Q1. The Average Age Of A Woman And Her Daughter Is 42 Years. The Ratio Of Their Ages Is 2 : 1 Respectively. What Is The Daughter’s Age?

Let the age of mother be M and that of her daughter be D Therefore, [M+D]/2 = 42 [M/D] =2/1 and Solving the above equations we get D = 28 yrs.

Q2. Two Trains Of Length 150 M And 200 M Respectively, Are Travelling In Opposite Directions At A Speed Of 54 Km/hr And 72 Km/hr. What Is The Total Time Taken By Them To Cross Each Other?

54 km/hr = 54 x (5/18) = 15m/s

72 km/hr = 72 x (5/18) = 20 m/s

Total distance = 150 + 200 = 350 m

Relative speed = 15 + 20 = 35 m/s

Total time = Total distance/Relative speed

= 350/35 = 10s

Q3. The Banker’s Discount On Rs.1800 At 12% Per Annum Is Equal To The True Discount On Rs.1872 For The Same Time At The Same Rate. Find The Time?

S.I on Rs.1800 = T.D on Rs.1872.

P.W on Rs.1872 is Rs.1800.

Rs.72 is S.I on Rs. 1800 at 12%.

Time = (100×72 / 12×1800)

= 1/3 year = 4 months.

Q4. Find Compound Interest On Rs. 7500 At 4% Per Annum For 2 Years, Compounded Annually?

Amount = Rs [7500x (1+4/100)²]

= Rs.(7500 x 26/25×26/25Rs)

= Rs.8112.

C.I = Rs (8112 – 7500)

= Rs.612.

Q5. A Cone Has Vertical Height And Slant Height As 15 Cm And 17 Cm Respectively. A Hemisphere With The Same Radius As The Cone Is Placed On The Face Of The Cone. What Is The Total Volume Of The Figure For

Using the Pythagoras theorem, the radius of the cone = √ (172 – 152) = 8 cm.

Volume of cone = (1/3) x π x r2 x h= (1/3) x π x 82 x 15

Radius of hemisphere=radius of cone

Volume of hemisphere = (2/3) x π x r3 = (2/3) x π x 83

Total volume of figure = Volume of cone + Volume of hemisphere = 4233.58 cu. cm.

Q6. What Is The Probability Of Finding A Red Face Card In A Deck Of Cards?

A deck of card has 52 cards of which 26 are black and the other 26 are red.

The number of face cards is 12, but only 6 of them are red.

So, required probability is 6/52 = 3/26

Q7. On Selling 100 Articles, A Shopkeeper Earns A Profit Amount Equal To The Selling Price Of 50 Articles. What Is The Profit Percentage Of The Shopkeeper?

Let SP of each article be Re. 1

So, SP of 100 articles =Rs. 100

Implies profit = Rs. 50

So, CP = 100 – 50 = Rs. 50

Therefore, profit percentage of the shopkeeper is 50/50 × 100 = 100%

Q8. The Speed Of A Boat In Still Water Is 15 Km/hr And The Rate Of Current Is 3 Km/hr. The Distance Travelled Downstream In 12 Minutes Is?

Speed Downstream = (15 + 3) km/hr

= 18 km/hr.

Distance travelled = (18 x 12/60) hrs

= 3.6km.

Q9. The Sum Of 3rd And 6th Term Of An A.p Is 2

T3 = a + 3d

T6 = a + 6d

T3 + T6 = 2a + 9d = 27

**Sum of first 10 terms of an AP is: **

S= (10/2) (2a+9d)

= 5 x 27

= 135

Q10. Even After Reducing The Marked Price Of A Tristor By Rs. 32, A Shopkeeper Makes A Profit Of 15%. If The Cost Price Be Rs. 320, What Percentage Of Profit Would He Have Made If He Had Sold The Tristor A

Let x be the marked price,

So x – 32 = 320 X 1.15

x = 400.

So required value is

400 = 320 (1 + profit/100),

So profit is 25%

Q11. The Product Of Two Numbers Is 192 And The Sum Of These Two Numbers Is 2

Let the number be x and (28 – x) = Then,

x (28 – x) = 192

‹=›x^{2} – 28x + 192 = 0.

‹=›(x – 16) (x – 12) = 0

‹=›x = 16 or x = 12.

Q12. A Can Contains A Mixture Of Two Liquids A And B In The Ratio 7:

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively

Quantity of An in mixture left

= (7x – 7/12 x 9) litres = (7x – 21/4) litres.

Quantity of B in mixture left

= (5x – 5/12 x 9) litres = (5x – 15/4) litres.

