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The average of five results is 46 and that of the first four is 45. The fifth result is?
5 * 46 – 4 * 45 = 50
5 * 46 – 4 * 45 = 50
Which of the following is the greatest?
5 > 4 and 1/2 > 1/4
5^{1/2} > 4^{1/4}
4^{1/4} cannot be the greatest. Hence 3^{1/4} also cannot be the greatest.
3^{7/10} = (3^{7})^{1/10} = (2187)^{1/10}
5^{1/2} = 5^{5/10} = (5^{5})^{1/10} = (3125)^{1/10}
6^{1/5} = 6^{2/10} = (6^{2})^{1/10} = (36)^{1/10}
As (3125)^{1/10} > (2187)^{1/10} > (36)^{1/10} , 5^{1/2} is the greatest.
5 > 4 and 1/2 > 1/4
5^{1/2} > 4^{1/4}
4^{1/4} cannot be the greatest. Hence 3^{1/4} also cannot be the greatest.
3^{7/10} = (3^{7})^{1/10} = (2187)^{1/10}
5^{1/2} = 5^{5/10} = (5^{5})^{1/10} = (3125)^{1/10}
6^{1/5} = 6^{2/10} = (6^{2})^{1/10} = (36)^{1/10}
As (3125)^{1/10} > (2187)^{1/10} > (36)^{1/10} , 5^{1/2} is the greatest.
A tank is filled in eight hours by three pipes A, B and C. Pipe A is twice as fast as pipe B, and B is twice as fast as C. How much time will pipe B alone take to fill the tank?
1/A + 1/B + 1/C = 1/8 (Given)
Also given that A = 2B and B = 2C
=> 1/2B + 1/B + 2/B = 1/8
=> (1 + 2 + 4)/2B = 1/8
=> 2B/7 = 8
=> B = 28 hours.
1/A + 1/B + 1/C = 1/8 (Given)
Also given that A = 2B and B = 2C
=> 1/2B + 1/B + 2/B = 1/8
=> (1 + 2 + 4)/2B = 1/8
=> 2B/7 = 8
=> B = 28 hours.
Amar, Bhavan and Chetan divide an amount of Rs.5600 among themselves in the ratio 3:6:5. If an amount of Rs.400 is deducted from each of their shares, what will be the new ratio of their shares of the amount?
Let the shares of Amar, Bhavan and Chetan be 3x, 6x and 5x respectively.
3x + 6x + 5x = 5600
14x = 5600 => x = 400
Required ratio = 3x – 400 : 6x – 400 : 5x – 400
= 3x – x : 6x – x : 5x – x
= 2x : 5x : 4x => 2:5:4
Let the shares of Amar, Bhavan and Chetan be 3x, 6x and 5x respectively.
3x + 6x + 5x = 5600
14x = 5600 => x = 400
Required ratio = 3x – 400 : 6x – 400 : 5x – 400
= 3x – x : 6x – x : 5x – x
= 2x : 5x : 4x => 2:5:4
A can finish a piece of work in 5 days. B can do it in 10 days. They work together for two days and then A goes away. In how many days will B finish the work?
2/5 + (2 + x)/10 = 1 => x = 4 days
2/5 + (2 + x)/10 = 1 => x = 4 days
The length of rectangle is thrice its breadth and its perimeter is 96 m, find the area of the rectangle?
2(3x + x) = 96
l = 36 b = 12
lb = 36 * 12 = 432
2(3x + x) = 96
l = 36 b = 12
lb = 36 * 12 = 432
Find the one which does not belong to that group ?
In each of the groups B4, D16, E25 and F36 the number is the square of the position value of the letter. This pattern is not followed in I91.
In each of the groups B4, D16, E25 and F36 the number is the square of the position value of the letter. This pattern is not followed in I91.
The average of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is:
Excluded number = (27 * 5) – (25 * 4) = 135 – 100 = 35.
Excluded number = (27 * 5) – (25 * 4) = 135 – 100 = 35.
When 2 is added to half of onethird of onefifth of a number, the result is onefifteenth of the number. Find the number?
Let the number be
2 + 1/2[1/3(a/5)] = a/15
=> 2 = a/30 => a = 60
Let the number be
2 + 1/2[1/3(a/5)] = a/15
=> 2 = a/30 => a = 60
In an office, totally there are 6400 employees and 65% of the total employees are males. 25% of the males in the office are atleast 50 years old. Find the number of males aged below 50 years?
Number of male employees = 6400 * 65/100 = 4160
Required number of male employees who are less than 50 years old = 4160 * (100 – 25)%
= 4160 * 75/100 = 3120.
Number of male employees = 6400 * 65/100 = 4160
Required number of male employees who are less than 50 years old = 4160 * (100 – 25)%
= 4160 * 75/100 = 3120.
A train crosses a platform of 150 m in 15 sec, same train crosses another platform of length 250 m in 20 sec. then find the length of the train?
Length of the train be ‘X’
X + 150/15 = X + 250/20
4X + 600 = 3X + 750
X = 150m
Length of the train be ‘X’
X + 150/15 = X + 250/20
4X + 600 = 3X + 750
X = 150m
A shopkeeper sells two articles at Rs.1000 each, making a profit of 20% on the first article and a loss of 20% on the second article. Find the net profit or loss that he makes?
SP of first article = 1000
Profit = 20%
CP = (SP)*[100/(100+P)] = 5000/6 = 2500/3
SP of second article = 1000
Loss = 20%
CP = (SP)*[100/(100L)] = 5000/4 = 1250
Total SP = 2000
Total CP = 2500/3 + 1250 = 6250/3
CP is more than SP, he makes a loss.
Loss = CPSP = (6250/3) 2000 = 250/3
Loss Percent = [(250/3)/(6250/3)]*100 =
0.04 * 100 = 4%
SP of first article = 1000
Profit = 20%
CP = (SP)*[100/(100+P)] = 5000/6 = 2500/3
SP of second article = 1000
Loss = 20%
CP = (SP)*[100/(100L)] = 5000/4 = 1250
Total SP = 2000
Total CP = 2500/3 + 1250 = 6250/3
CP is more than SP, he makes a loss.
Loss = CPSP = (6250/3) 2000 = 250/3
Loss Percent = [(250/3)/(6250/3)]*100 =
0.04 * 100 = 4%
Pipe A can fill a tank in 6 hours. Due to a leak at the bottom, it takes 9 hours for the pipe A to fill the tank. In what time can the leak alone empty the full tank?
Let the leak can empty the full tank in x hours 1/6 – 1/x = 1/9
=> 1/x = 1/6 – 1/9 = (3 – 2)/18 = 1/18
=> x = 18.
Let the leak can empty the full tank in x hours 1/6 – 1/x = 1/9
=> 1/x = 1/6 – 1/9 = (3 – 2)/18 = 1/18
=> x = 18.
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;
(B + C)’s 1 day work = 1/15;
(A + C)’s 1 day work = 1/20
Adding, we get: 2(A + B + C)’s 1 day work = (1/12 + 1/15 + 1/20) = 1/5
(A + B + C)’s 1 day work = 1/10
So, A, B and C together can complete the work in 10 days.
(A + B)’s 1 day work = 1/12;
(B + C)’s 1 day work = 1/15;
(A + C)’s 1 day work = 1/20
Adding, we get: 2(A + B + C)’s 1 day work = (1/12 + 1/15 + 1/20) = 1/5
(A + B + C)’s 1 day work = 1/10
So, A, B and C together can complete the work in 10 days.
Find the greatest 4digit number exactly divisible by 3, 4 and 5?
Greatest 4digit number is 9999. LCM of 3, 4 and 5 is 60. Dividing 9999 with 60, we get remainder 39. Thus, the required number is 9999 – 39 = 9960.
