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__START QUIZ__

__CASLET DI__**Directions :- Study the following data carefully and answer the questions given below it .**

**1. The total numbers of students who have taken all the three subjects are what per cent of the total number of students in the institute?**

**A15**

**B12.5**

**C13.6**

**D14.8**

**ENone of these**

**2.What is the total number of students in the institute who have taken only Cookery as their subject?**

**A3180**

**B3200**

**C3020**

**D3135**

**ENone of these**

**3.The total numbers of students who have taken only Social Work as their subject from the discipline of Arts are approximately what per cent of the total number of students taking the same subject from the discipline of Science?**

**A43**

**B48**

**C36**

**D28**

**E51**

**4.What is the total number of students who have taken Social Work and Physical Training from all the three disciplines?**

**A13855**

**B9861**

**C10281**

**D12555**

**ENone of the above**

**5.What is the respective ratio of the total number of students from the discipline of Commerce to those from the discipline of Science?**

**A9:8**

**B7:8**

**C8:9**

**D8:7**

**ENone of these**

**DIRECTION ( 6 - 10 )**

**Direction ( 11 - 15 )**

The table shows the number of students participated in different sections from different schools.

**11) A group of four students has to form from school C such that the group contains one student in each section and the remaining from any of the section. The number of ways in which this can be possible is 25920. Find the number of students in section III from school C?**

a) 15

b) 12

c) 10

d) 20

e) 18

**12) A group of two students has to form from school E such that the group contains at least one student in section III. The number of ways in which this can be possible is 195. A committee of five students is to be formed from school E such that the committee contains 2 students from section I, 1 student from section II and 2 students from section III. Find the number of ways in which this can be possible?**

a) 15920

b) 16478

c) 23420

d) 17820

e) None of these

**13) Find the total number of students in school A?**

**Statement I: **All the students in section II from school A is handshaking with each other and the total number of handshakes is 105.

**Statement II: **All the students in school A is handshaking with each other and the total number of handshakes is 780.

a) Only I

b) Only II

c) Both I and II

d) Either I or II

e) Neither I nor II

**14) In school B, number of students in section III is 75% of the number of students in section II.**

**Quantity I: **In committee P, five students is to be formed from school B such that the committee contains 1 student from each section and the remaining students from any of the section.

**Quantity II: **In committee Q, five students is to be formed from school B such that committee contains at least one student from section I.

a) Quantity I > Quantity II

b) Quantity I < Quantity II

c) Quantity I ≥ Quantity II

d) Quantity I ≤ Quantity II

e) Quantity I = Quantity II (or) Relationship cannot be determined

**15) A committee of three members is to be formed from each school B and school D. Number of possible ways of a committee contains three students in section III from school D is what percentage of the number of possible ways of committee contains three students in section I from school B? **

a) 180%

b) 380%

c) 160%

d) 280%

e) None of these

**Directions (16 – 20): Study the following information carefully and answer the given questions**

__SOLUTIONS:__

**Solutions—**

**Directions (11-15) :**

11) Answer: c)

No of students in section III from school C = x

Total number of students in school C = x + 8 + 12 = 20 + x

Number of possible ways = 8C1 * 12C1 * xC1 * (20 + x – 3) C1 = 25920

8 * 12 * x * (17+x) = 25920

17x + x2 = 270

= > x2 + 17x – 270 = 0

= > x2 – 10x + 27x – 270 = 0

= > x (x – 10) + 27 (x – 10) = 0

= > x = 10, – 27 (Eliminate the –Ve value)

Number of students in section III from school C = 10

12) Answer: d)

Number of students in section III from school E = y

Total number of students in school E = (y + 12 + 18)

= y + 30

Number of possible ways = yC1 * 30C1 + yC2 = 195

= > y * 30 + (y*(y-1)/(1 *2)) = 195

= > 30y + (y2 – y)/2 = 195

= > 60y + y2 – y = 390

= > y2 + 59y – 390 = 0

= > y2 + 65y – 6y – 390 = 0

= > y (y + 65) – 6 (y + 65) = 0

= > (y + 65) (y – 6) = 0

= > y = 6, -65 (Eliminate –Ve value)

