Simple Interest
1. A sum of Rs. 2665 is lent into two parts so that the interest on the first part for 8 years at 3% per annum may be equal to the interest on the second part for 3 years at 5% per annum. Find the second sum?
A. Rs.1678
B. Rs.1640
C. Rs.2341
D. None
Answer: Option B
Explanation:
(x*8*3)/100 = ((2665 - x)*3*5)/100
24x/100 = 39975/100 - 15x/100
39x = 39975 => x = 1025
Second sum = 2665 – 1025 = 1640
2. If A lends Rs.3500 to B at 10% per annum and B lends the same sum to C at 11.5% per annum then the gain of B in a period of 3 years is?
A. Rs.107.50
B. Rs.115.50
C. Rs.157.50
D. Rs.177.50
Answer: Option C
Explanation:
(3500*1.5*3)/100 => 157.50
3. If Rs.450 amount to Rs.540 in 4 years, what will it amount to in 6 years at the same rate % per annum?
A. Rs.575
B. Rs.675
C. Rs.585
D. Rs.685
Answer: Option C
Explanation:
90 = (450*4*R)/100
R = 5%
I = (450*6*5)/100 = 135
450 + 135 = 585
Height and Distance
1. 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.
2. 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.
3. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 173 m
B. 200 m
C. 273 m
D. 300 m
Answer: Option C
Explanation:
Let AB be the lighthouse and C and D be the positions of the ships.
Then, AB = 100 m, ACB = 30° and ADB = 45°.
[latex]\frac {AB}{AC}[/latex] = tan 30° = [latex]\frac {1}{\sqrt {3}}[/latex] => AC = AB x[latex]\sqrt {3}[/latex] = 100[latex]\sqrt {3}[/latex] m.
[latex]\frac {AB}{AD}[/latex] = tan 45° = 1 => AD = AB = 100 m.
CD = (AC + AD) = (100[latex]\sqrt {3}[/latex] + 100) m
= 100([latex]\sqrt {3}[/latex] + 1)
= (100 x 2.73) m
= 273 m.
Volume and Surface Area
1. The edges of a cuboid are 4 cm; 5 cm and 6 cm. Find its surface area?
A. 120 cm[latex]^ {2}[/latex]
B. 148 cm[latex]^ {2}[/latex]
C. 74 cm[latex]^ {2}[/latex]
D. 15 cm[latex]^ {2}[/latex]
Answer: Option B
Explanation:
2(4*5 + 5*6 + 4*6) = 148
2. The area of the floor of a room is 20 m[latex]^ {2}[/latex] that of a longer wall 15 m[latex]^ {2}[/latex] and of the shorter wall 12 m[latex]^ {2}[/latex], find the edge of the new cube?
A. 450 m[latex]^ {3}[/latex]
B. 100 m[latex]^ {2}[/latex]
C. 60 m[latex]^ {3}[/latex]
D. 400 m[latex]^ {3}[/latex]
Answer: Option C
Explanation:
lb = 20 ; lh = 15 ; fh = 12
(lbh)[latex]^ {2}[/latex] = 20 * 15 * 12 => lbh = 60 m[latex]^ {3}[/latex]
3. The area of a base of a cone is 30 cm[latex]^ {2}[/latex]. If the height of the cone is 6cm, find its volume?
A. 120 cm[latex]^ {3}[/latex]
B. 40 cm[latex]^ {3}[/latex]
C. 50 cm[latex]^ {3}[/latex]
D. 60 cm[latex]^ {3}[/latex]
Answer: Option D
Explanation:
πr[latex]^ {2}[/latex] = 30 h = 6
1/3 * 30 * 6 = 60.
Races and Games
1. A can run a kilometer race in 4 1/2 min while B can run same race in 5 min. How many meters start can A give B in a kilometer race, so that the race mat end in a dead heat?
A. 150 m
B. 125 m
C. 130 m
D. 100 m
Answer: Option D
Explanation:
A can give B (5 min - 4 1/2 min) = 30 sec start.
The distance covered by B in 5 min = 1000 m.
