A = P { 1 - r%}^n where A is the depreciated value after n years at the depreciation rate of r% , So , A = 20,00,000 { 1 - 20/100}^3 = 20,00,000 {4/5}^3 = 20,00,000 x 4/5x4/5x4/5 = 10,24,000 Depreciated value after 3 years will be Rs 10,24,000

The ball rebounces only 80% of orginal height i.e. it is going 20% less height, so by using formula A = P { 1 - r%}^n where A is final height after third bounce, P = original height achived, r% = rate of loss in height & n = number of bounces A = 20 {1 - 20/100}^3 = 20 x 4/5x4/5x4/5 = 10.24 m so after third bounce it will achieve a height of 10.24 m

CI=P[{1+R/100}^2-1]=9650[{1+8.3/100}^2-1]=1668.378=1668.38

Simple interest (SI)=(time×rate×principle)/100 Thus principle=(1240×100)/3×6 =6880

Using the formula for compound interest, CI = P { (1+r%)^n - 1 } where CCI = compond interest, P = princcipal, r% = rate, n = time in years. Putting the values 210 = P { (1 + 4%)^2 - 1 } = P { 1.0816 - 1 } = P x 0.0816 or P = 210/0.0816 = 2573.53 so principal = Rs 2573.53 Now SI = {PXrxt}/100 = { 2573.53 x 8 x 2 }/100 = 411.76 so simple interest will be Rs 411.76

A =P(1+%r)^n. Hence 1296=1089(1+%r )^2 i.e.sq.rt(1296/1089)= 36/33=12/11=1+%r, hence r= 100/11= 9.09%.

Amount/Principal =1323/1200 =(1+%r)^2. Hence sq.root(1323/1200)=1.05,which makes interest rate as 5%.

principal P (1+.05)^2 =1323 amount .Hence P= Rs.1200.

Let length and breadth of the rectangle be l and b. Given: l = 1.5b ------ (1) Perimeter of a rectangle P = 2(l+b) Using equation (1), P = 2(1.5b + b) = 2(2.5 b) P = 5b ------------ (2) Rs.60 - 1m can be fenced Rs.12,000 - (1/60)*12000 = 200meters So, perimeter of the rectangle is 200 meters ------------ (3) From (2), P = 5b Hence, 5b = 200 b = 40 m l = 1.5b = 1.5 * 40 = 60 m Length = 60m Breadth=40m

C.P= 1800 Sales tax= 8% He has to pay, 1800+1800*8/100= 1944Rs But he wants to pay 1800Rs(Including sales Tax) Let new C.P =x x + x*8/100=1800 X=1666.66 (New C.P after reduction) Reduction Percentage= (1800-1666.66)/1800*100= 7.40%

CP = (108/100)*(98/100)*(97/100)*(94/100)*Rs.3,500 = Rs. 3,377.67.

Let the distance be 4km. Time taken to cover 1/4th distance i.e 1/4*4km i.e 1km=1/10 hr. Time taken to cover next 1/4th distance i.e 1/4*4km i.e 1km=1/5 hr. Time taken to cover last 1/2 distance i.e 1/2*4km i.e 2km=2/10 hr. Total time taken=1/10+1/5+2/10=5/10=1/2hr speed=distance/time=4km/(1/2hr)=8km/hr

After allowing 10% discount price of vase = 1500-(10/100 *1500) = 1350 Now, the shopkeeper earns a profit of 25% on selling it at 1350 Cost price = 1350 * (100/125) = 1080

Let the distance be 40km. Time taken to cover 1/4th  i.e 10km=10km/10kmph=1hour

Time taken to cover next 1/4th i.e 10km=10km/5kmph=2hour. Time taken to cover remaing 1/2 i.e 20km = 20km/10kmph=2hour.

Average speed =total distance/total time=40/(1+2+2)=40/5 = 8 kmph.

Amount=P* (1+.05)^2 =1323 amount .Hence P= Rs.1200.

 let assume "x" meter be the trains length

When it pass an electric pole it only cover its own length, whereas when it cross a platform it has to cover distance equal to sum of its length and platform's length.

Then assuming it is moving in same speed we get

  x/18   =   (x+100)/30

or, 30x = 18x + 1800

or 12x = 1800

so x = 150

So the train's length is 150 meter.

As trains are moving in opposite direction

 So combined speed of the two trains = (50+40) km/hr = 90 km/hr = 25 m/s

Let length of train B=x metre

Total  length of two trains = (400+x) m. 

Time taken = 30s .

So, (400+x) = 25*30 = 750 . 

Thus, x = 750 - 400 = 350 .

Ans. The length of train B is 350 m .

Simple interest for 3 years=225. So interest for one year=225/3=75

When it is compounded, Interest for first year+Interest for second year=153

75+Interest for second year=153. Interest for second year=153-75=78

Difference interest for first and second year=78-75=3

This Rs3 extra interest is of Rs75(Interset of first year) for one year.

Rate of interest=3*100/75=4%

Now using the simple interest of one year

Principal=75*100/4=1875

Ans--Principal=Rs1875

Let, the distance be x km.

Time taken to reach office if speed is 40kmph=x/40 hr

Time taken to reach office if speed is 50kmph=x/50 hr

Since he reaches 10 min late if speed is 40kmph and 5 min early if speed is 50kmph. The difference in time is 15min 

15 min=15/60 hr =1/4 hr.

Now x/40-x/50=1/4

=> x/200 = 1/4   or x=200/4 =50km.

Ans. The required distance is 50 km . 

Present age of Pravash=10+4=14 years

Present age of Rajat=3*Pravash age=3*14=42 years

Present age of Basu=42-5=37 years

Basu age after 6 years=37+6=43 years.

After arranging letters of word TIGER in ascending order alphabetically, it will become EGIRT. So, no letter will remain in the same position.

Here it is given that P is mother of X

Also Q is husband of P .

Hence X is the daughter of Q or say Q is the father of X.

Here it is given that Riton is the mother of Hyder.

Prashanta is the husband of Riton. And Hyder is the wife of Parash. So Hyder is the daughter of Riton and Prashanta

So Prashanta is the father in law of Parash.

Position of Nilesh from right end=17

Position of sharmila from right end=14

Position of Rohit from right end=19

Position of Rohit from left end=20

Total number of children=20+19-1(Position of Rohit from both end-1)

Total children=38

The position will be as follows.

 Left end    Rohit                  Nilesh                            Sharmila      Right end

            20(L)19(R)  18(R)    17(R)   16(R) 15(R)        14(R)

Since we have to divide the boys and girls into equal section of either boys or girls, We have to first find the HCF of 408 and 312.

HCF of 408 and 312=24

Number of sections of boys=408/24=17

Number of section of girls=312/24=13

Total number of section=17+13=30

For finding the distance covered by all three men before they 'IN STEP' again , we will have to find the LCM of their steps.

LCM of 68,51,85=1020 cm.

Ans--1020 cm

Here the length of wire is sold in 1m=100cm

Lenghth of wire required=85 cm

To find the minimum length of wire that has to be purchased, we have to find the LCM of 100cm and 85 cm.

LCM of 100 and 85=1700 cm

Ans--1700 cm=17 m

To find when the bells will simultaneously again we have to find the LCM

LCM of 6,8,12,18=72seconds

So the bells will again ring simultaneously after 72 seconds

6 minutes=6*60=360 seconds

Number of times bells will ring in 6 minutes i.e 360 second=360/72=5 times

But  the bells rang simultaneously for first time  at 12 o'clock

So total number of times=5+1=6 times.

To find the least number of coconuts  we will have to find the LCM

LCM of 2,3,5=30

Ans--30 coconuts.

To find the time when both will again be at starting point we have to find the LCM

LCM of 10 minutes and 9 minutes=90 minutes

Ans--Both cyclist will be at starting point after 90 minutes.

To find the capacity we have to find the HCF of 27,33 and 45

HCF of 27,33 and 45=3

Ans--3 Litre

Total number of chocolates Priyanka had in the begining=30

Number of chocolates collected in 5 weeks=5*5=25

Total number of chocolates after 5 weeks=30+25=55

Weight of diferent varities of tea are 132kg,156kg,84kg.

To find the size of box we have to find the HCF of 132,156 and 84

HCF of 132,156 and 84=12kg

Ans--12kg

To find the time when all the three traffic lights will change simultaneously we will have to find the LCM of 48,72 and 108

LCM of 48,72 and 108=432 seconds=432/60 minutes=7minutes and 12 seconds

So the traffic lights will change simultaneously at 7:07:12 am

ANS--A

To find the number of books in each stack we have to find the HCF of 420 and 130

HCF of 420 and 130=10

Ans--10

C.P of 1000 items. = Rs. 1000*3 = Rs. 3000

S,P, of 1000 items = Rs. (700*4.5 + 300*2.5) = Rs. 3900

Profit = SP - CP = Rs. (3900 - 3000) = Rs. 900

Hence,  Avg. Profit  per item : Rs. 900/1000=.Re. 0.90 = 90 paise

Let, the original price and consumption be P1 and  C1 respectively .

So, original expenditure = P1*C1

New price P2=1.25*P1

Let new consumption be C2

Since expenditure remains the same so P2*C2=P1*C1

or 1.25*P1*C2=P1*C1

C2=P1*C1/1.25P1  i.e C2=4/5*C1

% change in consumption=(1-4/5)*100=1/5*100=20%

Ans--20%

From the given condition we can find

Total coins they have = 3*100 -50 = 250 (as 100 is average coins one has)

Now  50+a+b = 250 => a+b=200.----(i)

a - 15 = 1/3*(b + 15) . => 3a - b = 60 ... (ii)

Adding (i) and (ii), 4a = 260 .  => a = 65 .

 b = 200-a=200-65=135

So a:b= 65:135 = 13:27.

Ans--13:27

Let first and second numbers be a and b respectively.

Now b- 30%a = 3/4b

=> b-3/4b = 30/100a

=>b/4 = 3/10a

=>a/b = 5/6

Total distance covered by boat=1km

Total time taken=15 minutes=1/4 hour

Speed of boat=distance/time=1/(1/4)=4kmph

Let the speed of river Vr

Speed of boat=speed of boat in still water-speed of river

4=5-Vr

Vr=5-4=1kmph

Velocity of river=1kmph

Ans--B

Total Outcomes = 25C3

Favourable Outcomes = 10C1*15C2

Probability = Favourable/Total = 10C1*15C2/25C3 = 10*7*15/23*4*25 = 21/46

ANS--C

Favouralbe case = 10C1

Total cases=14C1

Probability=Favouralbe/Total = 10C1/14C1 = 10/14 = 5/7 

ANS--D

Total cases, n(S) = 52C2 = (52∗51) / (2∗1) =1326

One King and One queen case, n(E) = 4C1 * 4C1 = 4 ∗ 4 = 16

Probability = n(E) / n(S) = 16 / 1326 = 8 / 663

ANS--D

p ( getting an even number ) = 3 / 6 = 1 / 2

q ( not getting an even number ) = 3 / 6 = 1 / 2 

n ( number of times die is thrown ) = 100

Variance = n * p * q = ( 1 / 2 ) * ( 1 / 2 ) * 100 = 25

Therefore, the variance of the number of successes = 25

Let the speed of boat in still water is x.

