Height and Distance
1. The angle of elevation of the sun is 60°. Find the length of the shadow of a man who is 180 cm tall.
A. 127.27 cm
B. 103.92 cm
C. 311.77 cm
D. None of these
Answer: Option B
Explanation:
Let AB be the man and BC be his shadow
AB = 180
tan60° = AB/BC
[latex]\sqrt {3}[/latex] = 180/BC
BC = 103.92 cm
2. A ship is approaching an observation tower. If the time taken by the ship to change the angle of elevation from 30° to 45° is 10 minutes, then find the time the ship will take to cover the remaining distance and reach the observation tower assuming the ship to be traveling at a uniform speed.
A. 15 minutes 20 seconds
B. 13 minutes 40 seconds
C. 16 minutes 40 seconds
D. Cannot be determined
Answer: Option B
Explanation:
Let AB be the observation tower and h be its height.
Also, let the ship be at C when the angle of elevation is 30° and at D when the angle of elevation is 45°.
The time taken by the ship to travel from C to D is 10 minutes and we need to find out the time the ship will take to reach B from D.
tan30 = AB/CB = h/CB = [latex]\frac{1}{\sqrt {3}}[/latex]
CB = [latex]\sqrt {3}[/latex]*h
tan45 = AB/DB = h/DB = 1
DB = h
CD = CB – DB = ([latex]\sqrt {3}[/latex]h – h) = h([latex]\sqrt {3}[/latex] – 1)
Now, as h([latex]\sqrt {3}[/latex] – 1) distance is covered in 10 minutes, a distance of h is covered in = 13.66 minutes = 13 minutes 40 seconds
3. On the two sides of a road are two tall buildings exactly opposite to each other. The height of the taller building is 60 m. If the angle of elevation from the top of the smaller building to the top of the taller one is 30° and the angle of depression from top of the taller building to the foot of the smaller one is 30°, then find the height of the smaller building.
Answer: Option C
Explanation:
Let AB be the taller building of height 60m and CD be the smaller one of height h m.
tan30= DB/AB = DB/60 = [latex]\frac{1}{\sqrt {3}}[/latex]
DB = 34.64 m
tan30= AE/CE = AE/DB = [latex]\frac{1}{\sqrt {3}}[/latex] = AE/34.64
AE = 60 – h = 20
h = 40 m
Volume and Surface Area
1. Find the number of cubes of 2cm edge will be required to fill a cubical box of 2m edge.
A. 10[latex]^{5}[/latex]
B. 10[latex]^{3}[/latex]
C. 10[latex]^{4}[/latex]
D. 10[latex]^{6}[/latex]
Answer: Option D
Explanation:
Volume of the box = N*volume of the cubes … (N = number of cubes)
2[latex]^{3}[/latex] = N*(0.02)[latex]^{3}[/latex]
8 = N*(0.000008)
N = 10[latex]^{6}[/latex]
2. What is the mass of the man, if a boat of length 4m, width 1.5m floating on a river sinks by 0.015 m when the man gets on the boat?
A. 100kg
B. 90kg
C. 80kg
D. 70kg
Answer: Option B
Explanation:
Volume of water displaced = 4*1.5*0.015 = 0.09m[latex]^{3}[/latex]
Mass = volume of water displaced * density of water
Mass = 0.09 * 1000
Mass = 90kg
3. What is the ratio of the volumes of a right cylinder to that of a right cone having equal diameters?
A. 1:3
B. 3:1
C. 2:3
D. 3:2
Answer: Option B
Explanation:
a volume of cylinder : volume of cone
pr[latex]^{2}[/latex]h : 1/3 pr[latex]^{2}[/latex]h = 3 : 1
Logarithm
1. If [latex]log_{10}[/latex] 2 = 0.3010, the value of log10 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.
2. If [latex]log_{10}[/latex] 5 + [latex]log_{10}[/latex] (5x + 1) = [latex]log_{10}[/latex] (x + 5) + 1, then x is equal to:
Answer: Option B
Explanation:
[latex]log_{10}[/latex] 5 + [latex]log_{10}[/latex] (5x + 1) = [latex]log_{10}[/latex] (x + 5) + 1
[latex]log_{10}[/latex] 5 + [latex]log_{10}[/latex] (5x + 1) = [latex]log_{10}[/latex] (x + 5) + log10 10
[latex]log_{10}[/latex] [5 (5x + 1)] = [latex]log_{10}[/latex] [10(x + 5)]
5(5x + 1) = 10(x + 5)
5x + 1 = 2x + 10
3x = 9
x = 3.
