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Sections: English, Aptitude, Reasoning
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If FRAGRANCE is written as SBHSBODFG, how can IMPOSING be written?
Each letter in the word is moved one step forward and the first letter of the group so obtained is put at the end, to obtain the code.
Each letter in the word is moved one step forward and the first letter of the group so obtained is put at the end, to obtain the code.
In a certain code, COMPUTER is written as RFUVQNPC. How is MEDICINE written in the same code ?
The Letters of the word are written in a reverse order and each letter, except the first and the last one, is moved one step forward, to obtain the code.
The Letters of the word are written in a reverse order and each letter, except the first and the last one, is moved one step forward, to obtain the code.
If in a certain language FASHION is coded as FOIHSAN, how is PROBLEM coded in that code ?
The first and the last letters of the word remain as such and the remaining letters are written in a reverse order, to obtain the code.
The first and the last letters of the word remain as such and the remaining letters are written in a reverse order, to obtain the code.
If VICTORY is coded as YLFWRUB, how can SUCCESS be coded ?
Each letter of the word is moved three steps forward to obtain the code.
Each letter of the word is moved three steps forward to obtain the code.
A person who is the husband of my son’s sister is my
My son’s sister is my daughter. My daughter’s husband is my soninlaw.
My son’s sister is my daughter. My daughter’s husband is my soninlaw.
How is my grandmother’s only child’s husband’s mother related to me ?
My grandmother’s only child is my mother. My mother’s husband is my father. My father’s mother is my grandmother.
My grandmother’s only child is my mother. My mother’s husband is my father. My father’s mother is my grandmother.
How is my son’s son’s mother’s daughter related to me ?
My son’s son’s mother is my son’s wife. My son’s wife’s daughter is my granddaughter.
My son’s son’s mother is my son’s wife. My son’s wife’s daughter is my granddaughter.
How is Ravi’s mother’s father’s son related to Ravi’s father ?
Ravi’s mother’s father’s son is Ravi’s mother’s brother i.e. Ravi’s uncle. Ravi’s uncle is Ravi’s father’s brotherinlaw.
Ravi’s mother’s father’s son is Ravi’s mother’s brother i.e. Ravi’s uncle. Ravi’s uncle is Ravi’s father’s brotherinlaw.
Tinku, introducing a person to Rinku, said “He is the father of your sister’s son and he is also my mother’s husband”. How is Tinku’s father related to Rinku’s mother ?
He is the father of Rinku’s Sister’s son means he is Rinku’s Sister’s husband. He is also Tinku’s mother’s husband means Tinku’s mother is Rinku’s sister. Now, Tinku’s father is Rinku’s brotherinlaw. i.e. Tinku’s father is Rinku’s mother’s soninlaw.
He is the father of Rinku’s Sister’s son means he is Rinku’s Sister’s husband. He is also Tinku’s mother’s husband means Tinku’s mother is Rinku’s sister. Now, Tinku’s father is Rinku’s brotherinlaw. i.e. Tinku’s father is Rinku’s mother’s soninlaw.
In a row of trees, one tree is fifth from either end of the row. How many trees are there in the row?
Clearly, number of trees in the row = (4 + 1 + 4) = 9.
Clearly, number of trees in the row = (4 + 1 + 4) = 9.
In a queue, Amrita is 10th from the front while Mukul is 25th from behind and Mamta is just in the middle of the two. If there be 50 persons in the queue, what position does Mamta occupy from the front ?
Number of persons between Amrita and Mukul = 50 (10 + 25) = 15. Since Mamta lies in middle of these 15 persons,so Mamta’s position is 8th from Amrita i.e. 18th from the front.
Number of persons between Amrita and Mukul = 50 (10 + 25) = 15. Since Mamta lies in middle of these 15 persons,so Mamta’s position is 8th from Amrita i.e. 18th from the front.
Raman ranks sixteenth from the top and forty ninth from the bottom in a class. How many students are there in the class ?
