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4 + 9/10 + 3/11 + 7/15 

= 4 + 9/10 + 3/11 + 7/15 

=1617+90+154330=1861330

4 + 9/10 + 3/11 + 7/15 

= 4 + 9/10 + 3/11 + 7/15 

=1617+90+154330=1861330

We are GIVEN a 5 digit number

Let FIRST digit be 'X'

then 5th digit is '3X'

let 2ND digit be 'Y'

then 3rd digit is 'Y - 3'

and 4th digit is 'Y + 4'

then the no is '(X)(Y)(Y - 3)(Y + 4)(3X)'

from the above we can say 3X <= 9

so X<=3 and any of the digit in the number is <= 9

and also given that 3 pairs sum is 11... 

Also Y–3 ≥ 0 and Y+4 ≤ 9

⇒ Y ≥ 3 and Y ≤ 5 i.e. Y = 3,4 or 5

for x = 1, all the conditions won’t be satisfied.

For x = 2,

Let Y = 5

∴ The number is 25296

We are given a 5 digit number

Let first digit be 'X'

then 5th digit is '3X'

let 2nd digit be 'Y'

then 3rd digit is 'Y - 3'

and 4th digit is 'Y + 4'

then the no is '(X)(Y)(Y - 3)(Y + 4)(3X)'

from the above we can say 3X <= 9

so X<=3 and any of the digit in the number is <= 9

and also given that 3 pairs sum is 11... 

Also Y–3 ≥ 0 and Y+4 ≤ 9

⇒ Y ≥ 3 and Y ≤ 5 i.e. Y = 3,4 or 5

for x = 1, all the conditions won’t be satisfied.

For x = 2,

Let Y = 5

∴ The number is 25296

We have interior ANGLE as 40 degrees. 

HENCE LET n be the number of sides of that polygon, 

Then formula is 

⇒180 × (n-2) = 40 × n 

⇒180n – 360 = 40n 

⇒140n = 360 

Since, no INTEGER value of n is possible, hence such a polygon can not exist

We have interior angle as 40 degrees. 

Hence Let n be the number of sides of that polygon, 

Then formula is 

⇒180 × (n-2) = 40 × n 

⇒180n – 360 = 40n 

⇒140n = 360 

Since, no integer value of n is possible, hence such a polygon can not exist

Let’s find a number which is COMPLETELY divisible by 4, 6, 8, 12 and 16.

To find such number, we NEED to find the LCM of all these numbers

LCM of (4,6) , 8 , 12 , 16

= 12 , 8 , 12 , 16

= LCM of (12 , 8) , LCM of (12 , 16)

= 24 , 48

= LCM of (24 , 48)

= 48

HENCE 48 + 2 ie 50 will leave a remainder of 2 if divided by any of the following numbers : 4, 6, 8, 12 and 16

Let’s find a number which is completely divisible by 4, 6, 8, 12 and 16.

To find such number, we need to find the LCM of all these numbers

LCM of (4,6) , 8 , 12 , 16

= 12 , 8 , 12 , 16

= LCM of (12 , 8) , LCM of (12 , 16)

= 24 , 48

= LCM of (24 , 48)

= 48

Hence 48 + 2 ie 50 will leave a remainder of 2 if divided by any of the following numbers : 4, 6, 8, 12 and 16

<P>LET those 2 numbers be p and q . 

We are GIVEN that HCF (p, q) = 13 and

LCM (p, q) = 273. 

We know that product of numbers = HCF × LCM 

Hence, p × q = 13 × 273 

⇒ p × q = 13 × 13 × 3 × 7 

P cannot be 13 × 3 or 13 × 13 or 7 × 3 or 13 or 3 or 7 as these numbers do not come between 60 and 140. 

Hence we can say that p = 13 × 7 = 91

Let those 2 numbers be p and q . 

We are given that HCF (p, q) = 13 and

LCM (p, q) = 273. 

We know that product of numbers = HCF × LCM 

Hence, p × q = 13 × 273 

⇒ p × q = 13 × 13 × 3 × 7 

P cannot be 13 × 3 or 13 × 13 or 7 × 3 or 13 or 3 or 7 as these numbers do not come between 60 and 140. 