(7x – 21/4) / [(5x – 15/4) +9] = 7/9 = › 28x – 21/20x + 21 = 7/9 =› 252x – 189 = 140x + 147

=› 112x = 336 =’ x = 3.

So, the can contained 21 litres of A.

Q13. A Completes 50% Of The Work In 10 Days And Then Decides To Take Help From B And C. B Is Half As Efficient As A And Similarly C Is Half As Efficient As B. How Many More Days Will They Take To Complete

As A completes half of the work in 10 days, he/she will complete the work in 20 days.

As B is half as efficient as A and C is half as efficient as B,

They will complete the work in 1/20 + 1/40 + 1/80 = 7/8@So 7/80 of the work in a single day.

So, they can complete the entire work in 80/7 days

Therefore, they can complete the remaining 50% of the work in (1/2) x 80/7 = 40/7 days.

Q14. A Car Travels A Distance Of 45 Km At The Speed Of 15 Km/hr. It Covers The Next 50 Km Of Its Journey At The Speed Of 25km/hr And The Last 25 Km Of Its Journey At The Speed Of 15 Km/hr. What Is The Aver

We know, Average speed = Total distance travelled / Total time taken Average = [45+50+25]/ [3+2+ {25/15}] = 18 kmph

Q15. When A Commodity Is Sold For Rs.34.80, There Is A Loss Of 2%. What Is The Cost Price Of The Commodity?

C.P = Rs. (100 / 75×34.80)

= Rs.46.40.

Q16. Three Times The First Of Three Consecutive Odd Integers Is 3 More Than Twice The Third. The Third Integer Is?

Let the three numbers be x, x+2, x+4

Then 3x = 2(x+4) + 3

‹=›x = 11

Third integer = x + 4 = 15.

Q17. The Value Of Log_{2} 16 Is?

Let log_{2}16 = n.

Then, 2^{n} = 16 = 24

‹=› n=4.

Q18. There Are Two Sections A And B Of A Class, Consisting Of 36 And 44 Students Respectively. If The Average Weight Of Sections A Is 40 Kg And That Of Sections B Is 35 Kg. Find The Average Weight Of The W

Total weight of (36+44) Students = (36×40+44×35) Kg = 2980 kg.

Average weight of the whole class = (2980 / 80) = 37.25.

Q19. Aman Started A Business Investing Rs.70, 0

Aman : Rakhi : Sagar =(70,000 x 36):(1,05,000 x 30):(1,40,000 x 24)

=12: 15: 16.

Q20. How Many Different Words Can Be Formed From The Word Oracle So That The Vowels Always Come Together?

We will group the letters that need to come together (A & E) and consider them as a single letter. So, here the letters are O, R, C, L, and AE.

Number of ways A & E can be arranged is 2!

So, the total number of ways in which the words can be formed so that all vowels are together is 5! x 2! = 240 ways.

Q21. The Banker’s Gain On A Bill Due 1 Year Hence At 12% Per Annum Is Rs.

T.D = [B.G x 100 / R x T]

= Rs. (6 x 100 / 12 x 1)

= Rs.50.

Q22. The Ratio Of The Present Ages Of A And B 9:

A/b=9/5; i.e. 5a = 9B………. (i)

A-5/B-5 = 2/1; i.e. A-5 = 2B ………10

∴A – 2B = -5………………. (ii)

From equation (i) and (ii), A = 45, B = 25

∴ Average =45+25/2 = 70/2 = 35.

Q23. The Length Of A Rectangular Plot Is 20 Metres More Than Its Breadth. If The Cost Of Fencing The Plot

Let breadth = x metres

Then, length = (x + 20) metres.

Perimeter = (5300 / 26.50) m

= 200m

Q24. A Batsman Makes A Score Of 87 Runs In The 17th Inning And Thus Increases His Averages By

Let the average after 17th inning = x. Then, average after 16th inning = (x – 3)

Average =16 (x-3) +87

= 17x or x= (87-48)

= 39.

Q25. A Clock Is Set At 5 A.m. The Clock Loses 16 Minutes In 24 Hours. What Will Be The True Time When The Clock Indicates 10 P.m. On 4th Day?

Time from 5 a.m on a day to 10 p.m.on 4th day = 89 hours.

Now 23 hrs 44 min. of this clock = 24 hours of correct clock.

Therefore 356 / 15 hrs of this clock = 24 hours of correct clock.

89 hrs of this clock = (24 x 15/356 x 89) hrs

= 90 hrs

So, the correct time is 11 p.m.

Q26. A Man Buys A Cycle For Rs.1400 And Sells It At A Loss Of 15%. What Is The Selling Price Of The Cycle?