Speed upstream=(x-3) kmph

Speed downstream=(x+3) kmph

Since distance is constant

(x-3)*2=(x+3)*1.5

on solving we get x=21 kmph

ANS--C

ANS--B

We know that sum of all angles of a triangle=180

So, 94+41+3rd angle=111+x+3rd angle=180

Since 3rd angle is common so 94+41=111+x

Hence x=24

ANS--D

Angle AMC=180-(56+78)=46

Angle DMB=46

Angle MBD=180-(62+46)=72

ANS--B

Let angle B=x

Then Angle A=2x

2x+x=90

3x=90 i.e x=90/3=30

Angle B=30

Angle A=2*30=60

Ans.  D 

As both the trains are moving in same direction the relative speed = 30-20=10m/s

Total distance = (200 + 300) m = 500 m .

Hence, time taken = 500/10 s = 50 s .

ANS--A

Let the daughter's share is x

 sons share is 2*x=2x

mother's share is 4times of son's share, then mothers share is 4*2x=8x

fatrher's share is 3times than of mother's share, thus father's share is 3*8x=24x

thus the ratio of father's, mother's, son's, daughter's shares are 24x:8x:2x:x=24:8:2:1

Taking, no. of cold drinks bottles = x, no. of water bottles becomes 2/3*x .

ATP, x + 30 = 4*(2/3*x) .  => 5x = 90 .  => x = 18 . 

So, 1/3*x = 6 . 

Hence, The reqd. no. of cold drink bottles is 6 

Ans. C

Diff, between CI and SI = Dif. between the Amounts = Rs.1000[(1 + 10/100)^4 - (1 + 10*4/100)] = Rs.1000[1.4641 - 1.4] = Rs. 64.10 .

ANS--A

CP for the manufacturer for 1kg wheat = Rs. x.

ATP, his SP for 0.9 kg wheat = Rs. (114/100)*x .

So, SP of 1 kg wheat = Rs. (1.14/0.9)*x = Rs.1.2667x .

Hence, profit% = [(SP - CP)/CP]*100 = 26.67% .

ANS--B

Cost Price = 12500

Marked up = 12500*10/100 = 1250

Toal Marked up price = 12500+1250 = 13750

Discount = 13750*5/100 = 687.50

Sales Price after discount = 13750-687.50 = 13062.50

Profit = sales Price - Cost Price

=> 13062.50-12500 = 562.50

% Profit = 562.50/12500*100 = 4.5%

ANS--C

Average speed = total distance/total time

Total time = 6h + 4h = 10h

distance 1 = 30 X 6 = 180 miles

distance 2 = 45 X 4 = 180 miles

Total distance = distance 1 + distance 2

Total distance = 180 + 180 = 360 miles

Average speed = total distance/total time

Average speed = 360/10= 36 miles per hour

ANS--D

Part of cistern filled in 1 min = (1/16 + 1/12 - 1/8) = 1/48. 

Hence, total  time to fill cistern=48 mins. 

ANS--C

Length of bigger rectangle=30 cm

Breadth of bigger rectangle=15 cm

Area of bigger rectangle=L*B=30*15=450  sq cm

Length of smaller unshaded rectangle=30-8=22cm

Breadth of smaller unshaded rectangle=15-4=11 cm

Area of smaller unshaded rectangle=22*11=242 sq cm

Area of shaded portion =Area of bigger rectangle - Area of smaller rectangle

Area of shaded portion=450-242=208 sq cm

ANS--D

First let us extend the cut portion to form arectangle.

Area of that rectangle=20*15=300 sq m

The cut portion will be a triangle having base=20-8=12 m

and its height=15-10=5m

Area of this triangle=12*5/2=30 sq m

Area of given figure=300-30=270 sq m

Ans.  C .

Ratio of area = Suare of ratio of sides .  => 9:25

So, area of smaller square = [9/(9+25) of 304] sq. cm = 81 sq. cm. 

Hence, side length of smaller square rt.81 = 9 cm .

ANS--B

Let, total volume be x ltrs. It remains same throughout. But, change in liquor is (18 - 15)% = 3% = 3x/100

ATP, 3x/100 = (18/100)*8 .  => x = 48 litres

ANS--B

Age of new friend = [(24*5) - {21*4)+(4*4)}] = (120 - 100) yrs = 20 yrs.

In given problem RMO make a right angle triangle

Where Distance between RM = 15km and MO = 9 km

Then (OR)2 =(RM)2 - (MO)2

i.e. (OR)2 = (15*15) - (9*9)

=> (OR)2 = 225 - 81

=> (OR)2 = 144

=> OR = 12km

T's paternal grandfather is P

T's paternal grandmother is R

P and R are husband and wife

Perimeter=sum of the sides Let X be a variable. The given ratio is 13:12:5 Therefore the sum of the sides=13x+12x+5x Perimeter=30x Given perimeter=450 30x=450 x=15m The sides are 12*15,13*15,5*15 It is a right angle triangle since 13square equal to sum of the square of 12 and 5. Area =1/2*b*h 1/2*12*15*5*15=6750sq.m

ANS--A

Total outcomes = 2×2×2×2= 16

Getting at least one tail means one or more tails. P(at least 1 tail) = 1- P(no tail) = 1- 1/16 = 15/16

ANS--C

Total number of balls=2+3+4=9

Number of ways of selecting 3 balls=9C3

Since we have to draw 3 balls and they should be of same colours. So we can take out black or white balls.

This can be done in 3C3 and 4C3 ways

probability=(3C3+4C3)/9C3=5/84

ANS--C

Circumradius = 2/3 rd of height = (2/3)*(√3/2) of each side = R .

Hence, each side = √3R. 

Ans. BD = DC = (a/√3)

Join AD. AD = 2r = 2*(a/√3) . Join OB and OC. OBDC is a rhombus. OB = OC = r = (a/√3). 

ANS--A

Mr.A has x children while Mrs.B has (x+1) children. After they got married, totally they have 10 children. But it is not given how many children do they have after they got married. Let they be 'y'. Then, we get, x+(x+1)+y=10 i.e. 2x+y=9. Here, if y can take only odd values, otherwise, x will not be a positive integer for any even value of y. Then there are following cases: 1) for y=1, x=4 and x+1=5, 2) for y=3, x=3 and x+1=4 3) for y=5, x=2 and x+1=3 4) for y=7, x=1 and x+1=2 5) for y=9, x=0 and x+1=1. As the children of same parents cannot quarrel, only y can quarrel with x and (x+1) and x can quarrel with (x+1). Then from above five cases, we have to consider the case which will give maximum value. It is the the 2nd case. That is, the product xy+y(x+1)+x(x+1) will be maximum for y=3, x=3 and (x+1)=4, that is 33. Hence 33 is the answer.

ANS--A

Original amt. of solute in 800 g solution = 40/100*800 = 320 g.

After 100 ppt, 220 g remains in 700 g of solution.

Hence, new solute percentage is (220/700)*100 = 31.4%

ANS--D

When sister was born father was of 28 years and when brother was born sister was of 4 years, so when the brother was born father was of 28 + 4 = 32 years of age.

Since, the average formula is:

(sum of student age) / 22 = 21

sum of student age = 22 x 21

Now the average increase by one when a teacher age included

(sum of student age + teacher's age) / (22 + 1) = 21 + 1

(sum of student age + teacher's age) = 22 x 23

teacher's age = 22 x 23 - sum of student age = 22 x 23 - 22 x 21 = 22 (23 - 21) = 22 x 22 = 44 years.

So, the age of teacher is 44 years.

ANS--B

If each wager is for half the money remaining, each win lead to multiplying the amount by 1.5 times and each loss multiplying the amount by 0.5 times.

the person won three times and lost three times. So we need to multiply the initial amount by 1.5 thrice and the loss by 0.5 times thrice in any order.

So the final amount will be Rs. 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5) = Rs 27 (ina ll caes)

The loss will be 64-27 =37 

ANS--A

Let me explain this question with an easy example. If 100 is increased by 20% it will become 120. Now if 120 is decreased by 20% u will get 120-24 = 96 a lesser value. So, to get same value of 100 the % increase of decrease can not be same. There decrease in % will always be less than increase in % Now, according to question, if x = y the value will not come as 1 million, it will be less. Therefore answer will be x>y 

ANS--C

relative speed = (68-8)km/s=  60km/s= 60*5/18 m/s= 50/3 m/s

total distance= 150m     (Train's length)

time taken to pass a man = 150*3/50 = 9s

ANS--D

Rs. 1 get double in 4 Years i.e. become Rs. 2

In next 4 years get doubled and become Rs. 4

In next 4 years get doubled and become Rs. 8

Hence in 4+4+4 = 12 years amount will become 8 times.

ANS--D

The man completes 40/60 = 2/3 part of work in 40 hrs.

So, his son completes (1 - 2/3) = 1/3 part of work in 40 hrs.

Hence, the son can alone complete the job in 40*3 = 120 hrs

ANS--C

Time taken for  walking one side = 45/2 mins = 22.5 mins.

So, time taken for riding one side = (33-22.5) mins = 10.5 mins. 

Hence, time taken for riding both ways = 10.5*2 mins = 21 mins

ANS--A

Let the length of one side be 2 cm

Then surface area=6*2*2=24 sq cm

After 50% increase the side becomes 3 cm

Then surface area=6*3*3=54

Percentage increase=(54-24)/24*100=30/24*100=125%

Present sum of ages of Subho and Sudip = (32*2 + 3*2) years = 70 years.

Again, present sum of ages of Subho, Sudip and Subir = 30*3 years = 90 years.

Hence, age of Subir = (90-70) years = 20 years

ANS--B

Profit=SP-CP=60-50=10

Gain %=Profit/CP*100=10/50*100=20%

ANS-A

 Let speed of train= 1/x and speed of car = 1/y Time = distance / speed Therefore, 120x + 480y = 8....... equation 1 200x + 400y = 25/3... equation 2 Solving 1/x = 60 and 1/y = 80 Therefore ratio 60:80 = 3:4

Let, time taken by Ashish be x hrs.

So, his speed = 40/x km/h and that of Tushar = 40/(x+2) km/h .

ATP, 40/[2*(40/x+2)] = x - 1 . Or, x + 2 = 2x - 2 .  Or, x = 4 . 

Hence, Tushar's speed = 40/(x+2)  = 40/6 km/h = 6.67 km/h

ANS-C

Speed of the train = (142/6)*(18/5) = 84.9 km/h (approx) 

ANS--A

red marbles=24

probability to pick red =3/7

total marbles= X= 24x7/3=56

blue marbles= 56-24=32

Difference in no. of red and blue=8

ANS--D

total marbles 15

White=5

Probability choosing white=5/15

addition of white=X

5+X/15+X=3/5

X=10

ANS--A

Total number of coins in piggy bank = 10 + 20 +10 = 40.

