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Aptitude

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Downstream/Upstream:

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

If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then:

Speed downstream = (u + v) km/hr.

Speed upstream = (u - v) km/hr.

If the speed downstream is a km/hr and the speed upstream is b km/hr, then:

Speed in still water = 1/2*(a + b) km/hr.

Rate of stream = 1/2*(a - b) km/hr.

Profit & Loss

• Profit, P = SP – CP; SP>CP

• Loss, L = CP – SP; CP>SP

• P% = (P/CP) x 100

• L% = (L/CP) x 100

• SP = {(100 + P%)/100} x CP

• SP = {(100 – L%)/100} x CP

• CP = {100/(100 + P%)} x SP

• CP = {100/(100 – L%)} x SP

• Discount = MP – SP

• SP = MP -Discou

• For false weight, profit percentage will be P% = (True weight – false weight/ false weight) x 100.

• When there are two successful profits say m% and n%, then the net percentage profit equals to (m+n+mn)/100

• When the profit is m% and loss is n%, then the net % profit or loss will be: (m-n-mn)/100

• If a product is sold at m% profit and then again sold at n% profit then the actual cost price of the product will be: CP = [100 x 100 x P/(100+m)(100+n)]. In case of loss, CP =[100 x 100 x P/(100-m)(100-n)]

• If P% and L% are equal then, P = L and %loss = P^2/100

Divisiblty Rule

Divisible by 7

To check multiply the number by 5 or 2 and than that number add(if mul by 5) else if mul by 2(subtract) through left number

eg : 532 div by 7

Step 1- mul 2x5=10 Step 2- neglect the unit digit and add to left number 53|2+10=63

check number divisible if yes than original number is also divisible.

Similary multiply it by 2 and instead addition subtract it.

Divisible by 17

Multiply unit digit by 5 and substract that multiplied value with rest digit if this multile than the number is divisible by 17.

eg : 272 iv by 17

Step 1-

mul 2*5=10 **Step 2- **

27-10=17 17 is divisble by 17 hence 272 is divisble by 17.

Age Base Questions

Blood Relation

Clock

1. Minute Spaces

The face or dial of clock is a circle whose circumference is divided into 60 equal parts, named minute spaces.

2. Hour hand and minute hand, A clock has two hands. The smaller hand is called the hour hand or short hand and the larger one is called minute hand or long hand.

3. In 60 minutes, minute hand gains 55 minute spaces over the hour hand.

(In 60 minutes, hour hand will move 5 minute spaces while the minute hand will move 60 minute spaces. In effect the space gain of minute hand with respect to hour hand will be 60 - 5 = 55 minutes.)

4. Both the hands of a clock coincide once in every hour.

5. The hands of a clock are in the same straight line when they are coincident or opposite to each other.

6. When the two hands of a clock are at right angles, they are 15 minute spaces apart.

7. When the hands of a clock are in opposite directions, they are 30 minute spaces apart.

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

9. Angle traced by minute hand in 60 min. = 360°.

10. Theta( aka Degree) = 30 X H- (11/2) X M

11. Reflax Angle = 360 - Theta

Probability

AP & GP

Number System

TYPES OF NUMBERS :

1. Natural numbers (N) = 1, 2, 3, . . . . 2. Whole numbers (W) = 0, 1, 2, 3, . . . . 3. Intezers (Z) = −∞ . . . −2, −1, 0, 1, 2, 3, . . . 4. Rational numbers (Q) = The numbers of the form p⁄q where q ≠ 0. Eg: 1⁄5 , 0.46, 0.333333 they are terminating, repeating. 5. Irrational numbers (I) = The numbers of the form x1⁄n ≠ Intezer. Also π and e also irrational numbers they are non-terminating, non-repeating.

Other types of numbers: a. Even numbers : Numbers which are exactly divisible by 2. These numbers are in the format of 2n. b. Odd numbers: Numbers which gives remainder 1 when divided by 2. These numbers are in the format of 2n ± 1. c. Prime numbers : The numbers which are divisible by 1 and the number itself are primes. The least prime is 2. d. Composite numbers : The numbers of which are divisible by more than 2 numbers.

