GATE General Aptitude - Number System
Exam Duration: 45 Mins Total Questions : 30
There are four prime numbers written in ascending order. The product of the first three is 1001 and that of the last three is 2431. Find the first prime number
- (a)
7
- (b)
11
- (c)
3
- (d)
5
Let the four prime numbers be a,b,c,d. abc=1001 and bcd=2431
\(\frac { abc }{ bcd } =\frac { 1001 }{ 2431 } =\frac { 7 }{ 17 } \therefore \frac { a }{ d } =\frac { 7 }{ 17 } \Rightarrow a=7\)
What is the remainder when \({ 7 }5^{ 3 }\) is divided by 4?
- (a)
1
- (b)
1
- (c)
3
- (d)
2
(c) The remainder when 75 is divided by 4 is 3
Threfore The remainder when 3x3x3 is divided by 4 is 3.
So, the reqired remainder is 3
A man is typing numbers 5 to 700 on his computer. How many times does he press the key of his computer?
- (a)
988
- (b)
1988
- (c)
1888
- (d)
1990
One digit numbers from 5 to 9 = 5
Two digit numbers from 10 to 99 = 90
Three digit numbers from 100 to 700 = 601
Total number of digits that are to be used
=(5x1)+(90x2)+(601x3)=5+180+1803=1988
What least number must to be added to 15763 so that it is exactly divisible by 18?
- (a)
5
- (b)
18
- (c)
12
- (d)
13
the least number that should be subtracted = 1
What least number must be subtracted from 178669 ao that it is exactly divisible by 36?
- (a)
18
- (b)
1
- (c)
36
- (d)
5
The least number that should be subtracted = 1
Find the numbers of prime factors contained in the product of \({ 23 }^{ 11 }\times { 7 }^{ 4 }\times { 3 }^{ 5 }\)
- (a)
25
- (b)
17
- (c)
30
- (d)
20
Number of prime factors=11+4+5=20
Two numbers when divided by a certain divisor give remainders 350and 20 respectively and when their sum is divided by the same divisor, the remainder is 1.5 find the divisor.
- (a)
30
- (b)
40
- (c)
35
- (d)
45
In such questions, the divisor is r1+r2-r3
Therefore Divisor = 35+20-15=40
A certain number when successively dividedby 5 and 7 leaves the remainder, if the same number is divided by 35?
- (a)
21
- (b)
23
- (c)
24
- (d)
29
X=7Y+4
n=5(7y+4) +3=35y+23
y=1,n=58
Therefore when 58 is divided by 35 the remainder is 23
What is the difference between the greatest and th smallest five digit numbers formed using the digits 7,0,3,4 and 2 without repeating the digits?
- (a)
65672
- (b)
53978
- (c)
54672
- (d)
65978
The greatest five-digit number that can be formed using the given digits =74320
The smallest five-digit number that can be formed using the given digits=20342
Therfore The required difference = 74320-20342 = 53978
A boy had to do a multiplication.instead of taking 35 as one of the multipliers, he took 53. As a result the product went up by 540.What is the new product?
- (a)
1590
- (b)
1450
- (c)
1550
- (d)
1420
Let the number that the boy wanted to mutiply be x
he was expected to find the value of 35 x>30x35x=450
x=30
If both \({ 5 }^{ 2 }\quad \)and \( { 7 }^{ 3 }\)are factors of the number \(p\times { 4 }^{ 3 }\times { 7 }^{ 2 }\times { 13 }^{ 5 }\), then what is the least possible value of p?
- (a)
175
- (b)
145
- (c)
190
- (d)
156
If \({ 5 }^{ 2 }\) is a factor of the given number. The number does not have a power or multiple of 11 as its factor. Hence, P should include \({ 5}^{ 2 }\), \({ 7 }^{ 3 }\) is a factor of the given number. The number has \({ 7 }^{ 2 }\). Hence p should \({ 7 }^{ 1 }\)to make it \({ 7 }^{ 3 }\)
Therefore P should be atleast \({ 5 }^{ 2 }\)x 7
=175 to have \({ 5 }^{ 2 }\) and \({ 7 }^{ 3 }\) as its factors
A worker earns rs.40 the first day and spends rs 25 on the second day. He earns rs 40 on the third day and spends rs 25 on the fourth day and so on. On which day would he have rs 100
- (a)
10
- (b)
9
- (c)
8
- (d)
11
His savings for every 2 days = rs 40-rs 25=rs 15
His savings in 8 days = rs 60
On the 9th day he earns rs 40
On the 9th day he has rs60+rs40=rs100
A shepherd has 324 sheeps, 396 goats and 428 cows. He wants to arrange them into groups of equal sie without mixing. Find the least number of groups required to arrange them.