S.P = 85% of Rs.1400

= Rs. (85/100×1400)

Rs.1190.

Q27. A Car Travels A Distance Of 170 Km In 2 Hours Partly At A Speed Of 100 Km/h And Partly At 50 Km/h. The Distance Travelled At A Speed Of 50 Km/h Is?

Suppose he covers x km at 100 kmph

So he covers 170-x at 50 kmph

So {X/100}+{170-X}/50=2

Solving this equation, we get x = 140.

So he covers 30km at 50 kmph.

Q28. A Rectangular Parking Space Is Marked Out By Painting Three Of Its Sides. If The Length Of The Unpainted Side Is 9 Feet, And The Sum Of The Lengths Of The Painted Sides Is 37 Feet, Then What Is The Ar

Clearly, we have l=9 and l+2b=37

Area = (l x b)

= (9 x 14) sq.ft = 126 sq.ft.

Q29. 20 Boys And 32 Girls Form A Group For Social Work. During Their Membership Drive Same No. Of Boys And Girls Joined The Group. How Many Members Does The Group Have Now, If The Ratio Of Boys To Girls Is

Let x be the new boys as well as girls, Therefore

[20+X]/ [32+X] =3/4

Solving this we get x = 16

So total will be 36 + 48 = 84.

Q30. Suganya And Surya Are Partners In A Business. Suganya Invests Rs. 35,000 For 8 Months And Surya Invests Rs.42, 000 For 10 Months. Out Of A Profit Of Rs.31, 57

Ratio of their shares = (35000×8): (42000×10)

= 2: 3.

Suganya’s share = Rs. (31570 ×2/5)

= Rs.12628.

Q31. Find The Compound Interest On Rs.16, 000 At 20% Per Annum For 9 Months, Compounded Quarterly?

Principal = Rs.16, 000;

Time=9 months = 3 quarters;

Amount = Rs. [16000x (1+5/100)³]

= [16000×21/20×21/20×21/20]

= Rs.18522.

C.I = Rs. (18522 – 16000)

= Rs.2522.

Q32. A Vessel Is Filled With Liquid, 3 Parts Of Which Are Water And 5 Parts Of Syrup. How Much Of The Mixture Must Be Drawn Off And Replaced With Water So That The Mixture May Be Half Water And Half Syrup?

Suppose the vessel initially contains 8 litres of liquid. Let x litters of this liquid be replaced with water.

Quantity of water in new mixture = (3 – 3x/8 + x) litres.

Quantity of syrup in new mixture = (5 – 5x/8) litres.

(3 – 3x/8 + x) = (5 – 5x/8) = 5x + 24 = 40 – 5x

=› 10x = 16 =› x = 8/5

So, part of the mixture replaced = (8/5 x 1/8) = 1/5.

Q33. An Accurate Clock Shows 8 O’clock In The Morning. Through How Many Degrees Will The Hour Hand Rotate When The Clock Shows 2 O’clock In The Afternoon?

Angle traced by hour hand in

5 hrs 10 min. = (360/12 x 6) °

= 180°.

Q34. What Decimal Of An Hour Is A Second?

Required decimal = 1/ 60 x 60

= 1/ 3600

= .00027.

Q35. If Log 2 = 0.30103, The Number Of Digits In 5^{20} Is?

Log 5^{20} =20 log 5

=20 × [log (10/2)]

=20 (log 10 – log 2)

=20 (1 – 0.3010)

=20×0.6990

=13.9800.

Characteristics = 13.

Q36. A Can Do A Work In 60 Days And B Can Do The Same Work In 40 Days. They Work Together For 12 Days And Then ‘a’ Goes Away. In How Many Days Will ‘b’ Finish The Remaining Work?

Work done by A and B in 12 days is = 12 * 5/120 = 1/2

Therefore Remaining work = 1- 1/2 = 1/2work

B does 1/40 work in one day

Therefore B does 1/2 work in 40*1/2 =20days

Q37. The Price Of Sugar Is Increased By 25%.find By How Much Percent The Consumption Of Sugar Be Decreased So As Not To Increase The Expenditure?

Using the formula to calculate % decrease as [R/(100+R)]x100 where R = percentage increase in price, we get

Required % decrease in consumption = 25/125 MULTIPLIED 100 = 20%.

Q38. A Boat Can Travel With A Speed Of 13 Km/hr In Still Water. If The Speed Of The Stream Is 4 Km/hr. Find The Time Taken By The Boat To Go 68 Km Downstream?

Speed Downstream = (13 + 4) km/hr

= 17 km/hr.

Time taken to travel 68 km downstream = (68 / 17) hrs

= 4 hrs.