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Answer :
Let present age of the elder person =x and
present age of the younger person =x−16
(x−6)=3(x−16−6)
⇒x−6=3x−66
⇒2x=60
⇒x=60/2=30.
Answer :
Let the present age the son =x
Then, present age of the father =3x+3
Given that, three years hence, father's age will be 1010 years more than twice the age of the son
⇒(3x+3+3)=2(x+3)+10
⇒x=10
Father's present age
=3x+3=3×10+3=33.
Answer :
Let age of the son before 88 years =x
Then, age of Kamal before 88 years ago =4x
After 88 years, Kamal will be twice as old as his son
⇒4x+16=2(x+16)
⇒x=8
Present age of Kamal
=4x+8=4×8+8=40.
Answer :
average age of 36 students in a group is 14
Sum of the ages of 36 students = 36 × 14
When teacher's age is included to it, the average increases by one
=> average = 15
Sum of the ages of 36 students and the teacher = 37 × 15
Hence teachers age
= 37 × 15 - 36 × 14
= 37 × 15 - 14(37 - 1)
= 37 × 15 - 37 × 14 + 14
= 37(15 - 14) + 14
= 37 + 14
= 51.
Answer :
Sum of 5 numbers = 5 × 27
Sum of 4 numbers after excluding one number = 4 × 25
Excluded number
= 5 × 27 - 4 × 25
= 135 - 100 = 35.
Answer :
Average of 6 numbers = x
=> Sum of 6 numbers = 6x
Average of the 3 numbers = y
=> Sum of these 3 numbers = 3y
Average of the remaining 3 numbers = z
=> Sum of the remaining 3 numbers = 3z
Now we know that 6x = 3y + 3z
=> 2x = y + z.
Answer :
Total age of husband and wife (at the time of their marriage) = 2 × 23 = 46
Total age of husband and wife after 5 years + Age of the 1 year old child
= 46 + 5 + 5 + 1 = 57
Average age of the family = 57/3 = 19.
Answer :
Let the rate along with the current is xx km/hr
x+3/2=5
⇒x+3=10
⇒x=7 kmph.
Answer :
Speed downstream =96/8 = 12 kmph
Speed of current = 4 km/hr
Speed of the boatman in still water = 12-4 = 8 kmph
Speed upstream = 8-4 = 4 kmph
Time taken to cover 8 km upstream =8/4 = 2 hours.
Answer :
Given that first day of a normal year was Friday
Odd days of the mentioned year = 1 (Since it is an ordinary year)
Hence First day of the next year = (Friday + 1 Odd day) = Saturday
Therefore, last day of the mentioned year = Friday.
Question 11. If 1st October Is Sunday, Then 1st November Will Be?
Answer :
Given that 1st October is Sunday
Number of days in October = 31
31 days = 3 odd days
(As we can reduce multiples of 7 from odd days which will not change anything)
Hence 1st November = (Sunday + 3 odd days) = Wednesday.
Question 12. 1.12.91 Is The First Sunday. Which Is The Fourth Tuesday Of December 91?
Answer :
Given that 1.12.91 is the first Sunday
Hence we can assume that 3.12.91 is the first Tuesday
If we add 7 days to 3.12.91, we will get second Tuesday
If we add 14 days to 3.12.91, we will get third Tuesday
If we add 21 days to 3.12.91, we will get fourth Tuesday
=> fourth Tuesday = (3.12.91 + 21 days) = 24.12.91.
Answer :
Given that seventh day of a month is three days earlier than Friday
=> Seventh day is Tuesday
=> 14th is Tuesday
=> 19th is Sunday.
Answer :
Let the required number of rounds be x
More radius, less rounds(Indirect proportion)
Hence we can write as
(radius) 14 : 20 :: xx : 70
⇒14×70=20x
⇒14×7=2x
⇒x=7×7=49.
Answer :
Given that fort had provision for 500 persons for 27 days
Hence, after 3 days, the remaining food is sufficient for 500 persons for 24 days
Remaining persons after 3 days = 500 + 300 = 800
Assume that after 3 days,the remaining food is sufficient for 800 persons for xx days
More men, Less days (Indirect Proportion)
(men) 500 : 800 :: xx : 24
⇒500×24=800x
⇒5×24=8x
⇒x=5×3=15.
Answer :
A hostel had provisions for 250 men for 40 days
If 50 men leaves the hostel, remaining men = 250 - 50 = 200
We need to find out how long the food will last for these 200 men.
Let the required number of days = xx days
More men, Less days (Indirect Proportion)
(men) 250 : 200 :: xx : 40
⇒250×40=200x
⇒5×40=4x
⇒x=5×10=50.
Answer :
LCM of 6, 9, 15 and 18 = 90
Required Number = (90k + 4) which is a multiple of 7
Put k = 1. We get number as (90 × 1) + 4 = 94. But this is not a multiple of 7
Put k = 2. We get number as (90 × 2) + 4 = 184. But this is not a multiple of 7
Put k = 3. We get number as (90 × 3) + 4 = 274. But this is not a multiple of 7
Put k = 4. We get number as (90 × 4) + 4 = 364. This is a multiple of 7
Hence 364 is the answer.