Greatest 4digit number is 9999. LCM of 3, 4 and 5 is 60. Dividing 9999 with 60, we get remainder 39. Thus, the required number is 9999 – 39 = 9960.
A boy rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. His average speed for the entire trip is approximately?
Total distance traveled = 10 + 12 = 22 km /hr.
Total time taken = 10/12 + 12/10 = 61/30 hrs.
Average speed = 22 * 30/61 = 10.8 km/hr.
Total distance traveled = 10 + 12 = 22 km /hr.
Total time taken = 10/12 + 12/10 = 61/30 hrs.
Average speed = 22 * 30/61 = 10.8 km/hr.
The average age of three boys is 15 years and their ages are in proportion 3:5:7. What is the age in years of the youngest boy?
3x + 5x + 7x = 45
x =3
3x = 9
3x + 5x + 7x = 45
x =3
3x = 9
Difference between two numbers is 5, six times of the smaller lacks by 6 from the four times of the greater. Find the numbers?
x – y = 5
4x – 6y = 6
x = 12 y = 7
x – y = 5
4x – 6y = 6
x = 12 y = 7
(6 – 2 * 3 + 6) / (2 + 6 * 3 – 4) = ?
(6 – 2 * 3 + 6) / (2 + 6 * 3 – 4) = (6 – 6 + 6) / (2 + 18 – 4)
= 6/16 = 3/8
(6 – 2 * 3 + 6) / (2 + 6 * 3 – 4) = (6 – 6 + 6) / (2 + 18 – 4)
= 6/16 = 3/8
The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average of the team?
Let the average of the whole team be x years.
11x – (26 + 29) = 9(x – 1)
= 11x – 9x = 46
= 2x = 46 => x = 23
So, average age of the team is 23 years.
Let the average of the whole team be x years.
11x – (26 + 29) = 9(x – 1)
= 11x – 9x = 46
= 2x = 46 => x = 23
So, average age of the team is 23 years.
31.2 * 14.5 * 9.6 = ?
? = 4343.04
? = 4343.04
A man whose speed is 4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph, find his average speed for the total journey?
M = 45
S = 1.5
DS = 6
US = 3
AS = (2 * 6 * 3) /9 = 4
M = 45
S = 1.5
DS = 6
US = 3
AS = (2 * 6 * 3) /9 = 4
A train 150 m long running at 72 kmph crosses a platform in 25 sec. What is the length of the platform?
D = 72 * 5/18 = 25 = 500 – 150 = 350
D = 72 * 5/18 = 25 = 500 – 150 = 350
Find the greatest number that, while dividing 47, 215 and 365, gives the same remainder in each case?
Calculate the differences, taking two numbers at a time as follows:
(21547) = 168
(365215) = 150
(36547) = 318
HCF of 168, 150 and 318 we get 6, which is the greatest number, which while dividing 47, 215 and 365 gives the same remainder in each cases is 5.
Calculate the differences, taking two numbers at a time as follows:
(21547) = 168
(365215) = 150
(36547) = 318
HCF of 168, 150 and 318 we get 6, which is the greatest number, which while dividing 47, 215 and 365 gives the same remainder in each cases is 5.
David invested certain amount in three different schemes. A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B?
Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.
Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.
Two men Amar and Bhuvan have the ratio of their monthly incomes as 6 : 5. The ratio of their monthly expenditures is 3 : 2. If Bhuvan saves onefourth of his income, find the ratio of their monthly savings?
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.
What is the length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high?
d^{2} = 12^{2} + 4^{2} + 3^{2} = 13
d^{2} = 12^{2} + 4^{2} + 3^{2} = 13
The spherical ball of lead 3 cm in diameter is melted and recast into 3 spherical balls. The diameters of two of these are 1 ½ cm and 2 cm respectively. The diameter of third ball is?
4/3 π * 3 * 3 * 3 = 4/3 π[(3/2)^{3} + 2^{3} + r^{3}]
r = 1.25
d = 2.5
4/3 π * 3 * 3 * 3 = 4/3 π[(3/2)^{3} + 2^{3} + r^{3}]
r = 1.25
d = 2.5
In a division sum, the remainder is 0. As student mistook the divisor by 12 instead of 21 and obtained 35 as quotient. What is the correct quotient?
Number = (12 * 35) = 420
Correct quotient = 420/21 = 20
Number = (12 * 35) = 420
Correct quotient = 420/21 = 20
The denominator of a fraction is 1 less than twice the numerator. If the numerator and denominator are both increased by 1, the fraction becomes 3/5. Find the fraction?
Let the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.
d = 2n – 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n – 1) + 3 => n = 5
d = 2n – 1 => d = 9
Hence the fraction is : 5/9.
Let the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.
d = 2n – 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n – 1) + 3 => n = 5
d = 2n – 1 => d = 9
Hence the fraction is : 5/9.
Sum of two numbers is 80. Greater number exceeds by 5 from four times of the smaller. Find the numbers?
x + y = 80
x – 4y = 5
x = 65 y = 15
x + y = 80
x – 4y = 5
x = 65 y = 15
Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full is?
(A + B)’s 1 hour work = (1/12 + 1/15) = 3/20
(A + C)’s 1 hour work = (1/12 + 1/20) = 2/15
Part filled in 2 hrs = (3/20 + 2/15) = 17/60
Part filled in 6 hrs = 3 * 17/60 = 17/20
Remaining part = 1 – 17/20 = 3/20
Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour.
Total time taken to fill the tank = (6 + 1) = 7 hrs.
(A + B)’s 1 hour work = (1/12 + 1/15) = 3/20
(A + C)’s 1 hour work = (1/12 + 1/20) = 2/15
Part filled in 2 hrs = (3/20 + 2/15) = 17/60
Part filled in 6 hrs = 3 * 17/60 = 17/20
Remaining part = 1 – 17/20 = 3/20
Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour.
Total time taken to fill the tank = (6 + 1) = 7 hrs.
The ratio between the present ages of P and Q is 6:7. If Q is 4 years old than P, what will be the ratio of the ages of P and Q after 4 years?
Let P’s age and Q’s age be 6x and 7x years respectively.
Then, 7x – 6x = 4 => x = 4
Required ratio = (6x + 4) : (7x + 4)
28 : 32 = 7:8
Let P’s age and Q’s age be 6x and 7x years respectively.
Then, 7x – 6x = 4 => x = 4
Required ratio = (6x + 4) : (7x + 4)
28 : 32 = 7:8
A, B, C enter into a partnership investing Rs. 35,000, Rs. 45,000 and Rs. 55,000 respectively. The respective shares of A, B, C in annual profit of Rs. 40,500 are:
A:B:C = 35000 : 45000 : 55000 = 7:9:11
A’s share = 40500 * 7/27 = Rs. 10500
B’s share = 40500 * 9/27 = Rs. 13500
C’s share = 40500 * 11/27 = Rs. 16500
A:B:C = 35000 : 45000 : 55000 = 7:9:11
A’s share = 40500 * 7/27 = Rs. 10500
B’s share = 40500 * 9/27 = Rs. 13500
C’s share = 40500 * 11/27 = Rs. 16500
If (4^{61} + 4^{62} + 4^{63} + 4^{64}) is divisible by ?, then ? =
4^{61} + 4^{62} + 4^{63} + 4^{64}
=> 4^{61}(1 + 4 + 16 + 64)
4^{61}(85) = 17 (5 * 4^{61}) So it is divisible by 17.
4^{61} + 4^{62} + 4^{63} + 4^{64}
=> 4^{61}(1 + 4 + 16 + 64)
4^{61}(85) = 17 (5 * 4^{61}) So it is divisible by 17.
