**Question 1. A Man, His Wife And Daughter Worked In A Graden. The Man Worked For 3 Days, His Wife For 2 Days And Daughter For 4 Days. The Ratio Of Daily Wages For Man To Women Is 5 : 4 And The Ratio For Man To Daughter Is 5 : 3. If Their Total Earnings Is Mounted To Rs. 105, Then Find The Daily Wage Of The Daughter?****Answer :**Assume that the daily wages of man, women and daughter are Rs 5x, Rs. 4x, Rs 3x respectively.

Multiply (no. of days) with (assumed daily wage) of each person to calculate the value of x.

[3 x (5x)] + [2 x (4x)] + [4 x (3x)] = 105

[15x + 8x + 12x] = 105

35x = 105

x = 3

Hence, man’s daily wage = 5x = 5 x 3 = Rs. 15

Wife’s daily wage = 4x = 4 x 3 = Rs. 12

Daughter’s daily wage = 3x = 3 x 3 = Rs. 9.

**Question 2. Amit, Raju And Ram Agree To Pay Their Total Electricity Bill In The Proportion 3 : 4 : 5. Amit Pays First Day’s Bill Of Rs. 50, Raju Pays Second Day’s Bill Of Rs. 55 And Ram Pays Third Day’s Bill Of Rs. 75. How Much Amount Should Amit Pay To Settle The Accounts?****Answer :**Toatal bill paid by Amit, Raju and Ram = ( 50 + 55 +75 ) = Rs. 180

Let amount paid by Amit, Raju and Ram be Rs. 3x, 4x and 5x respectively.

Therefore, (3x + 4x + 5x ) = 180

12x = 180

x = 15

Therefore, amount paid by,

Amit = Rs. 45

Raju = Rs. 60

Ram = Rs. 75

But actually as given in the question, Amit pays Rs. 50, Raju pays Rs. 55 and Ram pays Rs. 80. Hence, Amit pays Rs. 5 less than the actual amount to be paid. Hence he needs to pay Rs. 5 to Raju settle the amount.

**Question 3. Salaries Of Ram And Sham Are In The Ratio Of 4 : 5. If The Salary Of Each Is Increased By Rs. 5000, Then The New Ratio Becomes 50 : 60. What Is Sham’s Present Salary?****Answer :**Assume original salaries of Ram and Sham as 4x and 5x respectively.

Therefore,

(4x + 5000)/= 50

(5x + 5000) 60

60 (4x + 5000) = 50 (5x + 5000)

10 x = 50,000

5x = 25, 000

Sham’s present salary = 5x + 5000 = 25,000 + 5000

Sham’s present salary = Rs. 30,000.

**Question 4. Two Dice Are Tossed.the Probability That The Total Score Is A Prime Number Is?****Answer :**Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }

n(E) = 15.

P(E) = n(E)/n(S) = 15/36 = 5/12.

**Question 5. Find The Lcm Of Following Three Fractions:36/,48/,72/?****Answer :**Numerators = 36, 48 and 72.

72 is largest number among them. 72 is not divisible by 36 or 48

Start with table of 72.

72 x 2 = 144 = divisible by 72, 36 and 48

? LCM of numerators = 144

Denominators = 225, 150 and 65

We can see that they can be divided by 5.

On dividing by 5 we get 45, 30 and 13

We cannot divide further.

So, HCF = GCD = 5

LCM of fraction =144/5.

**Question 6. Find The Largest Number To Divide All The Three Numbers Leaving The Remainders 4, 3, And 15 Respectively At The End?****Answer :**Here greatest number that can divide means the HCF

Remainders are different so simply subtract remainders from numbers

17 – 4 = 13; 42 – 3 = 39; 93 – 15 = 78

Now let’s find HCF of 13, 39 and 78

By direct observation we can see that all numbers are divisible by 13.

? HCF = 13 = required greatest number.

**Question 7. The Two Given Numbers A And B Are In The Ratio 5:6 Such That Their Lcm Is 480. Find Their Hcf?****Answer :**Let K be common factor. So 2 numbers are 5K and 6K

Also K is the greatest common factor (HCF) as 5 and 6 have no other common factor

? 5K x 6K = 480 x K

K = 16 = HCF.

**Question 8. A Wall Is 4.5 Meters Long And 3.5 Meters High. Find The Number Of Maximum Sized Wallpaper Squares, If The Wall Has To Be Covered With Only The Square Wall Paper Pieces Of Same Size?****Answer :**Wall can be covered only by using square sized wallpaper pieces.

Different sized squares are not allowed.