Clearly, number of students in the class = (15 + 1 + 48) = 64.
Clearly, number of students in the class = (15 + 1 + 48) = 64.
Sanjeev ranks seventh from the top and twenty eight from the bottom in a class. How many students are there in the class ?
Clearly, number of students in the class = (6 + 1 + 27) = 34.
Clearly, number of students in the class = (6 + 1 + 27) = 34.
If Atul finds that he is twelfth from the right in a line of boys and fourth from the left, how many boys should be added to the line such that there are 28 boys in the line ?
Clearly, number of boys in the line = (11 + 1 + 3) = 15. Number of boys to be added = 28 – 15 = 13.
Clearly, number of boys in the line = (11 + 1 + 3) = 15. Number of boys to be added = 28 – 15 = 13.
Statement:
Statement:
Statement:
Statement:
If * stands for ‘addition’, / stands for ‘subtraction’, + stands for ‘multiplication’, and – stands for ‘division’, then 20 * 8 / 8 – 4 + 2 = ?
Using the correct symbols, we have: Given expression = 20 + 8 – 8 / 4 * 2 = 20 + 8 – 2 * 2 = 20 + 8 – 4 = 24.
Using the correct symbols, we have: Given expression = 20 + 8 – 8 / 4 * 2 = 20 + 8 – 2 * 2 = 20 + 8 – 4 = 24.
If – means *, * means +, + means / and / means , then 40 * 12 + 3 – 6 / 60 = ?
Using the correct symbols, we have : Given expression = 40 + 12 / 3 * 6 – 60 = 40 + 4 * 6 – 60 = 40 + 24 – 60 = 4
Using the correct symbols, we have : Given expression = 40 + 12 / 3 * 6 – 60 = 40 + 4 * 6 – 60 = 40 + 24 – 60 = 4
If + means /, * means , / means * and – means +, then 8 + 6 * 4 / 3 – 4 = ?
Using the correct symbols, we have: Given expression = 8 / 6 – 4 * 3 + 4 = 4 / 3 – 4 * 3 + 4 = 4 / 3 – 12 + 4 = – 20 / 3
Using the correct symbols, we have: Given expression = 8 / 6 – 4 * 3 + 4 = 4 / 3 – 4 * 3 + 4 = 4 / 3 – 12 + 4 = – 20 / 3
If * means /, – means *, / means + and + means , then (3 – 15 / 19) * 8 + 6 = ?
Using the correct symbols, we have: Given expression = (3 * 15 + 19) / 8 – 6 = (45 + 19) / 8 – 6 = 64 / 8 – 6 = 8 – 6 = 2
Using the correct symbols, we have: Given expression = (3 * 15 + 19) / 8 – 6 = (45 + 19) / 8 – 6 = 64 / 8 – 6 = 8 – 6 = 2
If + means *, / means , * means / and – means +, what will be the value of 4 + 11 / 5 – 55 = ?
Using the correct symbols, we have: Given expression = 4 * 11 – 5 + 55 = 44 – 5 + 55 = 94
Using the correct symbols, we have: Given expression = 4 * 11 – 5 + 55 = 44 – 5 + 55 = 94
Each of these questions are based on the information given below:
Who is sitting third to the right of O ?
If B shifts to the place of E, E shifts to the place of Q, and Q shifts to the place of B, then who will be the second to the left of the person opposite to O ?
Which of the following pair is diagonally opposite to each other ?
If O and P, A and E and B and Q interchange their positions, then who will be the second person to the right of the person who is opposite to the person second of the right of P ?
In the original arrangement who is sitting just opposite to N ?
A woman says, “If you reverse my own age, the figures represent my husband’s age. He is, of course, senior to me and the difference between our ages is oneeleventh of their sum.” The woman’s age is
Let x and y be the ten’s and unit’s digits respectively of the numeral denoting the woman’s age.
Then, woman’s age = (10X + y) years; husband’s age = (10y + x) years.
