Solution: 

The given equations are:

2x+3y+z=55 …(i); x + z – y=4 …(ii); y -x + z =12 …(iii)

Subtracting (ii) from (i), we get: x+4y=51  …(iv)

Subtracting (iii) from (i), we get: 3x+2y=43  …(v)

Multiplying (v) by 2 and subtracting (iv) from it, we get: 5x=35 or x=7.

Putting x=7 in (iv), we get: 4y=44 or y=11.

Putting x=7,y=11 in (i), we get: z=8

Solution:

L.C.M. of 5, 6, 7, 8 = 840.

Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

Required number = (840 x 2 + 3) = 1683.

Solution:

Let the average for 17 innings is x runs

Total runs in 17 innings = 17x

Total runs in 18 innings = 17x + 150

Average of 18 innings = 17x + 150/18

: . 17x + 150/18 = x + 6 — > x = 42

Thus, average after 18 innings = 42

Solution:

= 45 km/hr.

Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr.

x – 5 = 45          x = 50 km/hr.

Solution:

Amount of work P can do in 1 day = 1/15

Amount of work Q can do in 1 day = 1/20

Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60

Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15

Fraction of work left = 1 – 7/15= 8/15

Solution:

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 – x).

28x – 22x = 350800 – (13900 x 22)

6x = 45000

x = 7500.

So, sum invested in Scheme B = Rs. (13900 – 7500) = Rs. 6400.

Solution:

(a) Since log_a a = 1, so log₁₀ 10 = 1.

(b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3

log (2 + 3)  log (2 x 3)

(c) Since log_a 1 = 0, so log₁₀ 1 = 0.

(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.

So, (b) is incorrect.

Solution:

All 2 digit numbers divisible by 3 are :

12, 51, 18, 21, …, 99.

This is an A.P. with a = 12 and d = 3.

Let it contain n terms. Then,

12 + (n – 1) x 3 = 99 or n = 30.

\Required sum = 30/2 x (12+99) = 1665.

Solution: 

Let the two parallel sides of the trapezium be a em and b em.

Then, a – b = 4

And, (1/2) x (a + b) x 19 = 475  ó      (a + b) =((475 x 2)/19) ó a + b = 50

Solving (i) and (ii), we get: a = 27, b = 23.

So, the two parallel sides are 27 cm and 23 cm

Solution:

Let the number of keepers be x then,

Total number of heads =(50 + 45 + 8 + x)= (103 + x).

Total number of feet = (45 + 8) x 4 + (50 + x) x 2 =(312 +2x).

(312 + 2x)-(103 + x) =224ó x =15.

Hence, number of keepers =15.

Solution:

Let the first discount be x%

Then,87.5% of (100-x)% of 150= 105

87.5/100*(100-x)/100*450=150=>105=>100-x=(105*100*100)/(150*87.5)=80

x=(100-80)=20

first discount = 20%

Solution:

N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Solution:

Solution:

Solution:

Suppose commodity X will cost 40 paise more than Y after z years.

Then, (4.20 + 0.40z) – (6.30 + 0.15z) = 0.40

0.25z = 0.40 + 2.10

X will cost 40 paise more than Y 10 years after 2001 i.e., 2011.

Solution:

Relative speed = (120 + 80) km/hr

Let the length of the other train be x metres.

x + 270 = 500

x = 230.

Solution:

Each of the numbers except 81 is a prime number

Solution:

S.I. on Rs. (110 – 10) for a certain time = Rs. 10.

S.I. on Rs. 100 for double the time = Rs. 20.

T.D. on Rs. 120 = Rs. (120 – 100) = Rs. 20.

Solution:

Let the required distance be x km

Difference in the time taken at two speeds=1 min =1/2 hr

Hence x/5-x/6=1/5<=>6x-5x=6

x=6

Hence, the required distance is 6 km

Solution:

3^x – y = 27 = 3³        x – y = 3 ….(i)

3^x + y = 243 = 3⁵        x + y = 5 ….(ii)

On solving (i) and (ii), we get x = 4.

Solution: 

Let the length of the train be x meters,

Then, the train covers x meters in 8 seconds and (x + 180) meters in 20 sec

x/8=(x+180)/20 ó 20x = 8 (x + 180)       <=>      x = 120.

Length of the train = 120 m.

Speed of the train = (120/8) m / sec = m / sec = (15 *18/5) kmph = 54 km

Solution:

Total weight increased = (8 x 2.5) kg = 20 kg.

Weight of new person = (65 + 20) kg = 85 kg.

Solution:

Let the required number of revolutions made by larger wheel be x.

Then, More cogs, Less revolutions (Indirect Proportion)

14 : 6 :: 21 : x        14 x x = 6 x 21

x = 9.

Solution:

Relative speed of the trains = (72 – 54) km/hr = 18 km/hr

= (18 * 5/18) m/sec = 5 m/sec.