A money lender finds that due to a fall in the annual rate of interest from 8% to 7 3/4 % his yearly income diminishes by Rs. 61.50, his capital is?
Let the capital be Rs. x. Then,
(x * 8 * 1)/100 – (x * 31/4 * 1/100) = 61.50
32x – 31x = 6150 * 4
x = 24,600.
Let the capital be Rs. x. Then,
(x * 8 * 1)/100 – (x * 31/4 * 1/100) = 61.50
32x – 31x = 6150 * 4
x = 24,600.
Find the nearest to 25268 which is exactly divisible by 467?
The sum of the two digits of a number is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number?
Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 10 —– (1)
(10Q + P) – (10P + Q) = 54
9(Q – P) = 54
(Q – P) = 6 —– (2)
Solve (1) and (2) P = 2 and Q = 8
The required number is = 28
Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 10 —– (1)
(10Q + P) – (10P + Q) = 54
9(Q – P) = 54
(Q – P) = 6 —– (2)
Solve (1) and (2) P = 2 and Q = 8
The required number is = 28
50 men can complete a work in 65 days.Five days after started the work, 20 men left the group. In how many days can the remaining work be completed?
After 5 days, the following situation prevails.
50 men can complete the work in 60 days.
30 men can complete the work in ? days.
M_{1} D_{1} = M_{2} D_{2}
=> 50 * 60 = 30 * D_{2}
=> D_{2} = (50 * 60)/30 = 100 days.
After 5 days, the following situation prevails.
50 men can complete the work in 60 days.
30 men can complete the work in ? days.
M_{1} D_{1} = M_{2} D_{2}
=> 50 * 60 = 30 * D_{2}
=> D_{2} = (50 * 60)/30 = 100 days.
A rectangular field 30 m long and 20 m broad. How much deep it should be dug so that from the earth taken out, a platform can be formed which is 8 m long, 5.5 m broad and 1.5 m high where as the earth taken out is increase by 10/5?
30 * 20 * x = (8 * 5.5 * 1.5)/2
30 * 20 * x = (8 * 5.5 * 1.5)/2
(0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / (1.6 * 1.6 * 1.6 + 1.8 * 1.8 * 1.8) = ?
(0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / (1.6 * 1.6 * 1.6 + 1.8 * 1.8 * 1.8)
= (0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / 2^{3}(0.8 * 0.8 * 0.8 + 0.9 * 0.9 * 0.9) = 1/8 = 0.125
(0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / (1.6 * 1.6 * 1.6 + 1.8 * 1.8 * 1.8)
= (0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / 2^{3}(0.8 * 0.8 * 0.8 + 0.9 * 0.9 * 0.9) = 1/8 = 0.125
A man can row a boat at 20 kmph in still water. If the speed of the stream is 6 kmph, what is the time taken to row a distance of 60 km downstream?
Speed downstream = 20 + 6 = 26 kmph.
Time required to cover 60 km downstream = d/s = 60/26 = 30/13 hours.
Speed downstream = 20 + 6 = 26 kmph.
Time required to cover 60 km downstream = d/s = 60/26 = 30/13 hours.
The smallest number which when divided by 20, 25, 35 and 40 leaves a remainder of 14, 19, 29 and 34 respectively is?
LCM = 1400
1400 – 6 = 1394
LCM = 1400
1400 – 6 = 1394
Rajan borrowed Rs.4000 at 5% p.a compound interest. After 2 years, he repaid Rs.2210 and after 2 more year, the balance with interest. What was the total amount that he paid as interest?
4000
200 — I
200
10 — II
—————
4410
2210
——–
2000
110 — III
110
5.50 — IV
———–
2425.50
2210
———–
4635.50
4000
———
635.50
4000
200 — I
200
10 — II
—————
4410
2210
——–
2000
110 — III
110
5.50 — IV
———–
2425.50
2210
———–
4635.50
4000
———
635.50
The average runs scored by a batsman in 20 matches is 40. In the next 10 matches the batsman scored an average of 13 runs. Find his average in all the 30 matches?
Total score of the batsman in 20 matches = 800.
Total score of the batsman in the next 10 matches = 130.
Total score of the batsman in the 30 matches = 930.
Average score of the batsman = 930/30 = 31.
Total score of the batsman in 20 matches = 800.
Total score of the batsman in the next 10 matches = 130.
Total score of the batsman in the 30 matches = 930.
Average score of the batsman = 930/30 = 31.
Kiran travels from A to B by car and returns from B to A by cycle in 7 hours. If he travels both ways by car he saves 3 hours. What is the time taken to cover both ways by cycle?
Let the time taken to cover from A to B in car and cycle be x hours and y hours respectively.
x + y = 7 — (1) ; 2x = 4 — (2)
solving both the equations, we get y = 5
So, time taken to cover both ways by cycle = 2y hours = 10 hours.
Let the time taken to cover from A to B in car and cycle be x hours and y hours respectively.
x + y = 7 — (1) ; 2x = 4 — (2)
solving both the equations, we get y = 5
So, time taken to cover both ways by cycle = 2y hours = 10 hours.
The area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm is?
(5 * 3.5)/2 = 8.75
(5 * 3.5)/2 = 8.75
David invested certain amount in three different schemes. A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B?
Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.
Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.
How many meters of carpet 50cm, wide will be required to cover the floor of a room 30m * 20m?
50/100 * x = 30 * 20 => x = 1200
50/100 * x = 30 * 20 => x = 1200
3 men, 4 women and 6 children can complete a work in 7 days. A woman does double the work a man does and a child does half the work a man does. How many women alone can complete this work in 7 days?
Let 1 woman’s 1 day work = x.
Then, 1 man’s 1 day work = x/2 and 1 child’s 1 day work x/4.
So, (3x/2 + 4x + + 6x/4) = 1/7
28x/4 = 1/7 => x = 1/49
1 woman alone can complete the work in 49 days.
So, to complete the work in 7 days, number of women required = 49/7 = 7.
Let 1 woman’s 1 day work = x.
Then, 1 man’s 1 day work = x/2 and 1 child’s 1 day work x/4.
So, (3x/2 + 4x + + 6x/4) = 1/7
28x/4 = 1/7 => x = 1/49
1 woman alone can complete the work in 49 days.
So, to complete the work in 7 days, number of women required = 49/7 = 7.
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Find the H.C.F of 12, 16, 18 and 24?
How much 60% of 50 is greater than 40% of 30?
(60/100) * 50 – (40/100) * 30
30 – 12 = 18
(60/100) * 50 – (40/100) * 30
30 – 12 = 18
A jogger running at 9 km/hr along side a railway track is 240 m ahead of the engine of a 120 m long train running at 45 km/hr in the same direction. In how much time will the train pass the jogger?
Speed of train relative to jogger = 45 – 9 = 36 km/hr.
= 36 * 5/18 = 10 m/sec.
Distance to be covered = 240 + 120 = 360 m.
Time taken = 360/10 = 36 sec.
Speed of train relative to jogger = 45 – 9 = 36 km/hr.
= 36 * 5/18 = 10 m/sec.
Distance to be covered = 240 + 120 = 360 m.
Time taken = 360/10 = 36 sec.
Two trains of length 100 m and 200 m are 100 m apart. They start moving towards each other on parallel tracks, at speeds 54 kmph and 72 kmph. After how much time will the trains meet?
They are moving in opposite directions, relative speed is equal to the sum of their speeds.
Relative speed = (54 + 72)*5/18 = 7*5 = 35 mps.
The time required = d/s = 100/35 = 20/7 sec.
They are moving in opposite directions, relative speed is equal to the sum of their speeds.
Relative speed = (54 + 72)*5/18 = 7*5 = 35 mps.
The time required = d/s = 100/35 = 20/7 sec.
What should be the least number to be added to the 51234 number to make it divisible by 9?