3. If log 2 = 0.30103, the number of digits in 2[latex]^{64}[/latex] is:
Answer: Option C
Explanation:
log (2[latex]^{64}[/latex]) = 64 x log 2
= (64 x 0.30103)
= 19.26592
Its characteristic is 19.
Hence, then number of digits in 2[latex]^{64}[/latex]6 is 20.
Races and Games
1. If in a race of 80m, A covers the distance in 20 seconds and B in 25 seconds, then A beats B by:
A. 20m
B. 16m
C. 11m
D. 10m
Answer: Option B
Explanation:
The difference in the timing of A and B is 5 seconds. Hence, A beats B by 5 seconds.
The distance covered by B in 5 seconds = (80 * 5) / 25 = 16m
Hence, A beats B by 16m.
2. If X can run 48m and Y 42m, then in a race of 1km, X beats Y by:
A. 140m
B. 125m
C. 100m
D. 110m
Answer: Option B
Explanation:
When X runs 48m, Y runs 42m.
Hence, when X runs 1000m, Y runs (1000 * 42) / 48 = 875m
X beats Y by 125m.
3. In a race of 800m, if the ratio of the speeds of two participants P and R is 3:4, and P has a start of 120m, then R beats P by:
A. 100m
B. 80m
C. 90m
D. 110m
Answer: Option B
Explanation:
As P has a start of 120m, P has to cover 680m while R has to cover 800m.
Now, when P covers 3m, R covers 4m.
Hence, when R covers 800m, P covers (800 * 3) / 4 = 600m
Therefore, R beats P by 80m.
Chain Rule
1. A wheel that has 6 cogs has meshed with a larger wheel of 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions made by the larger wheel is:
Answer: Option B
Explanation:
Let the required number of revolutions made by larger wheel be x.
Then, More cogs, less revolutions (Indirect Proportion)
14 : 6 :: 21 : x => 14 x x = 6 x 21
x = [latex]\frac {6 \times 21}{14}[/latex]
x = 9.
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.
3. In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal?
Answer: Option B
Explanation:
There is a meal for 200 children. 150 children have taken the meal.
The remaining meal is to be catered to 50 children.
Now, 200 children = 120 men.
50 children = [latex]\frac {120 \times 50}{2500}[/latex] = 30 men.
Permutation and Combination
1. A boy has nine trousers and 12 shirts. In how many different ways can he select a trouser and a shirt?
A. 21
B. 12
C. 9
D. 108
E. 101
Answer: Option D
Explanation:
The boy can select one trouser in nine ways.
The boy can select one shirt in 12 ways.
The number of ways in which he can select one trouser and one shirt is 9 * 12 = 108 ways.
2. Using all the letters of the word "THURSDAY", how many different words can be formed?
A. 8
B. 8!
C. 7!
D. 7
E. 6!
Answer: Option B
Explanation:
Total number of letters = 8
Using these letters the number of 8 letters words formed is [latex]^8P_ {8}[/latex] = 8!.
3. Using all the letters of the word "NOKIA", how many words can be formed, which begin with N and end with A?
A. 3
B. 6
C. 24
D. 120
E. 12
Answer: Option B
Explanation:
There are five letters in the given word.
Consider 5 blanks
The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways.
The number of words = 3! = 6.
Surds and Indices
1. The value of [(10)[latex]^{150}[/latex] ÷ (10)[latex]^{146}[/latex]]
A. 1000
B. 10000
C. 100000
D. 10[latex]^{6}[/latex]
Answer: Option B
Explanation:
(10)[latex]^{150}[/latex] ÷ (10)[latex]^{146}[/latex] = [latex]\frac {(10)^{150}}{(10)^{146}}[/latex]
= 10[latex]^{150- 146}[/latex]
= 10[latex]^{4}[/latex]
= 10000.
