2 x 200 = 800

=> x + 400 = 800

Therefore x = 400

Solving (i) and (ii) we get x = 400, y = 200.

so Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.

3(B)Explanation:

a^2 + b^2+c^2 = 2(a-b-c)-3

a^2 + b^2+c^2-2(a-b-c)+3=0

a^2 + b^2+c^2-2a+2b+2c+1+1+1=0

(a^2+1-2a)+ (b^2+1+2b)+(c^2+1+2c)= 0

(a-1)^2+(b+1)^2+(c+1)^2= 0

so a = 1, b = -1, c = -1

so 2a - 3b + 4c = 2x1 - ((3x(-1)) + 4x(-1)

= 2 +3 - 4

= 5 - 4

= 1

4(C)Explanation :

Let x buckets of water be required ,if the capacity of the bucket is reduced to two-fifth

More buckets, less capacity (Indirect proportion)

Hence we can write as

Capacity 1 : 2/5 } :: x : 25

1×25 = 2/5×x

25 = 2x/5

x=25×5/2 = 62.5

i.e., 62.5 buckets are needed

5(C)Explanation:

Let C's share = Rs. x

Then, B's share = Rs. x/4 , A's share = Rs.(2/3 x x/4) = Rs. x/6

so, x/6 + x/4 + x = 1360

=>17x/12 = 1360

=> x =1360 x 12/17 = Rs. 960

Hence, B's share = Rs. (960/4) = Rs. 240.

6( A)Explanation :

Let the number of notes of each denomination be x

Then, x + 5x + 20x = 312

=> 26x = 312

=> x = 312/26 = 12

the total number of notes that he has = 3x = 3 × 12 = 36

7(D)Explanation:

b-[b-(a+b)-{b-(b-a+b)}+2a]

=b-[b-a-b-{b-(2b-a)}+2a]

=b-[-a-{b-2b+a}+2a]

=b-[-a-{-b+a}+2a]

=b-[-a+b-a+2a]

=b-[-2a+b+2a]

=b-b

=0

8(D)Explanation :

Let n be the total number of children.

More children, less notebooks(Indirect proportion)

If the number children = n, number of books each child will get = n/8

If the number children = n/2, number of books each child will get = 16

Hence we can write as

children n : n/2 } :: 16 : n/8

n×n/8 = n/2×16

n/8 = 16/2 = 8

n = 8×8 = 64

Then, total number of notebooks those were distributed = n×n/8 = 64×64/8 = 64×8=512

9(C)Explanation:

= 2(3)-3(5)+(3*5) + 2(5)-3(3)+(5*3)

= 6-15+15+10-9+15 = 31-9 = 22

10(A)Explanation :

Let savings in National Savings Certificate = x

and savings in Public Provident Fund = (180000 - x)

1/3 x = 1/2(180000-x)

2x = 3(180000-x)

2x = 540000-3x

5x = 540000

x=540000/5 = 108000

Savings in Public Provident Fund = (180000 - 108000) = 72000

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

### Set C

**1. 10 students started to eat biscuits. Among them, 1 ate 8 biscuits, 2 of them ate 6 biscuits each, 3 of them ate 4 biscuits each and 4 of them ate 2 biscuits each. The average number of biscuits, each of them ate, was:**

A. 4

B. 6

C. 8

D. 3

**2. The mean of 11 numbers is 35. If the mean of first 6 numbers is 32 and that of the last 6 numbers is 37, find the sixth number.**

A. 27

B. 28

C. 29

D. 30

**3. In a 20 over match, the required run rate to win is 7.2. If the run rate is 6 at the end of the 15th over, the required run rate to win the match is:**

A. 1.2

B. 13.2

C. 10.8

D. 12

**4. The average of 9 integers arranged in a certain order is 32. If the average of the first four integers is 26 and that of the last four integers is 28, then the fifth integer is:**

A. 76

B. 72

C. 68

D. 58

**5. A batsman has a certain average of runs for 11 innings. In the 12th innings, he makes a score of 90 runs, thereby increasing his average by 5. His average after the 12th innings is:**

A. 30

B. 35

C. 40

D. 45

**6.The average monthly income of A and B is Rs. 5050. The average monthly income of B and C is Rs. 6250 and the average monthly income of A and C is Rs. 5200. What is the monthly income of A?**

A. 2000

B. 3000

C. 4000

D. 5000

**7. In Kiran's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kiran and he thinks that Kiran's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Kiran?**

A. 70 kg

B. 69 kg

C. 61 kg

D. 67 kg

**8.A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?**

A. 28 4/7 years

B. 31 5/7 years

C. 32 1/7 years

D. None of these

**9. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, what is the average marks of all the students?**

A. 53.23

B. 54.68

C. 51.33

D. 50

**10.A student needed to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What is the value of x?**

A. 12

B. 5

C. 7

D. 9

Total number of biscuits = 1*8+2*6+3*4+4*2= 8 + 12 + 12 + 8 = 40

Average number of biscuits = 40/10= 4

The sixth number = 6*32 + 6 * 37-11*35

= 192 + 222 - 385

= 29

Total runs required to win the match = 20 * 7.2 = 144

Runs scored in 15 over = 15 * 6 = 90

Runs required in 5 overs = 144 - 90 = 54

Required run rate = 54/5 = 10.8

Fifth integer = 9*32 - 4*26 - 4*8

= 288 - 104 - 112

= 72

Let avrage run of 1 innings = x

Total runs of 11 innings = 11x

Score of 12th innings = 90

Average run of 12 innings = x=5

Total runs of 12 innings = 12(x+5)