The least number to be added to the numbers to make them divisible by 9 is equal to the difference of the least multiple of 9 greater than the sum of the digits and sum of the digits.
Sum of digits = 15.
Nearest multiple of 9 greater than sum of digits = 18.
Hence 3 has to be added.
The least number to be added to the numbers to make them divisible by 9 is equal to the difference of the least multiple of 9 greater than the sum of the digits and sum of the digits.
Sum of digits = 15.
Nearest multiple of 9 greater than sum of digits = 18.
Hence 3 has to be added.
Product of two coprime numbers is 117. Their L.C.M should be:
H.C.F of coprime numbers is 1.
So, L.C.M = 117/1 = 117.
H.C.F of coprime numbers is 1.
So, L.C.M = 117/1 = 117.
10 camels cost as much as 24 horses, 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants. If the cost of 10 elephants is Rs.170000, find the cost of a camel?
Cost of the camel = P
10 camels = 24 horses
16 horses = 4 oxen
6 oxen = 4 elephants
10 elephants = Rs.170000
P = Rs.[(24 * 4 * 4 * 170000)/(10 * 16 * 6 * 10)]
P = Rs.(65280000/9600) => P = Rs.6800
Cost of the camel = P
10 camels = 24 horses
16 horses = 4 oxen
6 oxen = 4 elephants
10 elephants = Rs.170000
P = Rs.[(24 * 4 * 4 * 170000)/(10 * 16 * 6 * 10)]
P = Rs.(65280000/9600) => P = Rs.6800
In a partnership between A, B and C. A’s capital is Rs.5000. If his share of a profit of Rs.800 is Rs.200 and C’s share is Rs.130, what is B’s capital?
200 + 130 = 330
800 – 330 = 470
200 — 5000
470 — ? => 11750
200 + 130 = 330
800 – 330 = 470
200 — 5000
470 — ? => 11750
The average of 9 observations was 9, that of the 1^{st} of 5 being 10 and that of the last 5 being 8. What was the 5^{th} observation?
1 to 9 = 9 * 9 = 81
1 to 5 = 5 * 10 = 50
5 to 9 = 5 * 8 = 40
5^{th} = 50 + 40 = 90 – 81 = 9
1 to 9 = 9 * 9 = 81
1 to 5 = 5 * 10 = 50
5 to 9 = 5 * 8 = 40
5^{th} = 50 + 40 = 90 – 81 = 9
The difference between a number and its threefifth is 50. What is the number?
Let the number be x. Then,
x – 3/5 x = 50 => 2/5 x = 50
x = (50 * 5)/2 = 125.
Let the number be x. Then,
x – 3/5 x = 50 => 2/5 x = 50
x = (50 * 5)/2 = 125.
Albert buys 4 horses and 9 cows for Rs. 13,400. If he sells the horses at 10% profit and the cows at 20% profit, then he earns a total profit of Rs. 1880. The cost of a horse is:
Let C.P. of each horse be Rs. x and C.P. of each cow be Rs. y.
Then, 4x + 9y = 13400 — (i)
And, 10% of 4x + 20% of 9y = 1880
2/5 x + 9/5 y = 1880 => 2x + 9y = 9400 — (ii)
Solving (i) and (ii), we get : x = 2000 and y = 600.
Cost price of each horse = Rs. 2000.
Let C.P. of each horse be Rs. x and C.P. of each cow be Rs. y.
Then, 4x + 9y = 13400 — (i)
And, 10% of 4x + 20% of 9y = 1880
2/5 x + 9/5 y = 1880 => 2x + 9y = 9400 — (ii)
Solving (i) and (ii), we get : x = 2000 and y = 600.
Cost price of each horse = Rs. 2000.
The length of a rectangular floor is more than its breadth by 200%. If Rs. 324 is required to paint the floor at the rate of Rs. 3 per sq m, then what would be the length of the floor?
Let the length and the breadth of the floor be l m and b m respectively.
l = b + 200% of b = l + 2b = 3b
Area of the floor = 324/3 = 108 sq m
l b = 108 i.e., l * l/3 = 108
l^{2} = 324 => l = 18.
Let the length and the breadth of the floor be l m and b m respectively.
l = b + 200% of b = l + 2b = 3b
Area of the floor = 324/3 = 108 sq m
l b = 108 i.e., l * l/3 = 108
l^{2} = 324 => l = 18.
When 5% is lost in grinding wheat, a country has to import 20 million bags; but when only 2% is lost, it has to import only 15 million bags. Find the quantity of wheat, which grows in the country?
5% – 2% = 3%
3% — 5
100% — ? => 166 2/3
5% – 2% = 3%
3% — 5
100% — ? => 166 2/3
Eight men, ten women and six boys together can complete a piece of work in eight days. In how many days can 20 women complete the same work if 20 men can complete it in 12 days?
Let the number of units which can be completed by each man, each women and each boy be m/day, w/day and b/day respectively.
Work = 8(8m + 10w +6b)units = 12(20m)
10w +6b = 22m
b is unknown.
We cannot find the relation between m and w.
We cannot answer the question.
Let the number of units which can be completed by each man, each women and each boy be m/day, w/day and b/day respectively.
Work = 8(8m + 10w +6b)units = 12(20m)
10w +6b = 22m
b is unknown.
We cannot find the relation between m and w.
We cannot answer the question.
A and B began business with Rs.3000 and Rs.4000 after 8 months, A withdraws Rs.1000 and B advances Rs.1000 more. At the end of the year, their profits amounted to Rs.630 find the share of A.
(3*8 + 2*4):(4*8 + 5*4)
8:13
8/21 * 630 = 240
(3*8 + 2*4):(4*8 + 5*4)
8:13
8/21 * 630 = 240
(64 + 9 + 9) / (2 * 20 + 1) = ?
(64 + 9 + 9) / (2 * 20 + 1) = 82/41 = 2
(64 + 9 + 9) / (2 * 20 + 1) = 82/41 = 2
30 men can do a work in 40 days. When should 20 men leave the work so that the entire work is completed in 40 days after they leave the work?
Total work to be done = 30 * 40 = 1200
Let 20 men leave the work after ‘P’ days, so that the remaining work is completed in 40 days after they leave the work.
40P + (20 * 40) = 1200
40P = 400 => P = 10 days
Total work to be done = 30 * 40 = 1200
Let 20 men leave the work after ‘P’ days, so that the remaining work is completed in 40 days after they leave the work.
40P + (20 * 40) = 1200
40P = 400 => P = 10 days
How many times the keys of a typewriter have to be pressed in order to write first 400 counting numbers?
1 to 9 = 9 * 1 = 9
10 to 99 = 90 * 2 = 180
100 to 400 = 301 * 3 = 903
———–
1092
1 to 9 = 9 * 1 = 9
10 to 99 = 90 * 2 = 180
100 to 400 = 301 * 3 = 903
———–
1092
The H.C.F of two numbers is 8. Which of the following can never be their L.C.M?
H.C.F of two numbers divides their L.C.M exactly. Clearly, 8 is not a factor of 60.
H.C.F of two numbers divides their L.C.M exactly. Clearly, 8 is not a factor of 60.
The subduplicate ratio of 1:4 is
√1:√4 = 1:2
√1:√4 = 1:2
Two men Amar and Bhavan have the ratio of their monthly incomes as 6:5. The ratio of their monthly expenditures is 3:2. If Bhavan saves onefourth of his income, find the ratio of their monthly savings?
Let the monthly incomes of Amar and Bhavan be 6x and 5x respectively.
Let the monthly expenditures of Amar and Bhavan be 3y and 2y respectively.