IMPORTANT RULES RELATED TO EVEN AND ODD NUMBERS:

odd ± odd = even;

even ± even = even;

even ± odd = odd

odd × odd = odd;

even × even = even;

even × odd = even.

odd(any number) = odd

even(any number) = even

BODMAS

1. (a - b)2 = (a2 + b2 - 2ab)

2. (a + b)2 = (a2 + b2 + 2ab)

3. (a + b) (a – b) = (a2 – b2 )

4. (a3 + b3) = (a + b) (a2 – ab + b2)

5. (a3 - b3) = (a - b) (a2 – ab + b2)

6. (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)

7. (a3 + b3 + c3 – 3abc) = (a + b + c) (ar>2 + b2 + c2 – ab – bc – ac)

Quick Tips and Tricks

1. Virnaculum (Bar ‘¬¬–’): If a given expression contains a virnaculum (bar), then first simplify the given expression under virnaculum and then follow the rule of ‘BODMAS’.

Step1: When simplifying the given expressions, first brackets are to be removed in the order: ‘––’, ‘( )’, ‘{ }’, ‘[ ]’

Step2: The operations are to be performed strictly in the order: Division, Multiplication, Addition and Subtraction

2. ‘BODMAS Rule’: This rule is used to solve and find out the value of given expressions by performing the operations in a correct sequence.

BODMAS is the shortcut used to remember the procedure of simplification.

B: Bracket

O: Order (i.e. power, square root etc.)

D: Division (Left to right)

M: Multiplication (Left to right)

A: Addition (Left to right)

S: Subtraction (Left to right)

Step1: When simplifying the given expressions, first brackets are to be removed in the order: ( ), { }, [ ]

Step2: The operations are to be performed strictly in the order: Division, Multiplication, Addition and Subtraction

Mensuration

Alligation or Mixture

**Important formula : **

• final mixture = initial misture(1-replaced amount/total amount)^n

Series

Important Formula

• Sum of Square of n natural number= n(n+1)(2n+1)/2

• Sum of n natural number=n(n+1)/2

• Sum in ap= nx(a+l)/2

Average

• Avg of first n odd number = n

• Avg of first n even number = n+1

• Avg of n naturat number = (n+1)/2

• Avg of number * total number = Sum

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Formulas and Tricks

Important Links

• Population Based Question

• Ratio And Proportion

• Code/Decode

• Log

• Amcat Verbal Question

• Negative Remainder

• Eligibility test

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Practice Questions

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1) What is the LCM of 147/64 and 30/44?

1. 735/2

2. 735/704

3. 3/704

4. 3/735

Solution

Option A

Find the LCM of the numerators.

LCM (147, 30) = 1470

Step 2: Find the HCF of denominators.

HCF (64, 44) = 4

So, the LCM of 147/64 and 30/44 is (LCM of Numerators) / (HCF of Denominators) = 1470 / 4 = 735/2

2) A man rows a boat at a speed of 15 mph in still water. Find the speed of the river if it takes her 4 hours 30 minutes to row a boat to a place 30 miles away and return.

A) 5 mph

B) 10 mph

C) 12 mph

D) 20 mph

Solution

Option C

Let the speed of the river=x mph, then

Time taken row 30 miles upstream and 30 miles downstream = 30/(15-x) + 30/(15+x) = 9/2

= 10/(15-x) + 10/(15+x) = 3/2

= 2[10(15+x) + 10(15-x)] = 3(15-x)²

= 300 + 20x + 300 – 20x = 675 -3x²

x² = 25 or x=5

3) Working 5 hours a day, A can Complete a work in 8 days and working 6 hours a day, B can complete the same work in 10 days. Working 8 hours a day, they can jointly complete the work in how many days?

A) 3 days

B) 4 days

C) 4.5 days

D) 5.4 days

Solution

option A

Working 5 hours a day, A can complete the work in 8 days = 5*8 = 40 hours

Working 6 hours a day, B can complete the work in 10 days = 6*10 = 60 hours

(A+B)’s 1 hour’s work = (1/40+1/60)

=(3+2)/120

= 1/24

Hence, A and B can complete the work in 24 hours which is equal to 3 days.

4) A mixture of 40 liters of milk and water contains 10% water. How much water should be added to this so that water may be 20% in the new mixture?

A) 6.5 liters

B) 5 liters

C) 4 liters

D) 7.5 liters

Solution

Option B

A mixture of 40 liters of milk = 36 liters of Milk and 4 liters of water = 90:10 ratio

Now the new mixture should be in the ratio of 80:20

Hence 80% is equivalent to 36 liters (no addition of milk is done)

100% is (36/80)*100 = 45 liters of milk is present in the new mixture

Thus water shall be added= (45 – 36 – 4) = 5 liters of water

**or**

5) Four different electronic devices make a beep after every 30 minutes, 1 hour, 3/2 hour and 1 hour 45 minutes respectively. All the devices beeped together at 12 noon. They will again beep together at:

A) 12 midnight

B) 3 a.m

C) 6 a.m

D) 9 a.m

Solution

Option D

Four different devices beep after every 30 mins, 60 mins, 90 mins and 105 mins.