2. (0.04)[latex]^{-1.5}[/latex] = ?
A. 25
B. 125
C. 250
D. 625
Answer: Option B
Explanation:
(0.04)[latex]^{-1.5}[/latex] =([latex]\frac {4}{100})^{-1.5}[/latex]
= ([latex]\frac {1}{25})^{-(3/2)}[/latex]
= (25)[latex]^{3/2}[/latex]
= (5[latex]^{2}[/latex])[latex]^{3/2}[/latex]
= (5)[latex]^{2\times {(3/2)}}[/latex]
= 5[latex]^{3}[/latex]
= 125.
3. ([latex]\frac {x^{b}}{x^{c}})^{b + c - a}[/latex] . ([latex]\frac {x^{c}}{x^{a}})^{c + a - b}[/latex] . ([latex]\frac {x^{a}}{x^{b}})^{a + b - c}[/latex] = ?
A. x[latex]^{abc}[/latex]
B. 1
C. x[latex]^{ab + bc + ca}[/latex]
D. x[latex]^{a + b + c}[/latex]
Answer: Option B
Explanation:
Given Exp.
= x[latex]^{(b - c)(b + c - a)}[/latex] . x[latex]^{(c - a)(c + a - b)}[/latex] . x[latex]^{(a - b)(a + b - c)}[/latex]
= x[latex]^{(b - c)(b + c) - a(b - c)}[/latex] . x[latex]^{(c - a)(c + a) - b(c - a)}[/latex]
. x[latex]^{(a - b)(a + b) - c(a - b)}[/latex]
= x[latex]^{(b^{2} - c^{2} + c^{2} - a^{2} + a^{2} - b^{2})}[/latex] . x[latex]^{-a(b - c) - b(c - a) - c(a - b)}[/latex]
= ([latex]x^{0} \times x^{0}[/latex])
= (1 x 1) = 1.
Pipes and Cistern
1. Two pipes A and B can separately fill a tank in 2 minutes and 15 minutes respectively. Both the pipes are opened together but 4 minutes after the start the pipe A is turned off. How much time will it take to fill the tank?
A. 9 min
B. 10 min
C. 11 min
D. 12 min
Answer: Option B
Explanation:
4/12 + x/15 = 1
x = 10
2. A cistern is normally filled in 8 hours but takes two hours longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in?
A. 16 hrs
B. 20 hrs
C. 40 hrs
D. 25 hrs
Answer: Option C
Explanation:
1/8 - 1/x = 1/10
x = 40
3. A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 liters per minute. When the tank is full in the inlet is opened and due to the leak, the tank is emptied in 8 hours. The capacity of the tank is?
A. 5260 liters
B. 5760 liters
C. 5846 liters
D. 6970 liters
Answer: Option B
Explanation:
1/x - 1/6 = -1/8
x = 24 hrs
24 * 60 * 4 = 5760
Boats and Streams
1. A man can row his boat with the stream at 6 km/h and against the stream in 4 km/h. The man's rate is?
A. 1 kmph
B. 5 kmph
C. 8 kmph
D. 3 kmph
Answer: Option A
Explanation:
DS = 6
US = 4
S =?
S = (6 - 4)/2 = 1 kmph
2. A man can row 6 kmph in still water. When the river is running at 1.2 kmph, it takes him 1 hour to row to a place and black. How far is the place?
A. 3.12 km
B. 2.88 km
C. 3 km
D. 2 km
Answer: Option B
Explanation:
M = 6
S = 1.2
DS = 6 + 1.2 = 7.2
US = 6 - 1.2 = 4.8
x/7.2 + x/4.8 = 1
x = 2.88
3. The current of a stream at 1 kmph. A motorboat goes 35 km upstream and back to the starting point in 12 hours. The speed of the motorboat in still water is?
A. 6 kmph
B. 7 kmph
C. 8.5 kmph
D. 8 kmph
Answer: Option A
Explanation:
S = 1
M = x
DS = x + 1
US = x - 1
35/(x + 1) + 35/(x - 1) = 12
x = 6
Partnership
1. A and B entered into a partnership investing Rs.25000 and Rs.30000 respectively. After 4 months C also joined the business with an investment of Rs.35000. What is the share of C in an annual profit of Rs.47000?
