Simple Interest
1. What will be the ratio of simple interest earned by a certain amount at the same rate of interest for 6 years and that for 9 years?
A. 1 : 3
B. 1 : 4
C. 2 : 3
D. Data inadequate
E. None of these
Answer: Option C
Explanation:
Let the principal be P and rate of interest be R%.
Required ratio =[latex]\frac{\frac{P\times R\times 6}{100}}{\frac{P\times R\times 9}{100}}[/latex]
=[latex]\frac{ P\times R\times 6}{ P\times R\times 9}[/latex] = [latex]\frac{6}{9}[/latex] = 2 : 3.
2. A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:
A. 5%
B. 7%
C. 7[latex]\frac{1}8{}[/latex]%
D. 10%
Answer: Option D
Explanation:
Let the rate be R% p.a.
Then, ([latex]\frac{5000 \times R \times 2}{100}[/latex]) + ([latex]\frac{3000 \times R \times 4}{100}[/latex]) = 2200.
100R + 120R = 2200
R = [latex]\frac{2200}{220}[/latex] = 10.
Rate = 10%.
3. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?
A. Rs. 4462.50
B. Rs. 8032.50
C. Rs. 8900
D. Rs. 8925
E. None of these
Answer: Option D
Explanation:
Principal
= Rs. [latex]\frac{100 \times 4016.25}{9 \times 5}[/latex]
= Rs. [latex]\frac{401625}{45}[/latex]
= Rs. 8925.
Height and Distance.
1. From a point P on level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:
A. 149 m
B. 156 m
C. 173 m
D. 200 m
Answer: Option C
Explanation:
Let AB be the tower.
Then, APB = 30° and AB = 100 m.
[latex]\frac{AB}{AP}[/latex] = tan 30° = [latex]\frac{1}{\sqrt 3}[/latex]
AP = (AB x [latex]\sqrt 3[/latex]) m
= 100 [latex]\sqrt 3[/latex] m
= (100 x 1.73) m
= 173 m.
2. The angle of elevation of the sun, when the length of the shadow of a tree 3 times the height of the tree, is:
A. 30°
B. 45°
C. 60°
D. 90°
Answer: Option A
Explanation:
Let AB be the tree and AC be its shadow.
Then, [latex]\frac{AC}{AB}[/latex] = [latex] \sqrt 3[/latex] = cot = [latex] \sqrt 3[/latex]
= 30°.
3. The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:
A. 2.3 m
B. 4.6 m
C. 7.8 m
D. 9.2 m
Answer: Option D
Explanation:
Let AB be the wall and BC be the ladder.
Then, ACB = 60° and AC = 4.6 m.
[latex]\frac{AC}{BC}[/latex] = cos 60° = [latex]\frac{1}{2}[/latex]
BC = 2 x AC
= (2 x 4.6) m
= 9.2 m.
Volume and Surface Area
1. 50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m3, then the rise in the water level in the tank will be:
A. 20 cm
B. 25 cm
C. 35 cm
D. 50 cm
Answer: Option B
Explanation:
Total volume of water displaced = (4 x 50) m[latex]^{3}[/latex] = 200 m[latex]^{3}[/latex].
Rise in water level = [latex]\frac{200}{40 \times 20}[/latex] m 0.25 m = 25 cm.
2. In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of the ground is:
A. 75 cu. m
B. 750 cu. m
C. 7500 cu. m
D. 75000 cu. m
Answer: Option B
Explanation:
1 hectare = 10,000 m[latex]^{2}[/latex]
So, Area = (1.5 x 10000) m[latex]^{2}[/latex] = 15000 m[latex]^{2}[/latex].
Depth = [latex]\frac{5}{100}[/latex] m = [latex]\frac{1}{20}[/latex] m.
Volume = (Area x Depth) = 15000 x [latex]\frac{1}{20}[/latex] m[latex]^{3}[/latex] = 750 m[latex]^{3}[/latex].