Distance covered in 30 sec = (1000 * 30)/300 = 100 m.
A can give B 100m start.
2. In a game of billiards, A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100?
Answer: Option C
Explanation:
A Scores 60 while B scores 40 and C scores 30.
The number of points that C scores when B scores 100 = (100 * 30)/40 = 25 * 3 = 75.
In a game of 100 points, B gives (100 - 75) = 25 points to C.
Simplification
1. 9000 + 16 2/3 % of ? = 10500
A. 1500
B. 1750
C. 9000
D. 7500
E. None of these
Answer: Option C
Explanation:
9000 + 16 2/3 % of ? = 10500 => 9000 + 50/3 % of ? = 10500
50/(3 * 100) of ? = 1500 => ? = 1500 * 6
? = 9000
2. 0.003 * ? * 0.0003 = 0.00000027
A. 9
B. 3
C. 0.3
D. 0.03
E. None of these
Answer: Option C
Explanation:
0.003 * ? * 0.0003 = 0.00000027
3/1000 * ? * 3/10000 = 3/1000 * 3/1000 * 3/100
? = 3/10 = 0.3
Time and Distance
1. A train 240 m in length crosses a telegraph post in 16 seconds. The speed of the train is?
A. 50 kmph
B. 52 kmph
C. 54 kmph
D. 56 kmph
Answer: Option C
Explanation:
S = 240/16 * 18/5 = 54 kmph
2. Two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively. In what time will they cross each other completely?
A. 10 sec
B. 11 sec
C. 12 sec
D. 14 sec
Answer: Option C
Explanation:
D = 250 m + 250 m = 500 m
RS = 80 + 70 = 150 * 5/18 = 125/3
T = 500 * 3/125 = 12 sec
Chain Rule
1. 20 women take 20 minutes to roll 20 chapattis. How many chapattis can 400 women roll in 400 minutes?
A. 20
B. 2000
C. 4000
D. 8000
Answer: Option D.
Explanation:
Men = M; Days = D; Time/Hours = T; Work = W
M1D1T1W2 = M2D2T2W1
Note that - W2 is on left side and W1 is on right side
∴ 20 women x 20 minutes x ? = 400 women x 400 minutes x 20 chapattis
∴ ? = 8000 chapattis = Are rolled by 400 women in 400 minutes.
2. A building is under construction and the task of paving the blocks is given to a group of men. 40 men can finish the given task in 96 days, working 9 hours/day. If 48 men take up the assignment and commit to finishing it in 45 days, how many hours will they need to work per day?
Answer: Option A
Explanation:
Men = M; Days = D; Time/Hours = T; Work = W
M1D1T1W2 = M2D2T2W1
Note that - W2 is on left side and W1 is on right side
Take work done = 1
∴ 40 men x 96 days x 9 hours x 1 = 48 men x 45 days x ? hours x 1
∴ ? = 16 hours = 48 men need to work these many hours per day
Permutation and Combination
Surds and Indices
1. (17[latex]^ {14}[/latex] * 17[latex]^ {16}[/latex]) / 17[latex]^ {-8}[/latex] = ?
A. 17[latex]^ {28 }[/latex]
B. 17[latex]^ {24 }[/latex]
C. 17[latex]^ {26 }[/latex]
D. 17[latex]^ {22 }[/latex]
E. None of these
Answer: Option D
Explanation:
? = 17[latex]^ {14+16-8}[/latex] = 17[latex]^ {22}[/latex]
2. (3[latex]^ {-3 }[/latex] * 9[latex]^ {5/2}[/latex]) / (27[latex]^ {2/3}[/latex] * 3[latex]^ {-4}[/latex]) = ?
