29 + Interview Questions in Essar Aptitude Interview Questions in Essar Aptitude Tutorial

1.	Ratio Of Ashok's Age To Pradeep's Age Is 4 : 3. Ashok Will Be 26 Years Old After 6 Years. How Old Is Pradeep Now ?
Answer» Given A/p= 4/3 Also A = 26 after 6 years, so his present AGE = 20YEARS, SUBSTITUTING we get P = 15 years. Given A/p= 4/3 Also A = 26 after 6 years, so his present age = 20years, Substituting we get P = 15 years.

1.

Ratio Of Ashok's Age To Pradeep's Age Is 4 : 3. Ashok Will Be 26 Years Old After 6 Years. How Old Is Pradeep Now ?

Answer»

Given A/p= 4/3 Also A = 26 after 6 years,

so his present AGE = 20YEARS,

SUBSTITUTING we get P = 15 years.

Given A/p= 4/3 Also A = 26 after 6 years,

so his present age = 20years,

Substituting we get P = 15 years.

Discussion

2.	The Incomes Of Chanda And Kim Are In The Ratio 9 : 4 And Their Expenditures Are In The Ratio 7 : 3. If Each Saves Rs. 2,000, Then Chanda's Expenditure Is
Answer» Let the incomes of Chanda and Kim be 9X and expenditures be 7y and 3y respectively. SINCE = Income – EXPENDITURE, we get 9x – 7y = 2000 and 4x – 3y = 2000. SOLVING, we get, x = 8000 and y = 10000. So Chanda’s expenditure = 7y = 7 × 10000 = Rs. 70,000. Let the incomes of Chanda and Kim be 9x and expenditures be 7y and 3y respectively. Since = Income – Expenditure, we get 9x – 7y = 2000 and 4x – 3y = 2000. Solving, we get, x = 8000 and y = 10000. So Chanda’s expenditure = 7y = 7 × 10000 = Rs. 70,000.

2.

The Incomes Of Chanda And Kim Are In The Ratio 9 : 4 And Their Expenditures Are In The Ratio 7 : 3. If Each Saves Rs. 2,000, Then Chanda's Expenditure Is

Answer»

Let the incomes of Chanda and Kim be 9X and expenditures be 7y and 3y respectively.

SINCE = Income – EXPENDITURE,

we get 9x – 7y = 2000 and 4x – 3y = 2000.

SOLVING, we get, x = 8000 and y = 10000.

So Chanda’s expenditure = 7y = 7 × 10000 = Rs. 70,000.

Let the incomes of Chanda and Kim be 9x and expenditures be 7y and 3y respectively.

Since = Income – Expenditure,

we get 9x – 7y = 2000 and 4x – 3y = 2000.

Solving, we get, x = 8000 and y = 10000.

So Chanda’s expenditure = 7y = 7 × 10000 = Rs. 70,000.

Discussion

3.	Solve 2 3 10 39 178 885 ?
Answer» The LOGIC is ×1+1, ×2+4, ×3+9, ×4+16, ×5+25,…. So following the logic we GET 178 is wrong INSTEAD it should be 172. The logic is ×1+1, ×2+4, ×3+9, ×4+16, ×5+25,…. So following the logic we get 178 is wrong instead it should be 172.

3.

Solve 2 3 10 39 178 885 ?

Answer»

The LOGIC is ×1+1, ×2+4, ×3+9, ×4+16, ×5+25,….

So following the logic we GET 178 is wrong INSTEAD it should be 172.

The logic is ×1+1, ×2+4, ×3+9, ×4+16, ×5+25,….

So following the logic we get 178 is wrong instead it should be 172.

Discussion

4.	A And B Can Do A Piece Of Work In 30 Days, While B And C Can Do The Same Work In 24 Days And C And A In 20 Days. They All Work Together For 10 Days When B And C Leave. How Many Days More Will A Take To Finish The Work ?
Answer» (A &AMP; B)'s 1 day work = 1/30 (B & C)'s 1 day work = 1/24 (C & A)'s 1 day work = 1/20 so 2 (A + B + C)'s 1 day's work = (1/30+1/24+1/20) = 15/120 = 1/8 => (A + B + C)'s 1 day's work = 1/16 Work done by A, B and C in 10 days = (101/16) = 5/8 so LEFT work = (1?5/8)=3/8 A's 1 day's work (1/16?1/24)=1/48 => 1/48 part of work is done by A = 1 day. So, 3/8 part of work will be done by A = (48?3/8) = 63 = 18 days. (A & B)'s 1 day work = 1/30 (B & C)'s 1 day work = 1/24 (C & A)'s 1 day work = 1/20 so 2 (A + B + C)'s 1 day's work = (1/30+1/24+1/20) = 15/120 = 1/8 => (A + B + C)'s 1 day's work = 1/16 Work done by A, B and C in 10 days = (101/16) = 5/8 so left work = (1?5/8)=3/8 A's 1 day's work (1/16?1/24)=1/48 => 1/48 part of work is done by A = 1 day. So, 3/8 part of work will be done by A = (48?3/8) = 63 = 18 days.

4.

A And B Can Do A Piece Of Work In 30 Days, While B And C Can Do The Same Work In 24 Days And C And A In 20 Days. They All Work Together For 10 Days When B And C Leave. How Many Days More Will A Take To Finish The Work ?

Answer»

(A &AMP; B)'s 1 day work = 1/30

(B & C)'s 1 day work = 1/24

(C & A)'s 1 day work = 1/20

so 2 (A + B + C)'s 1 day's work = (1/30+1/24+1/20) = 15/120 = 1/8

=> (A + B + C)'s 1 day's work = 1/16

Work done by A, B and C in 10 days = (10*1/16) = 5/8

so LEFT work = (1?5/8)=3/8

A's 1 day's work (1/16?1/24)=1/48

=> 1/48 part of work is done by A = 1 day.