Length = 4.5 m = 450 cm;

Height = 3.5 m = 350 cm

Maximum square size possible means HCF of 350 and 450

We can see that 350 and 450 can be divided by 50.

On dividing by 50, we get 7 and 9.

Since we cannot divide further,

HCF = 50 = size of side of square

Number of squares =Wall area/=450 x 350/= 63

Square area=50 x 50.

**Question 9. Find The Largest Number Of 4-digits Divisible By 12, 15 And 18?****Answer :**Required largest number must be divisible by the L.C.M. of 12, 15 and 18

L.C.M. of 12, 15 and 18

12 = 2 × 2 × 3

15 =5 × 3

18 = 2 × 3 × 3

L.C.M. = 180

Now divide 9999 by 180, we get remainder as 99

The required largest number = (9999 – 99) =9900

Number 9900 is exactly divisible by 180.

**Question 10. A Trader Mixes 26 Kg Of Rice At Rs. 20 Per Kg With 30 Kg Of Rice Of Other Variety At Rs. 36 Per Kg And Sells The Mixture At Rs. 30 Per Kg. His Profit Percent Is?****Answer :**C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

Gain =(80/1600*100) % = 5%.

**Question 11. By Selling 45 Lemons For Rs 40, A Man Loses 20%. How Many Should He Sell For Rs 24 To Gain 20% In The Transaction ?****Answer :**Let S.P. of 45 lemons be Rs. x.

Then, 80 : 40 = 120 : x or x = 40×120/80= 60

For Rs.60, lemons sold = 45

For Rs.24, lemons sold =4560×24= 18.

**Question 12. If Books Bought At Prices Ranging From Rs. 200 To Rs. 350 Are Sold At Prices Ranging From Rs. 300 To Rs. 425, What Is The Greatest Possible Profit That Might Be Made In Selling Eight Books ?****Answer :**Least Cost Price = Rs. (200 * 8) = Rs. 1600.

Greatest Selling Price = Rs. (425 * 8) = Rs. 3400.

Required profit = Rs. (3400 – 1600) = Rs. 1800.

**Question 13. If The Cost Price Is 25% Of Selling Price. Then What Is The Profit Percent?****Answer :**Let the S.P = 100

then C.P. = 25

Profit = 75

Profit% = (75/25) * 100 = 300%.

**Question 14. A Man Buys Oranges At Rs 5 A Dozen And An Equal Number At Rs 4 A Dozen. He Sells Them At Rs 5.50 A Dozen And Makes A Profit Of Rs 50. How Many Oranges Does He Buy?****Answer :**Cost Price of 2 dozen oranges Rs. (5 + 4) = Rs. 9.

Sell price of 2 dozen oranges = Rs. 11.

If profit is Rs 2, oranges bought = 2 dozen.

If profit is Rs. 50, oranges bought = (2/2) * 50 dozens = 50 dozens.

**Question 15. The Sum Of All 3 Digit Numbers Divisible By 3 Is?****Answer :****All 3 digit numbers divisible by 3 are :**102, 105, 108, 111, …, 999.

This is an A.P. with first element ‘a’ as

102 and difference ‘d’ as 3.

Let it contains n terms. Then,

102 + (n – 1) x

^{3}= 999102 + 3n-3 = 999

3n = 900 or n = 300

Sum of AP = n/2 [2*a + (n-1)*d]

Required sum = 300/2[2*102 + 299*3] = 165150.

**Question 16. The Speed Of A Car Increases By 2 Kms After Every One Hour. If The Distance Travelling In The First One Hour Was 35 Kms. What Was The Total Distance Travelled In 12 Hours?****Answer :**Total distance travelled in 12 hours =(35+37+39+…..upto 12 terms)

This is an A.P with first term, a=35, number of terms,

n= 12,d=2.

Required distance = 12/2[2 x 35+{12-1) x 2]

=6(70+23)

= 552 kms.

**Question 17. A Man Walking At The Rate Of 5 Km/hr Crosses A Bridge In 15 Minutes. The Length Of The Bridge (in Metres) Is?****Answer :**speed= (5×5/18)m/sec=25/18 m/sec.

Distance covered in 15 minutes= (25/18 x 15 x 60)m= 1250 m.

**Question 18. A Man On Tour Travels First 160 Km At 64 Km/hr And The Next 160 Km At 80 Km/hr. The Average Speed For The First 320 Km Of The Tour Is?****Answer :**Total time taken = (160/64 + 160/8)hrs

= 9/2 hrs.

Average speed = (320 x 2/9) km.hr

= 71.11 km/hr.

**Question 19. If The Product And H.c.f. Of Two Numbers Are 4107 And 37 Respectively, Then Find The Greater Number?****Answer :**4107 is the square of 37.