Logarithm
1. If log 27 = 1.431, then the value of log 9 is:
A. 0.934
B. 0.945
C. 0.954
D. 0.958
Answer: Option C
Explanation:
log 27 = 1.431
log (3[latex]^{3}[/latex] ) = 1.431
3 log 3 = 1.431
log 3 = 0.477
log 9 = log(3[latex]^{2}[/latex] ) = 2 log 3 = (2 x 0.477) = 0.954.
2. If [latex]log_{10}[/latex] 2 = 0.3010, the value of [latex]log_{10}[/latex] 80 is:
A. 1.6020
B. 1.9030
C. 3.9030
D. None of these
Answer: Option B
Explanation:
[latex]log_{10}[/latex] 80 = [latex]log_{10}[/latex] (8 x 10)
= [latex]log_{10}[/latex] 8 + [latex]log_{10}[/latex] 10
= [latex]log_{10}[/latex] (2[latex]^{3}[/latex] ) + 1
= 3 [latex]log_{10}[/latex] 2 + 1
= (3 x 0.3010) + 1
= 1.9030.
Races and Games
1. A runs 1[latex]\frac {2}{3}[/latex] times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?
A. 200 m
B. 300 m
C. 270 m
D. 160 m
Answer: Option A
Explanation:
The ratio of the speeds of A and B = [latex]\frac {5}{3}[/latex] : 1 = 5 : 3.
Thus, in the race of 5 m, A gains 2m over B.
2m is gained by A in a race of 5m.
80m will be gained by A in race of ([latex]\frac {5}{2}[/latex] x 80) m = 200 m.
Winning post is 200 m away from the starting point.
2. In a race of 200 m, A can beat B by 31 m and C by 18 m. In a race of 350 m, C will beat B by:
A. 22.75 m
B. 25 m
C. 19.5 m
D. 7[latex]\frac {4}{7}[/latex]m
Answer: Option B
Explanation:
A : B = 200 : 169.
A : C = 200 : 182.
[latex]\frac {C}{B}[/latex] = ([latex]\frac {C}{A}[/latex] x [latex]\frac {A}{B}[/latex]) = ([latex]\frac {182}{200}[/latex] x [latex]\frac {200}{169}[/latex]) = 182 : 169.
When C covers 182 m, B covers 169 m.
When C covers 350 m, B covers ([latex]\frac {169}{182}[/latex] x 350) m = 325 m.
Therefore, C beats B by (350 - 325) m = 25 m.
Simplification
1. If a - b = 3 and a[latex]^{2}[/latex] + b[latex]^{2}[/latex] = 29, find the value of ab.
Answer: Option A
Explanation:
2ab = (a[latex]^{2}[/latex] + b[latex]^{2}[/latex]) - (a - b)[latex]^{2}[/latex]
= 29 - 9 = 20
ab = 10.
2. To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifths of its present?
A. 10
B. 35
C. 62.5
D. Cannot be determined
E. None of these
Answer: Option C
Explanation:
Let the capacity of 1 bucket = x.
Then, the capacity of tank = 25x.
New capacity of bucket = [latex]\frac {2}{5}[/latex]x
Required number of buckets = [latex]\frac {25 \times x}{(2 \times x/5)}[/latex]
= 25x x [latex]\frac {5}{2 \times x}[/latex]
= [latex]\frac {125}{2}[/latex]
= 62.5
Time and Distance
1. The ratio between the speeds of two trains is 7:8. If the second train runs 400 km in 4 hours, then the speed of the first train is:
A. 70 km/hr
B. 75 km/hr
C. 84 km/hr
D. 87.5 km/hr
Answer: Option D
Explanation:
Let the speed of two trains be 7x and 8x km/hr.
Then, 8x = [latex]\frac {400}{4}[/latex] = 100
x = [latex]\frac {100}{8}[/latex] = 12.5
Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.
2. In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:
A. 5 kmph
B. 6 kmph
C. 6.25 kmph
D. 7.5 kmph
Answer: Option A
Explanation:
Let Abhay's speed be x km/hr.
Then, [latex]\frac {30}{x}[/latex] - [latex]\frac {30}{2x}[/latex] = 3
6x = 30
x = 5 km/hr.