- (a)
285
- (b)
287
- (c)
284
- (d)
289
HCF of (324,396,42)=4
So, the number of groups required
=\(\frac { 324 }{ 4 } +\frac { 396 }{ 4 } +\frac { 428 }{ 4 } \) =81+99+107=287
A number 475 is divided into two parts in such a way that the LCM and HCF of the two parts is 2250 and 25 respectively. Find the two numbers.
- (a)
250,225
- (b)
200,275
- (c)
175,300
- (d)
240,235
Let the two numbers be 25x and 25y
25x+25y=475
x+y=19 ...(i)
Also, the LCM of 25 xand 25 y=25xy
25xy=2250
xy=90 ...(ii)
\({ (x-y) }^{ 2 }={ (x+y) }^{ 2 }-4xy\\ { 19 }^{ 2 }={ (x+y) }^{ 2 }-4(90)\)
Solving x-y=1 ...(iii)
Using Eqs. (i) and (iii),
x=10 and y=9
Therefore The two numbers are 250 and 225.
A traffic lights at three different road crossingschange after every 50 s, 75 s and 100 s respectively. If thet start changing simultaneously at 10 am, after how much time will they change again simultaneously?
- (a)
6 min
- (b)
5 min
- (c)
3 min
- (d)
8 min
LCM of 50, 75 and 100 = 300 s
Therefore Traffic lights will change simultaneously after 300 s = 5 min
Find the unit digit in the expansion of \({ 42 }^{ 23 }\)
- (a)
4
- (b)
8
- (c)
6
- (d)
2
The unit digit comes from the unit place.So, we have to find the unit digit is the expansion of \({ 2 }^{ 23 }\)
Now, \({ 2 }^{ 1 }=2,{ 2 }^{ 2 }=4,{ 2 }^{ 3 }=8,{ 2 }^{ 4 }16,\\ { 2 }^{ 5 }=32,{ 2 }^{ 6 }=64......\)
We see that the unit digit is \({ 2 }^{ 1 }\) is the same as \({ 2 }^{ 5 }\)
Therefore The unit digit in \({ 2 }^{ 23 }\) is the same as \({ 2 }^{ 3 }\)=8
A worker was engaged for a certain number of days and was promised to be paid rs1755. Heremained absent for some days and was paid rs 1365 only. Waht were his daily wages?
- (a)
182
- (b)
195
- (c)
185
- (d)
192
The daily wages of the worker is the HCF of 1755 and 1365.
The daily wages of the worker is Rs.195
A and B had some candies each. If A gave 5 candies to B will have twice as many as A. Instead of that, if B were to give 5 candies to A, then they will both have the same number of candies. How amny did A have?
- (a)
28
- (b)
21
- (c)
26
- (d)
25
Let A have a candies and B have b candies.
Then , b+5=2(a-5)
b-5=a+5
On solving, we get a = 25
there was 240 multiple choice questions in an entrance examination of amedical college. Aa candiate was given 2 marks for every correct answer and penalised 1/2 mark for every wrong answer. The candidate scored 160 marks in the paper having answered all the questions. Find the number of correct answers given by the candidate.
- (a)
108
- (b)
112
- (c)
116
- (d)
104
Let the number of wrong answers marked = w
Let the number of right answers marked = r
r+w=240
2r-w/2=160
On solving, we get r=112
Find the LCM of \(\frac { 4 }{ 3 } ,\frac { 8 }{ 9 } ,\frac { 3 }{ 5 } \)
- (a)
20
- (b)
24
- (c)
1/24
- (d)
1/20
\(LCM\quad of\quad fraction=\frac { LCM\quad of\quad numerators }{ HCF\quad of\quad denomenators } \\ =\frac { LCM\quad of\quad 4,8,3 }{ HCF\quad of\quad 3,9,5 } =\frac { 24 }{ 1 } =24\)
Find the HCF of 40 and 64.