Therefore (10y + x) (10X + y) = (1/11) (10y + x + 10x + y)
(9y9x) = (1/11)(11y + 11x) = (x + y) 10x = 8y x = (4/5)y
Clearly, y should be a singledigit multiple of 5, which is 5.
So, x = 4, y = 5.
Hence, woman’s age = 10x + y = 45 years.
Let x and y be the ten’s and unit’s digits respectively of the numeral denoting the woman’s age.
Then, woman’s age = (10X + y) years; husband’s age = (10y + x) years.
Therefore (10y + x) (10X + y) = (1/11) (10y + x + 10x + y)
(9y9x) = (1/11)(11y + 11x) = (x + y) 10x = 8y x = (4/5)y
Clearly, y should be a singledigit multiple of 5, which is 5.
So, x = 4, y = 5.
Hence, woman’s age = 10x + y = 45 years.
A is 3 years older to B and 3 years younger to C, while B and D are twins. How many years older is C to D?
Explanation : Since B and D are twins, so B = D.
Now, A = B + 3 and A = C – 3.
Thus, B + 3 = C – 3 D + 3 = C3 C – D = 6.
Explanation : Since B and D are twins, so B = D.
Now, A = B + 3 and A = C – 3.
Thus, B + 3 = C – 3 D + 3 = C3 C – D = 6.
There are deer and peacocks in a zoo. By counting heads they are 80. The number of their legs is 200. How many peacocks are there ?
Explanation:
Let x and y be the number of deer and peacocks in the zoo respectively. Then,
x + y = 80 …(i) and
4x + 2y = 200 or 2x + y = 100 …(ii)
Solving (i) and (ii), we get) x = 20, y = 60.
Explanation:
Let x and y be the number of deer and peacocks in the zoo respectively. Then,
x + y = 80 …(i) and
4x + 2y = 200 or 2x + y = 100 …(ii)
Solving (i) and (ii), we get) x = 20, y = 60.
If you write down all the numbers from 1 to 100, then how many times do you write 3 ?
Explantion :
Clearly, from 1 to 100, there are ten numbers with 3 as the unit’s digit 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten’s digit – 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.
So, required number = 10 + 10 = 20.
Explantion :
Clearly, from 1 to 100, there are ten numbers with 3 as the unit’s digit 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten’s digit – 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.
So, required number = 10 + 10 = 20.
A pineapple costs Rs. 7 each. A watermelon costs Rs. 5 each. X spends Rs. 38 on these fruits. The number of pineapples purchased is
EXplanation:
Let the number of pineapples and watermelons be x and y respectively.
Then, 7x+5y=38 or 5y=(387x) or y=
Clearly, yis a whole number, only when (387x) is divisible by 5.
This happen when x=4.
EXplanation:
Let the number of pineapples and watermelons be x and y respectively.
Then, 7x+5y=38 or 5y=(387x) or y=
Clearly, yis a whole number, only when (387x) is divisible by 5.
This happen when x=4.
The lateeral surface area of cuboid length 12 m, breadth 8 m and height 6m.
Cuboid lateral surface = 2h(l+b)
= 2 × 6 (20) = 240 m^{2}
Cuboid lateral surface = 2h(l+b)
= 2 × 6 (20) = 240 m^{2}
The radius of a cone is 14 m, slant height is 20 m. Find the curved surface area?
Cone curved surface area = πrl
22/7 × 14 × 120 = 44 × 20 = 880 m^{2}
Cone curved surface area = πrl
22/7 × 14 × 120 = 44 × 20 = 880 m^{2}
If the radius of sphere is doubled, then its volume is increased by:
radius = r. Original volume = 4/3 πr^{3}
New radius = 2r New volume = 4/3 π(2r)^{3} = (32πr^{3})/3
increase in volume = ( (32πr^{3})/3 × 3/(4πr^{3}) × 100 ) % = 700%
radius = r. Original volume = 4/3 πr^{3}
New radius = 2r New volume = 4/3 π(2r)^{3} = (32πr^{3})/3
increase in volume = ( (32πr^{3})/3 × 3/(4πr^{3}) × 100 ) % = 700%
Steve traveled the first 2 hours of his journey at 40 mph and the remaining 3 hours of his journey at 80 mph. What is his average speed for the entire journey?
