15/25 + 10/x = 1 => x = 25
1/25 + 1/25 = 2/25
25/2 = 12 1/2 days.
30 * 25 = 750
425
-----------
325
25 + 7.50 = 32.5
325/32.5 = 10.
[(523 + 27) * (187 - 35) / (424 + 16)] / [110/22 of (2 * 38)]
= [(550 * 152) / 440] / [(110/44) * 38]
= (5 * 38)/(5/2 * 38) = 2.
Speed downstream = (5+1)km/hr = 6 km/hr
Speed upstream = (5-1)km/hr = 4 km/hr
Let the required distance be x km.
Then x/6 + x/4 = 75/60 = 5/4 => (2x + 3x) =15 => x =3.
Required distance = 3 km.
A+B+C = 1170
4A = 3B = 2C = x
A:B:C = 1/4:1/3:1/2 = 3:4:6
6/13 * 1170 = Rs.540.
Speed = [300 / 18] m/sec = 50/3 m/sec.
Let the length of the platform be x meters.
Then, x + 300 / 39 = 50/3
3(x + 300) = 1950 è x = 350m.
r = 7 h = 3
2πr(h + r) = 2 * 22/7 * 7(10) = 440.
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
A scores 60 while B score 40 and C scores 30.
The number of points that C scores when B scores 100 = (100 * 30)/40 = 25 * 3 = 75.
In a game of 100 points, B gives (100 - 75) = 25 points to C.
25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000.
If both agree stating the same fact, either both of them speak truth of both speak false.
Probability = 3/5 * 4/7 + 2/5 * 3/7
= 12/35 + 6/35 = 18/35.
Let the five consecutive odd numbers of set p be 2n - 3, 2n - 1, 2n + 1, 2n + 3, 2n + 5.
Sum of these five numbers
= 2n - 3 + 2n - 1 + 2n + 1 + 2n + 3 + 2n + 5
= 10n + 5 = 435 => n = 43
Largest number of set p = 2(43) + 5 = 91
The largest number of set q = 91 + 45 = 136
=> The five numbers of set q are 132, 133, 134, 135, 136.
Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670.
Length of the train be ‘X’
X + 120/15 = X + 180/18
6X + 720 = 5X + 900
X = 180m .
The fractions considered are 8/15 9/13 6/11
To compare them we make the denominators the same. So the fractions are
(8 * 13 * 11)/2145, (9 * 15 * 11)/2145 and (6 * 15 * 13)/2145
1144/2145, 1485/2145 and 1170/2145
so in descending order the fractions will be
1485/2145, 1170/2145 and 1144/2145 i.e., 9/13 , 6/11 , 8/15.
W = 62% L = 38%
62% - 38% = 24%
24% -------- 288
62% -------- ? => 744.
3x * 2x = 3750 => x = 25
2(75 + 50) = 250 m
250 * 1/2 = Rs.125.
2(l + b) = 50 => l + b = 25
l – b = 5
l = 15 b = 10.
Let the required percentage be x%.
3/20 = x% of 12/25 => 3/20 = x/100 * 12/25
=> 12x = (300 * 25)/20 => x = (25 * 25)/20
=> x = 31.25%.
HCF of 384, 1296 = 48
48 * 48 * x = 384 * 1296
x = 216.
The circumference of the circle is equal to the permeter of the rectangle.
Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * 3.5
=> x = 1
Therefore l = 6 cm and b = 5 cm Area of the rectangle = 6 * 5 = 30 cm2.
2x2 + 6x - 3x - 9 = 0
2x(x + 3) - 3(x + 3) = 0
(x + 3)(2x - 3) = 0
=> x = -3 or x = 3/2.
Amount = [1600 * (1 + 5/(2 * 100)2 + 1600 * (1 + 5/(2 * 100)]
= [1600 * 41/40(41/40 + 1)
= [(1600 * 41 * 81)/(40 * 40)] = Rs. 3321.
C.I. = 3321 - 3200 = Rs. 121.
Three consecutive numbers can be taken as (P - 1), P, (P + 1).
So, (P - 1) + P + (P + 1) = 102
3P = 102 => P = 34.
The lowest of the three = (P - 1) = 34 - 1 = 33.