Chain Rule
1. If the cost of x meters of wire is d rupees, then what is the cost of y meters of wire at the same rate?
A. Rs.[latex]\frac {xy}{d}[/latex]
B. Rs. (xd)
C. Rs. (yd)
D. Rs. [latex]\frac {yd}{x}[/latex]
Answer: Option D
Explanation:
Cost of x metres = Rs. d.
Cost of 1 metre = Rs. [latex]\frac {d}{x}[/latex]
Cost of y metres = Rs. ([latex]\frac {d}{x}[/latex] . y) = Rs. [latex]\frac {yd}{x}[/latex] .
2. If a quarter kg of potato costs 60 paise, how many paise will 200 gm cost?
A. 48 paise
B. 54 paise
C. 56 paise
D. 72 paise
Answer: Option A
Explanation:
Let the required weight be x kg.
Less weight, Less cost (Direct Proportion)
250 : 200 :: 60 : x = 250 x x = (200 x 60)
x= [latex]\frac {(200 \times 60)}{250}[/latex]
x = 48.
Permutation and Combination
1. In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
A. 360
B. 480
C. 720
D. 5040
E. None of these
Answer: Option C
Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
2. In how many ways can the letters of the word 'LEADER' be arranged?
A. 72
B. 144
C. 360
D. 720
E. None of these
Answer: Option C
Explanation:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
The required number of ways = [latex]\frac {6!}{(1!)(2!)(1!)(1!)(1!)}[/latex] = 360.
Surds and Indices
1. If 5[latex]^{a}[/latex] = 3125, then the value of 5[latex]^{(a - 3)}[/latex] is:
A. 25
B. 125
C. 625
D. 1625
Answer: Option A
Explanation:
5[latex]^{a}[/latex] = 3125 = 5[latex]^{a}[/latex] = 5[latex]^{5}[/latex]
a = 5.
5[latex]^{(a-3)}[/latex] = 5[latex]^{(5 - 3)}[/latex] = 5[latex]^{2}[/latex] = 25.
2. If m and n are whole numbers such that m[latex]^{n}[/latex] = 121, the value of (m - 1)[latex]^{n + 1}[/latex] is:
A. 1
B. 10
C. 121
D. 1000
Answer: Option D
Explanation:
We know that 11[latex]^{2}[/latex] = 121.
Putting m = 11 and n = 2, we get:
(m - 1)[latex]^{n + 1}[/latex] = (11 - 1)[latex]^{(2 + 1)}[/latex] = 103 = 1000.
Pipes and Cistern
1. Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P, Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?
A. [latex]\frac {5}{11}[/latex]
B. [latex]\frac {6}{11}[/latex]
C. [latex]\frac {7}{11}[/latex]
D. [latex]\frac {8}{11}[/latex]
Answer: Option B
Explanation:
Part filled by (A + B + C) in 3 minutes = 3 ([latex]\frac {1}{30} + \frac {1}{20} + \frac {1}{10}[/latex]) = 3 x [latex]\frac {11}{60}[/latex] = [latex]\frac {11}{20}[/latex] .
Part filled by C in 3 minutes = [latex]\frac {3}{10}[/latex] .
Required ratio =([latex]\frac {3}{10}[/latex] x [latex]\frac {20}{11}[/latex]) = [latex]\frac {6}{11}[/latex].
2. A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
A. 20 hours
B. 25 hours
C. 35 hours
D. Cannot be determined
E. None of these
Answer: Option C
Explanation:
Suppose pipe A alone takes x hours to fill the tank.
Then, pipes B and C will take [latex]\frac {x}{2}[/latex]and [latex]\frac {x}{4}[/latex] hours respectively to fill the tank.
[latex]\frac {1}{x}[/latex] + [latex]\frac {2}{x}[/latex] + [latex]\frac {4}{x}[/latex] = [latex]\frac {1}{5}[/latex]
[latex]\frac {7}{x}[/latex] = [latex]\frac {1}{5}[/latex]
x = 35 hrs.
Boats and Streams
1. A boat running downstream covers a distance of 16 km in 2 hours while for covering the same distance upstream, it takes 4 hours. What is the speed of the boat in still water?
A. 4 km/hr
B. 6 km/hr
C. 8 km/hr
D. Data inadequate
Answer: Option B
Explanation:
Rate downstream = [latex]\frac {16}{2}[/latex] kmph = 8 kmph.
