**Question 1. The Average Score Of A Cricketer For 13 Matches Is 42 Runs. If His Average Score For The First 5 Matches Is 54, Then What Was His Average Score (in Runs) For Last 8 Matches?****Answer :**Total Score = Average * Number of matches

Total score of 13 matches = 13 × 42 = 546

Total score of first 5 matches = 5 × 54 = 270

Therefore, total score of last 8 matches = 546 – 270 = 276

Average = 276/8 = 34.5

**Question 2. Simple Interest On A Sum Of Money For 4 Years At 7 P.c.p.a Is Rs. 3584/-. What Would Be The Compound Interest (compounded Annually) On The Same Amount Of Money For 2 Years At 4p.c.p.a?****Answer :**Let P be the sum

Therefore,

3584 = (P * 7 * 4) / 100 => P = 12800/-

CI = 12800(1 + (4/100))2 – 12800

=> 13844.48 – 12800 = 1044.48/-

**Question 3. 6*136/8+132 / 628/16-26.25=?****Answer :**6*136/8+132 / 628/16-26.25

⇒ 234/13=18

**Question 4. (1331)1/3/ 275 *52 = ?****Answer :**(1331)1/3/275 *(5)2

= (113)1/3 / 11*25 *(25)

=1

**Question 5. (43)2 + 841 = (?)2 + 1465?****Answer :**Using BODMAS

2 = (43)2 + 841 – 1465

= 1849 + 841 – 1465 = 1225

Therefore, = √1225 = 35

**Question 6. (216)4 ÷ (36)4 × (6)5 = (6) ?****Answer :**(6) = (216)4 ÷ (36)4 × (6)5 =

2164/364 * 65 =

612 / 68 * 65=

69 Therefore, = 9

**Question 7. (1097.63 + 2197.36 – 2607.24) ÷ 3.5 = ?****Answer :**(1097.63 + 2197.36 – 2607.24) ÷ 3.5 = 687.75 ÷ 3.5

= 196.5

**Question 8. A Certain Number Of Capsules Were Purchased For Rs. 216/-. 15 More Capsules Could Have Been Purchased In The Same Amount If Each Capsule Was Cheaper By Rs. 10/-. What Was The Number Of Capsules Purchased?****Answer :**Let x be the cost of each capsule and y be the no of capsules

Therefore, x = 216/y and x – 10 = 216/(y+15)

Putting the value of x from eq. 1 into eq. 2, we get

216/y – 216/(y+15) = 10

y2 +15y – 324 = 0

Solving for y, we get y = 12

**Question 9. What Will Come In Place Of Question Mark (?)in The Given Question?****Answer :**Here the difference between each term is in the form of 13+1, 33+1, 53+1, 73+1 and so on

4 6 34 160 504 1234

+2 +28 +126 +344 +730

13 + 1 33 + 1 53 + 1 73 + 1 93 + 1

So = 160

**Question 10. A And B Are Two Numbers. 6 Times Square Of B Is 540 More Than The Square Of A. If The Respective Ratio Between A And B Is 3:2. What Is The Value Of B?****Answer :**We are given that

6B2 = A2 + 540 and A : B = 3 : 2

Put the value of A from 2nd equation into the first equation, we get

6B2 = (1.5B)2 + 540

→ (6 – 2.25)B2 = 540 → 3.75B2 = 540 → B2 = 144 → B = 12

**Question 11. Pipe A Can Fill A Tank In 20 Minutes And Pipe B In 30 Minutes Respectively. Pipe C Can Empty The Same In 40 Minutes. If All The Three Pipes Are Opened Together, Find The Time Taken To Fill The Tank?****Answer :**Net part filled in 1 hour = ( 1/20 – 1/30 – 1/40) = 7 / 120

Therefore tank will be filled in = 120/7 = 17 1/7 minutes.

**Question 12. Cost Of 5 Mangoes + Cost Of 4 Oranges = Cost Of 7 Mangoes + Cost Of 1 Orange. Find The Ratio Of Cost Of Mango To Cost Of Orange ?****Answer :**⇒ 5 mangoes + 4 oranges = 7 mangoes + 1 orange

⇒4 orange – 1 orange = 7 mangoes – 5 mangoes

⇒ 3 orange = 2 mangoes

Hence Required Ratio is

⇒ Cost of mango: cost of orange = 3 : 2

**Question 13. There Are 6 Red Shoes & 4 Green Shoes. If Two Of Red Shoes Are Drawn Randomly What Is The Probability Of Getting Red Shoes?****Answer :**Total number of shoes=6 + 4=10;

Let S be the sample space.