A. 3
B. 9
C. 81
D. 21
E. None of these
Answer: Option C
Explanation:
(3[latex]^ {-3}[/latex] * 9[latex]^ {5/2}[/latex]) / (27[latex]^ {2/3}[/latex] * 3[latex]^ {-4}[/latex]) = [3[latex]^ {-3}[/latex] *([latex] (3^{2})^{5/2})[/latex]] / [([latex](3^{3})^{2/3})[/latex] * 3[latex]^ {-4}[/latex]]
= (3[latex]^ {-3}[/latex] * 3[latex]^ {5}[/latex]) / (3[latex]^ {2 }[/latex]* 3[latex]^ {-4}[/latex]) = (3[latex]^ {-3+5}[/latex])/(3[latex]^ {2-4}[/latex]) = 3[latex]^ {2}[/latex]/3[latex]^ {-2}[/latex] = 3[latex]^ {2 }[/latex]* 3[latex]^ {2}[/latex] = 81.
Pipes and Cistern
1. Pipe P can fill a cement tank in 24 hours. However, there is a leakage at the bottom of the tank due to which the Pipe P takes 12 hrs longer and fills it in 36 hours. How long would this leakage take to empty the entire tank?
A. 48 hours
B. 60 hours
C. 72 hours
D. 96 hours
Answer: Option C.
Explanation:
Normally, Pipe P can fill the cement tank in 24 hours. So in 1 hour it fills =[latex]\frac {1}{24}[/latex] part of tank
Due to leak, Pipe P can fill the tank in 36 hours. So in 1 hour it actually fills =[latex]\frac {1}{36}[/latex] part of tub
∴ Water removed by the leak in 1 hour =[latex]\frac {1}{24}[/latex] - [latex]\frac {1}{36}[/latex] = [latex]\frac {1}{72}[/latex]
∴ Leak empties the tank in 72 hours.
2. Pipe A takes 38 hours to fill a tank alone while Pipe B takes half the time taken by Pipe A to fill the same tank alone. The leakage C can empty the entire tank in 133 hours. How many hours will it take to fill the tank if Pipe A and B are opened together and leakage C is also allowed to empty the tank simultaneously?
A. 14 hours
B. 19 hours
C. 28 hours
D. 95 hours
Answer: Option A
Explanation:
Tank filled by A in 1 hour = [latex]\frac {1}{38}[/latex]
Tank filled by B in 1 hour (A takes 38 hrs, so B would take 19 hrs) =[latex]\frac {1}{19}[/latex]
Tank emptied by leakage C in 1 hour =[latex]\frac {1}{133}[/latex]
Tank filled by all 3 openings together in 1 hour =[latex]\frac {1}{38}[/latex]+[latex]\frac {1}{19}[/latex] -[latex]\frac {1}{133}[/latex]
Making denominators common,
Tank filled by all 3 pipes together in 1 hour =[latex]\frac {3.5}{133}[/latex]+[latex]\frac {7}{133}[/latex] -[latex]\frac {1}{133}[/latex] =[latex]\frac {9.5}{133}[/latex]=[latex]\frac {1}{14}[/latex]
So the entire tank is full in [latex]\frac {14}{1}[/latex] = 14 hours -------------> Inverse or reciprocal of [latex]\frac {1}{14}[/latex]
Boats and Streams
1. A motorist can go downstream at 18 km/hr and upstream at 10 km/hr. Find the speed of the stream and the speed of the motorist in still waters.
A. Motorist = 8 km/hr ; Stream = 28 km/hr
B. Motorist = 10 km/hr ; Stream = 5 km/hr
C. Motorist = 14 km/hr ; Stream = 4 km/hr
D. Motorist = 28 km/hr ; Stream = 8 km/hr
Answer: Option C.
Explanation:
Man's/Boat's Speed = X
Stream/Current/River Speed = Y
∴ Downstream speed = X + Y
Upstream speed = X - Y
X+Y = 18 km/hr and X-Y = 10 km/hr
Adding them we get,
X+Y+X-Y = 28 km/hr
∴ X = 14 km/hr = Speed of Motorist
Y=18-14 = 4 km/hr = Speed of stream.
2. A boatman rows his boat 48 km upstream and same distance downstream. The boat takes 12 hrs and 8 hrs to go upstream and downstream respectively. Find the speed of the boat in stagnant water and the speed of the stream respectively.
