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.