LCM of 30,60,90 and 105 is 1260.

Which means the devices beep together after every 1260 mins = 1260/60 = 21 hours

Hence 12 noon + 21 hours = 9 a.m

6) In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. In the same race, A can beat C by?

A) 21 m

B) 26 m

C) 28 m

D) 29 m

Solution

Option C

When A travels 100 m, B travels 75 m. Hence A:B = 100:75

When B travels 100 m, C travels 96 m. Hence B:C = 100:96

When B Travels 75 m, C travels (96 x 75)/100 = 72 m

Hence B:C = 75:72.

Therefore, A:B:C = 100:75:72.

So, when A Travels 100 m, C travels 72 m.

Therefore, A beat C by 28 m

7) In an examination, 70% of students passed in physics, 65% in chemistry, 27% failed in both subjects. The percentage of students who passed is?

A) 62

B) 68

C) 66

D) 69

Solution

Option A

70% students passed in physics = 30% failed in Physics.

65% students passed in Chemistry = 35% failed in Chemistry

Percentage of students failed in both subject = 27%.

Percentage of students failed = 35 + 30 – 27

= 38%.

Percentage of students passed = 100 – 38% = 62%

8) If 15 oxen or 20 cows can eat the grass of a field in 80 days, then in how many days will 6 oxen and 2 cows eat the same grass?

A) 40

B) 60

C)100

D) 160

Solution

Option D

15 oxen take 80 days so, 6 oxen take x days

x = 15*80/6 = 200 days

20 oxen also take 80 day. So, 2 cows take y days

y = 20*80/2 = 800 days

Together work will be done in 800*200/(800+200) = 160 days

9) Simplify {(3 * 2.333 + 2)/3} / (1/10 of 100 + 4.8181)

A) 297/10377

B) 188/121

C) 21/34

D) 33/163

Solution

Solution: Option D

10) In how many ways can 6 lottery tickets be distributed among 4 different people if all of the four different people can get any number of tickets?

A) 6C4

B) 6P4

C) 4^6

D) 6^4

Solution

C is correct because all 4 can get 4 ticket one by one

11) A salesman has the liberty to sell a hair dryer in his store at a price between Rs. 300 and Rs. 700. Profit earned by selling the hair dryer for Rs. 650 is twice the loss incurred when it is sold for Rs. 350. What is the cost price of the hair dryer?

A) 550

B) 450

C) 350

D) 150

Solution

Option B

Going through the options,

Taking Cost Price as Rs 450.

Profit = 650 – 450 = 200

Loss = 350 – 450 = 100

Clearly profit is twice the loss incurred.

Hence, Rs 450 is the correct option.

12) Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?

A) 7 hours 30 minutes

B) 8 hours

C) 8 hours 15 minutes

D) 8 hours 25 minutes

Solution

Option C

In 1hr, Ronald types = 32/6 pages and Elan types = 40/5 pages

If they work together they will type = 32/6 + 40/5 = 40/3 pages in 1 hr

Time needed  to complete the assignment is = (3 x 110)/40 = 33/4 = 8hrs 15mins

Hence, the time required is 8 hrs 15 mins.

13) A television manufacturing company has decided to increase the sale to beat the economic slowdown. It decides to reduce the price of television sets by 25% as a result of which the sales increased by 20%. What is the effect on the total revenue of the company?

A) Decreased by 20%

B) Increased by 20%

C) Increased by 10%

D) Decreased by 10%

Solution

Option D

Let the initial Price = Rs.100 and initial sales = 100

So, the initial revenue = Rs. 10000

Now, the price is reduced to 25% which is equal to Rs.75 and Sales is increased by 20% which is equal to 120.

Now new revenue = 120 x 75 = Rs. 9000

Change in revenue = (10000 – 9000) = Rs.1000 decrease

% decrease = (1000/10000) x 100 = 10%

Hence, the correct option is decrease of 10%.

14) Ravi has a bag full of 10 Nestle and 5 Cadbury chocolates. Out of these, he draws two chocolates. What is the probability that he would get at least one Nestle chocolate?

A) 19/21

B) 3/7

C) 2/21

D) 1/3

Solution

Option A

Probability of getting atleast one nestle chocolate = [(10C1 x 5C1) + 10C2] / 15C2

[(10 x 5) + (10 x 9)/2] / [(15 x 14)/2] = 19/21.

Hence, the required probability is 19/21.

