Aptitude Problems on Numbers with Solutions [CAT, BANK]

1.6, 24, 60,120, 210
336 366 330 660

2.1, 5, 13, 25
51 55 41 34

3.0, 5, 8, 17
24 14 4 36

4.1, 8, 9, 64, 25 (Hint : Every successive terms are related)
336 144 125 216

5.8,24,12,36,18,54
17 27 15 22

6.71,76,69,74,67,72
63 65 67 68

7.5,9,16,29,54
103 105 101 99

8.1,2,4,10,16,40,64 (Successive terms are related)
100 200 150 95

PROBLEMS ON NUMBERS -> DESCRIPTION

In this section, questions involving a set of numbers are put in the form of a puzzle. You have to analyse the given conditions, assume the unknown the numbers and form equations accordingly, which on solving yield the unknown numbers.

AVERAGE -> IMPORTANT FACTS AND FORMULAE 

I. Average = [Sum of observations / Number of observations] 

II. Suppose a man covers a certain distance at x kmph and an equal distance at y kmph. Then, the average speed during the whole journey is [2xy / x + y] kmph.

Numbers -> IMPORTANT FACTS AND FORMULAE 

1. Natural Numbers : Counting numbers 1, 2, 3, 4, 5, .. are called natural numbers. 

II. Whole Numbers : All counting numbers together with zero form the set of whole numbers. 

Thus, I. 0 is the only whole number which is not a natural number.  

II. Every natural number is a whole number. 

III.Some Important Formulae :

 I. ( 1 + 2 + 3 + .....+ n) = n (n + 1 ) / 2 

II. (1 2 + 22 + 32 + ..... + n2) = n ( n + 1 ) (2n + 1) / 6 

III. (1 3 + 23 + 33 + ..... + n3) = n2 (n + 1)2 / 4 

PROBLEMS ON TRAINS -> IMPORTANT FORMULAE

1. a km/hr = [a * 5/18]m/s. 

2. a m/s = [a * 18/5] km/hr. 

3. Time taken by a trian of length l metres to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover l metres. 

4. Time taken by a train of length l metres to pass a stationary object of length b metres is the time taken by the train to cover (l + b) metres. 

5. Suppose two trains or two bodies are moving in the same direction at u m/s and v m/s, where u>v, then their relatives speed = (u - v) m/s. 

6. Suppose two trains or two bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s 

7. If two trains of length a metres and b metres are moving in opposite directions at u 

8. If two trains of length a metres and b metres are moving in the same direciton at u m/s and v m/s, then the time taken by the faster train to cross theslower train = (a + b)/(u - v) sec. 

9. If tow trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then (A’s speed) : (B’s speed) = (√b : √a).

CLOCKS -> IMPORTANT FORMULAE

The face or dial of a watch is a circle whose circumference is divided into 60 equal parts, called minute spaces. A clock has two hands, the smaller one is called the hour hand or short hand while the larger one is called the minute hand or long hand. 

I. In 60 minutes, the minute hand gains 55 minutes on the hour hand. 

II. In every hour, both the hands coincide once. 

III. The hands are in the same straight line when they are coincident or opposite to each other. 

IV. When the two hands are at right angles, they are 15 minute spaces apart. 

V. When the hands are in opposite directions, they are are 30 minute spaces apart. 

VI. Angle traced by hour hand in 12 hrs = 360°. 

VII. Angle traced by munute hand in 60 min. = 360°

Too Fast and Too Slow : If a watch or a clock indicates 8.15, when the correct time is 8, it is said to be 15 minutes too fast. 

On the other hand, if it indicates 7.45, when the correct time is 8, it is said to be 15 minutes too slow. 

BANKERS DISCOUNT -> IMPORTANT CONCEPTS

Bankers’ Discount : Suppose a merchant A buys goods worth, say Rs. 10,000 from another merchant B at a credit of say 5 months. Then, B prepares a bill, called the bill of exchange. A signs this bill and allows B to withdraw the amount from his bank account after exactly 5 months. 

The date exactly after 5 months is called nominally due date. Three days (known as grace days) are added to it to get a date, known as legally due date. 

Suppose B wants to have the money before the legally due date. Then he can have the money from the banker or a broker, who deducts S.I. on the face value (i.e., Rs. 10,000 in this case) for the period from the date on which the bill was discounted (i.e., paid by the banker) and the legally due date. This amount is known as Banker’s Dicount (B.D.) 

Thus, B.D. is the S.I. on the face value for the period from the date on which the bill was discounted and the legally due date. 

Banker’s Gain (B.G.) = (B.D.) - (T.D.) for the unexpired time.

Note : When the date of the bill is not given, grace days are not to be added. 

BANKERS DISCOUNT -> IMPORTANT FORMULAE

I. B.D. = S.I. on bill for unexpired time.

II. B.G. = (B.D.) - (T.D.) = S.I. on T.D. = (T.D.)² / R.W. 

III. T.D. = √P.W. * B.G. 

IV. B.D. = [Amount * Rate * Time / 100] 

V. T.D. = [Amount * Rate * Time / 100 + (Rate * Time)]

VI. Amount = [B.D. * T.D. / B.D. - T.D.] 

VII. T.D. = [B.G. * 100 / Rate * Time] 

PARTNERSHIP -> IMPORTANT FACTS AND FORMULAE

I. Partnership : When two or more than two persons run a business jointly, they are called partners and the deal is known as partnership. 