A. Rs.18000
B. Rs.15000
C. Rs.17000
D. Rs.14000
Answer: Option D
Explanation:
25*12: 30*12: 35*8
15: 18: 14
14/47 * 47000 = 14000
2. A and B enter into a partnership with capital as 7:9. At the end of 8 months, A withdraws. If they receive the profits in the ratio of 8:9 find how long B's capital was used?
A. 6 months
B. 8 months
C. 10 months
D. 7 months
Answer: Option D
Explanation:
7 * 8 : 9 * x = 8:9 => x= 7
3. A, B, C together started a business. A invested Rs.6000 for 5 months B invested Rs.3600 for 6 months and C Rs.7500 for 3 months. If they get a total profit of Rs.7410. Find the share of A?
A. Rs.3750
B. Rs.3000
C. Rs.3200
D. Rs.2160
Answer: Option B
Explanation:
60*5: 36*6: 75*3
100: 72: 75
100/247 * 7410 = 3000
Ratio and Proportion
1. What number has a 5:1 ratio to the number 10?
Answer: Option B
Explanation:
5:1 = x: 10
x = 50
2. If a: b = 7: 5, b: c = 9: 11, find a: b: c?
A. 63: 14: 55
B. 63: 45: 55
C. 14: 14: 15
D. 7: 14: 15
Answer: Option B
Explanation:
a: b = 7: 5
b: c = 9: 11
a: b: c = 63: 45: 55
3. The ratio of two numbers is 2:3 and the sum of their cubes is 945. The difference of number is?
Answer: Option A
Explanation:
2x , 3x
8x cube + 27x cube = 945
35x cube = 945
x cube = 27 => x = 3
Problems on H.C.F and L.C.M.
1. LCM of 1/3, 5/6, 5/4, 10/7 is:
A. 10/7
B. 10
C. 10/11
D. 11/10
Answer: Option B
Explanation:
LCM of numerators = 10
HCF of denominators = 1
=> 10/1 = 10
2. HCF of 3/16, 5/12, 7/8 is:
A. 2/47
B. 3/47
C. 1/48
D. 5/48
Answer: Option C
Explanation:
HCF of numerators = 1
LCM of denominators = 48
=> 1/48
3. The least square number which divides 8, 12 and 18 is?
A. 100
B. 121
C. 64
D. 144
Answer: Option D
Explanation:
LCM = 72
72 * 2 = 144
Banker’s Discount
1. 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.
2. The banker's gain on a bill due 1 year hence at 12% per annum is Rs. 6. The true discount is:
A. Rs. 72
B. Rs. 36
C. Rs. 54
D. Rs. 50
Answer: Option D
Explanation:
T.D. = [latex]\frac {B.G. \times 100}{R \times T}[/latex] = Rs. [latex]\frac {6 \times 100}{12 \times 1}[/latex] = Rs. 50.
3. The present worth of a certain bill due sometime hence is Rs. 800 and the true discount is Rs. 36. The banker's discount is:
A. Rs. 37
B. Rs. 37.62
C. Rs. 34.38
D. Rs. 38.98
Answer: Option B
Explanation:
B.G. =[latex]\frac {(T.D.)^{2}}{P.W.}[/latex] = Rs. [latex]\frac {36 \times 36}{800}[/latex] = Rs. 1.62
B.D. = (T.D. + B.G.) = Rs. (36 + 1.62) = Rs. 37.62
Compound Interest
1. Simple interest on a sum at 4% per annum for 2 years is Rs.80. The C.I. on the same sum for the same period is?
A. Rs.81.60
B. Rs.160
C. Rs.1081.60
D. Rs.99
Answer: Option A
Explanation:
SI = 40 + 40
CI = 40 + 40 + 1.6 = 81.6
2. The C.I. on a certain sum for 2 years Rs.41 and the simple interest is Rs.40. What is the rate percent?
Answer: Option B
Explanation:
SI = 20 + 20
CI = 20 + 21
20 ---- 1
100 ---- ? => 5%
3. A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to eight times itself?
Answer: Option C
Explanation:
100 ---- 200 ---- 4
400 ---- 4
800 ---- 4
4 + 4 + 4 = 12 years
Area
1. What is the area of the square field whose side of length 15 m?
A. 225 sq m
B. 220 sq m
C. 100 sq m
D. 30 sq m
Answer: Option A
Explanation:
15 * 15 = 225 sq m
2. If the perimeter of a rectangular garden is 600 m, its length when its breadth is 100 m is?
A. 650 m
B. 600 m
C. 200 m
D. 300 m
Answer: Option C
Explanation:
2(l + 100) = 600 => l = 200 m
3. A rectangular field has an area equal to 150 sq m and perimeter 50 m. Its length and breadth must be?
A. 12 m, 10 m
B. 13 m, 12 m
C. 14 m, 11 m
D. 15 m, 10 m
Answer: Option D
Explanation:
lb = 150
2(l + b) = 50 => l + b = 25
l – b = 5
l = 15 , b = 10
Time and Work
1. A can do a piece of work in 30 days. He works at it for 5 days and then B finishes it in 20 days. In what time can A and B together it?
A. 16 2/3 days
B. 13 1/3 days
C. 17 1/3 days
D. 16 1/2 days
Answer: Option B
Explanation:
5/30 + 20/x = 1
x = 24
1/30 + 1/24 = 3/40
40/3 = 13 1/3 days
2. A and B can do a piece of work in 3 days, B and C in 4 days, C and A in 6 days. How long will C take to do it?
A. 18 days
B. 20 days
C. 24 days
D. 30 days
Answer: Option C
Explanation:
2c = 1/4 + 1/6 – 1/3 = 1/12
c = 1/24 => 24 days
3. A can do a piece of work in 12 days. When he had worked for 2 days B joins him. If the complete work was finished in 8 days. In how many days B alone can finish the work?
A. 18 days
B. 12 days
C. 24 days
D. 10 days
Answer: Option A
Explanation:
8/12 + 6/x = 1
X = 18 days
Decimal Fraction
1. 3889 + 12.952 - ? = 3854.002
A. 47.095
B. 47.752
C. 47.932
D. 47.95
Answer: Option D
Explanation:
Let 3889 + 12.952 - x = 3854.002.
Then x = (3889 + 12.952) - 3854.002
= 3901.952 - 3854.002
= 47.95.
2. 0.04 x 0.0162 is equal to:
A. 6.48 x 10[latex]^{-3}[/latex]
B. 6.48 x 10[latex]^{-4}[/latex]
C. 6.48 x 10[latex]^{-5}[/latex]
D. 6.48 x 10[latex]^{-6}[/latex]
Answer: Option B
Explanation:
4 x 162 = 648. Sum of decimal places = 6.
So, 0.04 x 0.0162 = 0.000648 = 6.48 x 10[latex]^{-4}[/latex]
3. 0.002 x 0.5 = ?
A. 0.0001
B. 0.001
C. 0.01
D. 0.1
Answer: Option B
Explanation:
2 x 5 = 10.
Sum of decimal places = 4
0.002 x 0.5 = 0.001
Probability
1. Three 6 faced dice are thrown together. The probability that all the three show the same number on them is -.
A. 1/216
B. 1/36
C. 5/9
D. 5/12
E. 7/12
Answer: Option B
Explanation:
It all 3 numbers have to be same basically we want triplets. 111, 222, 333, 444, 555 and 666. Those are six in number. Further the three dice can fall in 6 * 6 * 6 = 216 ways.
Hence the probability is 6/216 = 1/36
2. Three 6 faced dice are thrown together. The probability that no two dice show the same number on them is -.
A. 7/12
B. 5/9
C. 1/36
D. 5/12
E. 8/9
Answer: Option B
Explanation:
No two dice show the same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways.
Thus 6 * 5 * 4 = 120 favourable cases.
The total cases are 6 * 6 * 6 = 216.
The probability = 120/216 = 5/9.
3. If a number is chosen at random from the set {1, 2, 3, ...., 100}, then the probability that the chosen number is a perfect cube is -.
A. 1/25
B. 1/2
C. 4/13
D. 1/10
E. 9/13
Answer: Option A
Explanation:
We have 1, 8, 27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is
4/100 = 1/25.