Savings of Bhavan every month = 1/4 (5x)
=(his income) – (his expenditure) = 5x – 2y
=> 5x = 20x – 8y => y = 15x/8
Ratio of savings of Amar and Bhavan
= 6x – 3y : 1/4 (5x) = 6x – 3(15x/8) : 5x/4
= 3x/8 : 5x/4 => 3:10
Let the monthly incomes of Amar and Bhavan be 6x and 5x respectively.
Let the monthly expenditures of Amar and Bhavan be 3y and 2y respectively.
Savings of Bhavan every month = 1/4 (5x)
=(his income) – (his expenditure) = 5x – 2y
=> 5x = 20x – 8y => y = 15x/8
Ratio of savings of Amar and Bhavan
= 6x – 3y : 1/4 (5x) = 6x – 3(15x/8) : 5x/4
= 3x/8 : 5x/4 => 3:10
The curved surface of a sphere is 64 π cm^{2}. Find its radius?
4 πr^{2 }= 64 => r = 4
4 πr^{2 }= 64 => r = 4
A train 540 meters long is running with a speed of 54 kmph. The time taken by it to cross a tunnel 180 meters long is?
D = 540 + 180 = 720
S = 54 * 5/18 = 15 mps
T = 720/15 = 48 sec
D = 540 + 180 = 720
S = 54 * 5/18 = 15 mps
T = 720/15 = 48 sec
A, B and C started a business with capitals of Rs. 8000, Rs. 10000 and Rs. 12000 respectively. At the end of the year, the profit share of B is Rs. 1500. The difference between the profit shares of A and C is?
Ratio of investments of A, B and C is 8000 : 10000 : 12000 = 4 : 5 : 6
And also given that, profit share of B is Rs. 1500
=> 5 parts out of 15 parts is Rs. 1500
Now, required difference is 6 – 4 = 2 parts
Required difference = 2/5 (1500) = Rs. 600
Ratio of investments of A, B and C is 8000 : 10000 : 12000 = 4 : 5 : 6
And also given that, profit share of B is Rs. 1500
=> 5 parts out of 15 parts is Rs. 1500
Now, required difference is 6 – 4 = 2 parts
Required difference = 2/5 (1500) = Rs. 600
Ravi invested certain amount for two rates of simple interests at 6% p.a. and 7% p.a. What is the ratio of Ravi’s investments if the interests from those investments are equal?
Let x be the investment of Ravi in 6% and y be in 7%
x(6)(n)/100 = y(7)(n)/100
=> x/y = 7/6
x : y = 7 : 6
Let x be the investment of Ravi in 6% and y be in 7%
x(6)(n)/100 = y(7)(n)/100
=> x/y = 7/6
x : y = 7 : 6
A motorcyclist goes from Bombay to Pune, a distance of 192 kms at an average of 32 kmph speed. Another man starts from Bombay by car 2 ½ hours after the first, and reaches Pune ½ hour earlier. What is the ratio of the speed of the motorcycle and the car?
T = 192/32 = 6 h
T = 6 – 3 = 3
Time Ratio = 6:3 = 2:1
Speed Ratio = 1:2
T = 192/32 = 6 h
T = 6 – 3 = 3
Time Ratio = 6:3 = 2:1
Speed Ratio = 1:2
(180/37 of 14.8) / (17/60.1 of 180.30) = ?
(180/37 of 14.8) / (17/60.1 of 180.30) = 72/51 = 24/17 = 1.41
(180/37 of 14.8) / (17/60.1 of 180.30) = 72/51 = 24/17 = 1.41
If 5% more is gained by selling an article for Rs. 350 than by selling it for Rs. 340, the cost of the article is:
Let C.P. be Rs. x.
Then, 5% of x = 350 – 340 = 10
x/20 = 10 => x = 200
Let C.P. be Rs. x.
Then, 5% of x = 350 – 340 = 10
x/20 = 10 => x = 200
The cost price of an article is 64% of the marked price. Calculate the gain percent after allowing a discount of 12%?
Let marked price = Rs. 100.
Then, C.P. = RS. 64, S.P. = Rs. 88
Gain % = 24/64 * 100 = 37.5%.
Let marked price = Rs. 100.
Then, C.P. = RS. 64, S.P. = Rs. 88
Gain % = 24/64 * 100 = 37.5%.
Thrice the square of a natural number decreased by 4 times the number is equal to 50 more than the number. The number is:
Let the number be x. Then,
3x^{2} – 4x = x + 50
3x^{2} – 5x – 50 = 0
(3x + 10)(x – 5) = 0
x = 5
Let the number be x. Then,
3x^{2} – 4x = x + 50
3x^{2} – 5x – 50 = 0
(3x + 10)(x – 5) = 0
x = 5
A candidate got 35% of the votes polled and he lost to his rival by 2250 votes. How many votes were cast?
35%———–L
65%———–W
——————
30%———2250
100%———? => 7500
35%———–L
65%———–W
——————
30%———2250
100%———? => 7500
Two pipes A and B can separately fill a tank in 2 minutes and 15 minutes respectively. Both the pipes are opened together but 4 minutes after the start the pipe A is turned off. How much time will it take to fill the tank?
4/12 + x/15 = 1
x = 10
4/12 + x/15 = 1
x = 10
Two third of three fifth of one fourth of a number is 24, what is 30% of that number?
x * 2/3 * 3/5 * 1/4 = 24
x = 240
240 * 30/100 = 72
x * 2/3 * 3/5 * 1/4 = 24
x = 240
240 * 30/100 = 72
A cylinder and a cone have a same height and same radius of the base. The ratio between the volumes of the cylinder and cone is?
Sum of two numbers is 15. Two times of the first exceeds by 5 from the three times of the other. Then the numbers will be?
x + y = 15
2x – 3y = 5
x = 10 y = 5
x + y = 15
2x – 3y = 5
x = 10 y = 5
Find the roots of the quadratic equation: 2x^{2} + 3x – 9 = 0?
2x^{2} + 6x – 3x – 9 = 0
2x(x + 3) – 3(x + 3) = 0
(x + 3)(2x – 3) = 0
=> x = 3 or x = 3/2.
2x^{2} + 6x – 3x – 9 = 0
2x(x + 3) – 3(x + 3) = 0
(x + 3)(2x – 3) = 0
=> x = 3 or x = 3/2.
Eight years ago, Ajay’s age was 4/3 times that of Vijay. Eight years hence, Ajay’s age will be 6/5 times that of Vijay. What is the present age of Ajay?
Let the present ages of Ajay and Vijay be ‘A’ and ‘V’ years respectively.
A – 8 = 4/3 (V – 8) and A + 8 = 6/5 (V + 8)
3/4(A – 8) = V – 8 and 5/6(A + 8) = V + 8
V = 3/4 (A – 8) + 8 = 5/6 (A + 8) – 8
=> 3/4 A – 6 + 8 = 5/6 A + 20/3 – 8
=> 10 – 20/3 = 10/12 A – 9/12 A
=> 10/3 = A/12 => A = 40.
Let the present ages of Ajay and Vijay be ‘A’ and ‘V’ years respectively.
A – 8 = 4/3 (V – 8) and A + 8 = 6/5 (V + 8)
3/4(A – 8) = V – 8 and 5/6(A + 8) = V + 8
V = 3/4 (A – 8) + 8 = 5/6 (A + 8) – 8
=> 3/4 A – 6 + 8 = 5/6 A + 20/3 – 8
=> 10 – 20/3 = 10/12 A – 9/12 A
=> 10/3 = A/12 => A = 40.
Two men can complete a piece of work in four days. Two women can complete the same work in eight days. Four boys can complete the same work in five days. If four men, eight women and 20 boys work together in how many days can the work be completed?
Two men take four days to complete the work four men would take (2 * 4)/4 = 2 days to complete it.
