100

150

150-------50

100-------? => 331/3%

X - (10/100) X = X * ?

? = 90%

Let the two-digit number be 10a + b

a = b + 2 --- (1)

10a + b = 7(a + b) => a = 2b

Substituting a = 2b in equation (1), we get

2b = b + 2 => b = 2

Hence the units digit is: 2.

Let the number be in the form of 10a + b

Number formed by interchanging a and b = 10b + a.

a + b = 13 --- (1)

10b + a = 10a + b - 45

45 = 9a - 9b => a - b = 5 --- (2)

Adding (1) and (2), we get

2a = 18 => a = 9 and b = 4

The number is: 94.

Let Mudit's present age be 'm' years.

m + 18 = 3(m - 4)

=> 2m = 30 => m = 15 years.

Let there be x pupils in the class.

Total increase in marks =(x*1/2)=x/2

Therefore x/2=(83-63) =x/2=20 =x=40

(?% /100) * 120 = 90

? = 75%

X=100 y=120

120------20

100-------? => 16 2/3%

2 kmph = ( 2 x5 /18)m/sec = 5 /9m/sec.

4 kmph = ( 4 x 518)m/sec = 10 /9m/sec.

Let the length of the train be x metres and its speed by y m/sec.

Then, (x /y-5/9) = 9 and (x/y-10/9 )= 10.

9y-5=xand 10 (9y-10)=9x

9y-x=5 and 90y-9x=100

on solving we get x=50

Length of the train is 50 m

(60/100) * 50 – (40/100) * 30

30 - 12 = 18

Let the costs of each kg of apples and each kg of rice be Rs.a and Rs.r respectively.

10a = 24r and 6 * 20.50 = 2r

a = 12/5 r and r = 61.5

a = 147.6

Required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5

= 590.4 + 184.5 + 102.5 = Rs.877.40

X * (120/100) --- = 2/15

Y * (75/100)

X/Y = 1/12

I II III

120 125 100

125----------120

100-----------? => 96%

Let the numbers be x, x + 2, x + 4, x + 6, x + 8 and x + 10.

Given (x + 2) + (x + 10) = 24

=> 2x + 12 = 24 => x = 6

The fourth number = x + 6 = 6 + 6 = 12.

Let the three digit numbers be 100a + 10b + c

a = b + 2

c = b - 2

a + b + c = 3b = 18 => b = 6

So a = 8 and b = 4

Hence the three digit number is: 864

Total number of votes polled = (1136 + 7636 + 11628) = 20400.

Required percentage = (11628 / 20400 x 100) % = 57%.

Let the five consecutive even numbers be 2(x - 2), 2(x - 1), 2x, 2(x + 1) and 2(x + 2)

Their sum = 10x = 440

x = 44 => 2(x - 2) = 84

Second least number of the other set = 2(84) - 121 = 47

This set has its least number as 46.

Sum of the numbers of this set = 46 + 47 + 48 + 49 + 50

= 48 - 2 + 48 - 1 + 48 + 48 + 1 + 48 + 2 => 5(48) = 240

35%-----------L

65%-----------W

30%----------2250

100%---------? => 7500

Let the number be

2 + 1/2[1/3(a/5)] = a/15

=> 2 = a/30 => a = 60

X * (110/100) * (95/100) = 1045

X * (11/10) * (1/100) = 11

X = 1000

Let the cost of each pen and pencil be 'p' and 'q' respectively.

16p + 8q = 352 --- (1)

4p + 4q = 96

8p + 8q = 192 --- (2)

(1) - (2) => 8p = 160

=> p = 20

100 * 100 = 10000

80 * 115 = 9200

10000-----------800

100-----------? => 8% decrease

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

(f + m + d)/3 = 35

=> f + m + d = 105 --- (1)

m + 15 = f + d

Substituting f + d as m + 15 in (1), we get

2m + 15 = 105

2m = 90 => m = 45 years.

Suppose he move 4 km downstream in x hours. Then,

Speed downstream =(4/x)km/hr

Speed up stream =(3/x)km/hr

48/4/x+48(3/x)=14 or x=1/2

So,speed downstream = 8 km/hr,speed upstream =6km/hr

Rate of the stream = 1/2(8-6)km/hr =1 km/hr.

Let the two-digit number be 10a + b

a + b = 12 --- (1)

If a>b, a - b = 6

If b>a, b - a = 6

If a - b = 6, adding it to equation (1), we get

2a = 18 => a =9

so b = 12 - a = 3

Number would be 93.

if b - a = 6, adding it to the equation (1), we get

2b = 18 => b = 9

a = 12 - b = 3.

Number would be 39.

Therefore, Number would be 39 or 93.

Let B = 100

A = 50

C * (150/100) = 50

3C = 100

C = 33.3 then 'C' Cheapest

Let the present ages of Anand and Bala be 'a' and 'b' respectively.

a - 10 = 1/3 (b - 10) --- (1)

b = a + 12

Substituting b = a + 12 in first equation,

a - 10 = 1/3 (a + 2) => 3a - 30 = a + 2

=> 2a = 32 => a = 16.