So, 3/8 part of work will be done by A = (48?3/8) = 6*3 = 18 days.

(A & B)'s 1 day work = 1/30

(B & C)'s 1 day work = 1/24

(C & A)'s 1 day work = 1/20

so 2 (A + B + C)'s 1 day's work = (1/30+1/24+1/20) = 15/120 = 1/8

=> (A + B + C)'s 1 day's work = 1/16

Work done by A, B and C in 10 days = (10*1/16) = 5/8

so left work = (1?5/8)=3/8

A's 1 day's work (1/16?1/24)=1/48

=> 1/48 part of work is done by A = 1 day.

So, 3/8 part of work will be done by A = (48?3/8) = 6*3 = 18 days.

Discussion

5.	X And Y Entered Into Partnership With Rs. 700 And Rs. 600 Respectively. After 3 Months X Withdrew 2/7 Of His Stock But After 3 Months, He Puts Back 3/5 Of What He Had Withdrawn. The Profit At The End Of The Year Is Rs. 726. How Much Of This Should X Receive ?
Answer» X’s profit : Y’s profit = 700 × 3 + 500 × 3 + 620 × 6 : 600 × 12 = 2,100 + 1,500 + 3,720 : 7,200 = 7,320 : 7,200 = 61 : 60 X’s share in the profit = 61/(60+61) × 726 = 366 X’s profit : Y’s profit = 700 × 3 + 500 × 3 + 620 × 6 : 600 × 12 = 2,100 + 1,500 + 3,720 : 7,200 = 7,320 : 7,200 = 61 : 60 X’s share in the profit = 61/(60+61) × 726 = 366

5.

X And Y Entered Into Partnership With Rs. 700 And Rs. 600 Respectively. After 3 Months X Withdrew 2/7 Of His Stock But After 3 Months, He Puts Back 3/5 Of What He Had Withdrawn. The Profit At The End Of The Year Is Rs. 726. How Much Of This Should X Receive ?

Answer»

X’s profit : Y’s profit

= 700 × 3 + 500 × 3 + 620 × 6 : 600 × 12

= 2,100 + 1,500 + 3,720 : 7,200

= 7,320 : 7,200

= 61 : 60

X’s share in the profit = 61/(60+61) × 726 = 366

X’s profit : Y’s profit

= 700 × 3 + 500 × 3 + 620 × 6 : 600 × 12

= 2,100 + 1,500 + 3,720 : 7,200

= 7,320 : 7,200

= 61 : 60

X’s share in the profit = 61/(60+61) × 726 = 366

Discussion

6.	X And Y Can Do A Piece Of Work In 20 Days And 12 Days Respectively. X Started The Work Alone And Then After 4 Days Y Joined Him Till The Completion Of The Work. How Long Did The Work Last ?
Answer» X ONE day work = 1/20 y one day work = 1/12 work done by x in 4 DAYS = 4 * 1/20 = 1/5 left work = (1-1/5) = 4/5 x and y one day work = (1/20 + 1/12) = 8/60 = 2/15 => time required to do 2/15 PART of work by x and y = 1 day so for WHOLE work = 1/(2/15) = 15/2 so for 4/5 part of work x and y will take =( 4/515/2 ) = 6 days. => How long did the work LAST = 4 day + 6 day = 10 days. X one day work = 1/20 y one day work = 1/12 work done by x in 4 days = 4 1/20 = 1/5 left work = (1-1/5) = 4/5 x and y one day work = (1/20 + 1/12) = 8/60 = 2/15 => time required to do 2/15 part of work by x and y = 1 day so for whole work = 1/(2/15) = 15/2 so for 4/5 part of work x and y will take =( 4/5*15/2 ) = 6 days. => How long did the work last = 4 day + 6 day = 10 days.

6.

X And Y Can Do A Piece Of Work In 20 Days And 12 Days Respectively. X Started The Work Alone And Then After 4 Days Y Joined Him Till The Completion Of The Work. How Long Did The Work Last ?

Answer»

X ONE day work = 1/20

y one day work = 1/12

work done by x in 4 DAYS = 4 * 1/20 = 1/5

left work = (1-1/5) = 4/5

x and y one day work = (1/20 + 1/12) = 8/60 = 2/15

=> time required to do 2/15 PART of work by x and y = 1 day

so for WHOLE work = 1/(2/15) = 15/2

so for 4/5 part of work x and y will take =( 4/5*15/2 ) = 6 days.

=> How long did the work LAST = 4 day + 6 day = 10 days.

X one day work = 1/20

y one day work = 1/12

work done by x in 4 days = 4 * 1/20 = 1/5

left work = (1-1/5) = 4/5

x and y one day work = (1/20 + 1/12) = 8/60 = 2/15

=> time required to do 2/15 part of work by x and y = 1 day

so for whole work = 1/(2/15) = 15/2

so for 4/5 part of work x and y will take =( 4/5*15/2 ) = 6 days.

=> How long did the work last = 4 day + 6 day = 10 days.

Discussion

7.	X Can Do 1/4 Of A Work In 10 Days, Y Can Do 40% Of The Work In 40 Days And Z Can Do 1/3 Of The Work In 13 Days. Who Will Complete The Work First ?
Answer» X can do 1/4 of work in = 10 days so x can do WHOLE work in = (10 x 4) = 40 days. Y can do (40% or 40/100)of work in = 40 days so Whole work can be done by Y = (40x100/40)= 100 days. Z can do 1/3 of work in = 13 days Whole work will be done by Z in (13 x 3) = 39 days. so compare x , y ,z work compare = y &GT; x > z so Z can complete the work first. x can do 1/4 of work in = 10 days so x can do whole work in = (10 x 4) = 40 days. Y can do (40% or 40/100)of work in = 40 days so Whole work can be done by Y = (40x100/40)= 100 days. Z can do 1/3 of work in = 13 days Whole work will be done by Z in (13 x 3) = 39 days. so compare x , y ,z work compare = y > x > z so Z can complete the work first.