Rate upstream = [latex]\frac {16}{4}[/latex] kmph = 4 kmph.
Speed in still water = [latex]\frac {1}{2}[/latex](8 + 4) kmph = 6 kmph.
2. A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph, the speed of the stream is:
A. 2 mph
B. 2.5 mph
C. 3 mph
D. 4 mph
Answer: Option A
Explanation:
Let the speed of the stream x mph. Then,
Speed downstream = (10 + x) mph,
Speed upstream = (10 - x) mph.
[latex]\frac {36}{(10 - x)}[/latex] - [latex]\frac {36}{(10 + x)}[/latex] = [latex]\frac {90}{60}[/latex]
72x x 60 = 90 (100 - x[latex]^{2}[/latex])
x[latex]^{2}[/latex] + 48x - 100 = 0
(x+ 50)(x - 2) = 0
x = 2 mph.
Numbers
1. 1397 x 1397 = ?
A. 1951609
B. 1981709
C. 18362619
D. 2031719
E. None of these
Answer: Option A
Explanation:
1397 x 1397 = (1397)[latex]^{2}[/latex]
= (1400 - 3)[latex]^{2}[/latex]
= (1400)[latex]^{2}[/latex] + (3)[latex]^{2}[/latex] - (2 x 1400 x 3)
= 1960000 + 9 - 8400
= 1960009 - 8400
= 1951609.
2. 72519 x 9999 = ?
A. 725117481
B. 674217481
C. 685126481
D. 696217481
E. None of these
Answer: Option A
Explanation:
72519 x 9999 = 72519 x (10000 - 1)
= 72519 x 10000 - 72519 x 1
= 725190000 - 72519
= 725117481.
Partnership
1. A, B and C enter into a partnership in the ratio [latex]\frac {7}{4}[/latex] :[latex]\frac {4}{3}[/latex] : [latex]\frac {6}{5}[/latex] . After 4 months, A increases his share 50%. If the total profit at the end of one year be Rs. 21,600, then B's share in the profit is:
A. Rs. 2100
B. Rs. 2400
C. Rs. 3600
D. Rs. 4000
Answer: Option D
Explanation:
Ratio of initial investments = [latex]\frac {7}{2}[/latex] : [latex]\frac {4}{3}[/latex] : [latex]\frac {6}{5}[/latex] = 105 : 40 : 36.
Let the initial investments be 105x, 40x and 36x.
A : B : C = (105x x 4 + [latex]\frac {150}{100}[/latex] x 105x x 8) : (40x x 12) : (36x x 12)
= 1680x : 480x : 432x = 35 : 10 : 9.
Hence, B's share = Rs. (21600 x [latex]\frac {10}{54}[/latex]) = Rs. 4000.
2. A and B started a business in partnership investing Rs. 20,000 and Rs. 15,000 respectively. After six months, C joined them with Rs. 20,000. What will be B's share in the total profit of Rs. 25,000 earned at the end of 2 years from the starting of the business?
A. Rs. 7500
B. Rs. 9000
C. Rs. 9500
D. Rs. 10,000
Answer: Option A
Explanation:
A : B : C = (20,000 x 24) : (15,000 x 24) : (20,000 x 18) = 4 : 3 : 3.
B's share = Rs. (25000 x [latex]\frac {3}{10}[/latex]) = Rs. 7,500.
Ratio and Proportion
1. In a mixture 60 liters, the ratio of milk and water 2: 1. If this ratio is to be 1: 2, then the quantity of water to be further added is:
A. 20 liters
B. 30 liters
C. 40 liters
D. 60 liters
Answer: Option D
Explanation:
Quantity of milk = (60 x [latex]\frac {2}{3}[/latex] litres = 40 litres.
Quantity of water in it = (60- 40) litres = 20 litres.
New ratio = 1 : 2
Let quantity of water to be added further be x liters.
Then, milk : water = [latex]\frac {40}{20 + x}[/latex].
Now, [latex]\frac {40}{20 + x}[/latex] = [latex]\frac {1}{2}[/latex]
20 + x = 80
x = 60.