So let two numbers be 37x and 37y.

37x × 37y = 4107

xy = 3

3 is the product of (1 and 3)

x = 1 and y = 3

37x = 37 × 1 =37

37y = 37 × 3 = 111

Greater number = 111.

**Question 20. The Traffic Lights At Three Different Road Crossings Change After Every 40 Sec, 72 Sec And 108 Sec Respectively. If They All Change Simultaneously At 5 : 20 : 00 Hours, Then Find The Time At Which They Will Change Simultaneously?****Answer :**Traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively.

Therefore, find the L.C.M. of 40, 72 and 108.

L.C.M. of 40, 72 and 108 = 1080

The traffic lights will change again after 1080 seconds = 18 min

The next simultaneous change takes place at 5 : 38 : 00 hrs.

**Question 21. John, Smith And Kate Start At Same Time, Same Point And In Same Direction To Run Around A Circular Ground. John Completes A Round In 250 Seconds, Smith In 300 Seconds And Kate In 150 Seconds. Find After What Time Will They Meet Again At The Starting Point?****Answer :**L.C.M. of 250, 300 and 150 = 1500 sec

Dividing 1500 by 60 we get 25, which mean 25 minutes.

John, Smith and Kate meet after 25 minutes.

**Question 22. The Product Of Two Numbers Is 2028 And Their H.c.f. Is 13. The Number Of Such Pairs Is?****Answer :**Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

=>ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

**Question 23. Three Number Are In The Ratio Of 3 : 4 : 5 And Their L.c.m. Is 2400. Their H.c.f. Is?****Answer :**Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

**Question 24. The Sum Of Two Numbers Is 528 And Their H.c.f Is 33. The Number Of Pairs Of Numbers Satisfying The Above Condition Is?****Answer :**Let the required numbers be 33a and 33b.

Then 33a +33b= 528 => a+b = 16.

Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9).

Therefore, Required numbers are ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9)

The number of such pairs is 4.

**Question 25. The L.c.m Of Two Numbers Is 495 And Their H.c.f Is 5. If The Sum Of The Numbers Is 100, Then Their Difference Is?****Answer :**Let the numbers be x and (100-x).

Then,x(100-x)=5*495

=> x

^{2}-100x+2475=0=> (x-55) (x-45) = 0

=> x = 55 or x = 45

The numbers are 45 and 55

Required difference = (55-45) = 10.

**Question 26. A Rectangular Courtyard 3.78 Meters Long 5.25 Meters Wide Is To Be Paved Exactly With Square Tiles, All Of The Same Size. What Is The Largest Size Of The Tile Which Could Be Used For The Purpose?****Answer :**3.78 meters =378 cm = 2 × 3 × 3 × 3 × 7

5.25 meters=525 cm = 5 × 5 × 3 × 7

Hence common factors are 3 and 7

Hence LCM = 3 × 7 = 21

Hence largest size of square tiles that can be paved exactly with square tiles is 21 cm.

**Question 27. What Is The Probability Of Getting 53 Mondays In A Leap Year?****Answer :**1 year = 365 days . A leap year has 366 days

A year has 52 weeks. Hence there will be 52 Sundays for sure.

52 weeks = 52 x 7 = 364days

366 – 364 = 2 days

In a leap year there will be 52 Sundays and 2 days will be left.

**These 2 days can be:**1. Sunday, Monday

2. Monday, Tuesday

3. Tuesday, Wednesday

4. Wednesday, Thursday

5. Thursday, Friday

6. Friday, Saturday

7. Saturday, Sunday

Of these total 7 outcomes, the favourable outcomes are 2.

Hence the probability of getting 53 days = 2/7.

**Question 28. A Man Bought An Article And Sold It At A Gain Of 5 %. If He Had Bought It At 5% Less And Sold It For Re 1 Less, He Would Have Made A Profit Of 10%. The C.p. Of The Article Was?****Answer :**Let original Cost price is x

Its Selling price = (105/100) * x = 21x/20

New Cost price = (95/100) * x = 19x/20

New Selling price = (110/100 )* (19x/20 )= 209x/200

[(21x/20) – (209x/200)] = 1

=> x = 200.

**Question 29. Every Year Before The Festive Season,a Shopkeeper Increases The Price Of The Product By 35% And Then Introduce Two Successive Discount Of 10% And 15% Respectively.what Is Percentage Loss And Percentage Gain?****Answer :**Let cp= 100,

35 % increase in sp=135

10 % discount in 135((135*10)/100)=13.5

so 1st sp=(135-13.5)=121.5, again 15 % discount in 1st sp((121.5*15)/100)=18.225

2nd sp=(121.5-18.225)=103.275,

so finally cp=100,sp=103.275 ,gain by 3.27%.