7.

X Can Do 1/4 Of A Work In 10 Days, Y Can Do 40% Of The Work In 40 Days And Z Can Do 1/3 Of The Work In 13 Days. Who Will Complete The Work First ?

Answer»

X can do 1/4 of work in = 10 days

so x can do WHOLE work in = (10 x 4) = 40 days.

Y can do (40% or 40/100)of work in = 40 days

so Whole work can be done by Y = (40x100/40)= 100 days.

Z can do 1/3 of work in = 13 days

Whole work will be done by Z in (13 x 3) = 39 days.

so compare x , y ,z work compare = y &GT; x > z

so Z can complete the work first.

x can do 1/4 of work in = 10 days

so x can do whole work in = (10 x 4) = 40 days.

Y can do (40% or 40/100)of work in = 40 days

so Whole work can be done by Y = (40x100/40)= 100 days.

Z can do 1/3 of work in = 13 days

Whole work will be done by Z in (13 x 3) = 39 days.

so compare x , y ,z work compare = y > x > z

so Z can complete the work first.

Discussion

8.	A Man Sitting In A Train Travelling At The Rate Of 50 Km/hr Observes That It Takes 9 Sec For A Goods Train Travelling In The Opposite Direction To Pass Him. If The Goods Train Is 187.5 M Long, Find Its Speed.
Answer» LET REQUIRED SPEED be X. So,187.5/{ (x+50)5/18} =9 Let required speed be x. So,187.5/{ (x+50)5/18} =9

8.

A Man Sitting In A Train Travelling At The Rate Of 50 Km/hr Observes That It Takes 9 Sec For A Goods Train Travelling In The Opposite Direction To Pass Him. If The Goods Train Is 187.5 M Long, Find Its Speed.

Answer»

LET REQUIRED SPEED be X.

So,187.5/{ (x+50)*5/18} =9

Let required speed be x.

So,187.5/{ (x+50)*5/18} =9

Discussion

9.	A' And 'b' Complete A Work Togather In 8 Days.if 'a' Alone Can Do It In 12 Days.then How Many Day 'b' Will Take To Complete The Work?
Answer» A & B one day WORK = 1/8 A alone one day work = 1/12 B alone one day work = (1/8 - 1/12) = ( 3/24 - 2/24) => B one day work = 1/24 so B can complete the work in 24 DAYS. A & B one day work = 1/8 A alone one day work = 1/12 B alone one day work = (1/8 - 1/12) = ( 3/24 - 2/24) => B one day work = 1/24 so B can complete the work in 24 days.

9.

A' And 'b' Complete A Work Togather In 8 Days.if 'a' Alone Can Do It In 12 Days.then How Many Day 'b' Will Take To Complete The Work?

Answer»

A & B one day WORK = 1/8

A alone one day work = 1/12

B alone one day work = (1/8 - 1/12) = ( 3/24 - 2/24)

=> B one day work = 1/24

so B can complete the work in 24 DAYS.

A & B one day work = 1/8

A alone one day work = 1/12

B alone one day work = (1/8 - 1/12) = ( 3/24 - 2/24)

=> B one day work = 1/24

so B can complete the work in 24 days.

Discussion

10.	Find The Next Term Of The Following Series. 1, 1, 3, 9, 11, 121,?
Answer» 11=1, 1+2=3, 33=9, 9+2=11, 1111=121, 121+2=123 11=1, 1+2=3, 33=9, 9+2=11, 1111=121, 121+2=123

10.

Find The Next Term Of The Following Series. 1, 1, 3, 9, 11, 121,?

Answer»

1*1=1, 1+2=3, 3*3=9, 9+2=11, 11*11=121, 121+2=123

Discussion

11.	The Product Of Two Numbers Is 9375 And The Quotient, When The Larger One Is Divided By The Smaller, Is 15. The Sum Of The Numbers Is?
Answer» Let the numbers be X and y. Then, XY = 9375 and x/y=15 Xy/(x/y)=9375/15 y2 = 625. y = 25. x = 15Y = (15 x 25) = 375. Sum of the numbers = x + y = 375 + 25 = 400. Let the numbers be x and y. Then, xy = 9375 and x/y=15 Xy/(x/y)=9375/15 y2 = 625. y = 25. x = 15y = (15 x 25) = 375. Sum of the numbers = x + y = 375 + 25 = 400.

11.

The Product Of Two Numbers Is 9375 And The Quotient, When The Larger One Is Divided By The Smaller, Is 15. The Sum Of The Numbers Is?

Answer»

Let the numbers be X and y.

Then, XY = 9375 and x/y=15 Xy/(x/y)=9375/15 y2 = 625. y = 25. x = 15Y = (15 x 25) = 375.

Sum of the numbers = x + y = 375 + 25 = 400.

Let the numbers be x and y.

Then, xy = 9375 and x/y=15 Xy/(x/y)=9375/15 y2 = 625. y = 25. x = 15y = (15 x 25) = 375.

Sum of the numbers = x + y = 375 + 25 = 400.