Then, n(s) = Number of ways of drawing 2 shoes out of

10 =10C2 =(10 x 9)/(2 x 1) =45

Let E = Event of drawing 2 shoes which are red.

Therefore n(E) = Number of ways of drawing 2 red shoes out of

6 shoes.= 6C2 =(6 x 5)/(2 x 1) =15 P(E) = n(E)/n(S) = 1/3

**Question 14. To 15 Liters Of Water Containing 20% Alcohol, We Add 5 Liters Of Pure Water. What Is % Alcohol?****Answer :****Case 1:**Initial quantity of Water = 15 litres Quantity of Alcohol = 20% = (20/100) * 15 = 3 litres**Case 2:**Quantity of Water after adding 5 litres of water = 15 + 5 = 20 litres There of the % of alcohol in water = (3 /20) * 100 = 15%

**Question 15. A Certain Number Of Two Digits Is Three Times The Sum Of Its Digits. If 45 Be Added To It, The Digits Are Reversed. The Number Is?****Answer :**Let x and y are the digits of the number with x in ten’s place and y in one’s place.

So, 10x+y is the value of the number.

Given 10x + y = 3 (x + y)

⇒7x – 2y = 0 …………….(i)

And also given that,

10x + y + 45 = 10y + x (∵ digits are reversed⇒ x in one’s and y in ten’s place).

⇒ 9x – 9y = -45

⇒ x – y = -5 ………………(ii)

Put value of y from eq. (i) to eq. (ii)

⇒ x – (7x/2) = -5

⇒ x = 2

And y = 7

So the number is 27

**Question 16. Water Flows Into A Tank 200 M × 150 M Through A Rectangular Pipe Of 1.5m × 1.25 M @ 20 Kmph . In What Time (in Minutes) Will The Water Rise By 2 Metres?****Answer :****We know that, formula:**Volume of any Cuboid = Length × Breadth × Height

**According to the given question:**⇒ Volume of the water required in the tank = (200 × 150 × 2) m3 = 60000 m3 . . .

∴ Length of water column flown in1 min =(20 × 1000)/60 m = 1000/3 m

∴ Volume flown per minute = 1.5 × 1.25 × (1000/3) m3 = 625 m3 .

∴ Required time = (60000/625)min = 96min.

Hence, the required answer is 96 minutes.

**Question 17. A Boat Travels 20 Km Upstream In 4 Hours And 18 Km Downstream In 6 Hours. Find The Speed Of The Boat In Still Water?****Answer :**Rate downstream= 20 /4 = 5 kmph

Rate upstream= 18 / 6 = 3 kmph

Speed in still water = 1/2 (5 + 3) = 4 kmph.

**Question 18. A Person Is 80 Years Old In 490 And Only 70 Years Old In 500 In Which Year Is He Born ?****Answer :**Since we are given that a person is elder in 490 than in 500.

Hence, he must have been born in BC.

Hence, we can say that he must have been born in BC 490+80

= BC 570

**Question 19. Grass In Lawn Grows Equally Thick And In A Uniform Rate. It Takes 40 Days For 40 Cows And 60 Days For 30 Cows To Eat The Whole Of The Grass. How Many Cows Are Needed To Eat The Grass In 96 Days?****Answer :**Let quantity of grass initially = g

Let r be the rate at which grass grow in 1 day,

Let c be the quantity of grass a cow eat in 1 day

we can deduce from ‘. It takes 40 days for 40 cows and 60 days for 30 cows to eat the whole of the grass’ that

⇒ g + 40r = 40 × 40c = 1600c———–[1]

and

⇒ g + 60r = 60 × 30c = 1800c

⇒ g = 1800c – 60r—————-2

Putting this value of g in eqn [1]

we have

⇒ 1800c – 60r + 40r = 1600c

⇒ 200c = 20r

⇒ C = 0.1 r

⇒ r = 10c

Let m be the number of days required by 20 cows to eat the entire of the field = m

Then

We have the eqn as

⇒ g + 96r = 96nc

⇒ 96nc = 1800c – 60r + 96r (as we have g=1800c-60r from eqn 2)