Average speed of travel = Total distance travelledTotal time takenTotal distance travelledTotal time taken
Total distance traveled by Steve = Distance covered in the first 2 hours + distance covered in the next 3 hours.
Distance covered in the first 2 hours = speed * time = 40 * 2 = 80 miles.
Distance covered in the next 3 hours = speed * time = 80 * 3 = 240 miles.
Therefore, total distance covered = 80 + 240 = 320 miles.
Total time taken = 2 + 3 = 5 hours.
Hence, average speed = Total distance travelled
Total time taken=320/5
Total distance travelled
Total time taken=320/5 = 64 miles per hour.
Average speed of travel = Total distance travelledTotal time takenTotal distance travelledTotal time taken
Total distance traveled by Steve = Distance covered in the first 2 hours + distance covered in the next 3 hours.
Distance covered in the first 2 hours = speed * time = 40 * 2 = 80 miles.
Distance covered in the next 3 hours = speed * time = 80 * 3 = 240 miles.
Therefore, total distance covered = 80 + 240 = 320 miles.
Total time taken = 2 + 3 = 5 hours.
Hence, average speed = Total distance travelled
Total time taken=320/5
Total distance travelled
Total time taken=320/5 = 64 miles per hour.
Show that 4832718 is divisible by 11.
(Sum of digits at odd places) – (Sum of digits at even places)
= (8 + 7 + 3 + 4) – (1 + 2 + 8) = 11, which is divisible by 11.
Hence, 4832718 is divisible by 11.
(Sum of digits at odd places) – (Sum of digits at even places)
= (8 + 7 + 3 + 4) – (1 + 2 + 8) = 11, which is divisible by 11.
Hence, 4832718 is divisible by 11.
What least number must be added to 3000 to obtain a number exactly divisible by 19 ?
On dividing 3000 by 19, we get 17 as remainder.
Number to be added = (19 – 17) = 2.
On dividing 3000 by 19, we get 17 as remainder.
Number to be added = (19 – 17) = 2.
Find the number which is nearest to 3105 and is exactly divisible by 21.
On dividing 3105 by 21, we get 18 as remainder.
Number to be added to 3105 = (21 – 18) – 3.
Hence, required number = 3105 + 3 = 3108.
On dividing 3105 by 21, we get 18 as remainder.
Number to be added to 3105 = (21 – 18) – 3.
Hence, required number = 3105 + 3 = 3108.
Walking at the rate of 4 kmph a man cover certain distance in 2 hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in.
Distance = Speed × time
Here time = 2hr 45 min = 11/4 hr
Distance = 4×11/4=11 km
New Speed =16.5 kmph
Therefore time = DS=11/16.5=40 min
Distance = Speed × time
Here time = 2hr 45 min = 11/4 hr
Distance = 4×11/4=11 km
New Speed =16.5 kmph
Therefore time = DS=11/16.5=40 min
A train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 40 seconds. What is the length of the train in meters?
When a train overtakes another object such as a motorbike, whose length is negligible compared to the length of the train, then the distance travelled by the train while overtaking the motorbike is the same as the length of the train.
The length of the train:
= distance travelled by the train while overtaking the motorbike
= relative speed between the train and the motorbike × time taken
In this case, as both the objects i.e., the train and the motorbike are moving in the same direction, the relative speed between them = difference between their respective speeds = 100−64=36 kmph
Distance travelled by the train while overtaking the motorbike
=36 kmph ×40 seconds.
The final answer is given in meters and the speed is given in kmph and the time in seconds.
So let us convert the given speed from kmph to m/sec.