**Question 30. In A Scheme, A Pack Of Three Soaps With Mrp Rs.45 Is Available For Rs.42. If It Still Gives A Profit Of 5% To The Shopkeeper, Then The Cost Price Of The Pack Is ?****Answer :**Given M.P=45,S.P=42, Profit = 0.05

Let C.P=x , Then

Profit = (42-x)/x = 0.05

=> x = 40.

**Question 31. A Shopkeeper Sells One-third Of His Goods At A Profit Of 10%, Another One-third At A Profit Of 20%, And The Rest At A Loss Of 6%.what Is His Overall Profit Percentage?****Answer :**Let the shopkeeper buy 300g for Rs.300. Now he sells 100g for Rs.110, another 100g for Rs120, and the rest 100g for Rs94. sir

Therefore, the total amount he receives = Rs.110 + Rs.120 + Rs.94 = 324.

Therefore, the shopkeeper spends Rs.300 and gets back Rs.324.

Therefore, his profit percentage = 24/300×100 % = 8%.

**Question 32. By Mixing Two Qualities Of Pulses In The Ratio 2: 3 And Selling The Mixture At The Rate Of Rs 22 Per Kilogram, A Shopkeeper Makes A Profit Of 10 %. If The Cost Of The Smaller Quantity Be Rs 14 Per Kg, The Cost Per Kg Of The Larger Quantity Is?****Answer :**Cost Price of 5 kg = Rs.(14*2 + x*3) = (28 + 3x).

Sell price of 5 kg = Rs. (22×5) = Rs. 110.

[{110 – (28 + 3x)}/(28 + 3x) ]* 100 =10

[82-3x/28 + 3x]= 1 / 10

820 – 30x = 28 +3x

33x = 792

x = 24.

**Question 33. What Profit Percent Is Made By Selling An Article At A Certain Price, If By Selling At 2/3rd Of That Price, There Would Be A Loss Of 20%?****Answer :**SP2 = 2/3 SP1

CP = 100

SP2 = 80

2/3 SP1 = 80

SP1 = 120

100 — 20 => 20%.

**Question 34. Ajay Bought 15 Kg Of Dal At The Rate Of Rs 14.50 Per Kg And 10 Kg At The Rate Of Rs 13 Per Kg. He Mixed The Two And Sold The Mixture At The Rate Of Rs 15 Per Kg. What Was His Total Gain In This Transaction?****Answer :**Cost price of 25 kg = Rs. (15 x 14.50 + 10 x 13) = Rs. 347.50.

Sell price of 25 kg = Rs. (25 x 15) = Rs. 375.

profit = Rs. (375 — 347.50) = Rs. 27.50.

**Question 35. The Profit Earned By Selling An Article For Rs. 832 Is Equal To The Loss Incurred When The Same Article Is Sold For Rs. 448. What Should Be The Sale Price For Making 50% Profit ?****Answer :**Let C.P. = Rs. C.

Then, 832 – C = C – 448

2C = 1280 => C = 640

Required S.P. = 150% of Rs. 640 = 150/100 x 640 = Rs. 960.

**Question 36. A Man Buys An Item At Rs. 1200 And Sells It At The Loss Of 20 Percent. Then What Is The Selling Price Of That Item?****Answer :**Here always remember, when ever x% loss,

it means S.P. = (100 – x)% of C.P

when ever x% profit,

it means S.P. = (100 + x)% of C.P

So here will be (100 – x)% of C.P.

= 80% of 1200

= (80/100) * 1200

= 960.

**Question 37. ‘a’ Sold An Article To ‘b’ At A Profit Of 20%. ‘b’ Sold The Same Article To ‘c’ At A Loss Of 25% And ‘c’ Sold The Same Article To ‘d’ At A Profit Of 40%. If ‘d’ Paid Rs 252 For The Article, Then Find How Much Did ‘a’ Pay For It?****Answer :**Let the article costs ‘X’ to A

Cost price of B = 1.2X

Cost price of C = 0.75(1.2X) = 0.9X

Cost price of D = 1.4(0.9X) = 1.26X = 252

Amount paid by A for the article = Rs. 200.

**Question 38. A Shopkeeper Fixes The Marked Price Of An Item 35% Above Its Cost Price. The Percentage Of Discount Allowed To Gain 8% Is?****Answer :**Let the cost price = Rs 100

then, Marked price = Rs 135

Required gain = 8%,

So Selling price = Rs 108

Discount = 135 – 108 = 27

Discount% = (27/135)*100 = 20%.

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