Discussion

12.	Ravi's Salary Was Reduced By 25%.percentage Increase To Be Effected To Bring The Salary To The Original Level Is
Answer» Explanation: METHOD: 1 LET's assume Ravi SALARY = 100 It get reduced by 25% => Salary = 75 75(1 + P/100) = 100 1+ P/100 = 4/3 P = 100/3 = 33 1/3%. Method: 2 You can use directly formula i.e [(R100)/(100-R)]% Where 'R' is decresed % so put 25 at place of 'R' => [(25 100)/(100 - 25)]% => [(25 100)/75]% =>100/3% = 33 1/3% Explanation:* Method: 1 Let's assume Ravi salary = 100 It get reduced by 25% => Salary = 75 75(1 + P/100) = 100 1+ P/100 = 4/3 P = 100/3 = 33 1/3%. Method: 2 You can use directly formula i.e [(R100)/(100-R)]% Where 'R' is decresed % so put 25 at place of 'R' => [(25 100)/(100 - 25)]% => [(25 *100)/75]% =>100/3% = 33 1/3%

12.

Ravi's Salary Was Reduced By 25%.percentage Increase To Be Effected To Bring The Salary To The Original Level Is

Answer»

Explanation:

METHOD: 1

LET's assume Ravi SALARY = 100

It get reduced by 25% => Salary = 75

75(1 + P/100) = 100

1+ P/100 = 4/3

P = 100/3 = 33 1/3%.

Method: 2

You can use directly formula i.e

[(R*100)/(100-R)]% Where 'R' is decresed %

so put 25 at place of 'R'

=> [(25 * 100)/(100 - 25)]% => [(25 *100)/75]%

=>100/3% = 33 1/3%

Explanation:

Method: 1

Let's assume Ravi salary = 100

It get reduced by 25% => Salary = 75

75(1 + P/100) = 100

1+ P/100 = 4/3

P = 100/3 = 33 1/3%.

Method: 2

You can use directly formula i.e

[(R*100)/(100-R)]% Where 'R' is decresed %

so put 25 at place of 'R'

=> [(25 * 100)/(100 - 25)]% => [(25 *100)/75]%

=>100/3% = 33 1/3%

Discussion

13.	A Can Do A Work In 10 Days And B Can Do The Same Work In 15 Days. So How Many Days They Will Take To Finish The Same Work ?
Answer» First find the 1 DAY work of both (A &AMP; B) A 1 day's work = 1/10 and B 1 day's work = 1/15 So (A + B) 1 day's work = (1/10+1/15) = (3/30+2/30) = 5/30 = 1/6 So Both (A & B) together can finish work in 6 DAYS First find the 1 day work of both (A & B) A 1 day's work = 1/10 and B 1 day's work = 1/15 So (A + B) 1 day's work = (1/10+1/15) = (3/30+2/30) = 5/30 = 1/6 So Both (A & B) together can finish work in 6 days

13.

A Can Do A Work In 10 Days And B Can Do The Same Work In 15 Days. So How Many Days They Will Take To Finish The Same Work ?

Answer»

First find the 1 DAY work of both (A &AMP; B)

A 1 day's work = 1/10

and

B 1 day's work = 1/15

So (A + B) 1 day's work = (1/10+1/15)

= (3/30+2/30) = 5/30 = 1/6

So Both (A & B) together can finish work in 6 DAYS

First find the 1 day work of both (A & B)

A 1 day's work = 1/10

and

B 1 day's work = 1/15

So (A + B) 1 day's work = (1/10+1/15)

= (3/30+2/30) = 5/30 = 1/6

So Both (A & B) together can finish work in 6 days

Discussion

14.	A And B Can Finish A Piece Of Work In 20 Days .b And C In 30 Days And C And A In 40 Days. In How Many Days Will A Alone Finish The Job ?
Answer» FIND ONE day work for all three (A+B)'s 1 day work = 1/20 ----(1) (B+C)'s 1 day work = 1/30 ----(2) and (C+A)'s 1 day work = 1/40 ----(3) 2(A+B+C) 1 day work = (1/20 + 1/30 + 1/40) => (A+B+C) = (6+4+3)/2120 => (A+B+C) = 13/240 -----------(4) By eq. (2) and (4) A + 1/30 = 13/240 => A = 13/240 - 1/30 = (13-8)/240 = 1/48 then A's 1 day work = 1/48 so A alon can FINISH the job = 48 days Find one day work for all three (A+B)'s 1 day work = 1/20 ----(1) (B+C)'s 1 day work = 1/30 ----(2) and (C+A)'s 1 day work = 1/40 ----(3) 2(A+B+C) 1 day work = (1/20 + 1/30 + 1/40) => (A+B+C) = (6+4+3)/2120 => (A+B+C) = 13/240 -----------(4) By eq. (2) and (4) A + 1/30 = 13/240 => A = 13/240 - 1/30 = (13-8)/240 = 1/48 then A's 1 day work = 1/48 so A alon can finish the job = 48 days

14.

A And B Can Finish A Piece Of Work In 20 Days .b And C In 30 Days And C And A In 40 Days. In How Many Days Will A Alone Finish The Job ?