Time taken by the trains to cross each other

= Time taken to cover (100 + 120) m at 5 m /sec = (220/5) sec = 44 sec.

Solution:

Let the required weight be x kg.

Less weight, Less cost (Direct Proportion)

250 : 200 :: 60 : x       250 x x = (200 x 60)

x = 48.

Solution:

Let the total number of students be x .

Let A and B represent the sets of students who passed in English and Mathematics respectively

Then, number of students passed in one or both the subjects

= n (AÈB) =n (A)+n(B)- n(AÇB)=80% of x + 85% of x –75% of x

=[(80/100)x+(85/100)x-(75/100)x]=(90/100)x=(9/10)x

Students who failed in both the subjects = [x-(9x/10)]=x/10.

So, x/10=40 of x=400.

Hence, total number of students = 400

Solution:

Let the 3rd number be x, Then, second number = 3x

: . First number = 3x/2

: . x+3x + 3x/2 = (44*3)

— > 11x/2 = 132 — > x = 24

Largest number — > 3x

— > 3*24 = 72

Solution:

Solution:

Let breadth = x metres, length = 3x metres, height = H metres.

Area of the floor=(Total cost of carpeting)/(Rate/m²)=(270/5)m²=54m².

x* (3x/2) = 54 <=> x^2 = (54*2/3) = 36 <=> x = 6.

So, breadth = 6 m and length =(3/2)*6 = 9 m.

Now, papered area = (1720/10)m² = 172 m².

Area of 1 door and 2 windows = 8 m².

Total area of 4 walls = (172 + 8) m² = 180 m²

2*(9+ 6)* H = 180 <=> H = 180/30 = 6 m.

Solution:

Let original length = x and original breadth = y.

Solution:

Let radius of each be R and height of the cone be H.

Then, (4/3) ∏ R³ = (1/3) ∏ R²H (or) R/H = ¼ (or) 2R/H = 2/4 =1/2

\Required ratio = 1:2.

Solution:

Let the initial amount with Mr.Jones be Rs.x

Then,(1/2)[100-(3*20)]% of x=12000

ó (1/2)*(40/100)*(60/100)*x=12000

óx=((12000*25)/3)=100000

Solution:

S.I. = Rs. (15500 – 12500) = Rs. 3000.

Solution:

The differences between two successive terms from the beginning are 7, 5, 7, 5, 7, 5.

So, 40 is wrong.

Solution:

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o’clcok only).

So, in a day, the hands point in the opposite directions 22 times.

Solution:

Let the ten’s digit be x and unit’s digit be y.

Then, x + y = 15 and x – y = 3   or   y – x = 3.

Solving x + y = 15   and   x – y = 3, we get: x = 9, y = 6.

Solving x + y = 15   and   y – x = 3, we get: x = 6, y = 9.

So, the number is either 96 or 69.

Hence, the number cannot be determined.

Solution:

Let their investments be Rs. x for 14 months, Rs. y for 8 months and Rs. z for 7 months respectively.

Then, 14x : 8y : 7z = 5 : 7 : 8.

Solution:

Let the numbers be x and (15 – x).

Then, x² + (15 – x)² = 113           =>       x² + 225 + x² – 30x = 113

=>       2x² – 30x + 112 = 0                 =>              x² – 15x + 56 = 0

=>       (x – 7) (x – 8) = 0                         =>      x = 7 or x = 8.

So, the numbers are 7 and 8.

Solution:

Let no of passengers in the beginning be x

After first station no passengers=(x-x/3)+280=2x/3 +280

After second station no passengers =1/2(2x/3+280)+12

½(2x/3+280)+12=248

2x/3+280=2*236

2x/3=192

x=288

Solution:

S={HH,HT,TH,TT}

Let EE=event of getting one head

E={tt,ht,th}

P(e)=n(e)/n(s)=3/4

Solution:

Required percentage = [(20*100)/(100-20)]%=25%.

Solution:

Area of the park = (60 x 40) m² = 2400 m².

Area of the lawn = 2109 m².

Area of the crossroads = (2400 – 2109) m² = 291 m².

Let the width of the road be x metres. Then,

60x + 40x – x² = 291

x² – 100x + 291 = 0

(x – 97)(x – 3) = 0

x = 3.

Solution:

Let the present ages of Sameer and Anand be 5x years and 4x years respectively.

9(5x + 3) = 11(4x + 3)

45x + 27 = 44x + 33

45x – 44x = 33 – 27

x = 6.

Anand’s present age = 4x = 24 years.

Solution:

Ratio of times taken by A and B = 160 : 100 = 8 : 5.

Suppose B alone takes x days to do the job.

Then, 8 : 5 :: 12 : x = 8x = 5 x 12 =x = 7 1/2 days

Solution:

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.

Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).

14x : 21x : 28x

2 : 3 : 4.