A. 4 km/hr ; 3 km/hr
B. 5 km/hr ; 1 km/hr
C. 6 km/hr ; 4 km/hr
D. 12 km/hr ; 6 km/hr
E. Cannot be determined
Answer: Option B
Explanation:
Man's/Boat's Speed = X
Stream/Current/River Speed = Y
∴ Downstream speed = X + Y
Upstream speed = X - Y
Downstream Speed = [latex]\frac {Distance covered}{Time taken}[/latex] = [latex]\frac {48}{8}[/latex]= 6 km/hr
Upstream Speed = [latex]\frac {Distance covered}{Time taken}[/latex]= [latex]\frac {48}{12}[/latex]= 4 km/hr
X+Y = 6 km/hr and X-Y = 4 km/hr
Adding them we get,
X+Y+X-Y = 10 km/hr
∴ X=5 km/hr = Speed of Boat
Y=6-5 = 1 km/hr = Speed of stream
Numbers
1. Five consecutive numbers add up to 335. What will be the sum of largest and smallest number?
A. 134
B. 150
C. 174
D. 226
Answer: Option A
Explanation:
Let 5 consecutive numbers be N, N+1, N+2, N+3, N+4
Sum = N+(N+1)+(N+2)+(N+3)+(N+4) = 5N + 10 = 335
∴ N (Smallest Number) = 65; Largest Number = 65 + 4 = 69
∴ 65 + 69 = 134 = Sum of largest and smallest number.
2. We reverse a number and form a new one. The old number is 45 less than new number. The sum of the digits of the old number is 9. What is the new number?
A. 36
B. 54
C. 72
D. 81
E. None of the above
Answer: Option C
Explanation:
Let the two digits be X and Y.
Let the older number be A and newer one be B.
A = 10X + Y
∴ B = 10 Y + X
From given, B = 45 + A = 45 + 10X + Y
10Y + X = 45 + 10X + Y
So, 9Y - 9X = 45;
Y - X = 5 ----------------- (1)
X + Y = 9 ---------------- (2)
Solving (1) and (2), Y = 7; X = 2
So, A = 27; B = 72
Partnership
1. Polly and Jolly jointly buy a tractor. Jolly's investment is one-sixth the investment of Polly. In a profit of Rs. 91000 what will be Polly's share?
A. Rs. 50000
B. Rs. 66000
C. Rs. 72000
D. Rs. 78000
Answer: Option D
Explanation:
When the time period is the same,
The ratio of investment = Ratio of Profit
The ratio of investment of Polly to Jolly = 6:1 = Ratio of Profit
The share of Polly in profit =[latex]\frac {6}{6 + 1}[/latex] x 91000 = Rs. 78000
2. Three friends Mohan, Neeraj and Prateek make an equal investment in a business. Neeraj withdraws after just 5 months while Mohan withdraws 3 months after Neeraj. Only Pratik continues for a year. Find the ratio of profits of Mohan, Neeraj, and Pratik respectively, if Mohan's investment was Rs. 216,800/-?
A. 4 : 7 : 1
B. 5 : 8 : 12
C. 8 : 5 : 12
D. 12 : 5 : 8
Answer: Option C
Explanation:
When investment is same,
The ratio of time = Ratio of Profit
Let us denote Mohan by M, Neeraj by N and Prateek by P.
Ratio of time periods of M , N and P = 8 : 5 : 12 = Ratio of Profit.
Ratio and Proportion
1. Find the 3rd proportional to 9 and 72.
A. 8
B. 216
C. 576
D. 648
Answer: Option C
Explanation:
In a:b:c, 3rd proportional is c.
a:b:c can be written as a:b::b:c
a:b::b:c can be written as[latex]\frac {a}{b}[/latex] = [latex]\frac {b}{c}[/latex]=> b[latex]^ {2}[/latex] = ac
Here, a:b:c = 9:72:c
∴ 72 x 72 = 9 x c
∴ c = [latex]\frac {72 \times 72}{9}[/latex]= 576
2. Find the 4th proportional in 7, 23 and 217?
A. 66
B. 506
C. 713
D. 961
Answer: Option C
Explanation:
In a:b::c:d , 4th proportional is d.
a:b::c:d can be written as[latex]\frac {a}{b}[/latex] =[latex]\frac {c}{d}[/latex] => d = [latex]\frac {c \times b}{a}[/latex]
Here, a:b::c:d = 7:23::217:d
∴ d = [latex]\frac {23 \times 217}{7}[/latex]= 713
Problems on H.C.F and L.C.M.