Similarly four women would take two days to complete it and 20 children would take one day to complete it.
All the three groups working togerther will complete 1/2 + 1/2 + 1/1 work in a day
= 2 times the unit work in a day.
They will take 1/2 a day to complete it working together.
Two men take four days to complete the work four men would take (2 * 4)/4 = 2 days to complete it.
Similarly four women would take two days to complete it and 20 children would take one day to complete it.
All the three groups working togerther will complete 1/2 + 1/2 + 1/1 work in a day
= 2 times the unit work in a day.
They will take 1/2 a day to complete it working together.
If Re.1 amounts to Rs.9 over a period of 20 years. What is the rate of simple interest?
8 = (1*20*R)/100
R = 40%
8 = (1*20*R)/100
R = 40%
Two men Amar and Bhuvan have the ratio of their monthly incomes as 6 : 5. The ratio of their monthly expenditures is 3 : 2. If Bhuvan saves onefourth of his income, find the ratio of their monthly savings?
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.
A man rows his boat 85 km downstream and 45 km upstream, taking 2 1/2 hours each time. Find the speed of the stream?
Speed downstream = d/t = 85/(2 1/2) = 34 kmph
Speed upstream = d/t = 45/(2 1/2) = 18 kmph
The speed of the stream = (34 – 18)/2 = 8 kmph
Speed downstream = d/t = 85/(2 1/2) = 34 kmph
Speed upstream = d/t = 45/(2 1/2) = 18 kmph
The speed of the stream = (34 – 18)/2 = 8 kmph
A, B and C are partners in a business. Their capitals are respectively, Rs.5000, Rs.6000 and Rs.4000. A gets 30% of the total profit for managing the business. The remaining profit is divided among three in the ratio of their capitals. In the end of the year, the profit of A is Rs.200 more than the sum of the profits of B and C. Find the total profit.
A:B:C = 5:6:4
Let the total profit = 100 – 30 = 70
5/15 * 70 = 70/3
A share = 70/3 + 30 = 160/3
B + C share = 100 – 160/3 = 140/3
A(B+C) = 160/3 – 140/3 = 20/3
20/3 — 200
100 — ? => 3000
A:B:C = 5:6:4
Let the total profit = 100 – 30 = 70
5/15 * 70 = 70/3
A share = 70/3 + 30 = 160/3
B + C share = 100 – 160/3 = 140/3
A(B+C) = 160/3 – 140/3 = 20/3
20/3 — 200
100 — ? => 3000
A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they meet again at the starting point?
L.C.M of 252, 308 and 198 = 2772
So, A, B and C will again meet at the starting point in 2772 sec, i.e., 46 min 12 sec.
L.C.M of 252, 308 and 198 = 2772
So, A, B and C will again meet at the starting point in 2772 sec, i.e., 46 min 12 sec.
How much interest can a person get on Rs. 8200 at 17.5% p.a. simple interest for a period of two years and six months?
I = (8200 * 2.5 * 17.5)/100 = (8200 * 5 * 35)/(100 * 2 * 2) = Rs. 3587.50
I = (8200 * 2.5 * 17.5)/100 = (8200 * 5 * 35)/(100 * 2 * 2) = Rs. 3587.50
(112 * 5^{4}) = ?
(112 * 5^{4}) = 112 * (10/2)^{4}
= (112 * 10^{4})/24 = 1120000/16 = 70000
(112 * 5^{4}) = 112 * (10/2)^{4}
= (112 * 10^{4})/24 = 1120000/16 = 70000
In a pair of fractions, fraction A is twice the fraction B and the product of two fractions is 2/25. What is the value of fraction A?
A = 2B => B = 1/2 A, so, AB = 2/25
1/2 A^{2} = 2/25
A^{2} = 4/25
A = 2/5
A = 2B => B = 1/2 A, so, AB = 2/25
1/2 A^{2} = 2/25
A^{2} = 4/25
A = 2/5
The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?
Let the rate of interest be R% p.a.
4400{[1 + R/100]^{2} – 1} = 11193.60
[1 + R/100]^{2} = (44000 + 11193.60)/44000
[1 + R/100]^{2} = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]^{2} = 784/625 = (28/25)^{2}
1 + R/100 = 28/25
R/100 = 3/25
Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560
Let the rate of interest be R% p.a.
4400{[1 + R/100]^{2} – 1} = 11193.60
[1 + R/100]^{2} = (44000 + 11193.60)/44000
[1 + R/100]^{2} = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]^{2} = 784/625 = (28/25)^{2}
1 + R/100 = 28/25
R/100 = 3/25
Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560
If 12 men do a work in 80 days, in how many days will 16 men do it?
12 * 80 = 16 * x
x = 60 days
12 * 80 = 16 * x
x = 60 days
(0.9 * 0.9 – 0.8 * 0.8)/1.7 = ?
As numerator is of the form a^{2} – b^{2} = (a + b) (a – b), so the given expression simplifying becomes
(0.9^{2} – 0.8^{2})/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1
As numerator is of the form a^{2} – b^{2} = (a + b) (a – b), so the given expression simplifying becomes
(0.9^{2} – 0.8^{2})/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1
The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4?
Average after 11 innings = 36
Required number of runs
= (36 * 11) – (32 * 10) = 396 – 320 = 76.
Average after 11 innings = 36
Required number of runs
= (36 * 11) – (32 * 10) = 396 – 320 = 76.
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A sum of money becomes double itself in 8 years at simple interest. How many times will it become 10 years at the same rate?
P — 2P — 8 years
2 1/4 P — 10 years
P — 2P — 8 years
2 1/4 P — 10 years
The bus fare for two persons for travelling between Agra and Aligarh id fourthirds the train fare between the same places for one person. The total fare paid by 6 persons travelling by bus and 8 persons travelling by train between the two places is Rs.1512. Find the train fare between the two places for one person?
Let the train fare between the two places for one person be Rs.t
Bus fare between the two places for two persons Rs.4/3 t
=> 6/2 (4/3 t) + 8(t) = 1512
=> 12t = 1512 => t = 126.
Let the train fare between the two places for one person be Rs.t
Bus fare between the two places for two persons Rs.4/3 t
=> 6/2 (4/3 t) + 8(t) = 1512
=> 12t = 1512 => t = 126.
A bag contains five white and four red balls. Two balls are picked at random from the bag. What is the probability that they both are different color?
Two balls can be picked from nine balls in ⁹C₂ ways.
We select one white ball and one red ball from five white balls and four red balls. This can be done ⁵C₁ . ⁴C₁ ways.
The required probability = (5 * 4)/⁹C₂ = 20/36 = 5/9
Two balls can be picked from nine balls in ⁹C₂ ways.
We select one white ball and one red ball from five white balls and four red balls. This can be done ⁵C₁ . ⁴C₁ ways.
The required probability = (5 * 4)/⁹C₂ = 20/36 = 5/9
On dividing 2272 as well as 875 by 3digit number N, we get the same remainder. The sum of the digits of N is:
Clearly, (2272 – 875) = 1397, is exactly divisible by N.
Now. 1397 = 11 * 127
The required 3digit number is 127, the sum of whose digit is 10.
Clearly, (2272 – 875) = 1397, is exactly divisible by N.
Now. 1397 = 11 * 127
The required 3digit number is 127, the sum of whose digit is 10.
If Re.1 amounts to Rs.9 over a period of 20 years. What is the rate of simple interest?
8 = (1*20*R)/100
R = 40%
8 = (1*20*R)/100
R = 40%
The average age of a husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:
Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years.
Required average = 57/3 = 19 years.
Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years.
Required average = 57/3 = 19 years.
Two pipes A and B can fill a cistern in 37 1/2 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the pipe B is turned off after?