Answer»

FIND ONE day work for all three

(A+B)'s 1 day work = 1/20 ----(1)

(B+C)'s 1 day work = 1/30 ----(2)

and (C+A)'s 1 day work = 1/40 ----(3)

2(A+B+C) 1 day work = (1/20 + 1/30 + 1/40)

=> (A+B+C) = (6+4+3)/2*120

=> (A+B+C) = 13/240 -----------(4)

By eq. (2) and (4)

A + 1/30 = 13/240

=> A = 13/240 - 1/30 = (13-8)/240 = 1/48

then A's 1 day work = 1/48

so A alon can FINISH the job = 48 days

Find one day work for all three

(A+B)'s 1 day work = 1/20 ----(1)

(B+C)'s 1 day work = 1/30 ----(2)

and (C+A)'s 1 day work = 1/40 ----(3)

2(A+B+C) 1 day work = (1/20 + 1/30 + 1/40)

=> (A+B+C) = (6+4+3)/2*120

=> (A+B+C) = 13/240 -----------(4)

By eq. (2) and (4)

A + 1/30 = 13/240

=> A = 13/240 - 1/30 = (13-8)/240 = 1/48

then A's 1 day work = 1/48

so A alon can finish the job = 48 days

Discussion

15.	After Decreasing 24% In The Cost Price Of An Article,its Costs Rs.912. Find The Actual Cost Of An Article?
Answer» CP* (76/100) = 912 => CP = 912 * 100/76 CP= 12 * 100 => CP = 1200 COST PRICE of article = RS. 1200 CP* (76/100) = 912 => CP = 912 * 100/76 CP= 12 * 100 => CP = 1200 cost price of article = Rs. 1200

15.

After Decreasing 24% In The Cost Price Of An Article,its Costs Rs.912. Find The Actual Cost Of An Article?

Answer»

CP* (76/100) = 912 => CP = 912 * 100/76

CP= 12 * 100

=> CP = 1200

COST PRICE of article = RS. 1200

CP* (76/100) = 912 => CP = 912 * 100/76

CP= 12 * 100

=> CP = 1200

cost price of article = Rs. 1200

Discussion

16.	In A Bag, There Are Coins Of 25 P, 10 P And 5 P In The Ratio Of 1 : 2 : 3. If There Is Rs. 30 In All, How Many 5 P Coins Are There ?
Answer» Let the NUMBER of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively. Then, SUM of their values=Rs[(25x/100)+(102x)/100+(53x)/100]=Rs 60x/100 60x/100=30 x=(30100)/60=50 Hence, the number of 5 p coins = (3 x 50) = 150. Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively. Then, sum of their values=Rs[(25x/100)+(102x)/100+(53x)/100]=Rs 60x/100 60x/100=30 x=(30100)/60=50 Hence, the number of 5 p coins = (3 x 50) = 150.

16.

In A Bag, There Are Coins Of 25 P, 10 P And 5 P In The Ratio Of 1 : 2 : 3. If There Is Rs. 30 In All, How Many 5 P Coins Are There ?

Answer»

Let the NUMBER of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.

Then, SUM of their values=Rs[(25x/100)+(10*2x)/100+(5*3x)/100]=Rs 60x/100 60x/100=30

x=(30*100)/60=50 Hence, the number of 5 p coins = (3 x 50) = 150.

Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.

Then, sum of their values=Rs[(25x/100)+(10*2x)/100+(5*3x)/100]=Rs 60x/100 60x/100=30

x=(30*100)/60=50 Hence, the number of 5 p coins = (3 x 50) = 150.

Discussion

17.	Two Number Are In The Ratio 3 : 5. If 9 Is Subtracted From Each, The New Numbers Are In The Ratio 12 : 23. The Smaller Number Is?
Answer» Let the numbers be 3x and 5X. Then ,(3x-9)/(5x-9)=12/13 23(3x – 9) = 12(5x – 9) 9x = 99 x = 11. The smaller NUMBER = (3 x 11) = 33. Let the numbers be 3x and 5x. Then ,(3x-9)/(5x-9)=12/13 23(3x – 9) = 12(5x – 9) 9x = 99 x = 11. The smaller number = (3 x 11) = 33.

17.

Two Number Are In The Ratio 3 : 5. If 9 Is Subtracted From Each, The New Numbers Are In The Ratio 12 : 23. The Smaller Number Is?

Answer»

Let the numbers be 3x and 5X.

Then ,(3x-9)/(5x-9)=12/13 23(3x – 9) = 12(5x – 9) 9x = 99 x = 11.

The smaller NUMBER = (3 x 11) = 33.

Let the numbers be 3x and 5x.

Then ,(3x-9)/(5x-9)=12/13 23(3x – 9) = 12(5x – 9) 9x = 99 x = 11.

The smaller number = (3 x 11) = 33.

Discussion

18.	Two Trains Each 100 M Long, Moving In Opposite Directions, Cross Each Other In 8 Seconds. If One Is Moving Twice As Fast The Other, Then The Speed Of The Faster Train Is ?
Answer» SPEED of the faster train = 2x m/sec. RELATIVE speed of train = (x + 2x) m/sec = 3x m/sec. Total DISTANCE = (100 + 100)m = 200m 3x = 200/8 => 24x = 200 => x = 25/3 So speed of the faster train = 2 * 25/3 m/sec = 50/3 m/sec = 50/3 * 18/5 = 60 km/hr. speed of the faster train = 2x m/sec. Relative speed of train = (x + 2x) m/sec = 3x m/sec. Total distance = (100 + 100)m = 200m 3x = 200/8 => 24x = 200 => x = 25/3 So speed of the faster train = 2 * 25/3 m/sec = 50/3 m/sec = 50/3 * 18/5 = 60 km/hr.

18.

Two Trains Each 100 M Long, Moving In Opposite Directions, Cross Each Other In 8 Seconds. If One Is Moving Twice As Fast The Other, Then The Speed Of The Faster Train Is ?

Answer»

SPEED of the faster train = 2x m/sec.

RELATIVE speed of train = (x + 2x) m/sec = 3x m/sec.

Total DISTANCE = (100 + 100)m = 200m

3x = 200/8

=> 24x = 200 => x = 25/3

So speed of the faster train = 2 * 25/3 m/sec

= 50/3 m/sec

= 50/3 * 18/5 = 60 km/hr.

speed of the faster train = 2x m/sec.

Relative speed of train = (x + 2x) m/sec = 3x m/sec.

