d) Rs. 9600
a) Rs. 1200
b) Rs. 1140
c) Rs. 960
d) Rs. 1500
e) Rs. 840
1). B) Total of balls = 6 + 11 + 5 = 22
n(S) =22C2 = (21×22) / 2 = 231
Now, n(E) =6C1 x11C1 = 6 x 11 = 66
P(E) = n(E)/n(S) = 66/231 = 6/21 = 2/7
2). A)Let the radius of the cylinder be r and height be h.
Then, r + h = 19 …..(i)
Again, total surface area of cylinder = (2ᅲrh + 2ᅲr2)
Now, 2ᅲr(h + r) = 1672
or, 2ᅲr x 19 = 1672
or, 38ᅲr = 1672 , ᅲr = (1672/38) = 44m, r = (44 × 7) / 22 = 14
Height = 19 – 14 = 5m
Volume of cylinder = ᅲr2h = (22/7) x 14 x 14 x 5 =14m
= 22 × 2 × 14 × 5 = 3080m3
3). B)Let the speed of the boat in still water be x and that of the current be y.
Then, downstream speed = x + y and upstream speed = x – y
Now, downstream speed = 42 / [2 20/60] = (42 × 3) / 7 =18 km
x+y=18
Again, 3 : 18, 2 : 12
(As ratio of downstream to upstream is 2 : 3)
x – y = 12 Solving (i) and (ii), we get
(x+y=18) + (x – y =12) = 2x =30
x = 15 kmph
Hence speed of the boat 15 kmph
4). B)35 men complete the work in 16 days.
1 man completes the work in 16 x 35 days,
32 men complete the work in (16×35)/40 = 14 days.
Again, 20 women complete the same piece of work in 30 days.
1 woman completes the same piece of work in 20 × 30 days.
15 women can complete the work in (20×30)/15 = 40 days.
Ratio = 1/14 : 1/40 = 40 : 14 = 20 : 7
5). B)Let the cost price be x.
Then, loss = (x – 6800)
Again, profit = (7850 – x)
Now, (7850 – x) = (x – 6800)/2 or, 15700 – 2x = x – 6800
or, 3x = 15700 + 6800 = 22500
x = 22500/3 = 7500
Selling price = (7500×120)/100 = Rs. 9000
6). B)Total CP = 1500
Total SP = 1500 + 10% of 1500 = 1500 + 150 = 1650
CP of 1/3 of bananas = 1500/3 = Rs.500
SP of 1/3 of bananas at 25% loss
= 500 – [ (500 x25 / 100)] = 500 – 125 = 375
SP of the rest of bananas = 1650 – 375 = 1275
Now, CP of the test of bananas = 1500 – 500 = 1000
Profit on the rest of bananas = 1275 -1000 = 275
% of profit on the rest of bananas = (275/1000)×100 = 27.5%
7). A)(1/P) – (1/X) = (1/15)
Or, (1/6) – (1/X) = (1/15) (P = 6 hours)
Or, (1/X) = (1/6) – (1/15) = (10-4)/60 = 1/10
x = 10 hours
Now,
(1/Q) – (1/10) = (1/8) – (1/10) = (5-4)/40 = 1/40
Hence, Q fills the tank in 40 hours.
8). C)According to the question, A : B = 3 : 8
A : C = 1 : 4
B : A = 8 : 3
A : C = 1 : 4
8 : 3 : 12
Sum = 8x + 3x 12x = 23x
Now, 23x = 92
x = 4
Hence the present age of C = 12x = 12 x 4 = 48 years
9). A)Let the amount invested in scheme A be Rs.x and that in B be Rs. 3x.
Then, [(x × 4 × 8)/100] [(3x × 2 × 13) /100] = 1320
Or, (32x/100) + (78x/100) = 1320
110x/100 = 1320
x = (1320 x 100) / 110 = Rs. 1200
10). D)
Let the speed of Kim be x and that of Om be y.
Then, (400/x) – (400/y) = 1
Let 1/x = u and 1/y = v
400u — 400v = 1 …(i)
Again, (400/y) – (400/2y) = 3/2
400v – 200u = (3/2)
Or, 800v – 400u = 3 …(ii)
Solving (i) and (ii), we get
(400u – 400v =1) + (-400u + 800v) = 400v = 4
v = (4/400) = (1/100) km
y = 100 km
now, (400/x) – (400/100) = 1
or, (400/x) = 5
x = 80 kmph