12(x+5) - 11x = 90

12x + 60 - 11x = 90

x = 30

Thus, average run after the 12th innings = 30+5= 35

Let the monthly income of A = a

monthly income of B = b

monthly income of C = a

a+b=2×5050---------(Equation1)

b+c=2×6250---------(Equation2)

a+c=2×5200---------(Equation3)

(Equation 1) + (Equation 3) - (Equation 2)

=>a+b+a+c-(b+c)=(2×5050)+(2×5200)-(2×6250)

=> 2a = 2(5050 + 5200 - 6250)

=> a = 4000

=> Monthly income of A = 4000

Let Kiran's weight = x. Then

According to Kiran, 65 < x < 72 ------------(equation 1)

According to brother, 60 < x < 70 ------------(equation 2)

According to mother, x = 68 ------------(equation 3)

Given that equation 1,equation 2 and equation 3 are correct. By combining these equations,

we can write as

65<x=68

That is x = 66 or 67 or 68

average of different probable weights of Kiran = (66+67+68)/3=67

Required average = (67 x 2 + 35 x 2 + 6 x 3)/(2+2+3)

= (134 + 70 + 18)/7

= 222/7

= 31 5/7 years.

Average marks of batch1 = 50

Students in batch1 = 55

Total marks of batch1=55×50

Average marks of batch2 = 55

Students in batch2 = 60

Total marks of batch2=60×55

Average marks of batch3 = 60

Students in batch3 = 45

Total marks of batch3=45×60

Total students = 55 + 60 + 45 = 160

Average marks of all the students = ((55×50)+(60×55)+(45×60))/160

=(275+330+270)/16 = 875/16=54.68

(3+11+7+9+15+13+8+19+17+21+14+x)/12 =12

=>137+x/12=12

=> 137 + x = 144

=> x = 144 - 137 = 7

**ANSWER AND SOLUTION :**

**1(A)Explanation:**Total number of biscuits = 1*8+2*6+3*4+4*2= 8 + 12 + 12 + 8 = 40

Average number of biscuits = 40/10= 4

**2(C)Explanation:**The sixth number = 6*32 + 6 * 37-11*35

= 192 + 222 - 385

= 29

**3(C)Explanation:**Total runs required to win the match = 20 * 7.2 = 144

Runs scored in 15 over = 15 * 6 = 90

Runs required in 5 overs = 144 - 90 = 54

Required run rate = 54/5 = 10.8

**4(B)Explanation:**Fifth integer = 9*32 - 4*26 - 4*8

= 288 - 104 - 112

= 72

**5(B)Explanation:**Let avrage run of 1 innings = x

Total runs of 11 innings = 11x

Score of 12th innings = 90

Average run of 12 innings = x=5

Total runs of 12 innings = 12(x+5)

12(x+5) - 11x = 90

12x + 60 - 11x = 90

x = 30

Thus, average run after the 12th innings = 30+5= 35

**6(C)Explanation :**Let the monthly income of A = a

monthly income of B = b

monthly income of C = a

a+b=2×5050---------(Equation1)

b+c=2×6250---------(Equation2)

a+c=2×5200---------(Equation3)

(Equation 1) + (Equation 3) - (Equation 2)

=>a+b+a+c-(b+c)=(2×5050)+(2×5200)-(2×6250)

=> 2a = 2(5050 + 5200 - 6250)

=> a = 4000

=> Monthly income of A = 4000

**7 (D)Explanation :**Let Kiran's weight = x. Then

According to Kiran, 65 < x < 72 ------------(equation 1)

According to brother, 60 < x < 70 ------------(equation 2)

According to mother, x = 68 ------------(equation 3)

Given that equation 1,equation 2 and equation 3 are correct. By combining these equations,

we can write as

65<x=68

That is x = 66 or 67 or 68

average of different probable weights of Kiran = (66+67+68)/3=67

**8( B)Explanation:**Required average = (67 x 2 + 35 x 2 + 6 x 3)/(2+2+3)

= (134 + 70 + 18)/7

= 222/7

= 31 5/7 years.

**9(B)Explanation :**Average marks of batch1 = 50

Students in batch1 = 55

Total marks of batch1=55×50

Average marks of batch2 = 55

Students in batch2 = 60

Total marks of batch2=60×55

Average marks of batch3 = 60

Students in batch3 = 45

Total marks of batch3=45×60

Total students = 55 + 60 + 45 = 160

Average marks of all the students = ((55×50)+(60×55)+(45×60))/160

=(275+330+270)/16 = 875/16=54.68

**10(C)Explanation :**(3+11+7+9+15+13+8+19+17+21+14+x)/12 =12

=>137+x/12=12

=> 137 + x = 144

=> x = 144 - 137 = 7

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