A be event of number of Rs 5 coin = 10 P(A)    = (nuber of out comes favorable to A) / (Total number of possible out comes)            = 10/40 = 1/4

ANS--A

Weight of 11 people=11*48=528

Weight of 12 people=12*46=552

Weight of 12th man =552-528=24kg

Area of larger square=5*5=25

Area of smaller square=4*4=16

Area of shaded portion=25-16=9

Probability of point in shaded portion =9/25

ANS--A

 We know that the time period is proportional to the roo of length of a pendulam, 44% increase in legth means, it is now 144%, i.e, 1.44, root of that is 1.2, so, it os 20% increase.

ANS-A

 Total Mass of Solution is 800 g Ratio of Solute is 40℅ Hence, Mass of Solute is (40/100)x800 = 320 g Now, 100 g of the solute is precipitated. Hence, the remaining mass of Solution is 700 g And, the remaining mass of Solute is 220 g Now the Percentage of Solute in the Solution is (220/700)x100℅ = 31.4%

ANS--B

Apply the formula- v^2 = u^2 + 2gR (where u = 0)

V^2=0+2gR

V^2=2gR

V=Sqrt 2gR

ANS-C

 AngleDOB = [180 - (132 + 141)] = 93

ANS--D

Triangles DCQ and XCD are similar. [Since, <C is common, <X = 90 degree for both triangles]

DQ:DC = 2:1.  We know, ratio of areas = Square of the ratio of the sides

4:1

ANS--A

Total surface area of cylinder=2*3.14*R*H+2*3.14*7*7

or 224*3.14=2*3.14*R*H+2*3.14*7*7

Dividing both sides by 2*3.14

112=7*H+49

or 7*H=112-49

or H=63/7=9cm

ANS--D

Case 1 Probability of picking up a white marble = 5/15=1/3

Case-2  x white marbles were added.

Probability of selecting white marble = (5+x)/(15+x)=3/5 or 25+5x= 45+3x or x=10

ANS--B

Red=X

Green = Y

Yellow = Z

X= 2Y= 2x 3Z=6Z

Y=3Z

Probability of getting Green Marble: Y/X+Y+Z

3Z/6Z+3Z+Z=3/10

ANS--D

Draw the figure with given data and join the two centres A and B. Let, that line AB bisect PQ at O. Take, two rt. ang Tri. AOP and BOP. to find out AO and BO by Pyth. Theo. PO= 12 cm, AP = 13 cm and BO = 15 cm.  Thus, AO = 5 cm, BO = 9 cm. Hence, AB = (5 + 9) = 14 cm.

No of students taking neither tea nor coffee=600-(150+225-100)=600-275=325

Number of boxes damaged=700/35=20

Let total number of boxes be x

Then 2*x/100=20

x=20*100/2=1000 boxes

From the first statement T>Q

From the second statement  T>Q>P>S

Since R is remaining so R will be shortest

T>Q>P>S>R is the order of heights.

ANS--R is the shortest.

ANS--A

Increase in total age of 7 women=7*5=35

Age of women=35+25=60 years

ANS--A

A's 1 day work=1/12

B's 1 day work=1/18

Total work done by both A and B in one day=1/12+1/18=5/36

C's 1 day work=1/4-5/36=1/9

Time taken to complete the work by c alone=9 days

ANS--A

Portion of tank filled in 1 hour by pipe A=1/8

Portion of tank filled in 1 hour by pipe B=1/6

Portion of tank filled in 2 hours when A and B are opened alternately=1/8+1/6=7/24

Portion filled in 6(2*3) hours=7/24*3=21/24

Portion of tank empty=1-21/24=3/24=1/8

To fill this portion pipe A will take 1 hour

Total time taken=6+1=7 hours

ANS--A

Ratio given 4 years ago 19 : 6

Father age = 19x

Son age    =6x

ii)Difference 19x - 6x =26

                   13x       =26

                       x        =2

Therefore Their ages 4 years back in the ratio of 38 : 12

4 years hence (Present age) 38+4:12+4 = 42:16 = 21:8

ANS--D

Smita's position from left after interchange=14

From right this is 17th

Total number of girls=17+14-1=30

Sn = n/2 [2a+(n-1)d] to find the sum of n terms of an AP Where a= First term d = difference

 S20 = 20/2 [2a+ (20-1)d] = 10[2a+19d] = 20a+190d. ------(1)

S10 = 10/2 [2a+ (10-1)d] = 5[2a+9d] = 10a+45d. ------(2)

S30 = 30/2 [2a+ (30-1)d] = 15[2a+29d] = 30a+435d. -----(3)

Eqn (1) - (2) gives S20 - S10 = (20a+190d) - (10a+45d) = 10a+145d 3(S20-S10) = 3(10a+145d) = 30a+435d = S30

 (50 + a + b) + 50 = 3*100 .   => a + b = 200 ... (i)

Again, a - 15 = 1/3*(b + 15) .   => 3a - b = 60 ... (ii)

Adding, (i) and (ii), 4a = 260  .   => a = 65 .  So, b = 135 .

Hence, a:b = 65:135 = 13:27

ANS--D

(1+ r/100) = 23958/21780 [taking the 3rd year only]

r/100= 1/10

r=10

Hence, rate of interest is 10%

ANS--C

Let the length of train=x

(x+162)/(x+120) = 18/15 = 6/5 .

So, x = 5*162 - 6*120 = 90 . 

ANS--B

nitial water = 20% of 25 kg = 5 kg.

Let, water added be x kg.

ATP, 5 + x = 33.33% of (25 + x) .  => x = 5 .

R has more number of pages than P but less number of pages than Q.   Q>R>P

V has more number of pages than Q i.e V>Q 

but it does not have maximum number of pages and since R has 51 pages & T has 76 pages, the equations can be written as T>V>Q>R>P

Since exactly 2 books have less number of pages than P and let those books be x & y, so  P>x>y i.e T>V>Q>R>P>x>y

since Q & S does not have min pages, y cannot be S. So that leaves x as S and y as U. The eqn becomes T>V>Q>R>P>S>U

So the num of books with pages less than Q are 4

ANS--A

Daughter has the least share. Therefore the ratios of their shares are (3×4×2×1):(4×2×1):(2×1):1 = 24:8:2:1

Since tangents have equal length we have XP = XQ and AO = AR and BQ = BR

XB+BR = XB + BQ = XQ and similarly XA+AR = XP and we already have XQ = XP

Let d be the distance between tower and chimney.  Then tan 60 = h/d where h is height of tower.

Also tan 30 = (h-40)/d

Dividing 3 = h/(h-40) Or 3h-120 = h

h = 60 m.

Height is less than 60 hence meets pollution norms.

Use section formula for the 4 points as ratio 1:4, 2:3, 3:2, 4:1.

First point = (2, 6) and second point = (3, 5), III point = (4, 4) IV point = (5, 3)

If h is the height of the light house, and d the distance between the first ship and light house, then

tan 30 = h/d and tan 45 = h/(100-d)

Divide to get

1/sq rt 3 = (100-d)/d

d = 100 rt 3 - rt 3 d

= 173.2 - 1.732 d

d = 173.2/2.732 = 63.4

h = 100-d = 36.6 m

S7 = 7/2 (2a+6d) = 63 and S14 = 7(2a+13d) 

S14 = S7+161 = 224

2a+13d =32

2a+6d = 18 From S7

7d =14 and d =2. a = 9-3d=3

28th term= 3+27(2) = 57

the possibilities of sum 10 (considering both dice) is three-(6,4),(4,6),(5,5)

so, number of envent is -3

total number of events for 2 dice is-6*6=36

probability is= number of event of 10/ total number of event occour= 3/36=1/12

Area of rectangle = curved SA of cylinder

i.e. 880 = 2(3.14) (r)(40)

r = 7/2 = 3.5

ANS--C

We have cube of edge is 15 so radius of sphere half of the edge i.e 7.5 cm Then volume of sphere is= 4/3×pi×r^3 =4/3×22/7×7.5×7.5×7.5 = 1767.86 cm^3

ANS--B

The reqd. volume of water = [70*50*20 - 50*(5^3)] cm^3 = (70000 - 6250) cm^3 = 63750 cm^3

ANS--C

We have cube of edge is 15 cm so radius of sphere is half of edge i.e 7.5 cm Now we calculate volume = 4/3×pi×r^3 = 4/3×22/7×7.5×7.5×7.5 = 1767.86 cm^3

ANS--B

 We have (30 + x)(20 + x) = 5*(30*20) .  => x = 30 or -80.  -ve value to be ignored.

Hence, x = 30 m .

ANS--D

Area of the field = 4928/0.5 cm^2 = 9856 m^2 .

So, radius of the field = root(9856/pi) cm = 56 m.

Thus, circumference = 2*pi*56 = 352 m.

Hence, cost = Rs. 352*12 = Rs. 4224 .

ANS--C

Ratio of radii of large circle to small circle = 3:1 .  So, ratio of their areas = 3^2:1^2 = 9:1 .

Hence, unshaded area:shaded area = (9 - 1):1 = 8:1 .

BD = root.(AB^2 + AD^2) = root.(64 + 225) = root.289 = 17 cm.

So, radius of the circle or OD = 1/2 of BD = 8.5 cm .

Hence, area of the circle = 3.14*(8.5)^2 cm^2 = 226.865 cm^2 .

ANS--D

Reqd area = [35*25 - {2*pi/4*13^2)] = [875 - 265.57) cm^2 = 609.43 cm^2

Let nC(r-1) = 45, nCr = 120 and nC(r+1) = 210 => nCr / nC(r-1) = 120/45=8/3 and nC(r+1)/nCr = 210/120=7/4 nCr / nC(r-1)=[ ni /(r)i (n-r)i ]/[ni /(r-1)i (n-r+1)i] = (n-r+1)/r = 8/3 and (n-r)/(r+1) = 7/4 => 4n -11r =7… eq-1 and 3n - 11r =- 3… eq-2 Subtracting 2nd eqn. from the 1st, n = 10 and r = 3

To reach the path represented by the equation 6x - 7y+ 8=0 in the shortest possible time, the family has to travel through the path which is perpendicular to the path 6x - 7y + 8=0 and passes through the intersection of the paths whose equations are 2x - 3y - 4 =0 and 3 x + 4y -5 =0 To get the point of intersection of the two lines: 2x - 3y - 4 =0 and 3 x + 4y -5 =0, we need to solve the equation which will give the the point of intersection,O (x1,y1) 2x1 - 3y1 - 4 =0 ....line1 3x1 + 4y1 -5 =0......line 2 x1=31/17 and y1=(-2/17) The slope of the given line(line-3) is 6x - 7 y + 8=0 , m1=6/7 The slope of the required line(shortest path) will be m2= -7/6 ,ince the shortest path will be perpendicular to line 3 and product of slopes of perpendicular lines are -1 The equation of the required line is : (y - y1)/(x - x1)=- 7/6 ⇒ [y-(-2/17)]/(x -31/7)= (-7/6) ⇒ (17y+2)/(17x-31)=(-7/6) ⇒ 119x +102y= 205 Hence the equation of the path they should follow is 119x +102y- 205=0

Let the one end of the diameter be (x,y). The center (2,-5) is the midpoint of the diameter, whose other end is (-2,-3) By using the midpoint formula x+(-2)/2=2 and y+(-3)/2=-5 => x=6 and y = -7 Coordinates of the other end of the diameter is (6,-7)

ANS--B

Let the perpendicular hit the base at O. As we have A and B on either side of, we need to calculate, AB=OA+OB. Using trignometry we have Tan 45= 120/ OA, i.e. 1=120/OA, i.e.OA=120. Similarly, we have Tan 30=120/OB ,i.e=OB=120(1.732). Thus, the total length= AB=OA+OB= 120*2.732=327.84.