Total number of students in school E = 30 + 6 = 36

Required number of ways = 12C2 * 18C1 * 6C2

= (12*11 / 1*2) *18 *(6*5 / 1*2)

= 6*11*18*3*5

= 17820

13) Answer: b)

From I,

Number of students in section II from school A = n

Number of handshakes = nC2 = 105

= > [n*(n -1)] / (1 * 2) = 105

= > n2 – n = 210

= > n2 – n – 210 = 0

= > n2 – 15n + 14n – 210 = 0

= > (n – 15) (n – 14) = 0

= > n = 15, -14 (Eliminate –Ve value)

From statement I, We have to find only number of students in section II from school A. We can’t find the total number of students.

Hence, statement I alone is not sufficient to answer the given question.

From II,

Total number of students in school A = x

Number of handshakes = xC2 = 780

= > x * (x – 1)/1*2 = 780

= > x2 – x = 780*2

= > x2 – x – 1560 = 0

= > x2 – 40x + 39x – 1560 = 0

= > x (x – 40) + 39 (x – 40) = 0

= > (x – 40) (x + 39) = 0

= > x = 40, -39 (Eliminate –Ve value)

Total number of students in school A is 40.

Hence, Statement II alone is sufficient to answer the given question.

14) Answer: a)

Number of students in section II from school B = x

Number of students in section III from school B = 75/100 * x

= 3x/4

Total number of students in section II and III from school B = 20 – 6 = 14

= > x + 3x/4 = 14

= > 7x/4 = 14

= > x = 8

Number of students in section III from school B = 8 * 75/100 = 6

Quantity I:

Number of ways = 6C1*8C1*6C1*(20-3)C2

= 6 * 8 * 6 *(17*16/1*2)

= 6*8*6*17*8

= 39168

Quantity II: In committee Q, five students is to be formed from school B such that committee contains at least one student from section I.

Number of ways = 6C1*14C4 + 6C2*14C3 + 6C3*14C2 + 6C4*14C1 + 6C5

= 6006 + 5460 + 1820 + 210 + 6

= 13502

Hence, Quantity I > Quantity II

15) Answer: d)

Number of ways in school D = 8C3

= 8*7*6 / 1*2*3 = 56

Number of ways in school B = 6C3

= 6*5*4 / 1*2*3 = 20

Required percentage = (56/20) * 100 = 280%

**Directions (16 – 20):**

Ratio of number of Blue colour balls in bag A, B and C = 72: 120: 168 = 3: 5: 7

Let number of blue balls in bag A, B and C is 3x, 5x and 7x respectively.

Number of yellow balls in bag A = 3x * (5/3) = 5x

Total balls in bag A = (40 + 3x + 5x) = (40 + 8x)

Probability when two balls are selected from bag A and out of which one Red and other is Blue = [(40C1 * 3xC1)/(40 + 8x)C2] = 20/119

120x/(20 + 4x)(39 + 8x) = 20/119

3x/(5 + x)(39 + 8x) = 2/119

357x = 16x2 + 158x + 390

16x2 – 199x + 390 = 0

16x2 – 160x – 39x + 390 = 0

16x (x – 10) – 39 (x – 10) = 0

(16x – 39) (x – 10) = 0

= > x = 10, 39/16 (Eliminate the fraction value)

x = 10

Number of Blue balls in bag A = 3x = 30

Number of Yellow balls in bag A = 5x = 50

Total balls in bag A = 40 + 30 + 50 = 120

Number of Blue balls in bag B = 5x = 50

Number of Blue balls in bag C = 7x = 70

Let the number of yellow balls in bag B = b

Total balls in bag B = (60 + 50 + b) = (110 + b)

