Question 1]

A. Less than the number of huts existing at the beginning of 2001.

B. Less than the total number of huts destroyed by floods in 2001 and 2003.

C. Less than the total number of huts destroyed by floods in 2002 and 2003.

D. More than the total number of huts built in 2001 and 2002.

Explanation :

Let there be 2x buildings in 2001. Let us form a table :-

Huts destroyed in 2004 = 3.375x.

1st option : Total huts in 2001 = 2x.

2nd option: Total huts destroyed in 2001 and 2003 = 3.25x.

Total huts destroyed in 2002 and 2003 : 3.75x. 3.375 < 3.75. Hence this is true.

Huts built in 2001 and 2002 = 5x.

Question 2]

A. 27

B. 4

C. 5

D. 9

Explanation :

Let the 2nd piece length be x.

1st piece length is 3x.

3rd piece length is 3x – 23.

Sum of all the pieces is 40.

3x + x + 3x – 23 = 40.

7x = 63.

x = 9.

Lengths of the pieces are 27, 9, 4.

Question 3 ]

A. 4

B. 7

C. 14

D. 15

Explanation :

Since I live X floors above the ground floor and it takes me 30 s per floor to walk and 2 s per floor to ride, it takes 30X s to walk down and 2X s to ride down after waiting 420 s

=> 30X = 2X + 420.

=> X = 15.

Question 4]

A. 7

B. 8

C. 15

D. 16

Explanation :

Let a train from Chennai to Jammu starts on Day 1 and it reaches jammu on day 8 at 12:01 Hrs.

The train which starts from Jammu on Day 8 at 12:00 will cross this Day 1 train (from Chennai) at approx. 12:00:30 Hrs and this train will reach Chennai on day 15 at 12:01 Hrs.

So it will cross the train which started from Chennai at 12:00 Hrs on day 15 (in between 12:00 Hrs and 12:01 Hrs).

So, it will cross a total of 15 trains.

Question 5]

A. 32

B. 84

C. 64

D. Can't be determined

Explanation :

Let ∠DAC = x, Then

∠ACD= x

∠CDB= 2x

∠DBC= 2x

=> ∠ACD + ∠CDB + ∠BCE = 180.

∠CDB = 180 - 96 - x = 84 - x.

∠CDB + ∠CDB + ∠DBC = 180

=> 84 - x + 2x + 2x = 180.

=> 3x = 96.

=> x = 32.

=> 2x = 64.

Hence, ∠DBC = 640.

Question 6]

A. 1:√3

B. 1:√2

C. 1:2√3

D. 1:2

Explanation :

Let the radius of the circle be ‘R’ and ∠ODC=∠ADE= θ.

If OM is drawn perpendicular to DC,

DM = R* Cos θ

OM = R* Sin θ

Length of rectangle ABCD,

AB= CD= 2DM = 2R* Cos θ

AD=BC= 2OM=2R* Sin θ

Area of rectangle= AB * BC = 2R* Cos θ * 2R* Sin θ = 2R² Sin 2θ

Area of circle = πR²

According to the question,

=> πR² : 2R²Sin 2θ = π:√3

=> Sin 2θ =√3/2 =Sin 600.

θ =300.

In triangle ADE, tan θ=AE/AD.

AE:AD=tan 300= 1:√3.

Question 7]

A. 100

B. 150

C. 80

D. 60

Explanation :

<OCT = 90 (Tangent and Radius)

=> <OCA = 90 - 50 (<ACT) = 400

Triangle AOC is isosceles ( 2 sides are radius of same circle )

<OAC = <OCA = 400

<CAT = 1800 - 500 (<ACT) - 300 (<CTA) = 1000

so <BAC = 1800 - 1000 (<CAT) = 800

<OAB = <BAC - <OAC = 800 - 400.

∆BOA is isosceles ( 2 sides are radius of same circle )

<OBA = <OAB = 400

<BOA = 1800 - (400 + 400) = 1000.

Question 8]

A. 3

B. 3.5

C. 4

D. 4.5

Explanation :

B & K overtake A at the same instant. So all the 3 have travelled the same distance so far.

Distance = Time * Speed.

Let time taken by A be x.

Since B starts 2hrs later, time taken by B to reach the same point is x – 2.

Distance travelled by A and B are same, and speed of A & B are 30 and 40 respectively.