⇒ 96nc = 1800c + 36r

⇒ 96nc = 1800c+ 36 × 10c

⇒ 96nc = 1800c + 360c

⇒ 96nc = 2160c

Hence we have 96nc = 2160c

Dividing both the sides by 96c

We get

⇒ n = 22.5

**Question 20. A Man Buys Spirit At Rs. 60 Per Litre, Adds Water To It And Then Sells It At Rs. 75 Per Litre. What Is The Ratio Of Spirit To Water If His Profit In The Deal Is 37.5%?****Answer :**We have SP = 75 per litre

Profit Per cent = 37.5%

Hence

⇒ CP = SP × 100/(100 + 37.5)

⇒ CP = 7500/137.5 = 54.5454

Hence, to make the CP from 60 to 54.54,

Hence,

Required ratio is

= 54.54 : 60 – 54.54

= 54.54 : 5.454

= 10:1

**Question 21. In What Ratio Must Rice At Rs.9.30 Per Kg Be Mixed With Rice At Rs. 10.80 Per Kg So That The Mixture Be Worth Rs.10 Per Kg ?****Answer :**We need to mix rice varieties costing Rs 9.3 and Rs 10.8 per kg to make the mixture at Rs 10 /kg

To do so,

We need to add these varieties in the ratio

⇒10.8 – 10 : 10 – 9.3

= 0.8 : 0.7

= 8 : 7

**Question 22. Three Types Of Tea The A, B, C Costs Rs. 95/kg, 100/kg And70/kg Respectively. How Many Kgs Of Each Should Be Blended To Produce 100 Kg Of Mixture Worth Rs.90/kg, Given That The Quantities Of B And C Are Equal?****Answer :**Let the quantity of tea a = x

Let the quantity of tea b = y

Let the quantity of tea c = z

Given : y = z …… (i)

x + y + z = 100 ……. (ii)

Cost of Mixture = (Sum of Cost of individual items in a mixture / total quantity of mixture)

90 = (95x + 100y + 70z) / (x + y + z)

90x + 90y + 90z = 95x + 100y + 70z

5x + 10y = 20z

5x = 20y – 10y = 10y…………………………. (from Eq. (i), y =z )

x = 2y ……………….. (iii)

From Equation (i), (ii) & (iii)

x + y + z = 100

2y + y + y = 100

4y = 100

y = 25

**Question 23. A Man Buys 12 Litres Of Solution A Which Contains 20% Of The Liquid And The Rest Is Water. He Then Mixes It With 10 Litres Of Another Solution B With 30% Of Liquid. What Is The % Of Water In The New Mixture?****Answer :**% of Liquid in Solution A = 20%

⇒ % of Water in Solution A = 80%

Volume of Solution A = 12 Litres

Volume of Water in Solution A = (80/100) × 12 = 9.6 Litres

% of Liquid in Solution B = 30%

⇒ % of Water in Solution A = 70%

Volume of Solution B = 10 Litres

Volume of Water in Solution B = (70/100) × 10 = 7 Litres

If Solution A is mixed with Solution B

⇒Net Total Volume of Solution = (12 + 10) = 22 Ltrs

⇒Net Total Volume of Water in Solution = (9.6 + 7) = 16.6 Ltrs

% of Water in New Mixture = (Volume of Water in New Mixture / Volume of New Mixture) × 100

% of Water in New Mixture = (16.6 / 22) × 100 = 75.45%

**Question 24. A Fuel Dealer Mixes Two Brands Of Fuel Which Cost In The Ratio 2:3. A Solution Containing 30% Brand A And Remaining Brand B Yields Profit Of 10% When Sold At Rs 297. What Is Cost Of Brand B?****Answer :**Let quantity of solution sold be 1 litre

When the solution is sold for 297, the profit percent = 10%

Hence CP of of solution = SP × 100/(100+P%)

⇒CP = 297 × 100/(100+10)

⇒CP = 270

Ratio of quantities of brand A and brand B is 3:7

Let their costs be 2c and 3c.