II. Ratio of Division of Gains : 

(i) When investments of all the partners are for the same time, the gain or loss is distributed among the partners in the ratio of their investments. Suppose A and B invest Rs. x and Rs. y respectively for a year in a business, then at the end of the year : (A’s share of profit) : (B’s share of profit) = x : y. 

(ii) When investments are for different time periods, then equivalent capitals are calculated for a unit of time by taking (capital * number of units of time). Now, gain or loss is divided in the ratio of these capitals. Suppose A invests Rs. x for p months and B invests Rs. y for q months, then (A’s share of profit) : (B’s share of profit) = xp : yq. 

III. Working and Sleeping Partners : A partner who manages the business is known as working partner and the one who simply invests the money is a sleeping partner.

VOLUME AND SURFACE AREA -> IMPORTANT FACTS AND FORMULAE

I. CUBIOD 
Let length = l, breadth = b and height = h units. Then, 

1. Volume = (l x b x h) cubic units. 

2. Surface area = 2 (lb + bh + lh) 

II. CUBE 
Let each edge of a cube be of length a. Then, 

1. Volume = a³ cubic units. 

2. Surface area = 6a² sq. units. 

3. Diagonal = √3 a units. 

III. CYLINDER 
Let radius of base = r and Height (or length) = h Then, 

1. Volume = (∏r²h) cubic units.

 2. Curved surface area = (2∏rh) sq. units.

3. Total surface area = (2∏rh + 2∏r² sq. units) = 2∏r (h + r) sq. units. 

IV. CONE
Let radius of base = r and Height = h. Then, 

1. Slant height, l = √h² + r ² units. 

2. Volume = [1/3 ∏r²h] cubic units. 

3. Total surface area = (∏rl + ∏r²) sq.units. 

PROFIT AND LOSS -> IMPORTANT FACTS AND FORMULAE

Cost Price : The price at which an article is purchased, is called its cost price, abbreviated as C.P. 

Selling Price : The price at which an article is purchased, is called its cost price, abbreviated as C.P. 

Profit or Gain : The price at which an article is purchased, is called its cost price, abbreviated as C.P. 

Loss : If S.Pis less than C.P., the seller is said to have incurred a loss.

1. Gain = (S.P.) - (C.P.) 

2. Loss or gain is always reckoned on C.P.

 3. gain% = [Gain*100/C.P.] 

4. Loss = (C.P.) - (S.P.) 

5. Loss% = [Loss*100/C.P.] 

6. S.P. = (100+Gain%)/100 * C.P. 

7. S.P. = (100-Loss%)/100 * C.P. 

8. C.P. = 100/(100+Gain%) * S.P. 

9. C.P. = 100/(100-Loss%) * S.P. 

10. If an article is sold at a gain of say, 35%, then S.P. = 135% of C.P. 

11. If an article is sold at a loss of say, 35%, then S.P. = 65% of C.P.

TIME AND DISTANCE -> IMPORTANT FACTS AND FORMULAE

1. Speed = [Distance/Time], Time=[Distance/Speed], Distance = (Speed*Time) 

2. x km/hr = [x*5/18] m/sec. 

3. If the ratio of the speeds of A and B is a:b, then the ratio of the times taken by them to cover the same distance is 1/a : 1/b or b:a. 

4. x m/sec = [x*18/5] km/hr. 

5. Suppose a man covers a certain distance at x km/hr and an equal distance at y km/hr. then, the average speed during the whole journey is [2xy/x+y] km/hr.

BOATS AND STREAMS -> IMPORTANT FACTS AND FORMULAE 

I. In water, the direction along the stream is called downstream. And, the direction against the stream is called upstream. 

II. If the speed of a boat in still water is u km/ht and the speed of the stream is v km/hr, then : Speed downstream = (u + v) km/hr Speed upstream (u - v) km/hr.

III. If the speed downstream is a km/hr and the speed upstream is b km/hr, then : Speed in strill water = 1/2 (a + b) km/hr Rate of stream = 1/2 (a - b) km/hr  

Highest common factor (HCF)or Greatest common divisor (GCD)

HCF is the maximum divisor which divides all the given numbers exactly.  Let us say for 16, 24 there are several numbers i.e., 1, 2, 4, 8 divide them exactly. Of all these numbers 8 is maximum number so we could call 8 as HCF

Finding HCF:  

HCF can be found in two ways. Division Method and Factorization method.

Example: Find the HCF of 16, 24

Factorization Method: 

We need to write each number in its prime factorization format and take the prime numbers common to all given numbers and their minimum power.