30x = 40(x-2), solving, we get x = 8.

So A takes 8hrs and B takes 6 hrs, and they travel 30*8 = 240 kms.

Time taken by K to travel 240kms is 240/60 = 4hrs. (Speed of K is 60)

So, K starts 8 - 4 i.e 4 hrs. later than A.

Question 9]

A. 1 hour and 30 minutes

B. 2 hours

C. 2 hours and 30 minutes

D. 1 hr

Explanation :

Wind is from A to B. Hence on the forward journey, velocity is v – 50, and on return journey, velocity is v + 50, where v is the velocity of the plane.

Plane cruises from B to A, stays in A for 1hr, then starts the return journey.

B to A = time = 3000 / (v – 50)

1 hr in A.

A to B = time = 3000 / (v + 50)

Total time = 3000 / (v – 50) + 1 + 3000 / (v + 50).

We can see that the onward flight is at 8 AM from B and return flight reaches back B at 8PM. So, total time is 12hrs. So 3000 / (v – 50) + 1 + 3000 / (v + 50) = 12.

Substitute for ‘v’ from the options of the 2nd sub question, and we get v = 550 as the answer.

So, we now know v = 550, B to A time is 3000/v – 50 = 3000/500 = 6hrs. So B to A flight starts from B at 8 AM, and reaches A at 2PM (B’s local time). However A’s local time is 3PM then.

So, time difference is 1hr.

Question 10 ]

A. 6:15 am

B. 6:30 am

C. 6:45 am

D. 7:00 am

Explanation :

Basically, we have ABC forming a triangle with right angle at C.

(∠ABC = 600 and ∠BAC = 300, So, ∠ACB = 900)

We have AB = 500km. tan 60 = AC/CB. tan 60 = √3.

So, √3 = AC/CB => AC = √3 * CB.

Also in a right triangle, AC2 + CB2 = BA2. (Pythagoras theorem)

So, CB2 + (√3 * CB)2 = 5002

4(CB)2 = 5002

2CB = 500.

So CB = 250, AC = 250√3.

Now, the train travels along CB, at a speed of 50 kmph. CB distance is 250kms. So, time for train to reach C is 250/50 = 5hrs. Train starts at 8AM, and hence reaches C by 1PM.

So, Rahim needs to reach C minimum by 12:45PM. He has to travel a distance of 250√3 at a speed of 70kmph. This takes a time of 250√3/70 approximately is 6.2hrs

6.2 hrs is. 6hrs 12 minutes. He cannot start at 6:45, as he will reach only by 12:57.

He needs to reach by 12:45.

So, he can start by 6:30, so that he reaches before 12:45.

Question 11]

A. Rs. 1200

B. Rs. 1300

C. Rs. 1500

D. Rs. 2000

Explanation :

Interest on Rs.3000 at 5% per annum = ( 3000 × 5 × 1) / 100 = Rs. 150.

Let his additional investment at 8% = x.

Interest on Rs.x at 8% per annum = ( x × 8 × 1 ) / 100 = 2x/25.

To earn 6% per annum for the total, interest = (3000 + x) × 6 × 1/100.

=> 150 + 2x/25 = (3000 + x) × 6 × 1/100.

=> 15000 + 8x = (3000 + x) × 6.

=> 15000 + 8x = 18000 + 6x.

=> 2x = 3000.

=> x = 1500.

Question 12 ]

A. 2/5

B. 1/5

C. 3/5

D. None of these

Explanation :

Out of 6 subjects, one subject M has already been chosen. Ramaining 2 subjects are to be choden from rest (6-1) i.e 5 subjects.

Without restriction, no. of selections = 5C2.

With restriction, no. of selection of B from rest 4 subject = 4C1.

=> Probability

= Selections with restriction / Selections without restriction

= 4C1 / 5C2 .

= 2 / 5.

Hence, the probability of subject B being chosen is 2 / 5.

Question 13]

A. 0.86.

B. 0.91

C. 0.95

D. None of these

Explanation :

Let, A ≡ event that part A is non- defective.

B ≡ event that part B is non- defective.

A and B are independent events.

AB ≡ event that part A and part B both are non-defective so that instrument is not defective.

=> P(AB) = P(A) x P(B).

=> [ 1 - P(~A)] x [ 1 - P(~B)].

Here, P(~A)= 9/100 = probability of part A being defective

and, P(~B) = 5 / 100 = probability of part B being defective.