Total distance = (100 + 100)m = 200m

3x = 200/8

=> 24x = 200 => x = 25/3

So speed of the faster train = 2 * 25/3 m/sec

= 50/3 m/sec

= 50/3 * 18/5 = 60 km/hr.

Discussion

19.	A Bag Contains 4 White, 5 Red And 6 Blue Balls. Three Balls Are Drawn At Random From The Bag. The Probability That All Of Them Are Red, Is?
Answer» Let S be the sample space. Then, N(S)= number of ways of DRAWING 3 balls out of 15 = 15C3 = (15 X 14 x 13)/ (3 x 2 x 1) = 455. Let E = event of getting all the 3 red balls. n(E) = 5C3 = 5C2 = (5 x 4)/ (2 x 1)= 10. P(E) = n(E) / n(S)= 10/ 455 = 2/ 91 Let S be the sample space. Then, n(S)= number of ways of drawing 3 balls out of 15 = 15C3 = (15 x 14 x 13)/ (3 x 2 x 1) = 455. Let E = event of getting all the 3 red balls. n(E) = 5C3 = 5C2 = (5 x 4)/ (2 x 1)= 10. P(E) = n(E) / n(S)= 10/ 455 = 2/ 91

19.

A Bag Contains 4 White, 5 Red And 6 Blue Balls. Three Balls Are Drawn At Random From The Bag. The Probability That All Of Them Are Red, Is?

Answer»

Let S be the sample space. Then, N(S)= number of ways of DRAWING 3 balls out of 15 = 15C3 = (15 X 14 x 13)/ (3 x 2 x 1) = 455.

Let E = event of getting all the 3 red balls. n(E) = 5C3 = 5C2 = (5 x 4)/ (2 x 1)= 10.

P(E) = n(E) / n(S)= 10/ 455 = 2/ 91

Let S be the sample space. Then, n(S)= number of ways of drawing 3 balls out of 15 = 15C3 = (15 x 14 x 13)/ (3 x 2 x 1) = 455.

Let E = event of getting all the 3 red balls. n(E) = 5C3 = 5C2 = (5 x 4)/ (2 x 1)= 10.

P(E) = n(E) / n(S)= 10/ 455 = 2/ 91

Discussion

20.	Two Cards Are Drawn Together From A Pack Of 52 Cards. The Probability That One Is A Spade And One Is A Heart, Is ?
Answer» Let S be the SAMPLE space. Then, n(S) = 52C2 = 52 * 51/ (21)= 1326. Let E = EVENT of GETTING 1 SPADE and 1 heart. n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = (13C1 x 13C1) = (13 x 13) = 169. P(E) =n (E)/n(S) = 169/ 1326= 13/102 Let S be the sample space. Then, n(S) = 52C2 = 52 51/ (2*1)= 1326. Let E = event of getting 1 spade and 1 heart. n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = (13C1 x 13C1) = (13 x 13) = 169. P(E) =n (E)/n(S) = 169/ 1326= 13/102

20.

Two Cards Are Drawn Together From A Pack Of 52 Cards. The Probability That One Is A Spade And One Is A Heart, Is ?

Answer»

Let S be the SAMPLE space.

Then, n(S) = 52C2 = 52 * 51/ (2*1)= 1326.

Let E = EVENT of GETTING 1 SPADE and 1 heart. n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = (13C1 x 13C1)

= (13 x 13) = 169. P(E) =n (E)/n(S) = 169/ 1326= 13/102

Let S be the sample space.

Then, n(S) = 52C2 = 52 * 51/ (2*1)= 1326.

Let E = event of getting 1 spade and 1 heart. n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = (13C1 x 13C1)

= (13 x 13) = 169. P(E) =n (E)/n(S) = 169/ 1326= 13/102

Discussion

21.	One Card Is Drawn At Random From A Pack Of 52 Cards. What Is The Probability That The Card Drawn Is A Face Card ?
Answer» Clearly, there are 52 CARDS, out of which there are 12 FACE cards. P (getting a face card)=12/52=3/13 Clearly, there are 52 cards, out of which there are 12 face cards. P (getting a face card)=12/52=3/13

21.

One Card Is Drawn At Random From A Pack Of 52 Cards. What Is The Probability That The Card Drawn Is A Face Card ?

Answer»

Clearly, there are 52 CARDS, out of which there are 12 FACE cards. P (getting a face card)=12/52=3/13

Clearly, there are 52 cards, out of which there are 12 face cards. P (getting a face card)=12/52=3/13

Discussion

22.	A Thief Steals A Car At 2.30 P.m And Drives It At 60 Kmph. The Theft Is Discovered At 3 P.m And The Owner Sets Off In Another Car At 75 Kmph. When Will Be Overtake The Thief
Answer» As THEFT is discovered at 3:00pm but Thief STOLE the car at 2:30. This means thief COVERED some DISTANCE in this 30 min gap. Distance travelled by thief in 30 min = 60 * 1/2 = 30 km Owner Discovered Car at 3:00pm Now relative speed = (75-60)km/hr = 15km/hr Time needed to travel 30km by the speed of 15km/hr. Time at which owner MEETS thief = 30/15 = 2 hrs So After 2 hrs (i.e,at 5:00 pm) the owner will catch/overtake the thief As Theft is discovered at 3:00pm but Thief stole the car at 2:30. This means thief covered some distance in this 30 min gap. Distance travelled by thief in 30 min = 60 * 1/2 = 30 km Owner Discovered Car at 3:00pm Now relative speed = (75-60)km/hr = 15km/hr Time needed to travel 30km by the speed of 15km/hr. Time at which owner meets thief = 30/15 = 2 hrs So After 2 hrs (i.e,at 5:00 pm) the owner will catch/overtake the thief

22.

