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

Top 10,000+ Aptitude Memory Based Exercises

Questions : A trader marks his goods 40% above cost price and allows a discount of 25 %. The profit he makes, is :

(a) 10 %

(b) 2 %

(c) 5 %

(d) 15%

Answer: (c)

Let the cost price be Rs.100.

Marked price = Rs.140

S.P. = ${75 × 140}/100$ = Rs.105

Profit per cent = 5%

Using Rule 8,

Here, r = 40%, $r_1$ = 25%

Profit % = ${r × (100 - r_1)}/100 - r_1$

= ${40 × (100 - 25)}/100 - 25$

= ${40 × 75}/100 - 25$

= $3000/100$ - 25

= 30 - 25 = 5%

Question : 1

The lengths of a train and that of a platform are equal. If with a speed of 90 km/hr the train crosses the platform in one minute, then the length of the train (in metres) is

a) 600

b) 750

c) 900

d) 500

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

Question : 2

A train 'B' speeding with 100 kmph crosses another train C, running in the same direction, in 2 minutes. If the length of the train B and C be 150 metre and 250 metre respectively, what is the speed of the train C (in kmph)?

a) 88

b) 95

c) 110

d) 75

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.

Question : 3

The sum of two numbers is 36 and their H.C.F. is 4. How many pairs of such numbers are possible ?

a) 1

b) 3

c) 2

d) 4

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)

Question : 4

Arranging the following in descending order, we get $√^3{4}, √2, √^6{3}, √^4{5}$

a) $√2>√^6{3}>√^3{4}>√^4{5}$

b) $√^3{4}>√^4{5}>√2>√^6{3}$

c) $√^4{5}>√^3{4}>√^6{3}>√2$

d) $√^6{3}>√^4{5}>√^3{4}>√2$

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,$√^3{4}>√^4{5}>√2>√^6{3}$

Question : 5

The compound interest on a certain sum of money at 5% per annum for 2 years is Rs.246. The simple interest on the same sum for 3 years at 6% per annum is

a) Rs.432

b) Rs.435

c) Rs.430

d) Rs.450

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

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