ANS--D

Volume of cylindrical part = pi*3^2*5 = 45pi m^3

Volume of conical part = (1/3)*pi*3^2*2 = 6pi m^3

Hence, total volume = (45 + 6)pi m^3 = 51pi m^3 = 160.29 m^3

ANS--D

Total volime of two cones = volume of cylinder

=> (1/3)*pi*h*a^2 + (1/3)*pi*h*b^2 = pi * h* r^2

=> r = sqrt((a^2 + b^2)/3)

ANS--C

 (pi*r*l + pi*r^2):(pi*r^2) = 18:13 .  => r:l = 5:13 .  => r^2:(r^2 + h^2) = 25:169 .  => h^2:r^2 = 144:25 .

Hence, h:r = 12:5 

ANS--D

Total length of the wire = 66*3 cm = 198 cm .

So, circumference of the circle = 198 cm. So, radius of the circle = 198/(2*pi) cm = 31.5 cm

Hence, area of the circle = pi*(31.5)^2 cm^2 = 3118.5 cm^2

ANS--B

Volume of cylinder= pir^2h 22/7×7.5×7.5×75 13258.92 cm^3 Given that 1cm^3 is equal to 10 g or 100 kg so mass of the solid is 13258.92/100 132.589 or 132.59 

ANS--A

Surface Area of a Sphere = 4πr^2

Curved Surface Area of Cylinder = 2πrh

Thus, ratio of Surface Area of a Sphere : Curved Surface Area of Cylinder = 4πr2 : 2πrh

or 2r:h

or d=h

Ratio 1:1

ANS--C

The average speed is 8 miles per hour.

The total time taken to complete tw course  are 20 +25 = 45 minutes.

! course  = 3 miles. So 2 course = 2 X 3 = 6 miles.

45 minutes taken to complete miles  = 6 miles

1 minute = 6 / 45

60 minutes = (6 * 60) / 45 = 8 miles

ANS--A

Let us assume that the quantity of alcohol as 4x and the quantity of water as 3x.

then he said that 5 liters of water is added

4x/(3x+5)=4/5

20x=12x+20

8x=20

x=2.5

alcohol=4x

4*2.5=10 litres

ANS--C

The investment ratio of A&B is 3:2, 

then their profit ratio must be 3:2

but  out of 100% profit, 5% was given to charity, the remaining 95% be shared in the ratio  3:2

say profit as X

then A share  95/ 100 X * 3/5  = 855     =>  X = 1500

ANS--A

Length and bredth of rectangle is l cm and b cm

(l2 + b2)^(1/2) = 41.

Also, lb = 20.

(l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81

(l + b) = 9.

Perimeter = 2(l + b) = 18 cm.

ANS--B

The fruit content in both the fresh fruit and dry fruit is the same.

Given, fresh fruit has 68% water.so remaining 32% is fruit content. weight of fresh fruits is 100kg

Dry fruit has 20% water.so remaining 80% is fruit content.let weight if dry fruit be y kg.

fruit % in freshfruit = fruit% in dryfruit

(32/100) * 100 = (80/100 )* y

we get, y = 40 kg

ANS--D

Total Cost => Set up Cost 2800 + cost of Paper 7 ink 1600+ printing cost 3200 => 7600

Total Sales => 1500*5 =>7500

Let, money obtained from advertising = x

then Profit = sales - cost

=> 25% of 7500 = (7500+x) -7600

=> 1875 = -100 + x

=> x = 1975

ANS--B

Let, B1 : B2 : B3 = 3x : 4x : 5x Again, B1 : B2 : B3 = 5y : 4y : 3y Number of oranges remain constant in third basket as increase in oranges takes place only in first two baskets. Hence, 5x = 3y and, 3x : 4x : 5x → 9y /5 : 12y /5 : 15y /5 = 9y : 12y : 15y And, 5y : 4y : 3y → 25y : 20y : 15y Therefore, Increment in first basket = 16. Increment in second basket = 8. Thus, Required ratio = 16 /8 = 2 : 1

ANS--C

CI for the 4th yr = CI after 4 yrs - CI after 3 yrs = Amt. after 4 yrs - Amt. after 3 yrs = P*[(1.2)^4 - (1.2)^3] = 0.3456P .

SI for the 4th yr = SI for the 1st yr = 0.2P .

So, diff. betn CI and SI for 4th yr = 0.3456P - 0.2P = 0.1456P .

ATP, 0.1456P = Rs. 7280.  => P = Rs. 50,000 

ANS--A

Temp. on Wednesday = [40*3 - (41*3 - 42)] deg = 39 deg

ANS--A

Total speed of train and boy = (60+6)km/h = 18.33 m/s

Distance covered = 120 m

Hence, time taken = 120/18.33 s = 6.54 s

ANS--A

Sima:Ayesa= 21000:17500= 6:5 Total profit= Rs. 26400 So, Sima gets= 6×26400/(6+5)= 6×26400/11=6×2400=14400

ANS--D

For only women selected means we chose 2 from 5 so 5C2 and total is 12C2

so probability= 5C2/12C2=5/33

ANS--B

Let Bikash runs = x meters

Then, x*25/100 = 7

=> 25x =700

=> x = 28 meters

The length of race = 28+7 => 35 meters

ANS--D

Relative speed = 68 -x Distance traversed= 0.1km Time interval =1/36 hrs 1/36 = 0.1 /(68-x) 5/18 = 1/(68-x) X=32

Let,point C lie closer to point A and point B closer to D. Given that, AC=BD. Now,AB= AC+ BC and CD= BD+ BC. BC being common and AC= BD, we have AB = CD. Had the case been reverse,that's B been closer to A and C being closer to D, AB= AC- BC and CD=BD- BC. Then again, AB= CD

Rohit is standing facing sun in the evening,    i.e. he is facing west

he starts walking towards right   i.e. walking towards north

Then he take left turn   i.e. walking towards west

Then he take right turn    i.e. walking towards north

And then he take right turn  ie. walking towards east.

so, now he is walking towards east.

ANS--A

4 years ago the the ratio between the ages is 19:6. Ao the difference=19-6=13 The ratio 13:26=1:2 Therefore the ages four years back will be 19×2 and 6×2. That is 38 and 12. So the present ages are 38 + 4 and 12+4. That is 42 and 16. Therefore the ratio between the present ages are 42:16. That is 21:8

 Simple Intereest =(P x R x T)/100

                           240 = (P x 5 x 6)/100

​                          Hence P =Rs. 800

ANS--C

the total no. of employees at present is 5000

the no. increases @2% every year

so for the first year the no. of employees would be 5000+(2%*5000)=5100

during the second year the no. of employees would be 5100+(2%*5100)=5202

so total no. of employees after 2 years =5202

Water content = 25 (20%) = 5 kg.

If x kg is added then percent = (5+x)/(25+x) = 1/3

i.e. 15+3x = 25+x

x = 5 kg.

It "crosses" the platform. Means it travels the distance equal to platform pluse its own length. so, x = 280x3 + 280 = 1120m t =50 v = x/t = 1120/50 m/s to convert it into kmph multiply by 18/5 (remember this) v = 112/5x 18/5 = 896/25 = 36kmph approx

A, B, C  together can finish the work in 4 days

Hence the work is done in one day by A, B, C together is = 1/4 

It can get by adding one day work of all the individuals i.e. 1/4=(1/12+1/18+1/C)

                                                                                               1/4=5/36+1/C

                                                                                             1/4 - 5/36=1/C

                                                                                                1/C=1/9

Therefore   C can finish the work alone in 9 days.

As per given data, A can fill 1/8 tank in 1 hr and B can fill 1/6 in 1 hr. If A is opened first 1/8 + 1/6 + 1/8 + 1/6 + 1/8+1/6+1/8+1/6=1 in this fashion the tank will get full in 7 hrs.

ANS--D

 {(x -16)/x}*100 = x.

=> By solving, x = 80 or 20

so x=20

ANS--C

Let Cost Price = Rs. 100

Marked Price = 100+(100*30/100) = Rs. 130

Discount = 130*20/100 => Rs. 26

Sales Price = 130-26 => Rs.104

Profit = 104-100 => Rs. 4

% Profit => 4/100*100 => 4%

ANS--A

Sara can finish the work in 12 Days. In I day she does 1/12 parts of the work

Bula can finish the work in 18 Days. In I day she does 1/18 parts of the work

Kamal can finish the work in d Days. In I day she does 1/d parts of the work

They together can complete in 4 days. In I day they can do 1/4 parts of the work

 Hence 1/12+1/18+1/d=1/4 so 1/d= 1/4-1/12-1/18= 1/9

Hence Kamal can complete the work in 9 days

ANS--A

P+C+M = C+120

=> P+M = C+120-C

=> P+M = 120

Average = 120/2 => 60

Let Total Boxes = 100

Damaged Boxes = 100*2/100 => 2 

Loss per Damaged Box = 2*35 => Rs. 70

When loss is 70 there are total no of boxes =100

When loss is 700 there are total no of boxes =100 / 70*700 => 1000

If total population is 500, then 300 males and 200 females

Of these, 60 males and 50 females are graduates.

Hence 110/500 i.e. 22% are graudates.

Non graduates are 78%

P is the mother-in law of Z

Since K and P are immediate neighbours.

Total number of people=38+13=51

A can do work in 4 days. One day work of A is 1/4.

B can complete same work in 8 days.( as A is twice efficient as B) One day work of B is 1/8.

C can do same work in 16 days. (as B is twice efficient as C) One day work of C is 1/16.

If all of them work together then in one day they can complete (1/4+1/8+1/16= 7/16) of the work .

So they will take 16/7 days for them to complete work when all work together.

Relative velocity = v1+v2 = 60+v2 kmph

Distance = l1+l2 = 180+270 m = 0.45 km.