1. Find the LCM of following three fractions:
[latex]\frac {36}{225}[/latex] ,[latex]\frac {48}{150}[/latex] ,[latex]\frac {72}{65}[/latex]
A. [latex]\frac {72}{225}[/latex]
B. [latex]\frac {36}{65}[/latex]
C. [latex]\frac {144}{5}[/latex]
D. [latex]\frac {288}{5}[/latex]
Answer: Option C
Explanation:
LCM of fraction = [latex]\frac {LCM of numerators}{HCF of denominators}[/latex]
Numerators = 36, 48 and 72.
72 is largest number among them. 72 is not divisible by 36 or 48
Start with table of 72.
72 x 2 = 144 = divisible by 72, 36 and 48
∴ LCM of numerators = 144
Denominators = 225, 150 and 65
We can see that they can be divided by 5.
On dividing by 5 we get 45, 30 and 13
We cannot divide further.
So, HCF = GCD = 5
LCM of fraction = [latex]\frac {144}{5}[/latex]
2. Given : Three numbers 17, 42 and 93
Find the largest number to divide all the three numbers leaving the remainders 4, 3, and 15 respectively at the end?
Answer: Option A
Explanation:
Here greatest number that can divide means the HCF
Remainders are different so simply subtract remainders from numbers
17 - 4 = 13; 42 - 3 = 39; 93 - 15 = 78
Now let's find HCF of 13, 39 and 78
By direct observation we can see that all numbers are divisible by 13.
∴ HCF = 13 = required greatest number
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 gain of a certain sum due 2 years hence at 10% per annum is Rs. 24. The present worth is:
A. Rs. 480
B. Rs. 520
C. Rs. 600
D. Rs. 960
Answer: Option C
Explanation:
T.D. = [latex]\frac {B.G. \times 100}{Rate \times Time}[/latex]= Rs.[latex]\frac {24 \times 100}{10 \times 2}[/latex] = Rs. 120.
P.W. =[latex]\frac {100 \times T.D.}{Rate \times Time}[/latex]= Rs.[latex]\frac {100 \times 120}{10 \times 2}[/latex] = Rs. 600.
Compound Interest
1. Rohit says that, an amount became 4 times in 6 years. His friend wanted to know, how many years will it take for the amount to become 64 times, provided the rate of interest remains same?
A. 12 years
B. 16 years
C. 18 years
D. 24 years
Answer: Option C
Explanation:
Initial Amount = P.
P becomes 4 times in 6 years ∴ P → 4P in 6 years
∴ 4P → 4 x (4P) = 16P in next 6 years.
Continuing this, 16P → 4 x (16P) = 64P in next 6 years.
So total time to get to 64 times i.e. 64P = 6+6+6 = 18 years.
2. Sum: 6400
Time period: 6 months
The rate of interest: 25% compounded quarterly.
Find the interest earned.
A. Rs. 735.5
B. Rs. 815
C. Rs. 825
D. Rs. 855
Answer: Option C
Explanation:
Total Amount = P[latex]( 1 + \frac {R}{100})^{2}[/latex]
Principal = P = Rs. 6400
For quarterly compound interest, R = [latex]\frac {R}{4}[/latex] and n = 4n
Rate of interest, R = [latex]\frac {R}{4}[/latex] = [latex]\frac {25}{4}[/latex]
Time period, n = 4T = 4 x 6 months = 4 x[latex]\frac {1}{2}[/latex] years = 2 years
Total Amount = 6400 [latex]( 1 + \frac {25}{4 \times 100})^{2}[/latex] = 6400[latex](\frac {425}{400})^{2}[/latex] = 6400 [latex](\frac {17}{16})^{2}[/latex]
∴ Total Amount = Rs. 7225
Compound Interest = Amount - Principle = 7225 - 6400 = Rs. 825
Area
1. The length and breadth of a rectangular piece of fabric are in the ratio 7:4. The length of the fabric is 15 m more than that of the breadth. Find the area that this fabric would cover.