16=24$16=2^4$,  24=23×3$24=2^3\times 3$

Now HCF of 16, 24 = 23$2^3$ ( we must not consider 3 because 16 does not contain the prime factor 3)

Division Method: 

Important formulas: 

Formula 3: if a, b, c are the remainders in each case when A, B, C are divided by N then N = HCF (A-a, B-b, C-c)

LCM or Least common factor

LCM is defined as the least number which is divisible by all the given divisors.  Take 4,6 as two divisors which divide 12, 24, 36... perfectly with no remainder.  So 12, 24, 36 are called common multiples of 4 and 6.  In other words, 4 and 6 are factors of all these number.  Of all these common multiples, 12 is the least number.  So we can say 12 is Least common multiple of all the given numbers or LCM of 4, 6.

Finding LCM: 

There are two ways to find LCM.  First one is division method, second one is Factorization method.  

1. Division Method: LCM of 15, 18, 27

In division method we have to continue the division until the numbers in the last row become co - primes with each other.  So LCM = 3 x 3 x 5 x 2 x 3 =270

2. Factorization Method: 

Here we can write all the given numbers in their prime factorization format.

15 = 3 x 5

18 = 2×32$2\times 3^2$

27 = 33$3^3$

Now take all primes number the given numbers and write their maximum powers. So LCM of 15, 18, 27 = 2×33×5$2\times 3^3\times 5$= 270

Formula 1: If r is the remainder in each case when N is divided by x, y, z then the general format of the number is N= K x [LCM (x, y, z)] + r here K is a natural number

QUANTITATIVE APTITUDE

The largest square that can be inscribed in a right angled triangle ABC when one of its vertices lies on the hypotenuse of the triangle

Solution 1:

From the above diagram,  Δ$\mathrm{\Delta }$ABC and Δ$\mathrm{\Delta }$AFD are similar.
⇒Tanα=ADFD=ABBC$\Rightarrow Tan\alpha ={\displaystyle \frac{AD}{FD}}={\displaystyle \frac{AB}{BC}}$
⇒ADx=ab$\Rightarrow {\displaystyle \frac{AD}{x}}={\displaystyle \frac{a}{b}}$
⇒AD=xab$\Rightarrow AD={\displaystyle \frac{xa}{b}}$ - - - - - (1)
Also, Δ$\mathrm{\Delta }$ABC and Δ$\mathrm{\Delta }$EGC are similar.
⇒Tanα=GEEC=ab$\Rightarrow Tan\alpha ={\displaystyle \frac{GE}{EC}}={\displaystyle \frac{a}{b}}$
⇒xEC=ab$\Rightarrow {\displaystyle \frac{x}{EC}}={\displaystyle \frac{a}{b}}$
⇒EC=xba$\Rightarrow EC={\displaystyle \frac{xb}{a}}$ - - - - - (2)
We know that c= AD + x + EC
⇒c=xab+x+xba$\Rightarrow c={\displaystyle \frac{xa}{b}}+x+{\displaystyle \frac{xb}{a}}$
⇒c=x(ab+1+ba)$\Rightarrow c=x\left({\displaystyle \frac{a}{b}}+1+{\displaystyle \frac{b}{a}}\right)$
⇒c=x(a2+ab+b2ab)$\Rightarrow c=x\left({\displaystyle \frac{a^2+ab+b^2}{ab}}\right)$
⇒x=(abca2+ab+b2)$\Rightarrow x=\left({\displaystyle \frac{abc}{a^2+ab+b^2}}\right)$
Side of the square = abca2+b2+ab${\displaystyle \frac{abc}{a^2+b^2+ab}}$

Solution 2:

From the above diagram, drop a perpendicular to AC from vertex B. 

Area of Δ$\mathrm{\Delta }$ABC = 12×a×b=12×BF×c${\displaystyle \frac{1}{2}}\times a\times b={\displaystyle \frac{1}{2}}\times BF\times c$

∴BF=abc$\therefore BF={\displaystyle \frac{ab}{c}}$ - - - - - (1)

Now Δ$\mathrm{\Delta }$BDE and Δ$\mathrm{\Delta }$ABC are similar. 

⇒BGBF=DEAC$\Rightarrow {\displaystyle \frac{BG}{BF}}={\displaystyle \frac{DE}{AC}}$

⇒abc−xabc=xc$\Rightarrow {\displaystyle \frac{{\displaystyle \frac{ab}{c}}-x}{{\displaystyle \frac{ab}{c}}}}={\displaystyle \frac{x}{c}}$

⇒1−xcab=xc$\Rightarrow 1-{\displaystyle \frac{xc}{ab}}={\displaystyle \frac{x}{c}}$

⇒1=xcab+xc$\Rightarrow 1={\displaystyle \frac{xc}{ab}}+{\displaystyle \frac{x}{c}}$

⇒1=x(cab+1c)$\Rightarrow 1=x\left({\displaystyle \frac{c}{ab}}+{\displaystyle \frac{1}{c}}\right)$

⇒1=x(c2+ababc)$\Rightarrow 1=x\left({\displaystyle \frac{c^2+ab}{abc}}\right)$

⇒x=abcc2+ab$\Rightarrow x={\displaystyle \frac{abc}{c^2+ab}}$ = abca2+b2+ab