The quantity of water to be added = 60 liters.
2. Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40: 57. What is Sumit's salary?
A. Rs. 17,000
B. Rs. 20,000
C. Rs. 25,500
D. Rs. 38,000
Answer: Option D
Explanation:
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then,[latex]\frac {2x + 4000}{3x + 4000}[/latex] = [latex]\frac {40}{57}[/latex]
57(2x + 4000) = 40(3x + 4000)
6x = 68,000
3x = 34,000
Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.
Problems on H.C.F and L.C.M
1. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
A. 101
B. 107
C. 111
D. 185
Answer: Option C
Explanation:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
1369ab = 4107
ab = [latex]\frac { 4107}{1369}[/latex]
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
2. The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
A. 1677
B. 1683
C. 2523
D. 3363
Answer: Option B
Explanation:
L.C.M. of 5, 6, 7, 8 = 840.
The required number is of form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
3. 252 can be expressed as a product of primes as:
A. 2 x 2 x 3 x 3 x 7
B. 2 x 2 x 2 x 3 x 7
C. 3 x 3 x 3 x 3 x 7
D. 2 x 3 x 3 x 3 x 7
Answer: Option A
Explanation:
Clearly, 252 = 2 x 2 x 3 x 3 x 7.
Banker’s Discount
1. The banker's gain on a sum due 3 years hence at 12% per annum is Rs. 270. The banker's discount is:
A. Rs. 960
B. Rs. 840
C. Rs. 1020
D. Rs. 760
Answer: Option C
Explanation:
T.D. = [latex]\frac {B.G. \times 100}{R \times T}[/latex] = Rs. [latex]\frac {270 \times 100}{12 \times 3}[/latex] = Rs. 750.
B.D. = Rs.(750 + 270) = Rs. 1020.
2. The banker's discount of a certain sum of money is Rs. 72 and the true discount on the same sum for the same time is Rs. 60. The sum due is:
A. Rs. 360
B. Rs. 432
C. Rs. 540
D. Rs. 1080
Answer: Option A
Explanation:
Sum = [latex]\frac {B.D. \times T.D.}{B.D. - T.D.}[/latex] = Rs. [latex]\frac {72 \times 60}{72 - 60}[/latex] = Rs. [latex]\frac {72 \times 60}{12}[/latex]= Rs. 360.
Compound Interest
1. There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?
A. Rs. 2160
B. Rs. 3120
C. Rs. 3972
D. Rs. 6240
E. None of these
Answer: Option C
Explanation:
Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
R = [latex]\frac {100 \times 60}{100 \times 6}[/latex] = 10% p.a.
Now, P = Rs. 12000. T = 3 years and R = 10% p.a.
C.I. = Rs. (12000 x((1 + [latex]\frac {10}{100})^{3}[/latex] - 1))
= Rs. (12000 x [latex]\frac {331}{1000}[/latex])
= 3972.
2. Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit?
A. Rs. 8600
B. Rs. 8620
C. Rs. 8820
D. None of these
Answer: Option C
Explanation:
Amount
= Rs. (8000 x (1 + [latex]\frac {5}{100})^{2}[/latex])
= Rs. (8000 x [latex]\frac {21}{20}[/latex] x [latex]\frac {21}{20}[/latex])
= Rs. 8820.
Area
1. In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of the ground is:
A. 75 cu. m
B. 750 cu. m
C. 7500 cu. m
D. 75000 cu. m
Answer: Option B
Explanation:
1 hectare = 10,000 m[latex]^{2}[/latex]
So, Area = (1.5 x 10000) m[latex]^{2}[/latex] = 15000 m[latex]^{2}[/latex].
Depth = [latex]\frac {5}{100}[/latex] m = [latex]\frac {1}{20}[/latex] m.
Volume = (Area x Depth) = 15000 x [latex]\frac {1}{20}[/latex] m[latex]^{3}[/latex] = 750 m[latex]^{3}[/latex].
2. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:
A. 720
B. 900
C. 1200
D. 1800
Answer: Option C
Explanation:
2(15 + 12) x h = 2(15 x 12)
h = [latex]\frac {180}{27}[/latex]m = [latex]\frac {20}{3}[/latex]m.