⇒ 0.3×2c + 0.7×3c = 270

⇒ 2.7c = 270

⇒ c = 100

Hence cost of Brand B = 3c = 300

**Question 25. If 9 Men Working 7.5 Hours A Day Can Finish A Piece Of Work In 20 Days, Then How Many Days Will Be Taken By 12 Men, Working 6 Hours A Day To Finish The Work? It Is Being Given That 2 Men Of Latter Type Work As Much As 3 Men Of The Former Type?****Answer :**Work done by 9 men in a day = 1/20

∴Work done by 1 man in a day for 7.5 hours = (1/20) ÷ 9 = 1/180

∴ Work done by 1 man in an hour = (1/180) ÷ (7.5) = 1/1350

Given that 2 men of latter type are equal to 3 men of former type ⇒ 1 man of latter type = 1.5 men of former type.

∴ Work done by 1 man of latter type in an hour = 1.5 × (1/1350) = 1/900

∴ Work done by 1 man in 6 hours = 6 × (1/900) = 1/150

∴ Work done by 12 men in a day working 6 hours/day = 12 × (1/150) = 2/25

Days required to finish the work = 25/2 = 12.5 days.

**Question 26. A Man’s Basic Pay For A 40 Hours’ Week Is Rs. 200. Overtimes Is Paid At 25% Above The Basic Rate. In A Certain Week, He Worked Overtime And His Total Was Rs. 300. He Therefore, Worked For A Total Of (in Hours)?****Answer :**Basic pay per hour of normal work = 200/40 = Rs.5

Over time is paid at 25% more = (125/100) × 5 =Rs.6.25 per hour

Man was paid Rs.200 for his regular work and Rs.100 for over time (total Rs.300).

Number of regular hours = 40

Number of overtime hours at 6.25 per hour = 100 ÷ 6.25 = 16 hours.

Total number of hours worked = 40+16 = 56 hours.

**Question 27. A Retailer Buys A Radio For Rs.225. His Overhead Expenses Are Rs.15 And He Sells The Radio For Rs.300. What Is The Profit Percent Of The Retailer?****Answer :**Selling price = cost price + profit

Cost price = 225 + 15 = 240

i.e. 240 + X * 240 * 1/100 = 300 240 + 24X/ 10 = 300

(taking LCM) 2400 + 24X = 3000

24 X = 3000 – 2400

24X = 600

X = 600/ 24

X = 25

**Question 28. A Truck Covers A Distance Of 640 Km In 10 Hrs. A Car Covers The Same Distance In 8 Hrs. What Is The Respective Ratio Between The Speed Of The Truck And The Car?****Answer :**Both truck and car covers a distance = 640 km

Time taken by truck is 10 hrs.

∴ Speed of truck = 640/10 = 64 km/hr

Time taken by car is 8 hrs.

∴ Speed of car = 640/8 = 80 km/hr

∴ Ratio between speed of truck and car =

Speed of truck : speed of car = 64 : 80 = 4 : 5

**Question 29. Two Trains From The Points A And B Moving In Opposite Direction, At The Point They Meet The Second Train Travels 120 Kms More Than The First. The Speeds Are 50kmph And 60kmph Respectively Find The Distance Between A And B?****Answer :**We know that

Distance = time × speed

Suppose first train travelled the distance x kms from A.

Then second train will travel the distance (x + 120) kms from B.

Let they met each other after t time

∴ (x + 120)/60 = (x/50)

⇒ x = 600 kms

But the distances between two point is (x + x + 120)

= 2x +120

= 2× 600 + 120

= 1320 km

**Question 30. Every Day A Cyclist Meets A Train At A Particular Crossing. The Road Is Straight Before The Crossing And Both Are Traveling In The Same Direction. The Cyclist Travels With A Speed Of 10 Kmph. One Day The Cyclist Comes Early By 25 Min. And Meets The Train 5km Before The Crossing. What Is The Speed Of The Train?****Answer :**Cyclist speed is 10kmph and he’ll take 30 min to cover 5 kms.

If Cyclist come 25 min early he would be 5 min before crossing.

So it means train takes 5 min to reach crossing.

We have speed of train calculated as

⇒ Speed= distance/time

⇒ Speed = 5km/5 min

⇒ Speed = 1 km/min

Or 60 km/hr

**Question 31. The Average Speed Of A Bus From Koyambedu To Salem Is 57 Km Per Hour. The Bus Is Scheduled To Leave Koyambedu Bus Station At 10 Pm And Reach Salem At 4.35 Am On The Next Day. The Distance Between Salem And Koyambedu Bus Station Is 342 Km. On The Way In Between Koyambedu And Salem A Halt Is Scheduled Compulsorily. Find Out The Duration Of This Halt Scheduled?****Answer :**Given, distance between Salem and Koyambedu = 342 km.