Probability when two Blue balls are selected from bag B = 50C2/(110 + b)C2

= (49/447)

(50 * 49)/[(110 + b)(110 + b – 1)] = 49/447

50/[(110 + b) (109 + b)] = 1/447

22350 = 11990 + 110b + 109b + b2

22350 = 11990 + 219b + b2

b2 + 219b – 10360 = 0

b2 + 259b – 40b – 10360 = 0

(b + 259) (b – 40) = 0

= > b = 40, -259 (Eliminate –ve value)

b = 40

Number of yellow balls in bag B = b = 40

Total balls in bag B = 60 + 50 + 40 = 150

Let the number of yellow balls in bag C = c

Probability of selecting a Blue ball from bag B = 50/150 = 1/3

Total balls in bag C = (20 + 70 + c) = (90 + c)

Probability of selecting a Blue ball from bag C = 70/(90 + c)

According to the question:

70/(90 + c) = 7/12

10/(90 + c) = 1/12

120 = 90 + c

c = 30

Number of yellow balls in bag C = c = 30

Total balls in bag C = 20 + 70 + 30 = 120

16) Answer: b)

Probability of selecting two red balls from bag A = 40C2/120C2

= (20 * 39)/(60/119) = (13/119)

Probability of selecting one yellow ball from bag B = 40/150 = (4/15)

Probability of selecting one blue ball from bag C = 70/120 = (7/12)

Required probability = (13/119) * (4/15) * (7/12) = (13/765)

17) Answer: d)

Probability of selecting two Blue colour balls from bag A = 30C2/120C2 = (29/476)

Probability of selecting either a red colour ball or a blue colour ball from bag B

= (60/150) + (50/150) = (11/15)

Required difference = (11/15) – (29/476) = (5236 – 435)/7140

= (4801/7140)

18) Answer: a)

Probability of selecting one blue colour ball from bag A = (30/120) = (1/4)

Probability of selecting one blue colour ball from bag B = (50/150) = (1/3)

Difference = (1/3) – (1/4) = (1/12)

Probability of selecting one red colour ball from bag B = (60/150) = (2/5)

Probability of selecting one red colour ball from bag C = (20/120) = (1/6)

Difference = (2/5) – (1/6) = (7/30)

Required per cent = [{(7/30) – (1/12)}/(7/30)] * 100 = (450/7)%

19) Answer: e)

Probability that both the selected balls from bag B and C are of red colour

= (60/150) * (20/120) = (1/15)

Probability that both the selected balls from bag B and C are of blue colour

= (50/150) * (70/120) = (7/36)

Probability that both the selected balls from bag B and C are of yellow colour

= (40/150) * (30/120) = (1/15)

Total probability = (1/15) + (7/36) + (1/15) = (59/180)

20) Answer: b)

After transfer, number of red colour balls in bag B = 60 – [83(1/3)% of 60]

= 10

After transfer, number of yellow colour balls in bag B = (40 + p)

Total balls in bag B after transfer = 10 + 50 + (40 + p) = (100 + p)

After transfer, number of red colour balls in bag C = 20 + [83(1/3)% of 60]

= 70

After transfer, number of yellow colour balls in bag C = (30 – p)

Total balls in bag C after transfer = 70 + 70 + (30 – p) = (170 – p)

Probability of selecting a red colour ball from bag B = [10/(100 + p)]

Probability of selecting a blue colour ball from bag C = [70/(170 – p)]

According to the question:

20[10(170 – p) + 70(100 + p)] = 11(100 + p)(170 – p)

20[1700 – 10p + 7000 + 70p) = 11(17000 + 70p – p2)

174000 + 1200p = 187000 + 770p – 11p2

11p2 + 430p – 13000 = 0

11p2 – 220p + 650p – 13000 = 0

11p (p – 20) + 650 (p – 20) = 0

= > p = 20, – 650/11 (Eliminate –ve value)

p = 20

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