- (a)
4
- (b)
8
- (c)
2
- (d)
10
\(40)64(\\ \frac { 40 }{ 24)40(1 } \\ \quad \quad \frac { 24 }{ 16)24(1 } \\ \quad \quad \quad \quad \quad \frac { 16 }{ 8)16(2 } \\ \quad \quad \quad \quad \quad \quad \quad \frac { 16 }{ 0 } \\ \quad \quad \quad \quad \quad \quad \quad ----\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \therefore HCF=8\)
Find the HCF of \({ 2 }^{ 2 }\times { 3 }^{ 3 }\times { 5 }^{ 2 },{ 2 }^{ 3 }\times { 3 }^{ 2 }\times 5,{ 5 }^{ 2 }\times 7\)
- (a)
\({ 2 }^{ 2 }\times { 3 }^{ 2 }\times 5\times 7\)
- (b)
\({ 2 }^{ 2 }\times { 3 }^{ 3 }\times 5\)
- (c)
\({ 2 }^{ 3 }\times { 3 }^{ 2 }\times 7\)
- (d)
\({ 3 }^{ 2 }\times { 5 }^{ 2 }\times 7\)
\(HCF\quad of\quad fraction=\frac { HCF\quad of\quad numerator }{ HCF\quad of\quad denominator } \\ =\frac { HCF\quad of\quad 2,12,1 }{ LCM\quad of\quad 5,11,3 } \frac { 1 }{ 165 } \)
Find the smallest 4 digit number that is eactly divisible by 8,10 and 12.
- (a)
1080
- (b)
1100
- (c)
1050
- (d)
1120
The smallest four-digit number exactly divisible by 8,10 and 12 should also be divisible by the LCM of 8,10 and 12
LCM of 8,10 and 12=120
Smallest four-digit number=1000
120)1000(8
960/40
So, the required number = 1000+(120-40)
=1000+80=1080
Findthe HCF of \(\frac { 2 }{ 5 } ,\frac { 12 }{ 11 } ,\frac { 1 }{ 3 } \)
- (a)
\(\frac { 1 }{ 165 } \)
- (b)
\(\frac { 5 }{ 165 } \)
- (c)
\(\frac { 2 }{ 165 } \)
- (d)
\(\frac { 12 }{ 165 } \)
\(HCF\quad of\quad fraction=\frac { HCF\quad of\quad numerator }{ HCF\quad of\quad denominator } \\ =\frac { HCF\quad of\quad 2,12,1 }{ LCM\quad of\quad 5,11,3 } =\frac { 1 }{ 165 } \)
Find the greatest five digit number exactly divisible by 6,8, and 12.
- (a)
99924
- (b)
99988
- (c)
99984
- (d)
99972
The greatest five-digit number divisible by 6,8 and 12 should also be divisible by the LCM of 6,8 and 12.
LCM of 6,8 and 12=24
The greatest five-digit number = 99999\(24)99999(416\\ \quad \quad \quad \frac { 96 }{ 39\\ \frac { 24 }{ 159 } \\ \quad \frac { 144 }{ 15 } \\ \quad ---- } \)
So, the required number = 99999-15=99984
The number of boys in a class is three times the number of girls.Which one of the following numbers cannot represent the total number of children in the class?
- (a)
48
- (b)
44
- (c)
42
- (d)
40
The ratio of Boys :Girls =3:1
So, 42 is not possible as it is not divisible by
(3+1)=4
If 8+12=2,7+14=3, then 10+18=?
- (a)
10
- (b)
4
- (c)
6
- (d)
18
8+12=20\(\Rightarrow \)2+0=2
7+14=21\(\Rightarrow \)2+1=3
\(\therefore \)10+18=28\(\Rightarrow \)2+8=10
Which among the following is the least number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, when divided by 8 leaves a remainder of 7, when divided by 7 leaves a remainder of 6, when divided by 6 leaves a remainder of 5, when divided by 5 leaves a remainder of 5, when divided by 4 leaves a reaminder 3, when divided by 3 leaves a remainder of 2, when divided by 2 leves a reaminder of 1 ?
- (a)
7559
- (b)
839
- (c)
5039
- (d)
None of these
(10-9)=(9-8)=(8-7)=(7-6)=(6-5)
=(5-4)=(4-3)=(3-2)=1
LCM of 10,9,8,7,6,5,4,3,2,=2520
therefore Required number =2520-1=2519
The number zero (0) is surrounded by the same 2 digit number on both (left and right) sides: for example 25025, 67067 etc. The largest number that always divides such a number is
- (a)
7
- (b)
11
- (c)
13
- (d)
1001
The largest such number is 1001
If x,y and z are three integers such that x=y=16, y=z=20 and +x=22, then xy=
- (a)
880
- (b)
981
- (c)
900
- (d)
819
Given x+y=16
y+z=20
z+x=22
(i)+(ii)+(iii) \(\Rightarrow \)2(x+y+z)=58
\(\Rightarrow \)x+y+z=22
x+y+z=29
(iv)-(i) \(\Rightarrow \)z=13
(iv)-(ii)\(\Rightarrow \)x=9
(iv)-(iii)\(\Rightarrow \)y=7
\(\therefore \) xyz=13x9x7=819
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