0.45 = 10.8(60+v2)/3600   

60+v2 = 150 

v2= 90kmph

Speed of second train=90 kmph

let the two numbers be x and y.

y-0.3x=3y/4

y-3y/4=0.3x

y/4=0.3x

dividing both sides by y we get

1/4=0.3(x/y)

so x/y=10/12=5/6

so the rato=5:6

ANS--B

The number of visitors in January was 960.

The average number of visitors January-July is 870

∴ Total number of visitors January-July is 870 X 7= 6,090.

∴ Total number of visitors February-July =6,090-960= 5,130

Actual number of visitors in August was 827

∴ Total number of visitors February -August =5,130+827= 5,957

∴ Average number of visitors February-August= 5,957/7=851

We multiply the capital by number of months remained in the business.

For Sreekumar, 30000(6) +21000(6)= 306000

For Pete 30000(6)+18000(6) = 288000

For Ahmed 30000(6)+35000(6) = 390000

Hence ratio of profit share = 306:288:390 

Ahmed share = 390(88560)/984

=35100

ANS--B

Side of the square = 2(16+24)/4 = 20 cm.

Circumference of semi circle with radius 10 cm  = 3.14(10) = 31.4 cm=31 cm

ANS--D

In 1 hour P will fill up 1/12th of the tank and in 1 hour Q will fill up 1/18th of the tank.

In 2 days (1/12 + 1/18)= 5/36 parts of the tank will be filled.

In 14 days 5/36*7=35/36 part will be filled

The remaining portion is 1/36 which can be filled by P in 12/36 days i.e 0.33

Total days=14.33 days

ANS--C

Let the MRP be 100

After 20% discount S.P=80 and profit%=20

C.P=S.P/1.2=80/1.2

When 10% discount is given, S.P=90

profit=SP-CP=90-80/1.2=108-80/1.2=28/1.2

Profit%=Profit/cp*100

=(28/1.2)/(80/1.2)*100=28/80*100=35%

ANS--D

Let AB= x and AC=y 

Tan30 = x/2a => 1/√3= x/2a => x=2a/√3

Tan60=y/a => √3 =y/a => y= a√3

AB/BC = x/(y-x)  => (2a/√3)/(√3a-2a/√3) = 2:1 

ANS--D

Ratio between sides = square root of ratio between areas

= 3:1

ANS--A

l^2 + b^ = d^2, l^2 + b^2 = 41 and lb = 20.

(l+b)^2 = l^2+ b^2 + 2lb = 41 + 2x20 = 81, l+b = 9, 

P = 2(l+b) = 2x9 =18

Here,

Distance=1.56 x10^16m  

Speed of light =3 x10^8m/s  

Time =?    

speed= dist/time      

time=dist/speed

       =1.56 x10^-16 /3 x10^8

       =0.52 x10*-24sec

f length =l, then time for 162+l metres  is 18 seconds and time for 120+l is 15 sec

Speed is the same. Hence 162+l/18 = 120+l/15

Cross multiply and simplify :   810+5l = 720+6l

l = 90 metres.

ANS--B

Let the speed be 2v and 3v and distance be 3d and 7d

Time for first car=3d/2v

Time for second car=7d/3v

Ratio of time taken=3d/2v:7d/3v=9:14

4% of CP = 1080-1026 = 54 so CP= 1350

SP for 4 % profit = 1350+ 54 = 1404

P is the mother-in-Law of Z

U sits immediate right of the Q.., from left to right: P,T,R,W,U,Q,V,S..., as Q is from India his rigth is left as per answer.

T,W,Q,S are from India, and remaining people from Japan.

Distance between T and Z =[(Distance between T and U) +(Distance between U and Z) ] units

                                           =(7.5+8)units=15.5 units

1 unit=2 km then 15.5 units=15.5*2=31 km

F is paternal grandmother to Z

Sonal is at  5th to the left of Atul who is at 12th. So the posititon of Sonal is 7th from left and 15th from right. So, there are 21 persons in the row.

ANS--B

The probability is 1/36.

Any number can appear on first dice in 6 ways.

The same has to appear on second and third which is possible in one way only.

So required probability = 1*1/6*1/6 = 1/36.

ANS--A

2j+b+s = 47 -(i)

j+3b+s= 54 - (ii)

3j+b+3s = 98 - (iii)

(i) + (ii)-(iii) gives

3b-s=3 - (iv).

2*(ii)-(i) gives

5b+s=61-(v)

adding (iv) and (v)

8b =64=> b=8.

Age of ben is 8 years

ANS--B

A completes 1/24 and B completes 1/30 part of work in a day.IF A & B alone have worked for 10 days then part of work done by them would be = 10*(1/24 + 1/30) = 10* (9/120) = 3/4 of the work. Now remaining work has been completed by C.

Hence out of total Rs. 2000/- C will get money according to the ratio 25% of Rs.2000 = Rs 500/-

C will get Rs. 500/-

ANS--D

Father in law. 

ANS--B

The wheel moves at 140 rev/min. In 1 hour it is 140 X 60 rev. 

In I rev the wheel will travel 2πr cm = 2X 22/7X 35 cm = 2X 22/7X 35 X10-5 kms

In 1 hour it will travel    140 X 60 X2X 22/7X 35 X10-5 = 18,480 X 10-3 Kms = 18.48 Kms

ANS--A

Let the price of 1gm rice be Re.1

Therefore, price of 1000gm i.e, 1kg is Rs.1000 (this is the C.P of the manufacturer)

S.P of the manufacturer= 114/100 of Rs.1000 =Rs.1140

Now for calculating the net profit %, it should be kept in mind that the manufacturer is giving 900gm and not 1000gm.

So, C.P of the manufacturer should be taken as Rs.900 ( price of 1gm is Re.1)

Net profit % = (1140-900)/900 * 100

                     = 26.67%

a) Speed of cycle/speed of scooter=15/30=1/2=1:2

b) 5m/10km=5/10000=1:2000

c)50paise/Rs5=50/500=1:10

ANS--D

Let the distance =x

Speed upstream=5-1=4km/h

speed downstream=5+1=6km/h

Time taken upstream=x/4

Time taken downstream=x/6

Total time taken=x/4+x/6

Now x/4+x/6=75/60

(6x+4x)/24=5/4

10x=5/4*24

10x=30

x=3 km

ANS--C

Since it is a Right angle triangle

Area of triangle=1/2*16*12=96 sq m

Area of plot grazed=x/360*22/7*7*7+y/360*22/7*7*7+z/360*22/7*7*7

22/7*7*7(x+y+z)/360

Since x+y+z=180

So Area=22/7*7*7*180/360=77 sq m

Ratio=77:96

Perimeter of the square = 44

Or, 4*a = 44

Therefore, a = 11 (where a is the lenth of a side of square)

Hence, area of square is a2 = 11*11 = 121 cm2

According to question,

Circumference of circle = 44

Or,                                  = 44

Therefore, r = 44*7/44 = 7 cm

Radius(r) of the circle is 7 cm

Thus, area of the circle =  = 22*7*7/7 = 154 cm2

Therefore the difference between the area of square and circle is 154 – 121 = 33 cm2

ANS--B

According to question, 

Principle(P)=500

Compound Amount(C.A)=583.20

Time(T)=2 years

Now, applying the formula of compound amound, we have,

      C.A = P(1+R/100)2

Or, 583.20 =500(1+R/100)2

Or, 583.20/500 = (1+R/100)2

Or, (1.1664)1/2 = 1+R/100

Or, 1.08 – 1 = R/100

Or, 0.08 = R/100

      Now, cross multiplying,

      R = 8     (where R = Rate)

     Now, from question,

     Rate has been increased by 2%

     Previous rate as we found(R) = 8%

     And , the given time(T) is 1 year

     Therefore, the new rate of interest (Rn) = 8% + 2% = 10%

     So new Compound Amount with new Rate,

     C.A = P(1+R/100)T

            = 500(1+10/100)1

            = 500(110/100)1

            = 500(1.1)1

            = 500*1.1

            = 550

Therefore,  the new Compound Amount after 1 year with 10% Rate will be Rs.550           

ANS--B

Let the sum borrowed by A be x and by B be y

Interest paid by A = IA= x*4*5/100 and Interest paid by B = IB= y*5*4/100

Since interest paid by A is double that of B's; IA=2(IB) =>​ x/5 = 2y/5 =>​ x = 2y

x + y =30,000 =>​ ​ 2y + y = 30,000  =>​ y = 10,000. Therefore, x = 20,000

Sum borrowed by A is Rs. 20,000

ANS--B

Let, both the trains start at t=0 from their respective stations.

Given that they proceed towards each other.

Let T1 be the train starting from station A and T2 be the train from station B.

T1 has speed 16 kmph, and T2 has speed 21kmph.

Let us suppose that they meet each other after ‘x’ hours, also suppose that T2 has travelled 60 km more than T1.

Now,

Distance travelled by T1 = 16*x

Distance travelled by T2 = 21*x

So, according to the given condition,

21x – 16x = 60

Or,          5x             = 60

Or,          x               = 60/5

Or,          x               = 12

Now,

Distance travelled by T1 = 16*12 =192

Distance travelled by T2 = 21*12 = 252

Therefore,

The distance between station A and station B = Distance travelled by T1 + Distance travelled by T2

                                                                                          = 192+252

                                                                                          = 444 Km

ANS--B

Due to stoppage the bus travels (45-36) = 9 kms less.

Time taken for traveling 9 kms = 9/45*60= 12 mins

ANS--B

Marked Price of the laptop(M.P) = $1050,  Discount(D) = 10%,  Profit(P) = 25%

Now, Price after 10% discount = 1050 – 10*1050/100 = 1050 – 105 = 945 (dollar), So selling price (S.P) = $945

After giving discount profit is 25%

Now, P = (S.P – C.P)*100/C.P                  (C.P = cost price)

Cross multiplying,   Or,          25 * C.P = 100*S.P –100* C.P       Or,          125*C.P = 100*945                                         Or,          C.P = 94500/125 = 756

Cost price of the laptop was $756

ANS--A

Let Reena's age be x years and her daughter's be y years.  x=8y

Eight years from now, (x+8)/(y+8)= 10/3 => 3x+24 =10y+80 => 24y+24= 10y+80  => y=4. Therefore, x = 8*4=32

Reena's present age is 32 years 

ANS--C (Water)

ANS--D (1)

ANS--B([email protected])

ANS--C

ANS--B

   1_2_3_4_5_6_7_8_9_10_11_12_13_14_15_16_17_18

Before, 5th fm right  →   B                                     P →  6th from left

After                                P                                     B

Bimal is fifth from left and Prakash 6th from right. When they interchange Bimal is thirteenth from left. i.e. 13th from right = 6th from left => there are 13+6-1 =18 positions. Hence after interchange, Prakash is 14th from the right. 