Answer :
Since the numbers are co-prime, their HCF = 1
Product of first two numbers = 119
Product of last two numbers = 391
The middle number is common in both of these products. Hence, if we take HCF of 119 and 391, we get the common middle number.
HCF of 119 and 391 = 17
=> Middle Number = 17
First Number =119/17=7
Last Number =391/17=23
Sum of the three numbers = 7 + 17 + 23 = 47.
Answer :
LCM of 4, 8, 10, 12, 15 and 20 = 120
120 seconds = 2 minutes
Hence all the six bells will ring together in every 2 minutes
Hence, number of times they will ring together in 60 minutes =1+60/2=31.
Answer :
We have a 11 litre solution containing 42% of alcohol in the water.
=> Quantity of alcohol in the solution =11×42/100
Now 3 litre of water is added to the solution.
=> Total quantity of the new solution = 11 + 3 = 14
Percentage of alcohol in the new solution =11×42/100/14×100
=11×3/100=33%.
Question 21. ? + 3699 + 1985 - 2047 = 31111?
Answer :
Let x+3699+1985−2047=31111x+3699+1985−2047=31111
x=31111−3699−1985+2047=27474.
Question 22. What Is The Smallest 6 Digit Number Exactly Divisible By 111?
Answer :
Smallest 6 digit number = 100000
100000÷111 = 900, remainder = 100.
Hence 11 should be added to 100000 to get the smallest 6 digit number exactly divisible by 111
Therefore, smallest 6 digit number exactly divisible by 111= 100000 + 11 = 100011
Answer :
Here A's and B's capitals are there for equal time.Hence
A : B = 85000 : 15000
= 85 : 15
= 17 : 3.
Answer :
P : Q : R
=(25×1+35×2):(35×2+25×1)=(25×1+35×2):(35×2+25×1) :(30×3):(30×3)
=95:95:90=19:19:18
Answer :
Let total number of men = 100
Then,
20 men play football.
80 men are less than or equal to 50 years old.
Remaining 20 men are above 50 years old.
Number of football players above 50 years old
=20×20/100=4
Number of football players less than or equal to 50 years old
=20−4=16
Required percentage
=16/20×100=80%.
Answer :
He has 10 patterns of chairs and 8 patterns of tables
A chair can be selected in 10 ways.
A table can be selected in 8 ways.
Hence one chair and one table can be selected in 10×810×8 ways=80 ways.
Answer :
He can go in any of the 25 buses (25 ways).
Since he cannot come back in the same bus, he can return in 24 ways.
Total number of ways =25×24=600.
Answer :
Around a circle, 5 boys can be arranged in 4! ways.
Given that the boys and the girls alternate.
Hence there are 5 places for the girls. Therefore the girls can be arranged in 5! ways.
Total number of ways
=4!×5!=24×120=2880.
Answer :
Part filled by first pipe in 11 minute =1/2
Part filled by second pipe in 11 minute =1/6
Net part filled by pipe A and B in 11 minute
=1/2+1/6=2/3
i.e, pipe A and B together can fill the tank in 3/2 minutes =1.5minutes.
Answer :
Total number of balls, n(S) = 8 + 7 + 6 = 21
n(E) = Number of ways in which a ball can be selected which is neither yellow nor black
= 7 (∵ there are only 7 balls which are neither yellow nor black)
P(E) = n(E)/n(S)=7/21=1/3.
Answer :
Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces)
Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36
Let E = the event of getting two numbers whose product is even
= {(1,2), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,2), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,2),(5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Hence, n(E) = 27
P(E) = n(E)/n(S)=27/36=3/4.
Answer :
selling price =2880
labelled price =2880×100/120=2400
price at which he bought the computer
=2400−2400×15/100=2040.
Answer :
selling price =250
profit =25%
cost price =250×100125=200
Required ratio =200:250=4:5.
Answer :
B has a start of 30 metre
=> A has to run 220 metre and B has to run (220-30)=190 metre
Given that A takes 41 seconds to cover this 220 metre
B takes 44 seconds to cover 220 metre
=> B takes 44220 seconds to cover 1 metre
=> B takes 44/220×190=420×190 = 38 seconds to cover 190 metre
=> B takes 44/220 seconds to cover 1 metre
=> B takes 44/220×190=4/20×190 = 38 seconds to cover 190 metre
i.e., A beats B by (41-38)= 3 seconds.
Answer :
This means, B takes 4 seconds to run 40 metres
=> B takes 4/40=1/10 seconds to run 1 metre
=> B takes 1/10×1000=100 seconds to run 1000 metre.
Question 36. Find The Odd Man Out. 3576, 1784, 888, 440, 216, 105, 48?
Answer :
3576
(3576-8)/2 = 1784
(1784-8)/2 = 888
(888-8)/2 = 440
(440-8)/2 = 216
(216-8)/2 = 104
(104-8)/2 = 48
Hence, 105 is wrong. 104 should have come in place of 105.