Volume = 15 x 12 x [latex]\frac {20}{3}[/latex]m[latex]^{3}[/latex] = 1200 m[latex]^{3}[/latex].
Time and Work
1. A is thrice as good as workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:
A. 20 days
B. 22 [latex]\frac {1}{2}[/latex]days
C. 25 days
D. 30 days
Answer: Option B
Explanation:
The ratio of times taken by A and B = 1 : 3.
The time difference is (3 - 1) 2 days while B takes 3 days and A takes 1 day.
If the difference in time is 2 days, B takes 3 days.
If difference of time is 60 days, B takes ([latex]\frac {3}{2}[/latex] x 60) = 90 days.
So, A takes 30 days to do the work.
A's 1 day's work = [latex]\frac {1}{30}[/latex]
B's 1 day's work = [latex]\frac {1}{90}[/latex]
(A + B)it's 1 day's work = [latex]\frac {1}{30}[/latex] + [latex]\frac {1}{90}[/latex] = [latex]\frac {4}{90}[/latex] = [latex]\frac {2}{45}[/latex]
A and B together can do the work in [latex]\frac {45}{2}[/latex] = 22 [latex]\frac {1}{2}[/latex] days.
2. A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10 hours while machine R can print them in 12 hours. All the machines are started at 9 A.M. while machine P is closed at 11 A.M. and the remaining two machines complete work. Approximately at what time will the work (to print one lakh books) be finished?
A. 11:30 A.M.
B. 12 noon
C. 12:30 P.M.
D. 1:00 P.M.
Answer: Option D
Explanation:
(P + Q + R)'s 1 hour's work = ([latex]\frac {1}{8}[/latex] + [latex]\frac {1}{10}[/latex] + [latex]\frac {1}{12}[/latex] = [latex]\frac {37}{120}[/latex]
Work was done by P, Q, and R in 2 hours =([latex]\frac {37}{120}[/latex] x 2 )= [latex]\frac {37}{60}[/latex]
Remaining work = ( 1 - [latex]\frac {37}{60}[/latex]) = [latex]\frac {23}{60}[/latex].
(Q + R)'s 1 hour's work = ([latex]\frac {1}{10}[/latex] + [latex]\frac {1}{12}[/latex] ) = [latex]\frac {11}{60}
[/latex].
Now, [latex]\frac {11}{60}[/latex] work is done by Q and R in 1 hour.
So, [latex]\frac {23}{60}[/latex] work will be done by Q and R in ( [latex]\frac {60}{11}[/latex] x [latex]\frac {23}{60}[/latex]) = [latex]\frac {23}{11}[/latex] hours = 2 hours.
So, the work will be finished approximately 2 hours after 11 A.M., i.e., around 1 P.M.
Allegation or Mixture
1. A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A. 4 litres, 8 litres
B. 6 litres, 6 litres
C. 5 litres, 7 litres
D. 7 litres, 5 litres
Answer: Option B
Explanation:
Let the cost of 1 litre milk be Re. 1
Milk in 1 litre mix. in 1st can = [latex]\frac {3}{4}[/latex] litre, C.P. of 1 litre mix. in 1st can Re. [latex]\frac {3}{4}[/latex]
Milk in 1 litre mix. in 2nd can =[latex]\frac {1}{2}[/latex] litre, C.P. of 1 litre mix. in 2nd can Re. [latex]\frac {1}{2}[/latex]
Milk in 1 litre of final mix. = [latex]\frac {5}{8}[/latex] litre, Mean price = Re. [latex]\frac {5}{8}[/latex]
By the rule of alligation, we have:
Ratio of two mixtures = [latex]\frac {1}{8}[/latex] : [latex]\frac {1}{8}[/latex] = 1 : 1.
So, quantity of mixture taken from each can = ([latex]\frac {1}{2}[/latex] x 12) = 6 litres.
2. A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is:
A. 4%
B. 6[latex]\frac {1}{4}[/latex]%
C. 20%
D. 25%
Answer: Option C
Explanation:
Let C.P. of 1 litre milk be Re. 1
Then, S.P. of 1 litre of mixture = Re. 1, Gain = 25%.