A. 450 sq.m.
B. 700 sq.m.
C. 850 sq.m.
D. 900 sq.m.
Answer: Option B
Explanation:
Let the common factor be K
∴ Length = 7K and Breadth = 4K
Also, 4K+15m = 7K
∴ K = 5
∴ Area of Rectangular ground = Length x Breadth = 7K x 4K = 35 x 20 = 700 sq.m.
2. A rectangular piece of paper has an area 840 sq. cm. The length of the side is 48 cm. The
rectangle is stretched such that its area becomes 2 [latex]\frac {6}{7}[/latex] times original rectangle. The only dimension that change is the breadth of the rectangle. Find the perimeter of the new rectangle.
A. 1.69 m
B. 1.85 m
C. 1.96 m
D. 2.25 m
Answer: Option C
Explanation:
New area = 840 x 2 [latex]\frac {6}{7}[/latex] = 840 x [latex]\frac {20}{7}[/latex]= 2400 sq.cm
Area of Rectangle = length x breadth
New Breadth =[latex]\frac {New Area}{Length}[/latex] = [latex]\frac {2400}{48}[/latex] = 50 cm
Perimeter of new margin = 2(length + new breadth) = 2(48 + 50) = 196cm = 1.96m
3. Ratan takes turns to cycle around the circular garden and through the diameter of the garden on alternate days. His speed each day is 6m/min but when he cycles through the diameter, he takes 1 hr less to cross the park. Find the diameter of the park.
A. 1530 m
B. 1680 m
C. 1750 m
D. 3600 m
Answer: Option B
Explanation:
We know that, Time =[latex]\frac {Distance}{Speed}[/latex]
Speed = 60m/min = [latex]\frac {60}{60}[/latex] m/sec = 1 m/sec
We know that,
Time along the park (circumference) - Time along the diameter = 60 minutes
∴[latex]\frac {2πr}{1}[/latex] - [latex]\frac {2r}{1}[/latex] = 60 minutes
∴ 2r ([latex]\frac {22}{7}[/latex]- 1) = 3600 seconds
∴ r = 840m
∴ Diameter = 2 x radius = 1680m.
Time and Work
1. Amish is 5 times as efficient as Brijesh. Amish can finish the work 60 days earlier than Brijesh. If they work individually, how many days would they take to finish the work?
A. Amish = 10 days; Brijesh = 50 days
B. Amish = 15 days; Brijesh = 75 days
C. Amish = 20 days; Brijesh = 80 days
D. Amish = 20 days; Brijesh = 100 days
Answer: Option B
Explanation:
The speed of Work / Efficiency ∝ [latex]\frac {1}{Time needed}[/latex]
So more the efficiency, less time is needed to do the task.
Let us denote Amish by 'A' and Brijesh by 'B'
Hence, if A is 5 times faster than B, then B needs 5 times more time than A.
Let A need 'n' days to complete the work.
So B will need 5n days.
Also, 5n – n = 60 ------------------- Given
∴ n = 15 days = Days needed by A to complete work individually
Days needed by B to complete work individually = 5n = 5 x 15 = 75 days
2. Recollecting the progress of an assignment, a group of 24 men said that they finished a project together in 10 days. Find the number of days required to complete the same work if 30 men had worked on the same project.
A. 3 days
B. 6 days
C. 7.5 days
D. 8 days
Answer: Option D
Explanation:
24 men complete work in 10 days.
∴ In 1 day work done by 24 men = [latex]\frac {1}{10}[/latex]
In 1 day work done by 30 men =?
∴ ? x 24 = [latex]\frac {1}{10}[/latex]x 30
∴ ? =[latex]\frac {1}{8}[/latex] = Amount of work done by 30 men in one day
If A completes work in 'n' days, in 1 day he completes[latex]\frac {1}{n}[/latex] amount of work.