1 kmph = 5/18 m/sec
Therefore, 36 kmph = 36×5/18=10m/sec
Relative speed = 10 m/sec.
Time taken = 40 seconds.
Therefore, distance travelled =10×4010×40= 400 meters
When a train overtakes another object such as a motorbike, whose length is negligible compared to the length of the train, then the distance travelled by the train while overtaking the motorbike is the same as the length of the train.
The length of the train:
= distance travelled by the train while overtaking the motorbike
= relative speed between the train and the motorbike × time taken
In this case, as both the objects i.e., the train and the motorbike are moving in the same direction, the relative speed between them = difference between their respective speeds = 100−64=36 kmph
Distance travelled by the train while overtaking the motorbike
=36 kmph ×40 seconds.
The final answer is given in meters and the speed is given in kmph and the time in seconds.
So let us convert the given speed from kmph to m/sec.
1 kmph = 5/18 m/sec
Therefore, 36 kmph = 36×5/18=10m/sec
Relative speed = 10 m/sec.
Time taken = 40 seconds.
Therefore, distance travelled =10×4010×40= 400 meters
Ram covers a part of the journey at 20 kmph and the balance at 70 kmph taking total of 8 hours to cover the distance of 400 km. How many hours has he been driving at 20 kmph?
Let xx be the number of hours he travels at 20 kmph.
Distance covered in this x hours = 20x km
He would have therefore, travelled the balance (8–x) hours at 70 kmph.
Distance covered in this 8–x hours = (8−x)×70=(560–70x) km
The total distance travelled = 20x+560–70x=400
⇒160=50x
x=(16050) hours
= 3.2 hours
= 3 hours and 12 minutes
Let xx be the number of hours he travels at 20 kmph.
Distance covered in this x hours = 20x km
He would have therefore, travelled the balance (8–x) hours at 70 kmph.
Distance covered in this 8–x hours = (8−x)×70=(560–70x) km
The total distance travelled = 20x+560–70x=400
⇒160=50x
x=(16050) hours
= 3.2 hours
= 3 hours and 12 minutes
If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students is:
Required average
=(55 x 50 + 60 x 55 + 45 x 60)/(55 + 60 + 45)
=(2750 + 3300 + 2700)/160
=8750/160
= 54.68
Required average
=(55 x 50 + 60 x 55 + 45 x 60)/(55 + 60 + 45)
=(2750 + 3300 + 2700)/160
=8750/160
= 54.68
In a class, the average age of 30 boys is 13 years and the average of 20 girls is 12 years. what is the average age of the whole class?
Total age of 50 students
(30X 13+20 X 12) = 630
Average = 630/50 = 12.6 Years
Total age of 50 students
(30X 13+20 X 12) = 630
Average = 630/50 = 12.6 Years
The average of 5 consecutive numbers is n. if the next two number are also included. The average will?
The salary of a person was reduced by 10% .By what percent should his reduced salary be raised so as to bring it at par with his original salary ?
Let the original salary be Rs.100 . New salary = Rs.90.
Increase on 90=10 , Increase on 100=((10/90)*100)%
= (100/9)%
Let the original salary be Rs.100 . New salary = Rs.90.
Increase on 90=10 , Increase on 100=((10/90)*100)%
= (100/9)%
Raman`s salary was decreased by 50% and subsequently increased by 50%.How much percent does he lose?
Let the original salary = Rs.100
New final salary=150% of (50% of Rs.100)=
Rs.((150/100)*(50/100)*100)=Rs.75.
Decrease = 25%
Let the original salary = Rs.100
New final salary=150% of (50% of Rs.100)=
Rs.((150/100)*(50/100)*100)=Rs.75.
Decrease = 25%
Sixty five percent of a number is 21 less than four fifth of that number. What is the number ?
Let the number be x.
Then, 4*x/5 –(65% of x) = 21
4x/5 –65x/100 = 21
5 x = 2100
x = 140.