C.P. of 1 litre mixture = Re.([latex]\frac {100}{125}[/latex] x 1) = [latex]\frac {4}{5}[/latex]
By the rule of alligation, we have:
Ratio of milk to water = [latex]\frac {4}{5}[/latex] : [latex]\frac {1}{5}[/latex] = 4 : 1.
Hence, percentage of water in the mixture = ([latex]\frac {1}{5}[/latex] x 100) % = 20%.
3. The cost of Type 1 rice is Rs. 15 per kg and Type 2 rice is Rs. 20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is:
A. Rs. 18
B. Rs. 18.50
C. Rs. 19
D. Rs. 19.50
Answer: Option A
Explanation:
Let the price of the mixed variety be Rs. x per kg.
By rule of alligation, we have:
[latex]\frac {(20 - x)}{(x - 15)}[/latex] = [latex]\frac {2}{3}[/latex]
60 - 3x = 2x - 30
5x = 90
x = 18.
Decimal Fraction
1. [latex]\frac {4.2 \times 4.2 - 1.9 \times 1.9}{2.3 \times 6.1}[/latex] is equal to:
A. 0.5
B. 1.0
C. 20
D. 22
Answer: Option B
Explanation:
Given Expression = [latex]\frac {a^{2} - b^{2} }{(a + b)(a - b)}[/latex]) = ([latex]\frac {a^{2} - b^{2}} {a^{2} - b^{2}}[/latex]) = 1.
2. If [latex]\frac {144}{0.144}[/latex] = [latex]\frac {14.4}{x}[/latex], then the value of x is:
A. 0.0144
B. 1.44
C. 14.4
D. 144
Answer: Option A
Explanation:
[latex]\frac {144}{0.144}[/latex] = [latex]\frac {14.4}{x}[/latex]
[latex]\frac {144 \times 1000}{144}[/latex] =[latex]\frac {14.4}{x}[/latex]
x =[latex]\frac {14.4}{1000}[/latex] = 0.0144
Probability
1. What is the probability of getting a sum 9 from two throws of a dice?
A.[latex]\frac {1}{6}[/latex]
B.[latex]\frac {1}{8}[/latex]
C.[latex]\frac {1}{9}[/latex]
D.[latex]\frac {1}{12}[/latex]
Answer: Option C
Explanation:
In two throws of a dice, n(S) = (6 x 6) = 36.
Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.
P(E) = [latex]\frac {n(E)}{n(S)}[/latex] = [latex]\frac {4}{36}[/latex]
= [latex]\frac {1}{9}[/latex].
2. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
A.[latex]\frac {1}{2}[/latex]
B.[latex]\frac {3}{4}[/latex]
C.[latex]\frac {3}{8}[/latex]
D.[latex]\frac {5}{16}[/latex]
Answer: Option B
Explanation:
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(E) = 27.
P(E) = [latex]\frac {n(E)}{n(S)}[/latex] = [latex]\frac {27}{36}[/latex]
= [latex]\frac {3}{4}[/latex].
Average
1. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?
A. 23 years
B. 24 years
C. 25 years
D. None of these
Answer: Option A
Explanation:
Let the average age of the whole team by x years.
11x - (26 + 29) = 9(x -1)
11x - 9x = 46
2x = 46
x = 23.
So, average age of the team is 23 years.
2. The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:
A. 3500
B. 4000
C. 4050
D. 5000
Answer: Option B
Explanation:
Let P, Q and R represent their respective monthly incomes. Then, we have:
P + Q = (5050 x 2) = 10100 .... (i)
Q + R = (6250 x 2) = 12500 .... (ii)
P + R = (5200 x 2) = 10400 .... (iii)
Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 .... (iv)
Subtracting (ii) from (iv), we get P = 4000.
P's monthly income = Rs. 4000.