15) Anil and Ruhi started a business by investing Rs 2000 and Rs 2800 respectively. After 8 months, Anil added Rs 600 and Ruhi added Rs 400. At the same time Teena joined them with Rs 4200. Find the share of Teena if they get a profit of Rs 34,300 after a year.

A) Rs 7490

B) Rs 7350

C) Rs 8250

D) Rs 8530

E) None of these

Solution

Option B

Solution:
Share of Anil : Share of Ruhi : Share of Teena is
2000×8 + 2600×4 : 2800×8 + 3200×4 : 4200×4
33 : 44 : 21
so share of Teena = 21/(33+44+21) × 34300 = Rs 7350

16) A sum of Rs 7000 is deposited in two schemes. One part is deposited in Scheme A which offers 8% rate of interest. Remaining part is invested in Scheme B which offers 10% rate of interest compounded annually. If interest obtained in scheme A after 4 years is Rs 226 more than the interest obtained in scheme B after 2 years, find the part deposited in scheme B.

A) Rs 3200

B) Rs 4500

C) Rs 3800

D) Rs 3500

E) Rs 2800

Solution:

Option C

Solution:
Rs 3800
Solution:
(7000-x)*8*4/100 = x [ (1 + 10/100)2 – 1] + 226
70*8*4 – 32x/100 = 21x/100 + 226
2240 – 226 = 53x/100
2014 = 53x/100
So, x = Rs 3800

17) A work which is completed by 20 men in 8 days can be completed by 25 women 12 days. 16 men and 10 women start doing the work. After 3 days, they leave. If the remaining work is to be completed in 6 days by x number of men, find x.

A) 16

B) 18

C) 12

D) 10

E) 22

Solution:

Option A

Solution: 
20 men in 8 days so 16 men in 20 × 8/16 = 10 days and
25 women in 12 days so 10 women in 25 × 12/10 = 30 days
So in 3 days, they complete (1/10 + 1/30) × 3 = 2/5
So remaining work = 1 – 2/5 = 3/5
20 m 1 work in 8 days and x men 3/5 work in 6 days
So 20 × 8 × 3/5 = x × 6 × 1
So, x = 16 men

18) There are 140 tickets (numbered 1 to 140) in a bowl. Find the probability of choosing a ticket which bears multiple of either 3 or 7.

A) 3/5

B) 2/9

C) 1/8

D) 3/7

E) None of these

Solution:

Option D

Solution
Number of multiples of 3 in 140 = 140/3 = 46
Number of multiples of 7 in 140 = 140/7 = 20
Number of multiples of 3×7= 21 in 140 = 140/21 = 6
So required probability = (46+20 – 6)/140 = 60/140 = 3/7

19) Introducing a man, a woman said, "He is the only son of the mother of my mother." How is the woman related to the man?

A. Mother

B. Sister

C. Niece

D. Maternal aunt

Solution C

20) There are six persons A. B, C, D, E and F. C is the sister of F. B is the brother of E's husband. D is the father of A and grandfather of F. There are two fathers, three brothers and a mother in the group. Who is the mother ?

A) A

B) B

C) C

D) E

Solution:

E
Solution:
D is father of A and grandfather of F. So, A is father of F. 

Thus. D and A are the two fathers. C is the sister of F So. C is the daughter of A. 

Since there is only one mother, it is evident that E is the wife of A and hence the mother of C and F. 

So, B is brother of A There are three brothers. So. F is the brother of C. 

 Clearly, A is E's Husband.

21) Pointing to a girl in the photograph, Amar said, "Her mother's brother is the only son of my mother's father." How is the girl's mother related to Amar ?

A) Mother

B) Sister

C) Aunt

D) Grandmother

Solution:

C

Solution:
Only son of Amar's mother's father — Amar's maternal uncle. 

So, the girl's maternal uncle is Arnar's maternal uncle. Thus, the girl's mother is Amar's aunt. 

22) The total age of A and B is 12 years more than the total age of B and C. C is how many years younger than A ?

A) 12

B) 13

C) 14

D) 15

Solution:

A

Solution:
A+B=B+C+12
so
A=12

23) The ratio of the present ages of P and Q is 3 : 4. Five years ago, the ratio of their ages was 5 : 7. Find their present ages.

Solution:

P=(3*2*5)/1=30
Q=(4*2*5)/1=40

24) The average age of a group of 10 students is 15 years. When 5 more students join the group, the average age increase by 1 year. The average age of the new students is?

Solution:

(150+x)/15=16.

=)150+x=240

=x=90

Average of their ages=sum/number=)90/5=18

25) An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?