Let the number be x.
Then, 4*x/5 –(65% of x) = 21
4x/5 –65x/100 = 21
5 x = 2100
x = 140.
A man bought a horse and a car riage for Rs 3000.he sold the horse at a gain of 20% and the carriage at a loss of 10%,thereby gaining 2% on the whole.find the cost of the horse.
Let the C.p of the horse be Rs.x, then C.P of the carriage =Rs(3000x)
20% of x10% of(3000x)=2% of 3000
Let the C.p of the horse be Rs.x, then C.P of the carriage =Rs(3000x)
20% of x10% of(3000x)=2% of 3000
find the single discount equivalent to a series discount of 20% ,10% and 5%
let the marked price be Rs 100
then ,net S.P=95% of 90% of 80% of Rs 100
=Rs(95/100*90/100*80/100*100)=Rs68.40
let the marked price be Rs 100
then ,net S.P=95% of 90% of 80% of Rs 100
=Rs(95/100*90/100*80/100*100)=Rs68.40
After getting 2 successive discounts, a shirt with a list price of Rs 150 is available at Rs 105. If the second discount is 12.55,find the first discount.
Let the first discount be x%
Then,87.5% of (100x)% of 150= 105
Let the first discount be x%
Then,87.5% of (100x)% of 150= 105
In what ratio should the profit of Rs.8000 be divided if X starts a business with an investment of Rs. 20000, Y invests Rs.7500 for 4 months and Z invests Rs.15000 after 3 months from the start of the business.
Let the profit of X be P1, that of Y be P2 and of Z be P3.
P1:P2:P3 = 20000*12 : 7500*4 : 15000*9 = 240 : 30 : 135 = 80 : 10 : 45
= 16 : 2 : 9
Let the profit of X be P1, that of Y be P2 and of Z be P3.
P1:P2:P3 = 20000*12 : 7500*4 : 15000*9 = 240 : 30 : 135 = 80 : 10 : 45
= 16 : 2 : 9
A, B and C distribute Rs. 1,000 among them. A and C have Rs. 400, B and C Rs. 700. How much does C have?
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.
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
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
A purse contains 342 coins consisting of one rupees, 50 paise and 25 paise coins. If their values are in the ratio of 11 : 9 : 5 then find the number of 50 paise coins?
Let the value of one rupee, 50 paise and 25 paise be 11x, 9x, 5x respectively.
No. of 1 rupee coins = (11x / 1) =11x
No. of 50 paise coins = (9x / 0.5) = 18x
No. of 25 paise coins = (5x / 0.25) = 20x
11x + 18x + 9x = 342
38x = 342
x = 9
Therefore, no. of 1 rupee coins = 11 x 9 = 99 coins
No. of 50 paise coins = 18 x 9 = 162 coins
No. of 25 paise coins = 20 x 9 = 180 coins
Let the value of one rupee, 50 paise and 25 paise be 11x, 9x, 5x respectively.
No. of 1 rupee coins = (11x / 1) =11x
No. of 50 paise coins = (9x / 0.5) = 18x
No. of 25 paise coins = (5x / 0.25) = 20x
11x + 18x + 9x = 342
38x = 342
x = 9
Therefore, no. of 1 rupee coins = 11 x 9 = 99 coins
No. of 50 paise coins = 18 x 9 = 162 coins
No. of 25 paise coins = 20 x 9 = 180 coins
20 women can do a work in 9 days. After they have worked for 6 days. 6 more men join them. How many days will they take to complete the remaining work?
(20 * 16) women can complete the work in 1 day.
(20 * 16) women can complete the work in 1 day.
Ronald and Elan are working on an assignment. Ronald takes 6 hrs to type 32 pages on a computer, while Elan takes 5 hrs to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?
Number of pages typed by Ronald in 1 hour = 32/6 = 16/3
Number of pages typed by Ronald in 1 hour = 32/6 = 16/3
A can do a piece of work in 4 hours; B and C together can do it in 3 hours, which A and C together can do it in 2 hours. How long will B alone take to do it?