Hence we can say that p = 13 × 7 = 91

(5863 - √2704) × 0.5 = ?

(5863 - √2704) × 0.5 = X

⇒ (5863 – 52) × 0.5 = x

⇒ x = 2905.5

(5863 - √2704) × 0.5 = ?

(5863 - √2704) × 0.5 = x

⇒ (5863 – 52) × 0.5 = x

⇒ x = 2905.5

GIVEN EXPRESSION is- 

76% of 1285 = 35% of 1256 +?

⇒ (76/100) × 1285 = (35/100) × 1256 +?

⇒ 97660/100 = 43960/100 +?

⇒ ? = (97660/100) – (43960/100)

⇒ ? = 53700/100 = 537

Given expression is- 

76% of 1285 = 35% of 1256 +?

⇒ (76/100) × 1285 = (35/100) × 1256 +?

⇒ 97660/100 = 43960/100 +?

⇒ ? = (97660/100) – (43960/100)

⇒ ? = 53700/100 = 537

Finding the roots of given equations X2 – 14x + 45 = 0 x2 – 9 X – 5 x + 45 = 0 (x – 5) (x – 9) = 0

Finding the roots of given equations x2 – 14x + 45 = 0 x2 – 9 x – 5 x + 45 = 0 (x – 5) (x – 9) = 0

Given, distance between Salem and Koyambedu = 342 km. 

Speed of bus = 57 kmph 

Time = Distance/Speed 

⇒ Time needed by the bus to travel from Koyambedu to Salem = 342/57 hours = 6 hours. 

Time actually taken by the bus = 6 hours and 35 minutes.

Bus takes 35 minutes extra ⇒ The COMPULSORY HALT time of bus = 35 minutes

Given, distance between Salem and Koyambedu = 342 km. 

Speed of bus = 57 kmph 

Time = Distance/Speed 

⇒ Time needed by the bus to travel from Koyambedu to Salem = 342/57 hours = 6 hours. 

Time actually taken by the bus = 6 hours and 35 minutes.

Bus takes 35 minutes extra ⇒ The compulsory halt time of bus = 35 minutes

Cyclist SPEED is 10kmph and he'll take 30 MIN to cover 5 KMS. 

If Cyclist come 25 min early he would be 5 min before crossing. 

So it means train takes 5 min to reach crossing. 

We have speed of train calculated as 

⇒ Speed= distance/time 

⇒ Speed = 5km/5 min 

⇒ Speed = 1 km/min 

Or 60 km/hr

Cyclist speed is 10kmph and he'll take 30 min to cover 5 kms. 

If Cyclist come 25 min early he would be 5 min before crossing. 

So it means train takes 5 min to reach crossing. 

We have speed of train calculated as 

⇒ Speed= distance/time 

⇒ Speed = 5km/5 min 

⇒ Speed = 1 km/min 

Or 60 km/hr

We know that 

DISTANCE = TIME × speed 

Suppose FIRST TRAIN travelled the distance x kms from A. 

Then second train will travel the distance (x + 120) kms from B. 

Let they MET each other after t time 

∴ (x + 120)/60 = (x/50) 

⇒ x = 600 kms 

But the distances between two point is (x + x + 120) 

= 2x +120 

= 2× 600 + 120 

= 1320 km

We know that 

Distance = time × speed 

Suppose first train travelled the distance x kms from A. 

Then second train will travel the distance (x + 120) kms from B. 

Let they met each other after t time 

∴ (x + 120)/60 = (x/50) 

⇒ x = 600 kms 

But the distances between two point is (x + x + 120) 

= 2x +120 

= 2× 600 + 120 

= 1320 km

Both truck and car covers a distance = 640 km

Time taken by truck is 10 HRS.

∴ Speed of truck = 640/10 = 64 km/hr

Time taken by car is 8 hrs.

∴ Speed of car = 640/8 = 80 km/hr

∴ Ratio between speed of truck and car =

Speed of truck : speed of car = 64 : 80 = 4 : 5

Both truck and car covers a distance = 640 km

Time taken by truck is 10 hrs.

∴ Speed of truck = 640/10 = 64 km/hr

Time taken by car is 8 hrs.

