Aptitude Practice Questions
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, 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
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.
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.
The face or dial of clock is a circle whose circumference is divided into 60 equal parts, named minute spaces.
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.
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.)
Both the hands of a clock coincide once in every hour.
The hands of a clock are in the same straight line when they are coincident or opposite to each other.
When the two hands of a clock are at right angles, they are 15 minute spaces apart.
When the hands of a clock are in opposite directions, they are 30 minute spaces apart.
Angle traced by hour hand in 12 hrs = 360°
Angle traced by minute hand in 60 min. = 360°.
Theta( aka Degree) = 30 X H- (11/2) X M
Reflax Angle = 360 - Theta
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.
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
(a - b)2 = (a2 + b2 - 2ab)
(a + b)2 = (a2 + b2 + 2ab)
(a + b) (a – b) = (a2 – b2 )
(a3 + b3) = (a + b) (a2 – ab + b2)
(a3 - b3) = (a - b) (a2 – ab + b2)
(a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
(a3 + b3 + c3 – 3abc) = (a + b + c) (ar>2 + b2 + c2 – ab – bc – ac)
Quick Tips and Tricks
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
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
**Important formula : **
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
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
735/2
735/704
3/704
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
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
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.
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**
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
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
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%
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
A) 297/10377
B) 188/121
C) 21/34
D) 33/163
Solution
Solution: Option D
A) 6C4
B) 6P4
C) 4^6
D) 6^4
Solution
C is correct because all 4 can get 4 ticket one by one
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.
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.
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%.
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.
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
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
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
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
A. Mother
B. Sister
C. Niece
D. Maternal aunt
Solution C
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.
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.
A) 12
B) 13
C) 14
D) 15
Solution:
A
Solution:
A+B=B+C+12
so
A=12
Solution:
P=(3*2*5)/1=30
Q=(4*2*5)/1=40
Solution:
(150+x)/15=16.
=)150+x=240
=x=90
Average of their ages=sum/number=)90/5=18
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°
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°.
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.
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
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
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
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
192
10
38
2
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.
999980
999990
999984
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.
5000
4950
4980
4900
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
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.
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
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
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
Solution:
5
2, 3, 5, 8, 13, 20, 34
Solution:
first+second=third
follow this order you get **20** as a answer
196, 169, 144, 121, 100, 80
Solution:
80
First - Second=2
A. CMN
B. UJI
C. VIJ
D. IJT
Solution:
C
OLNNIE
ONILEN
NOILEN
LNOENI
ONNLIE
Solution:
1
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
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
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
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
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
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
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.
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
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.
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
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.
Solution:
Surface area of cube=6a^2
600=6*a^2
a=10
diagonal of cube=sqrt(3)*a
ans=sqrt(3)*10
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.
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
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
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.
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.
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
Solution:
(6!)/(2!)(2!)=180
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
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.
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
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
[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
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.
Solution:
circumference of an wheel=πd
=22/7×98
22×14
=308cm =1 revolution
distance covered
1540×100=154000
now,154000÷308
500 rotations
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%
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
Solution
P(A)=1/9
P(B)=1/6
P(C)=26/36=13/18
Apply GP
you get 2/5 ans