A. 154°

B. 170°

C. 160°

D. 180°

Solution:

We know that angle traced by hour hand in 12 hrs = 360°

From 8 to 2, there are 6 hours.

Angle traced by the hour hand in 6 hours =  6×360/12= 180°

26) At 3:40, the hour hand and the minute hand of a clock form an angle of:

A. 120°

B. 125°

C. 130°

D. 135°

Solution:

C
Solution: 
Angle traced by hour hand in 12 hrs. = 360°.

Angle traced by it in	11/3	hrs =	(360/12	x	11/3)°	= 110°.
Angle traced by minute hand in 60 min. = 360°.

Angle traced by it in 40 min. =	(360/60x40)°= 240°.
 Required angle (240 - 110)° = 130°.

27) A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

A. 3.6

B. 7.2

C. 8.4

D. 10

Solution:

B
Solution: 

Speed =600/(5 x 60)= 2 m/sec.
Converting m/sec to km/hr (see important formulas section)
   =	(2 x18/5)km/hr= 7.2 km/hr.

28) A train of length 110 meters is running at a speed of 60 kmph. In what time, it will pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?

A. 10

B. 8

C. 6

D. 4

Solution:

C

Opposite direction 
speed=60+6=66km/h
time=distance/speed=110/66=5/3 km/h
in m/s 5/3x18/5=6

29) Two trains of equal length, running in opposite directions, pass a pole in 18 and 12 seconds. The trains will cross each other in:

A. 14.4 seconds

B. 15.5 seconds

C. 18.8 seconds

D. 20.2 seconds

Solution :

A
Solution : 
Let length of each train be x meter.

Then, speed of 1st train = x/18m/sec

Speed of 2nd train = x/12 m/sec

Now,

When both trains cross each other, time taken

=2x/(x/18+x/12)=2x/(2x+3x)/36=(2x X 36)/5x=725=14.4seconds

30) What will come in place of question mark (?) in the following series?

767 495 359 291 257 ?

A. 230 B. 240 C. 250 D. 280 E. 260

Soluton :

B

Solution: 

797-495=272
495-359=136
So which means it is half of previous diffrences
272/2=136
291-257=32
34/2=17
So subtract 17
257-17=240

31) What will come in place of question mark (?) in the following series?

50 67 33 84 16 ?

A. 101 B. 109 C. 107 D. 103 E. 201

Solution:

A

Solution: 
17 is the gap once increase and than decrease follow this order you will get the answer

32) In each of the following number series given, one particular number is wrong. Find out that wrong number in each series.

A) 2 3 10 38 172

1. 192

2. 10

3. 38

4. 2

5. 3

Solution:

3

Solution: 
Logic is 2×1 + 1 = 3, 3 × 2 + 4 =10, 10 × 3 + 9 = 39, 39 × 4 + 16 = 172…. So in place of 38, it should be 39.

33) Find the greatest 6-digit number, which is a multiple of 12.

1. 999980

2. 999990

3. 999984

4. None of these

Solution:

Greatest six-digit number is 999999. Divide this number by 12 and get remainder as 3. Since the remainder is 3, if you subtract 3 from the number, the remaining number will be a multiple of 12. So the greatest such number will be 999999 – 3 =999996.

34) What is the sum of all natural numbers between 100 and 200 which are multiples of 3?

1. 5000

2. 4950

3. 4980

4. 4900

5. None of these

Solution:

Multiples of 3 between 100 and 200 are 102, 105, 108,… ,198.
Here, the first term = 102
last term = 198
Let the number of Multiples of 3 between 100 and 200 = n

W.K.T: Arithmetic Progression Formula:
an = a1 + (n - 1)d
Where, an = last term = 198
a1 = first term = 102
d = common difference = 105 - 102 = 3
---> 198 = 102 + (n - 1) * 3
---> 198 - 102 = (n - 1) * 3
---> 96 = (n - 1) * 3
---> (n - 1) = 96/3 = 32
---> n = 32 + 1
---> n = 33

Formula:
Sum of n terms = Sn = (n/2) * (a + l)
where n = number of elements = 33
a = first term = 102
l = last term = 198
Thus, using the above formula, Sum of all natural numbers between 100 and 200 which are multiples of 3 = (33/2) * (102 + 198)
= (33/2) * 300
= 33 * 150
= 4950

35) The sum of three numbers in an Arithmetic Progression is 45 and their product is 3000. What are the three numbers?

A. 5, 15, 25

B. 12, 15, 18

C. 10, 15, 20

D. -10, -15, -20

Solution:

Assuming that the numbers are (a – d), a, (a + d) and their sum is 45, we get the middle number as 15. Now, the product (a – d) (a + d) = 200. Solving, we get d = 5. Therefore, the numbers are 10, 15 and 20.

