model 6 sold in circular path Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (b)

Let the cost price of the bicycle for A be Rs.x

Cost price for B = selling price for A = 1.20x

Cost price for C = selling price for B = (1.25) (1.20x) = 1.5x

But 1.5x = 225 ⇒ x = $225/1.5$ = Rs.150

The cost price of the bicycle for A = Rs.150

Here, $r_1= 20%, r_2$ = 25%

C.P. for C = C.P. for A

$(1 + r_1/100)(1 + r_2/100)$

225 = C.P. for A

$(1 + 20/100)(1 + 25/100)$

C.P. for A = ${225 × 100 × 100}/{120 × 125}$ = Rs.150

Answer: (a)

Let the C.P. of A be Rs.x, then

$x × 110/100 × 120/100 = 264$

$x = {264 × 100 × 100}/{110 × 120}$ = Rs.200

Using Rule 15,

Here, $r_1 = 10%, r_2$ = 20%

C.P. for C = C.P. for A

$(1 + r_1/100)(1 + r_2/100)$

264 = C.P. for A

$(1 + 10/100)(1 + 20/100)$

C.P. for A = ${264 × 100 × 100}/{110 × 120}$ = Rs.200

Answer: (c)

Let the C.P. of A be Rs.x, then

${x × 125}/100 × 90/100$ = 675

$x = {675 × 100 × 100}/{125 × 90}$ = Rs.600

Using Rule 15,

Here, $r_1 = 25%, r_2$ = 10%, (Loss)

C.P. for C = C.P. for A

$(1 + r_1/100)(1 + r_2/100)$

675 = C.P. for A

$(1 + 25/100)(1 + 10/100)$

C.P. for A = ${675 × 100 × 100}/{125 × 90}$ = Rs.600

Answer: (a)

Let the C.P. for A be Rs.x, then

$x × 105/100 × 110/100$ = 2310

$x = {2310 × 100 × 100}/{105 × 110}$ = Rs.2000

Using Rule 15,

Here, $r_1 = 5%, r_2$ = 10%

C.P. for C = C.P. for A

$(1 + r_1/100)(1 + r_2/100)$

2310 = C.P. for A

$(1 + 5/100)(1 + 10/100)$

C.P. for A = ${2310 × 100 × 100}/{105 × 110}$ = 2000