Speed of bus = 57 kmph

Time = Distance/Speed

⇒ Time needed by the bus to travel from Koyambedu to Salem = 342/57 hours = 6 hours.

Time actually taken by the bus = 6 hours and 35 minutes.

Bus takes 35 minutes extra ⇒ The compulsory halt time of bus = 35 minutes

**Question 32. Find The Equation Whose Roots Are 9 And 5 ?****Answer :**Finding the roots of given equations x2 – 14x + 45 = 0 x2 – 9 x – 5 x + 45 = 0 (x – 5) (x – 9) = 0

**Question 33. What Will Come In Place Of Question Mark (?) In The Following Question?**

76% Of 1285 = 35% Of 1256 +?**Answer :**Given expression is-

76% of 1285 = 35% of 1256 +?

⇒ (76/100) × 1285 = (35/100) × 1256 +?

⇒ 97660/100 = 43960/100 +?

⇒ ? = (97660/100) – (43960/100)

⇒ ? = 53700/100 = 537

**Question 34. What Will Come In Place Of Question Mark In The Following Equation?****Answer :**(5863 – √2704) × 0.5 = ?

(5863 – √2704) × 0.5 = x

⇒ (5863 – 52) × 0.5 = x

⇒ x = 2905.5

**Question 35. Let 13 And 273 Are The Hcf And Lcm Of Two Numbers Respectively, And If One Of Them Is Less Than 140 And Greater Than 60. Then What Will Be That Number?****Answer :**Let those 2 numbers be p and q .

We are given that HCF (p, q) = 13 and

LCM (p, q) = 273.

We know that product of numbers = HCF × LCM

Hence, p × q = 13 × 273

⇒ p × q = 13 × 13 × 3 × 7

P cannot be 13 × 3 or 13 × 13 or 7 × 3 or 13 or 3 or 7 as these numbers do not come between 60 and 140.

Hence we can say that p = 13 × 7 = 91

**Question 36. The Least Number Which When Divided By 4, 6, 8, 12 And 16 Leaves A Remainder Of 2 In Each Case Is?****Answer :**Let’s find a number which is completely divisible by 4, 6, 8, 12 and 16.

To find such number, we need to find the LCM of all these numbers

LCM of (4,6) , 8 , 12 , 16

= 12 , 8 , 12 , 16

= LCM of (12 , 8) , LCM of (12 , 16)

= 24 , 48

= LCM of (24 , 48)

= 48

Hence 48 + 2 ie 50 will leave a remainder of 2 if divided by any of the following numbers : 4, 6, 8, 12 and 16

**Question 37. Find The Number Of Sides Of A Regular Convex Polygon Whose Interior Angle Is 40 Degrees?****Answer :**We have interior angle as 40 degrees.

Hence Let n be the number of sides of that polygon,

Then formula is

⇒180 × (n-2) = 40 × n

⇒180n – 360 = 40n

⇒140n = 360

Since, no integer value of n is possible, hence such a polygon can not exist

**Question 38. There Is A 5 Digit No. Sum Of 3 Pairs Of Digits Is Eleven Each. Last Digit Is 3 Times The First One. 3rd Digit Is 3 Less Than The Second. 4th Digit Is 4 More Than The Second One. Find The Number?****Answer :**We are given a 5 digit number

Let first digit be ‘X’

then 5th digit is ‘3X’

let 2nd digit be ‘Y’

then 3rd digit is ‘Y – 3’

and 4th digit is ‘Y + 4’

then the no is ‘(X)(Y)(Y – 3)(Y + 4)(3X)’

from the above we can say 3X <= 9

so X<=3 and any of the digit in the number is <= 9

and also given that 3 pairs sum is 11…

Also Y–3 ≥ 0 and Y+4 ≤ 9

⇒ Y ≥ 3 and Y ≤ 5 i.e. Y = 3,4 or 5

for x = 1, all the conditions won’t be satisfied.

For x = 2,

Let Y = 5

∴ The number is 25296

**Question 39. What Will Come In Place Of Question Mark In The Following Equation?**

4910+311+715= ?**Answer :**4 + 9/10 + 3/11 + 7/15

= 4 + 9/10 + 3/11 + 7/15

=1617+90+154330=1861330

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