36) if 1 is added to the denominator of a fraction, the fraction becomes 1/2. if 1 is added to the numerator of the fraction, the fraction becomes 1. find the fraction

Solution:

x/(y+1)=1/2

and

(x+1)/y=1

2x-y=1 ....eq (1)

x=y-1 ....eq (2)

by solving eq 1 and 2 we get

x=2 and y=3

37) After the division of a number successively by 3, 4 and 7, the remainder obtained is 2, 1 and 4 respectively. What will be remainder if 84 divide the same number?

A. 80

B. 75

C. 42

D. 53

Answer:

Option D

Solution: 
As the Number gives a remainder of 4 when it is divided by 7, then the number must be in form of (7x + 4)

The same gives remainder 1 when it is divided 4, so the number must be in the form of {4 × (7x + 4) + 1}

Also, the number when divided by 3 gives remainder 2, thus number must be in form of [3 × {4 × (7x + 4) + 1} + 2]

Now, On simplifying,
[3 × {4 × (7x + 4) + 1} + 2]
= 84x + 53
We get the final number 53 more than a multiple of 84 Hence, if the number is divided by 84,
The remainder will be 53

38) If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

A) 55/601

B) 601/55

C) 11/120

D) 120/11

Solution:

Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum =1a+1b = a+bab= 55600=11120

39) In a certain code language, ‘sea is deep’ means ‘213’, ‘sky is blue’ means ‘514’ and ‘sea looks blue’ means ‘264’

What number is the code for ‘sky’?

Solution:

5

40) In the following questions, one term in the number series is wrong. Find the wrong term.

2, 3, 5, 8, 13, 20, 34

Solution:

first+second=third
follow this order you get **20** as a answer

41) In the following question, one term in the number series is wrong. Find the wrong term.

196, 169, 144, 121, 100, 80

Solution:

80
First - Second=2

42) SCD, TEF, UGH, ____, WKL

A. CMN

B. UJI

C. VIJ

D. IJT

Solution:

C

43) In a certain code 'MISSIONS' is written as 'MSIISNOS'. How is 'ONLINE' written in that code?

1. OLNNIE

2. ONILEN

3. NOILEN

4. LNOENI

5. ONNLIE

Solution:

1

44) The population of a city A which is 68000 decreases at the rate of 1200/year. Population of city B which is 42000, increases at the rate of 800 per year. Find in how many years, the population of cities A and B are equal?

a. 9 years

b. 10 years

c. 13 years

d. 15 years

Solution:

Correct Option: (c)

We have to find the population of cities A and B after x years.

Step 1: Population of city A = 68000, decreases at the rate of 1200/year
68000 – 1200x

Step 2: Population of city B = 42000, increases at the rate of 800/year
42000 + 800x

Step 3: Find after how many population of cities A and B are equal.

Population of city A = Population of city B
68000 – 1200x = 42000 + 800x
68000 – 42000 = 1200x + 800x
26000 = 2000x
x = 13

45) A contractor pays Rs. 20 to a worker for each day and the worker forfeits Rs. 10 for each day if he is idle. At the end of 60 days, the worker gets Rs. 300. Find for how many days the worker was idle?

a. 28 days

b. 30 days

c. 34 days

d. 40 days

Solution:

Correct Option: (b)

Step 1: Number of days worked by the worker = 60 and he remained idle for x days. Therefore, number of days worked = (60 – x)

Step 2: Each day he was getting paid Rs. 20. Therefore, the payment received for working days = (60 – x) 20

Step 3: After subtracting the amount which he forfeited, he receives Rs. 300.

Therefore,
(60 – x) 20 – 10x = 300
1200 – 20x – 10x = 300
900 = 30x
x = 30 days

46) In a farm, along with 50 hens, there were 45 goats and 8 horses and some farmers. If total number of feet be 224 more than number of heads, then find the number of farmers.

a. 11

b. 15

c. 16

d. 18

Solution:

Correct Option: (b)

Let’s the number of farmers be y.

Step 1: Find number of heads
= (50 hens + 45 goats + 8 horses + y farmers)
= (103 + y)

Step 2: Number of feet
= [(Hens 2 × 50) + (45 × 4) + (8 × 4) + (y × 2)]
= [100 + 180 + 32 + 2y]
= 312 + 2y

Step 3: Find number of farmers
(312 + 2y) – (103 + y) = 224
312 + 2y – 103 – y = 224
y = 15

47) What is the difference in the place value of 5 in the numeral 754853?