=> P(AB) = [1- (9 / 100 ) ] x [1- (5 / 100)].

=> 0.91 x 0.95.

=> 0.86.

Hence, the required probability is 0.86.

Question 14]

A. 280

B. 140

C. 90

D. None of these

Explanation :

According to the question, W1 will not be a member if W2 is a member and vice versa.

Case I :- W1 is a member

4 men is selected out of 7 men and 3 women will be chosen out of 5-1 i.e 4 women.

The number of ways of selection is 7C4 x 4C3.

Case II :- W2 is not a member

4 men is selected out of 7 men and 3 women will be chosen out of 5-1 i.e 4 women.

The number of ways of selection is 7C4 x 4C3.

Hence, the required number of ways = 7C4 x 4C3 + 7C4 x 4C3 i.e 280.

Question 15]

A. 5426

B. 5005

C. 5000

D. 4500

Explanation :

Here,

(i) each combination should have 6 balls out of 10 balls.

(ii) each of the 10 balls may be included in any of the combination any number of times

Hence, the required number of ways is:-

=> 10+6-1C6.

=> 15C6.

=> (15 x 14 x 13 x 12 x 11 x 10) / (6 x 5 x 4 x 3 x 2).

=> 5005.

Question 16]

A. – 8 ≤ x ≤ 1

B. – 1 ≤ x ≤ 8

C. 1 ≤ x ≤ 8

D. – 8 ≤ x ≤ 8

Explanation :

Let, x1/3 = y. The equation becomes y2 + y – 2 ≤ 0, or y2 + y ≤ 2.

Now, look at the upper limit. We know that y2 + y ≤ 2.

Among the positive numbers, 1 is the biggest number that satisfies the equation.

No number greater than 1 can have y2 + y ≤ 2. So the upper limit is 1.

Maximum value of y is 1, hence maximum value of x1/3 is 1, and hence maximum value of x is 1.

This answer is there only for option 1. Hence that is the answer.

Question 17]

A. 1023

B. 2047

C. 4095

D. 8195

Explanation :

un+1 = 2*un+ 1.

u0 = 0 (given)

u1 = 2 * 0 + 1 = 1.

u2 = 2 * 1 + 1 = 3.

u3 = 2 * 3 + 1 = 7.

u4 = 2 * 7 + 1 = 15.

On observing the pattern, un= (2n) - 1.

Hence, u10= 210 - 1 = 1023.

Question 18]

A. Standard 75 bags, Deluxe 80 bags

B. Standard 100 bags, Deluxe 60 bags

C. Standard 60 bags, Deluxe 90 bags

D. Standard 50 bags, Deluxe 100 bags

Explanation :

Let 'x' be the number of standard bags and 'y' be the number of deluxe bags.

Thus, 4x + 5y ≤ 700 and 6x + 10y ≤ 1250

Among the choices, (3) and (4) do not satisfy the second equation.

Choice (2) is eliminated as, in order to maximize profits the number of deluxe bags should be higher than the number of standard bags because the profit margin is higher in a deluxe bag.

Question 19]

A. 15 hours

B. 16 hours

C. 12 hours

D. 10 hours

Explanation :

If A can complete work alone in a days

and

B can complete work alone in b days, then

=> 2/a = 5/b = 1/2 or 1/a +5/2b = 1/4 ........... (1)

and

=> 3/a +3/b= 1/2 - 1/20 = 9/20 or 1/a +1/b = 9/60 .... ( 2)

From (1) and (2), we get :-

b=15 hrs and a= 12 hrs

Hence (A) is the correct answer.

Question 20]

A. 33.33%

B. 31.33%

C. 66.66%

D. 25%

Explanation :

Let d size of container A is "x", then B's size will be "2x".

A is half full of wine = >x/2 .

So remaining "x/2" of A contains water .

B is quarter full of win => 2x/4 => x/2

So remaining => 2x - x/2 = 3x/2.

3x/2 of B contains water.

Totally C has A's content + B's Content = x + 2x = 3x .

Wine portion in C = x/2 of "A" + x/2 of "B"

=> x portion of wine

Water portion in C = x/2 of "A" + 3x/2 of "B"

=> 4x/2 => 2x portion of water

So portion of wine in C is x/3x =1/3 portion of wine

if v expressed in %, 1/3 * 100 = 33.33%.