Average
1. The average age of three boys is 15 years and their ages are in proportion 3:5:7. What is the age in years of the youngest boy?
Answer: Option B
Explanation:
3x + 5x + 7x = 45
x =3
3x = 9
2. The average of 10 numbers is calculated as 15. It is discovered later on that while calculating the average, one number namely 36 was wrongly read as 26. The correct average is?
A. 12.4
B. 14
C. 16
D. 18.6
Answer: Option C
Explanation:
10 * 15 + 36 – 26 = 160/10 = 16
Stocks and Share
1. A 6% stock yields 8%. The market value of the stock is:
A. Rs. 48
B. Rs. 75
C. Rs. 96
D. Rs. 133.33
Answer: Option B
Explanation:
For an income of Rs. 8, investment = Rs. 100.
For an income of Rs. 6, investment = Rs.([latex]\frac {100}{8}[/latex] x 6 )= Rs. 75.
Market value of Rs. 100 stock = Rs. 75.
2. A man invested Rs. 1552 in a stock at 97 to obtain an income of Rs. 128. The dividend from the stock is:
A. 7.5%
B. 8%
C. 9.7%
D. None of these
Answer: Option B
Explanation:
By investing Rs. 1552, income = Rs. 128.
By investing Rs. 97, income = Rs. ([latex]\frac {128}{1552}[/latex] x 97) = Rs. 8.
Dividend = 8%
3. A 12% stock yielding 10% is quoted at:
A. Rs. 83.33
B. Rs. 110
C. Rs. 112
D. Rs. 120
Answer: Option D
Explanation:
To earn Rs. 10, money invested = Rs. 100.
To earn Rs. 12, money invested = Rs. ([latex]\frac {100}{10}[/latex] x 12) = Rs. 120.
Market value of Rs. 100 stock = Rs. 120.
Square Root and Cube Root
1.[latex]\sqrt { 0.0169 \times ?}[/latex] = 1.3
A. 10
B. 100
C. 1000
D. None of these
Answer: Option B
Explanation:
Let[latex]\sqrt { 0.0169 \times x}[/latex] = 1.3.
Then, 0.0169x = (1.3)[latex]^{2}[/latex] = 1.69
x =[latex]\frac {1.69}{0.0169}[/latex] = 100
2. How many two-digit numbers satisfy this property.: The last digit (unit's digit) of the square of the two-digit number is 8?
A. 1
B. 2
C. 3
D. None of these
Answer: Option D
Explanation:
A number ending in 8 can never be a perfect square.
3. The least perfect square, which is divisible by each of 21, 36 and 66 is:
A. 213444
B. 214344
C. 214434
D. 231444
Answer: Option A
Explanation:
L.C.M. of 21, 36, 66 = 2772.
Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11
To make it a perfect square, it must be multiplied by 7 x 11.
So, required number = 2[latex]^{2}[/latex] x 3[latex]^{2}[/latex] x 7[latex]^{2}[/latex] x 11[latex]^{2}[/latex] = 213444
Problems on Ages
1. A person's present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old id the mother at present?
A. 32 years
B. 36 years
C. 40 years
D. 48 years
Answer: Option C
Explanation:
Let the mother's present age be x years. Then, the person's present age = 2/5 x years.
(2/5 x + 8) = 1/2 (x + 8)
2(2x + 40) = 5(x + 8) => x = 40
2. The sum of the present ages of father and his son is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, son's age will be:
A. 12 years
B. 14 years
C. 18 years
D. 20 years
Answer: Option D
Explanation:
Let the present ages of son and father be x and (60 - x) years respectively.
Then, (60 - x) - 6 = 5(x - 6)
6x = 84 => x = 14
Son's age after 6 years = (x + 6) = 20 years.
3. A father said to his son, "I was as old as you are at present at the time of your birth." If the father's age is 38 years now, the son's age five years back was:
A. 19 years
B. 14 years
C. 33 years
D. 38 years
Answer: Option B
Explanation:
Let the son's present age be x years.
Then, (38 - x) = x
2x = 38 => x = 19
Son's age 5 years back = (19 - 5) = 14 years.