A) 49500

B) 49950

C) 45000

D) 49940

Solution:

The Correct answer is (B)

Answer with explanation:

The digit 5 has two place values in the numeral, 5 * 105 = 50,000 and 5 * 101 = 50.

∴Required difference = 50000 - 50 = 49950

48) What is the compound interest on Rs. 2500 for 2 years at rate of interest 4% per annum?

A) Rs. 180

B) Rs. 204

C) Rs. 210

D) Rs. 220

Solution:

Option B
CI=P(1+r/100)^t

CI=2500*(1+4/100)^2
CI=2704
So the diffrenece is 204

48) Sohan started a business with a capital of Rs. 80000. After 6 months Mohan joined as a partner by investing Rs. 65000. After one year they earned total profit Rs. 20000. What is share of Sohan in the profit?

A) Rs. 5222.2

B) Rs. 5777.7

C) Rs. 6222.2

D) Rs. 6777.7

Solution:

Option B
80000*12/65000*6=32/13
113/32*20000=5777.7rs

49) If January 1, 1996, was Monday, what day of the week was January 1, 1997?

A) Thursday

B) Wednesday

C) Friday

D) Sunday

Solution:

The correct option is (B)

Explanation:

The year 1996 is divisible by 4, so it is a leap year with 2 odd days.

As per the question, the first day of the year 1996 was Monday, so the first day of the year 1997 must be two days after Monday. So, it was Wednesday.

50) The speed of a boat in still water is 5km/hr. If the speed of the boat against the stream is 3 km/hr, what is the speed of the stream?

A) 1.5 km/hr

B) 2 km/hr

C) 2.5 km/hr

D) 1 km/hr

Solution:

The correct answer is B

Answer with explanation:

Let the speed of stream = X km/hr

Speed of boat = 5 km/hr

Speed upstream = 3km/hr

Apply formula: Speed upstream = speed of boat - speed of stream

∴ 3 = 5 - X

X = 5 - 3 = 2 km/hr

51) How many times the hands of a clock coincide in a day?

A) 24

B) 22

C) 23

D) 21

Solution:

The Correct answer is (B)

Explanation:

The hands of a clock coincide only once between 11 O' clock and 1 O' clock, so in every 12 hours, the hands of a clock will coincide for 11 times.

∴ In a day or 24 hours, the hands of a clock will coincide for 22 (11+11) times.

52) A pipe can fill a tank in 6 hours and another pipe can empty the tank in 12 hours. If both the pipes are opened at the same time,the tank can be filled in

A) 10 hours

B) 12 hours

C) 14 hours

D) 16 hours

Solution:

Option B
Water enter in 1 hr=1/6
Water empty in 1 hr=1/12
net=1/6-1/12=1/12
or 12hr

53) A running man crosses a bridge of length 500 meters in 4 minutes. At what speed he is running?

A) 8.5 km/s

B) 7.5 km/s

C) 9.5 km/s

D) 6.5 km/s

Solution:

Option B
sec=4*60=240s
speed=500/240=25/12m/s
in km/s speed is =25/12*18/5=7.5km/s.

54) The surface area of a cube is 600 cm2. The length of its diagonal is

Solution:

Surface area of cube=6a^2
600=6*a^2
a=10
diagonal of cube=sqrt(3)*a
ans=sqrt(3)*10

55)

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A. 276

B. 299

C. 322

D. 345

Solution:

Answer: Option C

Explanation:

Clearly, the numbers are (23 x 13) and (23 x 14).

 Larger number = (23 x 14) = 322.

56)

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A. 4

B. 10

C. 15

D. 16

Solution:

Answer: Option D

Explanation:

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

In 30 minutes, they will toll together	30	+ 1 = 16 times.
2

57)

Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

A. 4

B. 5

C. 6

D. 8

Solution:

Answer: Option A

Explanation:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

  = H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

58) The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A. 9000

B. 9400

C. 9600

D. 9800

Solution

Answer: Option C

Explanation:

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

 Required number (9999 - 399) = 9600.

59) If 15 men can reap the crops of a field in 28 days, in how many days will 5 men reap it?

A) 50 days

B) 60 days

C) 84 days

D) 9.333 days

Solution:

Answer: C

Explanation:

Let 5 men can reap a field in x days
So, put the same quantities on the same side.
Men: Days
Now, Men and Days are inversely proportional to each other. If we increase the number of men fewer days will be required to complete the work.
Inversely proportional means Apti Chain Rule
Apti Chain Rule

i.e., 5: 15 = 28: x
Or, x = (28*15)/ 5
Or, x = 84 days
Hence, 5 men can reap a field in 84 days.