Let B be turned off after x minutes. Then, part filled by (A + B) in x min + part filled by A in (30 – x) min = 1.
x(2/75 + 1/45) + (30 x) 2/75 = 1
11x + 180 – 6x = 225 => x = 9
Let B be turned off after x minutes. Then, part filled by (A + B) in x min + part filled by A in (30 – x) min = 1.
x(2/75 + 1/45) + (30 x) 2/75 = 1
11x + 180 – 6x = 225 => x = 9
The difference between the compound interest compounded annually and simple interest for 2 years at 20% per annum is Rs.144. Find the principal?
P = 144(100/5)^{2} => P = 3600
P = 144(100/5)^{2} => P = 3600
Find the length of the wire required to go 15 times round a square field containing 69696 m^{2}.
a^{2} = 69696 => a = 264
4a = 1056
1056 * 15 = 15840
a^{2} = 69696 => a = 264
4a = 1056
1056 * 15 = 15840
The ratio of the radius of two circles is 1: 3, and then the ratio of their areas is?
r_{1}: r_{2} = 1: 3
Πr_{1}^{2}: Πr_{2}^{2}
r_{1}^{2}: r_{2}^{2 }= 1: 9
r_{1}: r_{2} = 1: 3
Πr_{1}^{2}: Πr_{2}^{2}
r_{1}^{2}: r_{2}^{2 }= 1: 9
A metallic sphere of radius 12 cm is melted and drawn into a wire, whose radius of cross section is 16 cm. What is the length of the wire?
Volume of the wire (in Cylindrical shape) is equal to the volume of the sphere.
π(16)^{2} * h = (4/3)π (12)^{3} => h = 9 cm
Volume of the wire (in Cylindrical shape) is equal to the volume of the sphere.
π(16)^{2} * h = (4/3)π (12)^{3} => h = 9 cm
A brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 25 m * 2 m * 0.75 m?
25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000
25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000
If the numerator of a fraction is increased by 20% and its denominator is diminished by 25% value of the fraction is 2/15. Find the original fraction.
X * (120/100)
————— = 2/15
Y * (75/100)
X/Y = 1/12
X * (120/100)
————— = 2/15
Y * (75/100)
X/Y = 1/12
A bag contains a total of 93 coins in the form of one rupee and 50 paise coins. If the total value of coins in the bag is Rs.56, find the number of 50 paise coins in the bag?
Let the number of one rupee coins in the bag be x.
Number of 50 paise coins in the bag is 93 – x.
Total value of coins
[100x + 50(93 – x)]paise = 5600 paise
=> x = 74
Let the number of one rupee coins in the bag be x.
Number of 50 paise coins in the bag is 93 – x.
Total value of coins
[100x + 50(93 – x)]paise = 5600 paise
=> x = 74
A and B enter into partnership with capital as 7:9. At the end of 8 months, A withdraws. If they receive the profits in the ratio of 8:9 find how long B’s capital was used?
7 * 8 : 9 * x = 8:9 => x= 7
7 * 8 : 9 * x = 8:9 => x= 7
A can do a piece of work in 21 days and B in 28 days. Together they started the work and B left after 4 days. In how many days can A alone do the remaining work?
Let A worked for x days.
x/21 + 4/28 = 1 => x/21 = 6/7 => x = 18
A worked for 18 days. So, A can complete the remaining work in 18 – 4 = 14 days.
Let A worked for x days.
x/21 + 4/28 = 1 => x/21 = 6/7 => x = 18
A worked for 18 days. So, A can complete the remaining work in 18 – 4 = 14 days.
Five men and nine women can do a piece of work in 10 days. Six men and twelve women can do the same work in 8 days. In how many days can three men and three women do the work?
(5m + 9w)10 = (6m + 12w)8
=> 50m + 90w = 48w + 96 w => 2m = 6w => 1m = 3w 5m + 9w = 5m + 3m = 8m
8 men can do the work in 10 days.
3m +3w = 3m + 1w = 4m
So, 4 men can do the work in (10 * 8)/4 = 20 days.
(5m + 9w)10 = (6m + 12w)8
=> 50m + 90w = 48w + 96 w => 2m = 6w => 1m = 3w 5m + 9w = 5m + 3m = 8m
8 men can do the work in 10 days.
3m +3w = 3m + 1w = 4m
So, 4 men can do the work in (10 * 8)/4 = 20 days.
The speed of a train is 90 kmph. What is the distance covered by it in 10 minutes?
90 * 10/60 = 15 kmph
90 * 10/60 = 15 kmph
Find the one which does not belong to that group ?
G^{1}F^{+3}I, Q^{2}O^{+3}R, L^{1}K^{+3}N, Y^{1}X^{+3}A and T^{1}S^{+3}V.
Except QOR, all other groups follows similar pattern.
G^{1}F^{+3}I, Q^{2}O^{+3}R, L^{1}K^{+3}N, Y^{1}X^{+3}A and T^{1}S^{+3}V.
Except QOR, all other groups follows similar pattern.
252 can be expressed as a product of primes as:
Clearly, 252 = 2 * 2 * 3 * 3 * 7
Clearly, 252 = 2 * 2 * 3 * 3 * 7
Find the smallest number which when divided by 13 and 16 leaves respective remainders of 2 and 5.
Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders) = (208) – (11) = 197.
Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders) = (208) – (11) = 197.
A pipe can fill a cistern in 20 minutes whereas the cistern when fill can be emptied by a leak in 28 minutes. When both pipes are opened, find when the cistern will be full?
1/20 – 1/28 = 1/70
70 minutes
1/20 – 1/28 = 1/70
70 minutes
(17^{14} * 17^{16}) / 17^{8} = ?
? = 17^{14+168} = 17^{22}
? = 17^{14+168} = 17^{22}
An officer was appointed on maximum daily wages on contract money of Rs. 4956. But on being absent for some days, he was paid only Rs. 3894. For how many days was he absent?
HCF of 4956, 3894 = 354
(4956 – 3894)/354 = 3
HCF of 4956, 3894 = 354
(4956 – 3894)/354 = 3
Dacid obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, Mathematics, Physics, Chemistry and Biology. What are his average marks?
Average = (76 + 65 + 82 + 67 + 85)/5 = 375/5 = 75.
Average = (76 + 65 + 82 + 67 + 85)/5 = 375/5 = 75.
A basket has 5 apples and 4 oranges. Three fruits are picked at random. The probability that at least 2 apples are picked is .
Total fruits = 9
Since there must be at least two apples,
(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42.
Total fruits = 9
Since there must be at least two apples,
(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42.
If a man lost 4% by selling oranges at the rate of 12 a rupee at how many a rupee must he sell them to gain 44%?
96% — 12
144% — ?
96/144 * 12 = 8
96% — 12
144% — ?
96/144 * 12 = 8
If 3 workers collect 48 kg of cotton in 4 days, how many kg of cotton will 9 workers collect in 2 days?
(3 * 4)/48 = (9 * 2)/ x
x = 72 kg
(3 * 4)/48 = (9 * 2)/ x
x = 72 kg
HCF of 3/16, 5/12, 7/8 is:
HCF of numerators = 1
LCM of denominators = 48
=> 1/48
HCF of numerators = 1
LCM of denominators = 48
=> 1/48
The sum of three numbers is 98. If the ratio of the first to the second is 2:3. And that of the second to the third is 5:8, then the second number is:
Let the three parts A, B, C.
Then, A:B = 2:3 and B:C = 5:8
= (5 * 3/5) : (8 * 3/5) = 3:24/5
A:B:C = 2:3:24/5
= 10:15:24 => B = 98 * 15/49 = 30.
Let the three parts A, B, C.