∴ Speed of car = 640/8 = 80 km/hr

∴ Ratio between speed of truck and car =

Speed of truck : speed of car = 64 : 80 = 4 : 5

Basic pay per hour of normal WORK = 200/40 = Rs.5

Over time is paid at 25% more = (125/100) × 5 =Rs.6.25 per hour

Man was paid Rs.200 for his REGULAR work and Rs.100 for over time (total Rs.300).

NUMBER of regular HOURS = 40

Number of overtime hours at 6.25 per hour = 100 ÷ 6.25 = 16 hours.

Total number of hours worked = 40+16 = 56 hours.

Basic pay per hour of normal work = 200/40 = Rs.5

Over time is paid at 25% more = (125/100) × 5 =Rs.6.25 per hour

Man was paid Rs.200 for his regular work and Rs.100 for over time (total Rs.300).

Number of regular hours = 40

Number of overtime hours at 6.25 per hour = 100 ÷ 6.25 = 16 hours.

Total number of hours worked = 40+16 = 56 hours.

Work done by 9 men in a day = 1/20 

∴Work done by 1 man in a day for 7.5 hours = (1/20) ÷ 9 = 1/180 

∴ Work done by 1 man in an hour = (1/180) ÷ (7.5) = 1/1350 

Given that 2 men of latter TYPE are equal to 3 men of former type ⇒ 1 man of latter type = 1.5 men of former type. 

∴ Work done by 1 man of latter type in an hour = 1.5 × (1/1350) = 1/900 

∴ Work done by 1 man in 6 hours = 6 × (1/900) = 1/150 

∴ Work done by 12 men in a day working 6 hours/day = 12 × (1/150) = 2/25 

Days REQUIRED to finish the work = 25/2 = 12.5 days.

Work done by 9 men in a day = 1/20 

∴Work done by 1 man in a day for 7.5 hours = (1/20) ÷ 9 = 1/180 

∴ Work done by 1 man in an hour = (1/180) ÷ (7.5) = 1/1350 

Given that 2 men of latter type are equal to 3 men of former type ⇒ 1 man of latter type = 1.5 men of former type. 

∴ Work done by 1 man of latter type in an hour = 1.5 × (1/1350) = 1/900 

∴ Work done by 1 man in 6 hours = 6 × (1/900) = 1/150 

∴ Work done by 12 men in a day working 6 hours/day = 12 × (1/150) = 2/25 

Days required to finish the work = 25/2 = 12.5 days.

Let quantity of solution sold be 1 litre

When the solution is sold for 297, the profit percent = 10%

HENCE CP of of solution = SP × 100/(100+P%)

⇒CP = 297 × 100/(100+10)

⇒CP = 270

RATIO of quantities of brand A and brand B is 3:7

Let their COSTS be 2C and 3c.

⇒ 0.3×2c + 0.7×3c = 270

⇒ 2.7c = 270

⇒ c = 100

Hence cost of Brand B = 3c = 300

Let quantity of solution sold be 1 litre

When the solution is sold for 297, the profit percent = 10%

Hence CP of of solution = SP × 100/(100+P%)

⇒CP = 297 × 100/(100+10)

⇒CP = 270

Ratio of quantities of brand A and brand B is 3:7

Let their costs be 2c and 3c.

⇒ 0.3×2c + 0.7×3c = 270

⇒ 2.7c = 270

⇒ c = 100

Hence cost of Brand B = 3c = 300

% of Liquid in SOLUTION A = 20%

⇒ % of Water in Solution A = 80%

Volume of Solution A = 12 Litres

Volume of Water in Solution A = (80/100) × 12 = 9.6 Litres

% of Liquid in Solution B = 30%

⇒ % of Water in Solution A = 70%

Volume of Solution B = 10 Litres

Volume of Water in Solution B = (70/100) × 10 = 7 Litres

If Solution A is MIXED with Solution B

⇒Net TOTAL Volume of Solution = (12 + 10) = 22 LTRS

⇒Net Total Volume of Water in Solution = (9.6 + 7) = 16.6 Ltrs

% of Water in New Mixture = (Volume of Water in New Mixture / Volume of New Mixture) × 100