Conversely, if A completes[latex]\frac {1}{n}[/latex] amount of work in 1 day, he completes entire work in 'n' days.
Allegation or Mixture
1. A shopkeeper mixes two varieties of pulses with price Rs 420/ Kg and Rs 520/ Kg. The price of the mixture he gets is Rs 480/ Kg. Find the ratio of quantities of the two types of pulses to be mixed?
A. 2:3
B. 2:1
C. 1:3
D. 4:1
Answer: Option A
Explanation:
Ratio of Quantities = 40 : 60 = 2:3
2. While creating the fragrance mix, a shopkeeper mixes some amount of Jasmine essence with 240 liters water. The price of the mixture is Rs. 275/liter. The price of Jasmine essence is Rs 325/liter. What is the quantity of Jasmine essence in the mix?
A. 720 litres
B. 1000 litres
C. 1200 litres
D. 1320 litres
Answer: Option D
Explanation:
Water is free so its price is Rs. 0
Ratio of Water to Essence = 50:275 = 2:11
Water is 2 parts. Hence, 2 parts of water = 240litres.
So 11 parts of essence = ?
∴ ? = [latex]\frac {11 \times 240}{2}[/latex] = 1320 litres
Decimal Fraction
1. 0.03 × 0.0124 = ?
A. 3.72 × 10[latex]^ {-6 }[/latex]
B. 3.72 × 10[latex]^ {-5}[/latex]
C. 3.72 × 10[latex]^ {-3}[/latex]
D. 3.72 × 10[latex]^ {-4}[/latex]
Answer: Option D
Explanation:
3 × 124 = 372
Sum of the decimal places in 0.03 and 0.0124 = 2 + 4 = 6
Hence, 0.03 × 0.0124 = 372 × 10[latex]^ {-6}[/latex] = 3.72 × 10[latex]^ {-4}[/latex]
2. 7212 + 15.231 - ? = 6879
A. 359.022
B. 362.02
C. 328.221
D. 348.231
Answer: Option D
Explanation:
= 7212 + 15.231 - 6879
= 348.231
Probability
1. A mother has 24 pairs of white socks and 18 pairs of grey socks for her twin children kept in a bag. In a hurry to send the children to school on time, she picks 3 socks randomly. What is the possibility of her picking up the matching pair?
A. [latex]\frac {1}{108}[/latex]
B. [latex]\frac {1}{36}[/latex]
C. [latex]\frac {7}{36}[/latex]
D. 1
Answer: Option D
Explanation:
Since a pair has 2 socks in it and there are just two colors - White and grey, when the mother chooses 3 socks, he will surely select 2 socks of the same color - be it grey or white.
So possibility = probability = 1
2. Rishabh studies with the help of flashcards. He has a set of 30 flashcards out of which 17 cards are white and rest are grey. 4 white and 5 grey cards are marked ENGLISH. Find the possibility of choosing a grey card or an "ENGLISH" Card randomly from the set.
A. [latex]\frac {9}{13}[/latex]
B. [latex]\frac {13}{30}[/latex]
C. [latex]\frac {17}{30}[/latex]
D. [latex]\frac {22}{30}[/latex]
Answer: Option C
Explanation:
Probability = [latex]\frac {What we want}{Total}[/latex]
OR = add AND = multiple
We want a grey card OR ENGLISH card
There are 30 - 17 = 13 grey cards
There are 4 + 5 = 9 ENGLISH cards
Total cards = 30
Also, 5 grey cards are ENGLISH cards.
So probability = [latex]\frac {13}{30}[/latex] +[latex]\frac {9}{30}[/latex]-[latex]\frac {5}{30}[/latex] =[latex]\frac {17}{30}[/latex]
Remember:
This subtraction is needed a grey card gets counted twice - once in 13 grey cards and once again in 9 ENGLISH cards.
Average
1. Kritika is on a 4 days trip with schoolmates. She had decided to keep her average expense for 4 days at Rs 80. However, at the end of the third day, she realized that her expenses on the first three days were Rs 80, Rs 90, Rs 110 and respectively. How much should she spend on 4th day to ensure that she meets her targeted average expense?