60) Find the logarithm of 1/256 to the base 2√2.

A) 16

B) 13/5

C) -16/3

D) 12

Solution:

Answer: C

Explanation:

Let log2√2 [1/256] = x

We know that loga y = x is similar to ax = y

So, we can write it as [1/256] = (2√2) x

Or, (2√2) x = [1/28]

Or, [21 * 21/2]x = 1/28

Or, 23x/2 = 2-8

Therefore, 3x/2 = -8

Hence, x = (-8 * 2)/ 3 = -16/3

61) In what ways the letters of the word "RUMOUR" can be arranged?

Solution:

(6!)/(2!)(2!)=180

62) Find the remainder when 1! + 2! + 3! +4! + 5! + ————– 1000! is divided by 8?

Solution:

You can see number less than 3! are not divisible by 8 so it decide your output
(1!+2!+3!)=9
9%8=1
1 is the answer

63) 106 x 106 - 94 x 94 = ?

A. 2400

B. 2000

C. 1904

D. 1906

E. None of these

Solution:

106 x 106 - 94 x 94	= (106)2 - (94)2
= (106 + 94)(106 - 94)    [Ref: (a2 - b2) = (a + b)(a - b)]
= (200 x 12)
= 2400.

64)

The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number ?

A. 240

B. 270

C. 295

D. 360

Answer: Option B

Explanation:

Let the smaller number be x. Then larger number = (x + 1365).

 x + 1365 = 6x + 15

 5x = 1350

 x = 270

Smaller number = 270

65) The sum of first 45 natural numbers is:

A. 1035

B. 1280

C. 2070

D. 2140

Answer: Option A

Explanation:

Let Sn =(1 + 2 + 3 + ... + 45). This is an A.P. in which a =1, d =1, n = 45.

Sn =	n	[2a + (n - 1)d]	=	45	x [2 x 1 + (45 - 1) x 1]	=		45	x 46		= (45 x 23)
2	2	2
= 45 x (20 + 3)

= 45 x 20 + 45 x 3

= 900 + 135

= 1035.

Shorcut Method:

Sn =	n(n + 1)	=	45(45 + 1)	= 1035.
2	2

66) Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take working together on two different computers to type an assignment of 110 pages ?

[A]7 hours 30 minutes

[B]8 hours

[C]8 hours 15 minutes

[D]8 hours 25 minutes

Solution)

C)
Ronald 1 hr work = 32/6=16/3
Elan 1 hr work = 40/5=8
Show both work in an hr=8+16/3=40/3
Show for 110 pages it will take 110/(40/3) or (110 x 3)/40=33/4hr 
Since: convert it into hr 4*8=32 1 left in 1 hr 60 min 60/4=15min

Show final answer is 8hr 15 mon

67) 4^x + 6^x = 9^x

Solution:

link to solution

4^x/4^x + 6^x/4^x = 9^x/4^x
Now,
1 + (3/2)^x=(3/2)^(2x)
Consider (3/2)^x=u
Then,
1 + u = u^2
Simplifying this

0 = u^2 -u -1

By solving we get
u = (1 + sqrt(5))/2

and this equal to
(1 + sqrt(5))/2 = (3/2)^x
Take log both side

and you get 1.187 approx value.

68) The diameter of a wheel is 98 cm. The number of revolutions in which it will have to cover a distance of 1540 m is

Solution:
circumference of an wheel=πd
=22/7×98
22×14
=308cm =1 revolution
distance covered
1540×100=154000
now,154000÷308
500 rotations

69) An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

A. 2%

B. 2.02%

C. 4%

D. 4.04%

E. None of these

Answer: Option D

100 cm is read as 102 cm.
∴ A1 = (100 x 100) cm2 and A2 (102 x 102) cm2
(A2 - A1) = [(102)2 - (100)2]
              = (102 + 100) x (102 - 100)
              = 404 cm2
∴ Percentage error
=(404100×100×100)%=4.04%

69) A copper wire when bent in the form of a square encloses an area of 484 sq cm if the same wire is bent in the form of a circle the area enclosed by it is

Area of the square=484cm 
side-22cm
perimeter=22*4=88cm
circumfrence of cirlce is 2*pi*r=88
r=14cm
area=pi*r*r=616cm^2

70) When two dice are rolled, what is the probability that the sum is either 5 or 8?

Solution

P(A)=1/9
P(B)=1/6
P(C)=26/36=13/18
Apply GP
you get 2/5 ans