Hence (A) is the correct answer.

**In a coastal village, every year floods destroy exactly half of the huts. After the flood water recedes, twice the number of huts destroyed are rebuilt. The floods occurred consecutively in the last three years namely 2001, 2002 and 2003. If floods are again expected in 2004, the number of huts expected to be destroyed is :-**A. Less than the number of huts existing at the beginning of 2001.

B. Less than the total number of huts destroyed by floods in 2001 and 2003.

C. Less than the total number of huts destroyed by floods in 2002 and 2003.

D. More than the total number of huts built in 2001 and 2002.

Explanation :

Let there be 2x buildings in 2001. Let us form a table :-

Huts destroyed in 2004 = 3.375x.

1st option : Total huts in 2001 = 2x.

2nd option: Total huts destroyed in 2001 and 2003 = 3.25x.

Total huts destroyed in 2002 and 2003 : 3.75x. 3.375 < 3.75. Hence this is true.

Huts built in 2001 and 2002 = 5x.

Question 2]

**A string of length 40 meters is divided into three parts of different lengths. The first part is three times the second part, and the last part is 23 meters smaller than the first part. Find the length of the largest part.**A. 27

B. 4

C. 5

D. 9

Explanation :

Let the 2nd piece length be x.

1st piece length is 3x.

3rd piece length is 3x – 23.

Sum of all the pieces is 40.

3x + x + 3x – 23 = 40.

7x = 63.

x = 9.

Lengths of the pieces are 27, 9, 4.

Question 3 ]

**I live X floors above the ground floor of a high-rise building. It takes me 30 s per floor to walk down the steps and 2 s per floor to ride the lift. What is X, if the time taken to walk down the steps to the ground floor is the same as to wait for the lift for 7 min and then ride down?**A. 4

B. 7

C. 14

D. 15

Explanation :

Since I live X floors above the ground floor and it takes me 30 s per floor to walk and 2 s per floor to ride, it takes 30X s to walk down and 2X s to ride down after waiting 420 s

=> 30X = 2X + 420.

=> X = 15.

Question 4]

**The North-South Express is a pair of trains between Jammu and Chennai. A train leaves Jammu for Chennai exactly at 12 noon everyday of the week. Similarly, there's a train that leaves Chennai bound for Jammu on everyday of the week exactly at 12 noon. The time required by a train to cover the distance between Chennai and Jammu is exactly 7 days and 1 minute. Find the number of trains from Chennai to Jammu which a train from Jammu to Chennai would cross in completing its journey, assuming all the trains are on time.**A. 7

B. 8

C. 15

D. 16

Explanation :

Let a train from Chennai to Jammu starts on Day 1 and it reaches jammu on day 8 at 12:01 Hrs.

The train which starts from Jammu on Day 8 at 12:00 will cross this Day 1 train (from Chennai) at approx. 12:00:30 Hrs and this train will reach Chennai on day 15 at 12:01 Hrs.

So it will cross the train which started from Chennai at 12:00 Hrs on day 15 (in between 12:00 Hrs and 12:01 Hrs).

So, it will cross a total of 15 trains.

Question 5]

**In the figure (not drawn to scale) given below, if AD = CD = BC and ∠ BCE = 960. How much is the ∠ DBC ?**A. 32

B. 84

C. 64

D. Can't be determined

Explanation :

Let ∠DAC = x, Then

∠ACD= x

∠CDB= 2x

∠DBC= 2x

=> ∠ACD + ∠CDB + ∠BCE = 180.

∠CDB = 180 - 96 - x = 84 - x.

∠CDB + ∠CDB + ∠DBC = 180

=> 84 - x + 2x + 2x = 180.

=> 3x = 96.

=> x = 32.

=> 2x = 64.

Hence, ∠DBC = 640.

Question 6]

**In the figure below (not drawn to scale), rectangle ABCD. Is inscribed in the circle with centre at O. The length of side AB is greater than that of side BC. The ratio of the area of the circle to the area of the rectangle ABCD is π:√3. The line segment DE intersects AB at E such that ∠ODC = ∠ADE.****What is the ratio AE : AD ?**A. 1:√3

B. 1:√2

C. 1:2√3

D. 1:2

Explanation :

Let the radius of the circle be ‘R’ and ∠ODC=∠ADE= θ.