A Thief Steals A Car At 2.30 P.m And Drives It At 60 Kmph. The Theft Is Discovered At 3 P.m And The Owner Sets Off In Another Car At 75 Kmph. When Will Be Overtake The Thief

Answer»

As THEFT is discovered at 3:00pm but Thief STOLE the car at 2:30.

This means thief COVERED some DISTANCE in this 30 min gap.

Distance travelled by thief in 30 min = 60 * 1/2 = 30 km

Owner Discovered Car at 3:00pm

Now relative speed = (75-60)km/hr = 15km/hr

Time needed to travel 30km by the speed of 15km/hr.

Time at which owner MEETS thief = 30/15 = 2 hrs

So After 2 hrs (i.e,at 5:00 pm) the owner will catch/overtake the thief

As Theft is discovered at 3:00pm but Thief stole the car at 2:30.

This means thief covered some distance in this 30 min gap.

Distance travelled by thief in 30 min = 60 * 1/2 = 30 km

Owner Discovered Car at 3:00pm

Now relative speed = (75-60)km/hr = 15km/hr

Time needed to travel 30km by the speed of 15km/hr.

Time at which owner meets thief = 30/15 = 2 hrs

So After 2 hrs (i.e,at 5:00 pm) the owner will catch/overtake the thief

Discussion

23.	The Salaries A, B, C Are In The Ratio 2 : 3 : 5. If The Increments Of 15%, 10% And 20% Are Allowed Respectively In Their Salaries, Then What Will Be New Ratio Of Their Salaries ?
Answer» Let A = 2k, B = 3k and C = 5k. A’s new salary = 115/ 100 of 2k =[115/ 100 x 2k]= 23k/ 10 B’s new salary = 110/ 100 of 3k = [110/ 100X 3k] = 33k/ 10 C’s new salary = 120/ 100 of 5k = [120/ 100 x 5k] = 6k New ratio = 23k : 33k : 6k = 23 : 33 : 60 Let A = 2k, B = 3k and C = 5k. A’s new salary = 115/ 100 of 2k =[115/ 100 x 2k]= 23k/ 10 B’s new salary = 110/ 100 of 3k = [110/ 100x 3k] = 33k/ 10 C’s new salary = 120/ 100 of 5k = [120/ 100 x 5k] = 6k New ratio = 23k : 33k : 6k = 23 : 33 : 60

23.

The Salaries A, B, C Are In The Ratio 2 : 3 : 5. If The Increments Of 15%, 10% And 20% Are Allowed Respectively In Their Salaries, Then What Will Be New Ratio Of Their Salaries ?

Answer»

Let A = 2k, B = 3k and C = 5k.

A’s new salary = 115/ 100 of 2k =[115/ 100 x 2k]= 23k/ 10

B’s new salary = 110/ 100 of 3k = [110/ 100X 3k] = 33k/ 10

C’s new salary = 120/ 100 of 5k = [120/ 100 x 5k] = 6k

New ratio = 23k : 33k : 6k = 23 : 33 : 60

Let A = 2k, B = 3k and C = 5k.

A’s new salary = 115/ 100 of 2k =[115/ 100 x 2k]= 23k/ 10

B’s new salary = 110/ 100 of 3k = [110/ 100x 3k] = 33k/ 10

C’s new salary = 120/ 100 of 5k = [120/ 100 x 5k] = 6k

New ratio = 23k : 33k : 6k = 23 : 33 : 60

Discussion

24.	Fresh Fruit Contains 68% Water And Dry Fruit Contains 20% Water. How Much Dry Fruit Can Be Obtained From 100 Kg?
Answer» 1. Given FRESH fruit has 68% water, =&GT; Remaining 32% will be fruit content. 2. Given DRY fruit has 20% water => Remaining 80% is fruit content. Here assume WEIGHT of dry fruit = x kg. "fruit content in both the fresh fruit and dry fruit is the same" Fruit % in fresh-fruit = fruit% in dry-fruit so (32/100) * 100 = (80/100 )* x => x = 40 kg. 1. Given fresh fruit has 68% water, => Remaining 32% will be fruit content. 2. Given Dry fruit has 20% water => Remaining 80% is fruit content. Here assume weight of dry fruit = x kg. "fruit content in both the fresh fruit and dry fruit is the same" Fruit % in fresh-fruit = fruit% in dry-fruit so (32/100) * 100 = (80/100 )* x => x = 40 kg.

24.

Fresh Fruit Contains 68% Water And Dry Fruit Contains 20% Water. How Much Dry Fruit Can Be Obtained From 100 Kg?

Answer»

1. Given FRESH fruit has 68% water,

=&GT; Remaining 32% will be fruit content.

2. Given DRY fruit has 20% water

=> Remaining 80% is fruit content.

Here assume WEIGHT of dry fruit = x kg.

"fruit content in both the fresh fruit and dry fruit is the same"

Fruit % in fresh-fruit = fruit% in dry-fruit

so (32/100) * 100 = (80/100 )* x

=> x = 40 kg.

1. Given fresh fruit has 68% water,

=> Remaining 32% will be fruit content.

2. Given Dry fruit has 20% water

=> Remaining 80% is fruit content.

Here assume weight of dry fruit = x kg.

"fruit content in both the fresh fruit and dry fruit is the same"

Fruit % in fresh-fruit = fruit% in dry-fruit

so (32/100) * 100 = (80/100 )* x

=> x = 40 kg.

Discussion

25.	The Fourth Proportional To 5, 8, 15 Is?
Answer» LET the fourth PROPORTIONAL to 5, 8, 15 be x. Then, 5 : 8 : 15 : x 5x = (8 x 15) x=(815)/5=24 Let the fourth proportional to 5, 8, 15 be x. Then, 5 : 8 : 15 : x 5x = (8 x 15) x=(815)/5=24

25.