Then, A:B = 2:3 and B:C = 5:8
= (5 * 3/5) : (8 * 3/5) = 3:24/5
A:B:C = 2:3:24/5
= 10:15:24 => B = 98 * 15/49 = 30.
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs. ) is?
Let the sum be Rs. x. Then,
[x (1 + 4/100)^{2} – x] = (676/625 x – x) = 51/625 x
S.I. = (x * 4 * 2)/100 = 2x/25
51x/625 – 2x/25 = 1 or x = 625.
Let the sum be Rs. x. Then,
[x (1 + 4/100)^{2} – x] = (676/625 x – x) = 51/625 x
S.I. = (x * 4 * 2)/100 = 2x/25
51x/625 – 2x/25 = 1 or x = 625.
Income and expenditure of a person are in the ratio 5:4. If the income of the person is Rs.18000, then find his savings?
Let the income and the expenditure of the person be Rs.5x and Rs.4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income – expenditure = 5x – 4x = x
So, savings = Rs.3600.
Let the income and the expenditure of the person be Rs.5x and Rs.4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income – expenditure = 5x – 4x = x
So, savings = Rs.3600.
Tanya’s grandfather was 8 times older to her 16 years ago. He would be 3 times of her age 8 years from now. Eight years ago, what was the ratio of Tanya’s age to that of her grandfather?
16 years ago, let T = x years and G = 8x years. After 8 years from now, T = (x + 16 + 8) years and G = (8x + 16 + 8) years.
8x + 24 = 3(x + 24) => 5x = 48
8 years ago, T/G = (x + 8)/(8x + 8) = (48/5 + 8) / (8 * 48/5 + 8) = 88/424 = 11/53
16 years ago, let T = x years and G = 8x years. After 8 years from now, T = (x + 16 + 8) years and G = (8x + 16 + 8) years.
8x + 24 = 3(x + 24) => 5x = 48
8 years ago, T/G = (x + 8)/(8x + 8) = (48/5 + 8) / (8 * 48/5 + 8) = 88/424 = 11/53
Two taps can separately fill a cistern 10 minutes and 15 minutes respectively and when the waste pipe is open, they can together fill it in 18 minutes. The waste pipe can empty the full cistern in?
1/10 + 1/15 – 1/x = 1/18
x = 9
1/10 + 1/15 – 1/x = 1/18
x = 9
A train speeds past a pole in 15 sec and a platform 100 m long in 25 sec, its length is?
Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 100)/25 = x/15 => x = 150 m.
Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 100)/25 = x/15 => x = 150 m.
The radius of the two circular fields is in the ratio 3: 5 the area of the first field is what percent less than the area of the second?
r = 3 πr^{2} = 9
r = 5 πr^{2} = 25
25 π – 16 π
100 — ? => 64%
r = 3 πr^{2} = 9
r = 5 πr^{2} = 25
25 π – 16 π
100 — ? => 64%
Sum of two numbers is 80. Greater number exceeds by 5 from four times of the smaller. Find the numbers?
x + y = 80
x – 4y = 5
x = 65 y = 15
x + y = 80
x – 4y = 5
x = 65 y = 15
Which one of the following is the least number of four digits divisible by 71?
1000/71 = 14 6/71
1000 + 71 – 6 = 1065
1000/71 = 14 6/71
1000 + 71 – 6 = 1065
Find the 37.5% of 976 =
37.5 % of 976
= 37.5/100 * 976 = 375/1000 * 976 = 3/8 * 976
= 3 * 122 = 366.
37.5 % of 976
= 37.5/100 * 976 = 375/1000 * 976 = 3/8 * 976
= 3 * 122 = 366.
Rajan got married 8 years ago. His present age is 6/5 times his age at the time of his marriage. Rajan’s sister was 10 years younger to him at the time of his marriage. The age of Rajan’s sister is:
Let Rajan’s present age be x years.
Then, his age at the time of marriage = (x – 8) years.
x = 6/5 (x – 8)
5x = 6x – 48 => x = 48
Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = 30 years.
Rajan’s sister’s present age = (30 + 8) = 38 years.
Let Rajan’s present age be x years.
Then, his age at the time of marriage = (x – 8) years.
x = 6/5 (x – 8)
5x = 6x – 48 => x = 48
Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = 30 years.
Rajan’s sister’s present age = (30 + 8) = 38 years.
Find the one which does not belong to that group ?
20 = 4^{2} + 4, 42 = 6^{2} + 6, 58 = 7^{2} + 9, 72 = 8^{2} + 8 and 90 = 9^{2} + 9.
20, 42, 72 and 90 can be expressed in n^{2} + n form but not 58.
20 = 4^{2} + 4, 42 = 6^{2} + 6, 58 = 7^{2} + 9, 72 = 8^{2} + 8 and 90 = 9^{2} + 9.
20, 42, 72 and 90 can be expressed in n^{2} + n form but not 58.
A and B can do a work in 4 hours and 12 hours respectively. A starts the work at 6AM and they work alternately for one hour each. When will the work be completed?
Work done by A and B in the first two hours, working alternately = First hour A + Second hour B = 1/4 + 1/12 = 1/3.
Total time required to complete the work = 2 * 3 = 6 days.
Work done by A and B in the first two hours, working alternately = First hour A + Second hour B = 1/4 + 1/12 = 1/3.
Total time required to complete the work = 2 * 3 = 6 days.
A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy
=>5(300/x) – 10x – 50 = 0 => 150 + x^{2} + 5x = 0
multiplying both sides by 1/10x
=> x^{2} + 15x – 10x – 150 = 0
=> x(x + 15) – 10(x + 15) = 0
=> x = 10 or 15
As x>0, x = 10.
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy
=>5(300/x) – 10x – 50 = 0 => 150 + x^{2} + 5x = 0
multiplying both sides by 1/10x
=> x^{2} + 15x – 10x – 150 = 0
=> x(x + 15) – 10(x + 15) = 0
=> x = 10 or 15
As x>0, x = 10.
A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is?
Suppose, first pipe alone takes x hours to fill the tank. Then, second and third pipes will take (x – 5) and (x – 9) hours respectively to fill the tank.
1/x + 1/(x – 5) = 1/(x – 9)
(2x – 5)(x – 9) = x(x – 5)
x^{2} – 18x + 45 = 0
(x 15)(x – 3) = 0 => x = 15
Suppose, first pipe alone takes x hours to fill the tank. Then, second and third pipes will take (x – 5) and (x – 9) hours respectively to fill the tank.
1/x + 1/(x – 5) = 1/(x – 9)
(2x – 5)(x – 9) = x(x – 5)
x^{2} – 18x + 45 = 0
(x 15)(x – 3) = 0 => x = 15
Find the one which does not belong to that group ?
41, 43, 47 and 53 are prime numbers, but not 57.
41, 43, 47 and 53 are prime numbers, but not 57.
Salaries of Ravi and Sumit are in the ratio 2:3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40:57. What is Sumit’s present salary?
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then, (2x + 4000)/(3x + 4000) = 40/57
6x = 68000 => 3x = 34000
Sumit’s present salary = (3x + 4000) = 34000 + 4000 = Rs. 38,000.
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then, (2x + 4000)/(3x + 4000) = 40/57
6x = 68000 => 3x = 34000
Sumit’s present salary = (3x + 4000) = 34000 + 4000 = Rs. 38,000.
A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?
Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 4^{5} – 1.
= 2^{10} – 1 = 1024 – 1 = 1023
Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 4^{5} – 1.
= 2^{10} – 1 = 1024 – 1 = 1023
The proportion of copper and zinc in the brass is 13:7. How much zinc will there be in 100 kg of brass?
7/20 * 100 = 35
7/20 * 100 = 35
A man can swim in still water at 4.5 km/h, but takes twice as long to swim upstream than downstream. The speed of the stream is?