If OM is drawn perpendicular to DC,

DM = R* Cos θ

OM = R* Sin θ

Length of rectangle ABCD,

AB= CD= 2DM = 2R* Cos θ

AD=BC= 2OM=2R* Sin θ

Area of rectangle= AB * BC = 2R* Cos θ * 2R* Sin θ = 2R² Sin 2θ

Area of circle = πR²

According to the question,

=> πR² : 2R²Sin 2θ = π:√3

=> Sin 2θ =√3/2 =Sin 600.

θ =300.

In triangle ADE, tan θ=AE/AD.

AE:AD=tan 300= 1:√3.

Question 7]

**In the figure below (not drawn to scale), A, B, and C are three points on a circle with center O. The chord BA is extended to a point T such that CT becomes tangent to the circle at point C. If ∠ATC = 300 and ∠ACT = 500, then the ∠BOA is**A. 100

B. 150

C. 80

D. 60

Explanation :

<OCT = 90 (Tangent and Radius)

=> <OCA = 90 - 50 (<ACT) = 400

Triangle AOC is isosceles ( 2 sides are radius of same circle )

<OAC = <OCA = 400

<CAT = 1800 - 500 (<ACT) - 300 (<CTA) = 1000

so <BAC = 1800 - 1000 (<CAT) = 800

<OAB = <BAC - <OAC = 800 - 400.

∆BOA is isosceles ( 2 sides are radius of same circle )

<OBA = <OAB = 400

<BOA = 1800 - (400 + 400) = 1000.

Question 8]

**Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start?**A. 3

B. 3.5

C. 4

D. 4.5

Explanation :

B & K overtake A at the same instant. So all the 3 have travelled the same distance so far.

Distance = Time * Speed.

Let time taken by A be x.

Since B starts 2hrs later, time taken by B to reach the same point is x – 2.

Distance travelled by A and B are same, and speed of A & B are 30 and 40 respectively.

30x = 40(x-2), solving, we get x = 8.

So A takes 8hrs and B takes 6 hrs, and they travel 30*8 = 240 kms.

Time taken by K to travel 240kms is 240/60 = 4hrs. (Speed of K is 60)

So, K starts 8 - 4 i.e 4 hrs. later than A.

Question 9]

**Cities A and B are in different time zones. A is located 3000 km east of B.****The table below describes the schedule of an airline operating non-stop flights between A and B. All the times indicated are local and on the same day. Assume that planes cruise at the same speed in both directions. However, the effective speed is influenced by a steady wind blowing from east to west at 50 km per hour. What is the time difference between A and B?**A. 1 hour and 30 minutes

B. 2 hours

C. 2 hours and 30 minutes

D. 1 hr

Explanation :

Wind is from A to B. Hence on the forward journey, velocity is v – 50, and on return journey, velocity is v + 50, where v is the velocity of the plane.

Plane cruises from B to A, stays in A for 1hr, then starts the return journey.

B to A = time = 3000 / (v – 50)

1 hr in A.

A to B = time = 3000 / (v + 50)

Total time = 3000 / (v – 50) + 1 + 3000 / (v + 50).

We can see that the onward flight is at 8 AM from B and return flight reaches back B at 8PM. So, total time is 12hrs. So 3000 / (v – 50) + 1 + 3000 / (v + 50) = 12.

Substitute for ‘v’ from the options of the 2nd sub question, and we get v = 550 as the answer.

So, we now know v = 550, B to A time is 3000/v – 50 = 3000/500 = 6hrs. So B to A flight starts from B at 8 AM, and reaches A at 2PM (B’s local time). However A’s local time is 3PM then.

So, time difference is 1hr.

Question 10 ]

**Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and still catch the train is closest to**A. 6:15 am

B. 6:30 am

C. 6:45 am

D. 7:00 am

Explanation :

Basically, we have ABC forming a triangle with right angle at C.

(∠ABC = 600 and ∠BAC = 300, So, ∠ACB = 900)

We have AB = 500km. tan 60 = AC/CB. tan 60 = √3.

So, √3 = AC/CB => AC = √3 * CB.

Also in a right triangle, AC2 + CB2 = BA2. (Pythagoras theorem)

So, CB2 + (√3 * CB)2 = 5002

4(CB)2 = 5002

2CB = 500.

So CB = 250, AC = 250√3.