A’s 1 hour work = 1/4;
A’s 1 hour work = 1/4;
A and B can do a work in 12 days, B and C in 15 days, C and A in 20 days. If A, B and C work together, they will complete the work in?
(A + B)’s 1 day work = 1/12;
(A + B)’s 1 day work = 1/12;
In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?
A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is:
If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 1200, find the compound interest on the same sum for the same period at the same rate.
Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200. . .
So principal=RS [100*1200]/3*5=RS 8000
Amount = Rs. 8000 x [1 +5/100]^3 – = Rs. 9261.
.. C.I. = Rs. (9261 – 8000) = Rs. 1261.
Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200. . .
So principal=RS [100*1200]/3*5=RS 8000
Amount = Rs. 8000 x [1 +5/100]^3 – = Rs. 9261.
.. C.I. = Rs. (9261 – 8000) = Rs. 1261.
In what time will Rs. 1000 become Rs. 1331 at 10% per annum compounded annually?
Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a. Let the time be n years. Then,
[ 1000 (1+ (10/100))^{n }] = 1331 or (11/10)^{n = }(1331/1000) = (11/10)^{3}
n = 3 years.
Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a. Let the time be n years. Then,
[ 1000 (1+ (10/100))^{n }] = 1331 or (11/10)^{n = }(1331/1000) = (11/10)^{3}
n = 3 years.
If Rs. 600 amounts to Rs. 683.20 in two years compounded annually, find the rate of interest per annum.
Principal = Rs. 500; Amount = Rs. 583.20; Time = 2 years.
Let the rate be R% per annum.. ‘Then,
[ 500 (1+(R/100)^{2 }] = 583.20 or [ 1+ (R/100)]^{2 = }5832/5000 = 11664/10000
[ 1+ (R/100)]^{2 }= (108/100)^{2} or 1 + (R/100) = 108/100 or R = 8
So, rate = 8% p.a.
Principal = Rs. 500; Amount = Rs. 583.20; Time = 2 years.
Let the rate be R% per annum.. ‘Then,
[ 500 (1+(R/100)^{2 }] = 583.20 or [ 1+ (R/100)]^{2 = }5832/5000 = 11664/10000
[ 1+ (R/100)]^{2 }= (108/100)^{2} or 1 + (R/100) = 108/100 or R = 8
So, rate = 8% p.a.
If the compound interest on a certain sum at 16 (2/3)% to 3 years is Rs.1270, find the simple interest on the same sum at the same rate and f or the same period.
Let the sum be Rs. x. Then,
C.I. = [ x * (1 + (( 50/(3*100))^{3} – x ] = ((343x / 216) – x) = 127x / 216
127x /216 = 1270 or x = (1270 * 216) / 127 = 2160.
Thus, the sum is Rs. 2160
S.I. = Rs ( 2160 * (50/3) * 3 * (1 /100 ) ) = Rs. 1080.
Let the sum be Rs. x. Then,
C.I. = [ x * (1 + (( 50/(3*100))^{3} – x ] = ((343x / 216) – x) = 127x / 216
127x /216 = 1270 or x = (1270 * 216) / 127 = 2160.
Thus, the sum is Rs. 2160
S.I. = Rs ( 2160 * (50/3) * 3 * (1 /100 ) ) = Rs. 1080.
A boy has nine trousers and 12 shirts. In how many different ways can he select a trouser and a shirt?
The boy can select one trouser in nine ways.
The boy can select one trouser in nine ways.
The least multiple of 25 which when divided by 16, 24 and 28 leaves remainder 7, 10, 13 respectively Is
The —————should understand that they would get a licence only if they have basic skills of driving.
The tunnel was so ……… and congested, that we became ………
Although he is a ……… person, he occasionally loses his ………
Centre should ________ministries whose functions ……….with the state ministries to save money, deliver efficiency and avoid duplication of work.