% of Water in New Mixture = (16.6 / 22) × 100 = 75.45%

% of Liquid in Solution A = 20%

⇒ % of Water in Solution A = 80%

Volume of Solution A = 12 Litres

Volume of Water in Solution A = (80/100) × 12 = 9.6 Litres

% of Liquid in Solution B = 30%

⇒ % of Water in Solution A = 70%

Volume of Solution B = 10 Litres

Volume of Water in Solution B = (70/100) × 10 = 7 Litres

If Solution A is mixed with Solution B

⇒Net Total Volume of Solution = (12 + 10) = 22 Ltrs

⇒Net Total Volume of Water in Solution = (9.6 + 7) = 16.6 Ltrs

% of Water in New Mixture = (Volume of Water in New Mixture / Volume of New Mixture) × 100

% of Water in New Mixture = (16.6 / 22) × 100 = 75.45%

Let the quantity of tea a = x

Let the quantity of tea b = y

Let the quantity of tea c = z

Given : y = z ...... (i)

x + y + z = 100 ....... (ii)

Cost of Mixture = (Sum of Cost of INDIVIDUAL ITEMS in a mixture / TOTAL quantity of mixture)

90 = (95x + 100y + 70z) / (x + y + z)

90x + 90y + 90z = 95x + 100y + 70z

5x + 10y = 20z

5x = 20Y - 10y = 10y............................... (from Eq. (i), y =z )

x = 2y .................... (iii)

From Equation (i), (ii) & (iii)

x + y + z = 100

2y + y + y = 100

4y = 100

y = 25

Let the quantity of tea a = x

Let the quantity of tea b = y

Let the quantity of tea c = z

Given : y = z ...... (i)

x + y + z = 100 ....... (ii)

Cost of Mixture = (Sum of Cost of individual items in a mixture / total quantity of mixture)

90 = (95x + 100y + 70z) / (x + y + z)

90x + 90y + 90z = 95x + 100y + 70z

5x + 10y = 20z

5x = 20y - 10y = 10y............................... (from Eq. (i), y =z )

x = 2y .................... (iii)

From Equation (i), (ii) & (iii)

x + y + z = 100

2y + y + y = 100

4y = 100

y = 25

We need to MIX rice varieties costing Rs 9.3 and Rs 10.8 PER kg to MAKE the mixture at Rs 10 /kg

To do so,

We need to ADD these varieties in the ratio

⇒10.8 - 10 : 10 - 9.3 

= 0.8 : 0.7

= 8 : 7

We need to mix rice varieties costing Rs 9.3 and Rs 10.8 per kg to make the mixture at Rs 10 /kg

To do so,

We need to add these varieties in the ratio

⇒10.8 - 10 : 10 - 9.3 

= 0.8 : 0.7

= 8 : 7

We have SP = 75 per LITRE 

Profit Per cent = 37.5% 

Hence 

⇒ CP = SP × 100/(100 + 37.5) 

⇒ CP = 7500/137.5 = 54.5454

Hence, to MAKE the CP from 60 to 54.54, 

Hence, 

REQUIRED RATIO is 

= 54.54 : 60 - 54.54 

= 54.54 : 5.454 

= 10:1

We have SP = 75 per litre 

Profit Per cent = 37.5% 

Hence 

⇒ CP = SP × 100/(100 + 37.5) 

⇒ CP = 7500/137.5 = 54.5454

Hence, to make the CP from 60 to 54.54, 

Hence, 

Required ratio is 

= 54.54 : 60 - 54.54 

= 54.54 : 5.454 

= 10:1

Let quantity of grass initially = g 

Let r be the rate at which grass grow in 1 day, 

Let c be the quantity of grass a cow EAT in 1 day

we can deduce from ‘. It takes 40 days for 40 COWS and 60 days for 30 cows to eat the whole of the grass’ that 

⇒ g + 40r = 40 × 40c = 1600c-----------[1] 

and

⇒ g + 60r = 60 × 30c = 1800c 

⇒ g = 1800c - 60r----------------2

Putting this value of g in eqn [1] 

we have 

⇒ 1800c – 60r + 40r = 1600c 

⇒ 200c = 20r 

⇒ C = 0.1 r 

⇒ r = 10c 

Let m be the number of days required by 20 cows to eat the entire of the field = m 