The Fourth Proportional To 5, 8, 15 Is?

Answer»

LET the fourth PROPORTIONAL to 5, 8, 15 be x.

Then, 5 : 8 : 15 : x 5x = (8 x 15) x=(8*15)/5=24

Let the fourth proportional to 5, 8, 15 be x.

Then, 5 : 8 : 15 : x 5x = (8 x 15) x=(8*15)/5=24

Discussion

26.	The Numerator Of A Fraction Is 4 Less Than Its Denominator. If The Numerator Is Decreased By 2 And The Denominator Is Increased By 1, Then The Denominator Becomes Eight Times The Numerator. Find The Fraction?
Answer» Original fraction = (x - 4)/x In case II, 8(x - 4 - 2) = x + 1 ⇒ 8X - 48 = x + 1 ⇒ 7x = 49 ⇒ x = 7 ∴Original fraction = (7 - 4)/7 = 3/7 Original fraction = (x - 4)/x In case II, 8(x - 4 - 2) = x + 1 ⇒ 8x - 48 = x + 1 ⇒ 7x = 49 ⇒ x = 7 ∴Original fraction = (7 - 4)/7 = 3/7

26.

The Numerator Of A Fraction Is 4 Less Than Its Denominator. If The Numerator Is Decreased By 2 And The Denominator Is Increased By 1, Then The Denominator Becomes Eight Times The Numerator. Find The Fraction?

Answer»

Original fraction = (x - 4)/x

In case II,

8(x - 4 - 2) = x + 1

⇒ 8X - 48 = x + 1

⇒ 7x = 49 ⇒ x = 7

∴Original fraction

= (7 - 4)/7 = 3/7

Original fraction = (x - 4)/x

In case II,

8(x - 4 - 2) = x + 1

⇒ 8x - 48 = x + 1

⇒ 7x = 49 ⇒ x = 7

∴Original fraction

= (7 - 4)/7 = 3/7

Discussion

27.	An Equation Of The Form Ax + By + C = 0 Where A ≠ 0, B ≠ 0, C = 0 Represents A Straight Line Which Passes Through?
Answer» Ax+by+C = 0 When c = 0 ax+by = 0 by = -ax ⇒ y = - ax/b when x = 0, y = 0 i.e., this line passes through the ORIGIN (0,0). Ax+by+c = 0 When c = 0 ax+by = 0 by = -ax ⇒ y = - ax/b when x = 0, y = 0 i.e., this line passes through the origin (0,0).

27.

An Equation Of The Form Ax + By + C = 0 Where A ≠ 0, B ≠ 0, C = 0 Represents A Straight Line Which Passes Through?

Answer»

Ax+by+C = 0

When c = 0

ax+by = 0

by = -ax ⇒ y = - ax/b

when x = 0, y = 0 i.e., this line passes through the ORIGIN (0,0).

Ax+by+c = 0

When c = 0

ax+by = 0

by = -ax ⇒ y = - ax/b

when x = 0, y = 0 i.e., this line passes through the origin (0,0).

Discussion

28.	A Number X When Divided By 289 Leaves 18 As A Remainder. The Same Number When Divided By 17 Leaves Y As A Remainder. The Value Of Y Is?
Answer» Here, the first divisor (289) is a multiple of second divisor (17) ∴ Required REMAINDER = Remainder obtained on dividing 18 by 17 = 1 Here, the first divisor (289) is a multiple of second divisor (17) ∴ Required remainder = Remainder obtained on dividing 18 by 17 = 1

28.

A Number X When Divided By 289 Leaves 18 As A Remainder. The Same Number When Divided By 17 Leaves Y As A Remainder. The Value Of Y Is?

Answer»

Here, the first divisor (289) is a multiple of second divisor (17)

∴ Required REMAINDER = Remainder obtained on dividing 18 by 17 = 1

Here, the first divisor (289) is a multiple of second divisor (17)

∴ Required remainder = Remainder obtained on dividing 18 by 17 = 1

Discussion

29.	If (4x - 3)/x + (4y - 3)/y + (4z - 3)/z = 0, Then The Value Of 1/x + 1/y + 1/z Is?
Answer» (4x - 3)/X + (4Y - 3)/y + (4Z - 3)/z = 0 => 4x/x - 3/x + 4y/y - 3/y + 4z/z - 3/z = 0 => 3/x + 3/y + 3/z = 4 + 4 + 4 = 12 => 1/x + 1/y + 1/z = 12/3 = 4 (4x - 3)/x + (4y - 3)/y + (4z - 3)/z = 0 => 4x/x - 3/x + 4y/y - 3/y + 4z/z - 3/z = 0 => 3/x + 3/y + 3/z = 4 + 4 + 4 = 12 => 1/x + 1/y + 1/z = 12/3 = 4

29.

If (4x - 3)/x + (4y - 3)/y + (4z - 3)/z = 0, Then The Value Of 1/x + 1/y + 1/z Is?

Answer»

(4x - 3)/X + (4Y - 3)/y + (4Z - 3)/z = 0

=> 4x/x - 3/x + 4y/y - 3/y + 4z/z - 3/z = 0

=> 3/x + 3/y + 3/z = 4 + 4 + 4 = 12

=> 1/x + 1/y + 1/z = 12/3 = 4

(4x - 3)/x + (4y - 3)/y + (4z - 3)/z = 0

=> 4x/x - 3/x + 4y/y - 3/y + 4z/z - 3/z = 0

=> 3/x + 3/y + 3/z = 4 + 4 + 4 = 12

=> 1/x + 1/y + 1/z = 12/3 = 4

Discussion

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