model 2 hiking & discounting Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

Let, length of train = length of platform = x metre

Speed of train = 90 kmph

= $({90 × 5}/18)$ m/sec.

= 25 m/sec.

Speed of train

= $\text"Length of train and platform"/ \text"Time taken in crossing"$

25 = ${2x}/60$

2x = 25 × 60

$x = {25 × 60}/2$ = 750 metre

Answer: (a)

Let the speed of train C be x kmph.

Relative speed of B

= (100 – x ) kmph.

Time taken in crossing

= $\text"Length of both trains"/ \text"Relative speed"$

$2/60 = {({150 + 250}/1000)}/{100 – x}$

$1/30 = 2/{5(100 – x)}$

$1/6 = 2/{100 – x}$

100 – x = 12

x = 100 – 12 = 88 kmph.

Answer: (b)

HCF of two numbers = 4.

Hence, the numbers can be given by 4x and 4y

where x and y are co-prime.

Then, 4x + 4y = 36

4 (x + y) = 36

x + y = 9

Possible pairs satisfying this condition are : (1, 8), (4, 5), (2, 7)

Answer: (b)

$√^3{4}, √2, √^6{3}, √^4{5}$

LCM of indices = LCM of 3,6, 4 and 2 = 12

$√^3{4}=(4)^{1/3}=(4)^{1/12}=√^12{4^4}=√^12{256}$

$√2=(2)^{1/2}=√^12{2^6}=√^12{64}$

$√^6{3}=√^12{3^2}=√^12{9}$

$√^4{5}=√^12{5^3}=√^12{125}$

Clearly,$√^3{4}>√^4{5}>√2>√^6{3}$

Answer: (a)

Using Rule 1,

C.I. = P$[(1 + R/100)^T - 1]$

246 = P$[(1 + 5/100)^2 - 1]$

246 = P$[(21/20)^2 - 1]$

246 = P$({441 - 400}/400)$

246 = ${41P}/400$

P = ${246 × 400}/41$ = Rs.2400

SI = $\text"Principal × Time × Rate"/100$

= ${2400 × 3 × 6}/100$ = Rs.432