Answer :
Interest earned in 3rd year = Rs. 1596
Interest earned in 2nd year = Rs. 1400
i.e, in 3rd year, Andrews gets additional interest of (Rs. 1596 - Rs. 1400) = Rs.196
This means, Rs.196 is the interest obtained for Rs.1400 for 1 year
R=100×SI/PT=100×196/1400×1=196/14=14%.
Answer :
Let number of ducks =d=d
Number of cows =c=c
Then, total number of legs =2d+4c=2d+4c
Total number of heads =d+c=d+c
Given that total number of legs is 2828 more than twice the number of heads
⇒2d+4c=28+2(d+c)⇒2c=28⇒c=14⇒2d+4c=28+2(d+c)⇒2c=28⇒c=14
i.e., total number of cows =14.
Answer :
Let initial number of men = initial number of women =x
2(x−8)=x
⇒2x−16=x
⇒x=16
Total number of men and women
=2x=2×16=32.
Answer :
Face value of each share = Rs.5
Total dividend received by John = 100×5×12/100 = Rs.60
Let market value of 100 shares = Rs.x
x×10/100=60
x = 600
ie, Market value of 100 shares = Rs.600
Hence, Market value of each share = Rs.6
Answer :
Since face value is not given, take it as Rs.100.
As it is an 8% stock, income (dividend) per stock = Rs.8
ie, For an income of Rs.8,amount of stock needed = Rs.100
For an income of Rs.800, amount of stock needed = 100×800/8=10000.
Answer :
market Value of Company X (his selling price) = Rs.30
Total shares sold = 4000
Amount he gets = Rs.(4000 × 30)
He invests this amount in ordinary shares of Company Y
Market Value of Company Y(His purchasing price) = 15
Number of shares of company Y which he purchases = 4000×30/15=8000.
Answer :
semi-annual dividend = 10×12/100 = Rs.1.2
Total semi-annual dividend = 2000 × 1.2 = Rs.2400
Total annual dividend = 2 × Rs.2400 = Rs.4800.
Answer :
Investment = Rs.333000
Since face value is not given, we can take it as Rs.100
and dividend per share = Rs.11/2
Market Value = 110 + 1 = 111
Number of shares purchased = 333000/111 = 3000
Total income = 3000×11/2 = Rs.16500.
Answer :
Assume that they meet xx hours after 88 a.m.
Then, train 1,1, starting from A, travels xx hours till the trains meet.
Distance travelled by train 11 in xx hours =60x km=60x km
Train 2,2, starting from B, travels (x−1)(x−1) hours till the trains meet.
Distance travelled by train 22 in (x−1)(x−1) hours =75(x−1) km=75(x−1) km
Total distance travelled
= Distance travelled by train 11 + Distance travelled by train 22
⇒330=60x+75(x−1)
⇒12x+15(x−1)=66
⇒12x+15x−15=66
⇒27x=66+15=81
⇒3x=9
⇒x=3.
Answer :
Time needed to travel 600 km =600/100=6 hour
Now we need to find out the number of stops in the 600600 km journey. Given that the train stops after every 7575 km.
600/75=8
It means, the train stops 7 times before 600 km and 11 time just after 600 km. Hence we need to take only 7 stops into consideration for the 600 km journey.
Hence, total stopping time in the 600 km journey
=7×3=21minutes
Total time needed to reach the destination
=6 hours +21 minutes
=6 hours 21 minutes.
Answer :
Let P takes x days to complete the work
Then Q takes x/2 days and R takes x/3 days to finish the work
Amount of work P does in 1 day = 1/x
Amount of work Q does in 1 day = 2/x
Amount of work R does in 1 day = 3/x
Amount of work P,Q and R do in 1 day = 1/x + 2/x + 3/x = 1/x (1 + 2 + 3) = 6/x
6/x = 2
=> x = 12
=> Q takes 12/2 days = 6 days to complete the work.
Answer :
Work done by Q in 1 day = 1/12
Work done by P in 1 day = 2 × (1/12) = 1/6
Work done by P and Q in 1 day = 1/12 + 1/6 =1/4
=> P and Q can finish the work in 4 days.
Answer :
Assume both trains meet xx hours after 77 a.m.
Distance covered by train starting from P in xx hours =20x km=20x km
Distance covered by train starting from Q in (x−1)(x−1) hours =25(x−1) km=25(x−1) km
Total distance =110 km=110 km
⇒20x+25(x−1)=110
⇒45x=135
⇒x=3
Hence, they meet 33 hours after 77 a.m.
i.e., they meet at 10 a.m.
Answer :
The train can cover a distance equal to its length in 1515 seconds.
The train can cover (a distance equal to its length + 100100 metre) in 2525 seconds.
Therefore, it can be concluded that the train can travel 100100 metre in (25−15)=10 seconds and therefore its speed is 100/10=10 m/s
Hence, length of the train =10×15=150 metre.
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