A. Rs.100
B. Rs.700
C. Rs.1600
D. Rs.2000
Let the amount be divided into 7x,4x,3x and 2x.
7x-2x = 500 5x=500 x = 100
Total amount = 7x+4x+3x+2x = 16x = Rs.1600
A. 1:4
B. 3:4
C. 1:3
D. 2:3
x = z+(20/100)z = 1.2z y = 1.8z x:y = 1.2z:1.8z = 12:18 = 2:3
A. 18
B. 15
C. 12
D. 5
2x/25 = 6/(x/3) 2x/25 = 18/x 2x*x = 18*25 x² = 9*25 x=3*5 x=15
A. 1:2
B. 3:5
C. 1:3
D. 3:4
72 should be divisible by addition of the above ratio.
Using trial and error method
For (a): – 1+2 = 3, 72 is divisible by 3
For (b): – 3+5 = 8, 72 is divisible by 8
For (a): – 1+3 = 4, 72 is divisible by 4
For (a): – 3+4 = 7, 72 is not divisible by 7
Therefore, the answer is option (d)
A. 2:3
B. 3:2
C. 3:4
D. 4:3
x+y = 750 y-x = 150 Solving,
y = 450
x = 300 x:y = 300:450 = 2:3
A. 3:2
B. 1:2
C. 1:5
D. 2:1
p = q+(50/100)q = 1.5q p:q = 1.5q:q = 1.5:1 = 3:2
A. 8
B. 6
C. 12
D. 18
Let the numbers be 3x,5x and 6x. (3x)²+(5x)²+(6x)²=280 9x²+25x²+36x²=280 70x²=280 x²=4 x=2 The greatest number is 6x=12.
A. 9:19
B. 19:9
C. 15:28
D. 28:15
Let A1 be the area of flat and A2 be that of the commercial space
Total cost = area * rate
Therefore, cost of flat = A1*4500 ; cost of commercial space = A2*9500
Both the above costs are same
A1*4500 = A2*9500
A1:A2 = 9500:4500 = 19:9
A. 44
B. 55
C. 66
D. 77
Let the numbers be a, b and c. a+b+c = 172 a:b = 8:11 b:c = 5:7 11a = 8b a = 8b/11 7b = 5c c = 7b/5 (8b/11)+b+(7b/5) = 172 40b+55b+77b = 172*55 172b = 172*55 b=55
A. Rs.18000
B. Rs.13000
C. Rs.15000
D. Rs.10000
4S+5D+3C=130000 3C=2D 3D=2S S=3D/2 [4*(3D/2)]+5D+2D = 130000 6D+5D+2D = 130000 13D = 130000 D = Rs.10000 S=3D/2 = Rs.15000