1/20 + 1/30 = 1/12
1 + 1/3 = 4/3
1 --- 12
4/3 --- ?
4/3 * 12 = 16 hrs.
HCF of 323, 437 = 19
323 * 437 = 19 * 19 * x
x = 391.
(P*5*15)/100 - (P*3.5*15)/100 = 144
75P/100 – 52.5P/100 = 144
22.5P = 144 * 100
=> P = Rs.640.
Area of the four walls = 2h(l + b)
Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) - (6 * 3) - 3(4 * 3) = 906 sq.ft.
Total cost = 906 * 5 = Rs. 4530.
Let the length of the train be x meters and its speed be y m/sec.
They, x / y = 15 => y = x/15
x + 100 / 25 = x / 15
x = 150 m.
The ratio of the times taken is 2:1.
The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1) = 3/1 = 3:1
Speed of the stream = 42/3 = 14 kmph.
1/2:1/3:1/4 = 6:4:3
Ram = 6/13 * 3250 = 1500
Shyam = 4/13 * 3250 = 1000
Mohan = 3/13 * 3250 = 750.
(x*3*1)/100 + [(4000 - x)*5*1]/100 = 144
3x/100 + 200 – 5x/100 = 144
2x/100 = 56 è x = 2800.
60(x + 1) = 64x
X = 15
60 * 16 = 960 km.
100 * 100 = 10000
80 * 110 = 8800
10000------- 1200
100 ------- ? = 12%.
D = 250 m + 250 m = 500 m
RS = 80 + 70 = 150 * 5/18 = 125/3
T = 500 * 3/125 = 12 sec.
Let the amount with R be Rs.r
r = 2/3 (total amount with P and Q)
r = 2/3(6000 - r) => 3r = 12000 - 2r
=> 5r = 12000 => r = 2400.
D = 100 + 150 = 250
S = 36 * 5/18 = 10 mps
T = 250/10 = 25 sec.
A + B = 60, A = 2B
2B + B = 60 => B = 20 then A = 40.
5 years, their ages will be 45 and 25.
Sum of their ages = 45 + 25 = 70.
1860 + 4/7 of 21.21 - 41.4 = 1860 + 4(3.03) - 41.4
= 1860 + 12.12 - 41.4 = 1872.12 - 41.4
= 1830.72.
The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.
Area of the triangle = 1/2 * 24 * 10 = 120 cm2.
Let x percent of 80 be 64.
80 * x/100 = 64 => x = (64 * 100)/80 => x = 80.
80% of 80 is 64.
Length = 6x
Breadth = 5x
Height = 4x in cm
Therefore, 2(6x × 5x + 5x × 4x + 6x × 4x) = 33300
148x2 = 33300
=> x2 = 33300/148 = 225
=> x = 15.
Therefore, Length = 90cm,
Breadth = 75cm,
Height = 60cm
90, 75 , 60 cm.
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.
2C + 3T = 1300 --- (1)
3C + 3T = 1200 --- (2)
Subtracting 2nd from 1st, we get
-C + T = 100 => T - C = 100.
6 1/4% = 1/16
x *15/16 * 15/16 * 15/16 = 21093
x = 25600.24.
2x.....3x
8x cube + 27x cube = 945
35x cube = 945
x cube = 27 => x = 3.
1 to 11 = 11 * 10.9 = 119.9
1 to 6 = 6 * 10.5 = 63
6 to 11 = 6 * 11.4 = 68.4
63 + 68.4 = 131.4 – 119.9 = 11.5
6th number = 11.5.
WC = 3:1
WT = 1:3
x ............3x
1/x – 1/3x = 1/10
x = 20/3
3/20 + 1/20 = 1/5 => 5 days.
5/30 + 20/x = 1
x = 24
1/30 + 1/24 = 3/40
40/3 = 13 1/3 days.
65 * 6 : 84 * 5 : 100 * 3
26:28:20
C share = 74000 * 95/100 = 7030 * 20/74 => 1900.
1/4 + 1/5 = 9/20
20/9 = 2 2/9
9/20 * 2 = 9/10 ---- 4 hours
WR = 1 - 9/10 = 1/10
1 h ---- 1/4
? ----- 1/10
2/5 * 60 = 24 = 4 hrs 24 min.