Many people________ genetically modified food but the reality is that all the food that we eat has been genetically modified naturally by thousands of years of ………..
Given that only seven per cent of the country’s labour force is in the organised sector, training options _____ for the unorganized sectors should also be _____ .
Indians will______ onefourth of total work force in the next five years but India needs to introspect whether its education system is _____ for these demographic dividends.
Rearrange the following sentences (A), (B), (C), (D), (E) and (F) in proper sequence to from a meaningful paragraph; then answer the questions given below them. (A) He felt that his honest ways were responsible for the poverty and starvation of his family. (B) Sixteen miles away from Mysore, there is a dense forest. (C) He told them how his honesty was useless and asked if he should try an alternative. (D) They told him that they would prefer starving to dishonesty. (E) Once there lived a poor but honest woodcutter in the forest. (F) So he wanted to discuss his feeling with his wife and children.
Question 1: Which of the following should be the SIXTH (LAST) sentence?
Question 2: Which of the following should be the FIRST sentence?
Question 3:Which of the following should be the FIFTH sentence?
Question 4 : Which of the following should be the SECOND sentence?
Question 5 : Which of the following should be the THIRD sentence?
The secretary ________ the society’s funds, ________ he was dismissed.
________ the broker had warned him that the stock was a ________ investment, he insisted on buying a thousand shares
Jamshedji Tata is …1… to be the pathfinder of modern industrial builders. He known as the grandfather of the Indian industry for his acumen and enthusiasm. Nobody else could have…2…. of the new industries started by Jamshedji at that time when industrial….3…. and revolution was yet to come to India.
Jamshedji’s father Nasarvanji Tata used to trade in jute with China and Britain. He started….4… from India. Jamshedji started a clothe mill in nagpur more than hundred years ago. At that time almost all the …5… used to come from Lancashire in England. What Jamshedji …6…. was praiseworthy.
Jamshedji ….7…. very well that an industrial revolution can only be brought in the country by setthing up iron and steel industry. ……8…… he did not live to see the industry he had in mind, he had done all …9…. work. In fact, he laid the ground work for it. He had planned the entire steel city now known as Jashedpur, complete with streets, roads, schools, parks, playgrounds, temples, mosques, churches, etc. His ….10.. was fulfilled by his sons, Sire Dorabji Tata and Sir Ratan Tata, When they started the Tata Iron & Steel Factory in 1907 just after three years of his death.
Question 9:
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I felt the wall of the tunnel shiver. The master alarm squealed through my earphones. Almost simultaneously, Jack yelled down to me that there was a warning light on. Fleeting but spectacular sights snapped into ans out of view, the snow, the shower of debris, the moon, looming close and big, the dazzling sunshine for once unfiltered by layers of air. The last twelve hours before reentry were particular bonechilling. During this period, I had to go up in to command module. Even after the fiery reentry splashing down in 81^{o} water in south pacific, we could still see our frosty breath inside the command module.
Question 1 : The word ‘Command Module’ used twice in the given passage indicates perhaps that it deals with
Question 2 : Which one of the following reasons would one consider as more as possible for the warning lights to be on?
Question 3 : The statement that the dazzling sunshine was “for once unfiltered by layers of air” means
In the world today we make health and end in itself. We have forgotten that health is really means to enable a person to do his work and do it well. a lot of modern medicine and this includes many patients as well as many physicians pays very little attention to health but very much attention to those who imagine that they are ill. Our great concern with health is shown by the medical columns in newspapers. the health articles in popular magazines and the popularity of television programmes and all those books on medicine. We talk about health all the time. Yet for the most part the only result is more people with imaginary illness. The healthy man should not be wasting time talking about health: he should be using health for work. The work does the work that good health possible.
Question 1 : Modern medicine is primarily concerned with
Question 2 : The passage suggests that
Question 3 : Talking about the health all time makes people
Question 4 : The passage tells us