Then 

We have the eqn as 

⇒ g + 96r = 96nc 

⇒ 96nc = 1800c - 60r + 96r (as we have g=1800c-60r from eqn 2) 

⇒ 96nc = 1800c + 36r 

⇒ 96nc = 1800c+ 36 × 10c 

⇒ 96nc = 1800c + 360c 

⇒ 96nc = 2160c 

HENCE we have 96nc = 2160c 

Dividing both the sides by 96C 

We get 

⇒ N = 22.5

Let quantity of grass initially = g 

Let r be the rate at which grass grow in 1 day, 

Let c be the quantity of grass a cow eat in 1 day

we can deduce from ‘. It takes 40 days for 40 cows and 60 days for 30 cows to eat the whole of the grass’ that 

⇒ g + 40r = 40 × 40c = 1600c-----------[1] 

and

⇒ g + 60r = 60 × 30c = 1800c 

⇒ g = 1800c - 60r----------------2

Putting this value of g in eqn [1] 

we have 

⇒ 1800c – 60r + 40r = 1600c 

⇒ 200c = 20r 

⇒ C = 0.1 r 

⇒ r = 10c 

Let m be the number of days required by 20 cows to eat the entire of the field = m 

Then 

We have the eqn as 

⇒ g + 96r = 96nc 

⇒ 96nc = 1800c - 60r + 96r (as we have g=1800c-60r from eqn 2) 

⇒ 96nc = 1800c + 36r 

⇒ 96nc = 1800c+ 36 × 10c 

⇒ 96nc = 1800c + 360c 

⇒ 96nc = 2160c 

Hence we have 96nc = 2160c 

Dividing both the sides by 96c 

We get 

⇒ n = 22.5

Since we are given that a PERSON is elder in 490 than in 500. 

Hence, he must have been born in BC. 

Hence, we can say that he must have been born in BC 490+80 

= BC 570

Since we are given that a person is elder in 490 than in 500. 

Hence, he must have been born in BC. 

Hence, we can say that he must have been born in BC 490+80 

= BC 570

We KNOW that, formula: 

Volume of any Cuboid = Length × Breadth × Height 

According to the given question: 

⇒ Volume of the water required in the tank = (200 × 150 × 2) m3 = 60000 m3 . . . 

∴ Length of water column flown in1 MIN =(20 × 1000)/60 m = 1000/3 m 

∴ Volume flown per minute = 1.5 × 1.25 × (1000/3) m3 = 625 m3 . 

∴ Required time = (60000/625)min = 96min. 

Hence, the required answer is 96 minutes.

We know that, formula: 

Volume of any Cuboid = Length × Breadth × Height 

According to the given question: 

⇒ Volume of the water required in the tank = (200 × 150 × 2) m3 = 60000 m3 . . . 

∴ Length of water column flown in1 min =(20 × 1000)/60 m = 1000/3 m 

∴ Volume flown per minute = 1.5 × 1.25 × (1000/3) m3 = 625 m3 . 

∴ Required time = (60000/625)min = 96min. 

Hence, the required answer is 96 minutes.

LET x and y are the digits of the number with x in ten’s place and y in one’s place. 

So, 10x+y is the value of the number. 

Given 10x + y = 3 (x + y) 

⇒7x – 2y = 0 …………….(i) 

And also given that, 

10x + y + 45 = 10y + x (∵ digits are reversed⇒ x in one’s and y in ten’s place). 

⇒ 9x – 9y = -45 

⇒ x – y = -5 ………………(ii) 

PUT value of y from eq. (i) to eq. (ii) 

⇒ x – (7x/2) = -5 

⇒ x = 2 

And y = 7

So the number is 27

Let x and y are the digits of the number with x in ten’s place and y in one’s place. 

So, 10x+y is the value of the number. 

Given 10x + y = 3 (x + y) 

⇒7x – 2y = 0 …………….(i) 

And also given that, 

10x + y + 45 = 10y + x (∵ digits are reversed⇒ x in one’s and y in ten’s place). 