ANS--A

ANS--C

ANS--A

ANS--A

ANS--C

ANS--D

ANS--A

ANS--B

Total dist. = 4*2*(22/7)*35 m =880 m 

Time = 15 mins

Speed = (880/15)*60/1000 k/hr = 3.52 km/hr

ANS--B

Total dist. = 4*2*(22/7)*35 m =880 m . 

Time = 15 mins.

So, speed = (880/15)*60/1000 k/hr = 3.52 km/hr

ANS--B

ANS--B

ANS--E

ANS--B

ANS--A

ANS--B

ANS--D

ANS--A

ANS--A

ANS--A

ANS--A

ANS--C

ANS--A

ANS--B

ANS--C

ANS--C

ANS--A

ANS--E

ANS--E

ANS--A

ANS--D

ANS--C

ANS--D

ANS--C

ANS--C

ANS--A

ANS--A

ANS--B

ANS--D

ANS--B

ANS--A

ANS--D

ANS--D

LET B=X

A=2/9X

C=3/4*2/9=X/6

A+B+C=2/9X+X+X/6=625

X=450

A'share=2/9*450=100

B'sahre=x=450

C'share=450/6=75

ANS--B

100*90/100*80/100=72%

Hence discount=100-72=28%

ANS--A

SI for 1 yr = Rs. 420 .  => P = Rs. 8400  .  => CI for 6 months = Rs 210  .

Thus, P for next 6 months = Rs. (8400 + 210) = Rs. 8610 .

Now, CI for next 6 months = Rs. 215.25 .

Hence, total CI for 1 year = Rs. (210 + 215.25) = Rs. 425.25 

ANS--D

(25249.54 - 23820.32)/23820.32*100 = 6

ANS--C

45/100×66+28/100×450 =297+126 =423 

ANS--C

(45000*9%) + (45000+9%)*9%

ANS--D

There are 4 significant digits

ANS--A

First applying, v = u + a.t,

Here v=0, u=16 and t=4 sec

we get a = -4 m/s^2

Again applying v^2 = u^2 - 2.a.s , we get, s = 32 m 

ANS--D

Information given is not adequate to provide answer so

ANS--D

Let N1, N2, N3 be numbers which when divided by a divisor D, give quotients Q1, Q2, Q3 and remainders R1, R2, R3… respectively.

Now, Remainder of  (N1X N2XN3)/D => Remainder of (R1XR2XR3)/D.

Therefore, Remainder of  (14X 21X1423X1425)/12 => (2X9X7X5)/12 => (18 X35)/12 => (6X9)/12 => 6

ANS--A

ANS--B

ANS--B

n(T)=35

n(C)=25

n(T and C)=35+25-50=10

n(T) only=35-10=25

ANS--C

2^(73) – 2^(72) – 2^(71)

= 2^(71) x {2^(2) – 2^(1) – 1}

=2^(71) x (4 – 2 – 1)

=2^(71) x (1)

=2^(71)

ANS--C

• The total number of pens are 18.
• There is only one colour that is neither green nor blue and it is yellow.
• The number of yellow pens is 7.
• The number of ways 1 pen can be picked from all the 18 pens is 18C1. So that is the denominator of the probability.
• The number of ways 1 pen can be picked from 7 yellow pens is 7C1 so that's the numerator of the probability.
• i) Therefore the the required probability is 7C1/18C1 = 7/18
• There are total (7yellow + 5green)=12 pens that are not blue.
• ii) Therefore the required probability is 12C1 / 18C1  = 12/18 = 2/3 

[ - ]

1) Total no of cards= 52 No of Aces = 4 P(ace) = 4/52 = 1/13. 2) P(not red) = 1 - P (red) Total no of cards =52 No of red cards = 26 P(not red)= 1 - (26/52) = 1 - (1/2) = 1/2

Please mention the time taken to return to original position. It is missing from the question.

ANS--B

Total value of share = 200 x 15 = Rs 3000

Total dividend received = Rs 300

So rate of dividend = (300/3000) x 100 = 10%

ANS--C

Each shares (of Rs 20/-) gives 8% dividend = 8/100*20 = Rs 8/5

The man can earn Rs  8/5 from 1 shares

؞ He can earn Rs 1,000 from 1/8/5*1000 = 5000/8 = 625 shares.

Now each Rs. 20 shares is selling at Rs 40. ؞ To buy 635 shares he need to invest Rs 40 * 625 = Rs 25,000

ANS--C

By using distance formula, Distance between two possible vertices = √{(x2-x1)^2 + (y2-y1)^2} = √{(14-8)^2 + (3-3)^2} = √ 6^2 = 6 cm

Let, the reqd. height be h m.

Here, h/2 = tan 60 = root3 . 

Hence, h = 2*root3 m = 2*1.732 m = 3.464 m

Let the height of the tower be x m and the separation distance between the cliff and the ower be y m

So tan 45 = 1 = 80/y. This gives y = 80 m

Also tan 30 = 0.577 = (80-x)/y = (80-x)/80 . This gives x = 33.84 metre

Shopkeeper buys at 20% discount or at Rs 4000. ST paid by him = 5% on Rs 4000 =Rs 200

Selling price Rs 5000. Sales tax paid by customer = 7% pn 5000 = Rs 350

So Value Added  Tax = 350 -200 = Rs 150

Area of triangle COA = 60 = (1/2) x 5 x AC. This gives AC = 24 cm

By proportionality, BD/AC = OD/OC

So BD = (OD/OC) x AC = (8/5) x 24 This gives BD =  38.4 cm

Area of triangle BOD = (1/2) x 8 x 38.4 = 153.6 sqare cm

ANS--B

Volume of the original hollow cylinder = 22/7 x 5 x (10 x 10 - 8 x 8)= 22/7 x 5 x 36 cubic cm.....(1)

Volume of one cone = 1/3 x 22/7 x 3 x 3 x 5 = 22/7 x 5 x 3 cubic cm.......(2)

Dividing (1) by (2) we get the number of cones = 36/3 = 12 

ANS--D

Let the number of children be x

So [(6800/x) - (6800(/(x +9)] = 153

This gives x(x+9) = 400 

Or soving the quadratic equatuib x = 16

ANS--B

 The compound interest in the first two successive years is Rs. 2000 and Rs. 2500 respectively, So, extra Rs. 500 interest is paid on the interest on first year, i.e., on Rs. 2000 for 1 year, therefore, rate of interest = 500 * 100 / (2000* 1) = 25%

ANS--B

Simple interest for 2 years will be 40000 x 2 x 15/100 = 12000

Amount due after two years at 15 % per annum = Rs 52000

Compund interest for the first six months for six-monthly compounding for the third year = 52000 x 0.05 = 2600

Amount after 6 months - 52000 + 2600 = 54600

Interest for the second six months = 54600 x 0.05 = 2730

So amount due after third year = 54600 + 2730 = 57330 

Radius of circle ‘p’ = 15

So Radius of circle Q = 15+ 7 = 22

Similarly Radius of circle R = 15 + 7 + 7 = 29

In Same way find areas of all circles individually as

Area of circle P = 22/7*15*15 = 706.5

Similarly area of circle Q And R are resp. 1519 and 2640.74 (with help of 3.14*r*r) 

So area of ring Q = Area of circle Q — area of circle P = 1519– 706.5 = 812.5 cm

Similarly area lf ring R = area of circle R —area of circle Q = 2640.74–1519=1121.74 centimetre

ANS--C

Let d be the distance between the building and the tower.

We get tan 60 = 30/d = 1.732. This gives d = 30/1.732 = 1.73

Let H be the height of the tower above the height of the building

Now we also get tan 45 = 1 = H/17.32 = 1 This gives the dimension H = 17.32 m

So total height of tower = 30 + 17.32 = 47.32 metre

ANS--C

To find th eheight of building, consider one triangle with 30 and other with 60 degree with top of building.

As per one triangle with 30 point, tan 30 = h/(50+y)

As per second triangle with 60 degrees , tan 60 = h/y

now substituting value of y and then solving above equation , the h comes out to be 43.3 metre

ANS--B

Market price = 2500

10% discount on the market price = 2500 * 10/100  = 250

Reduced market Price = 2500 - 250 = 2250

5% of Cash discount = 2250 * 5/100 = 112.50

New price for payment through cash = 2250 - 112.50 = 2137.50

10% Sales Tax on the price for cash payment = 2137.50 *10/100 = 213.75

FINAL amount to be paid by customer by CASH = 2137.50 + 213.75 = Rs. 2351.25 

ANS--D

ANS--B

Taking the respective distances x and y,

Here, cot 30 = x/25 and cot 45 = y/25 

Or, x = 25*root3 and y = 25

Hence, total width of the river = x + y = 25(root3 + 1) m = 25*2.732 m = 68.3 metre

ANS--B

an theta = 450/778.5 = 0.578 = 1/root3 (approx)

Hence, theta = 30 degree 

ANS--C

By simple division we get  Price in Rs per oz for W = 0.035, for X it is 0.03. for Y it is 0.28 and for Z it is 0.03 respectively.

So the least expensive brand is Y

ANS--B

N is the LCM of first 100 natural numbers.

Now 101 is a prime number, so 101 and N do not have any common factors. Hence their LCM = 101XN

102= 2X51, since 2<100 and 51<100, 102 divides N

103 is a prime number, no common factors between 103 and N

104= 52X2; since 2<100 and 52<100, 105 divides N 

105 = 21X5; since 5<100 and 21<100, 105 divides N 

Hence, LCM of first 105 natural number is NX101X103

ANS--E

Let the desired weight of the alloy containing 40 % gold be W

So    (0.40 W + 0.20 x 10) / (W + 100 ) = 0.32 

This gives W = 150 gms 

ANS--B

numbers 4569   4571  4576 4586  4603  4629  4666        

Difference        2         5      10       17       26      37

Difference             3          5          7        9        11

ANS--E

ANS--B

North is vertically upwards 10 KM, Coming vertically downwards 6 KM . So 4 KM to North from starting point

Then East is to the right 3 KM

Therefore he is SQRT(3^2 + 4^2)  or 5 KM 

ANS--A

           A ______30___________C____10_____B

Distance AC=30 Kms, AB=40 kms  Let speed of boat P be 'x' speed of boat Q be 'y' and speed of stream be 'z'

Relative speed of P upstream = x-y and speed of Q upstream z-y. Relative speed of P downstream = x+y and speed of Q upstream z+y.  Now P travels from A to B = 40 kms upstream in 10 hours and B to A downstream in 5 hours.

time =distance/speed Therfore, x-y= 40/10 =4..(i)  and x+y = 40/5=8 ...(ii) solving (i) and (ii) we get x= 6 and y=2

Speed of boat P = 6 kms/hr  and speed of stream = 2 kms/hr and Speed of boat Q = z-2= 30/5=6 => z= 8 kms/hr 

Now, P starts from A and Q starts from C  and Pand Q will meet at D after t hours.

t = 30/(Speed of P upstream+speed of Q downstream)  = 30/(4+10) = 15/7

In 15/7 hours P will travel 15/7*4= 60/7 kms upstream from A., = 40-60/7 =220/7 kms from B.