A. Rs. 40
B. Rs. 60
C. Rs. 90
D. Rs. 220
Answer: Option A
Average = [latex]\frac {Sum of Observations}{Number of Observations}[/latex]
Average = [latex]\frac {80+90+110+ 4th day}{4}[/latex] = 80
∴ 4th day expenditure = Rs. (320 - 280) = Rs 40
2. Find the average of natural numbers from 1 to 67?
A. 33.5
B. 34
C. 50
D. 67
Answer: Option B
Explanation:
Average of first n natural numbers = [latex]\frac {(n+1)}{2}[/latex]
Average of natural numbers from 1 to 65 = [latex]\frac {67+1}{2}[/latex]= 34
Stocks and Share
1. Find the market value of the stock if 6% yields 10%.
Answer: Option A
Explanation:
Let the investment be Rs.100 for an income of Rs.10
Therefore, for an income of Rs.6, the investment = 600/10 = Rs.60
2. What is the investment made if one invests in 15% stock at 50 and earns Rs.2000?
A. 8000
B. 7000
C. 5000
D. 6000
Answer: Option C
Explanation:
To earn Rs.15, investment = Rs.50.
Hence, to earn Rs.1500, investment = (1500*50)/15
= Rs.5000
Square Root and Cube Root
1. Prashant had some pebbles in his treasure. His friend said, that he could arrange all the 17424 pebbles in a two dimensional square matrix. How many rows did he make?
A. 132
B. 138
C. 141.2
D. 145.8
Answer: Option A
Explanation:
Since it is a square matrix, no. of rows = No. of columns
Let number of pebbles in each row be P
∴ Number of rows = P
Total Pebbles = P x P = P[latex]^{2}[/latex] = 17424
∴ P = 132 = No. of rows
2. What is 335[latex]^{2}[/latex]?
A. 112225
B. 120225
C. 122225
D. 150225
Answer: Option A
Explanation:
To find square of a number ending in 5
Example - 335[latex]^{2}[/latex]
335 ends with 5.
So 335[latex]^{2}[/latex] will have 5[latex]^{2}[/latex] = 25 at end
Now leave the last digit i.e. 5.
We have 33. Add 1 to it.
We get 33 + 1 = 34
33 x 34 = 1122
3352 = 112225
Problems on Ages
1. Raj's age is 4 times that of Dhiraj, his cousin. 3 years back, Raj was 5 times as old as Dhiraj. What is his present age?
A. 12 years
B. 16 years
C. 24 years
D. 48 years
Answer: Option D
Explanation:
Present day,
Let us consider Dhiraj's present age as K years
Raj's present age is 4 times Dhiraj's = 4K years
3 years ago,
Dhiraj's age = K-3 years
Raj's age = 4K-3 years.
That time, Raj's age = 5 times Dhiraj's age
∴ 4K-3 = 5(K-3)
4K-3 = 5K-15
∴ K = 12 years
Present age of Raj = 4K = 4 x 12 = 48 years.
2. Referring to his two sons Ram and Shyam, their father says - 12 years ago, Ram was 3 times as old as Shyam. 12 years later, the ratio of ages of Shyam and Ram will be 2:3. What is the present age of Ram?
A. 24 years
B. 36 years
C. 54 years
D. 144 years
Answer: Option B
Explanation:
Let us consider Shyam's present age as K years
12 years ago,
Shyam's age = K-12 years
So Ram's age is 3 times Shyam's = 3(K-12) = (3K-36) years → This age of Ram was 12 years ago
Hence, Ram's present age = (3K- 36) + 12 years = (3K-24) years
12 years later from present day,
Shyam's age = K+12 years
Ram's age = (3K-24) + 12 = (3K-12) years
Also,[latex]\frac{Shyam's age}{Ram's age}[/latex] =[latex]\frac{K+12}{3K-12}[/latex]= [latex]\frac{2}{3}[/latex]
∴ K = 20 years = Shyam's present age
∴ Ram's present age = 3K-24 = 3 x 20 - 24 = 36 years