⇒ 9x – 9y = -45 

⇒ x – y = -5 ………………(ii) 

Put value of y from eq. (i) to eq. (ii) 

⇒ x – (7x/2) = -5 

⇒ x = 2 

And y = 7

So the number is 27

⇒ 5 mangoes + 4 oranges = 7 mangoes + 1 orange 

⇒4 orange – 1 orange = 7 mangoes – 5 mangoes 

⇒ 3 orange = 2 mangoes 

Hence Required RATIO is 

⇒ Cost of mango: cost of orange = 3 : 2

⇒ 5 mangoes + 4 oranges = 7 mangoes + 1 orange 

⇒4 orange – 1 orange = 7 mangoes – 5 mangoes 

⇒ 3 orange = 2 mangoes 

Hence Required Ratio is 

⇒ Cost of mango: cost of orange = 3 : 2

We are given that

6B2 = A2 + 540 and A : B = 3 : 2

Put the VALUE of A from 2nd EQUATION into the first equation, we GET

6B2 = (1.5B)2 + 540

→ (6 – 2.25)B2 = 540 → 3.75B2 = 540 → B2 = 144 → B = 12

We are given that

6B2 = A2 + 540 and A : B = 3 : 2

Put the value of A from 2nd equation into the first equation, we get

6B2 = (1.5B)2 + 540

→ (6 – 2.25)B2 = 540 → 3.75B2 = 540 → B2 = 144 → B = 12

Here the DIFFERENCE between each term is in the FORM of 13+1, 33+1, 53+1, 73+1 and so on

4 6 34 160 504 1234

+2 +28 +126 +344 +730

13 + 1 33 + 1 53 + 1 73 + 1 93 + 1

So = 160

Here the difference between each term is in the form of 13+1, 33+1, 53+1, 73+1 and so on

4 6 34 160 504 1234

+2 +28 +126 +344 +730

13 + 1 33 + 1 53 + 1 73 + 1 93 + 1

So = 160

Let x be the cost of each capsule and y be the no of capsules

Therefore, x = 216/y and x - 10 = 216/(y+15)

Putting the VALUE of x from eq. 1 into eq. 2, we get

216/y - 216/(y+15) = 10

Y2 +15y - 324 = 0

Solving for y, we get y = 12

Let x be the cost of each capsule and y be the no of capsules

Therefore, x = 216/y and x - 10 = 216/(y+15)

Putting the value of x from eq. 1 into eq. 2, we get

216/y - 216/(y+15) = 10

y2 +15y - 324 = 0

Solving for y, we get y = 12

(1097.63 + 2197.36 - 2607.24) ÷ 3.5 = 687.75 ÷ 3.5

= 196.5

(1097.63 + 2197.36 - 2607.24) ÷ 3.5 = 687.75 ÷ 3.5

= 196.5

(6) = (216)4 ÷ (36)4 × (6)5 =

2164/364 * 65 =

612 / 68 * 65=

69 Therefore, = 9

(6) = (216)4 ÷ (36)4 × (6)5 =

2164/364 * 65 =

612 / 68 * 65=

69 Therefore, = 9

Using BODMAS

2 = (43)2 + 841 – 1465

= 1849 + 841 – 1465 = 1225

Therefore, = √1225 = 35

Using BODMAS

2 = (43)2 + 841 – 1465

= 1849 + 841 – 1465 = 1225

Therefore, = √1225 = 35

(1331)1/3/275 *(5)2

= (113)1/3 / 11*25 *(25)

=1 

(1331)1/3/275 *(5)2

= (113)1/3 / 11*25 *(25)

=1 

6*136/8+132 / 628/16-26.25

⇒ 234/13=18 

6*136/8+132 / 628/16-26.25

⇒ 234/13=18 

<P>Let P be the SUM

Therefore,

3584 = (P * 7 * 4) / 100 => P = 12800/-

CI = 12800(1 + (4/100))2 - 12800

=> 13844.48 - 12800 = 1044.48/-

Let P be the sum

Therefore,

3584 = (P * 7 * 4) / 100 => P = 12800/-

CI = 12800(1 + (4/100))2 - 12800

=> 13844.48 - 12800 = 1044.48/-