Therefore P and Q will meet at 220/7=  31.4 kms form B

ANS--D

CI for 1st yr = SI for 1 yr = Rs 1300/2 = Rs. 650 .

CI for 2nd yr only = Rs. [650(1 + 8/100)^1] = Rs. 702 .

Hence, total CI for 2 yrs = Rs. (650 + 702) = Rs. 1352

ANS--C

Let, SI for 1yr be Rs. x .

ATP, x(1 + 10/100)^1 = 420 - x  [i.e. the CI for 2nd yr only]

By solving, x = 200

Hence, reqd. SI = Rs. 400*3/2 = Rs. 600

ANS--C

Let the time taken by the nth tap to fill up the tank be tn. The efficiency of the nth tap is equal to the sum of the efficiencies of all previous taps. i.e. the time taken by the 3rd tap= sum of the time taken by 1st and 2nd tap.

t3= t2+t1.

t4= t3+t2+t1= (t2+t1) + (t2+t1) = 2(t2+t1)

t5= t4+t3+t2+t1= 2(t2+t1)+(t2+t1) + (t2+t1) = 4(t2+t1)= 2^2(t2+t1).

؞tn= 2^(n-3)(t2+t1). the 8th tap takes 48 hours to fill the tank. ؞ t8= 2^(8-3)(t2+t1) ⇒ 2^5(t2+t1)= 48 

⇒ 2^5(t2+t1) = 48

When tap number 1to9 are working together to fill the tank it will be equal to the time taken by 10th tap

2^(10-3)(t2+t1) = 2^7(t2+t1) = 2^2*t^5(t2+t1) = 4*48 = 192 

ANS--C

Let, no. of original employees be x and time taken by them be t . Later, y people join them.

ATP, 1st case: (x + 20).2/3*t = x.t  => Solving, x = 20

2nd case: (y + 20).1/3*t = 20.t  => Solving, y = 40 

Hence, reqd no. of employees is 40

ANS--C

Speed of A = 36 km/hr. Therefore, Distance traveled by A in 2 mins = 36/60*2= 12/10kms = 1200 meters

Speed of B = 15 m/s. Therefore, Distance traveled by B in 2 mins = 15*120= 1800 meters

Distanc ebetween them after 2 mins = 1800-1200 = 600 meters

ANS--A

If average age is increased by 2 years, total age is increased by 16 years

Total age of the two women = (35+45)+16 = 96 years Average of the two women is 96/2 = 48 years

ANS--E

In the initial solution let the volume of A be 3x and that of B be 2x. When 10 litres of solution is taken out, 6 liters of A and 4 liters of B is taken out. Then 10 liters of B is added.

Therefore, (3x-6)/(2x-4+10)=1/2 => 3x-6 = x + 3 => 2x=9 => x= 9/2 => 5x = 45/2 = 22.5

Volume of initial solution = 22.5 liters

ANS--C

William share: John's share = 7:9 Total amount deposited $ 72,000. William's share = $7/16*72,000= $31,500

John's share = $40,500.  William withdrew $ 7,500 and John withdrew $ x.

(31,500-7,500)/(40,500- x) = 4/5 => 24,000/(40,500-x) = 4/5 =>  x  = 40,500-30,000 = $10,500

ANS--C

40/100×620+35/100*4500+X=324×324/18 4×62+35×45+X=18×324 248+1575+X=5832 X=5832-1823=4009 

ANS--A

ANS--A

ANS--B

ANS--C

Depositing Rs 4000 in a recurring deposit each month for two years, which is equal to 24 months. So, n=24, PMT = 4000 and FV = 106000.

Now for Annuity FV = C*/i * ((1+i)^n - 1) here i is monthly interest rate, Calculating i=0.0085 or 0.85% per month, annualizing this returns = 10% pa

ANS--C

Maturity value of a recurring deposit is given by  M =  

R [ (1+i)n – 1] ——————– 1- (1+i) -1/3       Where, M = Maturity value R = Monthly Installment r = Rate of Interest (i) / 400 (Annual interest rate converted into quarterly rate) n = Number of Quarters (Assuming the compounding is done quarterly)                               

By putting the values given we get M= Rs21,244.00

Triangles COA and DOB are similar by A-A axiom [since, < COA = vert. opp. <BOD, < ACO = <BDO = 90]

Since, ratio of similar sides OC and OD is 5:8, the ratio of areas of triangle DOB: triangle COA = 64:25.

Hence, area of triangle DOB = (64/25)*60 = 153.6 square cm

ANS--C

et the dist. betn. the feet of the tower and the building be d m.

ATP, 30/d = tan 60 = root3  .  => d = 30/(root3) m = 10*root3 m = 17.32 m .

Since, angle of elevation is 45, the part of height of the tower excess to the height of the building = d = 17.32 m.

Hence, the total height of the tower = (30 + 17.32) m = 47.32 metre

Let d be the distance of the building from the foot of the hill

We can see from the sketch tan 30 = 50/d = 0.5773

That gives d = 86.61 m

Let H be the height of the hill

Again from the sketch tan 60 = H/86.61 = 1.7320

That determines the height of the hill = 1.732 x 86.61 = 150 metre

Area of the vertical walls = 2 x 40 x 7 = 560 square metre

Area of the dome = 22/7 x 5 x 40 = 628 .5 square metre

Cost of plastering = 5 x (560+628.5) = Rs 5942.50

ANS--A

A = P(1+r/n)^nt   where A=amount, P= principal, r= rate of interest in decimal, n= number of times interest compounded per year, t= time in years

A = 36000(1+0.16/2)^2*3/2 = 36000 (1.08)^3 = 36000*1.25972 = 45,349.63

Therefore, interest = Rs. A - P = 45,349.63 - 36000= Rs. 9,349.63

Area of the shaded region = [(pi/2)*20^2 + 2*(pi/2)*10^2] cm^2 = 300*pi cm^2  . Ans.

Perimeter of the shaded region = [4*pi*10 + pi*20 + 20] cm = [60pi + 20] centi metre

ANS--C

Joining the two centers divide the cords into two equal parts, with 12/2 = 6.5cm each.

Applying pythagorus theorem in each circle and triangle sqrt(10^2  - 6^2) +  sqrt(15^2  -  6^2)

After Calculating these all You will get  21.75 cm

ANS--A

We use the formula s = ut -1/2 a t^2

So s = 15 x 60 - 0.5 x 0.3 x 60 x 60 = 360 m

So distance from signal would be 400 - 360 =40 metre

• In x litres of 80% boric acid,
• water content is x/5 and boric is 4x/5.
• by adding 20 litres of water ratio is (4x/5)/{20 +(x/5)} = 60/40 .
• 2 X 4x/5 = 60+3x/5 or, (8x/5 ) - (3x/5) = 60 so x= 60 lit

The difference between the third generation computers and the fourth generation computers are as follows:

Third Generation

• They used integrated circuit (I.C) and large scale integration (LSI).
• They were smaller in size and called as mini computer.
• The operating speed was measured in nano seconds.
• Main memory was increased in the form of PROM and DRAM.
• Magnetic disks were used for secondary storage.
• They could understand large number of high level languages.

Fourth Generation:

• They used very large scale integration (VLSI) and microprocessor.
• They were small in size and called as micro computer.
• They could perform millions of calculations per second.
• Main memory was increased in the form of EPROM and SRAM.
• They have a large memory and high functional speed.
• Multiprocessing and multiprogramming OS (operating system) are used

You can just do copy then paste by the office ClipBoard

#Name error occurs when excel didn't recognise a text in Formula

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Total thickness of books = 5x20= 100mm ​Total thickness of papers = 5x0.016 = 0.08 ​Thickness of stack = 100+0.08 = 100.08 mm Thus 10.008 centimetre

from above we get m+3 = 5

therfore m=2

Distance covered in 25 min= 14 km Distance covered in 1 min= 14/25 km Distance covered in 60 min or 1 hr= 0.56×60 km Distance covered in 5 hr. = 33.6×5 km = 168 km

i shadow = 3.2/5.6 x 10.5 = 6 metre

ii length = 5.6/3.2 x 5 = ​8.75 m = 8m 75 centimetre

​

18 km = 1 cm

72km = 72/18 = 4 cm

Given that:  2kg sugar = 9*10^6 crystals

By Unitary Method 1 kg will have = 9*10^6 /2 = 45*10^5 crystals

1.)         5 kg sugar = 5*45*10^5 = 225*10^5 crystals

2.)         1.2 kg sugar = 1.2* 45*10^5 =  54*10^5 crystals

Let x be the length of model Then, 9/x=12/28 9×28/12= x x=21 centimetre

840 bottles in 6 hr. Then In 1 hr. 840/6=140 bottles then In 5hr. 140*5=700 bottles

Actual length of the bacteria is 5/50,000= 0.0001 cm (ii) Length after 20,000 enlargement 0.0001×2= 2 cm

11 as the number ending in 1 must have a digit having 1 in ones place

The volume of each cuboid = 5 x 5 x 2 (lbh) = 50 cu cm

​volume of cube = l x l x l = 10 x 10 x 10 = 1000 cu cm

no. Of cubes required = 100/50 = 20​

Value after one year =42000 - (8/100)×42000 =42000-3360 =38640 

In 2001 54000*(100/105)*(100/105)=48980 In 2005 54000*(105/100)*(105/100)=59535

For 2 year

A=P (1+r/100)^n

8000* 105/100  * 105/100

=8820

For 3 yr

A=P (1+r/100)^n

8000* 105/100  * 105/100 * 105/100 

=9261

interest paid during 3rd year=

amount for 3yr - amount for 2 yr

9261-8820

=441

C.P of VCR = C.P of TV = 8,000.  4% loss on VCR. S.P. of VCR =96% of Rs. 8,000 = Rs. 7,680.

8% gain on TV. S.P. of TV =108% of Rs. 8,000 =  Rs. 8640.

Total S.P. = 16,320 Total C.P. 8,000 + 8,000 = 16,000

Profit/Loss % = (S.P. - C.P.)/C.P/ X 100 = 2%

cost = 15500, repair cost = 450

total cost after repairing = 15950

profit on cost price= 15%

selling price = (15950x15/100)+15950 = 18342.50

Selling price of 80 items =2400+(16/100)×2400 =2400+384=2784Rs hence selling price of one item =2784/80=Rs 34.80

percentage of 169 out of 845 is 169/845 X 100=20%

% decrease on monday = (100-20)%=80%

100-(60+30) =100-90= 10℅ like other games,no of people like cricket = (60/100)×5000000=3000000(thirty lakh), no of people like football = (30/100)×5000000=1500000(fifteen lakh), no of people like other game =(10/100)×5000000=500000(fife lakh

Let they played Xmatch , (40/100 )×X=10 or X =(10×100)/40 = 25

answer is 72% of 25 = 18 are good in mathematics Therefore 25-18 = 7 are not good in mathematics

a) 15:30 = 1:2

(b) 1 km = 1000m. So 10 km = 10000 m. Ratio is 5:10000 = 1:2000

(c) 1 Re = 100 paise. So 5 Rs = 500 paise. Ratio is 50 : 500 = 1:100

20+30=50

The sequence is the squqre of odd numbers. Hence the answer is 11^2=121

A compiler is computer software that transforms computer code written in one programming language into another programming language.