Stocks and Share
1. A man invests some money partly in 9% stock at 96 and partly in 12% stock at 120. To obtain equal dividends from both, he must invest the money in the ratio:
A. 3 : 4
B. 3 : 5
C. 4 : 5
D. 16 : 15
Answer: Option D
Explanation:
For an income of Re. 1 in 9% stock at 96, investment = Rs.[latex]\frac {96}{9}[/latex] = Rs.[latex]\frac {32}{3}[/latex]
For an income Re. 1 in 12% stock at 120, investment = Rs.[latex]\frac {120}{12}[/latex] = Rs. 10.
Ratio of investments = [latex]\frac {32}{3}[/latex] : 10 = 32 : 30 = 16 : 15.
2. The market value of a 10.5% stock, in which an income of Rs. 756 is derived by investing Rs. 9000, brokerage being [latex]\frac {1}{4}[/latex] %, is:
A. Rs. 108.25
B. Rs. 112.20
C. Rs. 124.75
D. Rs. 125.25
Answer: Option C
Explanation:
For an income of Rs. 756, investment = Rs. 9000.
For an income of Rs.[latex]\frac {21}{2}[/latex], investment = Rs.([latex]\frac {9000}{756}[/latex] x [latex]\frac {21}{2}[/latex]) = Rs. 125.
For a Rs. 100 stock, investment = Rs. 125.
Market value of Rs. 100 stock = Rs. (125 - [latex]\frac {1}{4}[/latex]) = Rs. 124.75
Square Root and Cube Root
1. What should come in place of both x in the equation [latex]\frac {x}{\sqrt {128}}[/latex] = [latex]\frac {\sqrt {162}}{x}[/latex].
A. 12
B. 14
C. 144
D. 196
Answer: Option A
Explanation:
Let[latex]\frac {x}{\sqrt {128}}[/latex] = [latex]\frac {\sqrt {162}}{x}[/latex]
Then x[latex]^{2}[/latex] = [latex]\sqrt{128 \times 162}[/latex]
= [latex]\sqrt{64 \times 2 \times 18 \times 9}[/latex]
=[latex]\sqrt{82 \times 62 \times 32}[/latex]
=[latex]\sqrt{8 \times 6 \times 3}[/latex]
= [latex]\sqrt{144}[/latex].
x = [latex]\sqrt{144}[/latex] = 12.
2. If 3[latex]\sqrt{5}[/latex] + [latex]\sqrt{125}[/latex] = 17.88, then what will be the value of [latex]\sqrt{80}[/latex] + 6[latex]\sqrt{5}[/latex] ?
A. 13.41
B. 20.46
C. 21.66
D. 22.35
Answer: Option D
Explanation:
3[latex]\sqrt{5}[/latex]+ [latex]\sqrt{125}[/latex]= 17.88
3[latex]\sqrt{5}[/latex] + [latex]\sqrt{12 \times 5}[/latex] = 17.88
3[latex]\sqrt{5}[/latex] + 5[latex]\sqrt{5}[/latex] = 17.88
8[latex]\sqrt{5}[/latex] = 17.88
[latex]\sqrt{5}[/latex] = 2.235
= [latex]\sqrt{80}[/latex] + 6[latex]\sqrt{5}[/latex] = [latex]\sqrt{16 \times 5}[/latex] + 6[latex]\sqrt{5}[/latex]
= 4[latex]\sqrt{5}[/latex] + 6[latex]\sqrt{5}[/latex]
= 10[latex]\sqrt{5}[/latex] = (10 x 2.235) = 22.35
Problems on Ages
1. Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand's present age in years?
A. 24
B. 27
C. 40
D. Cannot be determined
E. None of these
Answer: Option A
Explanation:
Let the present ages of Sameer and Anand be 5x years and 4x years respectively.
Then,[latex]\frac {5x + 3}{4x + 3}[/latex] = [latex]\frac {11}{9}[/latex]
9(5x + 3) = 11(4x + 3)
45x + 27 = 44x + 33
45x - 44x = 33 - 27
x = 6.
Anand's present age = 4x = 24 years.
2. A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:
A. 14 years
B. 18 years
C. 20 years
D. 22 years
Answer: Option D
Explanation:
Let the son's present age be x years. Then, man's present age = (x + 24) years.
(x + 24) + 2 = 2(x + 2)
x + 26 = 2x + 4
x = 22.