Now, the train travels along CB, at a speed of 50 kmph. CB distance is 250kms. So, time for train to reach C is 250/50 = 5hrs. Train starts at 8AM, and hence reaches C by 1PM.

So, Rahim needs to reach C minimum by 12:45PM. He has to travel a distance of 250√3 at a speed of 70kmph. This takes a time of 250√3/70 approximately is 6.2hrs

6.2 hrs is. 6hrs 12 minutes. He cannot start at 6:45, as he will reach only by 12:57.

He needs to reach by 12:45.

So, he can start by 6:30, so that he reaches before 12:45.

Question 11]

**A man invests Rs. 3,000 at the rate of 5% per annum. How much more should he invest at the rate of 8%, so that he can earn a total of 6% per annum?**A. Rs. 1200

B. Rs. 1300

C. Rs. 1500

D. Rs. 2000

Explanation :

Interest on Rs.3000 at 5% per annum = ( 3000 × 5 × 1) / 100 = Rs. 150.

Let his additional investment at 8% = x.

Interest on Rs.x at 8% per annum = ( x × 8 × 1 ) / 100 = 2x/25.

To earn 6% per annum for the total, interest = (3000 + x) × 6 × 1/100.

=> 150 + 2x/25 = (3000 + x) × 6 × 1/100.

=> 15000 + 8x = (3000 + x) × 6.

=> 15000 + 8x = 18000 + 6x.

=> 2x = 3000.

=> x = 1500.

Question 12 ]

**A student has to select 3 subject out of 6 subjects M,B,H,U,L. If he has chosen M, what is the probability of B being chosen ?**A. 2/5

B. 1/5

C. 3/5

D. None of these

Explanation :

Out of 6 subjects, one subject M has already been chosen. Ramaining 2 subjects are to be choden from rest (6-1) i.e 5 subjects.

Without restriction, no. of selections = 5C2.

With restriction, no. of selection of B from rest 4 subject = 4C1.

=> Probability

= Selections with restriction / Selections without restriction

= 4C1 / 5C2 .

= 2 / 5.

Hence, the probability of subject B being chosen is 2 / 5.

Question 13]

**An instrument manufactured by a company consists of two parts A and B. In manufacturing part A, 9 out of 100 are likely to be defective and in manufacturing part B, 5 out of 100 are likely to be defective. Calculate the probability that the instrument will not be defective.**A. 0.86.

B. 0.91

C. 0.95

D. None of these

Explanation :

Let, A ≡ event that part A is non- defective.

B ≡ event that part B is non- defective.

A and B are independent events.

AB ≡ event that part A and part B both are non-defective so that instrument is not defective.

=> P(AB) = P(A) x P(B).

=> [ 1 - P(~A)] x [ 1 - P(~B)].

Here, P(~A)= 9/100 = probability of part A being defective

and, P(~B) = 5 / 100 = probability of part B being defective.

=> P(AB) = [1- (9 / 100 ) ] x [1- (5 / 100)].

=> 0.91 x 0.95.

=> 0.86.

Hence, the required probability is 0.86.

Question 14]

**There are 7 men and 5 women who are eligible for being included in a committee of 4 men and 3 women. In how many ways can the committee be formed if one women W1 refuses to be in the committee if another particular woman W2 be included?**A. 280

B. 140

C. 90

D. None of these

Explanation :

According to the question, W1 will not be a member if W2 is a member and vice versa.

Case I :- W1 is a member

4 men is selected out of 7 men and 3 women will be chosen out of 5-1 i.e 4 women.

The number of ways of selection is 7C4 x 4C3.

Case II :- W2 is not a member

4 men is selected out of 7 men and 3 women will be chosen out of 5-1 i.e 4 women.

The number of ways of selection is 7C4 x 4C3.

Hence, the required number of ways = 7C4 x 4C3 + 7C4 x 4C3 i.e 280.

Question 15]

**A box has 10 balls. Find how many ways 6 balls can be selected if each ball may be repeated any number of times.**A. 5426

B. 5005

C. 5000

D. 4500

Explanation :

Here,

(i) each combination should have 6 balls out of 10 balls.

(ii) each of the 10 balls may be included in any of the combination any number of times

Hence, the required number of ways is:-

=> 10+6-1C6.

=> 15C6.

=> (15 x 14 x 13 x 12 x 11 x 10) / (6 x 5 x 4 x 3 x 2).