Here A/B=3/7 and B/C =2/5

So A:B:C=6:14:35

No. of A officers=6/55*1375=150

ANS--A (Resolution)

Sum of ages of 25 students=25*12=300

Sum of ages of 25 students+ one teacher=26*13=338

Age of teacher=338-300=38  

Let CP=100, Then Marked Price=140

Disount=10%, Hence SP=140-14=126

So profit=126-26=26

When profit=26 then CP=100

When profit=325 the CP =325*100/26=Rs1250

Actual CP=RS1250

The given series is an AP

Here First term a=-15, Common diff d=9

Sum of 30 terms=30/2(2*(-15)+29*9)

            =15(-30+261)

           =15*231

           =3465

Let the age of son=x

Then age of father=4x

After 10 years

Son's age=x+10

Fathers age=4x+10

so 2(x+10)=4x+10

Solving we get x=5 years

Son's age=5 years

Focal length=10mm=10/1000 m=0.01m

Power =1/f =1/0.01=100D

Each medium has different refractive index. when light travels from one medium to another it bends due to difference in refractive index.

The amount of energy required to move an electron from valance band to conduction band is called Forbidden enegy gap.

The amount of enegy required to move an electron from valance band to conduction band is called forbidden energy gap

The three quantities with no dimension are

1.solid angle

2.coefficient of friction

3.refractive index

On analysis we found that next term can be found by multiplying the previous number by 2 and adding 11 to it

19=4*2+11

49=19*2+11

Next term=49*2+11=109 (Answer)

Area of parallelogram=Base*Altitude

Area=4.2*5 Cm^2=21 cm^2

ANS--B

X^2+1=0

If we square any number then it will be a positive number and when 1 is added to it the result cannot be zero

ANS--C

ANS--B

ANS--A

ANS--D

ANS--C(300)

ANS--D

Since intelligent student cannot be defined

ANS--B

n(A)=3

n(B)=3

n(A*B)=n(A)*n(B)=3*3=9

Let CP=100

Then Marked price=100*140/100=140

Selling price=140*80/100=112

Profit=112-100=12

Profit %=12/100*100=12%

Ratio of investments=12000:15000:18000   i.e 4:5:6

Share of profits will be in the ratio of investments

Ratio of profits=4:5:6

A' share =4/(4+5+6)*Total investment=3000

Total investment=3000*15/4= Rs11250

Total cost of AC=12500+800+300=13600

Selling price=13600*115/100=15640

ANS--c   2^n

Let Principal=P

Then Time=10 years

SImple Interest=2P-P=P

SI=P*R*T/100

P=P*R*10/100

R=10%

Total travel time=1hr 30 min=90 min

Time when it stopped=12*2=24 min

Total time when bus was moving=90-24=66 min

distance covered =40/60*66=44 Km

ANS

41472/8=5184

51884/9=576

576/8=72

72/9=8

8/8=1

The next term is 1

The portion of field reaped in 10 days by 20 workers=1/3

No. of workers after 10 days=20+20=40

20 workers can reap the remaining fiels in 20 days

40 workers will reap in 10 days

Total no of days=10+10=20 days

ANS--62 is wrong

ANS--207

963-927=36

927-855=72=36*2

855-747=108=36*3

747-603=144=36*4

603-423=180=36*5

427-36*6=207(ANS)

ANS--63

The sequence is of the form 2^n-1

2^6-1=63

ANS--24

The series is in form n^2*6

2^2*6=24

The terms are 15pq+15+9q+25p

=(15pq+9q)+(25p+15)

=3q(5p+3)+5(5p+3)

=(3q+5)(5p+3)  ANS

Perimeter=2*22/7*4.9/2=15.4cm

Total lenth used=25*15.4=385 cm

Principal=10000

Interest for 1st year=10000*0.04=400

Principal for second year=10000+400=10400

Interest for second year=10400*0.05=520

Principal for 3rd year=10400+520=10920

Interest for 3rd year=10920*0.06=655.20

Total interest for all 3 years=655.20+520+400=1575.20.

Let P be the principal

When it doubles in 5 years then SI=P

SI=P*R*T/100

P=P*R*5/100

R=20%

When it becomes 7 times the SI=6P

6P=P*20*T/100

T=30 years

Let Cp of 100Litres=Rs100

Then SP of 100 Litres=110

When 20 litres of water is added then amount of milk becomes 120 Litres

SP of 120 litre=120*1.1=132

Increase in profit=132-110=22%

Let the total distance=2X

Then half distance=X

X/21+X/24=10

8X+7X/168=10

15X/168=10

X=168*10/15

X=112

Total distance=2X=112*2=224 KM

Let speed of boat in still water=X

and speed of stream=Y

Upstream speed of boat=X-Y

Downstream speed of boat=X+Y

(X+Y)*2=4(X-Y)

X+Y=2(X-Y)

3Y=X

Putting X=3Y

(X+Y)*2=16

(3Y+Y)*2=16

8Y=16

Y=16/8=2

X=3Y=3*2=6KM/hr

Speed of boat in still water=6KM/hr

Time=2 yr 73 days=2.2 yr

Amount=30000(1+0.1)^2.2=36998.58

Interest=36998.58-30000=6998.58

Let P be the principal and R be the rate of interest

P(1+R/100)^3=8000

Squaring we get

P^2(1+R/100)^6=8000*8000

P(1+R/100)^6=8000*8000/P---(1)

Now P(1+R/100)^6=12167---(2)

Putting the value from (1) into (2)

8000*8000/P=12167

P=8000*8000/12167=5260

P=2400

Let R be rate of interest and T time period

SI=PRT/100

864=2400*T*T/100

T^2=864/24=36

T=6 years

Let he buys 3 litres of milk.

CP of 3 litres=6*3=18

The he adds 1/3 water so total milk becomes 3+1=4 litres

SP of 4 litres=4*7.2=28.8

Gain%=(28.8-18)/18*100

=10.8/18*100=60%

Let cost of 1cm=Rs1

Then CP=Rs33

SP=RS36

Profit %=3/33*100=100/11=9.09%

Ratio of capitals=25000:40000=5:8

Profit ratio=5:8

SP of 36 oranges=Rs1

CP of 36 oranges=36*100/99=1.01

For 8% gain

SP=1.01*108/100=1.09

Let cost of one Pen=X and cost of 1 pencil=Y

So 2X+5Y=240--(1)

4X+2Y=336--(2)

Dividing (2) by 2

2X+Y=118---(3)

Subtracting (3) from (1)

4Y=122

Y=122/4=30.5

Putting Y=30.5 in (3) and solving we get X=43.75

Cost of 1pen=Rs43.75

Cost of 1 pencil=Rs30.5

Let the numbers be 7x and 11x

Product of numbers=HCF*LCM

7x*11x=13*1001

x^2=13*1001/77

x^2=16

x=13

Larger number=11*13=143

LCM of 12,25,45,60=900

The smallest 5 digit number is 10000

So 10000/900=11.11

So smallest number=900*12=10800

The product of two prime numbers is always equal to its LCM

Hence LCM=493

Let X be 4th proportional

So 100*X=50*40

X=2000/100=20

Total amount Julie has=Rs250

Amount spent on sweater=Rs75

amount left=15

Amount spent on jacket=250-(75+15) =250-90=Rs160

ANS--B--Lactobacillus

CNG stands for Compressed Natural Gas

ANS--A-- Resolution

ANS--C--Superscript

The two main products of photosynthesis are Oxygen and Glucose

Polar star is the brightest star located in north and it remains stationary with respect to earth

The utensils are made of metals and hence it is coated balck to absorb maximum heat.

Metals have free electrons which makes it good conductor of heat and electricity

ANS--D--Zinc

Sound waves move in a medium by producing vibration.

Electrical power P=V*I

Switch is an electrical component of circuit which makes or breaks a circuit

We should wear light colured clothes in summer because light coloured clothes reflects most of light and we feel less heat.

The motion of an object which repeats itself in fixed interval of time is called periodic motion. The motion of a pendulum in clock and movement of earth around sun and on its own axis are examples of periodic motion.

The thickness of five rupee coin is 1.9 mm

The resistor opposes the flow of current, Hence it reduces the current if resistance is increased.

I=V/R

Current is inversely proportonal to resistance.

An inductor oppese sudden change of current in a circuit. It stores magnetic energy

capacitors are used to store electric energy.

Plasma is the fourth state of matter.

Relative density has no unit because it is the ratio of same units which gets cancelled.

Power=Work/Time

Trivial, Unimportant

First you should have good knowledge of grammar and literature. Thoroughly study the chapter that you are going to teach. Make some interesting stories in your teaching class. Create some humour in your teaching.

Function of mumcipal corporation are as follows.

1) providing water and electricity to residents

2)Providing roads and sewage facility

3) Registration of birth, death marriages etc

4)Providing education for kids

First war of indian independence

Raja ram mohan roy was the founder of bramha sabha and he played an important role in abolition of sati pratha.

Doctrine of lapse was a rule started by lord dalhousie according to which if a king died without a natural heir then the british will take over the control of that state.

Bronze is an alloy of copper , zinc and tin

Arya samaj was founded in 1875 by dayanand saraswati. He was a sanskrit scholar.The DAV group of schools are run by this samaj.

Bell metal is an alloy of copper and tin

SAARC was formed for collaboration among south asian countries.To grow trade and tourism among these countries.

Let the distance be X

Time taken uphill =X/10

Time taken downhill=X/36

Total time taken=X/10+X/36=23X/180

Average Speed=2X/(23X/180)=15.65KM/HR

n(A*B)=n(A)*n(B)=3*3=9

ANS--B

Womesh chandra bonerjee

The vice president of India

The king Ashoka

British naval force

Vijayalaya

Lord Mountbatten

ANS--A (0.1)

F=1/Power=1/0.5=2metre

Sun is the main source of solar energy

modified drum technology

Balochistan

gujrat

1907 at Jamshedpur

Wheat is Rabi crop

Hirakunad dam

Bajra

Limestone, clay,shale etc

kerala

karnataka

odisha

Assam

Mangrove

Bokaro and durgapur

puttalam

Up, punjab, Haryana, MP

Solar, Wind energy, Tidal energy, Geothermal

F=1/power=1/0.2=5m

Design of ship, submarine, boats

water and carbon dioxide

sabaceous gland

Hepatic artery

solid waste, liquid waste, organic waste