=> 5005.

Question 16]

**What values of x satisfy the equation x2/3 + x1/3 – 2 ≤ 0 ?**A. – 8 ≤ x ≤ 1

B. – 1 ≤ x ≤ 8

C. 1 ≤ x ≤ 8

D. – 8 ≤ x ≤ 8

Explanation :

Let, x1/3 = y. The equation becomes y2 + y – 2 ≤ 0, or y2 + y ≤ 2.

Now, look at the upper limit. We know that y2 + y ≤ 2.

Among the positive numbers, 1 is the biggest number that satisfies the equation.

No number greater than 1 can have y2 + y ≤ 2. So the upper limit is 1.

Maximum value of y is 1, hence maximum value of x1/3 is 1, and hence maximum value of x is 1.

This answer is there only for option 1. Hence that is the answer.

Question 17]

**Let un+1 = 2un + 1.( n = 1,2,3 .........) and u0 = 0, then u10 nearest to**A. 1023

B. 2047

C. 4095

D. 8195

Explanation :

un+1 = 2*un+ 1.

u0 = 0 (given)

u1 = 2 * 0 + 1 = 1.

u2 = 2 * 1 + 1 = 3.

u3 = 2 * 3 + 1 = 7.

u4 = 2 * 7 + 1 = 15.

On observing the pattern, un= (2n) - 1.

Hence, u10= 210 - 1 = 1023.

Question 18]

**A leather factory produces two kinds of bags, standard and deluxe. The profit margin is Rs 20 on a standard bag and Rs. 30 on deluxe bag. Every bag must be processed on machine A and machine B. The processing times per bag on the two machines are as follows :-**Time required( Hours/bag) | ||
---|---|---|

Machine A | Machine B | |

Standard bag | 4 | 6 |

Deluxe bag | 5 | 10 |

**The total time available on machine A is 700 hrs and on machine 1250 hours. Among the following production plan, which one meets the machine availability constraints and maximizes the profit?**A. Standard 75 bags, Deluxe 80 bags

B. Standard 100 bags, Deluxe 60 bags

C. Standard 60 bags, Deluxe 90 bags

D. Standard 50 bags, Deluxe 100 bags

Explanation :

Let 'x' be the number of standard bags and 'y' be the number of deluxe bags.

Thus, 4x + 5y ≤ 700 and 6x + 10y ≤ 1250

Among the choices, (3) and (4) do not satisfy the second equation.

Choice (2) is eliminated as, in order to maximize profits the number of deluxe bags should be higher than the number of standard bags because the profit margin is higher in a deluxe bag.

Question 19]

**Two workers A and B manufactured a batch of identical parts. A worked for 2 hours and B worked for 5 hours and they did half the job. Then they worked together for another 3 hours and they had to do (1/20)th of the job. How much time does B take to complete the job, if he worked alone?**A. 15 hours

B. 16 hours

C. 12 hours

D. 10 hours

Explanation :

If A can complete work alone in a days

and

B can complete work alone in b days, then

=> 2/a = 5/b = 1/2 or 1/a +5/2b = 1/4 ........... (1)

and

=> 3/a +3/b= 1/2 - 1/20 = 9/20 or 1/a +1/b = 9/60 .... ( 2)

From (1) and (2), we get :-

b=15 hrs and a= 12 hrs

Hence (A) is the correct answer.

Question 20]

**There are two containers on a table. A and B. A is half full of wine, while B,which is twice A's size, is one quarter full of wine. Both containers are filled with water and the contents are poured into a third container C. What portion of container C's mixture is wine?**A. 33.33%

B. 31.33%

C. 66.66%

D. 25%

Explanation :

Let d size of container A is "x", then B's size will be "2x".

A is half full of wine = >x/2 .

So remaining "x/2" of A contains water .

B is quarter full of win => 2x/4 => x/2

So remaining => 2x - x/2 = 3x/2.

3x/2 of B contains water.

Totally C has A's content + B's Content = x + 2x = 3x .

Wine portion in C = x/2 of "A" + x/2 of "B"

=> x portion of wine

Water portion in C = x/2 of "A" + 3x/2 of "B"

=> 4x/2 => 2x portion of water

So portion of wine in C is x/3x =1/3 portion of wine

if v expressed in %, 1/3 * 100 = 33.33%.

Hence (A) is the correct answer.

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